Introduction to Lorentz Geometry: Curves and Surfaces 2020028007, 2020028008, 9780367468644, 9781003031574

Table of contents :
Cover
Half Title
Title Page
Copyright Page
Contents
Preface of the Portuguese Version
Preface
Chapter 1: Welcome to Lorentz-Minkowski Space
1.1.
PSEUDO–EUCLIDEAN SPACES
1.1.1.
Defining Rnn
1.1.2.
The causal character of a vector in Rnn
1.2.
SUBSPACES OF Rnn
1.3.
CONTEXTUALIZATION IN SPECIAL RELATIVITY
1.4.
ISOMETRIES IN Rnn
1.5.
INVESTIGATING O1(2,R) AND O1(3,R)
1.5.1.
The group O1(2,R) in detail
1.5.2.
The group O1(3,R) in (a little less) detail
1.5.3.
Rotations and boosts
1.6.
CROSS PRODUCT IN Rnn
1.6.1.
Completing the toolbox
Chapter 2: Local Theory of Curves
2.1.
PARAMETRIZED CURVES IN Rnn
2.2.
CURVES IN THE PLANE
2.3.
CURVES IN SPACE
2.3.1.
The Frenet-Serret trihedron
2.3.2.
Geometric effects of curvature and torsion
2.3.3.
Curves with degenerate osculating plane
Chapter 3: Surfaces in Space
3.1.
BASIC TOPOLOGY OF SURFACES
3.2.
CAUSAL TYPE OF SURFACES, FIRST FUNDAMENTAL FORM
3.2.1.
Isometries between surfaces
3.3.
SECOND FUNDAMENTAL FORM AND CURVATURES
3.4.
THE DIAGONALIZATION PROBLEM
3.4.1.
Interpretations for curvatures
3.5.
CURVES IN A SURFACE
3.6.
GEODESICS, VARIATIONAL METHODS AND ENERGY
3.6.1.
Darboux-Ribaucour frame
3.6.2.
Christoffel symbols
3.6.3.
Critical points of the energy functional
3.7 THE FUNDAMENTAL THEOREM OF SURFACES
3.7.1.
The compatibility equations
Chapter 4: Abstract Surfaces and Further Topics
4.1.
PSEUDO-RIEMANNIAN METRICS
4.2.
RIEMANN’S CLASSIFICATION THEOREM
4.3.
SPLIT-COMPLEX NUMBERS AND CRITICAL SURFACES
4.3.1.
A brief introduction to split-complex numbers
4.3.2.
Bonnet rotations
4.3.3.
Enneper-Weierstrass representation formulas
4.4.
DIGRESSION: COMPLETENESS AND CAUSALITY
Appendix:
Some Results from Differential Calculus
Bibliography
Index

Citation preview

Introduction to Lorentz Geometry

Introduction to Lorentz Geometry Curves and Surfaces

Ivo Terek Couto Alexandre Lymberopoulos

First edition published 2021 by CRC Press 6000 Broken Sound Parkway NW, Suite 300, Boca Raton, FL 33487-2742 and by CRC Press 2 Park Square, Milton Park, Abingdon, Oxon, OX14 4RN © 2021 Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, LLC Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, access www.copyright.com or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. For works that are not available on CCC please contact [email protected] Trademark notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Library of Congress Cataloging-in-Publication Data Names: Couto, Ivo Terek, author. | Lymberopoulos, Alexandre, author. Title: Introduction to Lorentz geometry : curves and surfaces / Ivo Terek Couto, Alexandre Lymberopoulos. Other titles: Introdução à geometria Lorentziana. English. Description: First edition. | Boca Raton : C&H/CRC Press, 2020. | Translation of: Introdução à Geometria Lorentziana. | Includes bibliographical references and index. Identifiers: LCCN 2020028007 (print) | LCCN 2020028008 (ebook) | ISBN 9780367468644 (hardback) | ISBN 9781003031574 (ebook) Subjects: LCSH: Geometry, Differential. | Lorentz transformations. | Curves. | Surfaces. | Mathematical physics. Classification: LCC QA641 .C65 2020 (print) | LCC QA641 (ebook) | DDC 516.3/6--dc23 LC record available at https://lccn.loc.gov/2020028007 LC ebook record available at https://lccn.loc.gov/2020028008

ISBN: 978-0-367-46864-4 (hbk) ISBN: 978-1-003-03157-4 (ebk)

Contents Preface of the Portuguese Version

vii

Preface

ix

Chapter 1.1

1  Welcome to Lorentz-Minkowski Space PSEUDO–EUCLIDEAN SPACES

1.1.1 1.1.2

Defining

2

Rnν

2

The causal character of a vector in

Rnν

1.2

SUBSPACES OF Rnν

1.3

CONTEXTUALIZATION IN SPECIAL RELATIVITY

Rnν

ISOMETRIES IN

1.5

INVESTIGATING O1 (2, R) AND O1 (3, R)

19 29 43

1.5.1

The group O1 (2, R) in detail

43

1.5.2

The group O1 (3, R) in (a little less) detail

44

1.5.3

Rotations and boosts

47

CROSS PRODUCT IN

1.6.1

Chapter

3 4

1.4

1.6

1

Rnν

Completing the toolbox

2  Local Theory of Curves

53

57

63

2.1

PARAMETRIZED CURVES IN Rnν

64

2.2

CURVES IN THE PLANE

77

2.3

CURVES IN SPACE

97

2.3.1

The Frenet-Serret trihedron

98

2.3.2

Geometric effects of curvature and torsion

106

2.3.3

Curves with degenerate osculating plane

118

Chapter

3  Surfaces in Space

129

3.1

BASIC TOPOLOGY OF SURFACES

130

3.2

CAUSAL TYPE OF SURFACES, FIRST FUNDAMENTAL FORM

155

3.2.1

169

3.3

Isometries between surfaces

SECOND FUNDAMENTAL FORM AND CURVATURES

178

v

vi  Contents

3.4

THE DIAGONALIZATION PROBLEM

190

3.4.1

194

Interpretations for curvatures

3.5

CURVES IN A SURFACE

207

3.6

GEODESICS, VARIATIONAL METHODS AND ENERGY

219

3.6.1

Darboux-Ribaucour frame

222

3.6.2

Christoffel symbols

228

3.6.3

Critical points of the energy functional

232

3.7

Chapter

THE FUNDAMENTAL THEOREM OF SURFACES

250

3.7.1

The compatibility equations

250

4  Abstract Surfaces and Further Topics

259

4.1

PSEUDO-RIEMANNIAN METRICS

260

4.2

RIEMANN’S CLASSIFICATION THEOREM

280

4.3

SPLIT-COMPLEX NUMBERS AND CRITICAL SURFACES

286

4.3.1

A brief introduction to split-complex numbers

286

4.3.2

Bonnet rotations

298

4.3.3

Enneper-Weierstrass representation formulas

304

4.4

DIGRESSION: COMPLETENESS AND CAUSALITY

314

Some Results from Differential Calculus

325

Appendix



Bibliography

331

Index

335

Preface of the Portuguese Version This book has the goal of simultaneously approaching the Differential Geometry of curves and surfaces in two distinct ambient spaces: Euclidean and Lorentzian. The essential difference between them is the the way of measuring lengths, with the former being the one that we are already used to. What is new is the study of such objects in the Lorentzian space, making systematic comparisons between the two cases. Differential Geometry is an area of Mathematics which consists of the study of geometric properties of certain objects, by using techniques from Differential Calculus and Linear Algebra, occasionally employing more sophisticated tools. Lorentzian Differential Geometry, in turn, is interesting not only for Mathematics itself, but it reaches Physics as well, being the mathematical language of General Relativity. The main motivation for preparing the present text was to present, in a sufficiently self-contained (and, we hope, simple) way, part of the vast bibliography on the subject, which is scattered in scientific papers and articles, often with non-uniform language and notation. To begin our journey, we need to understand the “static” geometry of the Lorentzian space, studying the Linear Algebra of this ambient space. This is done in Chapter 1, which also presents results about the transformations that preserve this geometry. Such results are stated in parallel with results from Euclidean Geometry. Some interactions of the Lorentz-Minkowski space with the theory of Special Relativity also appear here. In Chapter 2, we study in detail the geometry of curves. To do so, we’ll combine what was covered in the previous chapter with results from single-variable Differential Calculus. The basic theory is the same, no matter the dimension of the ambient space, and we’ll develop this in the first section, leaving the particular cases of the plane and three-dimensional space for the next sections. To complete this introduction to Lorentzian Differential Geometry, we’ll do a systematic study of surfaces in three-dimensional spaces, developing simultaneously the theory in both Euclidean and Lorentzian spaces, emphasizing the differences between the two cases, whenever those occur. Here, we will need results from the previous chapters, and tools from Multivariable Calculus. In the end of the book, we present an appendix with the main results used. The fourth chapter has essentially three goals: showing the direction to which the previously seen results can be generalized and the modern way to approach Differential Geometry; establishing relations between certain surfaces and complex variables (or an equivalent, there defined); and, finally, briefly discussing the interpretation of General Relativity in this more abstract mathematical setting. Naturally, for this chapter, we expect more mathematical maturity, and some familiarity with the theory of complex variables and point-set topology. All the sections in this book are accompanied by many exercises, and there are over

vii

viii  Preface of the Portuguese Version

300 of them. Those that are more important for the development of the theory are indicated with a †, and many others are mentioned along the text. We do not expect the reader to find the time to work through all of the exercises, but we do expect all the statements to be at least read and understood. Throughout the text, we will follow the usual notations from mathematical literature, but to make the reading easier, we will adopt bold font for vectors in the ambient space, while Greek letters are reserved for curves, and Roman letters for fixed vectors or parametrized surfaces. When the parameter of a curve has any special meaning, some alteration in font will be made to emphasize such change. Lastly, several colleagues have contributed suggestions and constructive criticism to improve the presentation and content of the book. In particular, we would like to thank Prof. Antonio de Padua Franco Filho for the careful reading of the manuscript. We would also like to acknowledge FAPESP and CNPq for the partial support. And, to all, our sincere thanks.

São Paulo, June of 2018

Ivo Terek Couto Alexandre Lymberopoulos

Preface This is a translation of the Portuguese version originally published by the Brazilian Mathematical Society in 2018. The organization of the text in chapters and sections remains the same as in the original version. So, what has changed? During the translation, several small corrections were made here and there, but especially in statements of exercises which were not precise enough, or missing important details. And this time, there is a tentative solution manual available for instructors, at the following address: www.routledge.com/9780367468644 Still on the topic of exercises, we cannot miss the opportunity to emphasize that while the main goal of this book is indeed to introduce Differential Geometry of Curves and Surfaces to students, but with a (Lorentzian) twist, this text can nevertheless be used for teaching a “vanilla” course (as the reader should note by setting ν = 0 and eα = ηα = e M = 1 in everything to appear). If this path is chosen, most of Chapter 1 may be omitted, since a standard pre-requisite Linear Algebra class should have already covered most of the important results regarding positive-definite inner products. Section 1.4, however, is crucial (as it deals with the notion of isometry). Proceeding to Chapters 2 and 3, Subsection 2.3.3 and the first half of Section 3.4 are purely Lorentzian, and thus could be skipped. Topics in Chapter 4 could serve as ideas for student presentations in small classes, according to the instructor’s discretion. Lastly, we would like to thank Prof. Jan Lang, not only for finally convincing us to try and publish this English version of the text, but also for actually suggesting to work with CRC Press, whose outstanding support and professionalism were essential throughout the whole translation process.

Columbus, OH São Paulo, SP

Ivo Terek Couto Alexandre Lymberopoulos

ix

CHAPTER

1

Welcome to Lorentz-Minkowski Space

INTRODUCTION In this chapter we define the pseudo-Euclidean spaces Rnν and, in particular, the Lorentz-Minkowski space Ln = R1n , the ambient where the geometric objects studied throughout this book are contained. Beyond definition and terminology, we provide the main results from Linear Algebra derived from this new way to measure angles and lengths. In Section 1.1 we present the space Ln and the causal character of its elements, as well as a geometric interpretation of the position of vectors and their causal character when n = 2 or n = 3. In Section 1.2 the concept of orthogonality is introduced in this context and we extend the definition of causal character, previously made for vectors, to subspaces of Ln . We use a Sylvester-like criterion to determine the causal character of a subspace, and we establish a relation between the causal characters of a subspace and its orthogonal “complement” (the reason for the quotation marks used here will also be explained). We finish this section with a modified Gram-Schmidt process and establish conditions for the existence of orthonormal bases for subspaces of Rnν . Continuing into Section 1.3, we focus on a physical feature of L4 : it is the model of a spacetime free of gravity. We provide a physical interpretation of the causal character of vectors and discuss, in brief, causal and temporal cones, as well as the concepts of causal and chronological precedence. The new versions of the Cauchy-Schwarz and triangle inequalities are used to explain the Twin Paradox, a highlight in Special Relativity. Surprising results are revealed here, such as the fact that two lightlike vectors are Lorentzorthogonal if and only if they are parallel. To close this section, we state the AlexandrovZeeman Theorem, which classifies all the causal automorphisms of Ln , for n > 2, and a family of counterexamples to this theorem, when n = 2. Next, Section 1.4 covers the isometries of pseudo-Euclidean spaces. Those functions are essential to study curves and surfaces in terms of their geometric invariants. We study Lorentz transformations from the mathematical point of view, focusing on the group structure defined by the composition operation. To do this, we need the concepts of proper transformations and orthochronous transformations (they preserve, respectively, spatial and time orientations). The groups O1+↑ (2, R) and O1+↑ (3, R) play a major role in our theory. We give them special attention in Section 1.5, where their complete classification is given. With this in place, we turn our attention to two classes of isometries: pure rotations and Lorentz 1

2  Introduction to Lorentz Geometry: Curves and Surfaces

boosts, which are very important in Physics. In the end, we provide an alternative characterization of proper orthochronous Lorentz transformations. The final section in this chapter aims to extend the usual cross product operation in R3 to Rnν . We establish its basic properties and how it is related to the algebraic orientation of the space, pointing out the influence of the causal character of vectors on it. Lastly, some technical results are shown. They will be useful in the study of surface curvatures, to be studied in Chapter 3.

1.1

PSEUDO–EUCLIDEAN SPACES

1.1.1

Defining Rnν

Let n ≥ 2 and consider the bilinear form h·, ·iν : Rn × Rn → R given by: . h x, yiν = x1 y1 + · · · + xn−ν yn−ν − xn−ν+1 yn−ν+1 − · · · − xn yn ,

(†)

where x = ( x1 , . . . , xn ) and y = (y1 , . . . , yn ). Writing the canonical basis1 of Rn as can = (e1 , . . . , en ) , the matrix of h·, ·iν relative to such basis is   .

Idn−ν,ν = 

Idn−ν

0

0

− Idν

.

n ν 2 In short, we write Idn−ν,ν = (ηijν )i,j =1 , where ηij is analogous to the Kronecker’s delta

in this setting. Hence, if u, v ∈ Rn are column vectors, we write hu, viν = u> Idn−ν,ν v. The matrix Idn−ν,ν says that while h·, ·iν is not a positive-definite inner product in Rn , it is still symmetric and non-degenerate3 . Throughout the literature, there are some differences in terminology. For instance, in [54] a bilinear form is an inner product if it is bilinear, symmetric, and positive-definite; if not positive-definite, but still non-degenerate, it is called a scalar product. With this, we would say that h·, ·iν is just a scalar product. In general, we will refer to both cases as an inner product, Euclidean when positive-definite and pseudo-Euclidean when nondegenerate. Definition 1.1.1. The pseudo-Euclidean space of index ν, from now on denoted by Rnν , is the vector space Rn with its usual sum and scalar multiplication, equipped
with the . n n inner product h·, ·iν defined in (†). In short, we write Rν = R , h·, ·iν . The main . ambient space in this book is Ln = R1n , called the Lorentz-Minkowski space. Remark.

• The inner products of Rn ≡ R0n and Ln will be denoted by h·, ·i E and h·, ·i L , respectively. The product h·, ·i L is called Lorentz inner product, or Minkowski metric. • It is usual in the literature to define h·, ·iν with the ν minus signs in the first ν terms of the sum (†), instead of in the last ν terms. This is no more than an axis permutation. • The space L4 is the simplest model where the Special Relativity theory may be developed, that is, a gravity-free spacetime. Its points are usually called events. 1 This

notation will be consistent through this chapter.

2 δ = 1, if i = j, and 0 if i 6 = j. ij 3 That is, if h v, u i = 0 for all u,

then v = 0.

Welcome to Lorentz-Minkowski Space  3

Definition 1.1.2. The pseudo-Euclidean norm is the map k · kν : Rnν → R given by . p kvkν = |hv, viν |. We say that v is a unit vector if kvkν = 1. It is important to notice that “norm” in the above definition is an abuse of language, since k · kν is not a norm: there are non-zero vectors such that its pseudo-Euclidean norm vanishes and vectors for which the triangular inequality does not hold. As examples in L3 , we have √ k(1, 0, −1)k L = 0 and k(1, 1, −2)k L = 2 ≥ 0 = k(1, 0, −1)k L + k(0, 1, −1)k L . Despite these differences when compared to the Euclidean norm k · k E , the pseudoEuclidean norms satisfy kλvkν = |λ|kvkν for all v ∈ Rnν and λ ∈ R. 1.1.2

The causal character of a vector in Rnν

Definition 1.1.3 (Causal Character or Causal Type). A vector v ∈ Rnν is: (i) spacelike, if hv, viν > 0, or v = 0; (ii) timelike, if hv, viν < 0; (iii) lightlike, if hv, viν = 0 and v 6= 0. For any non-lightlike vector v 6= 0, we set its indicator ev to be the sign of hv, viν , that is, ev = 1 if v is spacelike, and ev = −1 if v is timelike. Remark.

• For ν = 1, the nomenclature comes from Physics. Such motivations will be further explained in Section 1.3. In that context, lightlike vectors and the vector 0 are called null vectors (we won’t use that convention). • For our convenience we will consider all the vectors in Rn to be spacelike. In particular, any non-zero vector in Rn has indicator 1. Example 1.1.4. In L3 we have: (1) (4, −1, 0) is spacelike; (2) (1, 2, 3) is timelike; (3) (3, 4, 5) is lightlike. Remark. The causal type of a vector remains the same, up to a change of sign in any of its entries. Furthermore, v and λv share the same causal type for any v ∈ Rnν and λ ∈ R \ {0}. It is important to notice that the trichotomy property of the usual order relation in R ensures that each vector in Rnν has one and only one of the three causal types above. To provide some geometric intuition for L2 and L3 (where we can actually draw something), we need to know the locus of all the vectors with a fixed causal type: (1) if v = ( x, y) ∈ L2 and c ∈ R, then hv, vi L = c ⇐⇒ x2 − y2 = c, which is the equation of:

• the bisectors of the coordinated axes, if c = 0;

4  Introduction to Lorentz Geometry: Curves and Surfaces

• a hyperbola with foci on the x-axis, if c > 0; • a hyperbola with foci on the y-axis, if c < 0;

Figure 1.1: Causal type loci in L2 . (2) if v = ( x, y, z) ∈ L3 and c ∈ R, then hv, vi L = c ⇐⇒ x2 + y2 − z2 = c, which is the equation of:

• a cone, if c = 0; • a one-sheeted hyperboloid, if c > 0; • a two-sheeted hyperboloid, if c < 0. z

timelike lightlike

spacelike y

x

Figure 1.2: Causal type loci in L3 . The set of all lightlike vectors is called the lightcone (centered at 0). In particular, the lightcone in L2 is just the bisectors of the axes. In Section 1.3 we will discuss some interpretations and applications of such concepts in Special Relativity.

1.2

SUBSPACES OF Rnν

The definition of a vector subspace does not take inner products into account. Thus, subspaces of L3 are precisely the trivial ones (the origin and L3 itself), as well as the straight lines and planes containing the 0 vector. We can define the causal type of subspaces in Ln and, as a first motivation, we recall from Linear Algebra the: Proposition 1.2.1. If V is a finite-dimensional vector space with a Euclidean inner product, then, for any subspace U ⊆ V, we have dim V = dim U + dim U ⊥

and

V = U ⊕ U⊥.

Welcome to Lorentz-Minkowski Space  5

We won’t bother proving this at this moment, as very soon we will see what happens for pseudo-Euclidean vector spaces, where in general just the equation for the dimensions holds. We have a proof for this fact in Proposition 1.2.14. In any case, this situation leads us to discuss orthogonality in Ln . As in Euclidean vector spaces, we have: Definition 1.2.2 (Orthogonality). (i) Two vectors u, v ∈ Rnν are pseudo-orthogonal if hu, viν = 0. In this case we write u ⊥ν v. (ii) An arbitrary basis B of Rnν is pseudo-orthogonal if its vectors are pairwise pseudoorthogonal. (iii) A pseudo-orthogonal basis B of Rnν is pseudo-orthonormal if kvkν = 1 for each v ∈ B. (iv) Let S ⊆ Rnν be any set. The subspace of Rnν pseudo-orthogonal to S is defined as . S⊥ = {v ∈ Rnν | hu, viν = 0, for all u ∈ S} . If ν = 1, we say that pseudo-orthogonal vectors are Lorentz–orthogonal. Remark.

• In the above definition, if S = {v} is a singleton, we just write v⊥ instead of {v}⊥ . • If u and v are orthogonal with respect to the usual Euclidean inner product, we write u ⊥ E v. Similarly, we write u ⊥ L v in Lorentz-Minkowski space. In any case, if the context is clear enough, we avoid any mention to the ambient space, simply writing u ⊥ v. • Given any basis B = (u1 , . . . , un ) of Ln , where un is timelike, then B is orthonormal if and only if hui , u j i L = ηij1 , for all 1 ≤ i, j ≤ n. To define the causal character of subspaces in Ln , we recall the following general definition: Definition 1.2.3. Let V be any real vector space and B : V × V → R a symmetric bilinear form. We say that (i) B is positive-definite if B(u, u) > 0, for all u 6= 0; (ii) B is negative-definite if − B is positive-definite; (iii) B is non-degenerate if B(u, v) = 0 for all v implies u = 0; (iv) B is indefinite if there exists u, v ∈ V such that B(u, u) > 0 and B(v, v) < 0. Definition 1.2.4 (Causal Character). A vector subspace {0} 6= U ⊆ Ln is: (i) spacelike, if h·, ·i L |U is positive-definite; (ii) timelike, if h·, ·i L |U is negative-definite, or indefinite and non-degenerate; (iii) lightlike, if h·, ·i L |U is degenerate.

6  Introduction to Lorentz Geometry: Curves and Surfaces

Remark.

• In this setting, the restriction of h·, ·i L to U being positive-definite implies that all vectors in U are spacelike, although being negative-definite means that U is a straight line. If degenerate, then U has lightlike vectors, but no timelike ones. See Exercise 1.2.2. • If dim U = 0 then, by definition, U is spacelike. • We define the causal character of an affine subspace as the causal type of its vector subspace counterpart. • If ν > 1, a subspace may not have a well-defined causal type. For instance, consider the subspace spanned by three orthogonal vectors in R42 , one of each causal type. Definition 1.2.5. Let U ⊆ Rnν be an m-dimensional subspace and B = (u1 , · · · , um ) a basis for U. The Gram matrix of h·, ·iν U with respect to B is 


. GU,B = hui , u j iν 1≤i,j≤m

h u1 , u1 i ν  .. = .

··· .. . ···

h u m , u1 i ν

 h u1 , u m i ν  .. . .

hum , um iν

If the context makes the basis or subspace clear enough we write GU,B just as GU or G. Lemma 1.2.6. Let B = (u1 , . . . , um ) and C = (v1 , . . . , vm ) be linearly independent . subsets of Rnν such that U = span B = span C. Then, GU,B = A> GU, C A, where A = ( aij )1≤i,j≤n ∈ Mat(n, R) is such that u j = ∑im=1 aij vi . In particular, det GU,B = (det A)2 det GU, C, and det GU,B 6= 0 if and only if det GU, C 6= 0. Proof: It suffices to note that: * m

h ui , u j iν =

m

∑ aki vk , ∑ a` j v`

k =1

`=1

+

m

= ν

∑

aki a` j hvk , v` iν ,

k,`=1

which is precisely the (i, j)-entry in the matrix A> GU, C A. The remaining claim follows from applying det to both sides of GU,B = A> GU, C A. Proposition 1.2.7. Let u1 , . . . , um ∈ Rnν be vectors such that the matrix (hui , u j iν )1≤i,j≤m is invertible. Then {u1 , . . . , um } is linearly independent. Proof: Let a1 , . . . , am ∈ R such that ∑im=1 ai ui = 0. Applying h·, u j iν to both sides we obtain ∑im=1 ai hui , u j iν = 0 for all 1 ≤ j ≤ m. In matrix form, this is written as: 

        · · · h u1 , u m i ν a1 0 a1 0  ..   ..    ..   ..  .. ..   .  =  .  =⇒  .  =  .  , . . h u m , u1 i ν · · · h u m , u m i ν am 0 am 0 {z } | invertible

h u1 , u1 i ν  ..  .

hence {u1 , . . . , um } is linearly independent.

Welcome to Lorentz-Minkowski Space  7

Remark. The converse of the above result does not hold. The Gram matrix of a light ray4 is the zero matrix (0), but if v 6= 0, then {v} is linearly independent. We will see soon that, under convenient conditions, Lemma 1.2.6 is a partial converse of the previous result (see Proposition 1.2.21). In what follows, we will study subspaces of L3 , in terms of the causal type of their vectors. After that, we will present some results relating orthogonal complements and causal characters, which hold in Ln . From Definition 1.2.4, the origin and Ln are spacelike and timelike, respectively. Proposition 1.2.8. Let r ⊆ Ln be a straight line passing through the origin. Then the causal type of r is the same of any vector giving its direction. Proof: See Exercise 1.2.3. Now we provide a criterion to know the causal character of a plane Π in L3 in terms of the coefficients of its general equation, written as Π : ax + by + cz = 0. If c = 0, the plane contains the vector e3 , hence it is timelike. If c 6= 0 and u = ( x, y, z) ∈ Π, we have − ax −by z= and c  − ax − by 2 hu, ui L = x + y − c 2 a 2ab b2 = x2 + y2 − 2 x2 − 2 xy − 2 y2 c c c     a2 2ab b2 2 = 1 − 2 x − 2 xy + 1 − 2 y2 . c c c 2



2

In matrix notation: a2 1 −
 c2 hu, ui L = x y   ab − 2 c 

 ab   c2   x ,  y b2 1− 2 c

−

and the matrix

 ab a2 − 2  1− 2 .  c c  GΠ =    ab b2 − 2 1− 2 c c 

 of Π. is the Gram matrix of h·, ·i L Π in the basis 1, 0, − bc , 0, 1, − ac To move on, we need the following: 

(‡)

Definition 1.2.9. A symmetric matrix A ∈ Mat(n, R) is (i) semi-positive (resp. positive-definite), if h Au, ui E ≥ 0 (resp. h Au, ui E > 0) for all u 6= 0. (ii) semi-negative (resp. negative-definite), if − A is semi-positive (resp. − A is positivedefinite); (iii) indefinite, if it is none of the above. 4A

straight line in the direction of a lightlike vector.

8  Introduction to Lorentz Geometry: Curves and Surfaces

The following criterion characterizes semi-positive and positive-definite matrices: Theorem 1.2.10 (Sylvester’s Criterion). A real symmetric matrix is positive-definite if n and only if all of its principal minors are positive. In other words, if A = ( aij )i,j =1 then
. k A is positive-definite if and only if ∆k = det ( aij )i,j=1 > 0 for all 1 ≤ k ≤ n. Proof: See [32, p. 439] or follow the steps in Exercise 1.2.8. Theorem 1.2.11. Let Π ⊆ L3 be a plane, Π : ax + by + cz = 0, such that c 6= 0. If GΠ is its Gram matrix, as in (‡), the following hold: (i) det GΠ > 0 ⇐⇒ Π is spacelike; (ii) det GΠ < 0 =⇒ Π is timelike; (iii) det GΠ = 0 ⇐⇒ Π is lightlike. Proof: Let us calculate det GΠ :     a2 b2 ab 2 det GΠ = 1 − 2 1− 2 − − 2 c c c 2 2 2 2 2 2 b a b a a b = 1− 2 − 2 + 4 − 4 c c c c  2  a + b2 = 1− c2  2 2 Also notice that tr GΠ = 2 − a c+2 b = 1 + det GΠ . Since GΠ is a 2 × 2 matrix, its characteristic polynomial is: 

cGΠ (t) = t2 − tr GΠ t + det GΠ

= t2 − (1 + det GΠ ) t + det GΠ , whose roots are 1 and det GΠ . Hence, if det GΠ > 0, all the eigenvalues of GΠ are positive and h·, ·i L |Π is positive-definite: Π is spacelike. If det GΠ < 0, h·, ·i L |Π is indefinite, then Π is timelike. Finally, if det GΠ = 0, h·, ·i L |Π is semi-positive, then Π is lightlike. We now establish a result that provides a geometric interpretation for causal character of planes. Theorem 1.2.12. Let Π ⊆ L3 be a plane of general equation Π : ax + by + cz = 0 and n = ( a, b, c) be its Euclidean normal vector. Then: (i) Π is spacelike ⇐⇒ n is timelike; (ii) Π is timelike ⇐⇒ n is spacelike; (iii) Π is lightlike ⇐⇒ n is lightlike. Proof: Suppose c = 0. Then, as before, Π is timelike and n is ( a, b, 0), a spacelike vector. Now, let c 6= 0 and GΠ be the Gram matrix of h·, ·i L Π . For each causal type:  2 2 (i) Π is spacelike ⇐⇒ 0 < det GΠ = 1 − a c+2 b

⇐⇒ hn, ni L = a2 + b2 − c2 < 0 ⇐⇒ n is timelike;

Welcome to Lorentz-Minkowski Space  9

(ii) Π is timelike ⇐⇒ 0 > det GΠ = 1 −



a2 + b2 c2



⇐⇒ hn, ni L = a2 + b2 − c2 > 0 ⇐⇒ n is spacelike; (iii) Π is lightlike ⇐⇒ 0 = det GΠ = 1 −



a2 + b2 c2



⇐⇒ hn, ni L = a2 + b2 − c2 = 0 ⇐⇒ n is lightlike.

Remark. In the statement we could replace “Euclidean normal” with “Lorentz-normal”. Why? Example 1.2.13. (1) The plane Π1 : 3x − 4y + 5z = 0 is lightlike, since its Euclidean normal vector, (3, −4, 5), is lightlike. (2) The plane Π2 : x − 2y + 5z = 0 is spacelike, since its Euclidean normal vector, (1, −2, 5), is timelike. (3) The line L : X (t) = t(1, 2, 3), t ∈ R is timelike, since (1, 2, 3) is a timelike vector. The following images correspond to the examples above:

Figure 1.3: Subspaces of L3 . Remark. Every lightlike plane is tangent to the lightcone, the spacelike planes are “outside” of that cone and timelike planes cut it in two “lightrays”, that is, lightlike lines. Those particular results about planes in L3 suggest the analysis of the Gram matrix of the restriction of h·, ·i L to subspaces of Ln , in order to achieve similar results in higher dimensions. We aim for such results in what follows. Proposition 1.2.14. Let U ⊆ Rnν be a vector subspace. Then dim U + dim U ⊥ = n

and U = (U ⊥ )⊥ .

10  Introduction to Lorentz Geometry: Curves and Surfaces

Proof: Let U ∗ be the dual space of U and consider the map Φ : Rnν → U ∗ given by Φ( x ) = h x, ·i U . This map is linear, surjective (since h·, ·iν is non-degenerate), and its kernel is U ⊥ . Since dim U = dim U ∗ , the dimension formula follows from the rank-nullity theorem. Said formula applied twice also says that dim U = dim(U ⊥ )⊥ , so U ⊆ (U ⊥ )⊥ implies U = (U ⊥ )⊥ . From this, we have the: Corollary 1.2.15. Let U ⊆ Rnν be a subspace. Then U ⊕ U ⊥ = Rnν if and only if U is non-degenerate (i.e., h·, ·iν U is non-degenerate). It also follows that U is non-degenerate if and only if U ⊥ is also non-degenerate. Proof: From dim(U + U ⊥ ) = dim U + dim U ⊥ − dim(U ∩ U ⊥ ) = n − dim(U ∩ U ⊥ ) it follows that U + U ⊥ = Rnν if and only if U ∩ U ⊥ = {0}, which in turn is equivalent to h·, ·iν U being non-degenerate. In the same way, we have the following relation between causal types and orthogonality: Corollary 1.2.16. A vector subspace U ⊆ Ln is spacelike if and only if U ⊥ is timelike. Proof: Assume that U is spacelike. So U is non-degenerate and we write Ln = U ⊕ U ⊥ . Then U ⊥ must necessarily contain a timelike vector because Ln does — more precisely, take a timelike v ∈ Ln and write v = x + y with x ∈ U and y ∈ U ⊥ , so that the relation h x, xi L + hy, yi L = hv, vi L < 0 with h x, xi L ≥ 0 forces y ∈ U ⊥ to be timelike. Conversely, assume now that U is timelike, and take u ∈ U timelike. Since U ⊥ ⊆ u⊥ , it suffices now to show that u⊥ is spacelike. We know again from Corollary 1.2.15 that u⊥ is not lightlike, and if we have v ∈ u⊥ timelike, the plane spanned by u and v in Ln has dimension 2 while being negative-definite, which is impossible. The following is a restatement of Corollary 1.2.15: Corollary 1.2.17. If U ⊆ Ln is a lightlike subspace, so is U ⊥ . Remark. In the definition of orthogonality we avoid saying “orthogonal complement” since, unlike what happens for the Euclidean case, for a subspace U ⊆ Ln , we do not necessarily have U + U ⊥ = Ln . As an example,  consider the subspace U ⊆ L3 given by  U = ( x, y, z) ∈ L3 | y = z . Then U ⊥ = span (0, 1, 1) ⊆ U. Sylvester’s criterion from Theorem 1.2.10 alone is not enough to decide the causal character of a subspace U ⊆ Ln in terms of the Gram matrix of h·, ·i L U . To remedy this we have the following stronger version of that criterion: Theorem 1.2.18 (Sylvester on steroids). Let A ∈ Mat(n, R) be a symmetric matrix and ∆k its k-th order leading principal minor determinant. Then: (i) A is positive-definite (resp. semi-positive) if and only if ∆k > 0 (resp. ∆k ≥ 0) for all 1 ≤ k ≤ n; (ii) A is negative-definite (resp. semi-negative) if and only if (−1)k ∆k > 0 (resp. (−1)k ∆k ≥ 0) for all 1 ≤ k ≤ n;

Welcome to Lorentz-Minkowski Space  11

(iii) A is indefinite if and only if one of the following holds:

• there exists an even number 1 ≤ k ≤ n such that ∆k < 0, or; • there exist distinct odd numbers 1 ≤ k, ` ≤ n ∆k > 0 and ∆` < 0. Proof: (i) This is exactly Theorem 1.2.10. (ii) It follows directly from applying Theorem 1.2.10 to − A, bearing in mind that the k-th leading principal minor of − A is (−1)k ∆k . (iii) Suppose that none of the given conditions hold, that is, all of the leading principal minors of even order are non-negative and all of the leading principal minors of odd order are simultaneously non-negative or simultaneously non-positive. If the odd minors are non-negative then, from item (i), A is semi-positive; if they are non-positive, from item (ii), A is semi-negative. For the converse, start supposing that first condition holds. By item (i), A is not semi-positive and, from item (ii), A can’t be semi-negative. Hence A is indefinite. If the second condition holds and A is semi-positive, minors of order k and ` are both positive, and if A is negative, both are negative. Hence A must be again indefinite.

The following example illustrates the above criterion to decide the causal type of a vector subspace U ⊆ Ln by analyzing the Gram matrix of h·, ·i L U . Example 1.2.19. Consider L4 . . (1) Let U = span{v1 , v2 , v3 }, where v1 = (2, 0, 0, 1), v2 = (0, 1, 2, 0) and v3 = (−1, −2, 0, 1). The Gram matrix of h·, ·i L U , relative to the basis (v1 , v2 , v3 ) is   3 0 −3 5 −2 , G= 0 −3 −2 4 whose leading principal minors are, respectively, ∆1 = 3, ∆2 = 15 and ∆3 = 3. By Sylvester’s Criterion, U is spacelike. . (2) For U = span{v1 , v2 , v3 }, with v1 = (1, −1, 1, 1), v2 = (0, 1, −1, −1) and v3 = (0, 0, 1, 1), the Gram matrix of h·, ·i L U , relative to the basis (v1 , v2 , v3 ) is   2 −1 0 G =  −1 1 0 . 0 0 0 The leading principal minors are ∆1 = 2, ∆2 = 1 and ∆3 = 0. Since there are no negative minors, G is neither indefinite nor negative-definite, hence U is not timelike, that is, it must be spacelike or lightlike. As ∆3 = 0, U is not positive-definite, so it must be lightlike.

12  Introduction to Lorentz Geometry: Curves and Surfaces

. (3) Finally, if U = span{v1 , v2 , v3 }, where v1 = (2, 1, 0, 1), v2 = (0, 1, 1, −1) and v3 = (1, 0, 0, −1), the Gram matrix of h·, ·i L U , relative to the basis (v1 , v2 , v3 ) is  4 2 3 G = 2 1 −1 . 3 −1 0 

Now ∆1 = 4, ∆2 = 0 and ∆3 = −25. Since ∆1 and ∆3 have opposite signs, G is indefinite and U is timelike. In particular, Corollaries 1.2.16 and 1.2.17 provide a generalization of Theorem 1.2.12 to hyperplanes in Ln : Theorem 1.2.20. Let Π ⊆ Ln be a hyperplane of general equation Π : a1 x1 + · · · + a n x n = 0 and n = ( a1 , . . . , an ) its Euclidean normal vector. Then (i) Π is spacelike ⇐⇒ n is timelike; (ii) Π is timelike ⇐⇒ n is spacelike; (iii) Π is lightlike ⇐⇒ n is lightlike; Remark. Just as in Theorem 1.2.12, we can replace “Euclidean normal” by “Lorentznormal”. Now we are ready to state a converse for Proposition 1.2.7 (p. 6). Proposition 1.2.21. Let {u1 , . . . , um } ⊆ Rnν be a linearly independent set of vectors whose span is non-degenerate. Then the Gram matrix of such vectors is invertible. Proof: Let B = (u1 , . . . , um ) and U = span B. From Proposition 1.2.15 (p. 10), since U is non-degenerate, we may write Rnν = U ⊕ U ⊥ . If C is any basis for U ⊥ , then the (ordered) union B ∪ C is a basis for Rnν , and the Gram matrix GRnν ,B∪ C is block-diagonal, and thus det GU,B det GU ⊥ , C = det GRnν ,B∪ C 6= 0 implies that det GU,B 6= 0, as required. Theorem 1.2.22. Let {0} 6= U ⊆ Rnν be a non-degenerate vector subspace. Then U has a Lorentz-orthonormal basis. Proof: By induction on k = dim U. From non-degeneracy of U, there exists v ∈ U such that hv, viν 6= 0, so v/kvkν is unit. With that in mind, it is enough to show that for any orthonormal subset S ⊆ U, with less than k vectors, we can add another vector in such a manner that the resulting set is also orthonormal. Since S spans a non-degenerate subspace, its orthogonal complement is non-degenerate and non-trivial. It then suffices to add a unit vector in S⊥ to S (and such a vector exists by the above argument).

Welcome to Lorentz-Minkowski Space  13

The previous result can also be stated and proved by an algorithm: Theorem 1.2.23 (Gram-Schmidt Orthogonalization Process). For any linearly independent vectors v1 , . . . , vk ∈ Rnν such that for all i ∈ {1, . . . , k} we have that span(v1 , . . . , vi ) is non-degenerate (i.e., not lightlike), there exists ve1 , . . . , vek ∈ Rnν pairwise Lorentzorthogonal such that   span v1 , . . . , vk = span ve1 , . . . , vek . Proof: Again by induction on k. For k = 1 just take ve1 = v1 . Now, assume that the result is valid for some k, that is, suppose that there exists ve1 , . . . , vek ∈ Rnν with the stated properties. None of the vei is lightlike. In fact, if at least one of them were lightlike, we would have vei orthogonal to all vectors, from ve1 to vg i −1 , and even to itself,  of the previous  so that span v1 , . . . , vi = span ve1 , . . . , vei would be degenerate, contradicting our assumption. So we can set: k hvk+1 , vei iν . v ] ve . k +1 = v k +1 − ∑ hvei , vei iν i i =1

  It is clear that span v1 , . . . , vk+1 = span ve1 , . . . , v ] k +1 , and if j ∈ {1, . . . , k }, we have: *

hv ] ej iν = k +1 , v

k

hv , ve iν vk+1 − ∑ k+1 i vei , vej hvei , vei iν i =1

+ ν

k

hvk+1 , vei iν hvei , vej iν hvei , vei iν i =1

= hvk+1 , vej iν − ∑ = hvk+1 , vej iν −

hvk+1 , vej iν hvej , vej iν hvej , vej iν

= hvk+1 , vej iν − hvk+1 , vej iν = 0.

Remark. The expression for v ] k +1 can also be written as k hvk+1 , vei iν . v ] vei , ei k +1 = v k +1 − ∑ e v kvei k2ν i =1

which may be more familiar, up to sign adjustments. Example 1.2.24. Consider the vectors v1 = (1, 2, 1, 0),

v2 = (0, 2, 0, −1) and v3 = (0, 0, 0, 1)

in L4 . Straightforward computations show that they are linearly independent, span{v1 } . and span{v1 , v2 } are spacelike, and U = span{v1 , v2 , v3 } is timelike. Theorem 1.2.23 then ensures the existence of a Lorentz-orthogonal basis for U from the given vectors. Taking ve1 = v1 , do:   4 2 2 −2 ve2 = (0, 2, 0, −1) − (1, 2, 1, 0) = − , , , −1 , 6 3 3 3

14  Introduction to Lorentz Geometry: Curves and Surfaces

and

0 1 ve3 = (0, 0, 0, 1) − (1, 2, 1, 0) − 6 1/3



2 2 −2 − , , , −1 3 3 3



= (2, −2, 2, 4).

The vectors ve1 , ve2 and ve3 are pairwise Lorentz–orthogonal, spanning the same hyperplane in L4 as the original vectors. Theorem 1.2.25. Let u1 , . . . , uν+1 ∈ Rnν be pairwise orthogonal lightlike vectors. Then (u1 , . . . , uν+1 ) is linearly dependent. Proof: Splitting Rnν as Rnν = Rn−ν × Rνν , let (ei )in=1 be the canonical basis of Rnν . Write ν

u j = x j + ∑ aij en−ν+i ,

1 ≤ j ≤ ν + 1,

i =1

according to the above split (with all the x j being spacelike and orthogonal to each timelike vector in the canonical basis). It follows that ν

0 = h ui , u j iν = h xi , x j iν −

ν

∑

∑ aki akj , 1 ≤ i, j ≤ ν + 1.

aki akj =⇒ h xi , x j iν =

k =1

k =1

Consider now the linear map A : Rν+1 → Rν , whose matrix in the canonical bases has the above aij as entries, and take b = (b1 , . . . , bν+1 ) 6= 0 in ker A (this is possible for dimension reasons). We have + * ν +1

ν +1

∑

bi x i ,

i =1

∑

ν +1

bj x j

=

j =1

∑

bi b j h x i , x j i ν

i,j=1

ν

ν +1

=

∑

ν

bi b j

i,j=1

∑ aki akj

k =1

>

= ( Ab) ( Ab) = 0. Since the linear combination ∑νj=+11 b j x j is spacelike, the preceding computation shows that ∑νj=+11 b j x j = 0. The entries in b also testify that (u1 , . . . , uν+1 ) is linearly dependent. In fact: ! ν +1

∑

ν +1

bj u j =

j =1

∑

j =1

ν

ν +1

i =1

j =1

bj x j + ∑

∑ bj aij

en−ν+i = 0,

as the second term also vanishes, since b ∈ ker A. In particular, we have the: Corollary 1.2.26. Two lightlike vectors in Ln are Lorentz-orthogonal if and only if they are proportional. Corollary 1.2.27. If U ⊆ Ln is a lightlike subspace, then dim(U ∩ U ⊥ ) = 1. Proof: Let u, v ∈ U ∩ U ⊥ . Then hu, vi L = 0, hence u and v are linearly dependent, and therefore dim(U ∩ U ⊥ ) ≤ 1. On the other hand, if U is lightlike, we have dim U > 0. In this way take a lightlike vector (non-zero, by definition) v ∈ U. Then hv, vi L = 0 and v ∈ U ⊥ , showing that dim(U ∩ U ⊥ ) ≥ 1. Remark. The result above does not hold in Rnν , with ν > 1. As an example, take any subspace containing two pairwise orthogonal and linearly independent lightlike vectors. Since its orthogonal complement contains itself, we have dim(U ∩ U ⊥ ) ≥ 2.

Welcome to Lorentz-Minkowski Space  15

Corollary 1.2.28. Let u, v ∈ Ln be linearly independent lightlike vectors. Then u + v and u − v are not lightlike and have distinct causal types. Proof: Note that hu ± v, u ± vi L = ±2hu, vi L 6= 0. In particular, any degenerate subspace U ⊆ Ln contains only one lightray. Proposition 1.2.29. Let U ⊆ Ln be a degenerate subspace. Then U admits an orthogonal basis. Proof: Let (v1 , . . . , vk ) ⊆ Ln be a basis of U such that vk is a lightlike vector (therefore the remaining vectors are spacelike). Applying the Gram-Schmidt process to {v1 , . . . , vk−1 } we get k − 1 pairwise orthogonal spacelike vectors. Such vectors are orthogonal to vk . In fact, for any (unit) spacelike vector u ∈ U such that hu, vk i L 6= 0, we may take λ ∈ R satisfying hu + λvk , u + λvk i L = 1 + 2λhu, vk i L < 0, a contradiction with the degeneracy of U. Remark. The preceding result still holds for subspaces of Rnν , but this requires a slightly more elaborate proof (see [31]). Lemma 1.2.30. Let U ⊆ Rnν to be a vector subspace. If U admits an orthonormal basis, then it is non-degenerate. Proof: Let (v1 , . . . , vm ) be an orthonormal basis for U and x ∈ U. Then x = ∑im=1 xi vi . Suppose that h x, v j iν = 0 for all j = 1, . . . , m. Then, since e j = hv j , v j iν ∈ {−1, 1}, we have e j x j = 0, leading to x j = 0 and x = 0. This means that U is non-degenerate. Lemma 1.2.31. If U ⊆ Ln is a lightlike hyperplane, then U ⊥ ⊆ U. Proof: Let u1 , . . . , un ∈ Ln be vectors such that U = span {u1 , . . . , un−1 } and with U ⊥ = span {un }. Since U is lightlike, it holds that dim(U ∩ U ⊥ ) = 1. Hence dim(U + U ⊥ ) = dim U + dim U ⊥ − dim(U ∩ U ⊥ ) = n − 1, as dim U = n − 1 and dim U ⊥ = 1. However, U + U ⊥ = span {u1 , . . . , un }, and from linear independence of {u1 , . . . , un−1 }, it follows that un ∈ U.

Figure 1.4: A lightlike vector contained in its own orthogonal “complement”.

16  Introduction to Lorentz Geometry: Curves and Surfaces

Remark. The above result no longer holds if dim U < n − 1 in Ln . Consider the following counter-example: let   U = span (0, 0, 1, 1), (1, 0, 0, 0) and U ⊥ = span (0, 1, 0, 0), (0, 0, 1, 1) in L4 . Note that (0, 1, 0, 0) 6∈ U. Lemma 1.2.32 (Orthonormal expansion). Let (u1 , . . . , un ) be a pseudo-orthonormal basis for Rnν . Then, every v ∈ Rnν can be written as n

v=

∑ eui hv, ui iν ui .

i =1

Proof: Start writing v = ∑nj=1 v j u j , for suitable v1 , . . . , vn ∈ R and suppose that the last ν vectors in the given basis are timelike. Applying h·, ui iν to the previous equality we have: * + n

hv, ui iν =

n

∑ v j u j , ui

j =1

= ν

∑ v j h u j , ui iν =

j =1

n

∑ v j ηijν = eui vi .

i =1

Hence, vi = eui hv, ui iν . Corollary 1.2.33. In Ln there are no pairwise orthogonal timelike vectors, as well as no timelike vectors orthogonal to lightlike ones. Furthermore, any orthogonal basis must contain precisely one timelike vector with the remaining ones being spacelike. Proof: See Exercise 1.2.13. Proposition 1.2.34. Let v ∈ Ln be a unit timelike vector. Then kvk E ≥ 1. Proof: Writing v = (v1 , . . . , vn ), we have:

hv, vi L = v21 + · · · + v2n−1 − v2n = −1 =⇒ v2n = 1 + v21 + · · · + v2n−1 . Therefore

  kvk2E = v21 + · · · + v2n = 1 + 2 v21 + · · · + v2n−1 ≥ 1.

Remark. The previous proposition says the “Euclidean eyes” see a timelike vector larger than “Lorentzian eyes” see it (kvk E ≥ kvk L ). In particular, this justifies us drawing a timelike unit vector, Lorentz-orthogonal to a spacelike plane, with Euclidean length greater than 1.

Exercises Exercise 1.2.1 (Warmup). Decide the causal type of the following vectors, lines, and planes of L3 : (a) u = (1, 3, 2); (b) v = (3, 0, −3);

Welcome to Lorentz-Minkowski Space  17

(c) r (t) = (−2, −1, 2) + t(0, 2, 3), t ∈ R; (d) r (t) = (10, 0, 0) + t(3, 2, −1), t ∈ R; (e) Π : 3x − 4y + 5z = 10; (f) Π : x = 5y. Recall that it is also usual in the literature to adopt the convention . h x, yi L = − x1 y1 + x2 y2 + x3 y3 . What would be the answers in that convention? Exercise† 1.2.2. Let {0} 6= U ⊆ Ln be a vector subspace. Show that (a) If h·, ·i L U is negative-definite, then U is a line. Hint. Show that if dim U ≥ 2, then h·, ·i L U cannot be negative-definite. For this, take {u, v} ⊆ U linearly independent, with timelike v. If u is spacelike or lightlike we are done. If u is timelike, write a “horizontal” (spacelike) vector by combining v and a convenient rescaling of u. (b) If U has a timelike vector, then U is non-degenerate. In particular, if h·, ·i L U is indefinite, it is also non-degenerate. Hint. If U is degenerate, take u, v ∈ U such that u is lightlike, v is timelike, and hu, vi L = 0. On one hand, note that h au + bv, au + bvi L ≤ 0 for all a, b ∈ R and, on the other hand, repeat the argument in the hint above to have a spacelike combination of u and v. Use this to show that: (c) U is spacelike if and only if all of its vectors are spacelike; (d) U is timelike if and only if has some timelike vector; (e) U is lightlike if it has some lightlike vector but no timelike vectors. Exercise 1.2.3. Prove Proposition 1.2.8 (p. 7). Exercise 1.2.4. Show that a subspace U ⊆ Ln is lightlike if and only if its intersection with the lightcone in Ln is precisely a light ray through the origin. Exercise 1.2.5. Show that if U ⊆ Ln is a 2-dimensional timelike subspace, then it intersects the lightcone of Ln in two lightrays passing through the origin. Exercise 1.2.6. Exhibit a Penrose basis for Ln , that is, a basis P = {u1 , . . . , un } such that each ui is lightlike and hui , u j i L = −1 for i 6= j. Write down the Gram matrix GLn ,P. Does this matrix depend on the Penrose basis P? Hint. Try to solve this for L2 and L3 first. Exercise† 1.2.7. Let U ⊆ Rnν be a vector subspace. An orthogonal projection of Rnν onto U is a linear operator that fixes U (pointwise) and collapses U ⊥ onto the set {0}.

18  Introduction to Lorentz Geometry: Curves and Surfaces

(a) Show that U is non–degenerate if and only if there exists an orthogonal projection of Rnν onto U, which is unique. If v ∈ Rnν , we write the orthogonal projection of v in U as projU v. (b) Show that, if dim U = k and {u1 , . . . , uk } is an orthonormal basis for U, then k

projU v =

hv, u iν

∑ hui , uii iν ui .

i =1

Exercise 1.2.8. This exercise is a guide to prove Sylvester’s Criterion. For this, let A ∈ Mat(n, R) be a symmetric matrix. (a) A is positive-definite if and only if all of its eigenvalues are positive. (b) If W ⊆ Rn is a k-dimensional subspace, then every subspace of Rn with dimension m > n − k has non-trivial intersection with W. Hint. Compute the dimension of the intersection. (c) If h Ax, xi E > 0 for all non–zero x in a k-dimensional subspace W ⊆ Rn , then A has at least k positive eigenvalues. (d) Deduce the criterion using induction in n. Exercise 1.2.9. Consider the following vectors in L4 v1 = (1, 0, 2, 0),

v2 = (0, 1, −1, 0) and v3 = (1, 2, 0, 0).

. The subspace U = span {v1 , v2 , v3 } is spacelike but   5 − 2 1
hvi , v j i L 1≤i,j≤3 = −2 2 2 1 2 5 has principal leading minors 5, 6 and 0. Proceeding like in Example 1.2.19 (p. 11), U would be lightlike. Explain this ostensible contradiction. Exercise 1.2.10. Consider in L5 the vectors v1 = (1, 2, 0, 0, 1),

v2 = (0, 0, 1, 1, 0) and v3 = (1, 0, 1, 0, 0).

(a) Show that such vectors are linearly independent. . (b) Show that span{v1 }, span{v1 , v2 } and U = span{v1 , v2 , v3 } are spacelike subspaces of L5 (hence non-degenerate). (c) Exhibit an orthonormal basis of L5 containing a basis for U. Exercise† 1.2.11. Write the proof of Theorem 1.2.25 (p. 14) in the particular case ν = 1. Exercise 1.2.12 (Triangles of light). Show that there are no pairwise linearly independent lightlike vectors u, v, w ∈ Ln such that u + v + w = 0. Generalize. Exercise† 1.2.13. In Ln , show that: (a) there are no pairwise orthogonal timelike vectors;

Welcome to Lorentz-Minkowski Space  19

(b) there are no timelike vectors orthogonal to lightlike ones; (c) any orthonormal basis for Ln has exactly one timelike vector and the other ones are spacelike. Hint. Use Lemma 1.2.32 (p. 16) and the item (a) above, assuming that all vectors in a orthonormal basis are spacelike.

1.3

CONTEXTUALIZATION IN SPECIAL RELATIVITY

In this section we give an interpretation of previous results in the setting of Special Relativity. We will focus the discussion on Lorentz-Minkowski space L4 , for it is natural to consider three dimensions for space and one for time. Fixing the inertial frame given by the canonical basis, write the coordinates of an event as p = ( x, y, z, t), so that

h p, pi L = x2 + y2 + z2 − t2 . In Physics, the expression above is usually written as x2 + y2 + z2 − (ct)2 , where c is the speed of light in vacuum. In Mathematics, we usually work with geometric units 5 , where c = 1. . Let p, q ∈ L4 be any given events. If v = q − p = (∆x, ∆y, ∆z, ∆t) is the vector joining p to q and ∆t 6= 0, the causal type of v helps us understand the interaction between both events. Note that

hv, vi L = (∆x )2 + (∆y)2 + (∆z)2 − (∆t)2 !  2  2  2 ∆x ∆y ∆z 2 = (∆t) + + −1 ∆t ∆t ∆t = (∆t)2 (kv ek2E − 1) = (∆t)2 (kv ek E + 1)(kv e k E − 1), . ∆y ∆z
3 where v e = ∆x ∆t , ∆t , ∆t ∈ R is the spatial velocity vector between the events p and q. In particular, observe that hv, vi L and kv ek E − 1 share the same sign. Hence, (1) if v is timelike, then ∆t 6= 0 and kv ek E < 1. When ∆t > 0, the event p may influence the event q, and ∆t < 0 says that event p may have been influenced by q — such influences may manifest themselves through propagation of material waves; (2) if v is lightlike and ∆t 6= 0, then kv ek E = 1, hence the influence of p over q (or the opposite, depending on the sign of ∆t as above), is realized as the propagation of an electromagnetic wave, such as a light signal emitted by p and received by q; (3) if v is spacelike, with ∆t 6= 0, there is no influence relation between p and q, since kv ek E > 1, that is, the speed required to move spatially from one event to another is greater than the speed of light. In other words, not even a photon or neutrino is fast enough to witness both events. 5 See

[54, p. 162] for details.

20  Introduction to Lorentz Geometry: Curves and Surfaces

q q

q

P

Figure 1.5: Picture of the situation above, omitting one spatial dimension. The above considerations lead to the following definition in Ln : Definition 1.3.1. A timelike or lightlike vector v = (v1 , . . . , vn ) ∈ Ln is future-directed (resp. past-directed) if vn > 0 (resp. vn < 0). Clearly v is future-directed if it is timelike or lightlike and hv, en i L < 0. Hence, we have: Definition 1.3.2. The lightcone centered in p ∈ Ln is the set . CL ( p) = {q ∈ Ln | q − p is lightlike}. The timecone centered in p ∈ Ln is .  CT ( p) = q ∈ Ln | q − p is timelike}. Each of the cones CL ( p) and CT ( p) may be split in two disjoint components, called future/past light/timecones, denoted by CL+ ( p), CL− ( p), CT+ ( p) and CT− ( p). For example, .  CL+ ( p) = q ∈ CL ( p) | q − p is lightlike and future-directed . In Physics books, when discussing Special Relativity, it is usual to consider, instead of L4 , an arbitrary four-dimensional vector space equipped with a pseudo-Euclidean inner product of index 1, which is the largest dimension of a subspace for which the restriction of the inner product to it is negative-definite (for example, Exercise 1.2.2 says that h·, ·i L has index 1). This is done to avoid the choice of a preferred frame of reference (like the canonical basis in L4 ). Results shown so far could be written in this setting with no extra effort. In Special Relativity, using either L4 or any vector space in the above conditions, we are modelling a spacetime free of gravity, in the vacuum. In General Relativity, to consider gravitation, the ambient spaces studied are no longer vector spaces, being so-called manifolds. In such an ambient, light rays are not necessarily straight lines. In this broader context it is possible to define, in a mathematically precise way, the future and past (chronological or absolute) of an event. Definition 1.3.3 (Informal). Let p be a point in spacetime. The chronological future of p and the absolute future (causal) of p are, respectively, . I + ( p) = { q in spacetime | there exists a future-directed timelike curve joining p and q} and . J ( p) = { q in spacetime | there exists a future-directed timelike or lightlike curve joining p and q} . +

The chronological past, I − ( p), and the absolute past, J − ( p), are defined similarly.

Welcome to Lorentz-Minkowski Space  21

We will study curves in Chapter 2, emphasizing the three-dimensional spaces R3 and L3 , where most of the theory in this text will be developed. The above definitions are valid in General Relativity as well as in Special Relativity. In the latter, we have I ± ( p) = CT± ( p) in L4 , allowing us to write the: Definition 1.3.4. Let S ⊆ Ln be any subset. The chronological future of S is the set . [ + I + (S) = CT ( p ) . p∈S

Similarly, one defines the chronological past of S, I − (S).

Figure 1.6: The future of a set. With those concepts in hand, we can define the chronological and causal orderings in Ln : Definition 1.3.5. Let p, q ∈ Ln . We say that p chronologically precedes q if q ∈ CT+ ( p), and this is denoted by p  q. Furthermore, p causally precedes q if q ∈ CT+ ( p) ∪ CL+ ( p), and this will be denoted by p 4 q. In the following, we present some basic properties of future and past sets in terms of chronological precedence. Proposition 1.3.6. Let p ∈ Ln . Then hu − p, v − pi L < 0 for all u, v ∈ CT+ ( p). Proof: It follows from the definition that u, v ∈ CT+ ( p) if and only if both u − p and v − p are in CT+ (0). Hence, without loss of generality, we may assume that p = 0. If u, v ∈ CT+ (0), recall that Ln = e⊥ n ⊕ span { en } and write u = x + aen

and

v = y + ben ,

for certain spacelike vectors x, y ∈ Ln and positive numbers a and b. We have

hu, ui L = h x + aen , x + aen i L < 0 =⇒ h x, xi L − a2 < 0, whence h x, xi L < a2 . Similarly hy, yi L < b2 . Since h·, ·i L e⊥ is a Euclidean inner product, n

the standard Cauchy-Schwarz inequality holds in e⊥ n : |h x, y i L | ≤ k x k L k y k L . Then,

hu, vi L = h x + aen , y + ben i = h x, yi − ab ≤ k xk L kyk L − ab < ab − ab = 0.

22  Introduction to Lorentz Geometry: Curves and Surfaces

Remark. The previous result extends to vectors in CT+ ( p) ∪ CL+ ( p), but the inequality is no longer strict. When does the equality hold? In the Lorentzian ambient space, the functions cosh and sinh play the same role that trigonometric functions do in the circle, when the inner product is Euclidean. You can recall some basic properties of such functions in Exercise 1.3.1. Now is a good moment to do so, if you are not familiar with them, since we will use those properties in the next proposition and in several of the following ones. Proposition 1.3.7. Let p ∈ Ln . Given any two vectors u, v ∈ CT ( p) such that hu − p, v − pi L < 0, then u and v are both in CT+ ( p) or in CT− ( p). Proof: Just as in the previous proposition, we can assume that p = 0. Also, suppose that u ∈ CT+ (0) and kuk L = kvk L = 1. Now our aim is to prove that v ∈ CT+ (0), assuming that hu, vi L < 0. If u and v are parallel, then v = u or v = −u and the assumption leads to v = u ∈ CT+ (0). If {u, v} is linearly independent, take an orthonormal basis {w1 , w2 } for the plane spanned by u and v, where w1 is spacelike and w2 is timelike. Exchanging signs of the wi if needed, we can assume that w2 is future-directed and hw1 , en i L ≤ 0. Write u = aw1 + bw2 for some a, b ∈ R. The function sinh : R → R is bijective and hence there exists a unique θ1 ∈ R such that a = sinh θ1 . Hence

hu, ui L = sinh2 θ1 − b2 = −1, and then b2 = cosh2 θ1 . Since u and w2 are future-directed, the previous proposition ensures that −b = hu, w2 i L < 0, that is, b > 0, whence u = sinh θ1 w1 + cosh θ1 w2 . The same argument shows that v = sinh θ2 w1 + e cosh θ2 w2 for some θ2 ∈ R and e ∈ {−1, 1}. Then we have

hu, vi L = sinh θ1 sinh θ2 − e cosh θ1 cosh θ2 = −e (cosh θ1 cosh θ2 − e sinh θ1 sinh θ2 ) = −e (cosh θ1 cosh(−eθ2 ) + sinh θ1 sinh(−eθ2 ))
= −e cosh θ1 − eθ2 , using that cosh is an even function and sinh is an odd one. Since cosh is a positive function, hu, vi L < 0 gives e = 1. In this way, we have:

hv, en i L = sinh θ2 hw1 , en i L + cosh θ2 hw2 , en i L ≤ cosh θ2 (hw1 , en i L + hw2 , en i L ) < 0, that is, v ∈ CT+ (0), as desired. Remark. This result cannot be extended like we did in the remark after Proposition 1.3.6. Can you find a counterexample? Proposition 1.3.8. Let p, q, r ∈ Ln such that p  q and q  r. Then p  r. Proof: It suffices to verify that r − p is timelike and future-directed:

• To see that r − p is timelike, just compute hr − p, r − pi L = hr − q + q − p, r − q + q − pi L = hr − q, r − qi L + 2hr − q, q − pi L + hq − p, q − pi L < 0, since p  q, q  r and, from Proposition 1.3.6, we have that hr − q, q − pi L < 0 as q − p, r − q ∈ CT+ (0).

Welcome to Lorentz-Minkowski Space  23

• To see that r − p is future-directed, use again that p  q and q  r, whence hr − p, en i L = hr − q, en i L + hq − p, en i L < 0.

The previous proposition still holds if we replace  by 4. See Exercise 1.3.4. The transition from Rn to Ln affects results depending on the inner product: some fail to hold, while others undergo drastic changes. Let us explore this, starting with the: Theorem 1.3.9 (Reverse Cauchy-Schwarz inequality). Let u, v ∈ Ln be timelike vectors. Then |hu, vi L | ≥ kuk L kvk L . Furthermore, equality holds if and only if u and v are parallel. Proof: Decompose Ln = span {u} ⊕ u⊥ . Write v = λu + u0 , for some λ ∈ R, and a spacelike u0 orthogonal to u. On one hand we have:

hv, vi L = λ2 hu, ui L + hu0 , u0 i L . On the other hand:

hu, vi2L = hu, λu + u0 i2L = λ2 hu, ui2L
= hv, vi L − hu0 , u0 i L hu, ui L

≥ hv, vi L hu, ui L > 0, using that u0 is spacelike and u is timelike. Taking square roots on both sides leads to |hu, vi L | ≥ kuk L kvk L , as desired. Finally, note that equality holds if and only if hu0 , u0 i L = 0, that is, if u0 = 0. In other words, this is the same as saying that u and v are parallel. Another way to prove the previous result is to adapt the proof given for the classical version, analyzing the discriminant of a certain quadratic polynomial. See how to do this on Exercise 1.3.6. Theorem 1.3.10 (Hyperbolic Angle). If u, v ∈ Ln are both future-directed or pastdirected timelike vectors, there is a unique real number ϕ ∈ [0, +∞[ such that the relation hu, vi L = −kuk L kvk L cosh ϕ holds. This ϕ is called the hyperbolic angle between the vectors u and v. Proof: We rewrite the reverse Cauchy-Schwarz inequality as:

|hu, vi L | ≥ 1. kuk L kvk L Since u and v point both to the future or to the past, Proposition 1.3.6 gives us that hu, vi L < 0, and then: hu, vi L − ≥ 1. kuk L kvk L The function cosh : [0, +∞[ → [1, +∞[, is bijective, so there always exists a unique ϕ ∈ [0, +∞[ such that: hu, vi L − = cosh ϕ. kuk L kvk L Reorganizing the expression leads to hu, vi L = −kuk L kvk L cosh ϕ.

24  Introduction to Lorentz Geometry: Curves and Surfaces

Theorem 1.3.11 (Reverse triangle inequality). Let u, v ∈ Ln be both future-directed or past-directed timelike vectors. Then

ku + vk L ≥ kuk L + kvk L . Proof: Note that u + v is also timelike, and points to the future or the past, along with u and v (this is a particular case of Exercise 1.3.3). The computation is straightforward:

ku + vk2L = −hu + v, u + vi L = − (hu, ui L + 2hu, vi L + hv, vi L ) = −hu, ui L + 2(−hu, vi L ) − hv, vi L = kuk2L + 2(−hu, vi L ) + kvk2L ≥ kuk2L + 2kuk L kvk L + kvk2L = (kuk L + kvk L )2 , leading to ku + vk L ≥ kuk L + kvk L . Example 1.3.12 (The Twins “Paradox”). In Special Relativity, the reverse triangle inequality is used to explain the famous Twin Paradox. Natalia and Leticia are twins at the age of 8. Natalia is a fearless explorer and decides to start a journey through the galaxy, despite the vehement disapproval of her parents and sister. She travels in her ship at about 95% of the speed of light, for 5 years (according to her ship’s calendar). After that, she gets bored and decides to return, at the same speed, arriving home another 5 years later. There, on Earth, Natalia (18 years old) meets her sister Leticia, now married and 40 years old. How could this be? If v is a future-directed timelike vector, connecting the events p and q, kvk L is interpreted as the proper time experienced by an observer traveling from p to q, following v. We can model this situation using the plane L2 . The following picture shows the worldlines of Natalia and Leticia:

(0, 2∆t)

t

Leticia

Natalia

(0, ∆t)

(∆x, ∆t) 5 x

(0, 0)

Figure 1.7: The worldlines of the sisters. The reverse triangle inequality says that Natalia’s path is shorter that Leticia’s, taking less proper time to be traveled. In particular, the figure helps to effectively calculate the difference between their ages. In geometric units (where c = 1), the speed of Natalia’s ship is 0.95. With this, the point (∆x, ∆t) satisfies ∆x2 − ∆t2 = −25

and

∆x = 0.95, ∆t

Welcome to Lorentz-Minkowski Space  25

whence:

(0.95∆t)2 − ∆t2 = −25 =⇒ ∆t = √

5 1 − 0.952

≈ 16.

The symmetry gives k(0, 32)k L = 32, hence Leticia’s age is 8 + 32 = 40 years. The reader might ask why this is a paradox. We could revert the analysis above and consider Natalia’s ship as the reference and, in this situation, the one moving at 95% of the speed of light would be Leticia. After Natalia’s return, the ages would be swapped in the above calculations. In fact, despite the argumentation in the previous paragraph, this is not a paradox, since we cannot suppose that Natalia’s ship is an inertial frame of reference, since it is subject to accelerations during the travel (at least at the departure, the arrival and in the return maneuver). Proposition 1.3.13. Let v ∈ Ln . Then hv, vi L = hv, vi E cos 2θ, where θ is the Euclidean angle between v and the hyperplane e⊥ n (defined as the complement of the Euclidean angle between v and en ). Proof: Given v ∈ Ln , note the vector Idn−1,1 v is just the reflection of v about the ⊥ hyperplane e⊥ n . If θ is the Euclidean angle between v and the plane en , 2θ is the Euclidean angle between v and Idn−1,1 v. So:

hv, vi L = hv, Idn−1,1 vi E = kvk E k Idn−1,1 vk E cos 2θ = kvk E kvk E cos 2θ = kvk2E cos 2θ = hv, vi E cos 2θ.

Some consequences of this Proposition are explored in Exercises 1.3.8 and 1.3.9. To wrap up this section, we introduce a class of transformations in Ln , whose importance is revealed in Special Relativity: they are the ones that preserve the causal precedence . Definition 1.3.14 (Causal automorphism). A map F : Ln → Ln is a causal automorphism if F is bijective and F, as well as F −1 , preserve , that is: x  y ⇐⇒ F ( x)  F (y) and

x  y ⇐⇒ F −1 ( x)  F −1 (y).

Remark. We do not assume linearity for causal automorphisms, and not even continuity. Those properties will be verified soon and it turns out that preserving  is equivalent to preserving the relation < defined by saying that p < q if and only if q ∈ CL+ ( p), see Exercise 1.3.10. Example 1.3.15. “Some” examples of causal automorphisms: (1) Positive homotheties: for λ > 0, Hλ : Ln → Ln given by Hλ ( x) = λx; (2) Translations: fixed a ∈ Ln , Ta : Ln → Ln given by Ta ( x) = x + a; (3) Orthochronous Lorentz transformations: linear maps Λ : Ln → Ln satisfying hΛx, Λyi L = h x, yi L , for any x, y ∈ Ln , such that the set of future-directed timelike vectors is fixed. Remark. Lorentz transformations play a crucial role in the differential geometry in Lorentzian ambient spaces. They will get full attention in Section 1.4.

26  Introduction to Lorentz Geometry: Curves and Surfaces

Finally, we present the surprising result establishing that every causal automorphism is a composition of the ones listed in the previous example: Theorem 1.3.16 (Alexandrov-Zeeman). Let n ≥ 3 be a fixed integer and F : Ln → Ln be a causal automorphism. There exist a positive number c ∈ R>0 , a vector a ∈ Ln , and an orthochronous Lorentz transformation Λ : Ln → Ln such that F ( x) = cΛ( x) + a, for all x ∈ Ln . Remark. In particular, causal automorphisms are, up to a translation, linear (hence continuous). The proof of this result is beyond the scope of this book, but can be found in [52, Section 1.6]. Example 1.3.17. The assumption n ≥ 3 in Theorem 1.3.16 is necessary. If n = 2, . . consider L2 with lightlike coordinates (u, v) given by u = x − y and v = x + y (this coordinate change takes the canonical axes on the light rays in the plane), such that 2

2

h( x, y), ( x, y)i L = x − y =



v+u 2

2



−

v−u 2

2

= uv.

Furthermore, if ( x, y) is timelike and future-directed, the condition y > 0 is rewritten as v > u. If h : R → R is any increasing diffeomorphism, define Fh : L2 → L2 by
Fh (u, v) = h(u), h(v) . In terms of the original variables x and y, Fh corresponds6 to   . h( x + y) + h( x − y) h( x + y) − h( x − y) Gh ( x, y) = . , 2 2 We claim that Fh (hence Gh ) is a causal automorphism. Let (u1 , v1 ), (u2 , v2 ) ∈ L2 be vectors such that (u1 , v1 )  (u2 , v2 ). We must see that Fh (u1 , v1 )  Fh (u2 , v2 ). Our assumptions are that (u2 − u1 )(v2 − v1 ) < 0 and v2 − v1 > u2 − u1 . First, we see that
Fh (u2 , v2 ) − Fh (u1 , v1 ) = h(u2 ) − h(u1 ), h(v2 ) − h(v1 ) is timelike. The Mean Value Theorem gives u∗ between u1 and u2 , and v∗ between v1 and v2 such that

h(u2 ) − h(u1 ) h(v2 ) − h(v1 ) = h0 (u∗ )h0 (v∗ ) (u2 − u1 )(v2 − v1 ) < 0. {z }| {z } | >0

0. Since h is increasing, it follows that h ( v2 ) − h ( v1 ) > 0 > h ( u2 ) − h ( u1 ), whence Fh (u1 , v1 )  Fh (u2 , v2 ). Finally, notice that Fh−1 = Fh−1 and h−1 is also increasing, so that we can repeat the argument with h−1 playing the role of h above, leading us to conclude that Fh is a causal automorphism, for any function h satisfying the given conditions. But we can choose h 6 More

. precisely, if ϕ : L2 → L2 is given by ϕ( x, y) = ( x − y, x + y), we consider Gh = ϕ−1 ◦ Fh ◦ ϕ.

Welcome to Lorentz-Minkowski Space  27

such that Gh is not of the form stated in Theorem 1.3.16. For example, if h(t) = sinh t, we have Gh ( x, y) = (sinh x cosh y, cosh x sinh y), which is not even an affine map. The surjectiveness of h is also crucial: if h(t) = et , then the image of Gh ( x, y) = (ex cosh y, ex sinh y) is contained in a spacelike sector of the plane L2 . Remark.

• Why is this argument invalid for L3 ? • “Baby” Alexandrov-Zeeman: if we consider the pathological case when n = 1, we would have a map F : L1 → L1 that is a causal automorphism if and only if it is monotonically increasing and surjective. So, there is a gap in theorem just for the case n = 2. The physical meaning of this result is the following: we can recover the linear structure of Lorentz-Minkowski space from the causal relation between its events. The first version of Theorem 1.3.16, proved by Zeeman in 1964, was a seminal work that motivated the classification of transformations in spacetimes satisfying properties analogous to 4, such as invariance of the light cone (see [1]) or invariance of the set of unit timelike vectors (see [22]).

Exercises Exercise† 1.3.1 (Review). Let ϕ ∈ R. The hyperbolic cosine and the hyperbolic sine of ϕ are defined by . e ϕ + e− ϕ cosh ϕ = 2

and

. e ϕ − e− ϕ sinh ϕ = . 2

Check the following properties: (a) cosh is an even function and sinh is an odd function. Furthermore, sinh is bijective. (b) cosh ϕ ≥ 1, cosh2 ϕ − sinh2 ϕ = 1 and sinh ϕ ≤ cosh ϕ for all ϕ ∈ R. (c) cosh and sinh are differentiable, with cosh0 = sinh and sinh0 = cosh. (d) sinh( ϕ1 + ϕ2 ) = sinh ϕ1 cosh ϕ2 + sinh ϕ2 cosh ϕ1 and cosh( ϕ1 + ϕ2 ) = cosh ϕ1 cosh ϕ2 + sinh ϕ1 sinh ϕ2 , for all ϕ1 , ϕ2 ∈ R. (e) Replace ϕ2 by − ϕ2 in (d) and use item (a) to write up formulas for sinh( ϕ1 − ϕ2 ) and cosh( ϕ1 − ϕ2 ). (f) Make ϕ = ϕ1 = ϕ2 in (d) to state formulas for cosh(2ϕ) and sinh(2ϕ). Exercise 1.3.2. Recall that a subset C ⊆ Ln is convex if for any u, v ∈ C we have [u, v] ⊆ C, where .  [u, v] = (1 − t)u + tv | 0 ≤ t ≤ 1 is the straight line segment joining u to v. Show that, for any p ∈ Ln , CT+ ( p) are CT− ( p) are convex sets.

28  Introduction to Lorentz Geometry: Curves and Surfaces

Exercise 1.3.3. Let p ∈ Ln and u1 , . . . , uk ∈ CT+ ( p). For any λ1 , . . . , λk > 0, show that   k k + ∑ λ i u i ∈ CT ∑ λ i p . i =1

i =1

Exercise 1.3.4. Show that 4 is a transitive relation, that is, if p 4 q and q 4 r, then p 4 r. Exercise 1.3.5. A subset C ⊆ Ln is 4-convex if it satisfies the following: if p, q ∈ C and r ∈ Ln are such that p 4 r 4 q, then r ∈ C. Also, given S ⊆ Ln , the 4-convex hull of S, denoted by H4 (S), is the smallest subset 4-convex of Ln containing S, that is: . \ 0 S ⊆ Ln | S0 is 4-convex and S ⊆ S0 . H4 (S) = (a) Show that arbitrary intersections of 4-convex sets are 4-convex. This says that H4 (S) is well-defined, for any set S ⊆ Ln . (b) Show that H4 (S) = {r ∈ Ln | there exist p, q ∈ S such that p 4 r 4 q}. Hint. Show that the set in the right-hand side is 4-convex and contained in H4 (S). Remark. It is possible to define a relation 4, called observable causality between regions in Ln and prove results analogous to Theorem 1.3.16 in this setting. The definition of 4-convex hull presented here is one of the first tools to achieve that. For details, see [38]. Exercise 1.3.6 (Reverse Cauchy-Schwarz inequality). In this exercise we propose an adaptation for the proof of the standard Cauchy-Schwarz inequality, but for timelike vectors. Let u, v ∈ Ln be timelike vectors. (a) If u and v are parallel to each other, prove directly that |hu, vi L | = kuk L kvk L , mimicking the argument in Theorem 1.3.9 (p. 23). (b) If u and v are linearly independent, they span a timelike plane. Hence u + tv may assume any causal type as we vary t ∈ R. Using this, analyze the discriminant of . p(t) = hu + tv, u + tvi L and conclude that |hu, vi L | > kuk L kvk L . Remark. The inequality is strict here, due to Exercise 1.2.5 (p. 17). Why? (c) Can you repeat that argument when one of the vectors is lightlike? What if both are? In this setting, is there a necessary and sufficient condition for equality? Discuss. Exercise 1.3.7 (Lorentz factor). Let p, q ∈ Ln be two events and . v = q − p = (∆x1 , . . . , ∆xn−1 , ∆t) the vector joining p and q. Suppose that v is timelike and future-directed (in such a way that p influences q). Show that the hyperbolic angle ϕ between v and en is determined by 1 . γ = cosh ϕ = q , 1 − kv ek2E . ∆xn−1
1 where v e = ∆x ∈ Rn−1 is the spatial velocity vector associated with v. Also ∆t , . . . , ∆t show that kv ek E = tanh ϕ.

Welcome to Lorentz-Minkowski Space  29

Remark. The number γ is known as the Lorentz factor and it is used in Special Relativity for calculations involving phenomena like time dilation and length contraction, and formulas for relativistic energy and relativistic linear momentum of particles traveling at speeds close to the speed of light. p Exercise 1.3.8. Given v ∈ Ln , show that kvk L = kvk E | cos 2θ |, where θ is the Euclidean angle between v and the hyperplane e⊥ n. Exercise 1.3.9. Let v ∈ Ln . Then kvk L ≤ kvk E and equality holds if and only if v is horizontal or vertical. Remark. We use “vertical” and “horizontal” in the Euclidean sense we’re used to. More precisely, u is vertical if u k en and horizontal if u ⊥ en . It does not matter whether you use ⊥ E or ⊥ L in this case. Exercise 1.3.10. Recall that in Ln we say that p < q if q ∈ CL+ ( p). The relations  and < may be expressed in terms of each other. For this problem, given x, y ∈ Ln , you may assume the following equivalences:

• x < y ⇐⇒ x  6 y and y  z implies x  z, for any z ∈ Ln . • x  y ⇐⇒ x < 6 y and there is z ∈ Ln with x < z < y. Show that a bijective map F : Ln → Ln is a causal automorphism if and only if F and F −1 preserve 0 , a1 , a2 ∈ Ln and Λ1 and Λ2 are orthochronous Lorentz transformations such that c1 Λ1 ( x ) + a1 = c2 Λ2 ( x ) + a2 , for all x ∈ Ln , show that c1 = c2 , a1 = a2 and Λ1 = Λ2 .

1.4

ISOMETRIES IN Rnν

The Euclidean norm k · k E induces a natural distance function in Rn and, from the geometric point of view, functions that preserve that distance are of great interest. We would like to repeat such study in the Lorentzian ambient, but k · k L does not induce a distance in Ln , since k · k L is not, in the strict sense, a norm. Definition 1.4.1. A Euclidean isometry in Rn is a map F : Rn → Rn such that k F ( x) − F (y)k E = k x − yk E , for any x, y ∈ Rn . The set of all Euclidean isometries in Rn is denoted by E(n, R). Remark. Euclidean isometries are also called rigid motions in Rn .

30  Introduction to Lorentz Geometry: Curves and Surfaces

To translate this to the Lorentzian ambient just replacing E by L in k · k, would bring complications, due to the absolute value in the definition of k · k L . In other words, this direct translation would not be sensitive to changes in causal type (for instance, switching the axes in L2 would fit the bill, but axes of different causal types are geometrically and physically very distinct). However, note that F is a Euclidean isometry if and only if

h F ( x) − F (y), F ( x) − F (y)i E = h x − y, x − yi E , for any vectors x, y ∈ Rn . We use this equivalence as a starting point to write the following generalization: Definition 1.4.2 (Pseudo-Euclidean Isometry). A pseudo-Euclidean isometry in Rnν is a map F : Rnν → Rnν such that

h F ( x) − F (y), F ( x) − F (y)iν = h x − y, x − yiν , for any vectors x, y ∈ Rnν . The set of all pseudo-Euclidean isometries in Rnν is denoted by Eν (n, R). Remark. For ν = 1 the pseudo-Euclidean isometries are called Poincaré transformations. The set E1 (n, R) is often written in the literature as P(n, R). To study such isometries it is also useful to study functions that preserve h·, ·iν : Definition 1.4.3 (Pseudo-orthogonal Transformations). A linear map Λ : Rnν → Rnν is a pseudo-orthogonal transformation if

hΛx, Λyiν = h x, yiν , for any vectors x, y ∈ Rnν . The set of all pseudo-orthogonal transformations in Rnν is denoted by Oν (n, R). Remark. If ν = 1, the pseudo-orthogonal transformations are called Lorentz transformations. It follows directly from the definition that every pseudo-orthogonal transformation is a pseudo-Euclidean isometry. We will freely identify a linear transformation with its matricial representation in the canonical basis, and the vectors of Rnν with column matrices. With that in mind, here are some examples: Example 1.4.4. In Rn : (1) Translations. For each a ∈ Rn , the map Ta : Rn → Rn given by Ta ( x) = x + a is an isometry. Notice that Ta is bijective and ( Ta )−1 = T− a . (2) Rotations. In R2 , given θ ∈ [0, 2π [, the map Rθ : R2 → R2 given by Rθ ( x, y) = ( x cos θ − y sin θ, x sin θ + y cos θ ) is an orthogonal transformation. The action of Rθ is to rotate ( x, y) counter-clockwise around the origin by an angle of θ radians. It is useful to note that:    cos θ − sin θ x Rθ ( x, y) = sin θ cos θ y

Welcome to Lorentz-Minkowski Space  31

y Rθ ( x, y)

111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 ( x, y) 000000000 111111111 000000 111111 θ 000000000 111111111 000000 111111 000000000 111111111 000000 111111 000000000 111111111 000000 111111 000000000 111111111 000000 111111 000000000 111111111 x Figure 1.8: Counter-clockwise rotation of an angle θ. (3) Rotations around coordinate axes. In R3 , given θ ∈ [0, 2π [, one can consider rotations in the form       1 0 0 cos θ 0 − sin θ cos θ − sin θ 0  sin θ cos θ 0 ,  0 1 0  and 0 cos θ − sin θ  , 0 sin θ cos θ sin θ 0 cos θ 0 0 1 which are all orthogonal. (4) Rotations around several axes. In R2n , the block matrix  cos θ1 − sin θ1  sin θ1 cos θ1   .. ..  . .   0 0 0 0

take angles θ1 , . . . , θn ∈ [0, 2π [ and consider

··· ··· .. .

0 0 .. .

··· ···

cos θn sin θn

0 0 .. .



   ,  − sin θn  cos θn

which is another orthogonal transformation. (5) Generalized reflections. In Rn , given e1 , · · · , en ∈ {−1, 1}, the map R : Rn → Rn given by . R ( x 1 , · · · , x n ) = ( e1 x 1 , · · · , e n x n ) is an orthogonal transformation. In particular, if ei = −1 for a single index i, then R is the reflection on the hyperplane xi = 0. When ei = −1 for all i, we have R = − IdRn , the so-called antipodal map in Rn . (6) Axes permutations. In Rn , considering a permutation7 σ ∈ Sn , we have that Σ : Rn → Rn given by . Σ ( x 1 , · · · , x n ) = ( x σ (1) , · · · , x σ ( n ) ) is an orthogonal transformation. 7 That

is, a bijection σ : {1, · · · , n} → {1, · · · , n}. The set of all such bijections, endowed with the composition operation, is a group and denoted by Sn (permutations of n letters). For further details, see [23].

32  Introduction to Lorentz Geometry: Curves and Surfaces

In analogy to the examples above, we have: Example 1.4.5. In Ln : (1) Translations, as before. (2) Hyperbolic rotations. Given any ϕ ∈ R, the map Rhϕ : L2 → L2 given by . Rhϕ ( x, y) = ( x cosh ϕ + y sinh ϕ, x sinh ϕ + y cosh ϕ) is a Lorentz transformation. As before, one can see Rhϕ in the form Rhϕ ( x, y)

y



=

cosh ϕ sinh ϕ sinh ϕ cosh ϕ

  x . y

y

Rhϕ ( x1 , y1 )

( x1 , y1 ) Rhϕ ( x2 , y2 )

( x2 , y2 ) x

(a)

1111111 0000000 0000000 1111111 Rhϕ ( x, y) 0000000 1111111 0000000 1111111 ( x, y) 0000000 1111111 0000000 1111111 00 11 0000000 1111111 00 11 0000000 1111111 00 11 0000000 1111111 00 11 ϕ/2 0000000 1111111 00 11 0000000 1111111 00 11 0000000 1111111 00 11 0000000 1111111 00 11 0000000 1111111 00 11

x

(b)

Figure 1.9: Hyperbolic rotation by angle ϕ > 0.  Notice that Rhϕ preserves each branch of x ∈ L2 | h x, xi L = ±1 . See Exercise 1.4.2. (3) Rotations around coordinate axes. Here we must take causal types into account: given θ ∈ [0, 2π [ and ϕ ∈ R, we can consider, among many others, the maps given by       cos θ − sin θ 0 cosh ϕ 0 sinh ϕ 1 0 0  sin θ cos θ 0 ,  0 1 0  and 0 cosh ϕ sinh ϕ  . 0 0 1 sinh ϕ 0 cosh ϕ 0 sinh ϕ cosh ϕ (4) Rotations around several axes. In L2n , consider angles θ1 , · · · , θn−1 ∈ [0, 2π [ and ϕ ∈ R and the block matrix   cos θ1 − sin θ1 · · · 0 0  sin θ1 cos θ1 · · · 0 0     .. .. . ..  , . . .  . . . . .     0 0 · · · cosh ϕ sinh ϕ  0 0 · · · sinh ϕ cosh ϕ which is a Lorentz transformation. Is it a Lorentz transformation if the block with hyperbolic functions is not the last one?

Welcome to Lorentz-Minkowski Space  33

(5) Generalized reflections, as before. (6) Spacelike axis permutations. In Ln , considering σ ∈ Sn−1 , Σ L : Ln → Ln given by
. Σ L ( x 1 , · · · , x n ) = x σ (1) , · · · , x σ ( n −1) , x n is a Lorentz transformation. Would Σ L still be a Lorentz transformation if the shuffling acted on xn ? Remark. From now on, the following natural identifications will be adopted: O0 (n, R) ≡ O(n, R),

E0 (n, R) ≡ E(n, R) and

Idn−0,0 ≡ Idn .

Lemma 1.4.6. Let A, B ∈ Mat(n, R) such that x> Ay = x> By, for any vectors x, y ∈ Rn . Then A = B. Proof: Take x = ei and y = e j , whence aij = ei> Ae j = ei> Be j = bij , for any i and j. Therefore A = B. Proposition 1.4.7. Let Λ : Rnν → Rnν be a linear map. Then Λ is pseudo-orthogonal if and only if Λ> Idn−ν,ν Λ = Idn−ν,ν . Proof: Suppose that Λ ∈ Oν (n, R). Then, given any vectors x, y ∈ Rnν , it holds that hΛx, Λyiν = h x, yiν . In matrix notation this is written as

(Λx)> Idn−ν,ν (Λy) = x> Idn−ν,ν y =⇒ x> Λ> Idn−ν,ν Λy = x> Idn−ν,ν y. Since x and y are arbitrary, the previous lemma says that Λ> Idn−ν,ν Λ = Idn−ν,ν . The converse is now clear. Corollary 1.4.8. Let Λ ∈ Oν (n, R). Then det Λ ∈ {−1, 1}. In particular, Λ is a linear isomorphism and Λ−1 is always defined. Proposition 1.4.9. The set Oν (n, R) is a group, if endowed with the usual matrix multiplication. Furthermore, Oν (n, R) is closed under matrix transposition. Thus, Oν (n, R) is called a pseudo-orthogonal group. In particular, O(n, R) and O1 (n, R) are respectively called the orthogonal group and the Lorentz group. Proof: The verification that these sets are in fact groups is left for Exercise 1.4.3. To check closure under transposition, consider Λ ∈ Oν (n, R): inverting the expression Λ> Idn−ν,ν Λ = Idn−ν,ν we achieve Λ−1 Idn−ν,ν (Λ> )−1 = Idn−ν,ν . Solving for Idn−ν,ν in the left-hand side, we have: Idn−ν,ν = Λ Idn−ν,ν (Λ> ) = (Λ> )> Idn−ν,ν Λ> , showing that Λ> ∈ Oν (n, R). Corollary 1.4.10. If Λ ∈ Oν (n, R) then its columns (as well as its rows) form an orthonormal basis of Rnν .

34  Introduction to Lorentz Geometry: Curves and Surfaces

Proof: Considering the previous proposition, it suffices to verify the statement for the columns of Λ. In fact, noting that the i-th column of Λ is Λei we have

hΛei , Λe j iν = hei , e j iν . The result follows from the fact that the canonical basis is orthonormal relative to h·, ·iν , for any ν. Remark. In other words, the result above shows that every Λ ∈ Oν (n, R) maps orthonormal bases to orthonormal bases. Note that the composition of elements of Oν (n, R) with translations produce elements of Eν (n, R). Now we prove that every map in Eν (n, R) is such a composition. Proposition 1.4.11. Let F ∈ Eν (n, R). If F (0) = 0, then F ∈ Oν (n, R). Proof: A polarization identity for h·, ·iν is

h x, yiν =

1 (h x, xiν + hy, yiν − h x − y, x − yiν ) , 2

for any vectors x, y ∈ Rnν . Applying this to F ( x) and F (y) instead of x and y, we have that:

h F ( x), F (y)iν =

1 (h F ( x), F ( x)iν + h F (y), F (y)iν − h F ( x) − F (y), F ( x) − F (y)iν ) , 2

Since F (0) = 0, it follows that

h F ( x), F ( x)iν = h F ( x) − 0, F ( x) − 0iν = h x − 0, x − 0iν = h x, xiν , and the same holds for y. Hence, the comparison of these two formulas gives h F ( x), F (y)iν = h x, yiν . The linearity of F then follows from Exercise 1.4.4. Theorem 1.4.12. Let F ∈ Eν (n, R). Then there are a unique a ∈ Rnν and Λ ∈ Oν (n, R) such that F = Ta ◦ Λ. Proof: We know that T− F(0) ◦ F ∈ Eν (n, R) is an isometry taking 0 into 0, hence T− F(0) ◦ F = Λ ∈ Oν (n, R). Thus F = TF(0) ◦ Λ. To check the uniqueness, suppose that Ta1 ◦ Λ1 = Ta2 ◦ Λ2 for some a1 , a2 ∈ Rnν and Λ1 , Λ2 ∈ Oν (n, R). Evaluating at 0 we obtain a1 = Ta1 (0) = Ta1 (Λ1 (0)) = Ta2 (Λ2 (0)) = Ta2 (0) = a2 . Since Ta1 is bijective, Ta1 ◦ Λ1 = Ta1 ◦ Λ2 implies Λ1 = Λ2 . Remark. In the notation above, Λ and Ta are called, respectively, the linear part and the affine part of F. Corollary 1.4.13. Every F ∈ Eν (n, R) is bijective. It is opportune to note that Eν (n, R) is also a group, when equipped with the composition operation. See Exercise 1.4.5. It is called the Euclidean group when ν = 0, and the Poincaré group when ν = 1. The following result depends on some Differential Calculus concepts. See Appendix A if necessary. Corollary 1.4.14. Let F ∈ Eν (n, R). Then F is differentiable and, for each p ∈ Rnν , DF ( p) ∈ Oν (n, R).

Welcome to Lorentz-Minkowski Space  35

Proof: Notice that F = Ta ◦ Λ, for some a ∈ Rnν and Λ ∈ Oν (n, R). Hence F is differentiable as a composition of differentiable maps. Moreover, we have: DF ( p) = D ( Ta ◦ Λ)( p) = DTa (Λ( p)) ◦ DΛ( p) = idRnν ◦Λ = Λ ∈ Oν (n, R).

Proposition 1.4.15. Let p, q ∈ Rnν , and (v1 , · · · , vn ), (w1 , · · · , wn ) be two bases of Rnν such that hvi , v j iν = hwi , w j iν for all 1 ≤ i, j ≤ n. Then there exists a unique F ∈ Eν (n, R) such that F ( p) = q and DF ( p)(vi ) = wi for all 1 ≤ i ≤ n. Proof: Since F = Ta ◦ Λ, for some vector a ∈ Rnν and some linear map Λ ∈ Oν (n, R), it suffices to exhibit a and Λ. The hypothesis hvi , v j iν = hwi , w j iν ensures that the unique linear map Λ characterized by Λvi = wi , for all i, is in Oν (n, R). Finally, the condition . F ( p) = q forces a = q − Λp. Beyond the fact that Λ ∈ Oν (n, R) implies det Λ = ±1 (see Corollary 1.4.8), we have that if Λ = (λij )1≤i,j≤n ∈ O1 (n, R), then |λnn | ≥ 1. This follows from Corollary 1.4.10, by observing that the column Λen is a unit timelike vector. On the other hand, for C = (cij )1≤i,j≤n ∈ O(n, R) it holds that |cnn | ≤ 1. Definition 1.4.16. The group . SOν (n, R) = {Λ ∈ Oν (n, R) | det Λ = 1} is called the special pseudo-orthogonal group. Remark.

• When ν = 1, we have the special Lorentz group. We conveniently write, as before, SO0 (n, R) = SO(n, R) when it is convenient. • An element of SO1 (n, R) is called a proper Lorentz transformation. A Poincaré transformation is also called proper when its linear part is proper. Now, let B = (u1 , . . . , un ) be an orthonormal basis for Ln , where un is timelike. Also, let B be the matrix with the vectors ui as its columns. Then, we have: Definition 1.4.17. An orthonormal basis B = (u1 , . . . , un ) of Ln is future-oriented if un is a future-directed vector. It is past-oriented if un is past-directed. Definition 1.4.18. A Lorentz transformation Λ ∈ O1 (n, R) is orthochronous (that is, preserves time orientation) if, for any future-oriented orthonormal basis B, the (ordered) basis formed by the columns of ΛB, denoted by ΛB, is also future-oriented. The set of orthochronous Lorentz transformations is denoted by O1↑ (n, R). Remark. A Poincaré transformation is orthochronous if its linear part is orthochronous. Proposition 1.4.19. The set O1↑ (n, R) is a group. Proof: Let B be a future-oriented orthonormal basis of Ln .

• Let Λ1 , Λ2 ∈ O1↑ (n, R). Then Λ2 B is future-oriented since Λ2 ∈ O1↑ (n, R). Hence (Λ1 Λ2 ) B = Λ1 (Λ2 B) is also future-oriented, since Λ1 ∈ O1↑ (n, R). In this way Λ1 Λ2 ∈ O1↑ (n, R).

36  Introduction to Lorentz Geometry: Curves and Surfaces

• Idn B = B, so it is clear that Idn ∈ O1↑ (n, R). • Let Λ ∈ O1↑ (n, R). Then B = Λ−1 (ΛB). Since ΛB and B are both futureoriented, Λ−1 preserves time orientation, that is, Λ−1 ∈ O1↑ (n, R). Previously we noted that if Λ = (λij )1≤i,j≤n ∈ O1 (n, R), then |λnn | ≥ 1, and we could have λnn being positive or negative. It is usual in the literature to declare a Lorentz transformation Λ to be orthochronous if λnn ≥ 1. The equivalence between this and our definition comes in the: Theorem 1.4.20 (Characterization of O1↑ (n, R)). Let Λ = (λij )1≤i,j≤n be a Lorentz transformation. Then: Λ ∈ O1↑ (n, R) ⇐⇒ λnn ≥ 1 Proof: Start supposing that Λ is orthochronous. Then Λen is future-directed, as well as en , and hence λnn ≥ 1. Conversely, suppose that λnn ≥ 1. Let (u1 , . . . , un ) be a future-directed orthonormal basis of Ln . Then un is timelike and future-directed, the same holding for Λun . We know that Λun is timelike, since Λ ∈ O1 (n, R). It remains to check that its last coordinate is positive. Computing the product (recall that Λ> ei is the i-th row of Λ), we see that such coordinate is

hΛ> en , un i E = hΛ> en , Idn−1,1 un i L . Since λnn ≥ 1, it follows that Λ> en is future-directed. Furthermore, the fact that un is future-directed implies that Idn−1,1 un is past-directed. Hence

hΛ> en , Idn−1,1 un i L > 0, as desired. Remark. The previous result could be proved using topological arguments: since Ln is finite-dimensional, Λ is continuous; det Λ 6= 0 ensures the non-singularity of Λ. We know that Λ−1 is also linear, hence Λ−1 is continuous. So Λ is a homeomorphism. The timecone, CT (0), has two connected components, that are preserved (if Λ is orthochronous) or switched (if Λ is not orthochronous). The group O1 (n, R) admits a partition in subsets according to the signs of λnn and det Λ or, equivalently, the signs of λnn and the determinant of the spatial part of Λ. We formalize this recalling the splitting Rnν = Rn−ν × Rνν , used in the proof of Theorem 1.2.25 (p. 14). With it we write Λ ∈ Oν (n, R) as blocks:

Λ=

ΛS

B

C

ΛT

! ,

where ΛS ∈ Mat(n − ν, R) and Λ T ∈ Mat(ν, R) are, respectively, the spatial and temporal parts of Λ. Since Λ is non-singular and preserves causal types, one sees (by composing with appropriate projections) that both ΛS and Λ T are also non-singular. We have:

Welcome to Lorentz-Minkowski Space  37

Theorem 1.4.21. Let 0 < ν < n and Λ ∈ Oν (n, R). Then det ΛS = det Λ T det Λ. Proof: As always let can = (ei )in=1 be the canonical basis of Rnν . Consider
also the n orthonormal basis of Rν formed by the columns of Λ, B = Λe1 , . . . , Λen . Suppose that Λ = (λij )1≤i,j≤n . We “delete” the block B, defining a linear map T : Rnν → Rnν by ( T (Λe j ) =

Λe j ,

if 1 ≤ j ≤ n − ν and if n − ν < j ≤ n.

∑in=n−ν+1 λij ei ,

One easily sees that [Λ]can,B = Idn and

[ T ] B,can =

ΛS

0

C

ΛT

! .

Now, we compute the matrix [ T ] B. The expression T (Λe j ) = Λe j , which holds for 1 ≤ j ≤ n − ν, says that the left upper and lower blocks of [ T ] B are, respectively, Idn−ν and 0. To compute the determinant of [ T ] B by blocks, we need to know the last ν entries . of T (Λe j ) in the basis B, for any n − ν < j ≤ n. Letting ek = eek , Lemma 1.2.32 (p. 16) gives n

n

∑

T (Λe j ) =

i = n − ν +1 n n

∑

=

∑

λij ei =

n

∑ ek hei , Λek iν Λek

λij

i = n − ν +1

k =1

n

∑ ∑ ek λij λ`k hei , e` iν Λek

i =n−ν+1 k=1 `=1 n

=

∑

k =1

n

∑

n

∑ ek λij λ`k ηiν`

! Λek .

i =n−ν+1 `=1

The desired last ν entries correspond to n − ν < k ≤ n and, in this setting, the entries in the right lower block of [ T ] B are given by n

∑

n

∑

n

∑

−λij λ`k (−δi` ) =

i =n−ν+1 `=1

λij λik ,

i = n − ν +1

which are the entries of the product of Λ> T and Λ T , up to renaming indexes, if necessary. Then  

Idn−ν

[T ]B =  

0

∗

Λ> T ΛT

 .

In particular, det T = (det Λ T )2 . Furthermore:   [ TΛ] B = [ T ] B,can [Λ]can,B = 

ΛS C

0 ΛT

  .

Hence

(det Λ T )2 det Λ = det T det Λ = det( TΛ) = det Λ T det ΛS , or det ΛS = det Λ T det Λ, as desired.

38  Introduction to Lorentz Geometry: Curves and Surfaces

With this we can, as stated above, index the elements of Oν (n, R) according to the signs of det ΛS and det Λ T , obtaining a partition of Oν (n, R): . O+↑ ν ( n, R) = . O+↓ ν ( n, R) = . O−↑ ν ( n, R) = . O−↓ ν ( n, R) =

{Λ ∈ Oν (n, R) | det ΛS > 0 and det Λ T > 0} {Λ ∈ Oν (n, R) | det ΛS > 0 and det Λ T < 0} {Λ ∈ Oν (n, R) | det ΛS < 0 and det Λ T > 0} {Λ ∈ Oν (n, R) | det ΛS < 0 and det Λ T < 0}

The elements of O+• ν ( n, R) are said to preserve space orientation, while a matrix •↑ in Oν (n, R) preserves time orientation (also called orthochronous). If det Λ > 0, then Λ preserves the algebraic orientation of Rnν . On the other hand, if Λ ∈ O1 (n, R) and det ΛS > 0, Λ preserves spatial orientation of Rnν , that is, it preserves the orientation of its spacelike subspaces. The previous theorem shows that if det Λ T > 0, then both orientation preservation concepts are equivalent. One can show that O+↑ ν ( n, R) is a (normal) subgroup of Oν ( n, R) and, if ν = 1, then +↑ O1 (n, R) is called the special orthochronous Lorentz group. It is clear, in this case, that . O1+↑ (n, R) = O1↑ (n, R) ∩ SO1 (n, R) is an intersection of subgroups of O1 (n, R), hence a group itself. The relevance of O+↑ ν ( n, R) is highlighted when we use the four principal representatives . τ +↑ = Idn ,

. τ +↓ = Idn−1,1 ,

. . τ −↑ = Id1,n−1 and τ −↓ = diag(−1, 1, . . . , 1, −1),

in the following: −↑ −↓ Proposition 1.4.22. The sets O+↓ ν ( n, R), Oν ( n, R) and Oν ( n, R) are cosets of +↓ −↑ −↓ O+↑ ν ( n, R), whose representatives are τ , τ , and τ . In other words, +↓ O+↓ · O+↑ ν ( n, R) = τ ν ( n, R) −↑ O−↑ · O+↑ ν ( n, R) = τ ν ( n, R) −↓ O−↓ · O+↑ ν ( n, R) = τ ν ( n, R).

Proof: It suffices to note how the four principal representatives act on elements of Oν (n, R) by left multiplication (for instance, τ −↑ reverts just the sign of first row, τ +↓ reverts the sign of last row, etc.), and that squaring all of the principal representatives we get the identity. See Exercise 1.4.14 for more details. . Remark. The set G = {τ +↑ , τ +↓ , τ −↑ , τ −↓ }, equipped with matrix multiplication is a group. This is a straightforward computation (suggested in Exercise 1.4.15). This helps us derive a “sign rule” for indexes in the elements of G.

Exercises Exercise 1.4.1. Follow the notation in examples 1.4.4 and 1.4.5 (p. 30 and p. 32) and for θ, θ 0 ∈ [0, 2π [ and ϕ, ϕ0 ∈ R: (a) Compute Rθ (cos θ 0 , sin θ 0 ) and Rhϕ (sinh ϕ0 , cosh ϕ0 ). Give geometrical interpretations.

Welcome to Lorentz-Minkowski Space  39

(b) Compute Rθ ◦ Rθ 0 and Rhϕ ◦ Rhϕ0 . Use those expressions to write ( Rθ )−1 and ( Rhϕ )−1 in terms of Rθ and Rhϕ . What is the angle between u and Rθ (u), and what is the hyperbolic angle between u and Rhϕ (u), if u is a timelike vector? (c) Let Dθ and D hϕ be the matrices whose entries are the derivatives of the entries in Rθ and Rhϕ , respectively. Compute . J = Dθ ◦ ( R θ ) − 1

and

. J h = D hϕ ◦ ( Rhϕ )−1 .

Remark. Notice that Dθ is a rotation, while D hϕ is not. (d) Show that

h Jx, yi E = −h x, Jyi E and h J h x, yi L = −h x, J h yi L , for any vectors x, y ∈ R2ν . Remark. Maps with the above property are called skew-symmetric. You will face them again in Exercise 1.4.12. Exercise 1.4.2. Let ϕ > 0 and Rhϕ be the hyperbolic rotation defined in Example 1.4.5.  (a) Show that the map Rhϕ preserves each branch of the set x ∈ L2 | h x, xi L = ±1 , as indicated in Figure 1.9 (a) (p. 32). (b) Let θ ∈ R. Show that the region bounded by the upper branch of hyperbola given by x2 − y2 = −1 and the line segments joining

(sinh θ, cosh θ ) and (sinh(θ + ϕ), cosh(θ + ϕ)) to the origin (0, 0) has area ϕ/2 in R2 , according to Figure 1.9 (b). Hint. You may find it useful to solve this for θ = 0 first. Exercise† 1.4.3. Show that: (a) Idn ∈ Oν (n, R); (b) if Λ1 , Λ2 ∈ Oν (n, R), then Λ1 Λ2 ∈ Oν (n, R); (c) if Λ ∈ Oν (n, R), then Λ−1 ∈ Oν (n, R). Use this to prove the first part of Proposition 1.4.9 (p. 33), that is, Oν (n, R) is a group. Exercise† 1.4.4. In the text, we defined pseudo-orthogonal transformations as linear maps that preserve h·, ·iν . The word “linear” is not necessary in that definition. Prove that if Λ : Rnν → Rnν preserves h·, ·iν , then Λ is automatically linear. n

Hint. Use that Λ takes orthonormal bases into orthonormal bases, write v = ∑ ai ei , n

i =1

Λv = ∑ bi Λei and then prove that ai = bi for all i. i =1

Exercise 1.4.5. Show that the set Eν (n, R) is a group, when equipped with composition of functions.

40  Introduction to Lorentz Geometry: Curves and Surfaces

Exercise 1.4.6. Consider the points p = (1, 0, 3, 3) as the vectors v1 = (1, 0, 3, 0), w1 = v2 = (0, 2, 0, −3), w2 = v3 = (−1, 1, 0, 3), w3 = v4 = (0, 0, 6, 3), w4 =

and q = (−1, 2, 5, 6) in L4 , as well

(1, 0, 5, 4) (0, 2, −4, −5) (−1, 1, 4, 5) (0, 0, 14, 13).

Write out the unique Poincaré transformation F ∈ P(4, R) such that F ( p) = q and DF ( p)(vi ) = wi , for 1 ≤ i ≤ 4. Exercise† 1.4.7. Let B = (u1 , . . . , un ) and C = (v1 , . . . , vn ) be ordered bases of Ln . Show that, if GLn ,B = GLn , C, then the change of basis matrix between B and C defines a Lorentz transformation. Exercise 1.4.8. In the product Oν (n, R) × Rn , define the operation ∗ by . ( A, v) ∗ ( B, w) = ( AB, Aw + v). With this operation, we write the product as Oν (n, R) n Rn . (a) Prove that (Idn , 0) is the identity for the operation ∗ (which is not commutative). (b) Prove that ∗ is associative. (c) Exhibit the inverse of an element ( A, v) according to ∗. Remark. The previous items show that Oν (n, R) n Rn is a group, called an (outer) semi-direct product of Oν (n, R) and Rn . (d) Prove that {Idn } × Rn is a normal subgroup of Oν (n, R) n Rn and that Oν (n, R) n Rn ∼ = Oν (n, R). {Idn } × Rn Hint. Show that the projection on Oν (n, R) is a group epimorphism and compute its kernel. (e) Define Φ : Oν (n, R) n Rn → Eν (n, R) by Φ( A, v) = Tv ◦ A. Show that Φ is a group isomorphism. Exercise† 1.4.9. Let Λ ∈ O1 (n, R) be a Lorentz transformation. (a) Show that for any non-lightlike eigenvector, its associated eigenvalue is 1 or −1. (b) Show that the product of any two eigenvalues associated to linearly independent lightlike eigenvectors is 1. (c) Let U ⊆ Ln be an eigenspace of Λ that contains a non-lightlike eigenvector. Show that any other eigenspace is Lorentz-orthogonal to U. Hint. Use item (a). (d) If U ⊆ Ln is a vector subspace, show that U is Λ-invariant if and only if U ⊥ is Λ-invariant.

Welcome to Lorentz-Minkowski Space  41

Exercise 1.4.10 (Householder reflections). Let H ⊆ Ln be a hyperplane. Suppose that H is not lightlike and take a vector n normal to H. Write Ln = H ⊕ span{n} (see Exercise 1.2.7, p. 17). Then each v ∈ Ln is written uniquely as v H + λn, for some λ ∈ R and v H ∈ H, satisfying hv H , ni L = 0. Define the reflection relative to the hyperplane H as Rn : Ln → Ln , given by Rn (v) = v H − λn. (a) Show that Rn is a Lorentz transformation. Hint. Use Exercise 1.4.4 to skip some calculations. (b) Let u, v ∈ Ln be vectors such that hu, ui L = hv, vi L 6= 0. Show that it is possible to get u from v using one or two reflections. Hint. Deal with two different situations: if v − u is not a lightlike vector, compute Rv−u (v); if v − u is lightlike, show that v + u is not lightlike, and compute Rv+u (v). Exercise 1.4.11 (Cartan-Dieudonné). Let Λ ∈ O1 (n, R) be a Lorentz transformation. Then Λ is a composition of reflections relative to non-lightlike hyperplanes. Hint. Proceed by induction on n. For the induction step, take a non-lightlike vector v ∈ Ln , apply item (b) of Exercise 1.4.10 to take Λv to v using a reflection R (or composition of reflections). Then consider ( R ◦ Λ) v⊥ and use item (d) of Exercise 1.4.9. Remark.

• It is possible to show that Λ is the composition of at most n reflections. This result holds also in a more general context, see [15] for details. • On the other hand, this result is false in infinite dimensions. For instance, since Rn acts as the identity on H, the isometry −id cannot be written as a finite product of reflections. Exercise 1.4.12. A linear map T : Rnν → Rnν is skew-symmetric if, for any v, w ∈ Rnν , we have h Tv, wiν = −hv, Twiν . In addition to the maps presented in Exercise 1.4.1, the cross product (to be defined in Section 1.6) with a fixed vector of L3 also has this property. (a) Show that for any pseudo-orthonormal basis B = (u1 , u2 , . . . , un ) of Rnν , where the last ν vectors are timelike, the map T is written in blocks as:  

TS

 [T ]B = 

A

A> TT

 ,

where TS ∈ Mat(n − ν, R) and TT ∈ Mat(ν, R) are skew-symmetric blocks, and A ∈ Mat((n − ν) × ν, R). What is the trace of T? (b) Show that ker T = (Im T )⊥ and Im T = (ker T )⊥ . Hint. Prove the first identity directly and apply ⊥ on both sides for the second one. (c) If λ is an eigenvalue of T and v is an eigenvector associated to it, show that λ = 0 or v is lightlike (both can happen simultaneously).

42  Introduction to Lorentz Geometry: Curves and Surfaces

(d) If U ⊆ Rnν is a vector subspace, show that U is T-invariant if and only if U ⊥ is T-invariant. Exercise 1.4.13. Let T : Ln → Ln be a skew-symmetric linear map, as in Exercise 1.4.12 above. The map ET : Ln → Ln given by   1 1 2 2 ET = tr( T ) idLn − T 4π n is called the energy-momentum map associated to T. (a) Show that ET is self-adjoint according to h·, ·i L , that is,

h ET v, wi L = hv, ET wi L , for any v, w ∈ Ln . Show also that tr( ET ) = 0. (b) Show that if v is an eigenvector of T, then v is an eigenvector of ET (perhaps associated to another eigenvalue). “Conversely”, show that if v is an eigenvector of ET , then v is an eigenvector of T 2 . Remark. Maps with such property appear naturally in Physics. They are used to model electromagnetic fields — it is possible to define the electric (E) and the magnetic (B) fields associated to T. One can then study properties of T from E and B. More about that on [52]. Exercise† 1.4.14. Provide details for the proof of Proposition 1.4.22 (p. 38). . Exercise 1.4.15. Consider G = {τ +↑ , τ +↓ , τ −↑ , τ −↓ }, equipped with the matrix multiplication. (a) Show that G is a group and derive a “sign rule” for indices in the elements of G. (b) Show that G is isomorphic to the group of reflections in the diagonal of a square (which is a realization of the Klein group). Exercise 1.4.16 (Isogonal transformations). A linear map T : Rnν → Rnν is called isogonal if it preserves orthogonality, that is: if x, y ∈ Rnν are vectors such that h x, yiν = 0, then h Tx, Tyiν = 0. Show that: (a) In Rnν , homotheties are isogonal. In Rn orthogonal transformations are isogonal and, in Ln , Lorentz transformations are isogonal. In general, compositions of isogonal transformations are isogonal. (b) In Rn , every isogonal transformation is the composition of an orthogonal transformation and a homothety. Hint. If T is isogonal and T 6= 0, show that T is injective. See how T acts on an orthonormal basis, and define an orthogonal transformation C in the “opposite direction”. Show that C ◦ T is a homothety (using that C ◦ T is also isogonal). (c) In Ln , every injective isogonal transformation is the composition of a Lorentz transformation and a homothety. Hint. Adapt your proof for the item above, with care. (d) Provide a counterexample to the previous item when the transformation is not injective. Hint. What happens if the image of T is a light ray?

Welcome to Lorentz-Minkowski Space  43

INVESTIGATING O1 (2, R) AND O1 (3, R)

1.5 1.5.1

The group O1 (2, R) in detail

In order to fix the ideas studied so far, we’ll characterize the Lorentz transformations in L2 . For this end, we’ll use the properties of hyperbolic trigonometric functions mentioned in Exercise 1.3.1 (p. 27). Taking Proposition 1.4.22 (p. 38) into account, it suffices to describe O1+↑ (2, R) along with its cosets. If Λ = (λij )1≤i,j≤n ∈ O1+↑ (2, R), then its columns are Lorentz-orthonormal, that is,  2 2 = 1   λ11 − λ21 1 2 2 hΛei , Λe j i L = ηij =⇒ = −1 and, λ12 − λ22   λ11 λ12 − λ21 λ22 = 0 in addition, λ11 , λ22 ≥ 0. It follows, from the first two equations, that λ11 , λ22 ≥ 1. Therefore, there are unique t, s ∈ R≥0 such that λ11 = cosh t and λ22 = cosh s. Those equations also show that |λ21 | = sinh t and |λ12 | = sinh s. Since Λ is a proper transformation, it follows that cosh t cosh s − λ12 λ21 = 1, whence 0 ≤ cosh t cosh s − 1 = λ12 λ21 , meaning that λ12 and λ21 share the same sign or vanish simultaneously. Regardless of which sign this is, the third equation above gives 0 = cosh t sinh s − sinh t cosh s = sinh(s − t) =⇒ s = t. Hence, we have   cosh t sinh t Λ= sinh t cosh t

 or

 cosh t − sinh t , for some t > 0. − sinh t cosh t

In a more concise way, we write   cosh ϕ sinh ϕ Λ= , for some ϕ ∈ R. sinh ϕ cosh ϕ Such matrices clearly lie in O1+↑ (2, R), leading to O1+↑ (2, R)



=

  cosh ϕ sinh ϕ ϕ∈R . sinh ϕ cosh ϕ

This allows us to write the remaining cosets in O1 (2, R): in L2 , we have         1 0 1 0 −1 0 −1 0 +↑ +↓ −↑ −↓ τ = , τ = , τ = and τ = , 0 1 0 −1 0 1 0 −1 whence   cosh ϕ sinh ϕ ϕ∈R , =τ = − sinh ϕ − cosh ϕ    − cosh ϕ − sinh ϕ −↑ +↑ −↑ ϕ∈R , O1 (2, R) = τ · O1 (2, R) = sinh ϕ cosh ϕ    − cosh ϕ − sinh ϕ −↓ +↑ −↓ O1 (2, R) = τ · O1 (2, R) = ϕ∈R . − sinh ϕ − cosh ϕ O1+↓ (2, R)

+↓

· O1+↑ (2, R)



44  Introduction to Lorentz Geometry: Curves and Surfaces

It is important to notice that an analogous procedure leads to a similar characterization of SO(2, R). In this case, if C = (cij )1≤i,j≤2 satisfies  2 c11 + c221     c2 + c2 22 12  c11 c12 + c21 c22    c11 c22 − c12 c21

= = = =

1 1 , 0 1

there exists θ ∈ [0, 2π [ such that  C= 1.5.2

 cos θ − sin θ . sin θ cos θ

The group O1 (3, R) in (a little less) detail

Just repeating the strategy used in the previous section is not enough: there we described the elements using a single parameter, while now we’ll need three parameters to describe O1+↑ (3, R). We start with the: Proposition 1.5.1. Let Λ ∈ SO1 (3, R). Then 1 is an eigenvalue of Λ. Proof: It suffices to show that det(Id3 −Λ) = 0. We have: det(Id3 −Λ) = det(Id22,1 −Λ)

= det(Id2,1 Λ> Id2,1 Λ − Λ) = det(Id2,1 Λ> Id2,1 − Id3 ) det Λ = det(Id2,1 (Λ> − Id3 ) Id2,1 ) det Λ = det(Λ> − Id3 ) = − det(Id3 −Λ).

Remark. The previous result, and its proof, remain valid for any odd integer n. This ensures that every element of SO1 (3, R) fixes, pointwise, some line passing through the origin of L3 . Definition 1.5.2. Let Λ ∈ SO1 (3, R), Λ 6= Id3 , and v be an eigenvector of Λ associated to the eigenvalue 1. The map Λ is: (i) hyperbolic, if v is spacelike; (ii) elliptic, if v is timelike; (iii) parabolic, if v is lightlike. Remark.

• The elements of SO1 (3, R) fix pointwise exactly one direction. To wit, if there exist two distinct eigenvectors associated to 1, since det Λ = 1, we would have that Λ = Id3 (write the matrix of Λ in the basis of its eigenvectors). • The nomenclature above extends to Poincaré maps, considering linear parts.

Welcome to Lorentz-Minkowski Space  45

As before, in order to study O1 (3, R), it suffices to focus on O1+↑ (3, R). Consider Λ = (λij )1≤i,j≤3 ∈ O1+↑ (3, R) and let U be the line fixed by Λ. To justify the nomenclature in the above definition we start analyzing the  special cases when U is span {e1 } (hyperbolic), span {e3 } (elliptic), or span (0, 1, 1) (parabolic), that is, one line of each causal type. (1) U = span {e1 }: in this case we have Λe1 = e1 . Exercise 1.4.9 (p. 40) shows that Λe2 , Λe3 ∈ span {e2 , e3 }, and hence λ12 = λ13 = 0. So far, we can write   1 0 0 Λ = 0 λ22 λ23  0 λ32 λ33 but, since Λ ∈ O1+↑ (3, R), we have   λ22 λ23 ∈ O1+↑ (2, R). λ32 λ33 Therefore, there exists ϕ ∈ R such that   1 0 0 Λ = 0 cosh ϕ sinh ϕ  . 0 sinh ϕ cosh ϕ (2) U = span {e3 }: now we have Λe3 = e3 . As above, Λe1 and Λe2 lie in span {e1 , e2 }, thus λ31 = λ32 = 0. So far, the matrix has the form   λ11 λ12 0 Λ = λ21 λ22 0 , 0 0 1 but, since Λ ∈ O1+↑ (3, R), we have   λ11 λ12 ∈ SO(2, R). λ21 λ22 Therefore, there exists θ ∈ [0, 2π [ such that   cos θ − sin θ 0 Λ =  sin θ cos θ 0 . 0 0 1 (3) U = span {e2 + e3 }: we shall determine Λ ≡ [Λ]can indirectly, calculating the matrix [Λ] B = (θij )1≤i,j≤3 first, where B = (e1 , e2 , e2 + e3 ). Relative to this basis, the Lorentz product is given by

h( x1 , y1 , z1 ) B, ( x2 , y2 , z2 ) Bi L = x1 x2 + y1 y2 + y1 z2 + z1 y2 , for any ( x, y, z) B = xe1 + ye2 + z(e2 + e3 ). Since Λ(e2 + e3 ) = e2 + e3 , we see that the third column of [Λ] B is precisely the vector e3 . Furthermore, e1 ⊥ (e2 + e3 ) and, from Exercise 1.4.9 (p. 40), it follows that Λe1 ∈ (e2 + e3 )⊥ = span {e1 , e2 + e3 }. In other words, θ21 = 0.

46  Introduction to Lorentz Geometry: Curves and Surfaces

Expanding the determinant of [Λ] B from its already known third column, we have 1 = det[Λ] B = θ11 θ22 , that is, θ22 = 1/θ11 . So far, we have 

θ11

θ12

0



  −1 [Λ] B =  0 θ11 0 . θ31 θ32 1 In order to find out the remaining entries, we note that −1 1 = he2 , e2 + e3 i L = hΛe2 , Λ(e2 + e3 )i L = θ11 , 2 1 = h e2 , e2 i L = hΛe2 , Λe2 i L = θ12 + 1 + 2θ32 , 0 = h e1 , e2 i L = hΛe1 , Λe2 i L = θ12 + θ31 .

. Setting θ = θ31 we have



1  [ Λ ] B = 0 θ

 −θ 0  1 0 . −θ 2/2 1

In the canonical basis: 1 Λ = [Λ]can = [IdL3 ] B,can [Λ] B [IdL3 ]− B,can   −1   1 0 0 1 0 0 1 −θ 0     1 0 0 1 1 = 0 1 1 0 0 0 1 0 0 1 θ −θ 2/2 1   1 −θ θ   2 2 θ /2  . = θ 1 − θ /2 −θ 2/2 θ 1 + θ 2/2

Theorem 1.5.3. Let Λ ∈ O1+↑ (3, R). Then Λ is similar to one of the three matrices described above. Proof: Let U be the line fixed by Λ. One (and only one) of the following cases holds: (1) If Λ is hyperbolic, let u1 ∈ U be a unit vector. Extend {u1 } to an orthonormal basis (u1 , u2 , u3 ) of L3 , with u3 being timelike and future-directed8 . Let P ∈ O1↑ (3, R) be such that Pei = ui , for 1 ≤ i ≤ 3. Then P−1 ΛP ∈ O1+↑ (3, R) satisfies P−1 ΛPe1 = P−1 Λu1 = P−1 u1 = e1 . Hence there exists ϕ ∈ R such that   1 0 0 Λ = P 0 cosh ϕ sinh ϕ  P−1 . 0 sinh ϕ cosh ϕ +↑

fact, it turns out that the time direction of u3 is irrelevant, since O1 (3, R) is a normal subgroup of O1 (3, R). 8 In

Welcome to Lorentz-Minkowski Space  47

(2) If Λ is elliptic, let u3 ∈ U be a future-directed unit vector. Extend {u3 } to an orthonormal basis (u1 , u2 , u3 ) of L3 . As before, take P ∈ O1↑ (3, R) such that Pei = ui , for 1 ≤ i ≤ 3. Then P−1 ΛP ∈ O1+↑ (3, R) satisfies P−1 ΛPe3 = P−1 Λu3 = P−1 u3 = e3 . Then there exists θ ∈ [0, 2π [ such that   cos θ − sin θ 0 Λ = P  sin θ cos θ 0 P−1 . 0 0 1 (3) If Λ is parabolic, take u ∈ U of the form u = ( a, b, 1). Since a2 + b2 = 1, we have a = cos θ0 and b = sin θ0 for some θ0 ∈ [0, 2π [. Consider

  cos π2 − θ0
sin π2 − θ0
0 P = − sin π2 − θ0 cos π2 − θ0 0 . 0 0 1 Once more, P−1 ΛP ∈ O1+↑ (3, R) satisfies P−1 ΛP(e2 + e3 ) = P−1 Λu = P−1 u = e2 + e3 . Hence there exists θ ∈ R such that   1 −θ θ   θ 2/2  P−1 . Λ = P θ 1 − θ 2/2 −θ 2/2 θ 1 + θ 2/2

1.5.3

Rotations and boosts

Now we study two classes of Lorentz transformations that allow us to classify all of those which are proper and orthochronous. We start with the: Definition 1.5.4 (Pure Rotation). A pure rotation in Ln is a Lorentz transformation Λ = (λij )1≤i,j≤n ∈ O1+↑ (n, R) that is time fixing, that is, λnn = 1, and therefore is written as   0  ..  ΛS  .   ,  0    0 ··· 0 1 with ΛS ∈ SO(n − 1, R). If n = 4, every pure rotation is determined by a vector in R3 and an angle. More precisely: Theorem 1.5.5. For any transformation A ∈ SO(3, R) there exists a positive orthonormal basis B = (w1 , w2 , w3 ) and an angle θ ∈ [0, 2π [ such that   cos θ − sin θ 0 [ A] B =  sin θ cos θ 0 . 0 0 1 The direction of w3 is called the axis of rotation of A.

48  Introduction to Lorentz Geometry: Curves and Surfaces

Proof: First we show that 1 is an eigenvector of A: det(Id3 − A) = det( AA> − A)

= det A det( A> − Id3 )
= det ( A> − Id3 )> = det( A − Id3 ) = − det(Id3 − A). Hence, there exists a unit vector w3 , such that Aw3 = w3 . Let w1 , w2 ∈ R3 be vectors such that (w1 , w2 , w3 ) is a positive orthonormal basis of R3 . Relative to this basis, we have   a11 a12 0 [ A] B =  a21 a22 0 , 0 0 1 where ( aij )1≤i,j≤2 ∈ SO(2, R). Right after our study of O1 (2, R) we saw that there exists θ ∈ [0, 2π [ such that   cos θ − sin θ 0 [ A] B =  sin θ cos θ 0 , 0 0 1 as desired. With this in place, we proceed to consider a vector v ∈ Rn−1 as a spacelike vector in Ln = Rn−1 × R, identifying (v1 , . . . , vn−1 ) in Rn−1 with (v1 , . . . , vn−1 , 0) in Rn−1 × {0}. Then we have the: Definition 1.5.6 (Lorentz boost). A pure boost in the direction of a unit vector v ∈ Rn−1 is a Lorentz transformation Λ ∈ O1+↑ (n, R) that fixes pointwise the subspace v⊥ ⊆ Rn−1 . It is possible to describe such boosts, in arbitrary dimensions, in terms of v and a real parameter (the boost intensity). We’ll do this when n = 3, as an example. The difficulty in generalizing this only lies in the larger number of equations involved. The symmetry of such equations will be made clear in what follows. If Λ = (λij )1≤i,j≤3 is a pure boost in the direction of v = (v1 , v2 ) ∈ R2 then, by Exercise 1.4.9 (p. 40), the orthogonal complement (in L3 ) of v⊥ × {0} is Λ-invariant. This complement is precisely span {v, e3 }. In particular, Λv = γv + δe3

and Λ(−v2 , v1 , 0) = (−v2 , v1 , 0),

with γ2 − δ2 = 1, since hΛv, Λvi L = hv, vi L = 1. Writing out the entries in the above relations we have λ11 v1 + λ12 v2 = γv1

(1.5.1)

λ21 v1 + λ22 v2 = γv2

(1.5.2)

λ31 v1 + λ32 v2 = δ

(1.5.3)

λ11 v2 − λ12 v1 = v2

(1.5.4)

−λ21 v2 + λ22 v1 = v1 −λ31 v2 + λ32 v1 = 0.

(1.5.5)

Solving the system formed by Equations (1.5.1) and (1.5.4) we have λ11 = 1 + (γ − 1)v21

and λ12 = (γ − 1)v1 v2 .

(1.5.6)

Welcome to Lorentz-Minkowski Space  49

In the same way, Equations (1.5.2) and (1.5.5) lead to λ21 = (γ − 1)v1 v2

and λ22 = 1 + (γ − 1)v22 .

Hence we have γ = det ΛS > 0. Now, using that

hΛe3 , γv + δe3 i L = he3 , vi L = 0 and hΛe3 , (−v2 , v1 , 0)i L = he3 , (−v2 , v1 , 0)i L = 0 we get λ13 γv1 + λ23 γv2 − δλ33 = 0

(1.5.7)

−λ13 v2 + λ23 v1 = 0.

(1.5.8)

From Equations (1.5.7) and (1.5.8), we write λ13 and λ23 in terms of the remaining parameters: δλ33 v2 δλ33 v1 and λ23 = . λ13 = γ γ The relation hΛe3 , Λe3 i L = −1 can be read as     δλ33 v1 2 δλ33 v2 2 + − λ233 = −1. γ γ From this, using that v21 + v22 = 1 and λ33 > 0, we have λ33 = γ. Replacing v with −v if needed, we may suppose δ ≤ 0. Such a hypothesis is physically reasonable since, once Λe3 = δv + γe3 (check this), we have that an observer subject to a pure boost in the direction of v starts to see the event (0, 0, 1) (its immediate future) with a negative component (precisely δ) in the boost direction. See Figure 1.10 below:

e3

e3 Λe3

Λ

v

Λv

Figure 1.10: A Lorentz boost in the direction of v.

−1 + γ2 . From hΛei , Λe3 i L = 0 we have q
1 + (γ − 1)v21 δv1 + (γ − 1)v1 v2 δv2 − λ31 γ = 0 =⇒ λ31 = −v1 −1 + γ2 . p Similarly, λ32 = −v2 −1 + γ2 . Therefore, p   1 + (γ − 1)v21 ( γ − 1) v1 v2 − v1 −1 + γ2 p   Λ =  ( γ − 1) v1 v2 1 + (γ − 1)v22 −v2 −1 + γ2  . p p − v1 −1 + γ2 − v2 −1 + γ2 γ Then we can proceed with δ = −

p

50  Introduction to Lorentz Geometry: Curves and Surfaces

In arbitrary dimension, the expression of a boost in the direction of a unit vector v = (v1 , . . . , vn−1 ) is p   − v1 −1 + γ2   .. Idn−1 +(γ − 1)vv>  . .  . p B(v, γ) =   2 − v n −1 − 1 + γ    p p 2 2 γ − v 1 − 1 + γ · · · − v n −1 − 1 + γ The steps to check the previous claim are the following: 1. Write Λv ∈ span {v, en } in the form Λv = γv + δen , where we assume δ ≤ 0. 2. Take {u1 , . . . , un−2 }, a basis (orthonormal, for ease) of v⊥ . Solve the linear system given by Λui = ui , 1 ≤ i ≤ n − 2, and the entries of Λv = γv + δen to write the block Idn−1 +(γ − 1)vv> . By now we have γ = det ΛS > 0. 3. Solve the system given by the equations hΛu pi , Λen i L = 0, with 1 ≤ i ≤ n − 2 and hΛv, Λen i L = 0 to find out that λin = −vi −1 + γ2 , with 1 ≤ i ≤ n − 1. 4. Use hΛen , Λen i L = −1, kvk = 1 and λnn > 0 to get λnn = γ. 5. Finally, solve p the system given by hΛei , Λen i L = 0, with 1 ≤ i ≤ n − 1, to get λni = −vi −1 + γ2 . It is convenient to reparametrize the matrix above, choosing a hyperbolic angle ϕ ≥ 0 . such that γ = cosh ϕ, and then setting B∠ (v, ϕ) = B(v, γ). In this form, we can express some properties of such boosts in a more natural way. See Exercises 1.5.3, 1.5.4, 1.5.5, and 1.5.6. To close this section we state the following classification for the elements of O1+↑ (4, R): Theorem 1.5.7. Every Lorentz transformation Λ ∈ O1+↑ (4, R) has a unique decomposition as a product of a pure rotation R and a pure boost B: Λ = BR. Such decomposition can also be written in the reverse order: e Λ = R B, e is a pure boost in another direction. Furthermore, there exists a pair of pure where B rotations R1 and R2 , and a number ϕ ≥ 0 such that Λ = R1 B∠ (e1 , ϕ) R2 . The latter decomposition is also unique, except when Λ itself is already a pure rotation. The proof of this result relies on the representation of O1+↑ (4, R) in a group of complex matrices and on a particular decomposition of them. This exceeds the scope of this text and we refer to [33] for details.

Welcome to Lorentz-Minkowski Space  51

Exercises Exercise 1.5.1. We have seen that the map Λ : R → O1+↑ (2, R) given by:   . cosh θ sinh θ Λ(θ ) = sinh θ cosh θ is surjective. Complete the proof that Λ is a group isomorphism and conclude that any two matrices in O1+↑ (2, R) commute. Exercise 1.5.2. In Physics, it is usual to adopt the following convention for the product in L4 : . h(t1 , x1 , y1 , z1 ), (t2 , x2 , y2 , z2 )i L = t1 t2 − x1 x2 − y1 y2 − z1 z2 . Assume this convention only for this exercise. Consider . . > Herm(2, C) = { A ∈ Mat(2, C) | A† = A = A}, the set of all Hermitian complex matrices of order 2. (a) Let Φ : L4 → Mat(2, C) be given by Φ(t, x, y, z) =



 t + x y + iz . y − iz t − x

Check that Φ is linear, injective, and its image is Herm(2, C). (b) Show that hv, vi L = det Φ(v), for all v ∈ L4 . (c) Given M ∈ Mat(2, C), consider Ψ M : Herm(2, C) → Herm(2, C) defined by Ψ M ( A) = M† AM. Show that Ψ M is linear, its image in fact lies in Herm(2, C), and if det M ∈ S1 = {z ∈ C | |z| = 1}, then Ψ M preserves determinants. (d) For any M ∈ Mat(2, C) with det M ∈ S1 , let Λ M : L4 → L4 be the conjugation map given by Λ M = Φ−1 ◦ Ψ M ◦ Φ. Show that Λ M ∈ O1 (4, R). Remark.

• In this convention, the definitions of “spacelike” and “timelike” are swapped, but the Lorentz transformations are the same as in our convention. . • For 1 ≤ i ≤ 4, the matrices σi = Φ(ei ) are called Pauli matrices. • Not all Lorentz transformations in L4 can be obtained in this way. Can you find one of them? Exercise 1.5.3. Determine, as in the text, the general form of a given Lorentz boost Λ ∈ O1+↑ (4, R) in the direction of a unit vector v ∈ R3 . In particular, for v = e1 , and √ . v ∈ R≥0 determined by the relation γ = 1/ 1 − v2 , and ( x 0 , y0 , z0 , t0 ) = Λ( x, y, z, t), show that   x 0 = γ( x − vt)    y0 = y  z0 = z    t0 = γ(t − vx ), where γ is defined as in the text.

52  Introduction to Lorentz Geometry: Curves and Surfaces

Exercise 1.5.4. Let v ∈ Rn−1 be a unit vector and consider two Lorentz boosts B∠ (v, ϕ1 ), B∠ (v, ϕ2 ) ∈ O1+↑ (n, R) in the direction of v. Show that B∠ (v, ϕ1 ) B∠ (v, ϕ2 ) = B∠ (v, ϕ1 + ϕ2 ). In particular, conclude that B∠ (v, ϕ1 ) and B∠ (v, ϕ2 ) commute. Furthermore, show that B∠ (v, ϕ)−1 = B∠ (v, − ϕ). Hint. Use the matrix product definition and analyze four cases, thinking of the block division of a general boost. It is not as complicated as it seems. Exercise 1.5.5. If v, w ∈ Rn−1 are two given (spatial) unit vectors and B∠ (v, ϕ1 ), B∠ (w, ϕ2 ) ∈ O1+↑ (n, R) are the Lorentz boosts in the directions of v and w, respectively, then, in general, B∠ (v, ϕ1 ) and B∠ (w, ϕ2 ) do not commute. Give an example of this. Hint. You can find examples even when ϕ1 = ϕ2 . Exercise 1.5.6. In the same way we parametrized Lorentz boosts using an “angle” ϕ, it is usual to parametrize them in terms of the applied speed, v = tanh ϕ (see Exercise 1.3.7, p. 28). For a unit vector v ∈ Rn−1 , denote by Bsp (v, v) the Lorentz boost in the . direction of v with speed v (more precisely, Bsp (v, v) = B∠ (v, ϕ)). (a) Show that  Bsp (v, v1 ) Bsp (v, v2 ) = Bsp

v + v2 v, 1 1 + v1 v2

 .

Hint. Use Exercise 1.5.4 and the hyperbolic trigonometric identities. (b) Define in ]−1, 1[ the relativistic addition of speeds operation: . v + v2 v1 ⊕ v2 = 1 . 1 + v1 v2 Show that ]−1, 1[ equipped with this operation is an (abelian) group. Since we use geometric units, where the speed of light is c = 1, the numbers v1 , v2 ∈ ]−1, 1[ may be seen as speeds with direction. (c) Show that tanh : R → ]−1, 1[ is a group isomorphism. Exercise 1.5.7 (Margulis Invariant). Let F ∈ P(3, R) be a hyperbolic Poincaré transformation, written as F ( x) = Λx + w, with Λ ∈ O1+↑ (3, R) and w ∈ L3 . (a) Show that Λ has three positive eigenvectors, 1/λ < 1 < λ. Conclude from Exercise 1.4.9 (p. 40), the eigenspaces associated to the eigenvalues λ and 1/λ are light rays. (b) Let vλ , v1/λ be lightlike and future-directed eigenvectors associated to the eigenvalues λ and 1/λ, respectively, and v1 be a unit eigenvector associated to 1 such that the basis B = (vλ , v1 , v1/λ ) is positive. Show that F fixes a single affine straight line that is parallel to the eigenvector v1 , acting as a translation in this line. In other words, show that there exist p ∈ L3 and α F ∈ R such that F ( p + tv1 ) = p + tv1 + α F v1 , The number α F is the Margulis invariant of F.

for all t ∈ R.

Welcome to Lorentz-Minkowski Space  53

Hint. Solve a system for the coordinates of p in the basis B. Remark. The choice of v1 in a way that the basis is positive is necessary to make α F well-defined. Changing the sign of v1 would change the sign of α F . (c) Show that α F = hw, v1 i L . Use this to show that if F1 , F2 ∈ P(3, R) are hyperbolic and conjugated by an element of O1 (3, R), it holds that α F1 = α F2 , justifying the name “invariant”. (d) Show that for all n > 0, we have α Fn = nα F . (e) Discuss the orientation of the eigenbasis of Λ−1 in terms of the orientation of B. Use this to show that α Fn = nα F holds, even for n < 0.

1.6

CROSS PRODUCT IN Rnν

In a first Analytic Geometry class, one learns the concept of cross product between two vectors in space R3 : if u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 ), we have e1 e2 e3 u × v = u1 u2 u3 , v1 v2 v3 where the above determinant is formal, and used to find the components of the cross product relative to the canonical basis. This is applied, among other things, to understand orthogonality relations and to compute volumes. In this section, we will see how to extend this definition to pseudo-Euclidean spaces of arbitrary (finite) dimension, again emphasizing the particular features of the case n = 3. Definition 1.6.1. The cross product of v1 , . . . , vn−1 ∈ Rnν , according to h·, ·iν , is the . unique vector v1 × · · · × vn−1 = v ∈ Rnν such that

hv, xiν = det( x, v1 , · · · , vn−1 ), for all x ∈ Rnν . Remark.

• The existence and uniqueness of the cross product v1 × · · · × vn−1 are guaranteed by Riesz’s Lemma applied to the particular linear functional f : Rnν → R given by f ( x) = det( x, v1 , · · · , vn−1 ). • In Rn we will write × E , and in Ln we will write × L . In particular, for n = 3 we will write u × E v, and similarly in L3 . • It is also usual to use the symbol ∧ instead of ×. For example, one could write u ∧ L v for the Lorentzian cross product of u, v ∈ L3 . The above definition is not very efficient to explicitly compute cross products. The proposition below not only solves this issue, but also shows that the above definition is indeed an extension of the usual cross product in R3 :

54  Introduction to Lorentz Geometry: Curves and Surfaces

Proposition 1.6.2. Let B = (ui )in=1 be a positive and orthonormal basis for Rnν and . v j = ∑in=1 vij ui ∈ Rnν , for 1 ≤ j ≤ n − 1, be given vectors. Writing ei = eui for the indicators of the elements in B, we have that: e1 u 1 · · · e n u n v11 ··· vn1 v 1 × · · · × v n −1 = . .. . .. .. . . v1,n−1 · · · vn,n−1 Proof: For simplicity, write v = v1 × · · · × vn−1 . By Lemma 1.2.32 (p. 16), we have that v = ∑in=1 ei hv, ui iν ui , but from the definition of cross product, each component is given by v11 ··· vi−1,1 vi+1,1 ··· vn1 .. .. .. . .. .. hv, ui iν = (−1)i+1 ... . . . . . v1,n−1 · · · vi−1,n−1 vi+1,n−1 · · · vn,n−1 With this, it suffices to recognize the orthonormal expansion of v as the expansion of the (formal) determinant given in the statement of this result via the first row. Remark. In particular, if u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 ) are vectors in L3 , we have that e1 e2 − e3 u × L v = u1 u2 u3 . v1 v2 v3 Next, we will register a few general properties of the cross product, which directly follow from the usual properties of the determinant function: Proposition 1.6.3. Let v1 , . . . , vn−1 ∈ Rnν and λ ∈ R. Then the cross product is: (i) alternating, that is, switching two of its entries changes the sign of the result. In particular, for n = 3, the cross product is skew-symmetric (v1 × v2 = −v2 × v1 ); (ii) such that v1 × · · · × vn−1 = 0 if and only if {v1 , . . . , vn−1 } is linearly dependent; (iii) multilinear, that is, linear in each one of its entries. In particular, for n = 3, we have:

(v1 + λu) × v2 = v1 × v2 + λ(u × v2 ) v1 × (v2 + λu) = v1 × v2 + λ(v1 × u), for all u ∈ R3ν and λ ∈ R; (iv) orthogonal to each one of its entries:

hv1 × · · · × vn−1 , vi iν = 0, for each 1 ≤ i ≤ n − 1; (v) for n = 3, cyclic, that is, it satisfies the cyclic identity:

h v1 × v2 , v3 i ν = h v1 , v2 × v3 i ν .

Welcome to Lorentz-Minkowski Space  55

Remark.

• Recall that a permutation σ ∈ Sk may be decomposed as the composition of a sequence of permutations which only move two elements (these permutations are called transpositions). We say that sgn(σ) = 1 if this decomposition has an even number of transpositions, and −1 if it has an odd number of transpositions. Even though this decomposition in terms of transpositions is not unique, the parity of the number of transpositions is an invariant of σ, so that sgn(σ ) is indeed well-defined. For more details, see [23]. • With this terminology, saying that the cross product is alternating is equivalent to saying that given v1 , . . . , vn−1 ∈ Rnν , we have that vσ(1) × · · · × vσ(n−1) = sgn(σ ) v1 × · · · × vn−1 , for each σ ∈ Sn−1 . Some additional properties of the cross product for n = 3 are stated in Exercise 1.6.4. Furthermore, note that (visually), the Lorentzian product of v1 . . . , vn−1 ∈ Ln is the Euclidean product of these same vectors, but reflected on the hyperplane xn = 0. More precisely: Idn−1,1 (v1 × L · · · × L vn−1 ) = v1 × E · · · × E vn−1 . The verification of this fact is left as Exercise 1.6.5. Moreover, we may analyze the causal type of the cross product in terms of the causal type of the space spanned by the vectors who enter the product. The following proposition is nothing more than a reformulation of Theorem 1.2.20 (p. 12): Proposition 1.6.4. Let v1 , . . . , vn−1 ∈ Ln be linearly independent and consider the . hyperplane S = span {v1 , . . . , vn−1 }. Then: (i) v1 × · · · × vn−1 is spacelike ⇐⇒ S is timelike; (ii) v1 × · · · × vn−1 is timelike ⇐⇒ S is spacelike; (iii) v1 × · · · × vn−1 is lightlike ⇐⇒ S is lightlike. Proposition 1.6.5. Let u1 , . . . , un−1 , v1 , . . . , vn−1 ∈ Rnν . Then we have that
hu1 × · · · × un−1 , v1 × · · · × vn−1 iν = (−1)ν det (hui , v j iν )1≤i,j≤n−1 . Proof: If (ui )in=−11 or (v j )nj=−11 is linearly dependent, there is nothing to do. Suppose that both of them are linearly independent. Since both sides of the proposed equality are linear in each one of the 2n − 2 present variables, and both the cross product and the determinant function are alternate, we may assume without loss of generality that uk = eik and v` = e j` , where (ei )in=1 is the canonical basis for Rnν and 1 ≤ i1 < · · · < in−1 ≤ n and 1 ≤ j1 < · · · < jn−1 ≤ n. We will proceed with the analysis by cases, in terms of the indices i∗ and j∗ omitted in each of the (n − 1)-uples of indices being considered.

• If i∗ 6= j∗ , both sides equal zero. To wit, the left side equals hei∗ , e j∗ iν = 0 and the determinant on the right side has the i∗ -th row and j∗ -th column consisting only of zeros.

56  Introduction to Lorentz Geometry: Curves and Surfaces

• If 1 ≤ i∗ = j∗ ≤ n − ν, the left side is hei∗ , ei∗ iν = 1, while the right side is (−1)ν det Idn−1,ν = (−1)ν (−1)ν = 1. • If n − ν < i∗ = j∗ ≤ n, the left side is hei∗ , ei∗ iν = −1, while the right side is (−1)ν det Idn−1,ν−1 = (−1)ν (−1)ν−1 = −1. Corollary 1.6.6 (Lagrange’s Identities). Let u, v ∈ R3ν . Then:

ku × E vk2E = kuk2E kvk2E − hu, vi2E , hu × L v, u × L vi L = −hu, ui L hv, vi L + hu, vi2L . The orientation of the bases chosen for Rnν will be very important for the development of the theory to be presented in the following chapters. We recall that the canonical basis for Rnν is, by convention, positive. If v1 , . . . , vn−1 ∈ Rnν are linearly independent,
don’t span a lightlike hyperplane, and v = v1 × · · · × vn−1 , then B = v1 , . . . , vn−1 , v is a basis for Rnν and it is natural to wonder whether such basis is positive or negative. The answer comes from the sign of the determinant containing those vectors (be it on rows or columns). We have that det(v1 , . . . , vn−1 , v) = (−1)n−1 det(v, v1 , . . . , vn−1 ) = (−1)n−1 hv, viν and, thus, positivity of the basis B depends not only on the parity of n, but also on the causal character of v. Explicitly: if v is spacelike, B is positive if n is odd, and negative if n is even; if v is timelike, B is positive if n is even, and negative if n is odd. Check this for the canonical basis in R4 , for example. In particular, for n = 3, we may represent all cross products between vectors in the canonical basis by the following two diagrams: e1

×E e3

− e1

e1

×L

e2

×L

e2

− e3

e3

(a) In R3

(b) In L3

Figure 1.11: Summarizing cross products in R3ν . The cross products are obtained by following the arrows. For example, we have that e2 × E e3 = e1 and e1 × L e2 = −e3 . In R3 , following the opposite direction of the given arrows will yield results with the opposite sign (since × E is skew-symmetric), for example, e1 × E e3 = −e2 . But in L3 , this does not work due to the influence of causal types: note that e3 × L e2 = −e1 6= −(−e1 ) = e1 . The cross products which may not be obtained directly from the diagram 1.11b may be found out by using that × L is skew-symmetric. Remark. Note that the diagrams remain true if we replace the canonical basis by any positive and orthonormal basis, provided that in L3 the timelike vector is the last one (corresponding to e3 ).

Welcome to Lorentz-Minkowski Space  57

1.6.1

Completing the toolbox

We know that the trace and the determinant of a linear operator are invariant under change of basis. In Chapter 3, to define certain curvatures of a surface, we need to extend those concepts for bilinear maps. Restricting ourselves to orthonormal bases, we have the following results: Lemma 1.6.7. Let Z be any vector space and B : Rnν × Rnν → Z be a bilinear map. So, if (vi )in=1 and (wi )in=1 are orthonormal bases for Rnν , we have that n

∑ evi B ( v i , v i ) =

i =1

n

∑ ewi B ( w i , w i ).

i =1

The above quantity is then called the trace of B relative to h·, ·iν and it is denoted by trh·,·iν B. Proof: It is possible to show that both considered bases have necessarily ν timelike vectors (this result is known as Sylvester’s Law of Inertia, see [54]). With this, we may (reordering the bases if needed) assume that hvi , v j iν = hwi , w j iν = ηijν and that, in . particular, that ei = evi = ewi for 1 ≤ i ≤ n. Applying Lemma 1.2.32 (p. 16) successively, we have ! wj = whence

n

n

n

k =1

i =1

k =1

∑ e k h w j , v k i ν v k = ∑ ∑ ei e k h w j , v k i ν h v k , w i i ν

wi ,

n

∑ ei ek hw j , vk iν hvk , wi iν = δij ,

1 ≤ i, j ≤ n,

k =1

by linear independence of the (w j )nj=1 . With this, we compute n

n

n

n

i =1

i =1

j =1

k =1

∑ ei B ( v i , v i ) = ∑ B ∑ e j h v i , w j i ν w j , ∑ e k h v i , w k i ν w k n

=

∑

j,k =1

n

ek

∑ ei e j h w j , v i i ν h v i , w k i ν

!

! B(w j , wk )

i =1

n

=

∑

ek δjk B(w j , wk )

j,k =1 n

=

∑ ek B ( w k , w k ),

k =1

as wanted. Lemma 1.6.8. Let B : Rnν × Rnν → R be a bilinear functional. If (vi )in=1 and (wi )in=1 are orthonormal bases for Rnν , the matrices of B relative to the given bases are related by    

det B(vi , v j ) 1≤i,j≤n = det B(wi , w j ) 1≤i,j≤n . The above quantity is called the determinant of B relative to h·, ·iν , and it is denoted by deth·,·iν B.

58  Introduction to Lorentz Geometry: Curves and Surfaces

Proof: Write, for each j, v j = ∑im=1 aij wi . Since A = ( aij )1≤i,j≤n is the change of basis matrix between orthonormal bases, the Exercise 1.4.7 (p. 40) ensures that A ∈ Oν (n, R) and so | det A| = 1. This way, ! ! n

∑

B ( vi , v j ) = B

n

aki wk ,

k =1

∑

n

a` j w`

=

`=1

∑

n

aki

k =1

∑ B(wk , w` ) a` j

,

`=1

which, in matrix terms, is rewritten as

B(vi , v j ) 1≤i,j≤n = A> B(wi , w j ) 1≤i,j≤n A and, thus, det



B ( vi , v j )




1≤i,j≤n

= det



B ( wi , w j )




1≤i,j≤n

as wanted. Lemma 1.6.9. Let v1 , . . . , vn−1 ∈ Rnν be linearly independent, S be their linear span, and T : S → S be a linear map. Then: n −1

(i)

∑ (v1 × · · · × vi−1 × Tvi × vi+1 × · · · × vn−1 ) = (tr T ) v1 × · · · × vn−1 ;

i =1

(ii) Tv1 × · · · × Tvn−1 = det( T ) v1 × · · · × vn−1 . Proof: Write [ T ] B = ( aij )1≤i,j≤n−1 , where B = (v1 , . . . , vn−1 ). (i) We have that: n −1

∑ (v1 × · · · ×vi−1 × Tvi × vi+1 × · · · × vn−1 ) =

i =1

n −1

=

n −1

∑

v 1 × · · · × v i −1 ×

n −1

n −1

i =1

ji =1

i =1

= (∗)

=

∑ a ji i v ji

!

!

× v i +1 × · · · × v n −1

ji =1

∑ ∑ a ji i v1 × · · · × vi−1 × v ji × vi+1 × · · · × vn−1

!

n −1

∑ (aii v1 × · · · × vn−1 )

i =1

= tr([ T ] B) v1 × · · · × vn−1 = tr( T ) v1 × · · · × vn−1 , where in (∗) we have used that if ji 6= i, the cross product vanishes (one of the factors repeats).

Welcome to Lorentz-Minkowski Space  59

(ii) We will use (one) definition9 of determinant. We have: ! n −1

Tv1 × · · · × Tvn−1 =

∑

a i1 1 v i1

×···×

(∗)

=

n −1

n −1

i1 ,...,in−1 =1

j =1

∑

∑

∑

∏ ai j j

n −1

σ ∈ S n −1

=

∑

ain−1 ,n−1 vin−1

i n −1 =1

i1 =1

=

!

n −1

∏ aσ( j) j

! v i 1 × · · · × v i n −1

! v σ (1) × · · · × v σ ( n −1)

j =1

n −1

sgn(σ )

σ ∈ S n −1

∏ aσ( j) j

! v 1 × · · · × v n −1

j =1

= det([ T ]> B) v1 × · · · × vn−1 = det( T ) v1 × · · · × vn−1 , where in (∗) we have used that the summand vanishes in case there is any repetition of indices, so that the summation may be reindexed with permutations. More precisely, for each non-zero term in the sum there is a unique σ ∈ Sn−1 , namely, the one such that σ(k) = ik , for each 1 ≤ k ≤ n − 1.

Exercises Exercise 1.6.1. Consider the vectors u = (−1, 3, 1), v = (2, 1, 1), and w = (0, 1, 1). Prove that there is no vector x such that x × E u = v and x ⊥ E w. However, there is a unique vector x satisfying x × L u = v and x ⊥ L w. Determine this vector. Exercise 1.6.2. Suppose that u, v ∈ R3ν are linearly independent vectors. Let w ∈ R3ν be such that w × u = w × v = 0, where × is the cross product in the ambient space under consideration. Show that w = 0 (that is, the result is true in both R3 and L3 ). Exercise 1.6.3. Prove that if u, v ∈ R3 are vectors with hu, vi E = 0 and u × E v = 0, then u or v must vanish. Verify with a counter-example that the corresponding result in L3 is false. Exercise 1.6.4. Let u, v, w ∈ R3ν be any vectors. (a) Prove the double cross product identities: u × E (v × E w) = hu, wi E v − hu, vi E w u × L (v × L w) = hu, vi L w − hu, wi L v (b) Prove the Jacobi identities: u × E (v × E w) + v × E (w × E u) + w × E (u × E v) = 0 u × L (v × L w) + v × L (w × L u) + w × L (u × L v) = 0 9 For

more details, see [8]

60  Introduction to Lorentz Geometry: Curves and Surfaces

Hint. Use item (a). Exercise 1.6.5. Let v1 , . . . , vn−1 ∈ Rnν . Show that Idn−1,1 (v1 × L · · · × L vn−1 ) = v1 × E · · · × E vn−1 . Exercise† 1.6.6. Verify that the diagrams given in Figure 1.11 (p. 56) indeed work. Exercise 1.6.7. Let B : R3ν × R3ν → R3ν be a skew-symmetric bilinear map. Show that there is a unique linear map T : R3ν → R3ν such that T (v × w) = B(v, w), for all v, w ∈ R3ν . Exercise 1.6.8 (Representation of × E ). Given any x = ( x1 , x2 , x3 ) ∈ R3 \ {0}, we may see the cross product with x as a linear map T = x× E : R3 → R3 . (a) Write the matrix of T relative to the canonical basis and check that its characteristic polynomial is c T (t) = t3 + h x, xi E t. (b) Describe the eigenspace associated to 0 (i.e., ker T). Regarding R3 inside C3 and considering an eigenvector v ∈ C3 associated to the eigenvalue ik xk E , compute the cross products x × E Re(v) and x × E Im(v). Conclude that Re(v) and Im(v) are both orthogonal to x. Verify also that Re(v) and Im(v) are orthogonal. Hint. Proposition 1.6.5 (p. 55). Remark.

• The interpretation of the eigenvalue ik xk E is given as follows: the restriction of T to the normal plane to x, which is spanned by Re(v) and Im(v), acts as a rotation of 90◦ counterclockwise (inside the plane), followed by a dilation of factor k xk E . • One possible v is   v = − x1 x3 − x2 k xk E i, − x2 x3 + x1 k xk E i, x12 + x22 . Exercise 1.6.9 (Representation of × L ). Given any x = ( x1 , x2 , x3 ) ∈ L3 \ {0}, we may see the cross product with x as a linear map T = x× L : L3 → L3 . (a) Write the matrix of T relative to the canonical basis and check that its characteristic polynomial is c T (t) = t3 − h x, xi L t. (b) Suppose that x is spacelike. The plane which has x as its normal direction is timelike and intersects the lightcone of L3 in two light rays. Show that T is diagonalizable and that the directions of these light rays are eigenvectors of T. (c) Suppose that x is timelike. We have a situation similar to what was discussed in Exercise 1.6.8: if v ∈ C3 is an eigenvector associated to the eigenvalue ik xk L , compute x × L Re(v) and x × L Im(v). Conclude again that Re(v) and Im(v) are both Lorentzorthogonal to x. Verify also that Re(v) and Im(v) are Lorentz-orthogonal. Remark. If x is timelike, one possible v is   v = x1 x3 − x2 k xk L i, x2 x3 + x1 k xk L i, x12 + x22 . Exercise 1.6.10. Let ν ∈ {0, 1} and v1 , . . . , vn−1 ∈ Rnν .

Welcome to Lorentz-Minkowski Space  61

(a) If θ is the (Euclidean) angle between v1 × L · · · × L vn−1 (or v1 × E · · · × E vn−1 ) and the hyperplane e⊥ n , show that:

h v 1 × L · · · × L v n −1 , v 1 × L · · · × L v n −1 i L = = hv1 × E · · · × E vn−1 , v1 × E · · · × E vn−1 i E cos 2θ. (b) Show that

k v 1 × L · · · × L v n −1 k L ≤ k v 1 × E · · · × E v n −1 k E , with equality if and only if the vectors are linearly dependent, or the hyperplane spanned by v1 , . . . , vn−1 is horizontal or vertical. Exercise 1.6.11. Show that trh·,·i h·, ·i = n and that deth·,·i h·, ·i = (−1)ν . Exercise 1.6.12. Let T : Rnν → Rnν be a linear operator. Define a bilinear functional . BT : Rnν × Rnν → R by BT (v, w) = h Tv, wi. Show that trh·,·i BT = tr T and that deth·,·i BT = (−1)ν det T. Exercise 1.6.13. Let T1 , T2 : Rnν → Rnν be two linear operators and consider a bilinear functional B : Rnν × Rnν → R. Define the ( T1 , T2 )-pull-back of B . ( T1 , T2 )∗ B : Rnν × Rnν → R, by ( T1 , T2 )∗ B(v, w) = B( T1 v, T2 w). Show that deth·,·i (( T1 , T2 )∗ B) = det T1 det T2 deth·,·i B.

CHAPTER

2

Local Theory of Curves

INTRODUCTION In this chapter we start our study of curves in Rn and in Ln . In Section 2.1 we present the definition of a parametrized curve in Euclidean or Lorentzian spaces of arbitrary finite dimension and extend the concept of causal type, seen in Chapter 1, to curves in Ln . We provide a wide range of examples and introduce the concept of regularity for a curve. After that, we show that there are no lightlike or timelike closed curves in Ln , emphasizing the physical interpretation of this result in terms of the causality in Ln . In what follows, we study two quantities naturally associated to a curve: arclength (proper time, for timelike curves) and energy. We discuss the viability and geometric interpretation of unit speed reparametrizations, show an alternative reparametrization for lightlike curves (arc-photon), and close this section with a motivation for the definition of congruence between curves. From now on, to deal with Euclidean and Lorentzian ambient spaces simultaneously, we denote the scalar products simply by h·, ·i, omitting the subscript indexes E and L. In Section 2.2, we particularize the results seen in the previous section to curves in the plane R2ν . We start showing that each curve is locally the graph of some smooth function, and then we characterize all the lightlike curves in the plane. With this in place, we then turn our attention to unit speed curves, defining the Frenet-Serret frame for such curves and the notion of oriented curvature: the required invariant to classify the remaining curves. After that we generalize this construction for non-unit speed curves and derive a formula for κα (t). We provide interpretations for the sign of the curvature, using Taylor formulae, and discuss osculating circles. The end of this section comes with Theorem 2.2.12, which characterizes all the plane curves with constant curvature and Theorem 2.2.13, which shows that the curvature is the single geometric invariant necessary to recover the trace of a plane curve, up to an isometry of R2ν . Lastly, in Section 2.3, we study curves in space. We start with the Frenet-Serret frame, for a certain class of curves (to be called admissible), and we introduce the torsion of a curve which, together with the curvature, determines the trace of the curve, up to an isometry of the ambient (see Theorem 2.3.20, p. 112). In the following we show, through examples, how the Frenet-Serret frames carries geometric information of the curve, via curvature and torsion. For instance, in Proposition 2.3.11 (p. 106) we see that an admissible curve is planar if and only if its torsion vanishes. We also give conditions to the trace of an admissible curve to be contained in a sphere of the ambient space (Theorem 2.3.15, p. 108). We cannot neglect an important class of curves: the helices, which are characterized in terms of the ratio τα /κα (Lancret’s Theorem, p. 110). We conclude this chapter with the so-called Cartan frame for lightlike or semi-lightlike curves, introducing a single invariant, the pseudo-torsion, for this class of curves. We use this tool to achieve 63

64  Introduction to Lorentz Geometry: Curves and Surfaces

results analogous to the previous ones in this context, being aware of the interpretations in each case. In particular, we obtain a lightlike version of Lancret’s Theorem (p. 125), show that every semi-lightlike curve is planar, and state a version of the Fundamental Theorem of Curves (p. 125).

2.1

PARAMETRIZED CURVES IN Rnν Throughout this chapter, I ⊆ R will denote a (bounded or not) open interval.

Definition 2.1.1. A parametrized curve is a smooth map α : I → Rnν . Its image, α( I ), is the trace of α. For each t ∈ I, the derivative α0 (t) is the velocity vector of α in t. Remark. We use “smooth” meaning “infinitely differentiable”, or “of class C∞ ”. Despite that requirement in the definition, most of the results presented here are valid for C3 or C4 curves. Hence we’ll allow ourselves to deal only with “sufficiently differentiable” parametrizations.. In Chapter 1, we defined the causal type of vectors in Ln , and extended this definition to its subspaces (including affine ones). Now we’ll have this concept for curves in Ln . An affine subspace naturally associated to each point of a curve is its tangent line at that point, whose causal type is determined by any spanning vector. This leads to the: Definition 2.1.2. Let α : I → Rnν be a curve and t0 ∈ I. Then α is: (i) spacelike at t0 , if α0 (t0 ) is a spacelike vector; (ii) timelike at t0 , if α0 (t0 ) at a timelike vector; (iii) lightlike in t0 , if α0 (t0 ) at a lightlike vector. If the causal type of α0 (t) is the same for all t ∈ I, this will be defined as the causal type . of the curve α. If so, and α is not lightlike, we set the indicator or α as eα = eα0 (t) . Remark.

• The continuity of I 3 t 7→ hα0 (t), α0 (t)i ∈ R, implies that “being spacelike” and “being timelike” are open properties, that is, if α is spacelike or timelike at t0 , then α retains this causal type for all t in some open interval around t0 . • If α is spacelike in t0 and timelike in t1 , the Intermediate Value Theorem ensures that there exists t2 between t0 and t1 such that α is lightlike at t2 . Informally, the causal type of a curve can’t jump between space and time. • Our focus here is the local geometry of curves, hence the remarks above allow us to assume in some proofs, possibly shrinking the domain of the curve, that α has constant causal type. • If α is timelike in Ln , we say that it is future-directed or past-directed according to time orientation of velocity vectors α0 (t). Again, by continuity, a timelike curve is either future-directed or past-directed (i.e., it cannot jump from one time orientation to the other). Example 2.1.3. (1) Points: let p ∈ Ln and consider α : R → Ln given by α(t) = p. Then α0 (t) = 0, hence α is spacelike.

Local Theory of Curves  65

(2) Straight lines: the line passing through p ∈ Ln with direction v ∈ Ln admits a parametrization given by α(t) = p + tv, for t ∈ R. Then α0 (t) = v, and α has the same causal type of v. This agrees with the previous definition for the causal type of a straight line. (3) Euclidean circle: consider α : R → L2 , given by α(t) = (cos t, sin t). Thinking geometrically of its image, we see that α is:

• spacelike for t ∈ ]π/4 + kπ, 3π/4 + kπ [; • timelike for t ∈ ]−π/4 + kπ, π/4 + kπ [; • lightlike for t = π/4 + kπ/2, for all k ∈ Z.

Figure 2.1: The Euclidean unit circle in L2 .

√ (4) Consider the curve α : R → L3 given by α(t) = (−6t, 2 7t, t2 ). We have that √ α0 (t) = (−6, 2 7, 2t), and then: hα0 (t), α0 (t)i L = 64 − 4t2 . Analyzing the sign of this last expression we see that α is: • spacelike on ]−4, 4[; • timelike on ]−∞, −4[ ∪ ]4, +∞[; • lightlike at t = −4 and t = 4. (5) A Euclidean circle of radius r > 0, centered in p ∈ L3 . Let α : R → L3 , given by α(t) = p + (r cos t, r sin t, 0). Then α0 (t) = (−r sin t, r cos t, 0), hence hα0 (t), α0 (t)i L = r2 > 0, so α is spacelike. Note that α is a planar curve, contained in a spacelike plane. (6) A hyperbola centered in p ∈ L3 and radius r > 0. Let α : R → L3 be given by α(t) = p + (0, r cosh t, r sinh t). Now α0 (t) = (0, r sinh t, r cosh t), and hα0 (t), α0 (t)i L = −r2 < 0, so α is timelike. This is an example of a timelike curve lying in a timelike plane. (7) In analogy to the previous item, if p ∈ L3 and r > 0, the hyperbola α : R → L3 given by α(t) = p + (0, r sinh t, r cosh t) is spacelike, but lying in a timelike plane.

66  Introduction to Lorentz Geometry: Curves and Surfaces

(8) Euclidean Helix: consider α : R → L3 , given by α(t) = ( a cos t, a sin t, bt), a, b > 0. We give the causal type of this helix in terms of a and b.

Figure 2.2: A lightlike Euclidean helix in L3 . We have α0 (t) = (− a sin t, a cos t, b), and then

hα0 (t), α0 (t)i L = a2 − b2 = ( a + b)( a − b). It follows directly that α is:

• spacelike, if a > b; • timelike, if a < b; • lightlike, if a = b. Remark. In general, a planar curve in L3 doesn’t have the same causal type as the plane containing it. Take for example the x-axis and any plane containing it. One can obtain planes of each causal type by rotating this plane around the x-axis. Definition 2.1.4. Let α : I → Rnν to be a parametrized curve. We say that α is regular if α0 (t) 6= 0, for all t ∈ I. If there is t0 ∈ I such that α0 (t0 ) = 0, we say that t0 is a singular point of this parametrization. Remark. The regularity hypothesis is reasonable, since it excludes, among other situations, the constant curve whose image is a point (as it doesn’t fit our intuitive notion of a “curve”). Theorem 2.1.5. Let α : I → Ln be a curve. If α is lightlike or timelike, then α is regular. Proof: Write α(t) = ( x1 (t), . . . , xn (t)). Then x10 (t)2 + · · · + xn0 −1 (t)2 − xn0 (t)2 ≤ 0. If xn0 (t) = 0, then xi0 (t) = 0 for all 1 ≤ i ≤ n − 1, and α0 (t) = 0. Hence α is spacelike in t. Therefore xn0 (t) 6= 0, for all t ∈ I and α0 (t) 6= 0. It is sometimes interesting to consider curves defined in closed intervals [ a, b]. In this case the map α : [ a, b] → Rnν is smooth if it is the restriction of some smooth map defined in an open interval containing [ a, b].

Local Theory of Curves  67

Definition 2.1.6. A smooth map α : [ a, b] → Rnν is a closed parametrized curve if α and all of its derivatives agree in the boundary of its domain. In other words, for all k ≥ 0, we have α(k) ( a) = α(k) (b). Our first result for closed parametrized curves is: Theorem 2.1.7. In Ln , there are no closed lightlike or closed timelike parametrized curves. Proof: Let α : [ a, b] → Ln be a curve satisfying α( a) = α(b). Writing it in coordinates as α(t) = ( x1 (t), . . . , xn (t)), we shall see that α is spacelike (excluding the singular case) at some point. We have xn ( a) = xn (b) and, by Rolle’s Theorem, there exists t0 ∈ ] a, b[ such that xn0 (t0 ) = 0. Hence

hα0 (t0 ), α0 (t0 )i L = x10 (t0 )2 + · · · + xn0 −1 (t0 )2 ≥ 0. If hα0 (t0 ), α0 (t0 )i L = 0, then α0 (t0 ) = 0 and α is singular in t0 . If hα0 (t0 ), α0 (t0 )i L > 0, then α is spacelike in t0 . Remark. The existence of α0 (t) for all t ∈ [ a, b] is all we needed in the proof above. We didn’t use α(k) ( a) = α(k) (b), for all k ≥ 1. Give a geometric interpretation for this. In the setting of Special Relativity, future-directed timelike curves are the worldlines of particles with positive mass, while future-directed lightlike curves model the trajectory of massless particles, like photons. In this way, Theorem 2.1.7 says that it is impossible for an observer in Ln to go back to the past, before they are born, to kill their own grandfather. This means that the causality of Lorentz-Minkowski is “well-behaved”. Therefore, spacetime models that admit closed timelike or closed lightlike curves are not usually considered in the General Relativity context, but are of interest in Causality Theory (see, for example, [7]). Compare the above result with Exercise 1.2.12 (p. 18). Definition 2.1.8. Let α : I → Rnν be a parametrized curve and a, b ∈ I. The arclength of α between a and b is defined by Lba [α]

. =

Z b a

kα0 (t)kν dt.

When α is future-directed and timelike in Ln , its arclength is called proper time, denoted by tba [α]. Furthermore, an arclength function for α is a smooth function s : I → R written as s(t) = Ltt0 [α], for some t0 ∈ I. Remark. The proper time of a future-directed timelike parametrized curve can be seen as the time measured by a clock carried by an observer traveling along the curve. Another quantity associated to a curve, whose importance appears in Chapter 3, is the energy: Definition 2.1.9. Let α : I → Rnν be a parametrized curve and a, b ∈ I. The energy of α between a and b is defined by . 1 Eab [α] = 2

Z b a

hα0 (t), α0 (t)iν dt.

Remark. To avoid overloaded notation we don’t mention the ambient space of the curve in arclength and energy, since it is given in the definition of the curve a priori. We also omit the boundaries a and b when the integration is over the whole interval I.

68  Introduction to Lorentz Geometry: Curves and Surfaces

Example 2.1.10. To illustrate, we compute the arclengths of a given curve, seeing it as a curve in Euclidean space and then in Lorentz-Minkowski space. (1) Let α : R → R3 be given by α(t) = ( a cos t, a sin t, bt). Then, L2π 0 [α]

=

Z 2π p 0

a2 + b2 dt = 2π

p

a2 + b2 .

(2) If α : R → L3 is given by the expression above, we have Z 2π q q 2 − b2 | dt = 2π | a2 − b2 |. L2π [ α ] = | a 0 0

In particular, if | a| = |b| the curve is lightlike and its length vanishes. (3) Let β : R → R3 be given by β(t) = (e−t cos t, e−t sin t, e−t ). Then  Z 1 √ √  1 1 −t L0 [ β ] = . e 3 dt = 3 1 − e 0 (4) If β : R → L3 is given by the expression above, L10 [ β] =

Z 1 0

1 e−t dt = 1 − . e

Now let’s compute the energy of some curves: (5) Let γ : R → R2 be given by γ(t) = (cos t, sin t), then E02π [γ] =

Z 2π 0

1 dt = 2π.

(6) The same parametrization in L2 gives E02π [γ]

=−

Z 2π 0

cos(2t) dt = 0.

(You would expect that, right? See Figure 2.1.) (7) The hyperbola, parametrized by η : R → R2 , η(t) = (sinh t, cosh t) has its energy given by Z 2π sinh(4π ) cosh(2t) dt = . E02π [η] = 2 0 (8) In L2 , the same map η has energy E02π [η]

=

Z 2π 0

1 dt = 2π.

The energy and arclength of a parametrized curve are related: Proposition 2.1.11. Let α : I → Rnν be a parametrized curve with constant causal type. Given any a, b ∈ I we have q b L a [α] ≤ 2eα (b − a) Eab [α]. The equality holds if and only if α has constant speed.

Local Theory of Curves  69

Proof: Apply the Cauchy-Schwarz inequality for the (positive-definite) inner product given by Z b . hh f , gii = f ( x ) g( x ) dx, a p for the functions f (t) = eα hα0 (t), α0 (t)i and g(t) = 1: Lba [α] = hh f , gii

≤ k f kk gk Z b 1/2 Z b 1/2 0 0 = eα hα (t), α (t)i dt 1 dt a a q √ = 2eα Eab [α] b − a. The equality holds if and only if the functions f and g are linearly dependent on the function space or, equivalently, α has constant speed. Remark. The previous result is false when the causal type of the curve is not constant. We saw a counter-example not too long ago. Which one it is? Two parametrized curves may have the same image (trace). An example is an arc of the unit circle S1 = {( x, y) ∈ R2 | x2 + y2 = 1}, which can be parametrized by α(t) = (cos t, sin t), t ∈ ]−π/2, 3π/2[ and by the expression obtained in Exercise 2.1.8. This motivates the: Definition 2.1.12. Let α : I → Rnν be a parametrized curve. A reparametrization of α is another parametrized curve of the form β = α ◦ h, for some diffeomorphism h : J → I between the open intervals J and I. The function h is called a change of parameters. Remark. In the notation above:

• if β is a reparametrization of α by h, then α is a reparametrization of β by h−1 ; • the reparametrization is positive if h0 > 0; • β is regular if and only if α is regular; • β has the same causal type of α. The interesting objects to study in Differential Geometry of curves are those intrinsic to the trace of the curve, that is, those invariant under reparametrization. In particular, and as expected, the length of a parametrized curve is invariant under reparametrizations (see Exercise 2.1.13), but energy is not intrinsic (see Exercise 2.1.14). One must be careful when assigning such invariant properties to the image of the curve via some parametrization. For example, the curves α : [0, 2π ] → S1 and β : [0, 4π ] → S1 . given by α(t) = β(t) = (cos t, sin t) have the same image, but L[α] = 2π and L[ β] = 4π. What would prevent us from setting the latter value as the length of the unit circle, even knowing that its length is 2π? The images α([0, 2π ]) and β([0, 4π ]) are the same as sets, but geometrically different in the sense that α covers S1 once, while β covers it twice. The injectivity condition for the parametrization avoids artificial distortions in the “actual” length of a curve. There are reparametrizations of a given curve that carry strong geometric meaning and will simplify many expressions to be obtained in the following sections. To explore this, we need the following result:

70  Introduction to Lorentz Geometry: Curves and Surfaces

Proposition 2.1.13. Let u, v : I → Rnν be parametrized curves. Then: (i) f : I → R given by f (t) = hu(t), v(t)iν is smooth and f 0 (t) = hu0 (t), v(t)iν + hu(t), v0 (t)iν ; (ii) g : I → R given by g(t) = ku(t)kν is smooth in each t such that ku(t)kν 6= 0 and g0 (t) =

eu hu0 (t), u(t)iν ; ku(t)kν

(iii) If n = 3, the map w : I → R3ν given by w(t) = u(t) × v(t) is smooth and w 0 ( t ) = u 0 ( t ) × v ( t ) + u ( t ) × v 0 ( t ). Proof: Since u, v, h·, ·iν and × are smooth and the square root function is differentiable wherever it does not vanish, all of the compositions in the statement are smooth. We could achieve expressions for the derivatives expanding each object in terms of its coordinates. A more elegant way to prove this, allowing generalizations, is to use results provided in Appendix A as follows: (i) From bilinearity of h·, ·iν we have f 0 (t) = D f (t)(1) = D (h·, ·iν ◦ (u, v))(t)(1)

= D (h·, ·iν )(u(t), v(t)) ◦ D (u, v)(t)(1) = D (h·, ·iν )(u(t), v(t)) ◦ ( Du(t), Dv(t))(1) = D (h·, ·iν )(u(t), v(t))(u0 (t), v0 (t)) = hu0 (t), v(t)iν + hu(t), v0 (t)iν . (ii) Setting v = u in the above, g0 (t) =

d dt

q

eu hu0 (t), u(t)iν 2eu hu0 (t), u(t)iν eu hu(t), u(t)iν = p = . ku(t)kν 2 eu hu(t), u(t)iν

(iii) Analogously to the item (i), bilinearity of × gives w0 (t) = Dw(t)(1) = D (× ◦ (u, v))(t)(1)

= D (×)(u(t), v(t)) ◦ D (u, v)(t)(1) = D (×)(u(t), v(t))(u0 (t), v0 (t)) = u 0 ( t ) × v ( t ) + u ( t ) × v 0 ( t ).

Remark. Item (iii) above can be generalized to the cross product in Rnν using the formula for the total derivative of a multilinear map. Try to deduce such an expression. Corollary 2.1.14. Let u, v : I → Rnν be parametrized curves. If hu, viν is constant, then hu0 (t), v(t)iν = −hu(t), v0 (t)iν , for all t ∈ I. Corollary 2.1.15. Let u : I → Rnν be a parametrized curve. If hu, uiν is constant, then u0 (t) ⊥ u(t), for all t ∈ I (interpret this geometrically).

Local Theory of Curves  71

Definition 2.1.16. Let α : I → Rnν be a parametrized curve. We say that α has unit speed if kα0 (t)kν = 1 for all t ∈ I, and α is parametrized by arclength if, for all t0 ∈ I, the arclength function from t0 is given by s(t) = t − t0 . Remark. The concepts defined above are equivalent. We ask you to verify this in Exercise 2.1.16. Theorem 2.1.17. Let α : I → Rnν be a non-lightlike regular curve. Then α admits a unit speed reparametrization. More precisely, there exists an open interval J and a diffeomor. phism h : J → I such that e α = α ◦ h has unit speed. Proof: Let t0 ∈ I be fixed and s : I → R be the arclength function from t0 . From the Fundamental Theorem of Calculus, s0 (t) = kα0 (t)kν > 0. Therefore s is an increasing . diffeomorphism over its image J = s( I ) ⊆ R. Hence we consider its inverse map, h : J → I (also increasing). Setting e α = α ◦ h and identifying s = s(t), we have:

ke α0 (s)kν = kα0 (h(s))h0 (s)kν = kα0 (h(s))kν h0 (s) = s0 (h(s))h0 (s) = (s ◦ h)0 (s) = 1.

Remark.

• The relation e α = α ◦ h implies α(t) = e α(s(t)) for all t ∈ I. This latter expression is more commonly used. • For timelike curves, one may also write α(t) = e α(t(t)). • The above reparametrizations are not unique. In fact, they depend on the choice of t0 , but the change of parameters between two unit speed reparametrizations is necessarily an affine function. Details in Exercise 2.1.19. For lightlike curves it is impossible to obtain unit speed reparametrizations. Intuitively, the proper time in any arc of a photon’s worldline is zero, hence it can’t be used as a parameter. The best we get is: Theorem 2.1.18 (Arc-photon). Let α : I → Ln be a lightlike parametrized curve such that kα00 (t)k L 6= 0 for all t ∈ I. Then α admits an arc-photon reparametrization, that . is, there exists an open interval J and a diffeomorphism h : J → I such that e α = α◦h satisfies ke α00 (φ)k L = 1 for all φ ∈ J. Proof: Such a function h must verify e α(φ) = α(h(φ)). Differentiating this twice we get e α00 (φ) = α00 (h(φ))h0 (φ)2 + α0 (h(φ))h00 (φ). Since α is lightlike, α00 (t) is orthogonal to α0 (t) and the condition kα00 (t)k L 6= 0 tells us that α00 (t) is spacelike, for all t ∈ I. Hence, applying h·, ·i L to both sides of the above equation, we have 1 = hα00 (h(φ)), α00 (h(φ))i L h0 (φ)4 , 1/2 so that h0 (φ) = kα00 (h(φ))k− : a first order ODE for the function h. For fixed numbers L φ0 ∈ J and t0 ∈ I, one shows that this equation has a single solution satisfying h(φ0 ) = t0 . Once the existence of h is ensured, we use it to define e α with the desired properties.

Remark. As in the previous remark, arc-photon reparametrizations are not unique and the change of parameters between any two of them is affine. See Exercise 2.1.20.

72  Introduction to Lorentz Geometry: Curves and Surfaces

Example 2.1.19. 3 (1) In R3 the √ t, a sin t, bt), has the function √parametrized helix α : R → R , α(t) = ( a cos 2 2 s(t) = t a + b as an arclength. That is, t = s(t)/ a2 + b2 , hence       s s bs √ √ √ e α(s) = a cos , a sin , a2 + b2 a2 + b2 a2 + b2

is a unit speed reparametrization of α. (2) In L3 consider the parametrized curve α : ]0, +∞[ → L3 given by

√

α(t) =

10 3 (t − 2) sin t + 2t cos t, (2 − t ) cos t + 2t sin t, t 3 2

Then α0 (t) = (t2 cos t, t2 sin t,

√

e α (t) =

p 3

(

.

10t2 ) so that α is timelike amd

t( t ) = Hence t =

!

2

Z t 0

3ξ 2 dξ = t3 .

t(t). This way,

√ 3

t2

− 2) sin

√ 3

√ 3

t + 2 t cos

√ 3

t, (2 −

√ 3

t2 ) cos

√ 3

√ 3

t + 2 t sin

√ 3

√

10 t, t 3

!

is a proper time reparametrization of α. (3) Consider the Euclidean helix α : R → L3 , α(t) = (r cos t, r sin t, rt), where r > 0. Note that α is lightlike. We will obtain an arc-photon reparametrization of α: since α00 (t) = (−r cos t, −r sin t, 0) we have kα00 (t)k L = r. Then 1 φ h0 (φ) = √ =⇒ h(φ) = √ . r r Hence

 e α(φ) =

 r cos

φ √ r



 , r sin

φ √ r

 ,

√

 rφ .

To close this section, we introduce the following concept: Definition 2.1.20 (Congruence of curves). Let α : I → Rnν and β : J → Rnν be parametrized curves. We say that α and β are congruent if there is a reparametrization e α : J → Rnν of α and an isometry F ∈ Eν (n, R) such that β = F ◦ e α. Since any two open intervals are diffeomorphic, we may assume that both curves are defined in the same domain I. In the notation above this means e α = α. Congruence is one of the most important concepts in Geometry, since it allows the study of geometric quantities of an object, which do not depend on the position of the object in the ambient space. Proposition 2.1.21. Let α, β : I → Rnν be congruent parametrized curves. Then h β0 (t), β0 (t)iν = hα0 (t), α0 (t)iν for all t ∈ I.

Local Theory of Curves  73

Proof: There exists F ∈ Eν (n, R) such that β = F ◦ α. We already know that F = Ta ◦ Λ for some Λ ∈ Oν (n, R) and a ∈ Rnν . Then β0 (t) = Dβ(t)(1) = D ( F ◦ α)(t)(1) = DF (α(t)) ◦ Dα(t)(1) = Λα0 (t). Hence h β0 (t), β0 (t)iν = hα0 (t), α0 (t)iν . Corollary 2.1.22. Let α, β : I → Rnν be two congruent parametrized curves. Then we have that L[ β] = L[α] and E[ β] = E[α]. In terms of congruence, we raise a question: what is the minimal geometric information necessary to recover, up to a congruence, a curve in the plane or space? The answer relies on the concepts of curvature and torsion. Both are the subject of the following sections.

Exercises In the following exercises, “curve” stands for “parametrized curve”. Exercise 2.1.1. As in Example 2.1.3 (p. 64), discuss the causal type of: (a) the hyperbolic helix α : R → L3 given by α(t) = ( at, b cosh t, b sinh t), where a, b > 0; (b) the curve α : R → L4 given by α(t) = ( a0 , a1 t, a2 t2 , a3 t3 ). (c) the curve α : R>0 → L3 given by α(t) = (t, log t a , log tb ), where a, b > 0. Exercise 2.1.2. Let α : I → Rnν be a curve. (a) Let ν = 0 and p ∈ Rn not be contained in the image of α. Suppose that α(t0 ) is the point in the image of α which is closest to p. Show that α0 (t0 ) and α(t0 ) − p are orthogonal. (b) Let v ∈ Rnν . Suppose hα0 (t), viν = 0 for all t ∈ I and that there exists t0 ∈ I such that α(t0 ) is orthogonal to v. Show that α(t) is orthogonal to v for all t ∈ I. (c) Let ν = 0 and β : I → Rn be another curve. Show that if, for all t ∈ I, the line passing through α(t) and β(t) is orthogonal to α and β at t, then the length of the segment joining α(t) and β(t) is constant. Remark. The angle between two curves at a point is, by definition, the angle between the tangent lines to the curves at that point. Exercise 2.1.3. Show that if α : I → Rn is a curve whose coordinates are polynomials of order at most k, then the trace of α is contained in an affine subspace of Rn of dimension at most k. Hint. Taylor polynomials. Exercise 2.1.4. Let α : I → L3 be a regular closed curve (hence spacelike). Show that if the trace of α is contained in an affine plane Π, then Π is also spacelike. How can you generalize this result to Ln ? Hint. Apply, if necessary, a Poincaré transformation and suppose that Π : x = 0.

74  Introduction to Lorentz Geometry: Curves and Surfaces

Exercise 2.1.5. Compute the arclength and energy of the following curves: (a) α : ]0, 2π [ → R4 , α(t) = (cos t, sin t, cos t, sin t); (b) β : ]1, cosh(2)[ → L3 , β(t) = (cos t, sin t, t2 /2); ! √ √ 3 3 √ 2 3 2 2 t 3 t t2 − , 15t − t2 + . (c) γ : R → L3 , γ(t) = 12t − 3 2t , 9t + 2 2 2 2 Exercise 2.1.6. Let a, b ∈ R, with a > 0 and b < 0. Consider the logarithmic spiral α : R → R2 given by α(t) = ( aebt cos t, aebt sin t). Show that: (a) lim α0 (t) = 0; t→+∞

Z +∞

(b) 0

kα0 (t)k E dt < +∞. Evaluate this integral in terms of a and b, interpreting it

when b → 0− ; (c) α intersects, with a constant angle, all rays starting at the origin. Curves with such a property are called loxodromic.

Figure 2.3: The logarithmic spiral with a = 1 and b = −0.1. Exercise 2.1.7. Let a, b ∈ R, with a > 0 and b < 0. Consider the Lorentzian analogue of the logarithmic spiral, β : R → L2 , given by β(t) = ( aebt cosh t, aebt sinh t). (a) Discuss the causal type of β in terms of a and b. (b) Discuss the existence of lim β0 (t) in terms of a and b. t→+∞

Z +∞

(c) Show that 0

k β0 (t)k L dt < +∞. Evaluate this integral in terms of a and b, inter-

preting it when b → 0− . Exercise 2.1.8 (Rational parametrization for the circle). For each t ∈ R, consider the ray welling up from (0, −1) passing through (t, 0). Let γ(t) be the unique point in the interior of this ray contained in S1 = {( x, y) ∈ R2 | x2 + y2 = 1}. (a) Writing the expression for γ(t), show that this construction defines a smooth map γ : R → S1 ⊆ R2 whose image omits just the point (0, −1).

Local Theory of Curves  75

(b) Evaluate the limits lim γ(t) and lim γ(t). Interpret geometrically. t→+∞

Z +∞

(c) Verify that

−∞

t→−∞

kγ0 (t)k E dt = 2π.

Remark. The map γ is also known as stereographic projection from the circle, via the south pole. Exercise 2.1.9 (Rational parametrization for the hyperbola). For each t ∈ ]−1, 1[, consider the ray emanating from (0, −1) passing through (t, 0). Let γ(t) be the unique point in the interior of this ray contained in . H1 = {( x, y) ∈ L2 | x2 − y2 = −1 and y > 0}. (a) Writing the expression of γ(t), show that this construction defines a diffeomorphism γ : ]−1, 1[ → H1 ⊆ L2 . (b) Evaluate the limits lim γ(t) and lim γ(t). Interpret geometrically. t →1

t→−1

Exercise† 2.1.10. Let α : I → Rn be a curve. Show that given any a, b ∈ I such that a < b, we have Lba [α] ≥ kα(b) − α( a)k E , with the equality holding if and only if the trace of α [ a,b] is the line segment joining α( a) to α(b). Hint. Write α(b) − α( a) = use Cauchy-Schwarz.

Z b a

α0 (t) dt, apply h·, α(b) − α( a)i E on both sides of it and

Exercise† 2.1.11. Let α : I → Ln be a future-directed timelike curve and a, b ∈ I such that a < b. (a) Show that α(b) − α( a) is a future-directed timelike vector. Hint. Write α = ( β, xn ), where β : I → Rn−1 . Since α is a future-directed timelike 0 0 curve, we have the inequality k β

(t)k E < xn (t) for all t ∈ I. Evaluate the quantity Z b

k β(b) − β( a)k E = β0 (t) dt

. a

E

Remark.

• This shows that the chronological future of a point (informally defined in Chapter 1) coincides with its future timecone, that is, I + ( p) = CT+ ( p), for all p ∈ Ln . • The result established here holds even if we assume that α is only a differentiable curve, i.e., we don’t actually need α to be of class C1 . But in this case the hint is not helpful, since we can’t guarantee that β0 is integrable, so that another approach is needed. Can you think of any? (b) Show that tba [α] ≤ kα(b) − α( a)k L , with equality holding if and only if the trace of α [ a,b] is the line segment joining α( a) and α(b). Give a physical interpretation for this and compare with the explanation for the Twins Paradox, given in Chapter 1.

76  Introduction to Lorentz Geometry: Curves and Surfaces

Hint. Use the strategy employed in the previous exercise, with the timelike version of Cauchy-Schwarz inequality. Pay attention to the signs. Exercise 2.1.12. The item (a) from the previous exercise has a small generalization when n = 2. (a) Show that if α : I → R2ν is a curve and a, b ∈ I are such that α(b) 6= α( a), then there exists c between a and b such that α(b) − α( a) and α0 (c) are proportional. Hint. Cauchy’s Mean Value Theorem. (b) Conclude that given distinct points p, q ∈ L2 , the vector q − p has the same causal type of any constant causal type curve joining p and q. Exercise† 2.1.13. Prove that the arclength of a curve is invariant under reparametrizations, that is, if α : I → Rnν and β : J → Rnν are parametrized curves such that β = α ◦ h, where h : J → I is a diffeomorphism, then: Z J

k β0 (s)k ds =

Z I

kα0 (t)k dt.

Exercise† 2.1.14. We know that arclength is invariant under reparametrizations, but this doesn’t hold for the energy. Let α : ] a, b[ → Rnν be a smooth curve and k > 0. Define . αk : ] a/k, b/k[ → Rnν as αk (t) = α(kt). Evaluate the energy of αk and conclude that any non-lightlike curve admits a reparametrization whose energy has an arbitrarily large absolute value. (Intuitively, we go through the trace of α k times faster. Think of kinetic energy.) Exercise 2.1.15. Let α : I → Rn be a curve representing the trajectory of a point with mass m > 0, moving under the action of a conservative force field. In other words, there exists a smooth function, defined in some open subset of Rn , V : U ⊆ Rn → R, such that α( I ) ⊆ U, satisfying mα00 (t) = −∇V (α(t)) for all t ∈ I. The function V is called potential energy. Define the kinetic energy function as T : Rn → R as T (v) = mkvk2 /2. Show the energy conservation principle: along α the sum of kinetic and potential energies is a constant. Compare the kinetic energy with the energy E[α] defined in the text. Exercise 2.1.16. Let α : I → Rnν be a curve. Show that α has unit speed if and only if it is parametrized by arclength. Exercise 2.1.17. Reparametrize by arclength the following curves: (a) the cycloid γ : ]0, 2π [ → R2 , γ(t) = (t − sin t, 1 − cos t); (b) the upper half of Neil’s parabola (a semicubic), given by η : R → R2 , η(t) = (t2 , t3 ). Exercise 2.1.18. Let α : R → L3 be given by √ α(t) = (6t − t3 , 3 2t2 , 6t + t3 ). Check that α is lightlike and reparametrize it by arc-photon. Exercise† 2.1.19. Let α : I → Rnν be a curve and suppose that e α1 : J1 → Rnν and e α2 : J2 → Rnν are two unit speed reparametrizations of α, such that e α1 (s1 (t)) = e α2 (s2 (t)), for all t ∈ I. Show that s1 (t) = ±s2 (t) + a, for some a ∈ R. What is the geometric meaning of a?

Local Theory of Curves  77

Exercise 2.1.20. Let α : I → Ln be a lightlike curve, and suppose that e α1 : J1 → Ln and n e α2 : J2 → L are arc-photon reparametrizations of α, such that e α1 (φ1 (t)) = e α2 (φ2 (t)), for all t ∈ I. Show that φ1 (t) = ±φ2 (t) + a, for some a ∈ R. What is the geometric meaning of a? Exercise 2.1.21. Let α : I1 → R2 and β : I2 → R2 be regular curves that intersect transversally at p = α(t1∗ ) = β(t2∗ ), i.e., such that {α0 (t1∗ ), β0 (t2∗ )} is linearly independent, where t1∗ ∈ I1 and t2∗ ∈ I2 . Let v ∈ R2 be a unit vector and define, for each s ∈ R, the perturbations of α in the direction of v, αs : I1 → R2 , by setting αs (t) = α(t) + sv. Show that, for s sufficiently small, the traces of αs and β intersect near p. Hint. Use the Implicit Function Theorem for a convenient map F : R × I1 × I2 → R2 . Exercise 2.1.22. Let α : I → Rn be a regular curve and t0 ∈ I. Show that there exists an open interval J ⊆ I around t0 such that: (a) α J is injective; (b) there exist smooth maps F1 , . . . , Fn−1 : Rn → R whose image α( J ) is contained in the set { x ∈ Rn | Fi ( x) = 0 for all 1 ≤ i ≤ n − 1}. Hint. Use the Implicit Function Theorem first and, if needed, seek inspiration from Exercise A.3 (p. 329, Appendix A).

2.2

CURVES IN THE PLANE

We’ll begin the study of curves in R2ν by observing that every regular curve may be, at least locally, reparametrized as the graph of a real function. In L2 , the causal type of the curve allows us to specify which of the coordinate axes is the domain of such graph. Theorem 2.2.1. Let α : I → R2ν be a regular curve. For each t0 ∈ I there is an open interval J, u0 ∈ J and a reparametrization e α : J → R2ν of α such that α(t0 ) = e α(u0 ), of the form e α(u) = (u, f (u)) or e α(u) = ( f (u), u), for some smooth function f . Moreover, in L2 , we ensure that the first case is always possible if α is spacelike, while the second one is always possible if α is timelike. Proof: Write α(t) = ( x (t), y(t)). Since α is regular, we know that x 0 (t0 ) and y0 (t0 ) are not simultaneously zero. Suppose without loss of generality that x 0 (t0 ) 6= 0. By the Inverse Function Theorem, there is an open interval J where the inverse function . x −1 : J → x −1 ( J ) ⊆ I is well-defined and smooth. Define the curve e α = α ◦ x −1 , so . . that e α(u) = (u, y( x −1 (u)). Thus, f = y ◦ x −1 and u0 = x (t0 ) fit the bill. In L2 , if α is spacelike, we necessarily have that x 0 (t0 ) 6= 0, while if it is timelike, we have y0 (t0 ) 6= 0. In the plane, it is easy to determine all the lightlike curves: Proposition 2.2.2. Let α : I → L2 be a lightlike curve. Then the trace of α is contained in a line parallel to one of the light rays of L2 .

78  Introduction to Lorentz Geometry: Curves and Surfaces

Proof: Write α(t) = ( x (t), y(t)). Since α is lightlike, we have x 0 (t)2 − y0 (t)2 = 0, whence | x 0 (t)| = |y0 (t)|. Suppose without loss of generality that x 0 (t) = y0 (t). Then there is c ∈ R such that x (t) = y(t) + c, and so we have α(t) = (y(t) + c, y(t)) = (c, 0) + y(t)(1, 1), as wanted. In view of the above result, we may focus our attention on curves that are not lightlike, which we may assume parametrized with unit speed (and thus with constant causal character). The idea is, then, to associate to each point of the curve a positive orthonormal basis of R2ν . The trace of the curve in the plane is directly related to the way such basis changes along the curve. More precisely, every vector in R2ν may be written as a linear combinations of the elements in this basis, and in particular the derivatives of the basis vectors themselves, and thus the coefficients of such combinations will provide geometric information about the curve. This strategy will be used again in the next section when we discuss curves in space R3ν . Definition 2.2.3 (Frenet-Serret Dihedron). Let α : I → R2ν be a unit speed curve. The Frenet-Serret frame of α at s ∈ I is F = ( T α (s), N α (s)), . where T α (s) = α0 (s) is the tangent vector to α at s, and N α (s) is the normal vector to α at s, characterized as the unique vector that makes the basis F orthonormal and positive.

(a) In R2

(b) In L2

Figure 2.4: The Frenet-Serret dihedron in R2ν . If α(s) = ( x (s), y(s)), then the vector N α (s) may be obtained geometrically in the following way:

• in R2 , N α (s) is the counterclockwise rotation of angle π/2 of the vector T α (s), that is, N α (s) = (−y0 (s), x 0 (s)); • in L2 , N α (s) is the reflection in one of the light rays, which depend on the causal type of α. The reflection of T α (s) relative to any of these light rays produces a vector orthogonal to T α (s), and the axis of reflection is chosen to make the

Local Theory of Curves  79

obtained basis positive. If α is spacelike, we perform the reflection relative to the main light ray (y = x), and thus N α (s) = (y0 (s), x 0 (s)), while if α is timelike, we perform the reflection relative to the secondary light ray (y = − x), so that N α (s) = −(y0 (s), x 0 (s)). We may deduce a general expression for the normal vector, N α (s) = ( a(s), b(s)), just from the conditions

h T α (s), N α (s)i = 0 and

det( T α (s), N α (s)) = 1.

Such conditions yield the system: ( x 0 (s) a(s) + (−1)ν y0 (s)b(s) −y0 (s) a(s) + x 0 (s)b(s)

=0 . =1

Noting that eα = x 0 (s)2 + (−1)ν y0 (s)2 , the solutions to the system are a(s) = −eα (−1)ν y0 (s) and b(s) = eα x 0 (s),
that is, N α (s) = eα (−1)ν+1 y0 (s), x 0 (s) . Since h T α (s), T α (s)i = eα is constant, we have that h T 0α (s), T α (s)i = 0 and thus T 0α (s) is parallel to N α (s). We have the: Definition 2.2.4 (Curvature). Let α : I → R2ν be a unit speed curve. The oriented curvature of α is the function κα : I → R characterized by the relation T 0α (s) = κα (s) N α (s). The number κα (s) is called the curvature of α at s. Remark. Applying h·, N α (s)i to the relation T 0α (s) = κα (s) N α (s) and noting that h N α (s), N α (s)i = (−1)ν eα , we obtain κα (s) = (−1)ν eα h T 0α (s), N α (s)i. Proposition 2.2.5 (Frenet-Serret equations). Let α : I → R2ν be a unit speed curve. Then  0     0 κα (s) T α (s) T α (s) = N 0α (s) N α (s) (−1)ν+1 κα (s) 0 holds, for all s ∈ I. Proof: The first equation is the definition of the curvature of α. For the second one, note that h N α (s), N α (s)i is constant, and thus N 0α (s) is parallel to T α (s). By orthonormal expansion we have that N 0α (s) = eα h T α (s), N 0α (s)i T α (s). But now Corollary 2.1.14 gives us that κα (s) = (−1)ν eα h T 0α (s), N α (s)i = (−1)ν+1 eα h T α (s), N 0α (s)i and, finally, N 0α (s) = (−1)ν+1 κα (s) T α (s).

80  Introduction to Lorentz Geometry: Curves and Surfaces

Example 2.2.6. (1) Straight lines: let p, v ∈ R2ν , with kvk = 1, and α : R → R2ν the curve given by α(s) = p + sv. Clearly α has unit speed. We then have T α (s) = v and T 0α (s) = 0, so that κα ≡ 0. (2) Euclidean circles: let p = ( x p , y p ) ∈ R2 , r > 0, and define . S1 ( p, r ) = {( x, y) ∈ R2 | ( x − x p )2 + (y − y p )2 = r2 }. Consider the parametrized curve α : R → S1 ( p, r ) given by  s s  . , y p + r sin . α(s) = x p + r cos r r


We have that T α (s) = − sin rs , cos rs . It follows from this that the normal


vector is N α (s) = − cos rs , − sin rs . This way, we may compute the curvature of α:  s s  1 1 1 0 T α (s) = − cos , − sin = N α ( s ), r r r r r and we conclude that κα (s) = 1/r, for all s ∈ R. (3) Hyperbolic lines: let p = ( x p , y p ) ∈ L2 , r > 0, and define . H1 ( p, r ) = {( x, y) ∈ L2 | ( x − x p )2 − (y − y p )2 = −r2 and y > y p } and . H1− ( p, r ) = {( x, y) ∈ L2 | ( x − x p )2 − (y − y p )2 = −r2 and y < y p }. We’ll discuss here the set H1 ( p, r ). Consider α : R → H1 ( p, r ), the parametrized curve given by  s s  . α(s) = x p + r sinh , y p + r cosh . r r


We have that T α (s) = cosh rs , sinh rs and thus eα = 1. It follows from this


s s that N α (s) = sinh r , cosh r . This way, we may compute the curvature of α:  s 1 s  1 1 T 0α (s) = sinh , cosh = N α ( s ), r r r r r and we conclude that κα (s) = 1/r, for all s ∈ R. (4) de Sitter line: let p = ( x p , y p ) ∈ L2 , r > 0, and define . S11 ( p, r ) = {( x, y) ∈ L2 | ( x − x p )2 − (y − y p )2 = r2 }. Let’s parametrize the right branch of S11 ( p, r ) with α : R → S11 ( p, r ) given by  s s  . α(s) = x p + r cosh , y p + r sinh . r r


We have that T α (s) = sinh rs , cosh rs and thus eα = −1. It follows from this


that N α (s) = − sinh rs , − cosh rs . This way, we may compute the curvature of α:  s 1 s  1 1 0 T α (s) = cosh , sinh = − N α ( s ), r r r r r and we conclude that κα (s) = −1/r, for all s ∈ R.

Local Theory of Curves  81

(5) Catenary 1 : one unit speed parametrization for this curve is α : R → R2 given by p α(s) = (arcsinh s, 1 + s2 ).     1 s s 1 Then T α (s) = √ ,√ and hence N α (s) = − √ ,√ . 1 + s2 1 + s2 1 + s2 1 + s2 So   s 1 1 0 T α (s) = − = , N α ( s ), 2 3/2 2 3/2 1 + s2 (1 + s ) (1 + s ) whence κα (s) = 1/(1 + s2 ) > 0. (6) Cosine wave: a unit speed parametrization for the graph of the function cos, seen in L2 , is α : ]−1, 1[ → L2 given by p α(s) = (arcsin s, 1 − s2 ). We have that   s 1 , −√ T α (s) = √ 1 − s2 1 − s2

 and

N α (s) =

−√

s 1 − s2

,√

1 1 − s2

 ,

since eα = 1. Thus T 0α (s)



=

s 1 ,− 2 3/2 (1 − s ) (1 − s2 )3/2



=

s2

1 N α ( s ), −1

whence κα (s) = 1/(s2 − 1) < 0.

(a) The catenary in R2

(b) The cosine wave in L2

Figure 2.5: Catenaries (Euclidean and Lorentzian). Remark.

• To discuss both hyperbolic lines simultaneously we will indicate, when needed, H1 ( p, r ) by H1+ ( p, r ). This “separation” is made for hyperbolic lines but not for the de Sitter line, since these objects have natural generalizations in higher dimensions, but it turns out that that these de Sitter spaces will be connected. In other words, it won’t be divided into a “right branch” and a “left branch”. 1 The

name “catenary” comes from Latin cat¯ena, which means “chain”. Indeed, the shape of the curve is the one of a chain being held by its endpoints, under the action only of its own weight.

82  Introduction to Lorentz Geometry: Curves and Surfaces

• In analogy with S1 ( p, r ), p and r will also be called the center and the radius of S11 ( p, r ) and H1± ( p, r ). If p = 0 or r = 1, they will be omitted to simplify notation. The above examples illustrate that our definitions so far, despite being relatively simple, are not practical for the analysis of curves which do not have unit speed. Before we extend all the Frenet-Serret apparatus for such curves, consider α again with unit speed and assume that β is a reparametrization of α also with unit speed. It follows from Exercise 2.1.19 (p. 76) that β(s) = α(±s + a), for some a ∈ R. We have that T β (s) = ± T α (±s + a),

N β (s) = ± N α (±s + a) and

T 0β (s) = T 0α (±s + a),

which gives us that κ β (s) = ±κα (±s + a). To summarize, the curvature is sensitive to the change of direction realized by the curve (such change is indicated by the sign ±) and, particularly, we see that the curvature is invariant under positive reparametrizations. In the theory to follow, we will make a local study of curves and thus we may assume that they are regular and with constant causal type. When the curve does not have unit speed, we have the: Definition 2.2.7. Let α : I → R2ν be a regular curve which is not lightlike. Writing α(t) = e α(s(t)), where s is an arclength function for α, the tangent vector to α at t is defined by . T α (t) = T eα (s(t)), the normal vector to α at t is defined by . N α (t) = N eα (s(t)) and, lastly, the curvature of α at t is defined by . κα (t) = κeα (s(t)). Remark. The discussion preceding the above definition in fact says that the FrenetSerret apparatus for α does not depend on the arclength function chosen. Proposition 2.2.8. Let α : I → R2ν be a regular curve which is not lightlike. Then κα (t) =

det(α0 (t), α00 (t)) . kα0 (t)k3

Proof: Write α(t) = e α(s(t)) with e α having unit speed, and s being an arclength function for α. Differentiating such relation twice, we have that α0 (t) = s0 (t) T eα (s(t)) and α00 (t) = s00 (t) T eα (s(t)) + s0 (t)2 κeα (s(t)) N eα (s(t)). With this: det(α0 (t), α00 (t)) = det s0 (t) T eα (s(t)), s00 (t) T eα (s(t)) + s0 (t)2 κeα (s(t)) N eα (s(t))




= s0 (t)s00 (t) det( T eα (s(t)), T eα (s(t))) + s0 (t)3 κeα (s(t)) det( T eα (s(t)), N eα (s(t))) = s0 (t)3 κeα (s(t)).

Since s0 (t) = kα0 (t)k and κα (t) = κeα (s(t)), the result follows.

Local Theory of Curves  83

Remark.

• The sign of the curvature of α does not depend on the ambient plane where it lies (i.e., R2 or L2 ). • If kα0 (t)k = 1, the curvature at t is the (Euclidean) oriented area of the parallelogram spanned by α0 (t) and α00 (t). Example 2.2.9. Let a ∈ R, a 6= 0, and α : R → R2ν be the parametrization of a parabola, given by α(t) = (t, at2 ). Note that α does not have unit speed, nor a constant causal character. Since α0 (t) = (1, 2at) and α00 (t) = (0, 2a), we have that seen in L2 , α is

• spacelike for |t| < 1/2| a|; • timelike for |t| > 1/2| a|; • lightlike at t = 1/2| a| and t = −1/2| a|. Moreover:

2a 1 κα (t) = . 3/2 , if |t| 6= 2| a | 1 + (−1)ν 4a2 t2

See the behavior of the curvature in both ambient planes:

Figure 2.6: The curvatures of a parabola for a = 1 in both ambient planes: the bottom one in R2 and the upper one in L2 . Note that, when both are defined, the absolute value of the Lorentzian curvature is always greater than the absolute value of the Euclidean curvature. This is in fact a very general phenomenon valid for every curve, see Exercise 2.2.6. Moreover, in L2 we have that lim |κα (t)| = +∞. t→±1/2a

Corollary 2.2.10. Let α, β : I → R2ν be two regular, non-lightlike and congruent curves. Then |κα (t)| = |κ β (t)| for all t ∈ I. In particular, equality without the absolute values hold if the linear part of the isometry realizing the congruence preserves the orientation of the plane. Proof: Let F = Ta ◦ A ∈ Eν (2, R) be such that β = F ◦ α. We have previously seen that k β0 (t)k = kα0 (t)k. Moreover, the relation det( β0 (t), β00 (t)) = det( Aα0 (t), Aα00 (t)) = det A det(α0 (t), α00 (t)) holds. Since | det A| = 1, the result follows from the curvature expression given in Proposition 2.2.8.

84  Introduction to Lorentz Geometry: Curves and Surfaces

This result is particularly useful for a local analysis of the curve. With it, we may assume that the curve α : I → R2ν is not only parametrized with unit speed, but is also “well placed” in the plane. More precisely, Exercise 2.1.19 (p. 76) and Proposition 1.4.15 (p. 35) allow us to assume that 0 ∈ I, α(0) = 0 and that the Frenet-Serret frame at this point is “almost”2 the standard basis for R2ν . For a geometric interpretation of the sign of the curvature, assume that κα (0) 6= 0, and consider the Taylor formula: α(s) = sα0 (0) +

s2 00 α (0) + R ( s ), 2

where R(s)/s2 → 0 if s → 0. Reorganizing, we have that α(s) − sT α (0) =

s2 κ α (0) N α (0) + R ( s ). 2

The tangent line to α at s0 = 0 divides R2ν into two half-planes. Thus, the above expression says that for s small enough, the difference α(s) − sT α (0) points to the same side that N α (0) if κα (0) > 0, and points to the opposite side if κα (0) < 0. In any case, α(s) − sT α (0) points in the same direction as α00 (0), which gives the direction to which the curve deviates from its tangent on that point. This justifies the name acceleration vector usually given to α00 (0) (see Exercise 2.2.8). N α (0) T α (0) α(s) − sT α (0)

T α (0) α00 (0)

(a) κα (0) < 0

α00 (0) α(s) − sT α (0)

N α (0)

(b) κα (0) > 0

Figure 2.7: Interpretation of the local canonical form for a planar curve. Following this train of thought, we may approximate the curve in a neighborhood of α(s0 ) with another curve having constant curvature equal to κα (0), with the same tangent vector as α at s0 . We have seen in Example 2.2.6 that, for p ∈ R2ν and r > 0, S1 ( p, r ), S11 ( p, r ) and 1 H± ( p, r ) are curves in the plane with constant curvature equal to 1/r, up to sign. With this in mind, we are interested in curves of this type with the so-called radius of curvature equal to r = 1/|κα (s0 )|. Let’s find the center of curvature p: we know that α(s0 ) − p is normal to α at s0 , and so there is λ ∈ R such that α(s0 ) − p = λN α (s0 ). Up to a positive reparametrization, we may assume that κα (s0 ) > 0. Taking into account that we seek a curve with the same causal type as α at s0 , we have that

hα(s0 ) − p, α(s0 ) − pi = 2 If

(−1)ν eα 1 =⇒ |λ| = . 2 κ α ( s0 ) κ α ( s0 )

the curve is timelike, one needs to consider (e2 , −e1 ) instead.

Local Theory of Curves  85

To decide the sign of λ, we note that in R2 , the vectors α(s0 ) − p and N α (s0 ) point in opposite directions, while in L2 they point in the same direction. With r and p determined, we call this curve the osculating circle of α at s0 . This blatant abuse of terminology in L2 is justified by the fact that both S11 ( p, r ) and H1 ( p, r ) ∪ H1− ( p, r ) are the locus of points which are (Lorentzian) equidistant to a given fixed center, sharing the same geometric definition of a circle in R2 . This analogy will be further emphasized in Theorem 2.2.12 to come.

(a) Osculating circle of a parabola in R2

(b) Osculating de Sitter to a parabola in L2

Figure 2.8: Justifying the sign of λ. That is, we have λ = (−1)ν+1 /κα (s0 ), and thus p = α ( s0 ) +

(−1)ν N α ( s0 ). κ α ( s0 )

Moreover, note that p depends on s0 in the above construction. Allowing s0 to range over an interval where κα does not vanish, we obtain a new curve describing the motion of the centers of curvature of α. Such curve is called the evolute of α (see Exercise 2.2.2). Now, let’s show that the osculating circle is, among all circles tangent to α at α(s0 ), the one that better approximates it. Denote, for each r > 0, by Cr,ν the “circle” of radius r in R2ν with this property. The branch of Cr,ν containing α(s0 ) divides the plane R2ν into two connected components. Assuming that branch to be parametrized with positive curvature, its interior is the connected component of R2ν with the removed branch for which its normal vector at α(s0 ) points to (and the exterior is the remaining component).

Figure 2.9: Interior of H1 and of a branch of S11 in L2 .

86  Introduction to Lorentz Geometry: Curves and Surfaces

With this, let’s say that Cr,ν is
too curved (resp., slightly curved) at α(s0 ) if there is e > 0 such that α ]s0 − e, s0 + e[ does not intersect the interior (resp., exterior) of the branch of Cr,ν which is tangent to α at α(s0 ).

(a) Candidates for osculating circles in R2

(b) Candidates for osculating hyperbolas in L2

Figure 2.10: Circles Cr,ν too curved or slightly curved for a parabola in R2ν Proposition 2.2.11. With the notation of the above discussion, if r0 = 1/κα (s0 ), we have that Cr,ν is too curved at α(s0 ) if r < r0 , and slightly curved if r > r0 . Proof: Consider f : I → R given by . f (s) = hα(s) − α(s0 ) + r (−1)ν+1 N α (s0 ), α(s) − α(s0 ) + r (−1)ν+1 N α (s0 )i. Note that f (s0 ) = (−1)ν eα r2 and that f 0 (s) = 2h T α (s), α(s) − α(s0 ) + r (−1)ν+1 N α (s0 )i   f 00 (s) = 2 h T 0α (s), α(s) − α(s0 ) + r (−1)ν+1 N α (s0 )i + eα , whence f 0 (s0 ) = 0 and f 00 (s0 ) = 2eα (1 − r/r0 ). Let’s then classify the critical point s0 of f in terms of eα and r. If r < r0 and eα = 1, we have that s0 is a local minimum for f . Thus, in R2 , we have that f (s) ≥ f (s0 ) = r2 for every s sufficiently close to s0 . In L2 , we’ll have that f (s) ≥ f (s0 ) = −r2 . In the case where eα = −1, we have that s0 is a local maximum for f , so that f (s) ≤ −r2 for every s sufficiently close to s0 . In any case, we conclude that Cr,ν is too curved at α(s0 ). Similarly, if r > r0 , then Cr,ν is slightly curved at α(s0 ). Remark. The above result shows that the osculating circle is the one with greatest curvature among all the slightly curved circles Cr,ν . Equivalently, it is the one with smallest curvature among all the too curved circles Cr,ν .

Local Theory of Curves  87

Instead of only dealing with osculating circles, we could also consider approximations by polynomial curves. Despite having simple parametrizations, these curves have the disadvantage of not having constant curvature, which makes it more difficult to obtain a similar result to the above in this new setting. We will give expressions for osculating parabolas via Taylor expansions, and we’ll compute their curvatures. In Exercise 2.2.17 we’ll point out how to do this for cubics. We have two situations to consider:

• If α is spacelike, assuming that T α (0) = (1, 0) and N α (0) = (0, 1), we have that the osculating parabola of α at s0 = 0 is given by   κ α (0) 2 . β(s) = α(s) − R(s) = s, s . 2 By Example 2.2.9 we have κ α (0) κ β (s) = 1 + (−1)ν κα (0)2 s2 3/2 and, in particular, κ β (0) = κα (0), as expected. Moreover, if s is sufficiently small, we have that |κ β (s)| ≤ |κα (0)| if ν = 0, while |κ β (s)| ≥ |κα (0)| if ν = 1.

• If α is timelike, assuming that T α (0) = (0, 1) and N α (0) = (−1, 0), we have that the osculating parabola of α at s0 = 0 is given by   κ α (0) 2 . β(s) = α(s) − R(s) = − s ,s . 2 Similarly, we have that κ α (0) κ β (s) = 1 − κα (0)2 s2 3/2 and, as before, κ β (0) = κα (0). Now, for s sufficiently small, we have the inequality |κ β (s)| ≥ |κα (0)|. This indeed shows that curves with constant curvature are the most adequate ones for good local approximations. We use S1 ( p, r ), S11 ( p, r ) and H1± ( p, r ) as such models. The following theorem tells us that no other choice was possible: Theorem 2.2.12 (Curves with constant curvature). Let α : I → R2ν be a non-lightlike regular curve. Suppose that the curvature of α is a constant κ. Then: (i) if κ = 0, the trace of α is contained in a straight line; (ii) if κ 6= 0 in R2 , there is p such that the trace of α is contained in S1 ( p, 1/|κ |); (iii) if κ 6= 0 in L2 , there is p such that the trace of α is contained in

• H1± ( p, 1/|κ |), if α is spacelike; • S11 ( p, 1/|κ |), if α is timelike. Proof: We may assume that α has unit speed. From the Frenet-Serret dihedron it follows that:

• if κ = 0, then α00 (s) = T 0α (s) = 0, and so there exist p, v ∈ R2ν such that α(s) = p + sv, for all s ∈ I;

88  Introduction to Lorentz Geometry: Curves and Surfaces

• if κ 6= 0 we consider, motivated by evolutes, the candidate to center p(s) = α(s) +

(−1)ν N α ( s ), κ

which satisfies

(−1)ν (−1)ν+1 κT α (s) = 0, κ whence p(s) = p ∈ R2ν , for all s ∈ I. So: p0 (s) = T α (s) +

hα(s) − p, α(s) − pi =

(−1)ν eα , κ2

concluding the proof of the theorem.

Remark. The above results confirm yet another intuition for the curvature: it is a measure of how much a curve deviates from being a straight line. We know that in R2 , the shortest distance between two points is a line and, informally, the curve with this property is called a geodesic. This would also justify calling our oriented curvature the geodesic curvature instead. We will formally study geodesics in Chapter 3. Lastly, we’ll answer the question posed at the end of Section 2.1 for planar curves as a natural extension of the previous result when the curvature is not a constant. Theorem 2.2.13 (Fundamental Theorem of Plane Curves). Given p, v ∈ R2ν , with v unit, a continuous function κ : I → R and s0 ∈ I, there is a unique unit speed curve α : I → R2ν such that κα (s) = κ (s) for all s ∈ I, α(s0 ) = p, and α0 (s0 ) = v. Moreover, if two unit speed curves have the same curvature and causal type, they’re congruent. Proof: Write p = ( x0 , y0 ) and v = ( a, b). We have three situations to consider, depending on the ambient space and causal type of the curve:

• In R2 , define α : I → R2 by   Z s Z s . α ( s ) = x0 + cos(θ (ξ ) + θ0 ) dξ, y0 + sin(θ (ξ ) + θ0 ) dξ , s0

s0

Rξ

where θ (ξ ) = s κ (τ ) dτ, and θ0 is such that cos θ0 = a and sin θ0 = b (for 0 instance, θ0 = arctan(b/a) if a 6= 0, or ±π/2 else). Of course that α(s0 ) = p. And since α0 (s) = (cos(θ (s) + θ0 ), sin(θ (s) + θ0 )), we have that α has unit speed and, by construction, α0 (s0 ) = v. Thus N α (s) = (− sin(θ (s) + θ0 ), cos(θ (s) + θ0 )) gives us that κα (s) = θ 0 (s) = κ (s), for all s ∈ I. For uniqueness, suppose that β : I → R2 given by β(s) = ( xe(s), ye(s)) is another unit speed curve such that κ β (s) = κ (s), for all s ∈ I. Then β also satisfies the Frenet-Serret system: ( x 00 (s) = −κ (s)y0 (s) y00 (s) = κ (s) x 0 (s) for all s ∈ I. If β(s0 ) = p and β0 (s0 ) = v, the existence and uniqueness theorem from the theory of ordinary differential equations gives us that α = β.

Local Theory of Curves  89

• In L2 , to obtain a spacelike curve, define α : I → L2 by   Z s Z s . α ( s ) = x0 + cosh( ϕ(ξ ) + ϕ0 ) dξ, y0 + sinh( ϕ(ξ ) + ϕ0 ) dξ , s0

s0

Rξ where ϕ(ξ ) = s κ (τ ) dτ, and ϕ0 is such that cosh ϕ0 = a and sinh ϕ0 = b (here 0 we’re assuming without loss of generality that a > 0; the case a < 0 being similar). Again, we have that α(s0 ) = p. Since α0 (s) = (cosh( ϕ(s) + ϕ0 ), sinh( ϕ(s) + ϕ0 )), we have that α is spacelike, has unit speed and, by construction, satisfies α0 (s0 ) = v. Thus N α (s) = (sinh( ϕ(s) + ϕ0 ), cosh( ϕ(s) + ϕ0 )) gives us that κα (s) = ϕ0 (s) = κ (s), for all s ∈ I. For uniqueness, suppose that β : I → L2 given by β(s) = ( xe(s), ye(s)) is another unit speed spacelike curve such that κ β (s) = κ (s), for all s ∈ I. Then β also satisfies the Frenet-Serret system: (

x 00 (s) = κ (s)y0 (s) y00 (s) = κ (s) x 0 (s)

for all s ∈ I. As in the previous case, if β(s0 ) = p and β0 (s0 ) = v, the existence and uniqueness theorem from the theory of ordinary differential equations gives us that α = β.

• In L2 , to obtain a timelike curve, define α : I → L2 by   Z s Z s . cosh( ϕ(ξ ) + ϕ0 ) dξ , sinh( ϕ(ξ ) + ϕ0 ) dξ, y0 + α ( s ) = x0 + s0

s0

Rξ where ϕ(ξ ) = − s κ (τ ) dτ, and ϕ0 is such that sinh ϕ0 = a and cosh ϕ0 = b 0 (here we’re assuming without loss of generality that b > 0; the case b < 0 is similar). By the third time, we have α(s0 ) = p. Since α0 (s) = (sinh( ϕ(s) + ϕ0 ), cosh( ϕ(s) + ϕ0 )), we have that α is timelike, has unit speed and, by construction, satisfies α0 (s0 ) = v. Then N α (s) = (− cosh( ϕ(s) + ϕ0 ), − sinh( ϕ(s) + ϕ0 )) gives us that κα (s) = − ϕ0 (s) = κ (s), for all s ∈ I. The argument for uniqueness is the same as in the previous two cases, with the system: ( x 00 (s) = −κ (s)y0 (s) y00 (s) = −κ (s) x 0 (s). Lastly, assume that β, γ : I → R2ν are both unit speed curves with same causal type and curvature. Once chosen s0 ∈ I, Proposition 1.4.15 (p. 35) gives us a transformation F ∈ Eν (2, R) satisfying F ( β(s0 )) = γ(s0 ), DF ( β(s0 ))( T β (s0 )) = T γ (s0 ) and DF ( β(s0 ))( N β (s0 )) = N γ (s0 ). By the uniqueness argued in each case above, we have F ◦ β = γ.

90  Introduction to Lorentz Geometry: Curves and Surfaces

Exercises Exercise 2.2.1. Let α : I → R2ν be a unit speed curve. (a) Assume that ν = 0 and, for each s ∈ I, let θ (s) be the Euclidean angle between the vector (1, 0) and T α (s). Show that κα (s) = θ 0 (s). (b) Assume that ν = 1, that α is a future-directed timelike curve and, for each t ∈ I, let ϕ(t) be the hyperbolic angle between the vector (0, 1) and T α (t). Show that κ α (t) = − ϕ 0 (t). (c) State and prove a result similar to item (b) for spacelike curves in L2 . Exercise 2.2.2 (Evolutes). Let α : I → R2ν be a unit speed curve with non-vanishing curvature. We define the evolute of α, β = evα : I → R2ν as

(−1)ν . β(s) = α(s) + N α ( s ). κα (s) Observe that β does not necessarily have unit speed, despite its parameter being denoted by s. (a) Fix s0 ∈ I. Show that β is regular at s0 if and only if κα0 (s0 ) 6= 0. (b) Under the above assumptions, given s ∈ I, show that the tangent vector to the evolute at s is parallel to the normal vector to α at s. In particular, conclude that in L2 , β and α have opposite causal types. (c) ([63]) Suppose that ν = 1. Show that α and β never intersect. Hint. Assume that they do intersect and use item (a) from Exercise 2.1.12 to obtain a contradiction. Exercise 2.2.3 (Four singularities, [63]). Let α : [ a, b] → L2 be a closed curve. Show that the set of points on which α is lightlike is the union of at least four non-empty closed and disjoint subsets of the plane. Hint. Regard α inside R2 instead. Since α is closed, the Euclidean normal map N α : [ a, b] → S1 is surjective. Consider the inverse images of the intersections of the two light rays with the circle S1 . Exercise 2.2.4 ([59]). Let α : I → R2ν be a unit speed regular curve. For each n ≥ 1, let an , bn : I → R be the functions satisfying α ( n ) ( s ) = a n ( s ) T α ( s ) + bn ( s ) N α ( s ) , for all s ∈ I, where α(n) denotes the n-th derivative of α. Show that for all n ≥ 1 the generalized Frenet-Serret equations    0     a n +1 ( s ) an (s) an (s) 0 (−1)ν+1 κα (s) = 0 + bn + 1 ( s ) bn ( s ) bn ( s ) κα (s) 0 hold.

Local Theory of Curves  91

Exercise 2.2.5. Let a, b ∈ R>0 . (a) Obtain the values of t for which the curvature of the ellipse α : R → R2 given by α(t) = ( a cos t, b sin t) is maximum and minimum. Hint. Thinking geometrically, there’s a natural guess to be made. Check whether your computations support such a guess. Remark. We’ll say that s0 ∈ I is a vertex of α if κα0 (s0 ) = 0. The Four Vertex Theorem states that every closed, simple and convex curve in R2 has at least four vertices. For more details, see [17]. Moreover, there is a sharper version of this theorem without the convexity assumption, see [14]. (b) Repeat the discussion for the Lorentzian analogues, β1 , β2 : R → L2 given by β1 (t) = ( a cosh t, b sinh t) and

β2 (t) = ( a sinh t, b cosh t).

Do the curvatures still attain a maximum and a minimum value? Exercise† 2.2.6. Let α : I → R2ν be a non-lightlike regular curve. Denote by κα,E and κα,L the curvatures of α when seen in R2 and L2 , respectively. Show that |κα,E (t)| ≤ |κα,L (t)| for all t ∈ I. Moreover, give an example of a curve satisfying κα,L (t) < κα,E (t) for all t ∈ I. Exercise 2.2.7. Give an example of a regular curve α : I → L2 which is lightlike at a single instant t0 ∈ I, satisfying lim κα (t) = −∞ and

t→t0−

lim κα (t) = +∞.

t→t0+

Hint. Try some cubic similar to the one in Figure 2.4 (p. 78). Exercise† 2.2.8. Let α : I → R2ν be a non-lightlike regular curve. Show, for all t ∈ I, that k T 0α (t)k = k N 0α (t)k. In particular, check that if α has unit speed, then |κα (s)| = kα00 (s)k, for all s ∈ I. Exercise 2.2.9. It is usually convenient to also study curves as the set of zeros of a smooth function. For example, find a parametrized curve whose trace is the solution set of the equation x3 + y3 = 3axy, where a ∈ R. This curve is called the Folium of Descartes.

Figure 2.11: The Folium of Descartes for a = 1/3.

92  Introduction to Lorentz Geometry: Curves and Surfaces

Hint. Try y = tx, that is, take as the parameter t the tangent of the Euclidean angle formed between the x-axis and the position vector ( x, y). Exercise† 2.2.10. Exercise 2.1.22 (p. 77) says that a curve in R2 is locally represented as the solution set of an equation of the form F ( x, y) = 0, for some smooth function F : R2 → R. That is, α : I → R2 satisfies F (α(t)) = 0 for all t ∈ I (perhaps reducing the domain I). It is possible to express the curvature of α in terms of the function F. Show that
Hess F (α(t)) Rπ/2 ∇ F (α(t)) |κα (t)| = . k∇ F (α(t))k3 Hint.

• Recall (from Appendix A) that for a smooth function f : Rn → Rk , we identify the second total derivative D ( D f )( p) with the symmetric bilinear map given by n D2 f ( p)(v, w) = ∑i,j =1 vi w j Fxi x j ( p ). The Hessian of f at p is then defined by . 2 Hess f ( p)(v) = D f ( p)(v, v). • Write the Frenet-Serret frame of α in terms of the gradient of F and use that s0 (t)κα (t) = h T 0α (t), N α (t)i. Remark. The absolute values are necessary since we could work with − F instead of F; the correct sign is decided by analyzing whether ∇ F (α(t)) points in the same direction as α0 (t) or not. Exercise† 2.2.11. Note that the result seen in Exercise 2.1.22 (p. 77) is independent of the scalar product used, and thus it still holds in L2 . However, to repeat the discussion made in the previous exercise, we need the correct notion of a gradient in L2 . Bearing in mind that for a smooth function F : R2ν → R the gradient ∇ F ( p) is nothing more than the vector associated via h·, ·i E to the linear functional DF ( p) with Riesz’s Lemma, we define   ∂F . ∂F ∇ L F ( p) = ( p ), − ( p ) . ∂x ∂y With this, if α : I → L2 is a non-lightlike regular curve satisfying F (α(t)) = 0 for all t ∈ I, show that:
Hess F (α(t)) f∇ L F (α(t)) |κα (t)| = , k∇ L F (α(t))k3L where the “flip” operator f : L2 → L2 is given by f( x, y) = (y, x ). Will the causal type of α influence the correct choice of sign? Test the formula for S11 and H1 using F ( x, y) = x2 − y2 ± 1. Exercise 2.2.12 (Graphs). Let f : I → R be a smooth function and α, β : I → R2ν be given by α(t) = (t, f (t)) and β(t) = ( f (t), t). Determine, when possible, κα (t) and κ β (t). In particular, check that if t0 ∈ I is a critical point for f , we have α00 (t0 ) = (0, κα (t0 )) and β00 (t0 ) = (−κ β (t0 ), 0). Since every curve is locally given as a graph, interpret again the signs of the curvatures in terms of the concavity of f . Exercise† 2.2.13 (Polar Coordinates). Recall that a point ( x, y) ∈ R2 may also be represented by a pair (r, θ ) ∈ R≥0 × [0, 2π [, where r is the distance between ( x, y) and the origin, and θ is the (oriented) angle formed between e1 = (1, 0) and ( x, y). That is, we have: x = r cos θ and y = r sin θ.

Local Theory of Curves  93

Moreover, if ( x, y) 6= (0, 0), such representation is unique. The variables r and θ are called the polar coordinates of the pair ( x, y). y

11111 00000 ( x, y) 00000 11111 00000 11111 00000 11111 00000 11111 r 00000 11111 00000 11111 00000 11111 00000 11111 θ 00000 11111 00000 11111

x

Figure 2.12: Polar coordinates in R2 . Curves in the plane may be expressed in the form r = r (θ ), using θ as a parameter ranging over a certain interval [θ0 , θ1 ]. (a) Explicitly write the parametrization α(θ ) for a curve given in polar representation r = r (θ ). Verify that α0 (θ ) = Rθ (r 0 (θ ), r (θ )) and α00 (θ ) = Rθ (r 00 (θ ) − r (θ ), 2r 0 (θ )), where Rθ denotes a counterclockwise Euclidean rotation by θ, as in Example 1.4.4 (p. 30). (b) Show that the arclength of α between θ0 and θ1 is given by Z θ q 1 L[α] = r (θ )2 + r 0 (θ )2 dθ. θ0

(c) Assuming that α is regular, show that the curvature of α is given by κα (θ ) =

2r 0 (θ )2 − r (θ )r 00 (θ ) + r (θ )2 .
3/2 r ( θ )2 + r 0 ( θ )2

(d) Use the previous items to compute the length and the curvature of the cardioid3 defined by r = 1 − cos θ, for θ ∈ [0, 2π [. On which points is the curvature minimal?

Figure 2.13: Cardioid in the Euclidean plane. 3 The

name “cardioid” comes precisely from the heart-like shape the curve has.

94  Introduction to Lorentz Geometry: Curves and Surfaces

Exercise† 2.2.14 (Rindler Coordinates). Each point ( x, y) ∈ L2 with |y| < x may be represented by a pair (ρ, ϕ) ∈ R≥0 × R, in an analogous way to polar coordinates in R2 , by setting x = ρ cosh ϕ and y = ρ sinh ϕ. The variables ρ and ϕ are called the Rindler coordinates of the pair ( x, y).

y

( x, y) “ρ” “ϕ” x

Figure 2.14: Rindler coordinates in L2 . Curves in this region of the plane L2 , known as the Rindler wedge, may be expressed in the form ρ = ρ( ϕ), using ϕ as a parameter ranging over a certain interval [ ϕ0 , ϕ1 ]. (a) Explicitly write the parametrization α( ϕ) for a curve given in Rindler representation ρ = ρ( ϕ). Verify that α0 ( ϕ) = Rhϕ (ρ0 ( ϕ), ρ( ϕ)) and α00 ( ϕ) = Rhϕ (ρ00 ( ϕ) + ρ( ϕ), 2ρ0 ( ϕ)), where Rhϕ denotes a hyperbolic rotation by ϕ, as in Example 1.4.5 (p. 32). (b) Show that the arclength of α between ϕ0 and ϕ1 is given by Z ϕ q 1 L[α] = |ρ( ϕ)2 − ρ0 ( ϕ)2 | dϕ. ϕ0

(c) Assuming that α is regular and non-lightlike, show that the curvature of α is given by 2ρ0 ( ϕ)2 − ρ( ϕ)ρ00 ( ϕ) − ρ( ϕ)2 κα ( ϕ) = . 3/2 | ρ ( ϕ )2 − ρ 0 ( ϕ )2 | (d) One may consider a Lorentzian version of the cardioid defined in Exercise 2.2.13: use the previous items to compute the length and curvature of the curve defined by ρ = −1 + cosh ϕ, for ϕ ∈ R \ {0}. Are there points where the curvature is maximum or minimum? Compare with the cardioid.

Local Theory of Curves  95

Figure 2.15: Cardioid in the Lorentzian plane. Exercise† 2.2.15 (Curves in the complex plane). We may regard curves in R2 as maps α : I → C, written in the form α(t) = x (t) + iy(t). Suppose that α is regular. Observe that if z1 = x1 + iy1 and z2 = x2 + iy2 are identified with the vectors z1 = ( x1 , y1 ) and z2 = ( x2 , y2 ), we have that hz1 , z2 i E = Re(z1 z2 ). (a) Show that d dt



α0 (t) |α0 (t)|



= iκα (t)α0 (t).

(b) Show that κα (t) = −

Im(α0 (t)α00 (t)) , |α0 (t)|3

and use this to compute the curvature of the Archimedes Spiral, parametrized by α(t) = ( a + bt)eit . (c) Suppose that α : I → C is given by α(θ ) = r (θ )eiθ for some positive function r, and use item (b) to recover the formula for κα (θ ) seen in Exercise 2.2.13. Exercise† 2.2.16 (A glimpse of the future). In this exercise we will see how to repeat the results from the previous exercise for curves in L2 . Consider the set of split-complex numbers . C0 = { x + hy | x, y ∈ R and h2 = 1}, with operations similar to C. Split-complex numbers encode the geometry of L2 in the same way that the complex numbers encode the geometry of R2 . We will study C0 in more detail in Chapter 4, when we discuss surfaces with zero mean curvature in L3 . . Given w = x + hy ∈pC0 , define its split-conjugate as w = x − hy, and its split-complex . absolute value as |w| = | x2 − y2 |. We’ll denote the projections onto the coordinate axes, just like in C, by Re and Im. Note that identifying w1 = x1 + hy1 and w2 = x2 + hy2 with the vectors w1 = ( x1 , y1 ) and w2 = ( x2 , y2 ), we have that hw1 , w2 i L = Re(w1 w2 ). We may regard curves in L2 as maps α : I → C0 written as α(t) = x (t) + hy(t). Suppose that α is regular and non-lightlike. (a) Show that d dt



α0 (t) |α0 (t)|



= eα hκα (t)α0 (t).

96  Introduction to Lorentz Geometry: Curves and Surfaces

(b) Show that κα (t) = −

Im(α0 (t)α00 (t)) , |α0 (t)|3

and use this to compute the curvature of a Lorentzian version of the Archimedes Spiral. Defining, for t ∈ R, the split-complex exponential by . eht = cosh t + h sinh t, this curve may be expressed as α(t) = ( a + bt)eht . (c) Suppose that α : I → C0 has the form α( ϕ) = ρ( ϕ)ehϕ for some positive function ρ, and use item (b) to recover the formula for κα ( ϕ) seen in Exercise 2.2.14. Remark. “Naively” assume that the usual rules from Calculus hold in the split-complex setting. We will formalize all of this in due time. Exercise 2.2.17 (Osculating cubic). Let α : I → R2ν be a unit speed curve. Assume that 0 ∈ I and α(0) = 0. Write α(s) = sα0 (0) +

s2 00 s3 α (0) + α000 (0) + R(s), 2 6

where R(s)/s3 → 0 as s → 0. Define β : I → R2ν by setting β(s) = α(s) − R(s). (a) If α is spacelike, assume in addition that T α (0) = (1, 0) and N α (0) = (0, 1). Compute the curvature κ β (s) and compare it to κα (0). (b) If α is timelike, repeat the argument used in item (a) above, now assuming that T α (0) = (0, 1) and N α (0) = (−1, 0). Exercise 2.2.18. The definition of the evolute of a curve (given in Exercise 2.2.2) is also valid for curves with arbitrary parameter, whose curvatures never vanish. Consider the cycloid parametrized by α : ]0, 2π [ → R2 , α(t) = (t − sin t, 1 − cos t). Determine its evolute evα , and show that they are congruent (that is, show that evα is another cycloid).

Figure 2.16: The cycloid. Exercise 2.2.19. Determine, up to congruence, all the non-lightlike regular curves α : R → R2ν with the following property: for each s ∈ R there is Fs ∈ Eν (2, R) such that Fs (α(t)) = α(t + s), for all t ∈ R. Exercise† 2.2.20. Let α : I → L2 be a curve and consider again the “flip” operator f : L2 → L2 given by f( x, y) = (y, x ). (a) Show that α is regular if and only if f ◦ α is also regular. (b) Suppose that α has constant causal type. Show that α is spacelike (resp. timelike, lightlike) if and only if f ◦ α is timelike (resp. spacelike, lightlike).

Local Theory of Curves  97

(c) Assuming that α is non-lightlike and regular, show that κf◦α (t) = −κα (t) for all t ∈ I. Conclude that up to congruence and reparametrization, to every spacelike curve in L2 corresponds a unique timelike curve with the same curvature (and vice versa, since f = f−1 ). Exercise 2.2.21. Determine all the non-lightlike regular curves in R2ν such that: (a) all their tangent lines intercept in a fixed point; (b) all their normal lines intercept in a fixed point. Exercise 2.2.22. Determine a unit speed curve α : R → R2 such that α(0) = 0 and κα (s) =

1 , 1 + s2

for all s ∈ R. Exercise 2.2.23. Determine all the unit speed curves α : I → R2ν such that κα (s) =

1 , as + b

with a, b ∈ R, a 6= 0.

2.3

CURVES IN SPACE

In this section, our main goal is to establish a classification result similar to Theorem 2.2.13 for curves in space. However, this time the curvature is not enough and, to this end, we’ll introduce the concept of torsion. The first thing to be done is to define the Frenet-Serret frame for a curve in space, following the guidelines outlined in the previous section: associate a positive orthonormal basis to each point of the curve. The general construction, for a non-lightlike curve, begins considering the velocity vector, and then taking a vector orthogonal to it. But in space R3ν , there is now the possibility for the “natural candidate” to normal vector to be lightlike, making unfeasible the construction of the desired orthonormal basis. We must investigate a solution around this issue. In contrast with what was seen for planar curves, in L3 there are many lightlike curves that are not straight lines, which may not be neglected. It is possible to characterize those by using a single invariant (see Theorem 2.3.34, p. 125), but for this we need to introduce yet another new frame along the curve, mimicking the idea of the Frenet-Serret frame. In view of these situations, we have many cases to discuss, depending on the causal type of the tangent and normal vectors. Moreover, we must highlight the situations where, for all t, α00 (t) vanishes or is lightlike. The first situation is very simple: Proposition 2.3.1. The trace of a curve is a straight line if and only if the curve admits a reparametrization α : I → Rnν such that α00 (t) = 0 for all t ∈ I. The second situation will be addressed in Subsection 2.3.3. When one of these situations occurs in isolated points, it might not be possible to construct the desired frame in such points, and so we will not consider these (pathological) cases.

98  Introduction to Lorentz Geometry: Curves and Surfaces

2.3.1

The Frenet-Serret trihedron

Definition 2.3.2. Let α : I → R3ν be a parametrized curve. We’ll say that α is admissible 4 if:  (i) α is biregular, that is, α0 (t), α00 (t) is linearly independent for all t ∈ I;  (ii) both α0 (t) and span α0 (t), α00 (t) are not lightlike, for each t ∈ I. Remark.

• Every unit speed curve in R3 which is not a straight line is automatically admissible. This way, this definition is not a restriction for the development of the usual theory of curves developed in R3 , but really only a device to facilitate the simultaneous treatment of several possible cases in L3 . • Being admissible is an intrinsic property of a curve. See Exercise 2.3.1. As in the previous section, we will begin by  focusing on unit speed curves. Note that if α0 (s) and α00 (s) are orthogonal and span α0 (t), α00 (t) is not lightlike, then α00 (s) cannot be lightlike. With this: Definition 2.3.3 (Frenet-Serret Trihedron). Let α : I → R3ν be an admissible unit speed . curve. The tangent vector to α at s is T α (s) = α0 (s). The curvature of the curve α at . s is κα (s) = k T 0α (s)k. The assumptions on α ensure that κα (s) > 0 for all s ∈ I, so that the unit vector pointing in the same direction as T 0α (s) is well defined, allowing us to define the normal vector to α at s, by the relation T 0α (s) = κα (s) N α (s). Lastly, the binormal vector
to α at s is defined as the unique vector Bα (s) making the basis T α (s), N α (s), Bα (s) orthonormal and positive. Remark. The curvature here defined is fundamentally distinct from the oriented curvature defined in the previous section, as it does not encode orientation and has no sign. For a relation between them, see Exercise 2.3.3. To express in an easier way the binormal vector in terms of α, the following definition is useful: Definition 2.3.4. Let α : I → R3ν be a unit speed admissible curve. The coindicator of . α is defined as ηα = e N α (s) . For space curves, we have the advantage of being able to express the binormal vector in the form Bα (s) = λT α (s) × N α (s) for some λ ∈ R. Since we want the obtained basis to be positive, we have

det T α (s), N α (s), Bα (s) = 1 =⇒ λ det T α (s), N α (s), T α (s) × N α (s) = 1. It follows from the orientability analysis done in Section 1.6 that λ = (−1)ν eα ηα , and thus Bα (s) = (−1)ν eα ηα T α (s) × N α (s). Let’s see that such Bα (s) is indeed a unit vector. To wit, we have:   (∗) e 0 h Bα (s), Bα (s)i = h T α (s) × N α (s), T α (s) × N α (s)i = (−1)ν det α = (−1)ν eα ηα , 0 ηα where in (∗) we have used Proposition 1.6.5 (p. 55). With this we are almost ready to present the Frenet-Serret equations. Recall that the idea behind those equations is to write the derivatives T 0α (s), N 0α (s) and B0α (s) as linear combinations of the Frenet-Serret frame. We have the: 4 It

is reasonably usual in Mathematics to call any object “admissible” if it has the convenient properties for the development of the theory.

Local Theory of Curves  99

Definition 2.3.5 (Torsion). Let α : I → R3ν be a unit speed admissible curve. The torsion of α at s, denoted by τα (s), is the component of N 0α (s) in the direction of Bα (s). Theorem 2.3.6 (Frenet-Serret Equations). Let α : I → R3ν be a unit speed admissible curve. Then  0     0 κα (s) 0 T α (s) T α (s)  0     0 τα (s)  N α (s)  N α ( s )  =  − eα ηα κ α ( s ) B0α (s) Bα (s) 0 (−1)ν+1 eα τα (s) 0 holds, for all s ∈ I. Proof: The first equation is immediate by definition. For the remaining ones, the idea is to apply corollaries 2.1.14 and 2.1.15 (p. 70) for the orthonormal expansions of N 0α (s) and B0α (s). To begin, we have N 0α (s) = eα h N 0α (s), T α (s)i T α (s) + (−1)ν eα ηα h N 0α (s), Bα (s)i Bα (s) B0α (s) = eα h B0α (s), T α (s)i T α (s) + ηα h B0α (s), N α (s)i N α (s). From the first Frenet-Serret equation and the definition of torsion, we have that κα (s) = ηα h T 0α (s), N α (s)i and τα (s) = (−1)ν eα ηα h N 0α (s), Bα (s)i. To determine the component of N 0α (s) in the direction of T α (s), note that eα h N 0α (s), T α (s)i = −eα h N α (s), T 0α (s)i = −eα ηα κα (s). Thus, we have the second Frenet-Serret equation. To obtain the last one, note that
B0α (s) = (−1)ν eα ηα T 0α (s) × N α (s) + T α (s) × N 0α (s)

= (−1)ν eα ηα T α (s) × N 0α (s), so that h B0α (s), T α (s)i = 0. And, finally: ηα h B0α (s), N α (s)i = −ηα h Bα (s), N 0α (s)i = (−1)ν+1 eα τα (s).

Remark. It is usual in the literature for the torsion to be defined with the sign opposite to our definition above. We have made this choice for the coefficient matrix in R3 to be skew-symmetric. A geometric interpretation for the torsion will be given in Proposition 2.3.11 (p. 106) later. Example 2.3.7. (1) Consider the curve α : R → R3 given by   s s 4s α(s) = 3 cos , 3 sin , . 5 5 5 Note that α has unit speed, and thus: T α (s) =

1 s s  s s  3  −3 sin , 3 cos , 4 =⇒ T 0α (s) = − cos , − sin , 0 , 5 5 5 25 5 5

100  Introduction to Lorentz Geometry: Curves and Surfaces

whence we obtain s  3 s , N α (s) = − cos , − sin , 0 and κα (s) = 5 5 25 

for all s ∈ R. Proceeding, we have that: Bα (s) =

1 s s  4 sin , −4 cos , 3 , 5 5 5

and then it directly follows that τα (s) = 4/25, for all s ∈ R. (2) Consider the curve α : R → L3 given by   s 2s s α(s) = cosh √ , √ , sinh √ . 3 3 3 The derivatives of α are:   1 s s α0 (s) = T α (s) = √ sinh √ , 2, cosh √ 3 3 3   1 s s . α00 (s) = T 0α (s) = cosh √ , 0, sinh √ 3 3 3 Note that α is unit speed spacelike and admissible. To wit, if Π = span{α0 (s), α00 (s)} then the Gram matrix   1 0 GΠ = 0 1/9 is positive-definite. We then obtain that   s s 1 N α (s) = cosh √ , 0, sinh √ and κα (s) = , 3 3 3 for all s ∈ R. Proceeding, we have that:   s s −1 sinh √ , 1, 2 cosh √ , Bα (s) = − T α (s) × L N α (s) = √ 3 3 3 and then τα (s) = −2/3, for all s ∈ R. It follows from Exercise 2.1.19 (p. 76), that if β is a positive reparametrization of α, both with unit speed, then β(s) = α(s + a), for some a ∈ R. Hence κ β (s) = κα (s + a) and τβ (s) = τα (s + a) (see Exercise 2.3.4). With this in mind, we now extend the FrenetSerret apparatus for admissible curves not necessarily having unit speed. Definition 2.3.8. Let α : I → R3ν be an admissible curve and s be an arclength function for α. Write α(t) = e α(s(t)). The tangent, normal and binormal vectors to α at t are defined by . T α (t) = T eα (s(t)),

. N α (t) = N eα (s(t)) and

. Bα (t) = Beα (s(t))

and, finally, the curvature and the torsion of α at t are defined by . . κα (t) = κeα (s(t)) and τα (t) = τeα (s(t)).

Local Theory of Curves  101

Proposition 2.3.9. Let α : I → R3ν be an admissible curve. Given t ∈ I, the formulas
det α0 (t), α00 (t), α000 (t) kα0 (t) × α00 (t)k κα (t) = and τα (t) = kα0 (t)k3 kα0 (t) × α00 (t)k2 hold. Proof: For the curvature, we need two derivatives. As done in the previous section, we have: α0 (t) = s0 (t) T α (t) and α00 (t) = s00 (t) T α (t) + s0 (t)2 κα (t) N α (t). With this, we have

α0 (t) × α00 (t) = s0 (t) T α (t) × s00 (t) T α (t) + s0 (t)2 κα (t) N α (t)

= s 0 ( t )3 κ α ( t ) T α ( t ) × N α ( t ). Recalling that the curvature is always positive, that T α (t) × N α (t) is a unit vector, and that s0 (t) = kα0 (t)k, taking k · k of both sides of the above identity allows us to solve for κα (t) and obtain kα0 (t) × α00 (t)k κα (t) = . kα0 (t)k3 As for the torsion, we do not necessarily need α000 (t), but only its component in the direction of Bα (t), namely, s0 (t)3 κα (t)τα (t) Bα (t) (verify). This way (already discarding terms due to linear dependences), we have that:

det α0 (t), α00 (t), α000 (t) = det s0 (t) T α (t), s0 (t)2 κα (t) N α (t), s0 (t)3 κα (t)τα (t) Bα (t)
= s0 (t)6 κα (t)2 τα (t) det T α (t), N α (t), Bα (t)

= s0 (t)6 κα (t)2 τα (t), whence



det α0 (t), α00 (t), α000 (t) det α0 (t), α00 (t), α000 (t) τα (t) = = , kα0 (t)k6 κα (t)2 kα0 (t) × α00 (t)k2

by the expression previously obtained for κα (t). Remark. Note that if we had defined the torsion with the opposite sign, not only would we have an annoying negative sign in the above expression, but also in L3 we would have to consider the causal type of α as well. In other words, our definition has the advantage of directly incorporating everything in τα (t). Example 2.3.10 (Viviani’s window). Consider the curve α : [0, 4π ] → R3ν given by:    t α(t) = 1 + cos t, sin t, 2 sin . 2 The curve α is called Viviani’s window. This curve has an interesting property: consider the cylinder over the xy-plane, centered at (1, 0, 0), with radius 1, and the Euclidean sphere of radius 2 centered at the origin. Then the trace of α always lies in the intersection of the cylinder with the sphere.

102  Introduction to Lorentz Geometry: Curves and Surfaces

Figure 2.17: The trajectory of α. To determine its curvature and torsion we need the following derivatives:    t 0 α (t) = − sin t, cos t, cos , 2    1 t α00 (t) = − cos t, − sin t, − sin , and 2 2    1 t 000 α (t) = sin t, − cos t, − cos . 4 2
Since hα0 (t), α0 (t)i L = 1 − cos2 2t ≥ 0, we have that α is lightlike at t = 0, 2π, and 4π, and spacelike elsewhere. To apply the formulas in the previous proposition, we still need to check whether the curve is admissible: the first two components of α0 (t) and α00 (t) show that they are linearly independent, and Sylvester’s Criterion (Theorem 1.2.18, p. 10) shows that span{α0 (t), α00 (t)} is spacelike for t 6∈ {0, 2π, 4π }. Excluding these points, and using the cross product and norm in each ambient space, we have s
6 cos 2t 3 cos t + 13 κα,E (t) = and τα,E (t) = , 13 + 3 cos t (3 + cos t)3 √     3 t t κα,L (t) = csc , and τα,L (t) = cot 1 − cos t 2 2 where the second subscript index denotes the ambient space where the calculation was done. We may now compare the graphs of the curvatures and torsions: 1.1

0 .3

1.0

0 .2 0.9 0 .1 0.8 2 0.7

4

6

8

10

12

- 0 .1 - 0 .2

0.6

- 0 .3 2

4

6

8

10

12

(a) The curvature of α in R3

(b) The torsion of α in R3

25 10 20 5 15

2

10

4

6

8

10

12

-5 5 - 10 2

4

6

8

10

12

(c) The curvature of α in L3

(d) The torsion of α in L3

Figure 2.18: Curvature and torsion of Viviani’s window in R3ν .

Local Theory of Curves  103

Note that, if t0 = 0, 2π, or 4π (where the curve is lightlike), we have that lim κα,L (t) = lim τα,L (t) = ±∞.

t → t0

t→ p

(a) In R3

(b) In L3

Figure 2.19: Frenet-Serret frames for Viviani’s window in R3ν .

Exercises Exercise† 2.3.1. Show that the property of being admissible is invariant under reparametrization. More precisely, if α : I → R3ν is admissible and h : J → I is a diffeomorphism, show that α ◦ h : J → R3ν is also admissible. Exercise 2.3.2. Consider α : R → R3 given by  −1/t2 ), if t > 0,  (t, 0, e 2 α(t) = (t, e−1/t , 0), if t < 0 and   (0, 0, 0), if t = 0. Show that α is regular, but not biregular. Exercise 2.3.3. We may look at “copies” of R2 and L2 inside L3 , by identifying R2 with the plane xy, and L2 with the planes xz or yz. Consider the projections π E , π L : R3ν → R2ν given by π E ( x, y, z) = ( x, y) and π L ( x, y, z) = (y, z). Suppose that α : I → R3ν is a unit speed curve. (a) Show that if α(s) = ( x (s), y(s), 0) is regarded as a curve in R3 , then its curvature satisfies κα (s) = |κπE ◦α (s)| for all s ∈ I. (b) Show that if α(s) = (0, y(s), z(s)) is regarded as a curve in L3 , then its curvature satisfies κα (s) = |κπ L ◦α (s)| for all s ∈ I. Exercise† 2.3.4. Verify that the Frenet-Serret Trihedron, as well as curvature and torsion, are invariant under a positive reparametrization between unit speed curves.

104  Introduction to Lorentz Geometry: Curves and Surfaces

Exercise 2.3.5. Show that the curvature and torsion of an admissible curve are invariant under isometries whose linear part preserves the orientation of space. Exercise 2.3.6. A Laguerre transformation in L3 is a map H : L3 → L3 of the form H ( x) = aΛx + v for certain a ∈ R \ {0} and v ∈ L3 , where Λ ∈ O1 (3, R) is a Lorentz transformation. If α : I → L3 is an admissible curve, show that H ◦ α is also admissible, and express its curvature and torsion in terms of the curvature and torsion of α. Exercise 2.3.7 ([59]). Let α : I → R3 be a unit speed admissible curve, and for every n ≥ 1, let an , bn , cn : I → R be the functions satisfying α ( n ) ( s ) = a n ( s ) T α ( s ) + bn ( s ) N α ( s ) + c n ( s ) B α ( s ) , for all s ∈ I. Show that for every n ≥ 1 the generalized Frenet-Serret equations hold:        0 0 − eα ηα κ α ( s ) 0 an (s) an (s) a n +1 ( s )        0 (−1)ν+1 eα τα (s) bn (s) .  bn+1 (s) =  bn0 (s) + κα (s) c0n (s) c n +1 ( s ) cn (s) 0 τα (s) 0 Exercise† 2.3.8. Let α : I → R3ν be a unit speed admissible curve and β : I → R3ν its tangent indicatrix, that is, β(s) = T α (s) (which does not necessarily have unit speed). Assuming that β is also admissible, show that (−1)ν ηα κα (s)2 + τα (s)2 2 (a) κ β (s) = and; κ α ( s )2 (b) τβ (s) = eα ηα

κα (s)τα0 (s) − κα0 (s)τα (s) κα (s) |(−1)ν ηα κα (s)2 + τα (s)2 |

What happens when the tangent indicatrix collapses to a point? Exercise 2.3.9. Let α : I → R3ν be a unit speed admissible curve and β : I → R3ν its binormal indicatrix, that is, β(s) = Bα (s) (which does not necessarily have unit speed). Assuming that β is also admissible, show that (−1)ν ηα κα (s)2 + τα (s)2 2 (a) κ β (s) = and; τα (s)2 (b) τβ (s) = (−1)ν+1 ηα

κα (s)τα0 (s) − κα0 (s)τα (s) . τα (s) |(−1)ν ηα κα (s)2 + τα (s)2 |

What happens when the binormal indicatrix degenerates to a point? Exercise 2.3.10 (Darboux vector). If a rigid body moves along a curve α in R3 with unit speed, its motion consists of a translation and a rotation, both along the curve. The rotation is described by the so-called Darboux vector associated to α, denoted by ωα . We will study the situation in both ambient spaces simultaneously. Suppose that α : I → R3ν is admissible and has unit speed. (a) Knowing that ωα (s) × T α (s) = T 0α (s), ωα (s) × N α (s) = N 0α (s) and ωα (s) × Bα (s) = B0α (s), show that ωα (s) = (−1)ν eα ηα τα (s) T α (s) + ηα κα (s) Bα (s), for all s ∈ I.

Local Theory of Curves  105

(b) In R3 , show that T 0α (s) × E T 00α (s) = κα (s)2 ωα (s), for all s ∈ I. Hint. Exercise 1.6.4 (p. 59) may be useful. Exercise 2.3.11. Let α : I → R3ν be an admissible curve. We could have considered other moving frames along α, according to the following construction: for every s ∈ I, take U α (s) unit and orthogonal to T α (s), and then let V α (s) be the vector making the
basis T α (s), U α (s), V α (s) orthonormal and positive. Assume that the causal type of . U α (s) does not depend on s and let eU = eU α (s) . (a) Show, as done in the text, that V α (s) = (−1)ν eα eU T α (s) × U α (s), for all s ∈ I. (b) Define the associated curvatures ω1 , ω2 , ω3 : I → R as the functions such that T 0α (s) = ω3 (s)U α (s) − ω2 (s)V α (s), and ω1 (s) is the component of U 0α (s) in the direction of V α (s). Show that    0   0 ω3 ( s ) − ω2 ( s ) T α (s) T α (s)    0   0 ω1 ( s )  U α ( s )  , U α (s) =  −eα eU ω3 (s) V α (s) V 0α (s) (−1)ν eU ω2 (s) (−1)ν+1 eα ω1 (s) 0 for all s ∈ I. Note that in R3 the coefficient matrix is skew-symmetric. (c) We may consider the Darboux vector ω(s) associated to the frame, satisfying ω(s) × T α (s) = T 0α (s),

ω(s) × U α (s) = U 0α (s) and ω(s) × V α (s) = V 0α (s)

for all s ∈ I. Show that ω(s) = (−1)ν eα eU ω1 (s) T α (s) + (−1)ν eα eU ω2 (s)U α (s) + eU ω3 (s)V α (s). Remark. Note that in R3 the above expression becomes simply ω ( s ) = ω1 ( s ) T α ( s ) + ω2 ( s )U α ( s ) + ω3 ( s ) V α ( s ) , which explains the seemingly random indexing of the associated curvatures. Exercise 2.3.12. Show that for a cylindrical curve α : I → R3ν , given in the form α(t) = (cos t, sin t, z(t)), we have, whenever it makes sense, that: (−1)ν + z0 (t)2 + z00 (t)2 1/2 z0 (t) + z000 (t) κα (t) = and τ ( t ) = . α 3/2 |(−1)ν + z0 (t)2 + z00 (t)2 | |1 + (−1)ν z0 (t)2 | Exercise 2.3.13. Show that for a Lorentzian cylindrical curve α : I → L3 of the form α(t) = ( x (t), cosh t, sinh t) we have, whenever it makes sense, that: 1 − x 0 (t)2 + x 00 (t)2 1/2 x 0 (t) − x 000 (t) κα (t) = and τ ( t ) = . α 3/2 |1 − x 0 (t)2 + x 00 (t)2 | |−1 + x 0 (t)2 | Exercise 2.3.14. Let α : I → R3ν be an admissible curve. Denote by τα,E and τα,L the torsions of α when seen in R3 and L3 , respectively. Show that |τα,E (t)| ≤ |τα,L (t)| for all t ∈ I, and give an example showing that we cannot remove the absolute values. Also try to investigate if there is any relation between the curvatures in both ambient spaces in this context.

106  Introduction to Lorentz Geometry: Curves and Surfaces

2.3.2

Geometric effects of curvature and torsion

With the Frenet-Serret frame for admissible curves in hand, we may now obtain geometric information about such curves in R3ν . We’ll begin with a geometric interpretation for torsion: Proposition 2.3.11. Let α : I → R3ν be an admissible curve. Then the trace of α is contained in a plane if and only if τα (t) = 0, for all t ∈ I. Proof: We may assume, without loss of generality, that α has unit speed. Moreover, note that if α is admissible then the plane spanned by T α (s) and N α (s) is non-degenerate for all s ∈ I and thus the binormal vector Bα (s) cannot be lightlike. If there is a plane Π in R3ν containing the trace of α, then the plane spanned by T α (s) and N α (s) passing through α(s) does not depend on s and is, in fact, the plane Π. Indeed, let v ∈ R3ν be a unit vector normal to Π. Fixed s0 ∈ I, we have that hα(s) − α(s0 ), vi = 0, for all s ∈ I. Repeatedly differentiating such expression we have that

h T α (s), vi = 0 and h T 0α (s), vi = κα (s)h N α (s), vi = 0. Thus v and Bα (s) are collinear for all s ∈ I and so Bα (s) = ±v is a constant. From the last Frenet-Serret equation, it follows that 0 = B0α (s) = (−1)ν+1 eα τα (s) N α (s). Then τα (s) = 0 for all s ∈ I. Conversely, if τα (s) = 0 for all s ∈ I, then Bα (s) = B. Fix s0 ∈ I and consider the function f : I → R given by f (s) = hα(s) − α(s0 ), Bi. Geometrically, this function measures how much α deviates from the plane orthogonal to B that passes through α(s0 ). Note that f (s0 ) = 0 and f 0 (s) = h T α (s), Bi = 0, whence f identically vanishes and we conclude that the trace of α is contained in such a plane. In other words, just like the curvature measures how much the curve deviates from being a straight line, the torsion measures how much the curve deviates from being planar. Remark. The admissibility assumption is essential. Indeed, if the curve is not admissible, it is possible to extend the definition of torsion for this case, in such a way that the curve is not planar and has zero torsion (see, for example, the curve in Exercise 2.3.2, p. 103). Recall that in R2ν the curves with non-zero constant curvature were congruent to H1± ( p, r ) or S11 ( p, r ). Identifying R2ν with the planes xy and yz, according to whether ν = 0 or 1 (see again Exercise 2.3.3, p. 103), we have the following result for planar curves in space: S1 ( p, r ),

Proposition 2.3.12. Let α : I → R3ν be an admissible curve. Then κα (s) = κ is constant and τα = 0 if and only if α is congruent to a piece of S1 (1/κ ), H1± (1/κ ) or S11 (1/κ ). Proof: If α is congruent to a piece of one of the curves listed above, it follows from Exercises 2.3.3 and 2.3.5, that α is planar and has non-zero constant curvature. Conversely, consider c : I → R given by c(s) = α(s) +

eα ηα N α ( s ). κ

Since τα (s) = 0, we have that c0 (s) = T α (s) +

eα ηα (−eα ηα κT α (s)) = 0, κ

Local Theory of Curves  107

whence c(s) = p, for all s ∈ I. It follows that

hα(s) − p, α(s) − pi =

ηα . κ2

In R3 or if the trace of α is contained in a spacelike plane in L3 , this means that α is congruent to a piece of a Euclidean circle of radius 1/κ on the plane xy. If the trace of α is contained in a timelike plane in L3 , ηα says whether α is congruent to a piece of H1± (1/κ ) or S11 (1/κ ) in the plane yz. Remark. An observer with “Euclidean eyes” (such as ourselves) sees a Euclidean circle in a non-horizontal spacelike plane in L3 as an ellipse. Aiming to give another geometric interpretation for the torsion of an admissible curve α : I → R3ν , let’s consider the following order 3 Taylor expansion, assuming that 0 ∈ I and α(0) = 0: s2 s3 α(s) = sα0 (0) + α00 (0) + α000 (0) + R(s) 2 6   2   s s3 0 s3 2 κ α (0) + κ α (0) N α (0) = s − eα ηα κ α (0) T α (0) + 6 2 6

+

s3 κα (0)τα (0) Bα (0) + R(s), 6

where R(s)/s3 → 0 as s → 0. Before proceeding, a bit more terminology is in order: Definition 2.3.13. Let α : I → R3ν be an admissible curve and t0 ∈ I. (i) The osculating plane to α at t0 is the plane spanned by T α (t0 ) and N α (t0 ), passing through α(t0 ). (ii) The normal plane to α at t0 is the plane spanned by N α (t0 ) and Bα (t0 ), passing through α(t0 ). (iii) The rectifying plane to α at t0 is the plane spanned by T α (t0 ) and Bα (t0 ), passing through α(t0 ).
The coordinates of α(s) − R(s) relative to the frame F = T α (0), N α (0), Bα (0) are   2 s3 s3 0 s3 2 s α(s) − R(s) = s − eα ηα κα (0) , κα (0) + κα (0), κα (0)τα (0) . 6 2 6 6 F For s small enough, we then consider the projections of α(s) onto the normal and rectifying planes: B Α H0L

B Α H0L

T Α H0L

N Α H0L

(a) The curve α

(b) The normal plane

(c) The rectifying plane

Figure 2.20: An interpretation for τα (0) > 0.

108  Introduction to Lorentz Geometry: Curves and Surfaces

That is, when τα (0) > 0 we have that α crosses the osculating plane at α(0) in the same direction as the vector Bα (0) points to, and in the opposite direction if τα (0) < 0. For the next result, we’ll introduce the analogues in L3 of the Euclidean sphere, as the locus of points which are (Lorentzian) equidistant with a given center p. Definition 2.3.14. Let p ∈ L3 and r > 0. (i) The de Sitter space of center p and radius r is the set . S21 ( p, r ) = { x ∈ L3 | h x − p, x − pi L = r2 }. (ii) The hyperbolic plane of center p and radius r is the set H2 ( p, r ) ≡ H2+ ( p, r ) . = { x ∈ L3 | h x − p, x − pi L = −r2 and h x − p, e3 i L < 0}. As in the previous section, we also have . H2− ( p, r ) = { x ∈ L3 | h x − p, x − pi L = −r2 and h x − p, e3 i L > 0}. Remark. In order to keep consistency with the notation adopted in the previous section, the center or radius will be omitted according to whether p = 0 or r = 1. The name hyperbolic plane will also be used in Section 4.1, when we discuss some models for Hyperbolic Geometry.

(a) The de Sitter space S21

(b) The two-sheeted hyperboloid H2 ∪ H2−

Figure 2.21: The analogues to Euclidean spheres in L3 . Theorem 2.3.15. Let α : I → R3ν be a unit speed admissible curve with non-vanishing torsion. If the trace of α is contained in S2 ( p, r ), S21 ( p, r ), H2± ( p, r ), or CL ( p), for certain p ∈ R3ν and r > 0, then we have that

− eα ηα ηα α(s) − p = N α (s) + (−1)ν+1 κα (s) τα (s)



1 κα (s)

0 Bα (s)

Local Theory of Curves  109

and also ηα ±r = + (−1)ν eα ηα κ α ( s )2 2

1 τα (s)



1 κα (s)

 0 !2 ,

for all s ∈ I. On the other hand, if this last relation holds and κα (s) is non-constant, there is p ∈ R3ν such that the trace of α is contained in S2 ( p, r ), S21 ( p, r ), H2± ( p, r ) or C L ( p ). Proof: Let’s assume without loss of generality that p = 0. By orthonormal expansion, we have that α(s) = eα hα(s), T α (s)i T α (s) + ηα hα(s), N α (s)i N α (s) + (−1)ν eα ηα hα(s), Bα (s)i Bα (s),

for all s ∈ I, so that our task is finding out those coefficients. Well, since hα(s), α(s)i is constant, it follows that hα(s), T α (s)i = 0. Differentiating that again we obtain

h T α (s), T α (s)i + hα(s), T 0α (s)i = 0 =⇒ hα(s), N α (s)i = −

eα , κα (s)

by the first Frenet-Serret equation. Differentiating one last time and using the second Frenet-Serret equation together with the fact that T α (s) and N α (s) are orthogonal and the relation hα(s), T α (s)i = 0 already obtained, it follows that: eα hα(s), Bα (s)i = − τα (s)



1 κα (s)

0 .

Putting everything together, we finally obtain

− eα ηα ηα α(s) = N α (s) + (−1)ν+1 κα (s) τα (s)



1 κα (s)

0 B α ( s ),

and thus ηα + (−1)ν eα ηα hα(s), α(s)i = κ α ( s )2

1 τα (s)



1 κα (s)

 0 !2 ,

as wanted. As for the converse, let’s see that the natural candidate to center is, indeed, constant. We have:   0  d eα ηα 1 ν ηα α(s) + N α (s) + (−1) Bα (s) = ds κα (s) τα (s) κα (s)  0 1 eα ηα = T α ( s ) + eα ηα N α (s) + (−eα ηα κα (s) T α (s) + τα (s) Bα (s)) κα (s) κα (s)  0 !0  0 1 1 1 ν ν ηα + (−1) ηα Bα (s) + (−1) (−1)ν+1 eα τα (s) N α (s) τα (s) κα (s) τα (s) κα (s)  0 !0 ! eα τα (s) 1 1 = ηα + (−1)ν B α ( s ). κα (s) τα (s) κα (s) Now, it remains to use the assumption to conclude that the coefficient of Bα (s) in the above vanishes, which happens if and only if the term inside parenthesis vanishes. To

110  Introduction to Lorentz Geometry: Curves and Surfaces

. . wit, setting ρ(s) = κα (s)−1 and σ(s) = τα (s)−1 to simplify the notation, we have that this term equals eα ρ ( s ) + (−1)ν (σ0 (s)ρ0 (s) + σ(s)ρ00 (s)). σ(s) On the other hand, differentiating the constant expression from the assumption gives us that  d  0= ηα ρ(s)2 + (−1)ν eα ηα (σ (s)ρ0 (s))2 ds   eα ρ ( s ) = 2eα ηα ρ0 (s)σ(s) + (−1)ν (σ0 (s)ρ0 (s) + σ(s)ρ00 (s)) . σ(s) Since both σ (s) and ρ0 (s) are non-zero, the desired coefficient vanishes and thus α(s) − p =

ηα − eα ηα N α (s) + (−1)ν+1 κα (s) τα (s)



1 κα (s)

0 Bα (s)

for some p ∈ R3ν . So, ηα hα(s) − p, α(s) − pi = + (−1)ν eα ηα κ α ( s )2

1 τα (s)



1 κα (s)

 0 !2

is constant, as wanted. Before delivering what we have promised for this section, the Fundamental Theorem of Curves in Space, we’ll give one last example of application of the theory of curves developed so far, studying a very important class of curves: Definition 2.3.16 (Helices). A unit speed admissible curve α : I → R3ν is a helix if there is a vector v ∈ R3ν , v 6= 0, such that h T α (s), vi is a constant. Moreover, in L3 , we’ll say that the helix is: (i) hyperbolic, if v is spacelike; (ii) elliptic, if v is timelike; (iii) parabolic, if v is lightlike. The direction determined by the vector v is called the helical axis of α. Remark.

• In R3 we may assume that v is a unit vector, and thus h T α (s), vi E = cos θ (s) says that for a helix, the angle formed between T α (s) and v is constant. • In general, differentiating the identity h T α (s), vi = cte. says that the acceleration vector of α always lies in the orthogonal plane to v. We may characterize helices in terms of the ratio τα /κα : Theorem 2.3.17 (Lancret). Let α : I → R3ν be a unit speed admissible curve. Then α is a helix if and only if the ratio τα (s)/κα (s) is constant.

Local Theory of Curves  111

Proof: Suppose that α is a helix and that the helical axis is determined by v. Setting . c = h T α (s), vi we have

hκα (s) N α (s), vi = 0 =⇒ h N α (s), vi = 0, since κα (s) 6= 0. Differentiating again, we have

−eα ηα κα (s)c + τα (s)h Bα (s), vi = 0. Then it suffices to verify that h Bα (s), vi is a non-zero constant. To wit, d h Bα (s), vi = (−1)ν+1 eα τα (s)h N α (s), vi = 0. ds If h Bα (s), vi = 0 for all s, it follows that c = 0, and by orthonormal expansion we conclude that v = 0, contradicting the definition of a helix. Hence τα (s)/κα (s) is a constant. Conversely, suppose that τα (s) = cκα (s), for some c ∈ R. If c = 0, then the curve is planar and Bα (s) = B may be taken to be the vector v we seek. If c 6= 0, we seek a constant vector v = v1 ( s ) T α ( s ) + v2 ( s ) N α ( s ) + v3 ( s ) B α ( s ) such that h T α (s), vi is also a constant. Such condition is equivalent to v1 (s) = v1 being a constant. Differentiating the expression for v yields 0 = − eα ηα κ α ( s ) v2 ( s ) T α ( s )   + v1 κα (s) + v20 (s) + (−1)ν+1 eα cκα (s)v3 (s) N α (s)
+ cκα (s)v2 (s) + v30 (s) Bα (s). By linear independence, we have   0 = − eα ηα κ α ( s ) v2 ( s ) 0 = v1 κα (s) + v20 (s) + (−1)ν+1 eα cκα (s)v3 (s),   0 = cκα (s)v2 (s) + v30 (s), and thus

(−1)ν eα v1 . c This means that we have found a parametrization for the helical axis of α, with v1 as the real parameter. For concreteness, we may set v1 = 1 and take v2 (s) = 0 and v3 (s) =

(−1)ν . v = T α (s) + eα B α ( s ). c This satisfies our requirements. Remark.

• In particular, the proof above shows the existence of exactly one helical axis for each helix. • If α is a parabolic helix, then τα (s) = ±κα (s). The converse holds if ηα = 1. An alternative way to express helices is in terms of their tangent indicatrices:

112  Introduction to Lorentz Geometry: Curves and Surfaces

Definition 2.3.18. Let α : I → R3ν be an admissible curve. Its tangent indicatrix is the curve β : I → R3ν , given by β(t) = T α (t). Remark.

• Note that, even when α has unit speed, we cannot say the same about its tangent indicatrix. • Similarly, one may define the normal indicatrix and the binormal indicatrix of a curve.

Figure 2.22: The tangent indicatrices of Viviani’s window in both ambient spaces. With this definition, Exercise 2.3.8 (p. 104) gives us the: Proposition 2.3.19. Let α : I → R3ν be a unit speed admissible curve. Then α is a helix if and only if its tangent indicatrix has constant curvature. Remark. It also follows from Exercise 2.3.8 that the curvature of any planar tangent indicatrix is constant. Theorem 2.3.20 (Fundamental Theorem for Admissible Curves). Let κ, τ : I → R be given continuous functions, with κ ≥ 0, p0 ∈ R3ν , s0 ∈ I, and ( T 0 , N 0 , B0 ) be a positive orthonormal basis for R3ν . There is a unique unit speed admissible curve α : I → R3ν such that:

• α ( s0 ) = p0 ;

• T α ( s0 ), N α ( s0 ), B α ( s0 ) = T 0 , N 0 , B0 ;

• κα (s) = κ (s) and τα (s) = τ (s), for all s ∈ I. Proof: Consider the following Initial Value Problem (IVP) in R9 :      0 (s)  T 0 κ ( s ) 0 T ( s )         0 τ (s)  N (s)  N 0 (s) = −eT 0 e N 0 κ (s)  B0 (s) 0 (−1)ν+1 eT 0 τ (s) 0 B(s)  
 and T (s ), N (s ), B(s ) = T , N , B
. 0 0 0 0 0 0 By the theorem of existence and uniqueness of solutions
to systems of ordinary differential equations, there is a unique solution T (s), N (s), B(s) of this IVP. We claim that such

Local Theory of Curves  113

a solution forms a positive orthonormal basis for R3ν for all s ∈ I (and not only for s0 ). Indeed, we now consider a second IVP for the curve a : I → R6 : ( a 0 ( s ) = A ( s ) a ( s ),
a(s0 ) = eT 0 , e N 0 , (−1)ν eT 0 e N 0 , 0, 0, 0 where A(s) is the following matrix of coefficients: 0  0   0  −eT e N κ (s) 0 0   0 0 

0 0 0 κ (s) 0 (−1)ν+1 eT 0 τ (s)

0 0 0 0 0 τ (s)

2κ (s) −2eT 0 e N 0 κ (s) 0 0 (−1)ν+1 eT 0 τ (s) 0

0 0 0 τ (s) 0 −eT 0 e N 0 κ (s)

 0  2τ (s)  2(−1)ν+1 eT 0 τ (s) .  0   κ (s) 0

If the components of a(s) are all the scalar products5 between the frame vectors T (s), N (s), and B(s), forming the solution to the previous IVP, we conclude that the constant vector a0 = (eT 0 , e N 0 , (−1)ν eT 0 e N 0 , 0, 0, 0) is the unique solution for the given initial conditions, from where our claim follows. Defining then α : I → R3ν by . α ( s ) = p0 +

Z s s0

T (ξ ) dξ,

we see that α(s0 ) = p0 , α0 (s) = T (s), and α00 (s) = T 0 (s) = κ (s) N (s). Thus, α is a unit speed curve, eα = eT 0 , and span{α0 (s), α00 (s)} = span{ T (s), N (s)}, hence α is admissible. Moreover, κα (s) N α (s) = κ (s) N (s). Taking k · k of both sides and noting that the curvatures are positive, we obtain κα (s) = κ (s), and it follows from this not only that N α (s) = N (s) for all s ∈ I, but also that ηα = e N 0 . If T α (s) = T (s) and N α (s) = N (s) we also obtain that Bα (s) = B(s). Differentiating this, we have that (−1)ν+1 eα τα (s) N α (s) = (−1)ν+1 eT 0 τ (s) N (s). From the relations established so far, we get that τα (s) = τ (s), as wanted. Lastly, to verify the uniqueness of α, let β : I → R3ν by another unit speed admissible curve such that α(s0 ) = β(s0 ), with the same Frenet-Serret frame as α at s0 , with the same curvature and torsion as α for all s ∈ I. Since curvatures and torsions match, both ( T α (s), N α (s), Bα (s)) and ( T β (s), N β (s), B β (s)) are solutions to the same IVP, and in particular it follows that T α (s) = T β (s) for all s ∈ I, so that α and β differ by a constant. Then α(s0 ) = β(s0 ) ensures that such constant is zero, and so α = β as wanted. Remark. The proof of uniqueness in the theorem above may be simplified in R3 . See how to do this in Exercise 2.3.20. Corollary 2.3.21. Two unit speed admissible curves, with same curvature and torsion, whose Frenet-Serret frames have the same causal type, differ by a positive isometry of R3ν . Proof: Let α, β : I → R3ν be two curves as in the statement above. Fix any s0 ∈ I. Since both α and β share the same causal type, we use Proposition 1.4.15 (p. 35) to obtain F ∈ Eν (3, R) such that F (α(s0 )) = β(s0 ), DF (α(s0 ))( T α (s0 )) = T β (s0 ), and similarly for N and B. Such an isometry F is in fact positive (since its linear part takes a positive basis of the space into another positive basis), and hence preserves torsions, so that the curves F ◦ α and β are now in the setting of Theorem 2.3.20 above. We conclude that β = F ◦ α, as wanted. 5 More


precisely, a = h T, T i, h N, N i, h B, Bi, h T, N i, h T, Bi, h N, Bi .

114  Introduction to Lorentz Geometry: Curves and Surfaces

Remark. Given an admissible α : I → L3 not necessarily with unit speed, some care is needed to determine the causal type of its Frenet-Serret trihedron. For example, if α(t) = (t2 , sinh(t2 ), cosh(t2 )),

t > 0,

√ we have√that α0 (t) is always spacelike, while α00 (t) is spacelike for 0 < t < 2, lightlike √ for t = 2 and timelike for t > 2 (verify). Meanwhile, a unit speed reparametrization of α is √ √ √ e α(s) = (s/ 2, sinh(s/ 2), cosh(s/ 2)). Then e α0 (s) is always spacelike and e α00 (s) is always timelike. If it is not possible to concretely exhibit a unit speed reparametrization of the curve, one may analyze the Gram matrix of the osculating plane span{α0 (t), α00 (t)} to determine the causal type of the binormal vector. In this case the Gram matrix  2  4t 8t 8t 8 − 16t4 is indefinite, for all t > 0. Thus, the binormal vector is always spacelike and so the normal vector is always timelike. Lastly, Theorem 2.3.20 also allows us to efficiently conclude our study of admissible helices, when both curvature and torsion are independently constant: Corollary 2.3.22. A helix α : I → R3ν with both constant curvature and torsion is congruent, for a suitable choice of a, b ∈ R, to a piece of one and only one of the following standard helices:
• β1 (s) = a cos(s/c), a sin(s/c), bs/c ;
• β2 (s) = a cos(s/c), a sin(s/c), bs/c ;
• β3 (s) = bs/c, a cosh(s/c), a sinh(s/c) ;
• β4 (s) = bs/c, a sinh(s/c), a cosh(s/c) ;
• β5 (s) = as2 /2, a2 s3 /6, s + a2 s3 /6 ;
• β6 (s) = as2 /2, s − a2 s3 /6, − a2 s3 /6 , . √ where β1 is seen in R3 , the remaining ones in L3 , c = a2 + b2 for β1 and β4 , and p . c = | a2 − b2 | for β2 and β3 ; Proof: We’ll denote the curvature and torsion of α, respectively, simply by κ and τ. If the helix lies in R3 , it is congruent with β1 . Let’s then focus on the remaining helices in L3 . Recalling the proof of Lancret’s Theorem, we have that a vector determining the helical axis of α is eα κ B α ( s ), v = T α (s) − τ
whence hv, vi L = eα 1 − ηα κ 2 /τ 2 . According to Exercise 2.3.25, the causal type of the curves given in the statement of this result is, in general, determined by the constants a and b. Thus, a timelike helix is:

• hyperbolic if κ > |τ |, and thus congruent with β3 ; • elliptic if κ < |τ |, and thus congruent with β2 ; and

Local Theory of Curves  115

• parabolic if κ = |τ |, and thus congruent with β5 . Similarly, a spacelike helix with timelike normal is necessarily hyperbolic and thus congruent with β4 . Lastly, a spacelike helix with timelike binormal is:

• hyperbolic if κ < |τ |, and thus congruent with β3 ; • elliptic if κ > |τ |, and thus congruent with β2 , and • parabolic if κ = |τ |, and thus congruent with β6 .

Exercises Exercise 2.3.15. Let α : I → R3 be a unit speed admissible curve and s0 ∈ I. Assume . that kα(s)k ≤ R = kα(s0 )k for all s sufficiently close to s0 . Show that κα (s0 ) ≥ 1/R. Can you get a similar result in L3 ? Which complications do arise? Hint. Use that s0 is a local maximum for f : I → R given by f (s) = kα(s)k2E . Exercise 2.3.16. Determine all the admissible curves in R3ν such that: (a) all of their tangent lines intercept at a fixed point; (b) all or their normal lines intercept at a fixed point. Why didn’t we bother writing an item (c) here for the binormal vector? Exercise 2.3.17. Let α : I → R3ν be an admissible curve. Show that if all the osculating planes of α are parallel, then α is a planar curve. Exercise† 2.3.18. Show Proposition 2.3.19 (p. 112). Exercise 2.3.19. Let α : I → R3ν be a unit speed admissible curve. Show that α is a helix if and only if det( N α (s), N 0α (s), N 00α (s)) = 0, for all s ∈ I. Exercise† 2.3.20. In R3 , the part of Theorem 2.3.20 (p. 112) regarding the uniqueness of the curve may be proved in other ways. In this exercise we will see two of them. Let α, β : I → R3 be two unit speed admissible curves and s0 ∈ I be such that α(s0 ) = β(s0 ). Suppose that α and β have the same Frenet-Serret frame at s0 , that κα (s) = κ β (s), and τα (s) = τβ (s) for all s ∈ I. (a) Consider a deviation function D1 : I → R, given by . D1 (s) = h T α (s), T β (s)i E + h N α (s), N β (s)i E + h Bα (s), B β (s)i E . Show that, under the given assumptions, we have that D1 (s) = 3 for all s ∈ I. Argue then that each of the terms in the definition of D1 (s) has to be precisely 1, and use this to deduce that the Frenet-Serret frames of α and β actually coincide at all points. Finally, conclude that α = β. Hint. The Cauchy-Schwarz inequality may be useful.

116  Introduction to Lorentz Geometry: Curves and Surfaces

(b) An alternative argument to the one given in the above item considers a different deviation function D2 : I → R given by . D2 (s) = k T α (s) − T β (s)k2E + k N α (s) − N β (s)k2E + k Bα (s) − B β (s)k2E , and then proving that, under the given assumptions, we have D2 (s) = 0 for all s ∈ I. Use this to conclude again that α = β. Remark. Item (b) follows almost trivially from (a) by observing the relation D2 (s) = 6 − 2 D1 (s). Many other books just present our deviation D2 without mentioning our D1 , so it is still instructive to prove item (b) without this remark. Think a bit about the several reasons that make these arguments fail in L3 and understand why we had to use the theory of ordinary differential equations all the way through in the proof of Theorem 2.3.20. Exercise 2.3.21. Let a, b, c ∈ R. Show that if 2b2 = ±3ac, then the curve α : R → R3 given by α(t) = ( at, bt2 , ct3 ) is a helix. Is there a similar condition for α to be a helix when seen in L3 ? Exercise 2.3.22. Let α : I → L3 be a hyperbolic (resp. elliptic, parabolic) helix. If F ∈ P(3, R) is a Poincaré transformation, show that F ◦ α is also a hyperbolic (resp. elliptic, parabolic) helix. Exercise 2.3.23. Let α : I → R3ν be a unit speed admissible curve. Suppose that α is a non-parabolic helix with constant curvature. If v ∈ R3ν is a unit vector giving the direction of the helical axis, consider the projection α : I → R3ν onto the orthogonal plane to v, given by . α(s) = α(s) − ev hα(s), viv. Show that α has constant curvature and justify (in part) the terminology adopted in Definition 2.3.16 (p. 110). Exercise† 2.3.24. Determine admissible curves α1 , α2 : R → L3 with constant curvature and torsion verifying κ = |τ | and α1 (0) = α2 (0) = 0, such that:

(a) T α1 (0), N α1 (0), Bα1 (0) = e3 , e1 , e2 ;

(b) T α2 (0), N α2 (0), Bα2 (0) = e2 , e1 , −e3 . Exercise† 2.3.25. This exercise is a guide to deduce the correct choice of the coefficients a and b in the statement of Corollary 2.3.22 (p. 114). Let a, b ∈ R be non-zero. For simplicity, we’ll omit the subscript indices in all of the curvatures and torsions to follow.
(a) Show that the curve β1 : R → R3 given by β1 (s) = a cos(s/c), a sin(s/c), bs/c , . √ where c = a2 + b2 , has unit speed and curvature and torsion given by κ=

| a| a2 + b2

Also show that

| a| =

κ2

κ + τ2

and τ =

and b =

b . a2 + b2

κ2

τ . + τ2

Local Theory of Curves  117

(b) Assuming | a| 6= |b|, show that β2 , β3 : R → L3 given by   s  bs  s , a sin , , β2 (s) = a cos c c c  s  s  bs β3 ( s ) = , a cosh , a sinh , c c c . p where c = | a2 − b2 |, have opposite causal types, unit speed, are admissible, and have curvature and torsion given by κ=

| a2

| a| − b2 |

Also show that

| a| =

|κ 2

κ − τ2 |

and τ =

and b =

| a2

b . − b2 |

|κ 2

τ . − τ2 |


(c) Show that β4 : R → L3 given by β4 (s) = bs/c, a sinh(s/c), a cosh(s/c) , where . √ c = a2 + b2 , has unit speed, and curvature and torsion given by κ=

| a| + b2

and τ =

κ κ2 + τ2

and b =

a2

Also show that

| a| =

a2

−b . + b2

−τ . κ2 + τ2

(d) Show that β5 , β6 : R → L3 given by  β5 ( s ) =  β6 ( s ) =

as2 a2 s3 a2 s3 , ,s+ 2 6 6



as2 a2 s3 a2 s ,s− ,− 2 6 6

, 3 ,

have opposite causal types, unit speed, are admissible, and have curvature and torsion given by κ = | a| and τ = a. Remark. The previous exercise motivates the definitions for the curves β5 and β6 above. Exercise 2.3.26. Suppose that α : I → R3ν is a unit speed admissible curve with constant curvature and torsion. Determine α explicitly, by solving the Frenet-Serret system. Hint. Differentiate the second Frenet-Serret equation to obtain a second order ODE with constant coefficients involving only N 00α (s) and N α (s) (which is then solved componentwise). Knowing N α (s), integrating the first Frenet-Serret equation gives us T α (s). Then integrate again to get α(s).

118  Introduction to Lorentz Geometry: Curves and Surfaces

2.3.3

Curves with degenerate osculating plane

In this subsection we will assume that all the curves are biregular. In particular, we’re excluding the case where α is a light ray. Our curves are assumed to have unit speed or to be parametrized by arc-photon. To begin this discussion, we have the: Definition 2.3.23. A unit speed curve α : I → L3 is called semi-lightlike if span{α0 (s), α00 (s)} is degenerate (and, thus, α00 (s) is lightlike) for all s ∈ I . Remark.

• In view of this definition, we’ll allow the indicator eα and the coindicator ηα to be zero. This way, if α is lightlike, we have that (eα , ηα ) = (0, 1), while if α is semilightlike, we have (eα , ηα ) = (1, 0). This is done to treat both cases simultaneously. • Since the arclength parameter is denoted by s and the arc-photon parameter by φ, we will allow ourselves to simply omit the parameter when discussing results for both cases. The idea is, as before, to define a suitable frame adapted to the curve. In this case, an orthonormal frame would not carry geometric information about acceleration vector of the curve. We’ll start redefining the tangent and normal vectors: Definition 2.3.24. Let α : : I → L3 be lightlike or semi-lightlike. We define the tangent and the normal to the curve by . T α = α0

and

. N α = α00 ,

respectively.

w

To complete the basis we seek, we need to determine a third vector Bα (to be called again the binormal vector), such that basis ( T α , N α , Bα ) is positive at all points of the curve. In general, we may define the orientation of a basis (v, w) for a lightlike plane in terms of a choice of vector n which is Euclidean-normal to the plane. More precisely, let’s say that (v, w) is positive if (v, w, n) is a positive basis of L3 , for n future-directed. If v is lightlike and w is unit, we have that v × L w is lightlike and proportional to v. Writing v × L w = λv for some λ ∈ R, we analyze the sign of λ as follows:

L

× v

v

w

×E v

n

w

×

L

w

w

v n

v

v

×E w

(a) (v, w) positive (λ < 0)

(b) (v, w) negative (λ > 0)

Figure 2.23: Orientations for a lightlike plane.

Local Theory of Curves  119

This way, if (v, w) is positive, then λ < 0 and, similarly, if (v, w) is negative we have λ > 0. Back to the construction of the basis ( T α , N α , Bα ): we may assume, up to reparametrization, that the basis ( T α , N α ) is positive. In this case, to determine the vector Bα , which will be lightlike, we need to know the values of h T α , Bα i L and h N α , Bα i L as well. In view of the above definition, one of these values has to be 0 (for it to be orthogonal to the spacelike vector) and the other one −1 (so it is linearly independent with the lightlike vector). Which one will be zero and which one will be −1 naturally depends on the causal type of α. Choosing Bα such that h T α , Bα i L = −ηα and h N α , Bα i L = −eα we treat both cases simultaneously, and then we have the: Proposition 2.3.25. Let α : I → L3 be lightlike or semi-lightlike. The triple ( T α , N α , Bα ) is a positive basis for L3 . Proof: We must show that det( T α , N α , Bα ) > 0. Let’s treat here only the case eα = 0 and ηα = 1 (the other case is similar and we ask you to do it in Exercise 2.3.27).
Writing Bα (φ) relative to the basis T α (φ), N α (φ), T α (φ) × E N α (φ) , we see that the only component of Bα (φ) relevant for the required determinant is the one in the direction of T α (φ) × E N α (φ), and let’s say that it is µ(φ) T α (φ) × E N α (φ). Then we have det( T α (φ), N α (φ), Bα (φ)) = µ(φ) det( T α (φ), N α (φ), T α (φ) × E N α (φ)), | {z } >0

so that it only remains to check that µ(φ) > 0. The discussion illustrated by Figure 2.23 tells us that T α (φ) × L N α (φ) = λ(φ) T α (φ) for certain λ(φ) < 0 (since the basis ( T α (φ), N α (φ)) is assumed positive). Applying Id2,1 we obtain that
T α (φ) × E N α (φ) = Id2,1 T α (φ) × L N α (φ) = λ(φ) Id2,1 T α (φ) =⇒

=⇒ h T α (φ) × E N α (φ), Id2,1 T α (φ)i E < 0. Finally, as T α (φ) and N α (φ) are both Lorentz-orthogonal to T α (φ), we have that

−1 = h Bα (φ), T α (φ)i L = µ(φ)h T α (φ) × E N α (φ), Id2,1 T α (φ)i E , and so we conclude that µ(φ) > 0.
So, the frame T α , N α , Bα is called the Cartan frame of α. Geometrically, when the curve is lightlike, the situation is as follows: the vector N α (φ) is spacelike, and so its orthogonal complement is a timelike plane that cuts the lightcone in two light rays, one of them in the direction of T α (φ). Thus, the binormal vector will be in the direction of the remaining light ray in N α (φ)⊥ , being determined by the relation h Bα (φ), T α (φ)i L = −1. A similar interpretation holds when the curve is semi-lightlike. Recall that to define the Frenet-Serret equations when we studied admissible curves, we have used a tool which is no longer available here: the expression of a vector in terms of the trihedron via orthonormal expansion. We’ll remediate this in the following: Lemma 2.3.26. Let α : I → L3 and v ∈ L3 . Then: (i) if α is lightlike, we have that v = −hv, Bα (φ)i L T α (φ) + hv, N α (φ)i L N α (φ) − hv, T α (φ)i L Bα (φ), for all φ ∈ I;

120  Introduction to Lorentz Geometry: Curves and Surfaces

(ii) if α is semi-lightlike, we have that v = hv, T α (s)i L T α (s) − hv, Bα (s)i L N α (s) − hv, N α (s)i L Bα (s), for all s ∈ I. Remark. A possible mnemonic is: switch the position and sign only of the coefficients corresponding to lightlike directions. Proof: We’ll treat both cases at the same time, noting the relations eαn = eα , ηαn = ηα for all n ≥ 1, eα ηα = 0 and eα + ηα = 1. They follow from the only possibilities at hand being (eα , ηα ) = (1, 0) and (eα , ηα ) = (0, 1). Recall that we’re still assuming that ( T α , N α ) is positive. That being said, write v = aT α + bN α + cBα . Applying all the possible products in both sides of this expression, and already organizing everything in matrix form, we have      hv, T α i L eα 0 − ηα a hv, N α i L  =  0 ηα − eα   b  . hv, Bα i L − ηα − eα 0 c It follows from the observations just made that the inverse of the coefficient matrix exists, and it is itself, so that      a eα 0 − ηα hv, T α i L b =  0 ηα −eα  hv, N α i L  . c − ηα − eα 0 hv, Bα i L Particularizing, we conclude the lemma. Before applying the above lemma for the derivatives of the vectors in the Cartan frame, we have the: Definition 2.3.27. Let α : I → L3 be lightlike or semi-lightlike. The pseudo-torsion of . α is given by α = −h N 0α , Bα i L .

d

Remark. In the literature, this pseudo-torsion is also called the Cartan curvature of α. Theorem 2.3.28. Let α : I → L3 be lightlike or semi-lightlike. Then we have that  0    Tα 0 1 0 Tα  N 0α  = ηα α eα α   ηα Nα . 0 Bα eα ηα α − eα α Bα

d

d d

d

Remark. Explicitly, the coefficient matrices when α is lightlike or semi-lightlike are, respectively,     0 1 0 0 1 0  α (φ) 0 1 and 0 0 . α (s) 0 1 0 − α (s) α (φ) 0

d

d

d

d

Proof: The first equation is the definition of the normal vector. For the second equation, applying the Lemma 2.3.26 regarding N 0α as a column vector, we have:

Local Theory of Curves  121

 0  eα 0 − ηα h N α, T α iL ηα −eα  h N 0α , N α i L  N 0α =  0 − ηα − eα 0 h N 0α , Bα i L      eα 0 − ηα − ηα ηα α ηα − eα   0  =  eα α  , = 0 − ηα − eα 0 − α ηα 

d d

d

and so we obtain the second row of the coefficient matrix given in the statement. Similarly for B0α , we have:   0  eα 0 − ηα h Bα , T α i L ηα −eα  h B0α , N α i L  B0α =  0 − ηα − eα 0 h B0α , Bα i L      eα 0 − ηα eα eα ηα − eα   α  =  ηα α  , = 0 − ηα − eα 0 0 − eα α

d

d d

and so we obtain the last row. Example 2.3.29. Let r > 0 and consider the curve α : R → L3 given by       √ φ φ α(φ) = r cos √ , r sin √ , rφ . r r We have seen in Example 2.1.19 (p. 72) that α is lightlike and parametrized by arc-photon. We have       √ √ √ φ φ T α (φ) = α0 (φ) = − r sin √ , r cos √ , r and r r       φ φ N α (φ) = α00 (φ) = − cos √ , − sin √ , 0 . r r To compute Bα (φ), note that the cross product       √ √ √ φ φ T α (φ) × E N α (φ) = r sin √ , − r cos √ , r r r seen in L3 is lightlike and future-directed, so the basis ( T α (φ), N α (φ)) is already positive and no reparametrization for α is required. Moreover, in this case we have something very particular: T α (φ) × E N α (φ) is also Lorentz-orthogonal to N α (φ). This means that Bα (φ) must be a positive multiple of the product T α (φ) × E N α (φ). For the condition h Bα (φ), T α (φ)i L = −1 to be satisfied, it suffices to take       1 φ 1 φ 1 √ sin √ , − √ cos √ , √ . Bα (φ) = 2 r r 2 r r 2 r And so, we finally have:

dα ( φ ) =

−h N 0α (φ), Bα (φ)i L

1 = − sin2 2r



φ √ r



1 − cos2 2r



φ √ r



+0 = −

1 . 2r

122  Introduction to Lorentz Geometry: Curves and Surfaces

Figure 2.24: Cartan frame for α with r = 1/4. The next two results illustrate some of the differences between Theorem 2.3.30. The only planar lightlike curves in

L3

dα and τα .

are light rays.

Proof: Of course, light rays are planar and if α is not a light ray, it admits a reparametrization with arc-photon parameter. Thus, it suffices to check that if α : I → L3 is a lightlike curve parametrized with arc-photon and hα(φ) − p, vi L = 0 for all φ ∈ I, for certain p, v ∈ L3 , then v = 0. Indeed, differentiating that expression three times we obtain

h T α (φ), vi L = h N α (φ), vi L = dα (φ)h T α (φ), vi L + h Bα (φ), vi L = 0. By Lemma 2.3.26 it follows that v = 0. Example 2.3.31. Let f : I → R be a smooth function with second derivative strictly positive (i.e., a strictly convex function) and consider α : I → L3 given by α(s) = (s, f (s), f (s)) . One then sees that α is semi-lightlike with T α (s) = α0 (s) = (1, f 0 (s), f 0 (s)) and
N α (s) = α00 (s) = 0, f 00 (s), f 00 (s) . We have that T α (s) × E N α (s) = (0, − f 00 (s), f 00 (s)) is lightlike and future-directed, so that the basis ( T α (s), N α (s)) is positive. We seek for a lightlike vector Bα (s) = ( a(s), b(s), c(s)), orthogonal to T α (s) and such that h Bα (s), N α (s)i L = −1. Explicitly, we have:  2 2 2  =0  a(s) + b(s) − c(s) 0 a(s) + f (s)(b(s) − c(s)) = 0   00 f (s)(b(s) − c(s)) = −1.

Local Theory of Curves  123

Substituting the third equation into the second one, it follows that a(s) = f 0 (s)/ f 00 (s). With this, the first equation reads

(b(s) − c(s))(b(s) + c(s)) = b(s)2 − c(s)2 = −

f 0 ( s )2 f 0 ( s )2 =⇒ b ( s ) + c ( s ) = , f 00 (s) f 00 (s)2

after using the third equation again. We then obtain Bα (s) =

 1  0 0 2 0 2 2 f ( s ) , f ( s ) − 1, f ( s ) + 1 . 2 f 00 (s)

Finally, we have that

dα (s) = −h N 0α (s), Bα (s)iL =

f 000 (s) . f 00 (s)

In particular, note that α is contained in the plane Π : y − z = 0, but we may choose functions f for which the pseudo-torsion never vanishes. The above example already says that, in general, the pseudo-torsion of a semi-lightlike curve is not a measure of how much the curve deviates from being planar. In fact, the next theorem says that the situation is much more extreme: Theorem 2.3.32. Every semi-lightlike is planar and contained in some lightlike plane. Proof: If α : I → L3 is semi-lightlike, we seek elements p, v ∈ L3 , with v lightlike, such that hα(s) − p, vi L = 0 for all s ∈ I. If this is to happen, differentiating twice we obtain h N α (s), vi L = 0, and conclude that v must be parallel to N α (s) (two orthogonal lightlike vectors in L3 must be proportional). Motivated by this, we seek a smooth function λ : I → R such that v = λ(s) N α (s) is constant. This leads us to 0 = (λ0 (s) +

dα (s)λ(s)) N α (s),

for all s ∈ I. Define v in such a way, by taking  Z s λ(s) = exp − s0

dα (ξ ) dξ

 ,

for some fixed s0 ∈ I. By construction, v is constant and then we may take p = α(s0 ). That done, the justification that p and v satisfy all that was required proceeds as usual: consider f : I → R given by f (s) = hα(s) − α(s0 ), vi L . Clearly, we have f (s0 ) = 0 and f 0 (s) = h T α (s), vi L = 0, for all s ∈ I. In contrast to what happens for admissible curves, the sign of the pseudo-torsion does not influence how the curve crosses its osculating planes. Indeed, it α : I → L3 and assuming that 0 ∈ I and α(0) = 0, we have the Taylor formula α(φ) = φα0 (0) +

φ2 00 φ3 α (0) + α000 (0) + R(φ), 2 6


where R(φ)/φ3 → 0 as φ → 0. Reorganizing, in the frame F = T α (0), N α (0), Bα (0) , we have that the coordinates of α(φ) − R(φ) are  α(φ) − R(φ) =

φ+

d

φ3 φ2 φ3 ( 0 ) , , α 6 2 6

 . F

124  Introduction to Lorentz Geometry: Curves and Surfaces

Figure 2.25: Local canonical form for lightlike α. Projecting, no matter the sign of

dα (0), we have: B Α H0L

B Α H0L

N Α H0L

(a) Projection onto the normal plane

T Α H0L

(b) Projection in the rectifying plane

Figure 2.26: Projections onto the Cartan frame’s coordinate planes. We observe that here, even though the vectors in the Cartan frame are not pairwise orthogonal (no matter the ambient space), we have still represented as the above, since for qualitative effects, only their linear independence is relevant. Thus, we conclude that no matter the sign of the pseudo-torsion, the curve always crosses its own osculating planes in the direction of the binormal vectors. If α is semi-lightlike, in turn, we’ll have  2  s s3 α(s) − R(s) = s, + α (0) , 0 , 2 6 F

d

which would actually allow us to predict Theorem 2.3.32 above. The only relevant projection is γ : I → R2 given by  2  s s3 γ(s) = s, + α (0) , 2 6

d

and it would be natural to seek a relation between the curvature of γ at 0 and the

Local Theory of Curves  125

d

pseudo-torsion α (0). The issue, however, is that since T α (0) is spacelike, N α (0) is lightlike and they’re orthogonal, to study γ we would have to consider the product given . by hh( x1 , x2 ), (y1 , y2 )ii = x1 y1 . Yet another caution is needed: the expression det(γ0 (s), γ00 (s)) = 1+ kγ0 (s)k3

dα (0 ) s

may no longer be interpreted as the curvature of γ, since the ambient plane is degenerate. To wit, there is no reasonable notion of curvature in this setting, since every curve of the form (s, f (s)), where f is a smooth function, may be taken to the x-axis via the function F : R2 → R2 given by F ( x, y) = ( x, y − f ( x )). The derivative DF ( x, y) is a linear map, orthogonal relative to hh·, ·ii and, thus, F would be an “isometry” of this plane. In other words, all the graphs of smooth functions would be congruent. Since every spacelike curve may be parametrized as a graph over the x-axis and the lightlike curves are vertical lines, we see that the is no geometric invariant to associate to each curve here. Before moving on and presenting a version of the Fundamental Theorem of Curves for lightlike and semi-lightlike curves, we will complete the comparison with the results obtained in the previous subsection by considering lightlike and semi-lightlike helices. Definition 2.3.16 (p. 110) is extended without issues to this new setting. Since every semilightlike is planar, it is automatically a helix. For lightlike curves, we recover Lancret’s Theorem: Theorem 2.3.33 (Lancret, lightlike version). Let α : I → L3 be a lightlike curve. Then α is a helix if and only if α (φ) is a constant.

d

Proof: Assume that α is a helix and let v ∈ L3 be such that the h T α (φ), vi L = c ∈ R is constant. Differentiating this twice we directly obtain that

h N α ( φ ), v i L = dα ( φ ) c + h B α ( φ ), v i L = 0

d

for all φ ∈ I. We claim that c 6= 0 and that h Bα (φ), vi L is constant, whence α (φ) must also be a constant. To wit, if c = 0 then Lemma 2.3.26 gives that v = 0, contradicting the definition of a helix and, moreover, we have d h B α ( φ ), v i L = dφ

d

dα (φ)h N α (φ), viL = 0.

d

Conversely, assume that α (φ) = is a constant. If constant vector that clearly satisfies the required. And if

d = 0, then v d 6= 0, define

= Bα (φ) is a

1 . v = T α ( φ ) − B α ( φ ).

d

Indeed, we have that

d d N α (φ) = 0 and v is constant. Furthermore, h T α (φ), vi L = 1/d is a constant, as desired. Theorem 2.3.34 (Fundamental Theorem, second version). Let d : I → R be a contindv 1 = N α (φ) − dφ

uous function, p0 ∈ L3 , s0 , φ0 ∈ I and ( T 0 , N 0 , B0 ) a positive basis for L3 such that B0 is lightlike, and ( T 0 , N 0 ) is a positive basis for a lightlike plane. Then:

(i) if T 0 is lightlike, N 0 is unit, and h T 0 , B0 i L = −1, there is a unique lightlike curve α : I → L3 with arc-photon parameter such that

126  Introduction to Lorentz Geometry: Curves and Surfaces

• α(φ0 ) = p0 ; • ( T α (φ0 ), N α (φ0 ), Bα (φ0 )) = ( T 0 , N 0 , B0 ); •

dα (φ) = d(φ),for all φ ∈ I;

(ii) if T 0 is unit, N 0 is lightlike, and h N 0 , B0 i L = −1, there is a unique semi-lightlike curve α : I → L3 such that

• α ( s0 ) = p0 ; • ( T α (s0 ), N α (s0 ), Bα (s0 )) = ( T 0 , N 0 , B0 ); •

dα (s) = d(s) for all s ∈ I;

Proof: We will focus only on the first case. As in the proof of the previous Fundamental Theorem, consider the following IVP in R9 :      0 (φ)  T 0 1 0 T ( φ )    0      0 1  N ( φ )   N (φ) =  (φ)  B0 (φ) 0 (φ) 0 B(φ)    and T (φ ), N (φ ), B(φ )
= T , N , B
. 0 0 0 0 0 0

d

d

By the usual result regarding existence and uniqueness of solutions for systems of ordinary
differential equations, there is a unique solution T (φ), N (φ), B(φ) for the IVP. We claim that the solution satisfies the conditions given in the hypothesis for all φ ∈ I (and not only φ0 ), that is: T (φ) and B(φ) are always lightlike, N (φ) is unit spacelike and orthogonal to B(φ), and we still have h T (φ), B(φ)i L = −1. To wit, we consider the following IVP for the curve a : I → R6 : ( a 0 ( φ ) = A ( φ ) a ( φ ),
a(φ0 ) = 0, 1, 0, 0, −1, 0 where     A(φ) =    

d

0 0 0 (φ) 0 0

d

0 0 0 1 0 (φ)

0 2 0 2 (φ) 0 0 0 0 0 (φ) 1 0

d

d

 0 0 0 2   0 2 (φ) . 1 0   0 1  (φ) 0

d

d

If the components of a(φ) are the possible scalar products6 between the vectors T (φ), N (φ) and B(φ), solutions for the previous IVP, we conclude that the unique solution
with the given initial conditions is the constant vector a0 = 0, 1, 0, 0, −1, 0 , whence the claim follows. With this done, we define . α ( φ ) = p0 +

Z φ

T (ξ ) dξ.

φ0

Clearly we have α(φ0 ) = p0 and α0 (φ) = T (φ), whence α is lightlike. Differentiating again, we obtain α00 (φ) = N (φ), so that α has arc-photon parameter. Thus, we have that T α (φ) = T (φ) and N α (φ) = N (φ), while the positivity of all bases under discussion also ensures that Bα (φ) = B(φ) as well. 6 In


order, a = h T, T i L , h N, N i L , h B, Bi L , h T, N i L , h T, Bi L , h N, Bi L .

Local Theory of Curves  127

With this, differentiating N α (φ) = N (φ) gives us that

dα ( φ ) T α ( φ ) + B α ( φ ) = d ( φ ) T ( φ ) + B ( φ ), and from the relations established so far it follows that dα (φ) = d(φ) for all φ ∈ I.

The verification that such α is unique is exactly the same as the one done in the proof of the Fundamental Theorem for admissible curves. Corollary 2.3.35. Two curves, both lightlike or semi-lightlike whose osculating planes are positively oriented, and having the same pseudo-torsion, differ by a positive Poincaré transformation of L3 . Proof: Let α, β : I → R3ν be curves as in the statement above and t0 ∈ I. Since the bases for the osculating planes are positive, the Cartan frames for both curves at t0 satisfy the assumptions of Proposition 1.4.15 (p. 35), which gives us F ∈ P(3, R) such that F (α(t0 )) = β(t0 ), DF (α(t0 ))( T α (t0 )) = T β (t0 ), and similarly for N and B. Such F is in fact positive (since its linear part takes a positive basis into another positive basis), and thus preserves pseudo-torsions, so that the curves F ◦ α and β are now in the setting of Theorem 2.3.34 above. We conclude that β = F ◦ α, as wanted. Corollary 2.3.36. A lightlike helix α : I → L3 is congruent, for a suitable choice of r > 0, to a piece of one and only one of the following standard helices: √ √ √
• γ1 ( φ ) = rφ, r cosh(φ/ r ), r sinh(φ/ r ) ; √ √ √
• γ2 (φ) = r cos(φ/ r ), r sin(φ/ r ), rφ ;   φ3 φ φ2 φ3 φ • γ3 ( φ ) = − + , , − − . 4 3 2 4 3

d

Proof: We’ll denote the pseudo-torsion of α, which we know to be constant, just by . From the proof of the lightlike version of Lancret’s Theorem, we know that the helical axis of the curve, if 6= 0, is given by

d

1

d B α ( φ ), whence hv, vi L = 2/d, while we take v = Bα (φ) if d = 0 (and thus hv, vi L = 0). This v = T α (φ) −

way, we have that α is

• hyperbolic if

d > 0, and thus congruent with γ1;

d < 0, and thus congruent with γ2; parabolic if d = 0, and thus congruent with γ3 .

• elliptic if •

Exercises Exercise† 2.3.27. Show Proposition 2.3.25 in the case where the curve is semi-lightlike. Exercise 2.3.28 (More examples). Determine the Cartan frame and the pseudo-torsion for the following curves:

128  Introduction to Lorentz Geometry: Curves and Surfaces

(a) β : R → L3 given by  β(φ) =

√

 rφ, r cosh

φ √ r



 , r sinh

φ √ r

 ,

where r > 0. (b) γ : R → L3 , γ(s) = (−es , s, es ). √ √ (c) ζ : R>0 → L3 , ζ (s) = ( 2/2)(s − log(1/s), s + log(1/s), 2 log s). Exercise† 2.3.29. Prove the Fundamental Theorem for semi-lightlike curves. Exercise 2.3.30 ([44]). Show that every semi-lightlike curve α : I → L3 with constant pseudo-torsion α (s) = 6= 0 contained in the plane Π : y = z has the form   b ds b ds α(s) = ±s + a, 2 e + cs + d, 2 e + cs + d

d

d

d

d

for certain constants a, b, c, d ∈ R, perhaps up to reparametrization.

CHAPTER

3

Surfaces in Space

INTRODUCTION Now we turn our attention to surfaces in R3ν , the main goal of this text. In Section 3.1, we introduce the concept of regular surface and present a few examples, to then classify such surfaces locally as inverse images of regular values of functions from R3 to R and graphs of functions from R2 to R (which provides a larger class of examples). In what follows, we formalize the important concept of tangent plane, seen in Calculus courses, now in our new setting. Aiming to replicate the tools from Calculus in the theory of regular surfaces, we establish the notions of smooth function and differential (through tangent planes), illustrating those with several examples. We conclude this section by discussing the notion of orientation for surfaces, which intuitively says when a surface has an “inside” and an “outside” or, equivalently, when an inhabitant of the surface is capable to decide what is “left” and what is “right”. In Section 3.2, unlike the previous section, we start to consider the influence of the ambient scalar product on the surface, that is, we start to actually study the geometry of the surface. The first step is, naturally, to define the causal type of a surface from the one of its tangent planes, and present some criteria for deciding such a causal type. Next, we’ll see how to measure lengths, angles, and areas in a surface with its First Fundamental Form, which is nothing more than the restriction of the ambient scalar product to its tangent planes. We conclude the section by presenting the concept of isometry: the correct notion of equivalence, when discussing geometry of surfaces. In Chapter 2, to study the trace of a curve in R3ν we employed the invariants curvature and torsion, which depended not only on the tangent directions to the curve, but on its entire Frenet-Serret trihedron. In Section 3.3, to study the analogous situation for surfaces, it suffices to understand how the normal directions to the surface are changing, as this is directly related to the tangent planes. Such analysis is done by using the Gauss normal map and its differential, the Weingarten map. With those, we introduce the Second Fundamental Form and, consequently, the curvatures of a surface. We state Gauss’ Theorema Egregium (whose proof is postponed to Section 3.7), which says that the Gaussian curvature of a surface depends only on geometric measures realized on the surface, independently on the ambient where the surface lies (be it R3 or L3 ). One issue that does not occur in R3 , but appears in the study of surfaces in L3 , concerns the diagonalizability of the Weingarten map, which is always possible at least in the so-called umbilic points of the surface. All of this is discussed in Section 3.4, where we also provide a complete classification of all the surfaces for which all the points are umbilic. In what follows, we prove the existence of inertial parametrizations for any surface, which allows us to obtain visual interpretations for the sign of the Gaussian curvature in any given point. The mean curvature, in turn, is closely related to areas of 129

130  Introduction to Lorentz Geometry: Curves and Surfaces

regions in the surface, and we classify surfaces with zero mean curvature (called critical surfaces) as critical points of the area functional. At this point, we are ready to study curves in a surface, from the point of view of its inhabitants. This begins in Section 3.5. Two important classes of curves in a surface are the asymptotic curves and lines of curvature, which may be described by certain differential equations. In Section 3.6, our focus is on the curves on a surface which play the same role as straight lines in a plane: geodesics. For this, we introduce covariant derivatives of vector fields along curves. In a similar way to what was done in Chapter 2, we use a suitable frame (Darboux-Ribaucour Trihedron) to simultaneously study geodesics, asymptotic curves, and lines of curvature, via new invariants: geodesic curvature and geodesic torsion. Next, we introduce Christoffel symbols of a parametrization, which may be used to locally characterize geodesics in terms of a system of differential equations. Geodesics also arise naturally as solutions of a variational problem, expressed through these same differential equations. We conclude this chapter with Section 3.7, where we state and prove the Fundamental Theorem of Surfaces, completing the comparison with the local theory of curves developed in Chapter 2. Moreover, the propositions to be proved here provide a quick proof of Gauss’ Theorema Egregium, first stated in Section 3.3.

3.1

BASIC TOPOLOGY OF SURFACES

Initially, we present the definition of a regular surface, which does not take into account aspects relative to the ambient geometry (R3 or L3 ), and we study a few useful general facts for what follows. In particular, we’ll need to consider functions defined only on a surface, as well as suitable notions of continuity and differentiability. For this, it is necessary to say what will be the “open subsets of the surface”. Given a subset S ⊆ Rn , recall that A ⊆ S is an open subset of S if A is the intersection of S with an open subset of Rn . That being said, we need to answer the following question: what does it mean for a subset of Ln to be open? Recall that, in Rn , given p = ( p1 , . . . , pn ) and r > 0, we define the open ball of center p and radius r as  Br ( p) = ( x1 , . . . , xn ) ∈ Rn | ( x1 − p1 )2 + · · · + ( xn − pn )2 < r2 , and with this one defines what is an open subset of Rn , which, a priori, does not depend on the extra structure h·, ·i E or h·, ·i L chosen. An open ball may be expressed in terms of k · k E , but not of k · k L (which is not a norm). Not only that, but the natural attempt to define open balls in a similar way to what was done above but using h·, ·i L instead does not yield a good “collection of open sets” in Ln . Thus: we’ll take the open subsets of Ln to be the same ones as in Rn . Throughout this chapter, U ⊆ R2 will always denote an open subset of the plane. Definition 3.1.1 (Regular Surfaces). A regular surface in R3ν is a subset M ⊆ R3ν such that for all p ∈ M, there are open sets U ⊆ R2 and p ∈ V ⊆ M, and a mapping x : U → V such that: (i) x is smooth; (ii) x is a homeomorphism; (iii) Dx(u, v) has full rank for all (u, v) ∈ U.

Surfaces in Space  131

Under these conditions we say that, around p, x is a parametrization for M, (u, v) are local coordinates, and x−1 : x(U ) → U is a chart. Remark. When convenient, we will write just (U, x) instead of x : U → x(U ) ⊆ M.

x (U ) x M

U Figure 3.1: Illustrating the above definition. Definition 3.1.2. A smooth map x : U → R3ν is called a regular parametrized surface if Dx(u, v) has full rank for all (u, v) ∈ U. Remark. A regular parametrized surface is not necessarily injective, so that its image in space may have self-intersections.

Figure 3.2: A regular parametrization with self-intersections. In general, keeping the notation from the previous two definitions, we have that Dx(u, v) having full rank is equivalent to the vectors xu (u, v) and xv (u, v) (which are

132  Introduction to Lorentz Geometry: Curves and Surfaces

the columns of Dx(u, v)) being linearly independent. This, in turn, is equivalent to any of the following conditions:

• xu (u, v) × E xv (u, v) 6= 0; • xu (u, v) × L xv (u, v) 6= 0; • k xu (u, v) × E xv (u, v)k2E 6= 0; • k xu (u, v) × L xv (u, v)k2E 6= 0. Fixing u = u0 and letting v change (or similarly, letting u change and fixing v = v0 ), we have the so-called coordinate curves of x:

Figure 3.3: The coordinate curves of a parametrization. The vectors xu (u0 , v0 ) and xv (u0 , v0 ) are tangent to such coordinate curves, at the point x(u0 , v0 ). The relation between the concepts listed above is summarized in the following: Proposition 3.1.3. Let M be a regular surface, p ∈ M, and consider a bijective smooth map x : U → x(U ) ⊆ M such that p ∈ x(U ) and Dx(u, v) has full rank for all (u, v) ∈ U. Then x−1 is continuous, and thus x is indeed a parametrization for M around p. Remark. In other words, forcing all the parametrizations x in Definition 3.1.1 to be homeomorphisms is redundant in the presence of the remaining conditions. Proof: Explicitly write x(u, v) = ( x (u, v), y(u, v), z(u, v)), and take (u0 , v0 ) ∈ U. Let’s prove that x−1 is continuous in a neighborhood of (u0 , v0 ), so that global continuity follows from (u0 , v0 ) being arbitrary. Since Dx(u, v) has full rank, assume without loss of generality that ∂x ∂x ( u0 , v0 ) ( u0 , v0 ) ∂( x, y) ∂v 6= 0. (u0 , v0 ) = ∂u ∂(u, v) ∂y (u , v ) ∂y (u , v ) 0 0 0 0 ∂u ∂v So, if ϕ : U ⊆ R2 → R2 is given by ϕ(u, v) = ( x (u, v), y(u, v)), we have that: det Dϕ(u0 , v0 ) =

∂( x, y) (u0 , v0 ) 6= 0, ∂(u, v)

Surfaces in Space  133

and the Inverse Function Theorem gives us an open set (u0 , v0 ) ∈ V ⊆ U where the inverse ϕ−1 : ϕ(V ) → V exists and is smooth. Write ϕ−1 ( x, y) = (u( x, y), v( x, y)) and observe that ϕ = π ◦ x, where π is the projection in the first two coordinates. So, given ( x, y, z) ∈ x(V ), we have: ϕ−1 ◦ π ( x, y, z) = ϕ−1 ( x, y) = (u( x, y), v( x, y)) = x−1 ( x, y, z), whence x−1 x(V ) = ϕ−1 ◦ π is the composition of continuous functions, and hence continuous as well. Corollary 3.1.4. Let x : U → R3ν be a regular parametrized surface. If x is injective, then x(U ) is a regular surface. Example 3.1.5. (1) Planes: let p, w1 , w2 ∈ R3ν be such that {w1 , w2 } is linearly independent. Then x : R2 → R3ν given by x(u, v) = p + uw1 + vw2 is an injective regular parametrized surface. Thus, the plane Π passing through p and spanned by w1 and w2 is a regular surface. Since every plane in space has this form for suitable p, w1 and w2 , we see that all planes are regular surfaces. (2) Graphs: if f : U → R is a smooth function, then its graph . gr( f ) = {(u, v, f (u, v)) ∈ R3ν | (u, v) ∈ U } is a regular surface. To wit, the map x : U → R3ν given by x(u, v) = (u, v, f (u, v)) is an injective regular parametrized surface whose image is precisely gr( f ). A parametrization x of this form is called a Monge parametrization. We have similar results for graphs of functions defined in the other coordinate planes in space, x = 0 or y = 0. (3) Each lightcone CL ( p) in L3 is a regular surface. Recall that we have removed the cone vertex (in this case, p itself). Writing p = ( p1 , p2 , p3 ), we have that the cone is covered by the two parametrizations x± : R2 \ {( p1 , p2 )} → L3 , given by   q x± (u, v) = u, v, p3 ± (u − p1 )2 + (v − p2 )2 . (4) The helicoid: consider a circular helix whose axis is the z-axis, and join its points to the z-axis by horizontal lines. One possible parametrization for the obtained set is x : ]0, 1[ × R → R3ν , given by x(u, v) = (u cos v, u sin v, v). Clearly x is smooth and we have that: ∂x (u, v) = (cos v, sin v, 0) and ∂u

∂x (u, v) = (−u sin v, u cos v, 1), ∂v

whence x is a regular parametrized surface (to wit, the partial derivatives are linearly independent because of their last
components). Moreover, note that x is injective, whence the helicoid x ]0, 1[ × R is a regular surface.

134  Introduction to Lorentz Geometry: Curves and Surfaces

Figure 3.4: The helicoid. (5) Surfaces of Revolution: consider a smooth curve, regular and injective in the plane y = 0, α : I → R3ν , given by α(u) = ( f (u), 0, g(u)), and such that f (u) > 0 for all u ∈ I. This condition only says that the curve does not touch the z-axis. Rotating the curve around the z-axis, we obtain the revolution surface generated by α. The curves given by intersections of such surface with horizontal planes are called parallels, while the intersections with vertical planes passing through the origin are called meridians. One standard parametrization that misses only one of the surface’s meridians is x : I × ]0, 2π [ → R3ν , given by x(u, v) = ( f (u) cos v, f (u) sin v, g(u)). One may derive such expression for x in at least two ways: fixed u, we are parametrizing circles of radius f (u) inside horizontal planes of height g(u), or then directly applying the rotation around the z-axis to the curve α, as follows:      cos v − sin v 0 f (u) f (u) cos v  sin v cos v 0  0  =  f (u) sin v  . 0 0 1 g(u) g(u) In any case, since both f and g are smooth, so is x. We have: ∂x (u, v) = ( f 0 (u) cos v, f 0 (u) sin v, g0 (u)) and ∂u ∂x (u, v) = (− f (u) sin v, f (u) cos v, 0). ∂v With this: ∂x ∂x (u, v) × E (u, v) = (− f (u) g0 (u) cos v, − f (u) g0 (u) sin v, f 0 (u) f (u)), ∂u ∂v and so:

2

∂x

(u, v) × E ∂x (u, v) = f (u)2 ( f 0 (u)2 + g0 (u)2 ) 6= 0.

∂u

∂v E

This shows that x is a regular
parametrized surface. And clearly x is injective, so that the image x I × ]0, 2π [ is a regular surface.

Surfaces in Space  135

Figure 3.5: A curve and the revolution surface it generates. For this parametrization, the parallels are the coordinate curves of the form u = cte., while the meridians are the curves of the form v = cte.. Besides the previous examples, many others may be exhibited implicitly via functions satisfying certain properties. That is, it is not always necessary to exhibit parametrizations to show that a set is a regular surface. We start with the: Definition 3.1.6. Let Ω ⊆ R3ν be open and f : Ω → R be a smooth function. We say that q ∈ R3ν is a regular point if D f (q) is surjective1 and a critical value if D f (q) is not surjective. Furthermore, a number a ∈ R is called a regular value if f −1 ({ a}) is non-empty and consists only of regular points. Theorem 3.1.7 (Inverse image of a regular value). Let Ω ⊆ R3ν be open and f : Ω → R be smooth. If a ∈ R is a regular value for f , then f −1 ({ a}) is a regular surface in R3ν . Proof: Let p ∈ f −1 ({ a}). Assume without loss of generality that f z ( p) 6= 0 and consider the function ϕ : Ω → R3 given by ϕ( x, y, z) = ( x, y, f ( x, y, z)). Then: 1 0 0 0 1 0 ∂f det Dϕ( p) = ( p) 6= 0. = ∂f ∂z ∂f ∂f ( p) ( p) ( p) ∂x ∂y ∂z So, the Inverse Function Theorem yields an open set p ∈ V ⊆ Ω for which the inverse ϕ−1 : ϕ(V ) → V exists and is smooth. In view of the first two components of ϕ( x, y, z), we have that ϕ−1 ( x, y, z) = ( x, y, g( x, y, z)) for some smooth function g. Restricted to V ∩ f −1 ({ a}) (which is open in f −1 ({ a})), we have that:

( x, y, z) = ϕ−1 ◦ ϕ( x, y, z) = ϕ−1 ( x, y, f ( x, y, z)) = ϕ−1 ( x, y, a) = ( x, y, g( x, y, a)), whence z = g( x, y, a). This way, if π denotes the projection of R3 in the first two 1 In

this case, this is equivalent to saying that D f (q) is not the zero linear functional.

136  Introduction to Lorentz Geometry: Curves and Surfaces

components, we may define a parametrization x : π (V ∩ f −1 ({ a})) → V ∩ f −1 ({ a}) by x(u, v) = (u, v, g(u, v, a)). Note that x indeed takes values in f −1 ({ a}), since f (u, v, g(u, v, a)) = a. Furthermore, it is straightforward to check that x is smooth, injective, and has full rank (hence a homeomorphism). As p ∈ f −1 ({ a}) was arbitrary, we conclude that f −1 ({ a}) is a regular surface. Example 3.1.8. If p ∈ R3ν is a fixed point and the function f : R3ν → R is defined by f (q) = hq − p, q − pi, we have that D f (q) = 2hq − p, ·i, which is only the zero functional when q = p, once h·, ·i is non-degenerate. In particular, if r 6= 0, the following are regular surfaces:

• f −1 ({r2 }) = S2 ( p, r ), in R3 ; • f −1 ({r2 }) = S21 ( p, r ) and f −1 ({−r2 }) = H2+ ( p, r ) ∪ H2− ( p, r ), in L3 . Each connected component of f −1 ({−r2 }) is, in its own right, a regular surface (see Exercise 3.1.12).

(a) The de Sitter space S21

(b) The two-sheeted hyperboloid H2 ∪ H2−

Figure 3.6: The analogues to Euclidean spheres, in L3 . Just as R3 is the disjoint union of the origin with spheres centered at the origin with arbitrary positive radius, L3 is also a disjoint union of de Sitter spaces, hyperbolic planes, the lightcone, and the origin. More precisely, we have that:       [ [ [ 3 2 2 2 L = S1 (r ) ∪ H (r ) ∪ H− ( r ) ∪ C L ( 0 ) ∪ { 0 } . r >0

r >0

r >0

Figure 3.7: The structure of L3 in slices.

Surfaces in Space  137

Among the examples of regular surfaces seen so far, graphs of smooth functions might seem a quite restricted class of examples, but in fact, in a similar fashion to what was done for curves in the last chapter, it holds that every regular surface is locally the graph of a smooth function. This is made precise in: Proposition 3.1.9 (Local Graph). Let M ⊆ R3ν be a regular surface and p ∈ M. Then, there is a Monge parametrization for M around p. Proof: Take a parametrization (U, x) for M around p. Writing its components as x(u, v) = ( x (u, v), y(u, v), z(u, v)). By the regularity of x, we may assume without loss of generality that ∂( x, y) −1 ( x ( p)) 6= 0. ∂(u, v) Define ϕ : U → R2 by ϕ(u, v) = ( x (u, v), y(u, v)). So we have that:
∂( x, y) −1 det Dϕ x−1 ( p) = ( x ( p)) 6= 0, ∂(u, v) and the Inverse Function Theorem gives us an open set x−1 ( p) ∈ V ⊆ U where the inverse ϕ−1 : ϕ(V ) → V exists and is smooth, of the form ϕ−1 (s, t) = (u(s, t), v(s, t)). Observe that:

(s, t) = ϕ ◦ ϕ−1 (s, t) = ϕ(u(s, t), v(s, t)) = ( x (u(s, t), v(s, t)), y(u(s, t), v(s, t))). . Now consider the map y = x ◦ ϕ−1 : ϕ(V ) → x(V ) ⊆ M. Since ϕ V is a diffeomorphism, y is a parametrization for M around p. And lastly, let’s see that y is a Monge parametrization: y(s, t) = x ◦ ϕ−1 (s, t)

= x(u(s, t), v(s, t)) = ( x (u(s, t), v(s, t)), y(u(s, t), v(s, t)), z(u(s, t), v(s, t))) = (s, t, z(u(s, t), v(s, t))).

Remark. We will see in the next section that it is possible to more precise with this result in L3 by using the causal character of M, to be properly defined there. Regular parametrizations may be used to describe, in terms of coordinates in the plane, certain aspects of the surface. This is important because in the plane, we have available the necessary tools from Calculus and Linear Algebra to study the local geometry of the surface. On the other hand, we want descriptions made using different coordinates to be, in a certain sense, equivalent, and the geometric objects to be defined to be independent of the choice of parametrization. A first step towards that is the: Theorem 3.1.10. Let M ⊆ R3ν be a regular surface and (Ux , x) and (Uy , y) be two . parametrizations for M such that W = x(Ux ) ∩ y(Uy ) is non-empty. Then the change of coordinates2 x−1 ◦ y : y−1 (W ) → x−1 (W ) is smooth. 2 Also

called change of parameters.

138  Introduction to Lorentz Geometry: Curves and Surfaces

y(Uy ) x (U x ) W M

x

y

Ux

Uy

x −1 ◦ y x − 1 (W ) y − 1 (W )

Figure 3.8: Summarizing the situation below. Proof: Pick any point (s0 , t0 ) ∈ y−1 (W ). We will show that x−1 ◦ y is smooth in an open neighborhood of (s0 , t0 ). Then global smoothness follows from (s0 , t0 ) being arbitrary. Explicitly write x(u, v) = ( x1 (u, v), x2 (u, v), x3 (u, v)) and y(s, t) = (y1 (s, t), y2 (s, t), y3 (s, t)), and let (u0 , v0 ) = x−1 ◦ y(s0 , t0 ). By the regularity of x, we may assume without loss of generality that ∂ ( x1 , x2 ) (u0 , v0 ) 6= 0. ∂(u, v) Define ϕ : x−1 (W ) → R2 by ϕ(u, v) = ( x1 (u, v), x2 (u, v)), and note that det Dϕ(u0 , v0 ) =

∂ ( x1 , x2 ) (u0 , v0 ) 6= 0, ∂(u, v)

whence the Inverse Function Theorem gives us an open set (u0 , v0 ) ∈ V ⊆ x−1 (W ) where the inverse function ϕ−1 : ϕ(V ) → V exists and is smooth, say, written in components as ϕ−1 ( x1 , x2 ) = (u( x1 , x2 ), v( x1 , x2 )). If π is the projection of R3 in the first two components, note that y−1 (π −1 ( ϕ(V ))) is an open set containing (s0 , t0 ). Lastly, if (s, t) ∈ y−1 (π −1 ( ϕ(V ))), we have that ( x−1 ◦ y) y−1 (π −1 ( ϕ(V ))) (s, t) = ϕ−1 (π (y(s, t)))

= ϕ−1 (y1 (s, t), y2 (s, t)) = (u(y1 (s, t), y2 (s, t)), v(y1 (s, t), y2 (s, t))) is the composition of smooth maps, hence smooth as well, as wanted. Remark. It follows from the above result that the change of coordinates is a diffeomorphism. It suffices to apply this result to its inverse as well.

Surfaces in Space  139

We know, from Calculus, that the best linear approximation for a smooth function f : U ⊆ R2 → R is represented by the tangent plane to its graph. Considering the Monge parametrization x : U → gr( f ) previously seen, we have that the tangent plane to the graph of f at the point (u0 , v0 , f (u0 , v0 )) is spanned by the vectors     ∂f ∂x ∂f ∂x (u0 , v0 ) = 1, 0, (u0 , v0 ) and (u0 , v0 ) = 0, 1, (u0 , v0 ) . ∂u ∂u ∂v ∂v The next step in our discussion will be, motivated by this, to extend this notion to arbitrary regular surfaces, as follows: Lemma 3.1.11. Let M ⊆ R3ν be a regular surface and (Ux , x) and (Uy , y) be two parametrizations for M such that x(u0 , v0 ) = y(s0 , t0 ), for some (u0 , v0 ) ∈ Ux and (t0 , s0 ) ∈ Uy . Then Dx(u0 , v0 )(R2 ) = Dy(s0 , t0 )(R2 ). Proof: By symmetry, it suffices to show one of the inclusions. Restricting domains, it follows from Theorem 3.1.10 that ϕ = x−1 ◦ y is a diffeomorphism. Thus, differentiating y = x ◦ ϕ at (s0 , t0 ), we have that Dy(s0 , t0 ) = Dx( ϕ(s0 , t0 )) ◦ Dϕ(s0 , t0 ) = Dx(u0 , v0 ) ◦ Dϕ(s0 , t0 ), and so we conclude that Dy(s0 , t0 )(R2 ) ⊆ Dx(u0 , v0 )(R2 ), as wanted. Remark. With the above notation, if  Dϕ(s0 , t0 ) =

 a c , b d

applying the equality in display in the above proof to the vectors (1, 0) and (0, 1), respectively, gives us that ∂y ∂x ∂x (s0 , t0 ) = a (u0 , v0 ) + b (u0 , v0 ) and ∂s ∂u ∂v ∂y ∂x ∂x ( s0 , t0 ) = c ( u0 , v0 ) + d ( u0 , v0 ). ∂t ∂u ∂v In particular, observe that ∂y ∂y ∂x ∂x (s0 , t0 ) × (s0 , t0 ) = det Dϕ(s0 , t0 ) (u0 , v0 ) × (u0 , v0 ). ∂s ∂t ∂u ∂v This lemma allows us to write the: Definition 3.1.12 (Tangent Plane). Let M be a regular surface and p ∈ M. The tangent plane to M at p is defined by   ∂x ∂x . 2 ( u0 , v0 ), ( u0 , v0 ) , Tp M = Dx(u0 , v0 )(R ) = span ∂u ∂v where (U, x) is any parametrization for M around p, such that x(u0 , v0 ) = p. Remark.

• The tangent plane, as defined above, is a vector subspace of R3ν and thus passes through the origin. It is usual to represent the tangent plane Tp M “affinely”, passing through the point p instead (which will play the role of 0). Except in the cases where the distinction is extremely necessary, we will identify both planes.

140  Introduction to Lorentz Geometry: Curves and Surfaces

• We have seen that regular parametrized surfaces may, in general, have selfintersections, and in such points the tangent plane (as defined above) is not welldefined. This is because if p = x(u0 , v0 ) = x(u1 , v1 ), we cannot ensure that Dx(u0 , v0 )(R2 ) = Dx(u1 , v1 )(R2 ). In other words, we do not have a canonical choice to make here. In this case, we define the tangent plane to the parametriza. tion x at (u0 , v0 ) as T(u0 ,v0 ) x = Dx(u0 , v0 )(R2 ). . We say that Bx = { xu (u0 , v0 ), xv (u0 , v0 )} is the basis for Tp M associated to the parametrization x. We may also describe the tangent plane Tp M as the space of velocities of curves starting at p, whose traces lie in the surface M. To make this idea rigorous, we start with the: Lemma 3.1.13. Let M ⊆ R3ν be a regular surface, α : I → M be a curve, and (U, x) be a parametrization for M such that α( I ) ⊆ x(U ). Then there are unique smooth functions u, v : I ⊆ R → R such that α(t) = x(u(t), v(t)), for all t ∈ I. Proof: It suffices to consider x−1 ◦ α : I → U. Then: . (u(t), v(t)) = x−1 ◦ α(t) =⇒ α(t) = x(u(t), v(t)). Clearly u and v are smooth, by definition, and their uniqueness follows from x being a homeomorphism. Proposition 3.1.14. Let M ⊆ R3ν be a regular surface and p ∈ M. Then Tp M = {α0 (0) | α : ]−e, e[ → M such that α(0) = p}. Proof: On one hand, if α : ]−e, e[ → M is a curve such that α(0) = p and (U, x) is a parametrization for M with x(u0 , v0 ) = p, the previous lemma allows us to write α in the form α(t) = x(u(t), v(t)). Note that, in particular, we have (u(0), v(0)) = (u0 , v0 ). Differentiating at t = 0, we have that α 0 (0) = u 0 (0)

∂x ∂x ( u0 , v0 ) + v 0 (0) ( u0 , v0 ), ∂u ∂v

proving one of the inclusions. On the other hand, if x is again a parametrization as before, take an arbitrary tangent vector v p = axu (u0 , v0 ) + bxv (u0 , v0 ) ∈ Tp M. As U is open, there is e > 0 such that (u0 + ta, v0 + tb) ∈ U, for all t ∈ ]−e, e[, so that we have a well-defined smooth curve . α : ]−e, e[ → x(U ) ⊆ M, given by α(t) = x(u0 + ta, v0 + tb). Clearly α(0) = p and α0 (0) = v p , proving the remaining inclusion. Remark.

• Under the above conditions, we say that α realizes v p . When there is no risk of confusion, we omit p from v p . • This result is particularly interesting because it allows us to characterize the tangent plane without the explicit use of any parametrization. Proposition 3.1.15. Let Ω ⊆ R3 be open, f : Ω → R be smooth, and a ∈ R be a regular value for f . If M = f −1 ({ a}) and p ∈ M, then Tp M = ker D f ( p). Proof: See Exercise 3.1.7. Now we move on to the study of functions defined over surfaces.

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Definition 3.1.16. Let M ⊆ R3ν be a regular surface and f : M → Rk be a function. We’ll say that f is smooth if, for every parametrization (U, x) for M, the composition f ◦ x : U → Rk is smooth as a map between Euclidean spaces.

x (U )

f M Rk

x

f ◦x

U Figure 3.9: Illustrating the above definition. Lemma 3.1.17. Let M ⊆ R3ν be a regular surface, f : M → Rk be a function, and . (Ux , x) and (Uy , y) be two parametrizations for M such that W = x(Ux ) ∩ y(Uy ) 6= ∅. Then f ◦ x : x−1 (W ) → Rk is smooth if and only if f ◦ y : y−1 (W ) → Rk is as well. Proof: It suffices to observe that f ◦ x = ( f ◦ y) ◦ (y−1 ◦ x) and that the change of coordinates y−1 ◦ x : x−1 (W ) → y−1 (W ) is a diffeomorphism. Remark. In view of this, to verify whether a function f : M → Rk is smooth, it is not necessary to check the smoothness of f ◦ x for all parametrizations of M, but only for enough parametrizations to cover all of M (usually a small number). Example 3.1.18. (1) Let M ⊆ R3ν be a regular surface, f , g : M → Rk smooth functions, and λ ∈ R. Then f + g, λ f and h f , gi are smooth. We denote by C∞ ( M, Rk ) the real vector space consisting of all the smooth functions from M to Rk . When k = 1, we denote the real algebra C∞ ( M, R) simply by C∞ ( M ). (2) Let Ω ⊆ R3ν be open, F : Ω → Rk be a smooth function, and M ⊆ Ω be a regular surface. Then F M : M → Rk is smooth. (3) Let x : U → R3ν be an injective regular parametrized surface. Then its inverse, x−1 : x(U ) → U, is smooth.

142  Introduction to Lorentz Geometry: Curves and Surfaces

(4) Suppose that p0 , n ∈ R3ν are given, and Π is the plane orthogonal to n and passing through p0 . If M ⊆ R3ν is a regular surface, then the height function relative to Π, h : M → R given by h( p) = h p − p0 , ni, is smooth. In R3 , when knk E = 1, h measures the (signed) height of points in M relative to Π. (5) Let M ⊆ R3ν be a regular surface and p0 ∈ R3ν \ M be given. Then the distance function to p0 , f : M → R given by f ( p) = k p − p0 k E , is smooth. Such function is not necessarily smooth when we replace k · k E with k · k L . Why? Definition 3.1.19. Let M1 , M2 ⊆ R3ν be regular surfaces. A function f : M1 → M2 is smooth if, for all parametrizations (U1 , x1 ) and (U2 , x2 ) of M1 and M2 such that f ( x1 (U1 )) ⊆ x2 (U2 ), the local representation x2−1 ◦ f ◦ x1 : U1 → U2 is smooth as a map between Euclidean spaces. f x1 (U1 ) x1 (U1 )




f x2 (U2 ) M2

M1

x1

x2

x2−1 ◦ f ◦ x1 U1

U2

Figure 3.10: Summarizing the above definition. Moreover, we’ll say that f is a (i) diffeomorphism (between the surfaces) if f is smooth and bijective, with its inverse also smooth. (ii) local diffeomorphism if for every p ∈ M1 , there is an open subset U of M1 containing p such that the restriction f U : U → f (U ) is a diffeomorphism. Like in the case of functions f : M → Rk , we have the: Lemma 3.1.20. Let M1 and M2 be regular surfaces in R3ν , f : M1 → M2 be a function, and (Ux1 , x1 ) and (Ux2 , x2 ) be parametrizations for M1 and M2 , respectively, with f ( x1 (Ux1 )) ⊆ x2 (Ux2 ). If (Uy1 , y1 ) and (Uy2 , y2 ) are further parametrizations . for M1 and M2 with f (y1 (Uy1 )) ⊆ y2 (Uy2 ), such that W1 = x1 (Ux1 ) ∩ y1 (Uy1 ) and . W2 = x2 (Ux2 ) ∩ y2 (Uy2 ) are both non-empty, then: (i) f (W1 ) ⊆ W2 and (ii) the local expression x2−1 ◦ f ◦ x1 : x1−1 (W1 ) → x2−1 (W2 ) is smooth if and only if y2−1 ◦ f ◦ y1 : y1−1 (W1 ) → y2−1 (W2 ) is smooth as well.

Surfaces in Space  143

Proof: The seemingly overwhelming quantity of conditions assumed on the domains and images of the parametrizations considered are necessary only to ensure that all the relevant compositions all make sense. That being understood, the proof is entirely similar to the proof given for Lemma 3.1.17, and we ask you to do it in Exercise 3.1.4. Example 3.1.21. (1) For each p ∈ S2 \ {(0, 0, ±1)}, let f ( p) ∈ S1 × R be the intersection of the horizontal ray starting at the z-axis passing through p, with the cylinder S1 × R. The map f : S2 \ {(0, 0, ±1)} → S1 × R so defined is known as the Lambert cylindrical projection. Let’s see that this map is smooth, by considering the parametrizations x : ]0, π [ × ]0, 2π [ → S2 \ {(0, 0, ±1)} and e x : R × ]0, 2π [ → S1 × R given by . x(u, v) = (cos u cos v, cos u sin v, sin u) . e x(ue, ve) = (cos ve, sin ve, ue), which omit a single meridian from each surface.

p

f ( p)

Figure 3.11: The Lambert projection. We then see that f ( x(u, v)) = e x(sin u, v), and since the map taking (u, v) to (sin u, v) is smooth, we conclude that f restricted to the image of x is smooth. To verify that f is smooth along the meridian omitted by x, one repeats this argument by considering instead of x and e x, other parametrizations y and y e given, for example, by y(u, v) = x(u, v + π ) and y e(u, v) = e x(u, v + π ). Note that f is not a diffeomorphism, since it is not surjective. (2) The unit open Euclidean disk D = {( x, y, 1) ∈ R3ν | x2 + y2 < 1} is diffeomorphic to H2 . Given p ∈ D, let f ( p) be the intersection of H2 with the ray starting from the origin and passing through p. This defines a function f : D → H2 , and we claim it is a diffeomorphism. We consider the parametrizations x : ]0, 1[ × ]0, 2π [ → D and e x : R>0 × ]0, 2π [ → H2 given by . x(u, v) = (u cos v, u sin v, 1) . e x(ue, ve) = (sinh ue cos ve, sinh ue sin ve, cosh ue), omitting, respectively, a line segment in D and a meridian in H2 .

144  Introduction to Lorentz Geometry: Curves and Surfaces

H2 f ( p)

D p

Figure 3.12: A diffeomorphism between D and H2 . To determine f ( x(u, v)), we look for t > 0 such that tx(u, v) ∈ H2 . Such t must satisfy htx(u, v), tx(u, v)i L = t2 u2 − t2 = −1, √ whence t = 1/ 1 − u2 . Thus, we have   u u 1 f ( x(u, v)) = √ cos v, √ sin v, √ 1 − u2 1 − u2 1 − u2   u = ex arcsinh √ ,v . 1 − u2 √
Since the map taking (u, v) to arcsinh(u/ 1 − u2 ), v is a diffeomorphism, we conclude that f is a diffeomorphism between the images of x and e x. Repeating the argument from the previous example, taking new parametrizations to cover what was left out by x and e x, show that f is smooth and, in fact, a local diffeomorphism. Since f is bijective, it follows that f is a diffeomorphism. It is natural, in the process of transferring notions of Calculus to a surface, not only to define a notion of smoothness (as we just did), but also to define what would be the “derivative” of a smooth function between surfaces. For this end, we need the following: Lemma 3.1.22. Let M ⊆ R3ν be a regular surface, p ∈ M, v ∈ Tp M and f : M → Rk be a smooth function. If α : ]−e, e[ → M is a curve which realizes v, then ( f ◦ α)0 (0) depends only on p and v (but not on α). Proof: Let (U, x) be a parametrization for M around p, with p = x(u0 , v0 ) = α(0), such that x(U ) contains the trace of α (this is possible by reducing the domain of α, if necessary). Write the curve in coordinates as α(t) = x(u(t), v(t)), noting that (u(0), v(0)) = (u0 , v0 ) = x−1 ( p) depends only on p, and that u0 (0) and v0 (0) depend only on v, since ∂x ∂x v = u0 (0) ( x−1 ( p)) + v0 (0) ( x−1 ( p)). ∂u ∂v With this:

( f ◦ α ) 0 (0) = u 0 (0)

∂ ( f ◦ x ) −1 ∂ ( f ◦ x ) −1 ( x ( p)) + v0 (0) ( x ( p)) ∂u ∂v

depends only on p and v, as wanted.

Surfaces in Space  145

Remark.

• That the above calculation does not depend on the chosen parametrization is a consequence of the remark following Lemma 3.1.11. • It is usual, in the above notation, to abbreviate ∂f . ∂ ( f ◦ x ) −1 ( p) = ( x ( p)) and ∂u ∂u

∂f . ∂ ( f ◦ x ) −1 ( p) = ( x ( p)). ∂v ∂v

That is, once one has a parametrization, one may talk about partial derivatives of a function defined on a surface, which measure how a function changes along coordinate curves of this parametrization. This allows us to write the: Definition 3.1.23. Let M ⊆ R3ν be a regular surface and f : M → Rk be a smooth function. The differential of f at the point p ∈ M is the map d f p : Tp M → Rk given by d f p (v) = ( f ◦ α)0 (0), where α realizes v. Remark. In the previous proof, we have seen that ( f ◦ α)0 (0) is linear on the coordinates of v relative to the basis for Tp M associated to x, Bx , whence it follows that the map d f p is linear. When the image of such a function f : M1 → R3ν is contained in another surface M2 , Proposition 3.1.14 (p. 140) says that d f p (v) ∈ T f ( p) M2 for each v ∈ Tp M1 , that is, we have d f p : Tp M1 → T f ( p) M2 . Given parametrizations of M1 and M2 around p and f ( p), we may write the matrix of the linear operator d f p relative to the bases associated to these parametrizations. We have: Proposition 3.1.24. Let M1 , M2 ⊆ R3ν be regular surfaces, p ∈ M1 and f : M1 → M2 be a smooth function. Suppose that (U1 , x1 ) and (U2 , x2 ) are two regular parametrizations for M1 and M2 around p = x1 (u0 , v0 ) and f ( p) = x2 (ue0 , ve0 ), respectively, such that f ( x1 (U1 )) ⊆ x2 (U2 ). Then   d f p B ,B = Dψ f (u0 , v0 ), x1

x2

where ψ f = x2−1 ◦ f ◦ x1 is the local representation of f . Proof: Writing ψ f = (ψ1 , ψ2 ), it suffices to note that ∂ ( x2 ◦ ψ f ) ∂ ( f ◦ x1 ) ( u0 , v0 ) = ( u0 , v0 ) ∂u ∂u ∂ψ ∂x ∂ψ ∂x = 1 (u0 , v0 ) 2 (ue0 , ve0 ) + 2 (u0 , v0 ) 2 (ue0 , ve0 ), ∂u ∂ue ∂u ∂e v   and similarly to obtain the second column of d f p B ,B . 

d fp

∂x1 ( u0 , v0 ) ∂u



=

x1

x2

Example 3.1.25. Let’s compute the differentials of the functions seen in Example 3.1.18. (1) Let M be a regular surface, f , g ∈ C∞ ( M, Rk ) and λ ∈ R. Then, for each p ∈ M, we have that d( f + g) p = d f p + dg p , d(λ f ) p = λd f p and d(h f , gi) p = h g( p), d f p i + h f ( p), dg p i.

146  Introduction to Lorentz Geometry: Curves and Surfaces 3 k (2) Let Ω ⊆ R ν
be open, F : Ω → R a smooth function and M ⊆ Ω a regular surface. Then d F M p = DF ( p) T M , for each p ∈ M. In particular, d(id M ) p = idTp M . p

(3) If p0 , n ∈ R3ν are given and M ⊆ R3ν is a regular surface, we have seen that the height function h : M → R relative to the plane Π (orthogonal to n and passing through p0 ), given by h( p) = h p − p0 , ni, is smooth. We have that dh p = h·, ni, for each p ∈ M. (4) Let M ⊆ R3 be a regular surface and p0 ∈ R3 \ M. Then f : M → R, defined by f ( p) = k p − p0 k E , has differential given by d fp =

h p − p0 , ·i E . k p − p0 k E

Proposition 3.1.26 (Chain rule). Let M1 , M2 , M3 ⊆ R3ν be three regular surfaces, f : M1 → M2 and g : M2 → M3 be smooth functions. Then g ◦ f : M1 → M3 is smooth and, for each p ∈ M1 , we have that d( g ◦ f ) p = dg f ( p) ◦ d f p . Proof: To verify smoothness of g ◦ f , it suffices to write the local representation of this composition in terms of the local representations for f and g, which are smooth. Now, let v ∈ Tp M1 and α : ]−e, e[ → M1 be a curve realizing v. Noting that f ◦ α realizes the tangent vector d f p (v) ∈ T f ( p) M2 , we have that d( g ◦ f ) p (v) = (( g ◦ f ) ◦ α)0 (0) = ( g ◦ ( f ◦ α))0 (0) = dg f ( p) (d f p (v)), as wanted. Corollary 3.1.27. Let M1 , M2 ⊆ R3ν be regular surfaces and f : M1 → M2 be a diffeomorphism. Then, for each p ∈ M1 , d f p is a linear isomorphism, whose inverse is given by (d f p )−1 = d( f −1 ) f ( p) . Just like for functions between Euclidean spaces, the above corollary has the following local converse: Theorem 3.1.28 (Inverse Function Theorem). Let M1 , M2 ⊆ R3ν be two regular surfaces and f : M1 → M2 be a smooth function. If p0 ∈ M1 is such that d f p0 is a linear isomorphism, then there exists an open subset U of M1 containing p0 such that the restriction f U : U → f (U ) is a diffeomorphism. Proof: Take parametrizations (U1 , x1 ) around p0 = x1 (u0 , v0 ) and (U2 , x2 ) around f ( p0 ), such that f ( x1 (U1 )) ⊆ x2 (U2 ). Since d f p0 is an isomorphism, the chain rule gives that D ( x2−1 ◦ f ◦ x1 )(u0 , v0 ) is the composition of three isomorphisms, hence an isomorphism as well. The Inverse Function Theorem for Euclidean spaces yields an open −1 subset V ⊆ U1 containing (u0 , v0 ) such that ( x2 ◦ f ◦ x1 ) V : V → ( x2−1 ◦ f ◦ x1 )(V ) . is a diffeomorphism. With this in place, if U = x1 (V ), we have that U is open in M1 and f U : U → f (U ) is a diffeomorphism, whose inverse is given by the composition x1 ◦ ( x2−1 ◦ f ◦ x1 )−1 ◦ x2−1 : f (U ) → U. As an example of application of the Inverse Function Theorem for surfaces, we have the following result, of interesting geometric intuition:

Surfaces in Space  147

Proposition 3.1.29. Let M ⊆ R3ν be a regular surface and p0 ∈ M. Then there are open subsets U ⊆ M and V ⊆ p0 + Tp0 M, containing p0 and 0, respectively, and a diffeomorphism h : V → U such that h(q) − q is (Euclidean) normal to Tp0 M, for each q ∈ V. In other words, the surface is locally the graph of a function defined on its tangent plane.

q h(q)

n p0

p0 + Tp0 M

Figure 3.13: A surface as a graph over one tangent plane. Proof: For this proof, we will use only the Euclidean inner product. Let n ∈ R3ν be a unit normal vector to Tp0 M. Consider then the orthogonal projection f : M → R3 given by f ( p) = p − h p − p0 , nin. Since f ( p0 ) = p0 and h f ( p) − p0 , ni = 0 we have, in fact, f : M → p0 + Tp0 M. Clearly f is smooth and its differential d f p : Tp M → Tp0 M is given by d f p (v) = v − hv, nin. In particular, as n ∈ ( Tp0 M )⊥ , we have that d f p0 = idTp0 M , and so the Inverse Function Theorem yields open subsets U ⊆ M and V ⊆ p0 + Tp0 M containing p0 such that f U : U → V is a diffeomorphism, with inverse h : V → U. Such h fits the bill: to wit, if h(q) = q, then h(q) − q ∈ ( Tp0 M )⊥ , trivially. Else, writing q = f ( p) ∈ V, we have

hh(q) − q, q − p0 i = h p − f ( p), q − p0 i = 0 and hh(q) − q, ni = h p − f ( p), ni = h p − p0 , ni 6= 0, since p − f ( p) ∈ ( Tp0 M)⊥ , q − p0 ∈ Tp0 M, and the last term does not vanish, seeing that if h(q) 6= q, then p 6∈ p0 + Tp0 M. We conclude that h(q) − q ∈ ( Tp0 M )⊥ in this case as well, as wanted. We will conclude this section with a brief discussion regarding a notion of fundamental importance for the following sections: orientability. The basic idea is to define the

148  Introduction to Lorentz Geometry: Curves and Surfaces

orientation of a regular surface from the orientation of its tangent planes, in a similar fashion done for the orientation of lightlike planes, in Section 2.3 (Subsection 2.3.3, to be precise). The initial definition is motivated by the remark made after the proof of Lemma 3.1.11 (p. 139). The change of basis matrix between bases associated to different parametrizations is the Jacobian matrix of the change of coordinates itself, which we know to be a diffeomorphism. Our focus is then turned to the sign of the determinant of such matrix. We have the: Definition 3.1.30 (Orientability). Let M ⊆ R3ν be a regular surface. We’ll say that M is orientable if it is possible to obtain a collection O of parametrizations whose images together cover M, such that for any pair of parametrizations (Ui , xi ) (i = 1, 2) in O, with . W = x1 (U1 ) ∩ x2 (U2 ) 6= ∅, one has that det D ( x2−1 ◦ x1 )(u, v) > 0 for all (u, v) ∈ x1−1 (W ). We’ll also say that: (i) a parametrization (U, x) is compatible with O if det D (e x−1 ◦ x) > 0 for each e e e ) 6= ∅; parametrization (U, x) in O such that x(U ) ∩ e x (U (ii) the collection O is an orientation for M; (iii) M is non-orientable if it is not possible to obtain such an orientation O. Proposition 3.1.31. Let M ⊆ R3ν be a regular surface. If M may be covered with a single parametrization, or with two parametrizations whose images have connected intersection, then M is orientable. Proof: If M is covered by a single parametrization, then the orientation O will consist of this single parametrization only, and the only possible change of coordinates between parametrizations in O is the identity map, whose derivative (itself) has positive determinant. If M is covered by two parametrizations with connected intersection, say, (Ux , x) and (Uy , y), put W = x(Ux ) ∩ y(Uy ) and take (u0 , v0 ) ∈ x−1 (W ). By connectedness, the sign of det D (y−1 ◦ x)(u, v) is the same as the sign of det D (y−1 ◦ x)(u0 , v0 ), for each (u, v) ∈ x−1 (W ). We then have two possibilities:

• if det D (y−1 ◦ x)(u0 , v0 ) > 0, we may take O to be simply the collection formed by x and y; e e • if det D (y−1 ◦ x)(u0 , v0 ) < 0, replace x by (U, x) given by e x(v, u) = x(u, v), where 2 e = {(v, u) ∈ R | (u, v) ∈ Ux }, and take O to be the collection formed by e U x and y only.

Corollary 3.1.32. Surfaces of revolution and graphs of smooth functions defined in open subsets of the plane are orientable regular surfaces. To motivate one equivalence, which we’ll give soon, with the definition of orientability, note that if x : U → R3ν is an injective regular parametrized surface, then for each (u, v) ∈ U, the vector xu (u, v) × xv (u, v) is normal to Tx(u,v) x(U ). If the tangent planes

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to the parametrization are not lightlike, we have a well-defined map N : x(U ) → R3ν , which takes each point x(U ) to a unit normal vector, given by . xu (u, v) × xv (u, v) . N ( x(u, v)) = k xu (u, v) × xv (u, v)k Naturally, we would like to repeat this for arbitrary regular surfaces. Consider for now e e just the Euclidean inner product. If M ⊆ R3 is a regular surface, and (U, x) and (U, x) e are two parametrizations for M with x(U ) ∩ e x(U ) non-empty and connected, we have that xu (u, v) × xv (u, v) e x (ue, ve) × e xve(ue, ve) = ± ue k xu (u, v) × xv (u, v)k kexue(ue, ve) × exve(ue, ve)k e such that x(u, v) = e for each (u, v) ∈ U and (ue, ve) ∈ U x(ue, ve), where the sign ± is the sign of the determinant of the derivative of the coordinate change. To summarize, if M is orientable, it is possible to choose parametrizations which make this “patching up” work, that is, that all the above signs are positive. This yields a unit normal field N : M → R3 globally defined on the whole surface M, that is, a map N that associates to each point p ∈ M a unit vector N ( p) normal to Tp M. Since orientability is a notion which does not depend on the ambient product, we may indeed work only with the Euclidean inner product. For surfaces with non-degenerate tangent planes, the existence of a Euclidean unit normal field is equivalent to the existence of a Lorentzian one. Theorem 3.1.33. Let M ⊆ R3ν be a regular surface. Then M is orientable if and only if there is a smooth Euclidean unit normal field N : M → R3ν , defined on all of M. Proof: If M is orientable, the argument was sketched above: let O be an orientation for M, and for each p ∈ M, take a parametrization (U, x) for M around p, which is in O. If p = x(u0 , v0 ), define . x u ( u0 , v0 ) × x v ( u0 , v0 ) N ( p) = . k xu (u0 , v0 ) × xv (u0 , v0 )k This definition does not depend on the choice of parametrization in O, because all parametrizations there are pairwise compatible. Conversely, assume that there is a smooth Euclidean unit normal field N : M → R3ν defined on all of M. Consider all the parametrizations (U, x) such that x(U ) is connected. Then, for each (u, v) ∈ U we have that N ( x(u, v)) = ±

xu (u, v) × xv (u, v) . k xu (u, v) × xv (u, v)k

If suffices to take O as the collection of all the parametrizations of M with connected image for which the sign in the above formula is positive. Remark. If follows that if M ⊆ R3ν is a regular surface and p ∈ M is any point, there is an orientable open subset of M around p. Corollary 3.1.34. Let Ω ⊆ R3ν be open, f : Ω → R be smooth, and a ∈ R be a regular value for f . Then M = f −1 ({ a}) is orientable. Proof: It suffices to note that N : M → R3ν given by N ( p) =

∇ f ( p) k∇ f ( p)k E

is a smooth Euclidean unit normal field along M.

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Remark. A converse holds: if M is an orientable regular surface, then M is (globally) the inverse image of a regular value of a smooth function. For a proof in the case where M is compact, see [17, p. 130]. It is also interesting to analyze the relation between diffeomorphisms and orientations. Suppose that M1 , M2 ⊆ R3ν are regular surfaces, and that f : M1 → M2 is a diffeomor. phism. If M1 is orientable, say, with an orientation O1 , then the collection O2 = f ∗ O1 in M2 consisting of the parametrizations f ◦ x, for x in O1 , is an orientation for M2 , called the orientation induced by f . Note that if two surfaces are diffeomorphic, then necessarily both are orientable, or both are non-orientable. Definition 3.1.35. Let M ⊆ R3ν be a regular surface equipped with an orientation O and f : M → M be a diffeomorphism. We say that f preserves orientation if f ∗ O = O, and that it reverses orientation otherwise. Proposition 3.1.36. Let M ⊆ R3ν be a connected and orientable surface, and f : M → M be a diffeomorphism. Then: (i) f preserving or reversing orientation depends only on f itself, and not on the orientation chosen for M a priori; (ii) for each orientation O of M we have that det D (y−1 ◦ f ◦ x) > 0

or

det D (y−1 ◦ f ◦ x) < 0,

for any parametrizations x and y in O. Moreover, f preserves orientation in the first case, and reverses it in the second. Example 3.1.37. Consider the function f : S2 → S2 given by√f ( x, y, z) = (− x, y, z),
and the Monge parametrization x : B1 (0) → S2 , x(u, v) = u, v, 1 − u2 − v2 . We have that x−1 ◦ f ◦ x(u, v) = (−u, v), whose derivative has negative determinant. We conclude that f inverts orientation. From this point on, we will assume that all the surfaces we’ll work with are orientable. To get an idea for the sort of surface we leaving out of future discussions, let’s see a “nonexample”: Example 3.1.38 (Möbius Strip). The surface is constructed by considering a line segment of length `, which is twisted while it rotates along a circle of radius r > ` in such a way that opposite ends of this line segment are identified when it returns to its original position in the circle.

Figure 3.14: Building a Möbius strip. A parametrization which describes this motion, for each point in the line segment, is x(u, v) = α(u) + (v − 1/2) β(u),

Surfaces in Space  151

where α(u) = 2(cos u, sin u, 0) and β(u) = cos(u/2)α(u) + sin(u/2)e3 . In the curve β we use u/2 instead of u in the trigonometric functions to ensure that the line segment returns to its original position with the endpoints reversed. If we do not use u/2, the resulting surface would be orientable.

Figure 3.15: The image of x in R3 . Remark. By relaxing the definition of a regular surface, dropping the requirement that parametrizations are homeomorphisms onto its images (i.e., also allowing surfaces to have self-intersections), we may obtain more non-orientable surfaces, by mimicking the above construction. For instance, the Klein bottle arises from replacing the line segment used to construct a Möbius strip by a lemniscate:

(a) The Klein bottle, according to the above construction.

(b) Another realization of the Klein bottle.

Figure 3.16: Realizations of the Klein bottle. Both surfaces above are diffeomorphic and the second one justifies the name “bottle” in Klein bottle. For even more examples, see Chapter 11 in [27].

Exercises Exercise 3.1.1 (Localization). Let M ⊆ R3ν be a regular surface and p ∈ M. Show that we can always choose a parametrization x for M around p whose domain contains the origin in the plane, with x(0, 0) = p.

152  Introduction to Lorentz Geometry: Curves and Surfaces

Exercise† 3.1.2 (Stereographic Projection). Consider the unit sphere S2 = {( x, y, z) ∈ R3 | x2 + y2 + z2 = 1}. Let e3 = (0, 0, 1) be the north pole of S2 . For each (u, v) ∈ R2 the line in R3 joining (u, v, 0) to e3 intercepts S2 \ {e3 } in precisely one point St−1 (u, v). This determines a map St−1 : R2 → S2 \ {e3 }. (a) Write an expression for St−1 and show that it is a regular parametrized surface. (b) Write an expression for the inverse St : S2 \ {e3 } → R2 . Hint. Thinking geometrically and using a similar reasoning as in the previous item is easier than inverting the expression you found. (c) Rewrite the expression for St−1 identifying R2 ≡ C via (u, v) 7→ z = u + iv, in terms of z, z, Re(z) and Im(z). Exercise 3.1.3. Determine the values c ∈ R for which the set . M = {( x, y, z) ∈ R3ν | z(z + 4) = 3xy + c} is a regular surface. Exercise† 3.1.4. Prove Lemma 3.1.20 (p. 142). Exercise 3.1.5. (a) Let φ : R3ν → R3ν be a diffeomorphism and M ⊆ R3ν be a regular surface. Show that the image φ( M ) is also a regular surface, and that Tφ( p) φ( M ) = dφ p ( Tp M ) for all p ∈ M. (b) Use the previous item to show that if M = S2 , S21 , or H2 and f : M → R>0 is . smooth, then M ( f ) = { f ( p) p | p ∈ M } is a regular surface, diffeomorphic to M. Exercise† 3.1.6. (a) Let M ⊆ R3ν be a regular surface. Show that . Diff( M ) = { f : M → M | f is a diffeomorphism} equipped with the operation of function composition is a group. (b) Show that if M1 , M2 ⊆ R3ν are diffeomorphic regular surfaces, then we have that Diff( M1 ) ∼ = Diff( M2 ). Remark. In more general settings, we may consider actions of a subgroup G ⊆ Diff( M ) on M and, under suitable conditions, obtain new “quotient surfaces” M/G (orbit spaces). Exercise† 3.1.7. Show Proposition 3.1.15 (p. 140). Hint. Use Proposition 3.1.14, analyzing the behavior of the function along curves in M. Exercise 3.1.8. Let f : R → R be a smooth function, and consider the regular parametrized surface x : R>0 × R → R3ν given by x(u, v) = (u, v, u f (v/u)). Show that all the tangent planes to this surface pass through the origin 0.

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Exercise 3.1.9. Let x : ]0, 2π [ × R → R3 be a parametrized surface of the form x(θ, z) = (r (θ, z) cos θ, r (θ, z) sin θ, z), where r is a positive function. Show that x has rotational symmetry, that is, ∂r/∂θ = 0, if and only if all (Euclidean) normal lines to the image of x pass through the z-axis. Exercise 3.1.10. Similarly to what was done above, let x : R2 → L3 be a parametrized surface of the form x( ϕ, x ) = ( x, ρ( ϕ, x ) cosh ϕ, ρ( ϕ, x ) sinh ϕ), where ρ is a positive function. Show that ∂ρ/∂ϕ = 0 if and only if all (Lorentzian) normal lines to the image of x pass through the x-axis. Compare with the previous exercise. Exercise 3.1.11 (Horocycles). Let v ∈ L3 be a future-directed lightlike vector and . c < 0. The set Hv,c = { x ∈ H2 | h x, vi L = c} is called a horocycle of H2 , based on v. Let α : I → Hv,c ⊆ H2 have unit speed. (a) Show that s2 v + sw1 + w2 , 2c where w1 and w2 are unit and orthogonal vectors, with w1 spacelike, w2 timelike, w1 orthogonal to v, and hw2 , vi L = c. α(s) = −

Hint. Write α00 (s) as a combination of α(s), α0 (s) and v, and also assume that 0 ∈ I to simplify. It is not necessary to parametrize H2 to solve this exercise. (b) Conclude that α is semi-lightlike, with zero pseudo-torsion. Exercise 3.1.12. Show that if M ⊆ R3ν is a regular surface and S is open in M, then S is also a regular surface and Tp S = Tp M for all p ∈ S. Moreover, the inclusion ι : S ,→ M is smooth. Exercise† 3.1.13. Show Corollary 3.1.27 (p. 146). Exercise 3.1.14. For each p ∈ S2 \ {(0, 0, ±1)}, let f ( p) be the intersection of the ray passing through the origin and passing through p with the cylinder S1 × R. Show that f : S2 \ {(0, 0, ±1)} → S1 × R thus defined is a diffeomorphism. Exercise 3.1.15 (Lambert Projection for S21 ). For each p ∈ S21 , consider f ( p) ∈ S1 × R the intersection of the horizontal ray starting at the z-axis passing through p, with the cylinder S1 × R. Show that the map f : S21 → S1 × R so defined is a diffeomorphism. Exercise 3.1.16. Let M ⊆ R3 be a regular surface and p0 6∈ M. Show that the central projection through p0 , f : M → S2 given by . f ( p) =

p − p0 k p − p0 k E

is a local diffeomorphism if and only if p − p0 6∈ Tp M, for each p ∈ M. Hint. Compute d f p and observe what happens if you’re able to evaluate d f p ( p − p0 ). Exercise 3.1.17 (Instructive challenge). Show that 2 2 2 . M = {( x, y, z) ∈ R3 | ex + ey + ez = a}

is a regular surface if a > 3, which is diffeomorphic to the sphere S2 .

154  Introduction to Lorentz Geometry: Curves and Surfaces

Hint. Verifying that M is a regular surface (not containing the origin) is a straightforward application of Theorem 3.1.7 (p. 135). To see that M is diffeomorphic to S2 , follow the steps:

• for each ( x, y, z) 6= 0, show that the (smooth) function h : R → R given by 2 2 2 2 2 2 h(t) = et x + et y + et z is increasing and surjective onto the interval [3, +∞[; • conclude that the central projection through the origin φ : M → S2 is bijective. Use Exercise 3.1.16 above to conclude that φ is a local diffeomorphism (and hence global). Exercise 3.1.18. Here we have a version of Exercise 2.1.21 (p. 77) in R3 . Let α : I → R3 be a regular parametrized curve and x : U → R3 be a regular parametrized surface. Suppose that α and x intersect transversally at p = α(t0 ) = x(u0 , v0 ), i.e., such that {α0 (t0 ), xu (u0 , v0 ), xv (u0 , v0 )} is linearly independent, where t0 ∈ I and (u0 , v0 ) ∈ U. Let v ∈ R3 be a unit vector and take an arbitrary s ∈ R. (a) Define αs : I → R3 , by αs (t) = α(t) + sv. Show that for small enough s, the traces of αs and x intersect near p. (b) Define xs : U → R3 , by xs (u, v) = x(u, v) + sv. Show that for small enough s, the traces of xs and α intersect near p. Hint. Use the Implicit Function Theorem for a suitable function F : R × I × U → R3 . Exercise 3.1.19. Let M ⊆ R3ν be a regular surface and I an open interval of the real line. Say that a function F : M × I → Rk is smooth if for each parametrization (U, x) for M, the composition F ◦ ( x × Id I ) : U × I → Rk is smooth as a map between Euclidean spaces. Show that: (a) to verify whether such F is smooth, it suffices to check it for a collection of parametrizations whose images cover M; (b) if F, G : M × I → Rk are smooth and λ ∈ R, then F + G, λF and h F, G i are also smooth; (c) given p0 ∈ M and t0 ∈ I, the maps Ft0 : M → Rk and α p0 : I → Rk given by . . Ft0 ( p) = F ( p, t0 ) and α p0 (t) = F ( p0 , t) are smooth; (d) the differential of F, dF( p,t) : Tp M × R → Rk , given by . d dF( p,t) (v, a) = F (α(s), t + sa), ds s=0 where α is any curve in M realizing v, is well-defined (that is, it does not depend on the choice of α). Remark. The ideas presented in this exercise naturally generalize for functions defined in the cartesian product of two (or more) regular surfaces. Moreover, one can put a third surface as the codomain (instead of some Rk ). Can you give statements of results similar to the ones presented above in this setting? Exercise 3.1.20 (Inverse Function Theorem). Let M ⊆ R3ν be a regular surface, I an interval in the real line, and F : M × I → R3 be a smooth function (as in the previous exercise). Show that if ( p0 , t0 ) ∈ M × I is such that dF( p0 ,t0 ) is non-singular, there is an

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open subset V of M containing p0 , an open subset W of R3 containing f ( p0 , t0 ), and r > 0 such that the restriction F V ×]t −r,t +r[ : V × ]t0 − r, t0 + r [ → W 0

0

is a diffeomorphism. Exercise 3.1.21 (Implicit Function Theorem). Let M1 , M2 , M3 ⊆ R3ν be regular surfaces, pi ∈ Mi (i = 1, 2, 3) and f : M1 × M2 → M3 be a smooth function. Suppose that f ( p1 , p2 ) = p3 , and also that the partial differential d2 f ( p1 ,p2 ) : Tp2 M2 → Tp3 M3 defined . by d2 f ( p1 ,p2 ) (w) = d f ( p1 ,p2 ) (0, w) is non-singular. Show that there are neighborhoods V1 ⊆ M1 and V2 ⊆ M2 of p1 and p2 , and a smooth function ϕ : V1 → V2 such that f ( p, ϕ( p)) = p3 for each p ∈ V1 . Hint. “Pull” everything down and apply the “old” Implicit Function Theorem. Exercise 3.1.22. Let M ⊆ R3ν be a regular surface, and suppose that M can be covered by two parametrizations (Ux , x) and (Uy , y), such that W = x(Ux ) ∩ y(Uy ) has two connected components. Call W1 and W2 the two connected components of the inverse image x−1 (W ). Show that if det D (y−1 ◦ x) is positive on W1 and negative on W2 , then M is non-orientable. Exercise 3.1.23. Show that if M1 , M2 ⊆ R3ν are regular surfaces, M2 is orientable, and f : M1 → M2 is a local diffeomorphism, then M1 is orientable. Exercise 3.1.24. Show Proposition 3.1.36 (p. 150). Exercise 3.1.25. Consider the antipodal map A : R3ν → R3ν given by A( p) = − p. One may consider the restrictions A : M → M, for M = S2 , S21 or H2 ∪ H2− . Investigate whether A preserves or inverts orientation. Does the answer depend on the choice of M we take here? Exercise 3.1.26. Let M ⊆ R3ν be a orientable regular surface, and f : M → M be a diffeomorphism. Investigate whether f ◦ f preserves or reverses orientation.

3.2

CAUSAL TYPE OF SURFACES, FIRST FUNDAMENTAL FORM

We finally begin the study of the actual geometry of regular surfaces, and now the ambient space (R3 or L3 ) from which the surface will get its geometry matters. The notion of causal character seen so far for vectors and subspaces of Ln , and then generalized to curves, has a version for surfaces: Definition 3.2.1 (Causal Character). Let M ⊆ L3 be a regular surface. We say that: (i) M is spacelike, if for each p ∈ M, Tp M is a spacelike plane; (ii) M is timelike, if for each p ∈ M, Tp M is a timelike plane; (iii) M is lightlike, if for each p ∈ M, Tp M is a lightlike plane.

156  Introduction to Lorentz Geometry: Curves and Surfaces

Remark.

• We will have a situation similar to what happened for curves: by continuity, if Tp M is spacelike or timelike for some p ∈ M, then Tq M will have the same causal type for each q in some neighborhood of p in M. This way, we may possibly restrict our attention to surfaces with constant causal character, when needed. • Keeping consistency with the conventions adopted throughout the text so far, we will consider regular surfaces in R3 to be spacelike. • If M has no points p for which Tp M is lightlike, we’ll simply say that M is nondegenerate. • If x : U → R3ν is an injective and regular parametrized surface, we will abuse terminology and give the causal character of the regular surface x(U ) to the map x itself. If x is not injective, we may define the causal character of x in a similar fashion to what was done above, but considering the tangent planes to x itself, T(u,v) x. Naturally, parametrizations are one of the main tools we have available to decide the causal type of a surface. We will register the next definition, which may prove convenient in the future to present in an easier way some expressions related to the geometry of a surface: Definition 3.2.2 (Partial indicators). Let M ⊆ R3ν be a non-degenerate regular surface and (U, x) be a parametrization for M. The partial indicators of x are defined as the . . indicators of the partial derivatives of x: eu = exu (u,v) and ev = exv (u,v) . Remark.

• This time, from the start, we’ll allow these partial indicators to assume the value zero, in case one of the derivatives is lightlike. • If we denote our coordinates by (s, t) instead of (u, v), the partial indicators will be es and et , for instance. We know that the tangent planes to a regular surface are 2-dimensional vector subspaces of R3ν and so, in L3 , their causal type is determined by the causal type of their normal directions, whether they are Euclidean or Lorentzian. To summarize, we will register the: Proposition 3.2.3. Let M ⊆ L3 be a regular surface and (U, x) be a parametrization for M such that x(u0 , v0 ) = p ∈ M. Then: (i) Tp M is spacelike if and only if xu (u0 , v0 ) × xv (u0 , v0 ) is timelike; (ii) Tp M is timelike if and only if xu (u0 , v0 ) × xv (u0 , v0 ) is spacelike; (iii) Tp M is lightlike if and only if xu (u0 , v0 ) × xv (u0 , v0 ) is lightlike. With this in mind, we may particularize Proposition 3.1.9, seen previously, as follows: Proposition 3.2.4. Let M ⊆ L3 be a regular surface. (i) If M is spacelike or lightlike, then M is locally the graph of a smooth function whose domain is a subset of the plane z = 0.

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(ii) If M is timelike, then M is locally the graph of a smooth function whose domain is a subset of the plane x = 0 or the plane y = 0. Proof: Take any point p ∈ M and choose a parametrization (U, x) for M with x(u0 , v0 ) = p. From here on, the proof goes exactly like the proof of Proposition 3.1.9, with the only difference being that here we can use the extra information regarding the causal type of M to precisely pinpoint which of the submatrices of Dx(u0 , v0 ) will be non-singular, instead of making a generic assumption as before. Explicitly writing x(u, v) = ( x (u, v), y(u, v), z(u, v)) and noting that   ∂x ∂x ∂(y, z) ∂( x, z) ∂( x, y) ×E = ,− , , ∂u ∂v ∂(u, v) ∂(u, v) ∂(u, v) we have that if M is spacelike (resp., lightlike), then the above vector is timelike (resp., lightlike) at (u0 , v0 ), so that ∂( x, y) (u0 , v0 ) 6= 0, ∂(u, v) and so M admits a reparametrization around p of the form (s, t, f (s, t)) for some smooth function f . Similarly, if M is timelike, then this cross product in display is spacelike, whence ∂(y, z) ∂( x, z) (u0 , v0 ) 6= 0 or (u0 , v0 ) 6= 0, ∂(u, v) ∂(u, v) which would give us reparametrizations around p of the forms ( f (s, t), s, t) or (s, f (s, t), t) for some smooth function f , respectively. In L3 , the additional information given by the causal type of a surface may also impose strong restrictions on its topology: Proposition 3.2.5. There is no compact regular surface of constant causal type in L3 . Proof: Consider the projection π1 , π3 : M → R given, respectively, by π1 ( x, y, z) = x and π3 ( x, y, z) = z. Since M is compact and these functions are continuous, each one of them admits a maximum value in M, say, at p1 and p2 , respectively. At these points, we have, for each v = (v1 , v2 , v3 ) ∈ Tp1 M and w = (w1 , w2 , w3 ) ∈ Tp2 M, that 0 = d ( π 1 ) p1 ( v ) = v 1

and

0 = d ( π 3 ) p2 ( w ) = w3 ,

whence Tp1 M = e1⊥ and Tp2 M = e3⊥ are tangent planes to M with distinct causal types. The next result allows us to determine the causal type of a surface without using parametrizations: Theorem 3.2.6. Let Ω ⊆ L3 be open, f : Ω → R be a smooth function and a ∈ R be a regular value for f . If M = f −1 ({ a}), then: (i) M is spacelike if and only if ∇ f ( p) is always timelike; (ii) M is timelike if and only if ∇ f ( p) is always spacelike; (iii) M is lightlike if and only if ∇ f ( p) is always lightlike. Proof: We have previously seen that Tp M = ker D f ( p), for each p ∈ M. Since the gradient ∇ f ( p) is the vector equivalent to D f ( p) under h·, ·i E , we also have that the tangent plane at p is Tp M = (∇ f ( p))⊥ .

158  Introduction to Lorentz Geometry: Curves and Surfaces

Remark. Exercise 2.2.11 (p. 92) illustrates how the notion of “gradient” may depend on the ambient product chosen. Given Ω ⊆ Rnν open and f : Ω → R smooth, the Euclidean gradient of f is the vector ∇ E f ( p) associated to D f ( p) via h·, ·i E by Riesz’s Lemma. In the same way, the Lorentzian gradient of f is the vector ∇ L f ( p) associated to D f ( p) in a similar fashion using h·, ·i L instead. The notation ∇ E is immediately discarded, as it coincides with the differential operator ∇ seen in basic Calculus courses. On the other hand, in the Lorentzian gradient we have a sign change only in the timelike component:   ∂f ∂f ∂f ∇ L f ( p) = ( p ), · · · , ( p ), − ( p) . ∂x1 ∂xn−1 ∂xn This same idea may be used to define the gradient of functions defined only along surfaces (or in even more general settings): see Exercise 3.2.6. The previous theorem remains valid using ∇ L f ( p) instead of ∇ f ( p). Example 3.2.7 (Causal type of spheres in L3 ). We have seen that if p ∈ L3 and r > 0 are given, and f : L3 → R is given by f (q) = hq − p, q − pi L , then we have that D f (q) = 2hq − p, ·i L . This says that ∇ L f (q) = 2(q − p), and thus if q ∈ S21 ( p, r ), then

h∇ L f (q), ∇ L f (q)i L = 4r2 > 0, whence we conclude that S21 ( p, r ) is a timelike surface. Similarly, if q ∈ H2± ( p, r ), we have h∇ L f (q), ∇ L f (q)i L = −4r2 < 0, whence H2± ( p, r ) is a spacelike surface. To study the geometry of the surface, we would have many ways to define “length” and “angle” over it. Let’s use the previous existence of those notions in the ambient R3ν , motivating the following: Definition 3.2.8 (First Fundamental Form). Let M ⊆ R3ν be a regular surface and p ∈ M. The First Fundamental Form of M at p is the bilinear map I p : Tp M × Tp M → R . given by I p (v, w) = hv, wi. Remark.

• We will abbreviate I p (v, v) simply by I p (v), let alone when we decide to omit the point p as well. • In L3 , I p is also called the Minkowski First Fundamental Form of M at p. • The First Fundamental Form is a particular case of a more general concept called a pseudo-Riemannian metric. When M is spacelike, the metric is called Riemannian, and when M is timelike the metric is called Lorentzian. We will briefly discuss this further in Chapter 4, ahead. For more details see, for instance, [54]. Definition 3.2.9 (Components of the First Form). Let M ⊆ R3ν be a regular surface and (U, x) be a parametrization for M. The components of the First Fundamental Form relative to x are defined by   ∂x . E(u, v) = Ix(u,v) (u, v) , ∂u   ∂x ∂x . F (u, v) = Ix(u,v) (u, v), (u, v) and ∂u ∂v   ∂x . G (u, v) = Ix(u,v) (u, v) . ∂v

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Remark. . . . • It is also usual to write g11 = E, g12 = g21 = F and g22 = G, so that all the relevant information regarding the First Fundamental Form is encoded in the Gram matrix ( gij (u, v))1≤i,j≤2 of Ix(u,v) relative to the basis Bx . We will see shortly that this Gram matrix is non-singular precisely when M is non-degenerate, in which case we’ll denote the inverse matrix by ( gij )1≤i,j≤2 , with upper indices.

• A third way to represent the First Fundamental Form in coordinates is through differential notation: ds2 =

2

∑

gij dui du j = E(u, v) du2 + 2F (u, v) du dv + G (u, v) dv2 ,

i,j=1

where we identify u ↔ u1 and v ↔ u2 (do not confuse this with exponents). We will repeat such identifications in the future without many comments, as well as use whichever way of representing the First Fundamental Form is more convenient in the moment. We have seen in Chapter 1 how to use Sylvester’s Criterion to determine the causal type of subspaces of Ln . When dealing with surfaces, though, the next result makes our lives easier: Proposition 3.2.10. Let x : U → L3 be a regular parametrized surface. The causal type of x is decided by the sign of the determinant of its First Fundamental Form:
(i) x is spacelike if and only if det ( gij )1≤i,j≤2 > 0;
(ii) x is timelike if and only if det ( gij )1≤i,j≤2 < 0;
(iii) x is lightlike if and only if det ( gij )1≤i,j≤2 = 0. Proof: This follows directly from Proposition 3.2.3 (p. 156) by using Lagrange’s Identity in L3 : h xu , xu i L h xu , xv i L
= − det ( gij )1≤i,j≤2 . h xu × L xv , xu × L xv i L = − h xv , xu i L h xv , xv i L

In principle, we could ask ourselves what is the importance of the First Fundamental Form of a regular surface, since it is nothing more than the restriction of the ambient product to each tangent plane. To some extent, the answer lies in the question itself: the First Fundamental Form allows us to study the geometry of the surface without making reference to the ambient space, once the restriction of h·, ·i has been made. In other words, to study the intrinsic geometry of the surface it is not necessary to know how the ambient product acts on vectors which are not tangent. Let’s see a few notions which become intrinsic: Example 3.2.11 (Intrinsic concepts). Let M ⊆ R3ν be a regular surface, p ∈ M, (U, x) be a parametrization for M, and α : I → M be a curve in M. (1) If M is spacelike, the angle between two vectors u, v ∈ Tp M \ {0} is the number θ ∈ [0, 2π [ determined by the relation cos θ = q

I p (u, v) q . I p (u) I p (v)

160  Introduction to Lorentz Geometry: Curves and Surfaces

(2) If M is timelike in L3 , the hyperbolic angle between two timelike vectors u, v ∈ Tp M, both future-directed or past-directed, is the number ϕ ≥ 0 determined by the relation cosh ϕ = − q

I p (u, v) q . −I p ( u ) −I p ( v )

(3) The arclength of α is expressed just as Z q L[α] = |Iα(t) (α0 (t))| dt. I

This also justifies the differential notation adopted to express the First Fundamental Form in coordinates. Suppose that M ⊆ R3 and that the curve is written in coordinates as α(t) = x(u(t), v(t)). If s is an arclength function for α, omitting points of evaluation, we have that Iα(t) (α0 (t)) =



ds dt

2



=E

du dt

2

+ 2F

du dv +G dt dt



dv dt

2 .

A more rigorous interpretation of the symbols du and dv is the following: suppose that x(u0 , v0 ) = p, and take v = axu (u0 , v0 ) + bxv (u0 , v0 ) ∈ Tp M. Regarding . du and dv as the basis for Tp∗ M = ( Tp M)∗ (the so-called cotangent plane to M at p) dual to xu (u0 , v0 ) and xv (u0 , v0 ), and denoting the coefficients of the First Fundamental Form at this point by E0 , F0 G0 only, we have: I p (v) = I p ( axu (u0 , v0 ) + bxv (u0 , v0 ))

= a2 I p ( xu (u0 , v0 )) + 2abI p ( xu (u0 , v0 ), xv (u0 , v0 )) + b2 I p ( xv (u0 , v0 )) = E0 a2 + 2F0 ab + G0 b2 = E0 (du(v))2 + 2F0 du(v) dv(v) + G0 (dv(w))2 = E0 du2 (v) + 2F0 du(v) dv(v) + G0 dv2 (v) = ds2p (v). (4) The energy of α is expressed just as 1 E[α] = 2

Z I

Iα(t) (α0 (t)) dt.

This functional will have a crucial role in the study of geodesics that will be done in Section 3.6. (5) If R is an open subset of M such that the closure R is compact and R ⊆ x(U ), we define the area of R as Z q . A( R) = | det( gij (u, v))| du dv. x −1 ( R )

In Exercise 3.2.7 we ask you to show that A( R) does not depend on the choice of parametrization x. We extend the last example above to regions not necessarily contained in the image of a single parametrization. For example, when R is covered by two parametrizations, we have the:

Surfaces in Space  161

Definition 3.2.12. Let M ⊆ R3ν be a regular surface and (Ux , x) and (Uy , y) be two parametrizations for M. If R is an open subset of M such that the closure R is compact and R ⊆ x(Ux ) ∪ y(Uy ), we define the area of R as . A( R) =

Z

q x−1 ( R)∩Ux

−

Z

Z

| det( gij (u, v))| du dv + y−1 ( R)∩Uy q | det( gij (u, v))| du dv,

q

| det( gfij (ue, ve))| de u de v−

x−1 ( R∩ x(Ux )∩y(Uy ))

where ( gij )1≤i,j≤2 and ( gfij )1≤i,j≤2 denote the components of First Fundamental Form of x and y, respectively. Remark.

• It also follows from Exercise 3.2.7 that the third integral above may be replaced by Z q | det( gfij (ue, ve))| de u de v. y−1 ( R∩ x(Ux )∩y(Uy ))

• It is a known fact, whose proof is outside the scope of this text, that every regular surface in R3ν may be covered by three or fewer parametrizations. For more details, see [6]. Example 3.2.13. (1) Planes: we have seen that every plane is the image of some (injective) regular parametrized surface of the form x(u, v) = p + uw1 + vw2 , where p, w1 , w2 ∈ R3ν are given, and {w1 , w2 } is linearly independent. If the plane is not lightlike, there is no loss of generality in assuming that w1 and w2 are orthogonal and both unit vectors. Observe that eu = ew1 and ev = ew2 . Thus, we have that the First Fundamental Form of the plane in those coordinates is given by ds2 = ew1 du2 + ew2 dv2 . In the pathological case where the plane is lightlike, if w1 is unit (spacelike) and w2 is lightlike and orthogonal to w1 , we obtain just the degenerate ds2 = du2 . (2) When expressing the First Fundamental Form of a surface in coordinates, one must pay attention to the domain of the chosen parametrization. Maps which are not regular may introduce singularities in the metric’s coordinate expression which are artificial and unrelated to the geometry of the surface. For example, we have seen that H2 is a spacelike surface. Consider x : R × ]0, 2π [ → H2 given by x(u, v) = (sinh u cos v, sinh u sin v, cosh u). We have that 

( gij (u, v))1≤i,j≤2 =

 1 0 , 0 sinh2 u

or ds2 = du2 + sinh2 u dv2 .

The Gram determinant is just sinh2 u ≥ 0, which vanishes only for u = 0. This indeed shows that H2 is spacelike on the points for which u 6= 0, but it does not say that H2 is lightlike at x(0, v) = (0, 0, 1). This happens because the derivative xv (0, v) is the zero vector, and so x is not a valid parametrization, unless the points of the form (0, v) are removed from its domain.

162  Introduction to Lorentz Geometry: Curves and Surfaces

This phenomenon is not particular to L3 : for the sphere S2 , we may take y(u, v) = (cos u cos v, cos u sin v, sin u), and verify that in this case we have the same situation as above, with   1 0 ( gij (u, v))1≤i,j≤2 = , or ds2 = du2 + sin2 u dv2 . 0 sin2 u (3) Graphs. If f : U → R is smooth, we have seen that the Monge parametrization x : U → R3ν given by x(u, v) = (u, v, f (u, v)) is an injective and regular parametrized surface, so that gr( f ) = x(U ) is a surface regular. We have in L3 that   1 − f u2 − f u f v ( gij,L )1≤i,j≤2 = , − f u f v 1 − f v2
and thus det ( gij,L )1≤i,j≤2 = 1 − k∇ f k2E , where this Euclidean norm is (clearly) taken in R2 . Then we conclude that

• gr( f ) is spacelike if and only if k∇ f k E < 1; • gr( f ) is timelike if and only if k∇ f k E > 1; • gr( f ) is lightlike if and only if k∇ f k E = 1.

spacelike

∇f

lightlike timelike 1

Figure 3.17: Illustrating the classification criterion for graphs over the plane z = 0. The matrix of the First Fundamental Form induced by h·, ·i E , i.e., regarding the graph as a surface in R3 , is   1 + f u2 fu fv ( gij,E )1≤i,j≤2 = . f u f v 1 + f v2 (4) The cylinder of radius r > 0 S1 (r ) × R = {( x, y, z) ∈ R3ν | x2 + y2 = r2 }, may be parametrized (excluding a meridian) by x : R × ]0, 2π [ → R3ν given by x(u, v) = (r cos u, r sin u, v). Its First Fundamental Form is given in coordinates by ds2 = r2 du2 + (−1)ν dv2 . We see from this that the cylinder is spacelike in R3 and timelike in L3 .

Surfaces in Space  163

Figure 3.18: The cylinder S1 × R. (5) The lightcone CL (0) may be parametrized (excluding a light ray) by the map x : (R \ {0}) × ]0, 2π [ → CL (0) given by x(u, v) = (u cos v, u sin v, u). Its Minkowski First Fundamental Form is given by ds2 = du2 . That is, we see that CL (0) is a lightlike surface, as the name suggests. In R3 , its metric is given by ds2 = du2 + u2 dv2 . (6) Surfaces of revolution: we may generalize the previous two examples. We have previously seen that if α : I → R3ν is an injective, regular, and smooth curve, of the form α(u) = ( f (u), 0, g(u)) with f (u) > 0 for all u ∈ I, then x : I × ]0, 2π [ → R3ν given by x(u, v) = ( f (u) cos v, f (u) sin v, g(u)) is an injective and regular parametrized surface, whose image is the surface of revolution generated by α. Its First Fundamental Form is given by:  0  hα (u), α0 (u)i 0 ( gij (u, v))1≤i,j≤2 = , 0 f ( u )2 no matter which is the ambient space considered. We then see that in L3 , the causal type of x is the same one as α’s. When α is not lightlike, it is usual to consider a unit speed reparametrization before generating the surface of revolution, and this gives us the metric ds2 = eα du2 + f (u)2 dv2 . The relation with the causal type of α is to be expected, in the following geometric sense: the vector xv (u, v) (in fact, any tangent vector to a parallel of the surface) will always give us a spacelike direction, while xu (u, v) is the image of α0 (u) under a rotation around the z-axis, which is a Lorentz transformation. In the same way, if we rotate a curve contained in the plane x = 0 or y = 0 around the x-axis or y-axis, respectively, the corresponding surface of revolution will have tangent planes of all possible causal types. Indeed, any tangent vector to a parallel of the surface will complete one full rotation around the revolution axis. The fact that Euclidean

164  Introduction to Lorentz Geometry: Curves and Surfaces

rotations around both the x and y-axes are not Lorentz transformations is further evidence for this. However, hyperbolic rotations around both the x and y-axes are Lorentz transformations, which hints that in L3 we will have new types of surfaces of revolution. See Exercise 3.2.9.

Exercises Exercise 3.2.1. Consider the cylinder S1 × R = {( x, y, z) ∈ R3ν | x2 + y2 = 1}. (a) In R3 , find all the curves in the cylinder which make a constant angle with all of its generating lines (vertical lines). (b) In L3 , find all timelike curves in the cylinder which make a constant hyperbolic angle with all of its generating lines. Is the result different from what you got in (a)? Exercise 3.2.2. Fix 0 < α0 < π/2 and consider x : R>0 × ]0, 2π [ → L3 given by x(u, v) = (u cos v tan α0 , u sin v tan α0 , u). Show that x is an injective and regular parametrized surface, and discuss the causal type of x in terms of α0 . Make sketches of the image of x for α0 = π/6, π/4 and π/3. Exercise 3.2.3 (Tangent surfaces - I). Let α : I → R3ν be an admissible curve whose curvature never vanishes. Consider its tangent surface, x : I × R → R3ν , given by x(t, v) = α(t) + vα0 (t). Show that x restricted to U = {(t, v) ∈ I × R | v 6= 0} is a regular parametrized surface, discuss its causal type in terms of the causal type of the Frenet-Serret trihedron for α, and determine the tangent planes to x along its coordinate curves. Exercise 3.2.4 (Helicoids). Let a, b > 0 and consider the helix α : R → R3ν given by α(t) = ( a cos t, a sin t, bt). (a) For each t ∈ R, consider the line passing through α(t) and orthogonally crossing the z-axis. Obtain a parametrized surface x : R2 → R3ν , regular and injective, whose image is the union of these lines: this surface is called a helicoid (we have seen in the text a particular case of this, with a = b = 1). (b) In L3 , discuss the causal type of x in terms of a and b and find the lightlike helix in the helicoid which divides it into two regions: a spacelike one and a timelike one. Exercise 3.2.5 (Hyperbolic helicoids). Let’s study now the Lorentzian analogue of the situation in the previous exercise. Let a, b > 0 and consider the helix α : R → L3 given by α(t) = (bt, a cosh t, a sinh t). (a) For each t ∈ R, consider the line passing through α(t) and orthogonally crossing the x-axis. Obtain a parametrized surface x : R2 → L3 , regular and injective, whose image is the union of these lines. (b) Discuss the causal type of x in terms of a and b and find the lightlike helix in the image x(R2 ) which divides it into two regions, like in the previous exercise.

Surfaces in Space  165

Exercise† 3.2.6 (Surface gradient). (a) Let M ⊆ R3ν be a non-degenerate regular surface and f : M → R be a smooth function. For each p ∈ M, since d f p is a linear functional defined in Tp M, the non-degeneracy of M allows us to apply Riesz’s Lemma to obtain a tangent vector grad f ( p) ∈ Tp M satisfying d f p (v) = hgrad f ( p), vi,

for all v ∈ Tp M.

If (U, x) is a parametrization for M, show that

(grad f ) ◦ x =

f u G − f v F ∂x f v E − f u F ∂x + , 2 EG − F ∂u EG − F2 ∂v

where f u and f v stand for the partial derivatives of the composition f ◦ x. In particular, this shows that grad f : M → R3ν is smooth. (b) Use item (a) to verify that in R2 we have grad f ( x, y) = ∇ f ( x, y), and in L2 we have grad f ( x, y) = ∇ L f ( x, y). (c) Use item (a) to deduce an expression for the gradient of a function defined on R2ν in polar and Rindler coordinates (see Exercise 2.2.14, p. 94). More precisely, considering x(r, θ ) = (r cos θ, r sin θ ) in R2 \ {0} and y(ρ, ϕ) = (ρ cosh ϕ, ρ sinh ϕ) in the Rindler wedge of L2 , show that ∂f 1 ∂f er + e and ∂r r ∂θ θ ∂f 1 ∂f (grad f ) ◦ y = eρ − eϕ, ∂ρ ρ ∂ϕ

(grad f ) ◦ x =

where er , eθ , eρ and e ϕ denote the unit vectors in the direction of the respective derivatives of x and y. Hint. If you want to, you may regard R2 and L2 as coordinate planes inside L3 . Verify that dx2 + dy2 = dr2 + r2 dθ 2 and dx2 − dy2 = dρ2 − ρ2 dϕ2 . Exercise 3.2.7. Let M ⊂ R3ν be a regular surface, and (Ux , x) and (Uy , y) be parametrizations for M. Suppose that R is an open subset of M for which R is compact, R ⊆ x(Ux ) ∩ y(Uy ). Show that the area of R does not depend on the choice of parametrization, that is: Z Z q q | det( gij (u, v))| du dv = | det( gfij (ue, ve))| de u de v, x −1 ( R )

y −1 ( R )

where ( gij )1≤i,j≤2 and ( gfij )1≤i,j≤2 denote the components of the First Fundamental Forms of x and y, respectively. Exercise 3.2.8 (More graphs). Let f : U → R be a smooth function and x : U → L3 be a parametrized surface, given by x(u, v) = (u, f (u, v), v) or ( f (u, v), u, v). Show that x is:

• spacelike if and only if hf(∇ f ), f(∇ f )i L > 1; • timelike if and only if hf(∇ f ), f(∇ f )i L < 1; • lightlike if and only if hf(∇ f ), f(∇ f )i L = 1,

166  Introduction to Lorentz Geometry: Curves and Surfaces

where the “flip” operator f : L2 → L2 is given by f( x, y) = (y, x ). Make a drawing similar to Figure 3.17 (p. 162) to illustrate this criterion. Hint. Switching the first two coordinates in L3 is a Poincaré transformation, so you only need to do one of the cases. Exercise 3.2.9 (Surfaces of hyperbolic revolutions – I). Let α : I → L3 be a regular, smooth and injective curve, of the form α(u) = ( f (u), 0, g(u)), with g(u) > 0 for each u ∈ I. Consider the surface generated by the hyperbolic rotation of α around the x-axis, parametrized by the map x : I × R → x( I × R) ⊆ L3 given by x(u, v) = ( f (u), g(u) sinh v, g(u) cosh v). Show that x is an injective and regular parametrized surface whose causal type is the same causal type of α, and compute its Minkowski First Fundamental Form. What does the metric look like (in differential notation) when α has unit speed? Remark.

• Following the usual terminology, we will say that the coordinate curves u = cte. are parallels and the curves v = cte. are meridians. The parametrization x above omits one meridian of the surface. • One may also consider generating curves in other planes, and apply suitable hyperbolic rotations. How many possibilities do we have? Exercise† 3.2.10. Let x : U → R3ν be a non-degenerate regular parametrized surface. Define . xu (u, v) × xv (u, v) N (u, v) = k xu (u, v) × xv (u, v)k and use the indices E and L to distinguish between ambient spaces. v u u det( gij,E (u, v)) (a) Show that N L (u, v) = t Id ( N (u, v)). det( gij,L (u, v)) 2,1 E (b) If θ is the (Euclidean) angle between N E (u, v) and the plane z = 0, show that | det( gij,L )| = | cos 2θ | det( gij,E ). In particular, we obtain the inequality | det( gij,L )| ≤ det( gij,E ). Exercise 3.2.11. Let f : U ⊆ R2 → R3ν be a smooth function, where U is connected, and consider the graph gr( f ) = {(u, v, f (u, v)) ∈ R3ν | (u, v) ∈ U }. (a) Show that π : gr( f ) → U given by π ( x, y, f ( x, y)) = ( x, y) is a diffeomorphism. (b) Show that in R3 , π decreases areas: if R ⊆ gr( f ) is open, A( R) ≥ A(π ( R)). Also verify that the equality holds if and only if R is contained in a horizontal plane. (c) Show that in L3 , if gr( f ) is spacelike, then π increases areas: if R ⊆ gr( f ) is open, A( R) ≤ A(π ( R)). Verify again that equality holds if and only if R is contained in a horizontal plane. (d) Give counter-examples when gr( f ) is timelike in L3 .

Surfaces in Space  167

Exercise 3.2.12. Consider an isosceles triangle T in the plane, with base b > 0 and height h > 0, say, conveniently positioned inside L3 with its vertices at (−b/2, 0, 0), (b/2, 0, 0) and (0, h, 0). We know that the area of such a triangle is A( T ) = bh/2. Let θ ∈ [0, 2π ] be a fixed angle and consider   1 0 0 R = 0 cos θ − sin θ  . 0 sin θ cos θ We have that R is an orthogonal map (and thus preserves Euclidean areas), but it is not . a Lorentz transformation. Show that if Tθ = R( T ) is the slanted triangle, the Lorentzian area of Tθ is p bh | cos 2θ | A( Tθ ) = . 2 When is this area minimal?

R

(−b/2, 0, 0)

(−b/2, 0, 0)

(0, y(θ ), z(θ ))

θ

(0, h, 0) (b/2, 0, 0)

(b/2, 0, 0)

Figure 3.19: Illustrating the triangle Tθ . Exercise 3.2.13 (Girard’s Formula). Recall that a great circle in S2 is the intersection of S2 with a plane passing through the origin of R3 . A spherical triangle is the region in S2 bounded by three arcs of great circles. The goal of this exercise is to show Girard’s Formula, which gives the area of a spherical triangle in terms of its interior angles (more precisely, the angles formed between the tangent vectors to the great circles at the vertices of the triangle).

p1 ϕ1

ϕ3 p3

ϕ2 p2

Figure 3.20: A spherical triangle in S2 .

168  Introduction to Lorentz Geometry: Curves and Surfaces

(a) Show that A(S2 ) = 4π, and that the area of the fuse bounded by v = 0 and v = v0 equals 2v0 , where (u, v) ∈ ]−π/2, π/2[ × ]0, 2π [ are spherical coordinates for S2 . Remark. By the symmetry of S2 , this shows that the area of any fuse of amplitude v0 is 2v0 . This may be formalized using the notion of isometry, to be seen soon. (b) Let T ⊆ S2 be a spherical triangle whose vertices are p1 , p2 , p3 ∈ S2 , with interior angles ϕ1 , ϕ2 and ϕ3 , respectively (as in the previous figure). Show that A( T ) = ϕ1 + ϕ2 + ϕ3 − π. Hint. Use a combinatorics argument: for 1 ≤ i < j ≤ 3, let Cij be the great circle in S2 passing through pi and p j . Given two such great circles, two fuses of amplitudes ϕk are defined, where k is the index of the common point of the two great circles considered. If ∆k is the union of these two fuses, argue that e)), A(∆1 ) + A(∆2 ) + A(∆3 ) = A(S2 ) + 2( A( T ) + A( T e is the spherical triangle with vertices − p , − p and − p . Use item (a) and where T 1 2 3 conclude it. Remark.

• In spherical geometry, the sum of the interior angles of a spherical triangle is always larger than π. Moreover, “similar” spherical triangles are, in fact, “congruent”; • Considering a “hyperbolic triangle” T in H2 defined in the same fashion as above, and keeping the same notation, it holds that A( T ) = π − ( ϕ1 + ϕ2 + ϕ3 ) (Lambert’s Formula). (c) In general, a n-sided spherical polygon P, is the region in S2 bounded by n arcs of great circles. Show that n

A( P) =

∑ ϕi − (n − 2)π,

i =1

where ϕ1 , . . . , ϕn are the interior angles of P. Hint. Divide P into spherical triangles and use (b). Remark. Girard’s Formula is generalized to some spacelike surfaces other than the sphere. Such generalization is known as the Gauss-Bonnet Theorem. For more details see, for example, [17]. Exercise 3.2.14. (a) For each r > 1, show that the area of the “disk” . Dr = {( x, y, z) ∈ H2 | z < r } is A( Dr ) = 2π (r − 1). (b) For each r > 0, show that the area of the “strip” . Sr = {( x, y, z) ∈ S21 | |z| < r } is A(Sr ) = 4πr.

Surfaces in Space  169

Hint. Use parametrizations of revolution. Those will omit a single meridian, which has no area.

r

r 1

0

−r

0 (a) The region Dr .

(b) The region Sr .

Figure 3.21: “Disks” in H2 and S21 . Remark. Extending the definition of area to unbounded regions in a surface, possibly allowing improper integrals, we may conclude that A(H2 ) = A(S21 ) = +∞.

3.2.1

Isometries between surfaces

We will conclude this section by introducing a class of smooth functions between surfaces which is extremely important for everything we will do in the text from here on: Definition 3.2.14 (Isometry). Let M1 , M2 ⊆ R3ν be regular surfaces. (i) A diffeomorphism φ : M1 → M2 is called an isometry if given any p ∈ M1 and vectors v1 , v2 ∈ Tp M1 , we have that
I p (v1 , v2 ) = Iφ( p) dφ p (v1 ), dφ p (v2 ) . (ii) A smooth map φ : M1 → M2 is called a local isometry if for every p ∈ M1 there is an open subset U of M1 containing p such that the restriction φ U : U → φ(U ) is an isometry. (iii) We say that M1 and M2 are isometric (resp. locally isometric) if there is an isometry (resp. local isometry) between M1 and M2 . Remark.

• Above, we’re using I to denote the First
Fundamental Form of M1 on the left side of I p (v1 , v2 ) = Iφ( p) dφ p (v1 ), dφ p (v2 ) , and also to denote the First Fundamental Form of M2 on the right side. • We could also ask ourselves if a surface in R3 is isometric to another surface in L3 . In this setup, we consider the First Fundamental Forms in the above definition induced by the ambient spaces where each surface lies.

170  Introduction to Lorentz Geometry: Curves and Surfaces

Example 3.2.15. Let M1 , M2 ⊆ R3ν be regular surfaces in the same ambient space such that M2 = F ( M1 ), for some F ∈ Eν (3, R). Then M1 and M2 are isometric. Indeed, writing F = Ta ◦ A (with A ∈ Oν (3, R) and a ∈ R3 ) we will have, for each p ∈ M1 , that dFp = DF ( p) T M = A T M and thus p

p

1

1




IF( p) dFp (v), dFp (w) = IF( p) ( Av, Aw) = h Av, Awi = hv, wi = I p (v, w), for all v, w ∈ Tp M1 . Since F is a diffeomorphism, F is an isometry. In this situation, we say that M1 and M2 are congruent. That is, congruent surfaces are isometric. We will soon see that isometric surfaces are not always congruent, that is, there are isometric surfaces which “do not look like each other”. With this concept in hand, we see that the theory of surfaces is, to some extent, parallel to the theory of curves, focusing on the following question: when are two regular surfaces congruent? Seeking the answer, in the next sections we will see how to define certain geometric invariants of a surface, aiming towards a “Fundamental Theorem of Surfaces”, similar to the Fundamental Theorem of Curves. Example 3.2.16. We see from the definition that two isometric surfaces must necessarily have the same causal type. For example, the cylinder S1 (r ) × R is not isometric to itself when considered inside both ambients R3 and L3 . In particular, the identity map itself is not an isometry. Thus, when dealing with isometries, it is essential to know clearly which are the First Fundamental Forms being considered. Before we present a few examples, it will be convenient to register some relations between isometries and parametrizations. The next result follows immediately from the definition of isometry: Proposition 3.2.17. Let M1 , M2 ⊆ R3ν be regular surfaces and φ : M1 → M2 be a local isometry. If (U, x) is a parametrization for M1 and (U, φ ◦ x) is the corresponding parametrization for M2 , we have that gij (u, v) = geij (u, v), for all (u, v) ∈ U and 1 ≤ i, j ≤ 2, where the geij denote the coefficients of the First Fundamental Form of M2 relative to φ ◦ x. Remark. Or, in another suggestive notation: e(u, v), E(u, v) = E

F (u, v) = Fe(u, v) and

e(u, v). G (u, v) = G

As expected, isometries also preserve lengths and areas: Proposition 3.2.18. Let M1 , M2 ⊆ R3ν be regular surfaces and φ : M1 → M2 an isometry. Then: (i) if α : I → M1 is a curve in M1 , then φ ◦ α is a curve in M2 with the same causal type as α, satisfying L[α] = L[φ ◦ α] and E[α] = E[φ ◦ α]; (ii) if R is an open subset of M1 with R compact, then A( R) = A(φ( R)).

Surfaces in Space  171

Proof: (i) We have that L[φ ◦ α] =

= =

Z q ZI q ZI q I


|Iφ◦α(t) (φ ◦ α)0 (t) | dt
|Iφ◦α(t) dφα(t) (α0 (t)) | dt
|Iα(t) α0 (t) | dt = L[α],

and similarly for the energy. (ii) Suppose without loss of generality that R ⊆ x(U ) for some parametrization (U, x) of M. We have that φ( R) ⊆ φ( x(U )). Since φ ◦ x is a parametrization for M2 , φ( R) is open in M, and φ( R) is compact (since φ is, in particular, a diffeomorphism), we have: Z q | det gfij (u, v)| du dv A(φ( R)) = (φ◦ x)−1 (φ( R)) Z q = | det gij (u, v)| du dv = A( R). x −1 ( R )

It is also very convenient to relate in a more general way the coordinate expression of an isometry with the First Fundamental Forms of the given surfaces: Proposition 3.2.19. Let M1 , M2 ⊆ R3ν be regular surfaces and φ : M1 → M2 an isometry. Suppose that: e e e ); • (U, x) and (U, x) are parametrizations for M1 and M2 such that φ( x(U )) ⊆ e x (U . • for any (u, v) ∈ U and (ue, ve) = ex−1 ◦ φ ◦ x(u, v), the matrix representing the differential of φ, relative to the bases associated to the parametrizations (in the  .  correct points) is denoted by A = dφx(u,v) B ,B = ( ai j (u, v))1≤i,j≤2 ; x

e x

. . e = • G = ( gij (u, v))1≤i,j≤2 and G ( geij (ue, ve))1≤i,j≤2 are the Gram matrices of the First Fundamental Forms of x and e x. e Then G = A> GA. Proof: It suffices to compute each gij (u, v) by using all the given assumptions. Identify u ↔ u1 and v ↔ u2 , and similarly for the coordinates in e x. We have:   ∂x 1 2 ∂x 1 2 gij (u1 , u2 ) = Ix(u1 ,u2 ) ( u , u ), j ( u , u ) ∂ui ∂u      ∂x 1 2 ∂x 1 2 (u , u ) = Iφ◦ x(u1 ,u2 ) dφx(u1 ,u2 ) (u , u ) , dφx(u1 ,u2 ) ∂ui ∂u j ! 2 x 1 2 2 ` 1 2 ∂e x 1 2 k 1 2 ∂e = Iφ◦ x(u1 ,u2 ) ∑ a i (u , u ) k (ue , ue ), ∑ a j (u , u ) ` (ue , ue ) ∂ue ∂ue k =1 `=1   2 x 1 2 ∂e x 1 2 ∂e k 1 2 ` 1 2 = ∑ a i (u , u ) a j (u , u )Iφ◦ x(u1 ,u2 ) (ue , ue ), ` (ue , ue ) ∂uek ∂ue k,`=1 2

=

∑

k,`=1

aki (u1 , u2 ) a`j (u1 , u2 ) gek` (ue1 , ue2 ).

172  Introduction to Lorentz Geometry: Curves and Surfaces

e This last expression is precisely the entry in position (i, j) of the matrix A> GA. Remark. Compare the above result with Lemma 1.2.6 (p. 6). For non-degenerate surfaces, we have the following auxiliary result: Proposition 3.2.20. Let M1 , M2 ⊆ R3ν be two non-degenerate regular surfaces, and φ : M1 → M2 be a smooth map. If dφ p preserves the First Fundamental Forms for each p ∈ M1 , then φ is automatically a local isometry. Proof: Take p ∈ M1 and an orthonormal basis (v1 , v2 ) for Tp M1 . By assumption, (dφ p (v1 ), dφ p (v2 )) is an orthonormal subset of Tφ( p) M2 and, thus, is linearly independent. Hence dφ p is surjective, and since the dimension of the tangent planes is the same, it follows that dφ p is non-singular. As p was arbitrary, the Inverse Function Theorem says that φ is a local diffeomorphism, as wanted. The same calculations done in the proof of Proposition 3.2.19 and the above result also give us that: e → R3ν be injective and regular Proposition 3.2.21. Let x : U → R3ν and e x: U e be a diffeomorphism. Suppose that for all parametrized surfaces, and ϕ : U → U e = ( geij ( ϕ(u, v)))1≤i,j≤2 and A = Dϕ(u, v), (u, v) ∈ U, if G = ( gij (u, v))1≤i,j≤2 , G we have that e G = A> GA. . e ) is an isometry between x(U ) and xe(U e ). Then φ = e x ◦ ϕ ◦ x −1 : x (U ) → e x (U Let’s see how to use those results in practice: Example 3.2.22. (1) Lambert’s cylindrical projection, f : S2 \ {(0, 0, ±1)} → S1 × R, seen in Example 3.1.21 (p. 143) is not an isometry. Indeed, note that the First Fundamental Form of x is du2 + cos2 u dv2 and the one for e x is de u2 + de v2 . Since f ( x(u, v)) = e x(sin u, v), we let ue = sin u and ve = v, whence de u = cos u du and de v = dv. But de u2 + de v2 = cos2 u du2 + dv2 6= du2 + cos2 u dv2 . Despite this, note that f locally preserves areas, since  2    1 0 cos u 0 det = det . 0 1 0 cos2 u (2) The plane R2 is locally isometric to the cylinder S1 × R seen in R3 . The local isometry is F : R2 → S1 × R given by F (u, v) = (cos v, sin v, u). Considering the identity map as a parametrization for R2 along with the usual parametrization of revolution for the cylinder, e x(ue, ve) = (cos ve, sin ve, ue), we see that the local expression x(u, v)), whose derivative is for F is precisely the identity (to wit, F (idR2 (u, v)) = e non-singular, giving that F is a local diffeomorphism (but not global, as it is not injective). That F is a local isometry then follows from the First Fundamental Form for the given cylinder parametrization being de u2 + de v2 = du2 + dv2 .

Surfaces in Space  173

Figure 3.22: Local isometry between the plane and the cylinder. There is no global isometry between these surfaces because there is not even a diffeomorphism between them. The proof of this fact is beyond the scope of this text, but may be found in [42]. (3) The surface of revolution generated by the catenary α : R → R3 (given by α(u) = (cosh u, 0, u)) is called the catenoid, and it may be parametrized (excluding one meridian, as usual) by the map x : R × ]0, 2π [ → x(R × ]0, 2π [) ⊆ R3 given by x(u, v) = (cosh u cos v, cosh u sin v, u). The First Fundamental Form, in differential notation, is cosh2 u(du2 + dv2 ). Also consider again the helicoid, parametrized by e x : ]0, 2π [ × R → e x(]0, 2π [ × R), given by e x(ue, ve) = (ve cos ue, ve sin ue, ue), with First Fundamental Form in differential notation is (1 + ve2 )de u2 + de v2 . Define F : x(R × ]0, 2π [) → e x(]0, 2π [ × R) by F ( x(u, v)) = e x(u, sinh v). Let’s see that F is an isometry. To wit, the map (u, v) 7→ (v, sinh u) is a diffeomorphism, and so F is as well. To verify that F preserves First Fundamental Forms, let ue = v and ve = sinh u, so that de u = dv and de v = cosh u du. Hence,

(1 + ve2 )de u2 + de v2 = (1 + sinh2 u)dv2 + cosh2 u du2 = cosh2 u(du2 + dv2 ). Observe that F maps, respectively, meridians and parallels of the catenoid into lines and helices in the helicoid, as Figure 3.23 shows:

Figure 3.23: Isometry between parts of the catenoid and helicoid. We have seen that even though isometries preserve areas, there are functions which preserve areas but are not isometries (e.g., Lambert’s projection). We will see next that, while preserving areas is not enough to characterize isometries, preserving lengths will do the trick:

174  Introduction to Lorentz Geometry: Curves and Surfaces

Proposition 3.2.23. Let M1 , M2 ⊆ R3ν be two regular surfaces and φ : M1 → M2 be a (local) diffeomorphism whose differential preserves the causal type of tangent vectors. If, for every curve α : I → M1 in M1 we have that L[φ ◦ α] = L[α], then φ is a (local) isometry. Proof: Fix p ∈ M1 and v ∈ Tp M1 . Let α : ]−e, e[ → M1 be a curve in M1 realizing v. The assumption says that Z t 0

kα0 (u)k du =

Z t 0

k(φ ◦ α)0 (u)k du =

Z t 0

kdφα(u) (α0 (u))k du,

for all t ∈ ]−e, e[. Differentiating both sides with respect to t, and evaluating at t = 0, we obtain kvk = kdφ p (v)k. Since dφ p preserves causal types, it follows from this that I p (v) = Iφ( p) (dφ p (v)). Polarizing, we conclude that I p (v1 , v2 ) = Iφ( p) dφ p (v1 ), dφ p (v2 )




for all v1 , v2 ∈ Tp M1 . But since p was arbitrary and φ is a (local) diffeomorphism, we conclude that φ is a (local) isometry. Remark.

• The hypothesis of preservation of causal type is automatically satisfied when the ambient space considered is just R3 , but it is crucial in the general case. The “flip” operator f : L2 → L2 given by f( x, y) = (y, x ), which we have seen in a few exercises so far, is a witness for that. • A similar result holds replacing arclength by energy. See Exercise 3.2.20.

Exercises Exercise† 3.2.15. Show Proposition 3.2.17 (p. 170). Exercise 3.2.16. (a) Let M ⊆ R3ν be a regular surface. Show that . Iso( M) = {φ : M → M | φ is an isometry} is a subgroup of Diff( M) (see Exercise 3.1.6, p. 152). Is it a normal subgroup? (b) Show that if M1 , M2 ⊆ R3ν are isometric regular surfaces, then Iso( M1 ) ∼ = Iso( M2 ). Remark. . • We have then defined an action of Iso( M) on M by φ · p = φ( p).

• Intuitively, the “larger” is Iso( M), the simpler is the geometry of M, since we have many more “symmetries”. For example, we have already determined in Chapter 1 the groups Iso(R2 ) and Iso(L2 ).

Surfaces in Space  175

Exercise 3.2.17 (Isometries of S2 ). o n (a) Show that Iso(S2 ) = C S2 | C ∈ O(3, R) . Hint. If φ ∈ Iso(S2 ), define C : R3 → R3 by    k pk φ p , if p 6= 0 k pk C ( p) =  0, if p = 0. (b) Compute the stabilizers3 of (1, 0, 0) and (0, 0, 1) in Iso(S2 ). Exercise 3.2.18 (Isometries of H2 ). n o (a) Show that Iso(H2 ) = Λ H2 | Λ ∈ O1↑ (3, R) . Hint. To avoid issues with lightlike vectors in the analogue of the construction suggested in the hint of Exercise 3.2.17 above, we may directly explore the vector space structure of L3 , as follows: take p1 , p2 ∈ H2 such that { p1 , p2 , e3 } is linearly independent and linearly extend φ to Λ : L3 → L3 from those points. Note that φ(H2 ) = H2 ,

Λ T

2 e3 H

= dφe3

and dφe3 (ei ) ∈ Tφ(e3 ) H2 = φ(e3 )⊥ ,

for 1 ≤ i ≤ 2. Use these facts to show that hΛei , Λe j i L = ηij , for 1 ≤ i, j ≤ 3. Observe that by construction, Λ is orthochronous. Moreover Λ does not depend on the choice of p1 and p2 on the given conditions. (b) Compute the stabilizer of (0, 0, 1) in Iso(H2 ). Exercise 3.2.19 (Isometries of S21 ). n o (a) Show that Iso(S21 ) = Λ S2 | Λ ∈ O1 (3, R) . 1

Hint. Adapt what was done in item (a) of Exercise 3.2.18 above, using e1 instead of e3 . (b) Compute the stabilizer of (1, 0, 0) in Iso(S21 ). Exercise 3.2.20. Let M1 , M2 ⊆ R3ν be non-degenerate regular surfaces and consider a diffeomorphism φ : M1 → M2 . If φ is energy-preserving, i.e., for every curve α : I → M1 we have that E[φ ◦ α] = E[α], then φ is an isometry. Exercise† 3.2.21. Show Proposition 3.2.21 (p. 172). Exercise 3.2.22. Consider usual parametrizations for the cone and the plane, x : R>0 × ]0, 2π [ → R3 and e x : R>0 × ]0, 2π [ → R2 , given by x(u, v) = (u cos v, u sin v, u) and e x(ue, ve) = (ue cos ve, ue sin ve), respectively. Show √ that √the map F : x(R>0 × ]0, 2π [) → ex(R>0 × ]0, 2π [) given by F ( x(u, v)) = e x(u 2, v/ 2) is an isometry onto its image, and determine it. 3 Recall

that if G is a group acting (on the left) on a set X, the stabilizer of an element x ∈ X is . Gx = { g ∈ G | g · x = x }.

176  Introduction to Lorentz Geometry: Curves and Surfaces

Exercise 3.2.23. Consider the cylinder S1 × R = {( x, y, z) ∈ R3ν | x2 + y2 = 1}. Exhibit an isometry f : S1 × R → S1 × R with exactly two fixed points. Hint. There is an isometry that works no matter in which ambient space we consider the cylinder. Exercise 3.2.24. (a) Let α : R → R3 be a unit speed injective parametrized curve, of the particular form α(u) = ( x (u), y(u), 0). Define x : R2 → R3 by x(u, v) = ( x (u), y(u), v). Show that x is an injective and regular parametrized surface, and that x(R2 ) and R2 are isometric. (b) Suppose now that α is seen in L3 , has unit speed, and has the particular form α(u) = (0, y(u), z(u)). This time, define x : R2 → L3 by x(u, v) = (v, y(u), z(u)) instead. Show that x(R2 ) is isometric to R2 if α is spacelike, and isometric to L2 if α is timelike. Exercise 3.2.25. Let U ⊆ R2 be open and connected, f , g : U → R be two smooth functions, and consider φ : gr( f ) → gr( g) given by φ(u, v, f (u, v)) = (u, v, g(u, v)). (a) Show that if both graphs are seen inside the same ambient space and φ is an isometry, then f (u, v) = ± g(u, v) + c, for some constant c ∈ R, for all (u, v) ∈ U. (b) Show that if the graphs are seen in different ambient spaces, i.e., if gr( f ) ⊆ R3 and gr( g) ⊆ L3 , and φ is an isometry, then both f and g are constant. Remark. That is, item (b) says that the “direct projection” is an isometry between graphs in different ambient spaces if and only if both graphs are actually horizontal spacelike planes (and hence isometric to R2 ). A priori, we could have some other isometry between such graphs. Can you think of a concrete example? We already know that this is impossible if k∇ gk E ≥ 1. Exercise 3.2.26 (Tangent Surfaces – II). (a) Let α1 , α2 : I → R3ν be two unit speed admissible curves, and consider their tangent surfaces (as done in Exercise 3.2.3, p. 164), x1 , x2 : I × R → R3ν given by xi (s, v) = αi (s) + vαi0 (s),

i = 1, 2.

Fix (s0 , v0 ) ∈ I × R with v0 6= 0 and take a neighborhood V of (s0 , v0 ) for which x1 (V ) and x2 (V ) are both regular surfaces. Suppose that κα1 (s) = κα2 (s) 6= 0 for all s and that the causal type of the Frenet-Serret trihedrons for both curves is always the same. Show that the composition x1 ◦ x2−1 : x2 (V ) → x1 (V ) is an isometry. (b) Show that if α : I → R3ν is an admissible curve with non-zero curvature, and x : I × R → R3ν is its tangent surface, then for every (t0 , v0 ) ∈ I × R with v0 6= 0, there is a neighborhood V of (t0 , v0 ) such that x(V ) is a regular surface isometric to an open subset of R2 or L2 , depending on the causal type of the Frenet-Serret trihedron for α. Hint. Suppose without loss of generality that α has unit speed, and combine the Fundamental Theorem of Curves with item (a) above.

Surfaces in Space  177

Exercise 3.2.27. In Chapter 1, we mentioned that it was usual in the literature to consider the Lorentzian product defined with a negative sign in the first term instead . of the last one, as we have done. Let L2(+,−) = L2 , and L2(−,+) stand for the plane R2 equipped with the scalar product . h(u1 , u2 ), (v1 , v2 )i(−,+) = −u1 v1 + u2 v2 or, in other words, equipped with the First Fundamental Form −du2 + dv2 . Justify the force and ubiquity of the “flip” operator f : L2(+,−) → L2(+,−) given by f( x, y) = (y, x ), by showing that it is, in this setting, an isometry. Exercise 3.2.28 (Conformal mappings). Let M1 , M2 ⊆ R3ν be regular surfaces. A (local) diffeomorphism ψ : M1 → M2 is called (locally) conformal if there is a smooth function λ : M1 → R>0 such that, given p ∈ M1 and v, w ∈ Tp M1 , the relation
Iψ( p) dψ p (v), dψ p (w) = λ( p)I p (v, w) holds. The function λ is called the conformality coefficient of ψ. (a) Show that a conformal mapping must preserve the causal types of surfaces, angles between spacelike tangent vectors, and hyperbolic angles between tangent timelike vectors with the same time direction (in timelike surfaces). (b) Show that if a diffeomorphism between spacelike surfaces preserves angles between tangent vectors, then it is actually a conformal mapping. (c) Show that the stereographic projection seen in Exercise 3.1.2 (p. 152) is a conformal mapping. Hint. You may regard R2 inside R3 as the coordinate plane z = 0, as usual. Exercise 3.2.29. Let f : R2 → R2 be smooth and given by f ( x, y) = (u( x, y), v( x, y)). (a) Suppose that the functions u and v satisfy the Cauchy-Riemann equations u x = vy and uy = −v x . If . Q = {( x, y) ∈ R2 | u x ( x, y)2 + uy ( x, y)2 6= 0}, show that f is a locally conformal mapping from Q into R2 . (b) To obtain a result similar to item (a) in L2 , we will again resort to the analogies of C with the set of split-complex numbers C0 = { x + hy | x, y ∈ R and h2 = 1}, informally presented in Exercise 2.2.16 (p. 95). Suppose that the functions u and v satisfy the revised Cauchy-Riemann equations u x = vy and uy = v x (which motivate a notion of “split-holomorphicity”, as we will see in Chapter 4). Adopting the notation given in Exercise 3.2.27 above, and setting . Q+ = {( x, y) ∈ L2(+,−) | u x ( x, y)2 − uy ( x, y)2 > 0} and . Q− = {( x, y) ∈ L2(−,+) | u x ( x, y)2 − uy ( x, y)2 < 0}, show that f is a locally conformal map from Q+ into L2(+,−) , and from Q− into L2(+,−) . Hint. Do like we did for isometries: write du = u x dx + uy dy, similarly for dv, and compute du2 ± dv2 according to each ambient space.

178  Introduction to Lorentz Geometry: Curves and Surfaces

3.3

SECOND FUNDAMENTAL FORM AND CURVATURES

We have previously seen that the orientability of a regular surface M ⊆ R3ν is equivalent to the existence of a unit normal vector field, N : M → R3ν , smooth and defined on all of M. At this stage, we are ready to present a few avatars of the notion of “curvature” for surfaces. The motivation is simple: we wish to know how the surface M bends near a given point p ∈ M. Near enough to said point, it would be reasonable to turn our attention to the “linearization” of M at p, namely, the tangent plane Tp M. But knowing how Tp M changes with p is clearly equivalent to knowing how N ( p) changes with p. This indicates the crucial role that the map dN p will play in the definition of curvature. We start to formalize such ideas now: Definition 3.3.1 (Gauss map). Let M ⊆ R3ν be a non-degenerate regular surface. A Gauss (normal) map for M is a smooth field N : M → R3ν of unit vectors normal to M, that is, N ( p) ∈ ( Tp M )⊥ , for all p ∈ M. We denote by e M the indicator of the normal direction to M, namely, e M = 1 if M is timelike, and e M = −1 if M is spacelike. Remark. In general, we can precisely say what the codomain of the Gauss map is:

• if M ⊆ R3 , we have N : M → S2 ; • if M ⊆ L3 is spacelike, we have N : M → H2± ; • if M ⊆ L3 is timelike, we have N : M → S21 .

p

p

p

N

N N

N ( p)

N ( p)

N ( p)

(a) In R3 .

(b) In L3 , e M = −1.

(c) In L3 , e M = 1.

Figure 3.24: The Gauss map of a surface.

Surfaces in Space  179

We note that in R3 , Tp M and TN ( p) S2 are the same vector space, namely, the orthogonal complement of the line which has N ( p) as direction. The same remark holds in L3 , with S21 or H2± instead of S2 . With this, we may regard the differential of the Gauss map as a linear operator in Tp M. Definition 3.3.2 (Weingarten Map). Let M ⊆ R3ν be non-degenerate regular surface, and N be a Gauss map for M. The Weingarten map of M at p is the differential −dN p : Tp M → Tp M. Remark. The Weingarten map is also known as the shape operator of M. A naive justification for the negative sign in the above definition, seemingly artificial, is that it is meant to reduce the quantity of negative signs in future computations. Proposition 3.3.3. Let M ⊆ R3ν be a non-degenerate regular surface and N be a Gauss map for M. Then, for each p ∈ M, the Weingarten map −dN p is self-adjoint relative to the First Fundamental Form I p . Proof: Let (U, x) be a parametrization for M around p = x(u0 , v0 ). By definition of differential, we have:     ∂( N ◦ x) ∂x ∂( N ◦ x) ∂x and (u, v) = dN x(u,v) (u, v) (u, v) = dN x(u,v) (u, v) , ∂u ∂u ∂v ∂v for all (u, v) ∈ U. We also know that:     ∂x ∂x (u, v), N ( x(u, v)) = 0 = (u, v), N ( x(u, v)) ∂u ∂v for all (u, v) ∈ U. Then, differentiating the first relation with respect to v and the second one with respect to u we obtain:  2     ∂ x ∂x ∂x (u, v), N ( x(u, v)) + (u, v), dN x(u,v) (u, v) =0 ∂v∂u ∂u ∂v  2     ∂ x ∂x ∂x (u, v), N ( x(u, v)) + (u, v), dN x(u,v) (u, v) = 0, ∂u∂v ∂v ∂u whence it follows that:       ∂x ∂x ∂x ∂x (u, v), dN x(u,v) (u, v) = (u, v), dN x(u,v) (u, v) . ∂u ∂v ∂v ∂u Since xu (u0 , v0 ) and xv (u0 , v0 ) span Tp M, it follows from linearity, evaluating everything at (u0 , v0 ), that h−dN p (v), wi = hv, −dN p (w)i, for all v, w ∈ Tp M, as wanted. If M is a spacelike surface, then the First Fundamental Form is positive-definite, and so −dN p is diagonalizable, by the Real Spectral Theorem. If M is timelike, we cannot guarantee that this will happen. Definition 3.3.4. Let M ⊆ R3ν be a non-degenerate regular surface and N be a Gauss map for M. Given p ∈ M, if the Weingarten map −dN p is diagonalizable, its two eigenvalues κ1 ( p) and κ2 ( p) are called the principal curvatures of M at p. The associated (orthogonal) eigenvectors are called the principal directions of M at p.

180  Introduction to Lorentz Geometry: Curves and Surfaces

Remark. In the usual development of the theory done when considering only surfaces in R3 , it is usual to define the curvatures of M from the principal curvatures defined above, which will always exist. In our case, we need an alternative approach that includes timelike surfaces in L3 as well. Definition 3.3.5 (Second Fundamental Form). Let M ⊆ R3ν be a non-degenerate regular surface and N be a Gauss map for M. The Second Fundamental Form of M at p (associated to N) is the symmetric bilinear map II p : Tp M × Tp M → ( Tp M )⊥ defined by the relation hII p (v, w), N ( p)i = h−dN p (v), wi, for all v, w ∈ Tp M. Remark.

• As done for I p , we will abbreviate II p (v, v) to II p (v) only. Occasionally it will be convenient to omit the p as well. • In L3 , II p is also called the Minkowski Second Fundamental Form of M at p. Definition 3.3.6 (Components of the Second Form). Let M ⊆ R3ν be a regular surface, N be a Gauss map for M, and (U, x) be a parametrization for M. The components of the Second Fundamental Form relative to x are defined by . e(u, v) = h xuu (u, v), N ( x(u, v))i, . f (u, v) = h xuv (u, v), N ( x(u, v))i and . g(u, v) = h xvv (u, v), N ( x(u, v))i. Remark.

• The above definition is justified by noting that, identifying indices u ↔ 1 and v ↔ 2, we have
hIIx(u,v) xi (u, v), x j (u, v) , N ( x(u, v))i = h−dN x(u,v) ( xi (u, v)), x j (u, v)i = h xij (u, v), N ( x(u, v))i. • It is also usual to write ` ≡ h11 = e, m ≡ h12 = h21 = f and n = h22 = g, which allows us to gather all the necessary information about IIx(u,v) in the matrix (hij (u, v))1≤i,j≤2 . • It follows from the above considerations that IIx(u,v) ( xu (u, v)) = e M e(u, v) N ( x(u, v)), IIx(u,v) ( xu (u, v), xv (u, v)) = e M f (u, v) N ( x(u, v)) and IIx(u,v) ( xv (u, v)) = e M g(u, v) N ( x(u, v)). Lemma 3.3.7. Let M ⊆ R3ν be a non-degenerate regular surface, N be a Gauss map for M and (U, x) be a parametrization for M. Identifying the indices u ↔ 1 and v ↔ 2 and . omitting all points of evaluation, we have that if (hi j )1≤i,j≤2 = [−dN ] Bx , then we have that hi j = ∑2k=1 gik hkj .

Surfaces in Space  181

Proof: By definition of the matrix of a linear transformation, for each j = 1, 2 we have that −dN ( x j ) = ∑2i=1 hi j xi . Taking products with xk on both sides, we have that

h−dN ( x j ), xk i = ∑2i=1 hi j gik . It follows that h jk = h N ◦ x, x jk i = ∑2i=1 hi j gik . Since M is non-degenerate, we have the inverse matrix ( gij )1≤i,j≤2 . Hence, multiplying both sides by gk` , summing over k and renaming ` → i, we obtain precisely hi j = ∑2k=1 gik hkj , as wanted. With this basic language, we may return to our initial idea: looking at the Weingarten map of M at p. We know, from Linear Algebra, that the trace and the determinant of a linear operator are invariant under change of basis. The Second Fundamental Form is a vector-valued bilinear form so that, a priori, we wouldn’t have its trace and determinant available. Precisely to avoid this hindrance, we have seen in Lemmas 1.6.7 and 1.6.8 (p. 57) in Chapter 1 how to define the trace and determinant of a bilinear form relative to the ambient scalar product. But since this “metric determinant” was defined only for . scalar-valued bilinear forms, we consider instead e II p (v, w) = hII p (v, w), N ( p)i. Thus we may write the: Definition 3.3.8 (Mean and Gaussian curvatures). Let M ⊆ R3ν be a non-degenerate regular surface, and N be a Gauss map for M. The mean curvature vector and the Gaussian curvature of M at p are defined by: 1 . 1 H ( p) = trI p (II p ) = (ev1 II p (v1 ) + ev2 II p (v2 )) and 2 2
. K ( p) = (−1)ν detI p e II p = (−1)ν det (e II p (vi , v j ))1≤i,j≤2 , where {v1 , v2 } is any orthonormal basis for Tp M. Moreover, the mean curvature of M at p is the number H ( p) determined by the relation H ( p) = H ( p) N ( p). Remark.

• Note that replacing N by − N, the sign of the mean curvature is reversed, but not the sign of the Gaussian curvature, since the matrix whose determinant is computed has even order. • We recall that the presence of indicators in the definition of the mean curvature is indeed natural: without them, the quantity to be defined is not invariant under a change of orthonormal basis. • The coefficient (−1)ν in the definition of K, in turn, has the purpose of recovering the information about the causal type of the surface which is lost in L3 (but not in R3 ), when considering the scalar e II instead of II. Naturally, we need to know how to express those new objects in terms of coordinates. For that end, we need the following technical lemma: Lemma 3.3.9. Let M ⊆ R3ν be a non-degenerate regular surface, N be a Gauss map for M, and (U, x) a parametrization for M. Then, omitting points of evaluation, we define . xu E1 = k xu k

and

. xv − ( F/E) xu E2 = . k xv − ( F/E) xu k

182  Introduction to Lorentz Geometry: Curves and Surfaces

Then, we have that { E1 (u, v), E2 (u, v)} is an orthonormal basis for Tx(u,v) M satisfying:
e e e II E1 = 1 M N ◦ x, E
e M ( E f − Fe) II E1 , E2 = p N◦x E | EG − F2 |

e2 e M Eg − 2F f + F2 e/E II E2 = N ◦ x, EG − F2 where e1 and e2 stand for the indicators of E1 and E2 .  Proof: That E1 (u, v), E2 (u, v) is an orthonormal basis for Tx(u,v) M is nothing more than a direct consequence of the Gram-Schmidt process. Initially, we have:
II E1 =

1 e e e 1 e M eN ◦ x = 1 M N ◦ x. II( xu ) = 2 e1 E E k xu k

Next, noting that:   F F 2F2 F2 EG − F2 xv − xu , xv − xu = G − + 2E = , E E E E E we have that:   II xu , xv − EF xu e M ( f − EF e) e M ( E f − Fe) =p N◦x= p N ◦ x, II ( E1 , E2 ) = p 2 2 | EG − F | | EG − F | E | EG − F2 | and lastly, noting that e1 e2 e M = (−1)ν , we have:     II xv − EF xu
| E| F II E2 = II xv − xu

2 = E | EG − F2 |

xv − EF xu   e1 E 2F F2 e g− f + 2e N ◦ x = E (−1)ν e M ( EG − F2 ) M E
2 e2 e M Eg − 2F f + F e/E = N ◦ x. EG − F2

Proposition 3.3.10 (Local curvature expressions). Let M ⊆ R3ν be a non-degenerate regular surface, N be a Gauss map for M, and (U, x) a parametrization for M compatible with N. Then we have: eM e Eg − 2F f + Ge tr(−dN ) = M and 2 2 EG − F2 eg − f 2 K ◦ x = e M det(−dN ) = e M . EG − F2

H◦x=

Remark. N and x are compatible if N ◦ x =

xu × xv . k xu × xv k

Surfaces in Space  183

Proof: Let E1 and E2 be as in Lemma 3.3.9 above, and omit points of evaluation. Let’s deal with the mean curvature first. We have that: 1 H ◦ x = (e1 II ( E1 ) + e2 II ( E2 )) 2   1 e M e e M ( Eg − 2F f + F2 e/E) = + N◦x 2 E EG − F2 e Eg − 2F f + Ge = M N ◦ x. 2 EG − F2 It follows from the definition that: H◦x=

e M eG − 2 f F + Eg . 2 EG − F2

To check the relation between H ◦ x and tr(−dN ), we now use Lemma 3.3.7: tr(−dN ) = h11 + h22

= g11 h11 + g12 h21 + g21 h12 + g22 h22 F F E G e− f− f+ g = EG − F2 EG − F2 EG − F2 EG − F2 eG − 2F f + Eg = . EG − F2 For the Gaussian curvature, in turn, we have:   K ◦ x = (−1)ν e II( E1 )e II( E2 ) − e II( E1 , E2 )2  !2   e   e ( Eg − 2F f + F2 e/E)  ( E f − Fe) 2  p = (−1)ν  1 e − 2 E ( EG − F ) E | EG − F2 |   (−1)ν e1 e2 Eeg − 2Fe f + F2 e2 /E E2 f 2 − 2EFe f + F2 e2 = − E EG − F2 E2  2  eM E eg − E2 f 2 = EG − F2 E2 eg − f 2 = eM . EG − F2 The relation between K ◦ x and det(−dN ) also follows from Lemma 3.3.7, which essentially says that the matrix of −dN p is the product of the inverse matrix of I p with the matrix of e II p , from where it follows that
det (hij )1≤i,j≤2 eg − f 2
= det(−dN ) = , EG − F2 det ( gij )1≤i,j≤2 as wanted. Example 3.3.11. (1) Consider a plane Π ⊆ R3ν , non-degenerate, passing through a certain point p0 ∈ R3ν with a unit vector n ∈ R3ν giving the normal direction. A Gauss map in this case is simply N : Π → R3ν given by N ( p) = n. This way dN p is the zero operator for all p ∈ Π, whence we conclude that K = H ≡ 0. Note that in this case the Weingarten map is trivially diagonalizable, both principal curvatures vanish, and every direction is principal.

184  Introduction to Lorentz Geometry: Curves and Surfaces

(2) For the “spheres” S2 (r ), S21 (r ) and H2 (r ) with radius r > 0, a Gauss map is simply N given by N ( p) = p/r. Directly, we have that dN p (v) = v/r, for every vector v tangent at p, and thus 1 1 −dN p = − idTp S2 (r) =⇒ det(−dN p ) = 2 r r

and

1 1 tr(−dN p ) = − . 2 r

Hence, both S2 (r ) and S21 (r ) have constant and positive Gaussian curvature 1/r2 , and also constant mean curvature, equal to −1/r. The hyperbolic plane H2 (r ), in turn, has constant and negative Gaussian curvature −1/r2 , and mean curvature equal to 1/r. In these cases, the Weingarten map is again diagonalizable, with all the directions being principal. (3) Considering now a straight cylinder S1 (r ) × R with radius r > 0, and the projection π : R3ν → R3ν onto the first two components, we have that a Gauss map is N : S1 (r ) × R → S2ν , given by N ( p) = π ( p)/r. It follows from this that the derivative is given by dN p (v) = π (v)/r for every v ∈ Tp (S1 (r ) × R), since N is the composition of the restrictions of linear maps. Fixed p ∈ S1 (r ) × R, we may consider the orthonormal basis of Tp (S1 (r ) × R) formed by the vector u1 tangent to the cylinder and horizontal (take any of the two possible vectors here), and the vector u2 = (0, 0, 1). Relative to the basis B = (u1 , u2 ), we have that   −1/r 0 [−dN p ] B = =⇒ K ( p) = 0 and 0 0

H ( p) = −

1 , 2r

independently of the ambient space considered. This in particular illustrates that it is possible, when considering different ambient spaces, that surfaces which are not congruent might have the same curvatures. Another way to obtain the same conclusions is by doing coordinate computations, considering the parametrization x : ]0, 2π [ × R → R3ν given by x(u, v) = (r cos u, r sin u, v), and computing all the gij and hij . Note that even in L3 , with S1 (r ) × R being timelike, the Weingarten map is diagonalizable. The principal vectors are precisely u1 and u2 chosen above. (4) Let f : U ⊆ R2 → R be a smooth function, and consider the usual Monge parametrization for its graph: x : U → gr( f ) given by x(u, v) = (u, v, f (u, v)). Suppose that the graph of f is non-degenerate. Denoting the curvature with indices according to the ambient space, abbreviating the partial derivatives of f and omitting points of evaluation, we have that KE ◦ x =

2 f uu f vv − f uv (1 + f u2 + f v2 )2

and

HE ◦ x =

f uu (1 + f v2 ) − 2 f u f v f uv + f vv (1 + f u2 ) 2 (1 + f u2 + f v2 )3/2

in R3 , and KL ◦ x =

2 − f f f uv uu vv , (−1 + f u2 + f v2 )2

HL ◦ x =

f uu (−1 + f v2 ) − 2 f u f v f uv + f vv (−1 + f u2 ) 2 | − 1 + f u2 + f v2 |3/2

in L3 . We ask you to verify this in Exercise 3.3.4.

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(5) Suppose that α : I → R3ν is a smooth, regular, injective and non-degenerate curve of the form α(u) = ( f (u), 0, g(u)), for certain smooth functions f and g such that f (u) > 0 for all u ∈ I. This way, we may consider the revolution surface around the z-axis generated by α, which will also be non-degenerate. Considering the usual parametrization x : I × ]0, 2π [ → x( I × ]0, 2π [) ⊆ R3ν given by x(u, v) = ( f (u) cos v, f (u) sin v, g(u)), as in the previous example, we have that KE ◦ x =

g0 (− f 00 g0 + f 0 g00 ) f hα0 , α0 i2E

and

HE ◦ x =

g0 (hα0 , α0 i E − f f 00 ) + f f 0 g00 2 f hα0 , α0 i3/2 E

for R3 and KL ◦ x =

g0 ( f 00 g0 − f 0 g00 ) f hα0 , α0 i2L

and

HL ◦ x =

g0 (−hα0 , α0 i L + f f 00 ) − f f 0 g00 2 f |hα0 , α0 i L |3/2

for L3 . The verification of those formulas is left to Exercise 3.3.5. Note that these expressions undergo a great simplification when α has unit speed. (6) Consider again the helicoid, image of the regular and injective parametrized surface x : R × R → R3ν given by x(u, v) = (u cos v, u sin v, v). For u 6= ±1, x is nondegenerate and:     √ −1ν 2 0 1 0 |(−1) +u |  ( gij )1≤i,j≤2 = and (hij )1≤i,j≤2 =  √ −1ν 2 0 (−1)ν + u2 0 |(−1) +u |

and, thus, K ( x(u, v)) =

(−1)ν+1 |(−1)ν + u2 |2

and

H ( x(u, v)) = 0.

In possession of the relations of H and K with the trace and determinant of the Weingarten map, previously seen, Lemma 1.6.9 (p. 58) from Chapter 1 gives us the: Proposition 3.3.12. Let M ⊆ R3ν be a non-degenerate regular surface, with orientation given by the unit normal field N. For each p ∈ M, we have: dN p (v) × dN p (w) = e M K ( p) v × w dN p (v) × w + v × dN p (w) = −2e M H ( p) v × w, for all linearly independent v, w ∈ Tp M. Remark. For an interesting application of this last proposition, see Exercise 3.3.7. At this point, we have enough tools to raise the following natural question: are the mean and Gaussian curvatures invariant under congruences, really deserving the name of “curvatures”? The affirmative answer to this first question is now easy to obtain: Proposition 3.3.13. Let M1 , M2 ⊆ R3ν be non-degenerate regular surfaces such that there is F ∈ Eν (3, R) with M2 = F ( M1 ). Then, if K1 , K2 , H1 and H2 denote the Gaussian and mean curvatures of M1 and M2 , we have the relations K1 ( p) = K2 ( F ( p)) and H1 ( p) = H2 ( F ( p)), for all p ∈ M1 . Remark. The equality between the mean curvatures only holds indeed without the absolute value, once a convenient choice of a Gauss map for M2 has been made, “compatible” with F, in a sense to be made precise in the following proof.

186  Introduction to Lorentz Geometry: Curves and Surfaces

Proof: Suppose that F ∈ Eν (3, R) is written as F = Ta ◦ A, with A ∈ Oν (3, R) and . a ∈ R3ν . If N 1 is a Gauss map for M1 , then N 2 = A ◦ N 1 ◦ F −1 is a Gauss map for M2 . More precisely, if N 1 ( p) is normal to M1 at p, then A( N 1 ( p)) is normal to M2 at F ( p). Take linearly independent vectors v, w ∈ Tp M1 . Then Av, Aw ∈ Tp M2 are also linearly independent, and Proposition 3.3.12 now gives that: d( N 1 ) p (v) × d( N 1 ) p (w) = e M1 K1 ( p)v × w,

and

d( N 2 ) F( p) ( Av) × d( N 2 ) F( p) ( Aw) = e M2 K2 ( F ( p)) Av × Aw. Noting that d( N 2 ) F( p) = A ◦ d( N 1 ) p ◦ A−1 , the second equation in display reduces to A(d( N 1 ) p (v)) × A(d( N 1 ) p (w)) = e M2 K2 ( F ( p)) Av × Aw. Now, directly using Lemma 1.6.9 (p. 58) and canceling det A on both sides, it follows that d( N 1 ) p (v) × d( N 1 ) p (w) = e M2 K2 ( F ( p))v × w. Since congruent surfaces have the same causal type, we have that e M1 = e M2 , so that by direct comparison we obtain K1 ( p) = K2 ( F ( p)), as wanted. The reasoning for the mean curvature is similar: Proposition 3.3.12 gives us two more relations to be compared: d( N 1 ) p (v) × w + v × d( N 1 ) p (w) = −2e M1 H1 ( p)v × w,

and

d( N 2 ) F( p) ( Av) × Aw + Av × d( N 2 ) F( p) ( Aw) = −2e M2 H2 ( F ( p)) Av × Aw. The same remarks done for the Gaussian curvature simplify the second expression above to d( N 1 ) p (v) × w + v × d( N 1 ) p (w) = −2e M2 H2 ( F ( p))v × w, and M1 and M2 having the same causal type again allows us to conclude, by comparing, that H1 ( p) = H2 ( F ( p)). The above result raises a slightly subtler question: are the mean and Gaussian curvatures invariant under local isometries? We may focus our attention on local isometries instead of necessarily global ones, since the values of the curvatures at a given point are inherently local quantities, with coordinate expressions. This new question is fundamentally distinct than the previous one, since isometries between surfaces need not be restrictions of rigid motions defined on the ambient space. This observation justifies the usual terminology used in geometry: objects invariant under isometries (local or global) are called intrinsic to the surface. We have previously seen, though, that the mean curvature is not intrinsic to the surface (this might have been hinted at by the sign ambiguity in its definition): the plane and the cylinder are locally isometric, but the plane has zero mean curvature, while the cylinder does not. It remains to understand what happens with the Gaussian curvature. The answer is registered in one of the most beautiful theorems in all of Mathematics, established by Gauss himself in 1827: Theorem 3.3.14 (Theorema Egregium). Let M1 , M2 ⊆ R3ν be non-degenerate regular surfaces. If φ : M1 → M2 is a local isometry, and K1 and K2 denote the Gaussian curvatures of M1 and M2 , respectively, then K1 ( p) = K2 (φ( p)), for all p ∈ M1 . In other words, the Gaussian curvature of a surface is intrinsic to it.

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A more geometric interpretation: inhabitants of a surface M are able to determine the Gaussian curvature of M by only measuring angles, distances and ratios in M, without any reference to the “outside world”, the ambient space R3ν . The proof of this theorem will be presented on a more opportune moment ahead, but its idea basically consists of expressing the Gaussian curvature in terms of the First Fundamental Form only, but not the Second. As local isometries preserve the First Fundamental Form of a surface, they will also preserve any object which depends only on it. Namely, if (U, x) is a parametrization for a non-degenerate regular surface M, and x is orthogonal (that is, it satisfies F = 0), it is possible to show that ! ! ! p p ( | G |)u ( | E|)v −1 p p K◦x= p eu + ev . | EG | | E| | G | u v As expected, the formula when F 6= 0 is much more complicated and its practical usefulness is questionable. Gauss’ Theorema Egregium is one of the most powerful tools we have to decide when any given surfaces are not isometric. See an example of this idea in Exercise 3.3.9.

Exercises Exercise† 3.3.1 (Alternative expressions for K and H). Let M ⊆ R3ν be a nondegenerate regular surface. (a) Show that if p ∈ M and {v, w} is any basis for Tp M, then Gauss’ equation K ( p) =

hII p (v), II p (w)i − hII p (v, w), II p (w, v)i hv, vihw, wi − hv, wihw, vi

holds. (b) Show that if (U, x) is any parametrization for M, then H◦x=

1 2 ij g II( xi , x j ), 2 i,j∑ =1

where we identify u ↔ 1 and v ↔ 2. Exercise 3.3.2. Let M ⊆ R3ν be a non-degenerate regular surface, N be a Gauss map for M, and p ∈ M be any point. The Third Fundamental Form of M at p is the map III p : Tp M × Tp M → R given by . III p (v, w) = hdN p (v), dN p (w)i. Show that the relation III p − 2e M H ( p)e II p + e M K ( p)I p = 0 holds, for all p ∈ M. Thus III gives no new geometric information about M. Hint. Cayley-Hamilton.

188  Introduction to Lorentz Geometry: Curves and Surfaces

Exercise 3.3.3. Consider the Monge parametrization x : R2 → x(R2 ) ⊆ R3ν given by x(u, v) = (u, v, uv). (a) In R3 , show that K ( x(u, v)) < 0 for all (u, v) ∈ R2 , that K ( x(u, v)) depends only on the distance between x(u, v) and the z-axis, and that K ( x(u, v)) → 0 when such distance goes to +∞. (b) In L3 , show that K ( x(u, v)) < 0 for all (u, v) ∈ R2 wherever x is spacelike, and also that K ( x(u, v)) > 0 for all (u, v) ∈ R2 wherever x is timelike. In this latter case, K ( x(u, v)) also depends only on the Euclidean distance between x(u, v) and the z-axis, and K ( x(u, v)) → 0 when such distance goes to +∞. Exercise 3.3.4 (Graphs). Let f : U → R be a smooth function. Assuming that its graph is non-degenerate, and considering the usual Monge parametrization x, show that its curvatures are given by KE ◦ x =

2 f uu f vv − f uv (1 + f u2 + f v2 )2

and

HE ◦ x =

f uu (1 + f v2 ) − 2 f u f v f uv + f vv (1 + f u2 ) 2 (1 + f u2 + f v2 )3/2

in R3 , and: KL ◦ x =

2 − f f f uv uu vv 2 (−1 + f u + f v2 )2

and

HL ◦ x =

f uu (−1 + f v2 ) − 2 f u f v f uv + f vv (−1 + f u2 ) 2 | − 1 + f u2 + f v2 |3/2

in L3 . Exercise 3.3.5 (Surfaces of Revolution). Let α : I → R3ν be a smooth, regular, injective and non-degenerate curve of the form α(u) = ( f (u), 0, g(u)) for certain functions f and g, with f (u) > 0 for all u ∈ I. Assuming that α is not lightlike, consider the usual revolution parametrization x : I × ]0, 2π [ → x( I × ]0, 2π [) ⊆ R3ν given by x(u, v) = ( f (u) cos v, f (u) sin v, g(u)). Compute the coefficients of the Second Fundamental Form of x and show that KE ◦ x =

g0 (− f 00 g0 + f 0 g00 ) f hα0 , α0 i2E

and

HE ◦ x =

g0 (hα0 , α0 i E − f f 00 ) + f f 0 g00 2 f hα0 , α0 i3/2 E

in R3 and KL ◦ x =

g0 ( f 00 g0 − f 0 g00 ) f hα0 , α0 i2L

and

HL ◦ x =

g0 (−hα0 , α0 i L + f f 00 ) − f f 0 g00 2 f |hα0 , α0 i L |3/2

in L3 . Note that if α has unit speed, we have that: KE ◦ x = −

f 00 f

and K L ◦ x = −eα

f 00 , f

only. Exercise 3.3.6 (Surfaces of hyperbolic revolutions – II). Consider again a curve smooth, regular and injective curve α : I → L3 of the form α(u) = ( f (u), 0, g(u)), with g(u) > 0, and the associated surface of hyperbolic revolution around the x-axis, x : I × R → L3 given by: x(u, v) = ( f (u), g(u) sinh v, g(u) cosh v), as in Exercise 3.2.9 (p. 166), where we had asked you to show that the First Fundamental

Surfaces in Space  189

Form for this parametrization is given by ds2 = hα0 (u), α0 (u)i L du2 + g(u)2 dv2 . Suppose, in addition, that α is not lightlike. Compute the coefficients of the Minkowski Second Fundamental Form of x and show that K◦x=

f 0 ( f 00 g0 − f 0 g00 ) g hα0 , α0 i2L

and

H◦x=

− f 0 (hα0 , α0 i L + gg00 ) + gg0 f 00 . 2g |hα0 , α0 i L |3/2

Note that if α has unit speed, then we simply have K ◦ x = −eα g00 /g, in a similar fashion to what happened for usual revolution surfaces in R3 . Exercise 3.3.7 (Parallel Surfaces). Let x : U → R3ν be an injective regular parametrized surface. Define y : U → R3ν by . y(u, v) = x(u, v) + aN ( x(u, v)), where a ∈ R is fixed. For a small enough, y is also regular and has the same causal type as x (by continuity). (a) Show that yu × yv = (1 − 2e M aH + e M a2 K ) xu × xv , where H ≡ H ◦ x and K ≡ K ◦ x stand for the mean and Gaussian curvatures of x. (b) Show that the mean and Gaussian curvatures of y, Ha ≡ Ha ◦ y and Ka ≡ Ka ◦ y, are respectively given by Ha =

H − aK 1 − 2ae M H + a2 e M K

and Ka =

K . 1 − 2ae M H + a2 e M K

Hint. If N 1 = N is a Gauss map for x and N 2 is a Gauss map for y, it follows from the previous item that dN 1 ( xu ) = dN 2 (yu ), and similarly for derivatives with respect to v. Why? (c) Show that if the mean curvature of x is a constant c and y is a = e M /(2c) far from x, then the Gaussian curvature of y is constant and equals 4e M c2 . Exercise 3.3.8. Let M ⊆ R3ν be a non-degenerate regular surface, N be a Gauss map for M, and λ > 0. Consider the homothetic image λM = {λp ∈ R3ν | p ∈ M }, which we know to be a regular surface. Show that λM is also non-degenerate, and that its mean and Gaussian curvatures, Hλ and Kλ , are given by Hλ (λp) =

H ( p) λ

and Kλ (λp) =

K ( p) , λ2

where H and K denote the mean and Gaussian curvatures of M. Exercise 3.3.9. Give curves α, e α : I → R3 whose images are contained in the plane y = 0, which are congruent under an element of Eν (2, R) (acting only on the place y = 0), such that their associated surfaces of revolution are not isometric. Hint. Look for surfaces of revolution whose Gaussian curvatures have opposite signs — Theorema Egregium ensures that this will be a legitimate counter-example.

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3.4

THE DIAGONALIZATION PROBLEM

We have previously seen that if M ⊆ R3ν is a non-degenerate regular surface which is spacelike, and N is a Gauss map for M, it follows from the Spectral Theorem that the Weingarten map −dN p is always diagonalizable. Our goal now is to give conditions ensuring the possibility of diagonalization in the general case. We start recording the: Proposition 3.4.1. Let M ⊆ R3ν be a non-degenerate regular surface with diagonalizable Weingarten map. Then it holds that H ( p) = e M

κ1 ( p ) + κ2 ( p ) 2

and

K ( p ) = e M κ 1 ( p )κ 2 ( p ).

Remark. In R3 it is usual to define the mean and Gaussian curvatures by the above expressions. The expression for H also justifies the name “mean curvature”. A particular class of points for which the Weingarten map is diagonalizable are the so-called umbilic points: Definition 3.4.2. Let M ⊆ R3ν be a non-degenerate regular surface, and p ∈ M. We’ll say that p is umbilic if there is λ( p) ∈ R such that e II p (v, w) = λ( p)hv, wi, for all v, w ∈ Tp M. We’ll also say that M is totally umbilic if all its points are umbilic. Informally, a point is umbilic if the fundamental forms of M at said point are “linearly dependent”. Proposition 3.4.3. Let M ⊆ R3ν be a non-degenerate regular surface, and p ∈ M be an umbilic point. Then the Weingarten map at p is a multiple of the identity map. Proof: Just take v, w ∈ Tp M and compute

hλ( p)v, wi = e II p (v, w) = h−dN p (v), wi. The result follows from non-degeneracy of h·, ·i restricted to Tp M. The computations done in previous examples show that non-degenerate planes and the “spheres” are totally umbilic surfaces (in their respective ambient spaces). The next result states that this short list of examples is complete: Theorem 3.4.4 (Characterization of totally umbilic surfaces). Let M ⊆ R3ν be a connected and totally umbilic non-degenerate regular surface. Then M is contained in a plane of R3ν , or there are c ∈ R3ν and r > 0 such that: (i) if M ⊆ R3 , then M ⊆ S2 (c, r ); (ii) if M ⊆ L3 is spacelike, then M ⊆ H2 (c, r ) or M ⊆ H2− (c, r ); (iii) if M ⊆ L3 is timelike, then M ⊆ S21 (c, r ). Remark. In item (ii) above, what decides between H2 (c, r ) and H2− (c, r ) is the causal type of p − c for one (and hence all) p ∈ M, by continuity. The time orientation of the Gauss map is not relevant at all here (why?).

Surfaces in Space  191

Proof: Since M is totally umbilic, for every p ∈ M there is λ( p) ∈ R such that −dN p = λ( p) idTp M , and thus we have defined a map λ : M → R, which is automatically smooth by the previous expression. Consider an arbitrary parametrization (U, x), with U connected. Identifying N ≡ N ◦ x and similarly for λ, we have that: ( − N u (u, v) = −dN x(u,v) ( xu (u, v)) = λ(u, v) xu (u, v) − N v (u, v) = −dN x(u,v) ( xv (u, v)) = λ(u, v) xv (u, v), for all (u, v) ∈ U. Differentiating the first equality with respect to v and the second with respect to u, we obtain: ( − N uv (u, v) = λv (u, v) xu (u, v) + λ(u, v) xuv (u, v) − N vu (u, v) = λu (u, v) xv (u, v) + λ(u, v) xvu (u, v). Since second order partial derivatives commute, and xu (u, v) and xv (u, v) are linearly independent, it follows that: λv (u, v) xu (u, v) − λu (u, v) xv (u, v) = 0 =⇒ λu ≡ λv ≡ 0 em U, and by connectedness of U, we have that λ ◦ x is constant. As M is covered by parametrizations with connected domains, it follows that λ itself is locally constant. Now from the connectedness of M, we conclude that the function λ is constant.

• If λ = 0, dN p = 0 for all p ∈ M, so that the Gauss map N ( p) = N 0 is constant, and then M is contained in some plane in R3ν with N 0 giving the normal direction. • If λ 6= 0, define c : M → R3ν by 1 . c ( p ) = p + N ( p ). λ Let’s see that c is constant. To wit, we have that
1 1 dc p = idTp M + dN p = idTp M + − λ idTp M = 0 λ λ and M is connected. With this, note that e h p − c, p − ci = M2 , λ and the conclusion follows.

Figure 3.25: The totally umbilic surfaces in L3 .

192  Introduction to Lorentz Geometry: Curves and Surfaces

Back to the discussion of the possibility of diagonalization of the Weingarten map, we have the: Proposition 3.4.5. Let M ⊆ L3 be a non-degenerate regular surface, and p ∈ M such that the Weingarten map at p is diagonalizable. Then H ( p)2 − e M K ( p) ≥ 0, with equality holding if and only if p is umbilic. Proof: Directly, we have:  0≤

κ1 ( p ) − κ2 ( p ) 2

2

=

κ1 ( p)2 − 2κ1 ( p)κ2 ( p) + κ2 ( p)2 4

κ1 ( p)2 + 2κ1 ( p)κ2 ( p) + κ2 ( p)2 − κ1 ( p )κ2 ( p ) 4   κ1 ( p ) + κ2 ( p ) 2 = − κ1 ( p )κ2 ( p ) 2

=

= (e M H ( p))2 − e M K ( p) = H ( p)2 − e M K ( p). Equality holds if and only if κ1 ( p) = κ2 ( p), that is to say, if p is umbilic. The above proposition gives a necessary (but not sufficient) condition for −dN p to be diagonalizable, and emphasizes the importance of the quantity H ( p)2 − e M K ( p), which will be explored in the proof of the: Theorem 3.4.6 (Diagonalization in L3 ). Let M ⊆ L3 be a non-degenerate regular surface, N be a Gauss map for M, and p ∈ M. Then: (i) if H ( p)2 − e M K ( p) > 0, −dN p is diagonalizable; (ii) if H ( p)2 − e M K ( p) < 0, −dN p is not diagonalizable; (iii) if H ( p)2 − e M K ( p) = 0 and M is spacelike, then p is umbilic, and hence −dN p is diagonalizable. Remark. With the above notation, if H ( p)2 − e M K ( p) = 0 and M is timelike, the criterion is inconclusive and the Weingarten map may or may not be diagonalizable. In the following, we will see examples illustrating both situations. Proof: Consider the characteristic polynomial c(t) of −dN p , given by c(t) = t2 − tr(−dN p ) t + det(−dN p ) = t2 − 2e M H ( p)t + e M K ( p), whose discriminant is:

(−2e M H ( p))2 − 4(e M K ( p)) = 4( H ( p)2 − e M K ( p)). • If H ( p)2 − e M K ( p) > 0, then c(t) has two distinct roots, which are the eigenvalues of −dN p , which then admits two linearly independent eigenvectors (hence diagonalizable). • If H ( p)2 − e M K ( p) < 0, c(t) does not have any real roots. Thus −dN p has no real eigenvalues, and hence it is not diagonalizable.

Surfaces in Space  193

• Now assume that H ( p)2 − e M K ( p) = 0 and that M is spacelike, that is, that K ( p) = − H ( p)2 . From the expression given for the discriminant of c(t), it follows that − H ( p) is an eigenvalue of −dN p . So, there is a unit (spacelike) vector u1 ∈ Tp M such that dN p (u1 ) = H ( p)u1 . Consider then an orthogonal basis . B = (u1 , u2 ) of Tp M. Then:     H ( p) a dN p B = , where dN p (u2 ) = au1 + bu2 . 0 b It suffices to check that a = 0 and b = H ( p) to conclude the proof. Applying h·, u1 i L , we have: a = hdN p (u2 ), u1 i L = hu2 , dN p (u1 )i L = hu2 , H ( p)u1 i L = H ( p)hu2 , u1 i L = 0. On the other hand:

− H ( p)2 = K ( p) = − det(−dN p ) = − det(dN p ) = − H ( p)b, so that H ( p)b = H ( p)2 . If H ( p) = 0, then dN p is the zero map (hence diagonalizable). If H ( p) 6= 0, we obtain b = H ( p), as wanted. Note that in this case p is umbilic.

Observe that in the above proof, we do not have any control over the causal type of the eigenvector u1 in the last described situation when M is timelike. If u1 were lightlike, we could not consider the basis B to proceed with the argument. With this in mind, we extend the above result, with the due adaptations: Corollary 3.4.7. Let M ⊆ L3 be a timelike regular surface and p ∈ M be a point with H ( p)2 − K ( p) = 0. If −dN p has no lightlike eigenvectors, then it is diagonalizable and p is umbilic, with both principal curvatures equal to − H ( p). Now, let’s see the promised examples illustrating the situation where H ( p)2 = K ( p) and the Weingarten map may or may not be diagonalizable. Example 3.4.8. (1) We have previously seen that for the de Sitter space S21 , we had −dN p = −idTp (S2 ) , hence diagonalizable, with K = 1 and H = −1, so that H 2 − K = 0.

1

(2) Consider a lightlike curve α : I → L3 , with arc-photon parametrization. Recall that
T α (φ) = α0 (φ), N α (φ) = α00 (φ) and that if T α (φ), N α (φ) is positive, then Bα (φ) is the unique lightlike vector orthogonal to
N α (φ) with h T α (φ), Bα (φ)i L = −1, thus making the basis T α (φ), N α (φ), Bα (φ) positive as well. Moreover, we have Cartan’s equations  0     T α (φ) 0 1 0 T α (φ)  N 0α (φ) =  α (φ) 0 1  N α ( φ )  , 0 Bα (φ) 0 Bα (φ) α (φ) 0

d

where

dα (φ) is the pseudo-torsion of α.

d

Define the B-scroll associated to α, x : I × R → L3 by . x(φ, t) = α(φ) + tBα (φ).

194  Introduction to Lorentz Geometry: Curves and Surfaces

Restricting the domain of x enough, we may assume that its image is a regular surface. Let, for each φ:
. D (φ) = det T α (φ), N α (φ), Bα (φ) > 0. Computing the derivatives xφ (φ, t) = T α (φ) + t

dα ( φ ) N α ( φ )

and

xt (φ, t) = Bα (φ),

we immediately have that 

( gij (φ, t))1≤i,j≤2 =

t2

dα ( φ )2 −1

 −1 , 0

whence x is timelike. Furthermore, note that | det(( gij (φ, t))1≤i,j≤2 )| = 1, and then we directly obtain N ( x(φ, t)) = T α (φ) × L Bα (φ) + t

dα ( φ ) N α ( φ ) × L B α ( φ ).

Computing the second order derivatives

dα (φ)2 T α (φ) + (1 + td0α (φ)) N α (φ) + tdα (φ)Bα (φ), xφt (φ, t) = dα (φ) N α (φ) and

xφφ (φ, t) = t

xtt (φ, t) = 0, we obtain the Second Fundamental Form  (−1 − t 0α (φ) + t2 α (φ)3 ) D (φ) − (hij (φ, t))1≤i,j≤2 = − α (φ) D (φ)

d

d

d

dα ( φ ) D ( φ )  . 0

Thus, we have K ( x(φ, t)) =

dα ( φ )2 D ( φ )2

and

H ( x(φ, t)) =

dα ( φ ) D ( φ ).

Hence, we know that at each point x(φ, t), −dN x(φ,t) has only one eigenvalue, namely, α ( φ ) D ( φ ). Then, it suffices to see that there are points for which the associated eigenspace has dimension 1, so that the Weingarten maps at such points are not diagonalizable. Using Lemma 3.3.7 (p. 180), we have:   h i 0 α (φ) −dN x(φ,t) = D (φ) . 1 + t α (φ) Bx α (φ)

d

d

The kernel of the matrix



0

d

0 1 + t α (φ) 0

d



d clearly has dimension 1 whenever 1 + tdα (φ) 6= 0 (for instance, along the curve α itself, when t = 0). 3.4.1

Interpretations for curvatures

At this point we are ready to present some geometric interpretations for the mean and Gaussian curvatures of a non-degenerate regular surface in R3ν . Aiming towards this end, we need a special parametrization explicitly emphasizing the Second Fundamental Form of the surface. The existence of one such parametrization is given in the:

Surfaces in Space  195

Theorem 3.4.9 (Inertial coordinates). Let M ⊆ R3ν be a non-degenerate regular surface, N be a Gauss map for M, p ∈ M, and (w1 , w2 ) an orthonormal basis for Tp M such that (w1 , w2 , N ( p)) is a positive basis for R3ν . Then there is a parametrization (U, x) around p satisfying the following conditions: (i) (0, 0) ∈ U and x(0, 0) = p; (ii) gij (0, 0) = ewi δij , for 1 ≤ i, j ≤ 2; (iii) (∂gij /∂uk )(0, 0) = 0, for 1 ≤ i, j, k ≤ 2; (iv) up to second order: x ( u1 , u2 ) − p = u1 w1 + u2 w2 +

eM 2

2

∑

hij (0, 0)ui u j N ( p) + R(u1 , u2 ),

i,j=1

where R(u1 , u2 )/k(u1 , u2 )k2E → 0 if (u1 , u2 ) → (0, 0). We will say that a parametrization satisfying these four properties is inertial. Remark. Condition (iv) above is equivalent to x ( u1 , u2 ) − p = u1 w1 + u2 w2 +

1 2 i j u u II p (wi , w j ) + R(u1 , u2 ), 2 i,j∑ =1

with the same R(u1 , u2 ) of the original statement. Proof: The strategy is to start with an arbitrary parametrization around p and perform successive changes of coordinates until the end result satisfies all the needed requirements. That said, take an initial parametrization (U1 , x) around p. By Exercise 3.1 (p. 151), we may assume from the start that (0, 0) ∈ U1 and that x(0, 0) = p. Now, let A : R2 → R2 be the linear isomorphism defined by Aei = vi , where the . vectors v1 , v2 ∈ R2 are such that Dx(0, 0)(vi ) = wi , for 1 ≤ i ≤ 2. Let U2 = A−1 (U1 ) . and define e x = x ◦ A : U2 → e x(U2 ) = x(U1 ) ⊆ M. Clearly we still have e x(0, 0) = p, but now we have in addition that ∂e x (0, 0) = Dex(0, 0)(ei ) = Dx( A(0, 0)) ◦ DA(0, 0)(ei ) ∂ui = Dx(0, 0)( Aei ) = Dx(0, 0)(vi ) = wi . We need one last reparametrization. Consider the second order Taylor expansion, centered at (0, 0): e x ( u1 , u2 ) − p = u1 w1 + u2 w2 +

∂2 e x 1 2 e ( u1 , u2 ), (0, 0)ui u j + R ∑ 2 i,j=1 ∂ui ∂u j

e (u1 , u2 )/k(u1 , u2 )k2 → 0 if (u1 , u2 ) → (0, 0). But, applying orthonormal exwhere R E pansion for these second order partial derivatives, we obtain:  2  eM 2 i j ∂ e x 1 2 1 2 e x ( u , u ) − p = u w1 + u w2 + uu (0, 0), N ( p) N ( p)+ 2 i,j∑ ∂ui ∂u j =1  2  1 2 ∂ e x i j + ∑ ewk u u ∂ui ∂u j (0, 0), wk wk + Re (u1 , u2 ). 2 i,j,k =1

196  Introduction to Lorentz Geometry: Curves and Surfaces

Our goal, then, is to perform a change of coordinates to eliminate the triple summation above. For such end, consider ψ : U2 → R2 given (in matrix notation, for convenience) by   2  ew1 2 i j ∂ e x 1 (0, 0), w1  u + 2 ∑ u u ∂ui ∂u j   i,j=1 1 2 .  . ψ(u , u ) =     2 ∂2 e x  u 2 + ew2 i j (0, 0), w2  uu j i 2 i,j∑ ∂u ∂u =1 To simplify notation in what follows, abbreviate  2  ew k 2 i j ∂ e x k . k uu ue = u + (0, 0), wk . 2 i,j∑ ∂ui ∂u j =1 Clearly we have ψ(0, 0) = (0, 0), and also Dψ(0, 0) = Id2 , whence the Inverse Function Theorem provides open subsets U20 ⊆ U2 and U of R2 around (0, 0) such that ψ : U20 → U . x(U20 ) ⊆ M. Let’s is a diffeomorphism. Define, finally, x = e x U 0 ◦ ψ −1 : U → x (U ) = e 2 verify that x now satisfies all the requirements. To begin with, we have x(0, 0) = e x(ψ−1 (0, 0)) = e x(0, 0) = p, which shows condition (i). Moreover: ∂x (0, 0) = Dx(0, 0)(ei ) = D (ex ◦ ψ−1 )(0, 0)(ei ) ∂uei = Dex(ψ−1 (0, 0)) ◦ Dψ−1 (0, 0)(ei ) = Dex(0, 0) ◦ Id2 (ei )

= Dex(0, 0)(ei ) = wi , whence gij (0, 0) = ewi δij and we conclude (ii). Factoring out w1 and w2 in the Taylor formula for e x, and using the definition of the coefficients e hij of the Second Fundamental Form of e x, we have: x(ue1 , ue2 ) − p = e x(ψ−1 (ue1 , ue2 )) − p

= ex(u1 , u2 ) − p ! 2e ∂ x = ∑ ∑ ui u j ∂ui ∂u j (0, 0), wk wk + i,j=1 k =1   2 2 eM ∂ e x i j e ( u1 , u2 ) + (0, 0), N ( p) N ( p) + R uu i ∂u j 2 i,j∑ ∂u =1 2

ew uk + k 2

= ue1 w1 + ue2 w2 +

2



eM 2

2

∑

e ( u1 , u2 ). ui u j e hij (0, 0) N ( p) + R

i,j=1

Since uek = uk + Rk (u1 , u2 ) with Rk (u1 , u2 )/k(u1 , u2 )k2E → 0 as (u1 , u2 ) → (0, 0), we may group all the terms that go to zero fast and conclude that x(ue1 , ue2 ) − p = ue1 w1 + ue2 w2 +

eM 2

2

∑

i,j=1

uei uej e hij (0, 0) N ( p) + R(ue1 , ue2 ),

where R(ue1 , ue2 )/k(ue1 , ue2 )k2E → 0 if (ue1 , ue2 ) → (0, 0). Since the partial derivatives up to second order of x and e x agree at (0, 0), we have that hij (0, 0) = e II p (wi , w j ) = e hij (0, 0),

Surfaces in Space  197

so that we conclude (iv). To conclude (iii), we differentiate (iv) and observe that in general gij (ue1 , ue2 ) = ewi δij + Rij (u1 , u2 ) holds, where Rij (ue1 , ue2 )/k(ue1 , ue2 )k E → 0 if (ue1 , ue2 ) → (0, 0), whence it follows that all the possible first order partial derivatives of the gij evaluated at (0, 0) vanish. Renaming the parameters (ue1 , ue2 ) → (u1 , u2 ), we even obtain the conditions with the same notation as in the original statement. This theorem combined with Proposition 3.1.29 (p. 147) says that given a point p in a non-degenerate regular surface M ⊆ R3ν , we may regard M locally as a graph over Tp M, discarding terms of order 3 and higher, of the quadratic form eM 2

2

∑

hij (0, 0)ui u j .

i,j=1

Then, of course, the behavior of the surface, near p, will be controlled by the matrix (hij (0, 0))1≤i,j≤2 or, in other words, by the scalar version of the Second Fundamental Form e II p . With this in mind, we just need some last piece of terminology to start discussing geometric interpretations for the Gaussian curvature: Definition 3.4.10. Let M ⊆ R3ν be a non-degenerate regular surface and p ∈ M. We’ll say that p is: (i) elliptic if e II p is definite; (ii) hyperbolic if e II p is indefinite; (iii) parabolic if e II p 6= 0 is degenerate; (iv) planar if e II p = 0. If (U, x) is a parametrization, we see from the coordinate
expression for K ◦ x in terms of gij and hij that K ( x(u, v)) and det (hij (u, v))1≤i,j≤2 have the same sign in R3 , and opposite signs in L3 . The conclusion we make from this observation, by applying Sylvester’s Criterion for (hij (u, v))1≤i,j≤2 , is then recorded in: Proposition 3.4.11. Let M ⊆ R3ν be a non-degenerate regular surface and p ∈ M. Then: (i) in R3 , if K ( p) > 0 (resp. < 0), p is elliptic (resp. hyperbolic); (ii) in L3 , if K ( p) > 0 (resp. < 0), p is hyperbolic (resp. elliptic); (iii) if K ( p) = 0 and H ( p) 6= 0, p is parabolic; (iv) if K ( p) = H ( p) = 0, p é planar. The details of the proof are left for Exercise 3.4.5. Example 3.4.12. (1) If Π ⊆ R3ν is a non-degenerate plane, all the points of Π are planar (surprise?). (2) All the points of S2 and H2 are elliptic, while the points of S21 are hyperbolic.

198  Introduction to Lorentz Geometry: Curves and Surfaces

(3) All the points of S1 × R are parabolic. (4) All the points of the B-scroll over a lightlike curve in L3 (seen in Example 3.4.8 above, p. 193) are hyperbolic. (5) All the points in the helicoid seen in Example 3.3.11 (p. 183), where it is nondegenerate, are hyperbolic. We have the following situations which justify the terminology given in Definition 3.4.10 above: (I) if p is elliptic, e II p is positive or negative-definite and then M is approximated by an elliptic paraboloid:

p + Tp M p (II) if p is hyperbolic, e II p is indefinite and then M is approximated by a hyperbolic paraboloid:

p + Tp M

p

(III) if p is parabolic, e II p 6= 0 is degenerate and then M is approximated by a “parabolic cylinder”:

p

p + Tp M

(IV) if p is planar, e II p = 0 and M coincides with Tp M up to terms of order 3. In terms of the sign of the Gaussian curvature only, we also have the following interpretation, which follows from the discussion after the proof of Theorem 3.4.9:

Surfaces in Space  199

Proposition 3.4.13. Let M ⊆ R3 be a regular surface and p ∈ M. Then: (i) if K ( p) > 0, there is an open neighborhood of p in M entirely contained in one of the half-spaces determined by Tp M; (ii) if K ( p) < 0, then every open neighborhood of p in M has points in both half-spaces determined by Tp M. The same conclusion holds if M ⊆ L3 is non-degenerate, reversing the sign of K ( p). Remark. A priori, to compute the Gaussian curvature of a non-degenerate regular surface at a certain point, it is not necessary for the Gauss map to be defined in all of the surface, but only on a neighborhood of this point. That is, the orientability assumption made initially could have been dropped for this local analysis. However, it is a consequence of the above proposition that every regular surface M in R3 (resp. non-degenerate in L3 ) with strictly positive (resp. negative) Gaussian curvature in all points is automatically orientable. It suffices to consider, for each p ∈ M, the neighborhood Wp of p in M given by the above proposition, and take the unit normal vector N ( p) pointing towards the half-space determined by Tp M which contains Wp . This defines an orientation for M. This last proposition also allows us to visually verify that the Gaussian curvatures of and S21 are positive, while the curvature of H2 is negative. Unfortunately, this criterion is not decisive for parabolic and planar points (see Exercise 3.4.6). We may also relate the sign of K ( p) 6= 0 with dN p preserving or reversing orientation in Tp M, simply by recalling that K ( p) = e M det(dN p ), as follows: S2

• in R3 : K ( p) > 0 if and only if dN p preserves orientation; • in L3 : if M is timelike, K ( p) > 0 if and only if dN p preserves orientation; • in L3 : if M is spacelike, K ( p) < 0 if and only if dN p preserves orientation. To conclude the interpretations for the Gaussian curvature for now, we have the: Proposition 3.4.14. Let M ⊆ R3ν be a non-degenerate regular surface, N a Gauss map for M, and p ∈ M. Then A( N ( R)) |K ( p)| = lim . A( R)→0 A( R) p∈ R

Remark.

• The above limit may be formally understood by thinking of a classical e-δ definition, as follows: given e > 0, there is δ > 0 such that, if R is a region in M containing A( N ( R)) p with A( R) < δ, then A( R) − |K ( p)| < e. • In other words, K may be seen as an infinitesimal ratio of areas of images under the Gauss map over the original areas.

200  Introduction to Lorentz Geometry: Curves and Surfaces

Proof: Since A( R) → 0, we may evaluate the limit along regions all contained inside the image of a fixed parametrization (U, x) around p. We have that A( N ( R)) 1 lim = lim A( R)→0 A ( R )→ 0 A( R) A( R)

Z

1 = lim A( R)→0 A( R )

Z

p∈ R

( N ◦ x)−1 ( N ( R))

p∈ R

(1)

p∈ R

(2)

= lim

A( R)→0 p∈ R

x −1 ( R )

k( N ◦ x)u (u, v) × ( N ◦ x)v (u, v)k du dv

|K ( x(u, v))| k xu (u, v) × xv (u, v)k du dv

|K ( x(u R , v R ))| A( R) A( R)

(3)

= |K ( p)|,

where in (1) we use Proposition 3.3.12 (p. 185), in (2) the Mean Value Theorem for integrals and, in (3), that (u R , v R ) ∈ x−1 ( R), A( R) → 0 and p ∈ R together imply that x(u R , v R ) → p. The Gaussian curvature is not the only curvature that may be interpreted in terms of areas, as seen above. One of the most important interpretations for the mean curvature comes from the first variation of area for a non-degenerate parametrized regular surface. Definition 3.4.15. Let x : U → x(U ) ⊆ R3ν be a smooth mapping such that x U is a non-degenerate regular parametrized surface. An admissible variation of x, for t in the interval ]−e, e[, is a mapping xt : U → xt (U ) ⊆ R3ν , given by . xt (u, v) = x(u, v) + tV (u, v), ˚ where V : U → R3ν is a smooth map that vanishes on the boundary ∂U = U \ U. Remark.

• We may assume, in the above definition, that e > 0 is small enough so that each xt U is also regular and with the same causal type as x U . • In the above definition, we ask those mappings to be defined in the closure U, because we’ll need to analyze the curve x(∂U ) in R3ν . Definition 3.4.16. Let xt : U → xt (U ) ⊆ R3ν be an admissible variation of a smooth mapping x : U → x(U ) ⊆ R3ν . The area functional associated to this variation is . A(t) =

Z U

k xtu (u, v) × xtv (u, v)k du dv.

Theorem 3.4.17 (First Variation of Area). Let U ⊆ R2 be a bounded open set with regular boundary, x : U → x(U ) ⊆ R3ν a smooth map such that x U is a non-degenerate and injective regular parametrized surface. Then x U has zero mean curvature if and only if A0 (0) = 0, for every admissible variation xt of x. Proof: Initially, we compute A0 (0) for an arbitrary variation xt . We will omit the points . of evaluation (u, v) to simplify the notation. Also set e M = ex(U ) . Since U is compact and all the relevant functions here are smooth, we may differentiate under the integral sign to obtain Z d 0 t t A (0) = k xu × xv k du dv. U dt t =0

Surfaces in Space  201

By Proposition 2.1.13 (p. 70), we have D E
0 Z eM xtu × xtv t=0 , xu × xv A0 (0) = du dv. k xu × xv k U Directly, we see that


0 xtu × xtv t=0 = V u × xv + xu × V v ,

whence A0 (0) =

Z U

e M hV u × xv + xu × V v , N ◦ xi du dv.

The strategy now is to apply the Green-Stokes theorem (which is possible since U has . . regular boundary), considering P = h N ◦ x, V × xu i and Q = h N ◦ x, V × xv i. We have ∂Q ∂P − = h N ◦ x, V u × xv + xu × V v i − hV , ( N ◦ x)u × xv + xu × ( N ◦ x)v i ∂u ∂v = h N ◦ x, V u × xv + xu × V v i + 2e M hV , ( H ◦ x) xu × xv i, by Proposition 3.3.12 (p. 185). So, Stokes’ formula Z U

∂Q ∂P − du dv = ∂u ∂v

I

P du + Q dv ∂U

boils down to e M A0 (0) + 2e M

Z U

hV , ( H ◦ x) xu × xv i du dv = =

I

h N ◦ x, V × xu i du + h N ◦ x, V × xv i dv = 0, ∂U

. since V ∂U = 0. Setting dA = k xu × xv k du dv, we finally obtain 0

A (0) = −2

Z U

hV , H ◦ xi dA.

If H ◦ x = 0, then A0 (0) = 0 for every admissible variation. On the other hand, if H ◦ x 6= 0 (and it is necessarily always spacelike or timelike, since x is non-degenerate), it is possible to construct an admissible variation of x for which A0 (0) 6= 0: considering V = H ◦ x does not yield an admissible variation since H ◦ x does not necessarily vanish on ∂U (a priori it might not even be defined there). To remedy this, it suffices to multiply H ◦ x by a smooth function whose support is contained in U. It is not an easy task to compute an explicit expression for A00 (0) which, in more general settings, may be given in terms of certain geometric objects which are outside the scope of this text (see [3], for instance). From such expression, one may conclude that:

• in R3 , if H = 0 then x is a local minimum of the area functional; • in L3 , if H = 0 and x is spacelike, then x is a local maximum of the area functional; • in L3 , if H = 0 and x is timelike, then x is a local minimum of the area functional. This motivates the: Definition 3.4.18. Let M ⊆ R3ν be a non-degenerate regular surface. If its mean curvature vanishes, we say that the surface is critical. In particular, M is minimal if M ⊆ R3 or if it is timelike, and maximal if M ⊆ L3 is spacelike.

202  Introduction to Lorentz Geometry: Curves and Surfaces

The Gaussian and mean curvatures impose restrictions on the topology of the surface. One example of this is given in the: Proposition 3.4.19. Let M ⊆ R3 be a compact regular surface. Then there is a point p0 ∈ M such that K ( p0 ) > 0. Proof: Consider the (smooth) function f : M → R given by f ( p) = h p, pi. Since M . is compact, f has a global maximum at some point p0 ∈ M. If r = k p0 k > 0, note that d f p0 = 2h p0 , ·i = 0 tells us that Tp0 M = Tp0 S2 (r ). Moreover, if v ∈ Tp0 M is any tangent vector realized by a curve α : ]−e, e[ → M, and N ( p0 ) denotes the unit normal to M (and also to S2 (r )) at p0 , we have that
d2 00 f ( α ( t )) ≤ 0 =⇒ 2 h α ( 0 ) , p i + h v, v i ≤ 0, 0 dt2 t=0 which may be reorganized as:

hα00 (0), rN ( p0 )i ≤ −hv, vi =⇒ r hv, −dN p0 (v)i ≤ −hv, vi. This in particular holds for the unit principal directions v1 , v2 ∈ Tp0 M, satisfying −dN p0 (vi ) = κi ( p)vi for 1 ≤ i ≤ 2, whence rκi ( p) ≤ −1 =⇒ κi ( p) ≤ for 1 ≤ i ≤ 2. Thus K ( p0 ) = κ 1 ( p0 )κ 2 ( p0 ) ≥

−1 , r

1 > 0, r2

as wanted. Corollary 3.4.20. There is no minimal and compact surface in R3 . Proof: It suffices to note that in R3 , we always have that H ( p)2 − K ( p) ≥ 0. If H = 0 then K ( p) ≤ 0 for all p ∈ M. Suppose now that the surface M ⊆ R3ν is the graph of some smooth function f : U ⊆ R2 → R. By taking a Monge parametrization, for instance, we see that according to Exercise 3.3.4, M is critical if and only if f uu ((−1)ν + f v2 ) − 2 f u f v f uv + f vv ((−1)ν + f u2 ) = 0, which is a quasi-linear Partial Differential Equation (PDE). Clearly all affine (linear) functions are solutions to this equation, i.e., planes are the simplest examples for critical surfaces. If the domain of the function f is the whole plane, we say that f is entire (and that its graph is an entire graph). Demanding an entire graph to be critical is very restrictive, as the following theorem (whose proof uses techniques beyond the scope of this text) shows. Theorem 3.4.21 (Calabi-Bernstein). Let M ⊆ R3ν be the graph of an entire smooth function f : R2 → R, satisfying the critical surfaces equation above. If M ⊆ R3 or if M is spacelike in L3 , then f is affine and M is an entire plane.

Surfaces in Space  203

Example 3.4.22. In general, solving PDEs such as the above can be very complicated. It is natural, then, to look for solutions having certain types of symmetries. For example, suppose that f depends smoothly only on the distance √ between the point (u, v) and the origin, that is, that f has the form f ( u, v ) = g ( u2 + v2 ) for some single-variable . √ 2 smooth function g. Letting r = u + v2 and omitting points of evaluation, we compute the derivatives f u = ug0 (r )/r and f v = vg0 (r )/r, as well as the second order derivatives: f uu =

u2 00 v2 0 g ( r ) + g (r ), r2 r3

f uv =

uv 00 uv 0 v2 00 u2 0 g ( r ) − g ( r ) and f = g ( r ) + g (r ). vv r2 r3 r2 r3

A direct substitution leads us to the Ordinary Differential Equation g00 (r ) +

g0 (r )(1 + (−1)ν g0 (r )2 ) = 0. r

After order reduction, this is a Bernoulli equation, which can be solved by a well-known . algorithm: set h = g0 and rewrite the equation as h 0 (r ) +

h (r ) h (r )3 h ( r ) −2 (−1)ν + (−1)ν = 0 =⇒ h(r )−3 h0 (r ) + + = 0. r r r r

. Let w = h−2 , so that w0 = −2h−3 h0 , and so we may further rewrite the equation as rw0 (r ) − 2w(r ) = 2(−1)ν . Multiplying both sides by r −3 and integrating, we finally obtain w(r ) = (−1)ν+1 + cr2 , for some constant c ∈ R. Now we’ll start discussing the situation in each ambient space, in terms of the sign of the constant c:

• In R3 , the relation w = h−2 > 0 does not allow the possibility that c ≤ 0. Hence c > 0 we proceed to solve the equation. We have √ √  2 − 1 + cr log c cr 1 √ h (r ) = ± √ =⇒ g(r ) = ± + b, c cr2 − 1 for some b ∈ R, where r ranges over some adequate domain. We then obtain a family of catenoids.

• In L3 , we indeed have two situations according to the sign of the constant c. Recall that k∇ f (u, v)k E = | g0 (r )| controls the causal type of M. With this in mind, we see from h(r )−2 = 1 + cr2 that c > 0 if and only if M is spacelike, and c < 0 if and only if M is timelike (the case c = 0 corresponds to lightlike M and it is outside the scope of this discussion). If c > 0, we have h (r ) = ± p

1

√

1 + ( cr )2

√
1 =⇒ g(r ) = ± √ arcsinh cr + b c

and if c < 0, it follows that h (r ) = ± q for certain b ∈ R.

√  1 1 √ =⇒ g ( r ) = ± arcsin − cr + b, √ −c 1 − ( −cr )2

204  Introduction to Lorentz Geometry: Curves and Surfaces

For the surfaces obtained above to be graphs over   plane z = 0, the domains of √ the the solutions must be, respectively, R>0 and 0, 1/ −c .

Figure 3.26: Catenoids in R3 , L3 (spacelike) and L3 (timelike). The above calculations in particular show that the assumption of f being entire is essential in the Calabi-Bernstein theorem: the surfaces, in R3 and spacelike in L3 , found above, are graphs of non-affine functions defined in proper open subsets of R2 . Moreover, for spacelike surfaces in L3 , we have that g(0) = b and g0 (0) = ±1, that is, their graphs are asymptotic to the lightcone centered at (0, 0, b). Finally we observe that, in L3 , the solutions √ obtained √ are odd functions, and so may have their domains extended to R and −1/ −c, 1/ −c , respectively, with the planes kπ of the form z = √ + b (with integer k) are symmetry planes for the surface in the 2 −c second case. Another example of a situation where it is interesting to look for solutions of the critical surface PDE with certain symmetries is seen in Exercise 3.4.3, where we ask you to investigate the critical translation surfaces in R3ν .

Exercises Exercise 3.4.1. Let M1 , M2 ⊆ R3ν be non-degenerate regular surfaces, and F ∈ Eν (3, R) be such that F ( M1 ) = M2 . Show that if the Weingarten map of M1 is diagonalizable at p, then the Weingarten map of M2 is diagonalizable at F ( p). Exercise 3.4.2 (A factory of critical surfaces). Let α : I → R3ν and β : J → R3ν be two regular curves with {α0 (u), β0 (v)} linearly independent for all possibilities of u and v. Define x : I × J → L3 by setting x(u, v) = α(u) + β(v). (a) Show that x is a parametrized regular surface. (b) Show that the tangent planes along a fixed coordinate curve are all parallel to a single line. (c) In L3 , show that if α and β are lightlike, then x is timelike with zero mean curvature. Remark. Actually, the converse holds: every timelike surface with zero mean curvature is locally the sum of lightlike curves, i.e., around every point there is a parametrization as above. Furthermore, if the Gaussian curvature of M also vanishes, we have that hα0 (u), β0 (v)i L = c 6= 0 for all (u, v) ∈ I × J. See [13] for more details.

Surfaces in Space  205

Exercise 3.4.3 (Translation surfaces). Consider the particular Monge parametrization x : R2 → x(R2 ) ⊆ R3ν given by x(u, v) = (u, v, h(u) + `(v)), where h and ` are certain smooth functions. (a) In R3 , show that x is minimal if and only if h00 (u) `00 (v) = − =a 1 + h 0 ( u )2 1 + ` 0 ( v )2 for some constant a ∈ R. Solve the differential equations and conclude that all the minimal surfaces of this type are given by 1 1 log cos( au + b1 ) + c1 and `(v) = log cos(− av + b2 ) + c2 , a a when a 6= 0, and h(u) = b1 u + c1 and `(v) = b2 v + c2 if a = 0, for suitable constants b1 , b2 , c1 , c2 ∈ R. h(u) =

(b) In L3 , assuming that x is non-degenerate, show that x is critical if and only if h00 (u) `00 (v) = − =a 1 − h 0 ( u )2 1 − ` 0 ( v )2 for some constant a ∈ R. Solve the differential equations and conclude that all the critical surfaces of this type are given by 1 1 log cosh( au + b1 ) + c1 and `(v) = log cosh(− av + b2 ) + c2 , a a when a 6= 0, and h(u) = b1 u + c1 and `(v) = b2 v + c2 if a = 0, for suitable constants b1 , b2 , c1 , c2 ∈ R. h(u) =

Hint. Recall that arctanh x = (1/2) log((1 + x )/(1 − x )) for all x ∈ R. Exercise† 3.4.4. Show Corollary 3.4.7 (p. 193). Exercise† 3.4.5. Show Proposition 3.4.11 (p. 197). Exercise 3.4.6. Consider the Sherlock hat, image of the regular parametrized surface x : ]−1, 1[ × ]0, 2π [ → R3 given by
x(u, v) = (1 − u3 ) cos v, u, (1 − u3 ) sin v + 1 . Show that given a parabolic point x(0, v0 ) in the image of x, there are points x(u, v) in both half-spaces determined by T(0,v0 ) x.

Figure 3.27: “Elementary, my dear Watson!”

206  Introduction to Lorentz Geometry: Curves and Surfaces

Exercise 3.4.7. Let M ⊆ R3ν be a non-degenerate regular surface and (U, x) be a parametrization for M. (a) If N E and N L stand for the Euclidean and Lorentzian unit normal fields along M, respectively, and N E and N L have their last components with opposite signs, then show that hij,E = hij,L , where hij,E and hij,L denote the components of the Euclidean and Minkowski Second Fundamental Forms of x, respectively. (b) Conclude that the classification of points in a surface as elliptic, hyperbolic, parabolic or planar is independent of the ambient space where the surface is. Also conclude that if KE and K L denote the Gaussian curvatures of M equipped with the metrics from R3 and L3 , respectively, then KE ( p)K L ( p) ≤ 0, for every p ∈ M. Exercise 3.4.8. Consider the torus T2 in R3 generated by the revolution of the unit circle centered at (0, 2, 0) and contained in the plane x = 0, around the z-axis. Compute the Gaussian curvature of T2 and conclude that T2 has elliptic, hyperbolic and parabolic points. Sketch a picture. Exercise 3.4.9. Let M ⊆ R3ν be a non-degenerate regular surface and N be a Gauss map for M. (a) Show that if α : I → M is a regular curve in M such that N is constant along α, then all the points in α are parabolic or planar. (b) Show that if α, β : I → M are two regular curves in M, t0 ∈ I is such that . p = α(t0 ) = β(t0 ) and {α0 (t0 ), β0 (t0 )} are linearly independent, and N is constant along α and β, then p is planar. Exercise 3.4.10. Let M ⊆ R3ν be the graph of a smooth function f : U → R. Suppose that f depends smoothly only on the angle formed between the x-axis and the point (u, v) in the domain U, that is, that f (u, v) = g(arctan(v/u)) for a certain singlevariable smooth function g. √ 2 If r = u + v2 and θ = arctan(v/u), verify that f u = −vg0 (θ )/r2 , f v = ug0 (θ )/r2 , v2 00 2uv g ( θ ) + 4 g 0 ( θ ), 4 r r u2 0 v2 uv = − 4 g (θ ) + 4 g0 (θ ) − 4 g00 (θ ) r r r 2 u 2uv = 4 g00 (θ ) − 4 g0 (θ ), r r

f uu = f uv f vv

and

and use this to show that the critical surfaces equation boils down to g00 (θ ) = 0, in both ambient spaces. Determine, as done in Example 3.4.22 (p. 203), all the surfaces with this property.

Surfaces in Space  207

3.5

CURVES IN A SURFACE

In Chapter 2, we studied the theory of curves in space. At this point we may start investigating the relation between curves whose images are contained in a non-degenerate regular surface M ⊆ R3ν and the geometry of M itself. Naturally, the first step is to look for a relation between the curvature of curves in M with the curvatures of M itself. We start building this bridge by using the First and Second Fundamental Forms of M to introduce yet another notion of curvature. For each . p ∈ M, let Tp† M = {v ∈ Tp M | hv, vi 6= 0}. We have the: Definition 3.5.1 (Normal curvature). Let M ⊆ R3ν be a non-degenerate regular surface and N be a Gauss map for M. Given p ∈ M, the normal curvature of M at p is the function κn : Tp† M → R given by II p (v) . e κn (v) = . I p (v) Moreover, we say that a vector v ∈ Tp† M is asymptotic if κn (v) = 0. Remark.

• Note that for all v ∈ Tp† M and all non-zero λ ∈ R, we have that κn (λv) = κn (v), so that we may restrict ourselves to unit vectors and talk about the normal curvature at p along a general tangent direction (non-zero and non-degenerate) to M. • To justify the name “asymptotic vector” see Exercise 3.5.8. For a geometric interpretation, consider a unit speed admissible curve α : I → M in M. Using the first Frenet-Serret equation for α (that is, the definition of curvature of α), we have that

h−dN α(s) (α0 (s)), α0 (s)i κn (α (s)) = = eα h N (α(s)), α00 (s)i = eα κα (s)h N (α(s)), N α (s)i. hα0 (s), α0 (s)i 0

This way, if M ⊆ R3 , for each s ∈ I the angle θ (s) ∈ [0, 2π [ between N (α(s)) and N α (s) is well-defined, and the relation κn (α0 (s)) = κα (s) cos θ (s) holds. If s0 ∈ I is such that θ (s0 ) = 0 or π, then we have that |κn (α0 (s0 ))| = |κα (s0 )|. In general, we say that a normal section of M at p along a non-zero vector v ∈ Tp M is the intersection of M with the plane passing through p and spanned by N ( p) and v. That is, the above calculation tells us, in R3 , that if v is a unit vector, then the normal curvature of M at p along v and the curvature of the normal section of M at p along v are equal in absolute value, justifying the terminology adopted. If M ⊆ L3 is spacelike, then N (α(s)) is a timelike vector. If N α (s) is also timelike, with the same time direction as N (α(s)), we may consider the hyperbolic angle ϕ(s) between N (α(s)) and N α (s), and the relation κn (α0 (s)) = −κα (s) cosh ϕ(s) holds. Note that given s0 ∈ I, we have that ϕ(s0 ) = 0 if and only if κn (α0 (s0 )) = −κα (s0 ), in which case not only α works as a sort of normal section, but the vectors N (α(s)) and N α (s) are actually equal. Lastly, if M ⊆ L3 is timelike, then the causal type of α may be anything, and the interpretation to be given depends on the causal type of the Frenet-Serret trihedron of α. For instance, if N α (s) is spacelike and such that the plane it spans with N (α(s)) is also spacelike, the discussion goes exactly like what was done in R3 . Our conclusion from this analysis is the:

208  Introduction to Lorentz Geometry: Curves and Surfaces

Theorem 3.5.2 (Meusnier). Let M ⊆ R3ν be a non-degenerate regular surface. Then the curves in M which, at a point p ∈ M, have the same non-lightlike tangent line all have the same normal curvature at p. It is to be expected, when the Weingarten map of M is diagonalizable, that nice relations between the normal curvature and principal curvatures and direction will appear. Proposition 3.5.3 (Euler Formulas). Let M ⊆ R3ν be a non-degenerate regular surface, p ∈ M, and N be a Gauss map for M. Suppose that −dN p is diagonalizable and consider u1 , u2 ∈ Tp M the unit principal vectors associated to the principal curvatures κ1 ( p) and κ2 ( p), with u1 spacelike. Then: (i) if M is spacelike, every unit vector in Tp M may be uniquely written in the form v(θ ) = cos θ u1 + sin θ u2 for some θ ∈ [0, 2π [ and κn (θ ) ≡ κn (v(θ )) = κ1 ( p) cos2 θ + κ2 ( p) sin2 θ holds; (ii) if M is timelike, every unit spacelike vector in Tp M may be uniquely written in the form v( ϕ) = ± cosh ϕ u1 + sinh ϕ u2 and every unit timelike vector has the unique form w( ϕ) = sinh ϕ u1 ± cosh ϕ u2 for some ϕ ∈ R, where ± indicates exactly on which branch of the “unit hyperbola” the vector is. Then . κns.l. ( ϕ) = κn (v( ϕ)) = κ1 ( p) cosh2 ϕ − κ2 ( p) sinh2 ϕ, and . κnt.l. ( ϕ) = κn (w( ϕ)) = κ2 ( p) cosh2 ϕ − κ1 ( p) sinh2 ϕ hold. In particular, if p is umbilic, the normal curvature of M at p is constant.

u2 u2

u2 ϕ

θ

θ u1

(a) In R3 .

u1

u1

(b) In L3 , e M = −1.

(c) In L3 , e M = 1.

Figure 3.28: Unit vectors in Tp M. Proof: (i) If M is spacelike, we have κn (θ ) = h−dN p (cos θ u1 + sin θ u2 ) , cos θ u1 + sin θ u2 i

= hκ1 ( p) cos θ u1 + κ2 ( p) sin θ u2 , cos θ u1 + sin θ u2 i = κ1 ( p) cos2 θ + κ2 ( p) sin2 θ.

Surfaces in Space  209

(ii) If M is timelike, let’s do the first case: κns.l. ( ϕ) = h−dN p (± cosh ϕ u1 + sinh ϕ u2 ) , ± cosh ϕ u1 + sinh ϕ u2 i

= h±κ1 ( p) cosh ϕ u1 + κ2 ( p) sinh ϕ u2 , ± cosh ϕ u1 + sinh ϕ u2 i = κ1 ( p) cosh2 ϕ − κ2 ( p) sinh2 ϕ. The computation of κnt.l. ( ϕ) is similar, recalling that since w( ϕ) is timelike, the denominator in the definition of normal curvature is now −1.

Corollary 3.5.4. Let M ⊆ R3ν be a non-degenerate regular surface, p ∈ M, and N a Gauss map for M. If −dN p is diagonalizable, the critical points for the normal curvature of M at p are precisely the principal directions of M at p. And the normal curvatures along these directions at p are precisely the principal curvatures of M at p. Proof: Keep all notation from Proposition 3.5.3 above. Suppose that p is not umbilic, otherwise there is nothing to prove. If M is spacelike, we have κn (θ ) = κ1 ( p) cos2 θ + κ2 ( p) sin2 θ. To find the critical points of κn , note that κn0 (θ ) = (−κ1 ( p) + κ2 ( p)) sin 2θ vanishes if and only if θ = 0, π/2, π or 3π/2, and observe that κn (0) = κn (π ) = κ1 ( p) and κn (π/2) = κn (3π/2) = κ2 ( p). If M is timelike, let’s first make our search among spacelike vectors in the unit hyperbola, using κns.l. ( ϕ) = κ1 ( p) cosh2 ϕ − κ2 ( p) sinh2 ϕ. We have that
0 κns.l. ( ϕ) = (κ1 ( p) − κ2 ( p)) sinh 2ϕ vanishes if and only if ϕ = 0, and κns.l. (0) = κ1 ( p). The search among timelike vectors using the formula for κnt.l. ( ϕ) is similar, and one obtains κnt.l. (0) = κ2 ( p). Remark. In the above proof, the nature of the critical points is determined by the sign of the difference κ1 ( p) − κ2 ( p), and no critical point is an inflection point if p is not umbilic. Usually, the convention in R3 is to assume that κ1 ( p) ≤ κ2 ( p), by renaming the curvatures if necessary, but this is not possible in L3 since the principal vectors may have distinct causal types. The Euler formulas seen above may be used to give further interpretations of the mean curvature as an (either discrete or continuous) average of the normal curvatures, even when M is timelike and the “unit circle” in the tangent plane is no longer compact. See Exercises 3.5.1 and 3.5.2. We now introduce the first special type of curve in a surface, whose existence is intimately related to the diagonalization problem for the Weingarten map: Definition 3.5.5 (Lines of curvature). Let M ⊆ R3ν be a non-degenerate regular surface and N be a Gauss map for M. We’ll say that a curve α : I → M is a line of curvature of M if α0 (t) is an eigenvector of −dN α(t) , for all t ∈ I. It follows from what we have seen so far about the Weingarten map that given a non-umbilic point p ∈ M, there are at most two lines of curvature of M passing through p and, when M is spacelike, there are always exactly two. We also have a very concrete criterion to determine whether a given curve is or not a line of curvature, in terms of their coordinates relative to a parametrization of the surface:

210  Introduction to Lorentz Geometry: Curves and Surfaces

Proposition 3.5.6. Let M ⊆ R3ν be a non-degenerate regular surface, N be a Gauss map for M, and (U, x) be a parametrization for M. If α : I → M is given in coordinates by α(t) = x(u(t), v(t)), then α is a line of curvature of M if and only if: 0 2 ( v ) − u 0 v 0 ( u 0 )2 E = 0, para todo t ∈ I, F G e f g where u0 , v0 are evaluated in t, and the coefficients of the Fundamental Forms are evaluated in (u(t), v(t)). Remark.

• Clearly there is a pointwise version of the above result: if p = x(u0 , v0 ) and v = axu (u0 , v0 ) + bxv (u0 , v0 ), then v is a principal direction if and only if 2 b − ab a2 E F G = 0, e f g where the coefficients of the Fundamental Forms are evaluated in (u0 , v0 ).

• From here onwards, it will be convenient to denote derivatives taken with respect to t with dots, as usual in Physics. Proof: Let’s identify u ↔ u1 and v ↔ u2 , as usual. Omitting points of application, the chain rule gives us that α0 = ∑2i=1 u˙ i xi . If α is a line of curvature, there is a smooth λ : I → R with −dN (α0 ) = λα0 . We have that ! 2

−dN

∑ u˙ i xi

=

i =1

2

2

i =1

i,j=1

∑ u˙ i (−dN (xi )) = ∑

j

u˙ i h i x j ,

whence it follows (from linear independence) that 2

∑ u˙ i h i = λu˙ j , j

i =1

for 1 ≤ j ≤ 2. Using Lemma 3.3.7, this equality reads as 2

∑

u˙ i hik gkj = λu˙ j .

i,k =1

Back to index-free notation, this last expression is nothing more than the two following equations together:     f G − gF eG − f F 0 0 u + v0 λu = EG − F2 EG − F2     −eF + f E − f F + gE 0 0 λv = u + v0 . 2 2 EG − F EG − F Multiplying the first by v0 ( EG − F2 ), the second one by u0 ( EG − F2 ), and equating them, we have:

(eG − f F )u0 v0 + ( f G − gF )(v0 )2 = (−eF + f E)(u0 )2 + (− f F + gE)u0 v0 ,

Surfaces in Space  211

which simplifies as:

( f G − gF )(v0 )2 + (eG − gE)u0 v0 + (eF − f E)(u0 )2 = 0. This expression is precisely the Laplace expansion of the determinant given in the statement, through the first row. From these calculations, it is easy to see that the converse holds. To wit, all the steps are reversible since M is non-degenerate and thus EG − F2 6= 0. Remark. If the Weingarten map is diagonalizable, we know that the two principal directions are orthogonal at each point. Thus, the equation above gives us that a necessary and sufficient condition for the principal directions to be the coordinate curves of some parametrization x is that both F and f vanish. In this case, x is called a principal parametrization. Moreover, it follows from the general theory of Ordinary Differential Equations that if p ∈ M is not umbilic, there is a principal parametrization for M around p. Before we present examples, we’ll introduce the second special type of curve in a surface, more closely related to the normal curvature: Definition 3.5.7 (Asymptotic lines). Let M ⊆ R3ν be a non-degenerate regular surface. A non-lightlike regular curve α : I → M is called an asymptotic line of M if α0 (t) is an asymptotic vector for all t ∈ I. The first thing to investigate are conditions for the existence of asymptotic lines in a surface. Before that, the existence of asymptotic vectors in a given tangent plane Tp M is characterized by the following: Proposition 3.5.8. Let M ⊆ R3ν be a non-degenerate regular surface, N be a Gauss map for M, and p ∈ M. Suppose that −dN p is diagonalizable. Then there are asymptotic vectors in Tp† M if and only if (−1)ν K ( p) ≤ 0. Remark. In particular, given an asymptotic line α : I → M, if the Weingarten map of M is diagonalizable along α, it holds that (−1)ν K (α(t)) ≤ 0 for all t ∈ I. Proof: Observe that (−1)ν K ( p) ≤ 0 if and only if κ1 ( p) and κ2 ( p) have opposite signs when M is spacelike and equal signs when M is timelike. We then have the following cases to discuss: (i) If M is spacelike, the Euler formula is κn (θ ) = κ1 ( p) cos2 θ + κ2 ( p) sin2 θ. Suppose that (−1)ν K ( p) ≤ 0. If one of the principal curvatures vanishes, the corresponding principal direction is asymptotic. If both of the principal curvatures are non-zero, the equation κn (θ ) = 0 has solutions s κ ( p) θ = ± arctan − 1 . κ2 ( p ) Conversely, suppose that there is θ0 ∈ [0, π [ with κn (θ0 ) = 0. If θ0 = 0 or θ0 = π/2, then κ1 ( p) or κ2 ( p) vanishes. Otherwise κ1 ( p ) = − tan2 θ0 < 0. κ2 ( p )

212  Introduction to Lorentz Geometry: Curves and Surfaces

(ii) If M is timelike, it suffices to look for spacelike directions, for which the Euler formula is κns.l. ( ϕ) = κ1 ( p) cosh2 ϕ − κ2 ( p) sinh2 ϕ, as the situation for timelike directions is completely similar. Suppose that K ( p) ≥ 0. If κ2 ( p) = 0 it follows that κ1 ( p) = 0 as well, so that all directions are asymptotic. Otherwise, the equation κns.l. ( ϕ) = 0 has solution s κ1 ( p ) . κ2 ( p )

ϕ = ± arctanh

Conversely, suppose that there is ϕ0 ∈ R such that κns.l. ( ϕ0 ) = 0. If ϕ0 = 0, then κ1 ( p) = 0 and thus K ( p) ≥ 0. Otherwise, both the principal curvatures vanish, or neither of them do. If both vanish, the situation is trivial. If neither vanishes, we have κ1 ( p ) = tanh2 ϕ0 > 0. κ2 ( p )

As we have done for lines of curvature, let’s establish a criterion for identifying asymptotic lines in terms of their coordinates relative to a parametrization of the surface: Proposition 3.5.9. Let M ⊆ R3ν be a non-degenerate regular surface and (U, x) be a parametrization for M. If α : I → M is a regular curve given in coordinates by α(t) = x(u(t), v(t)), then α is an asymptotic line if and only if: e(u0 )2 + 2 f u0 v0 + g(v0 )2 = 0,

for all t ∈ I,

where the coefficients of the Second Fundamental Form are evaluated in (u(t), v(t)), and u0 , v0 are evaluated in t. Proof: As in the proof of Proposition 3.5.6 above, we have that α0 = ∑2i=1 u˙ i xi , and α is an asymptotic line if and only if h−dN (α0 ), α0 i = 0. However: * ! +

h−dN (α0 ), α0 i =

2

−dN

∑ u˙ i xi

i =1 2

=

∑

i,j=1 2

=

∑

* u˙ i u˙ j

2

∑

k =1

2

2

, ∑ u˙ j x j

=

j =1

+ hki xk , x j

i =1 j =1 2

=

2

∑ ∑ u˙ i u˙ j h−dN (xi ), x j i

∑

u˙ i u˙ j hki gkj

i,j,k=1

u˙ i u˙ j hij = e(u0 )2 + 2 f u0 v0 + g(v0 )2 .

i,j=1

Remark. Note that a necessary and sufficient condition for the coordinate curves of a parametrization x around a hyperbolic point p (which satisfies (−1)ν K ( p) ≤ 0) to be asymptotic lines is that both e and g vanish. To wit, if u0 = 0, the equation boils down to g(v0 )2 = 0, whence g = 0. Similarly, v0 = 0 implies that e = 0. On the other hand, if e = g = 0, the equation becomes f u0 v0 = 0, which is trivially satisfied by the coordinate curves. In this case x is called an asymptotic parametrization.

Surfaces in Space  213

At last, the long-awaited examples: Example 3.5.10. (1) Let’s consider again the cylinder with radius r > 0, S1 (r ) × R, parametrized by x : ]0, 2π [ × R → S1 (r ) × R ⊆ R3ν , given by x(u, v) = (r cos u, r sin u, v). Its lines of curvature are characterized by: 0 2 ( v ) − u 0 v 0 ( u 0 )2 ν = 0. r2 0 (− 1 ) −r 0 0 Expanding through the third row, one easily sees that the equation boils down to u0 v0 = 0, whence u0 = 0 or v0 = 0. So, in the first case, we have that u ≡ u0 and α(t) = (r cos u0 , r sin u0 , v(t)) is a vertical line in the cylinder, while in the second case we have that v ≡ v0 and α(t) = (r cos u(t), r sin u(t), v0 ), and α is a horizontal circle in the cylinder. The equation for asymptotic lines, in turn, is just −r (u0 )2 = 0, whence we conclude that the asymptotic lines in the cylinder are vertical lines. Observe that the conclusion was the same for both ambient spaces. (2) Consider the surface of revolution generated by an injective and unit speed curve α : I → R3ν , of the form α(u) = ( f (u), 0, g(u)), with f (u) > 0 for all u ∈ I, and the usual revolution parametrization x : I × ]0, 2π [ → R3ν , given by x(u, v) = ( f (u) cos v, f (u) sin v, g(u)). The equation of the lines of curvature, in both ambient spaces, is: 0 )2 0 v 0 ( u 0 )2 ( v − u eα 0 f 2 = 0 ⇐⇒ u0 v0 (eα f g0 + f 2 f 00 g0 − f 2 f 0 g00 ) = 0. − f 00 g0 + f 0 g00 0 f g0 There are at most two lines of curvature passing through each non-umbilic point. Clearly the coordinate curves are solutions to this equation and, thus, the meridians and parallels are the only lines of curvature passing through non-umbilic points of a surface of revolution. On the other hand, if x(u0 , v0 ) is umbilic, then eα f (u0 ) g0 (u0 ) + f (u0 )2 f 00 (u0 ) g0 (u0 ) − f (u0 )2 f 0 (u0 ) g00 (u0 ) = 0. The equation for the asymptotic lines is, in turn:

(− f 00 g0 + f 0 g00 )(u0 )2 + ( f g0 )(v0 )2 = 0, which is better studied in particular cases. (3) Consider the graph of a smooth function f : U ⊆ R2 → R, with the usual Monge parametrization x : U → R3ν , x(u, v) = (u, v, f (u, v)). The equation for the lines of curvature is: ( v 0 )2 −u0 v0 ( u 0 )2 1 + (−1)ν f 2 (−1)ν f u f v 1 + (−1)ν f 2 = 0. u v f uu f uv f vv Expanding this determinant gives us a non-linear ordinary differential equation, in general unmanageable. We can, however, study simpler cases, such as when f depends only on u or v.

214  Introduction to Lorentz Geometry: Curves and Surfaces

Suppose that f u = 0. With this we also have f uu = f uv = 0 and the equation boils down to f vv u0 v0 = 0, in both ambient spaces. If f vv = 0, all the curves will be lines of curvature (to wit, the graph of f is a plane). And if f vv 6= 0, we have u0 v0 = 0, whence u0 or v0 is zero, and the coordinate curves are the lines of curvature we’re looking for. The equation for the asymptotic lines, in both ambient spaces, is: f uu (u0 )2 + 2 f uv u0 v0 + f vv (v0 )2 = 0. Again, looking at the case where f u = 0, we have that the equation boils down to f vv (v0 )2 = 0. If f vv = 0, all the curves are asymptotic lines, and if f vv 6= 0, it follows that the coordinate curve v ≡ v0 is an asymptotic line. (4) Consider again the helicoid with the parametrization x : R × R≥0 → R3ν given by x(u, v) = (v cos u, v sin u, u). The equation for the lines of curvature is: ( v 0 )2 −u0 v0 (u0 )2 (−1)ν + v2 0 1 = 0. 0 1 0 In R3 , this becomes (v0 )2 − (u0 )2 (1 + v2 ) = 0, and then we see that v0 6= 0 (or else we would have u0 ≡ 0 as well, and the curve described by u and v would not be regular). Hence we may write: 1 du = ±√ , dv 1 + v2 using v as implicit parameter (more precisely, we invert v = v(t√ ) and use the chain rule). Integrating with respect to v, we obtain u(v) = ± ln |v + 1 + v2 | + c, where c ∈ R is a constant to be determined by prescribing a point in the surface on which the curve will pass through. We have to be careful when repeating this in L3 . To wit, in the strip 0 < v < 1 the helicoid is timelike and its Weingarten map is not diagonalizable. The equation for the lines of curvature in this case is (v0 )2 − (u0 )2 (−1 + v2 ) = 0, and a similar argument to the previous one (assuming v 6= 1 only) gives us that v0 6= 0, allowing us to use v as a parameter again. The equation: 

du dv

2

=

1 −1 + v2

is incompatible on 0 < v < 1. We’ll suppose then that v > 1, so that: p du 1 = ±√ =⇒ u(v) = ± ln |v + −1 + v2 | + c, dv −1 + v2 as before.

Surfaces in Space  215

Figure 3.29: The lines of curvature in a helicoid—the inner one in R3 and the outer one in L3 . The middle one is the horizon of causal type change in the surface. Observe that the Lorentzian line of curvature does not enter the timelike region of the surface, as expected. The analysis of the asymptotic lines is immediate, once the equation becomes just u0 v0 = 0, whence we conclude that the asymptotic lines are the coordinate curves, in both ambient spaces. Example 3.5.11 (“Funnel surface”). In the last example above, we have seen that the parametrization given for the helicoid is asymptotic. In general, this is not the case, but when we do have asymptotic directions, we may integrate the equation for the asymptotic lines to obtain an asymptotic parametrization. We will illustrate this procedure by producing an asymptotic parametrization to the graph M,pseen in R3 , of the radial function h : R2 \ {(0, 0)} → R, given by h( x, y) = log x2 + y2 . Instead of starting with the usual Monge parametrization, consider instead x : R>0 × ]0, 2π [ → M defined by x(r, θ ) = (r cos θ, r sin θ, log r ). For this parametrization, we obtain

−1 , e(r, θ ) = √ r 1 + r2

f =0

and

g(r, θ ) = √

r 1 + r2

,

whence the equation for the asymptotic lines boils down to

−

r 0 ( t )2 + r (t)θ 0 (t)2 = 0. r (t)

This leads us to two equations, θ 0 (t) = ±r 0 (t)/r (t), whence θ (t) + 2u = log r (t) and θ (t) + 2v = − log r (t), for certain constants of integration u and v, which will be the new parameters. Solving for u and v in terms of r and θ, we obtain a new parametrization

. y(u, v) = x eu−v , −u − v = eu−v cos(u + v), −eu−v sin(u + v), u − v , which is asymptotic. You may verify this in Exercise 3.5.10.

216  Introduction to Lorentz Geometry: Curves and Surfaces

Figure 3.30: The funnel surface. To geometrically understand the whole situation, it is interesting to take the point of view of an inhabitant of M: more precisely, if α : I → M is regular, we know that the velocity vector α0 (t) is always tangent to M at α(t), but in general we do not know anything about the acceleration vector α00 (t). Since M is non-degenerate, we always have an orthogonal direct sum decomposition R3ν = Tα(t) M ⊕ ( Tα(t) M )⊥ , which allows us to write . Dα0 α00 (t) = α00 (t)tan + α00 (t)nor = (t) + IIα(t) (α0 (t)), dt where ( Dα0 /dt)(t) ∈ Tα(t) M is the so-called covariant derivative of α0 at t, being exactly the part of the acceleration of α which is “detected” by an inhabitant of M. This way, asymptotic lines in M are the curves for which an inhabitant of M feels the full acceleration of motion. To wit, κn (α0 (t)) = 0 directly implies that IIα(t) (α0 (t)) = 0, so that α00 (t) = ( Dα0 /dt)(t) is tangent to M. Remark. We will see in the next section the dual notion, of a geodesic: the curves for which an inhabitant of M does not feel any acceleration, that is, curves satisfying Dα0 /dt = 0. In a general surface, geodesics play the role that straight lines play in a plane, and they are of fundamental importance in Differential Geometry. In particular, the above discussion tells us that if a line is contained in a surface, it is simultaneously asymptotic and geodesic (in fact, the converse also holds).

Exercises Exercise 3.5.1 (Averages). Let M ⊆ R3ν be a spacelike surface and p ∈ M. Adopting the notation from Proposition 3.5.3 (p. 208), show that: (a) H ( p) = (−1)ν

κn (θ ) + κn (θ + π/2) , for each θ ∈ [0, 2π [; 2

Surfaces in Space  217

(b) H ( p) =

(−1)ν m

m

.

∑ κn (θk ), where m ≥ 2 and θk = 2kπ/m;

k =1

Hint. First show that ∑m k =1 cos(2θk ) = 0. Use complex numbers and look at a certain geometric progression.

(−1)ν (c) H ( p) = 2π

Z 2π 0

κn (θ ) dθ.

Exercise 3.5.2 (More averages). Even though the functions cosh and sinh are not periodic, it is possible to obtain similar results to the ones in the previous exercise. Let M ⊆ L3 be a timelike surface and p ∈ M. Assuming that −dN p is diagonalizable and adopting the notation from Proposition 3.5.3, show that: 1 m s.l. κn ( ϕk ) + κnt.l. ( ϕk ), where m ≥ 2 and ϕ1 , . . . , ϕm ∈ R are arbitrary m k∑ =1 parameters;

(a) H ( p) = −

(b) H ( p) = −

1 2ϕ

Z ϕ 0

κns.l. (τ ) + κnt.l. (τ ) dτ, for all ϕ > 0.

Exercise 3.5.3. Show that if M ⊆ R3ν is a spacelike surface and p ∈ M is hyperbolic, then the principal directions at p bisect the asymptotic ones. What does the situation look like if M is timelike? Exercise 3.5.4. Show that the meridians of a torus in R3 are lines of curvature. Remark. You may consider, for concreteness, the torus from Exercise 3.4.8 (p. 206), to be more precise. Exercise 3.5.5. Consider the paraboloid M ⊆ R3 , graph of the smooth function f : R2 → R, f (u, v) = u2 + v2 . Show that the origin 0 is an umbilic point of M with the following property: for every non-zero v ∈ T0 M, there is a line of curvature α : I → M with α(0) = 0 and α0 (0) = v. Exercise 3.5.6. Let M1 , M2 ⊆ R3ν be non-degenerate regular surfaces, N 1 and N 2 be Gauss maps for M1 and M2 , and α : I → M1 ∩ M2 be a curve in both surfaces. Show that if h N 1 (α(t)), N 2 (α(t))i 6= ±1 independent of t, then α is a line of curvature for one of the surfaces if and only if it is for the other one as well. Hint. By symmetry, it suffices to show that if α is a line of curvature for M1 , then it is also one for M2 . Write d( N 2 )α(t) (α0 (t)) as a linear combination of α0 (t), N 1 (α(t)) and N 2 (α(t)) (the assumptions ensure that this is possible). Remark. This happens, for instance, if the surfaces are in R3 and the normal directions form a constant angle along α. Being more precise, the exercise treats the situation where these normal directions are not parallel (because if this is the case, the result trivially holds). Exercise 3.5.7. Let M ⊆ R3ν be a non-degenerate regular surface with non-vanishing Gaussian curvature, N be a Gauss map for M, and (U, x) a principal parametrization . for M. Show that y : U → R3ν defined by y(u, v) = N ( x(u, v)) is a regular parametrized surface whose First Fundamental Form coefficients are given by E(y) = κ12 E( x) , F (y) = 0 and G (y) = κ22 G ( x) , where κi ≡ κi ◦ x are the principal curvatures of M.

218  Introduction to Lorentz Geometry: Curves and Surfaces

Exercise† 3.5.8 (Dupin Indicatrix). Let M ⊆ R3ν be a non-degenerate regular sur. face and (U, x) be a parametrization for M around p = x(u0 , v0 ). Assume that p is non-planar and, when (−1)ν K ( p) ≥ 0, choose the normal N in such a way that h11 (u0 , v0 ), h22 (u0 , v0 ) ≥ 0. (a) We define the Dupin indicatrix of M at p as the conic in Tp M given by the equation e II p (v) = 1. Show that if p is an elliptic (resp. hyperbolic, parabolic) point, then the Dupin indicatrix of M at p is an ellipse (resp. hyperbola, pair of parallel lines). (b) Show that if p is hyperbolic, then the asymptotes of the Dupin indicatrix of M at p are given by the equation e II p (v) = 0, that is, the collection of asymptotic vectors of M at p. Exercise 3.5.9. Determine, in both ambient spaces, the lines of curvature and the asymptotic lines for the graph of the function h : R2 → R, h( x, y) = xy. Exercise 3.5.10. Verify that the parametrization y obtained in Example 3.5.11 (p. 215) is indeed asymptotic. Exercise 3.5.11. Let h : R≥0 → R be a smooth function. Show that the equation for the asymptotic lines of a: (a) polar parametrization x : U → R3ν of the form x(r, θ ) = (r cos θ, r sin θ, h(r )), is h00 (r (t))r 0 (t)2 + h0 (r (t))r (t)θ 0 (t)2 = 0, in both ambient spaces; (b) Rindler parametrization x : U → L3 of the form x(ρ, ϕ) = (h(ρ), ρ cosh ϕ, ρ sinh ϕ), is h00 (ρ(t))ρ0 (t)2 − h0 (ρ(t))ρ(t) ϕ0 (t)2 = 0. Exercise 3.5.12. Repeat the strategy adopted in Example 3.5.11 (p. 215) to deduce that   √ √ √ y(u, v) = e(u−v)/ 1−2α cos(u + v), −e(u−v)/ 1−2α sin(u + v), e2α(u−v)/ 1−2α is an asymptotic parametrization for the graph, in R3 , of the smooth function h : R2 → R given by h( x, y) = ( x2 + y2 )α , where 0 6= α < 1/2. Hint. Start with a polar parametrization and let the constants of integration obtained from the differential equation be denoted by 2u and 2v instead of u and v, to simplify the notation later. Exercise 3.5.13 (Beltrami-Enneper). Let M ⊆ R3ν be a non-degenerate regular surface and α : I → M be an admissible unit speed asymptotic. Show that the Gaussian curvature along α is given by K (α(s)) = (−1)ν+1 τα (s)2 for all s ∈ I. Hint. If α is asymptotic, then α00 (s) is always tangent to M and thus Tα(s) M is always spanned by T α (s) and N α (s). Moreover, up to sign Bα (s) is a unit normal to M along α. Compute τα in terms of e II p and K (α(s)) by using the Gauss equation given in Exercise 3.3.1 (p. 187).

Surfaces in Space  219

3.6

GEODESICS, VARIATIONAL METHODS AND ENERGY

Following the conclusion from the discussion in the end of the previous section, we’ll start registering a few definitions: Definition 3.6.1. Let M ⊆ R3ν be a non-degenerate regular surface and α : I → M be a regular curve. A vector field along α is a smooth map V : I → R3ν such that for all t ∈ I, V (t) ∈ Tα(t) M. We’re interested in knowing how vector fields change along α, but from the point of view of an inhabitant of M. This motivates the: Definition 3.6.2. Let M ⊆ R3ν be a non-degenerate regular surface, α : I → M be a regular curve and V : I → R3ν be a vector field along α. The covariant derivative of V along α is defined by DV . (t) = projT M V 0 (t) ≡ V 0 (t)tan . α(t) dt We’ll also say that V is parallel if DV /dt = 0. Remark.

• In this definition it is crucial that M is non-degenerate, or else we cannot consider projections in the tangent planes. • Some basic properties of the operator D/dt are listed in Exercise 3.6.1. Example 3.6.3. (1) Our simplest example of a regular surface is a plane. Let Π = { p ∈ R3ν | h p, ni = c} be a non-degenerate plane. For each p ∈ Π, we have that the tangent plane is Tp Π = {v ∈ R3ν | hv, ni = 0}. If α : I → Π is any regular curve and V : I → R3ν is a vector field along α, differentiating the relation hV (t), ni = 0, we obtain the relation hV 0 (t), ni = 0, whence V 0 (t) ∈ Tα(t) Π for all t ∈ I. In other words, we see that in this case DV ( t ) = V 0 ( t ), dt so that the covariant derivative may be seen as a generalization of the usual derivative, that now takes into account the geometry of the surface on where the base curve lies. (2) Now let M be S2 , S21 or H2 , on the adequate ambient spaces. Let α : I → M be a regular curve and V : I → R3ν be a vector field along α. Since a Gauss map for M is the position vector field itself, we have that DV hV 0 (t), α(t)i (t) = V 0 (t) − α(t) dt hα(t), α(t)i = V 0 (t) − e M hV 0 (t), α(t)iα(t)

= V 0 (t) + e M hV (t), α0 (t)iα(t), where the last equality follows from the fact that hV (t), α(t)i = 0 holds for all t ∈ I. In particular, if α has unit speed we conclude that Dα0 (s) = α00 (s) + e M eα α(s). ds

220  Introduction to Lorentz Geometry: Curves and Surfaces

In the most general case, when M is a non-degenerate regular surface, we have a correction factor given precisely by the Second Fundamental Form: V 0 (t) =


DV (t) + IIα(t) α0 (t), V (t) . dt

The main example of a vector field we should keep in mind for now, already mentioned in item (2) above, is the velocity field of α, α0 : I → R3ν itself. Definition 3.6.4 (Geodesics). Let M ⊆ R3ν be a non-degenerate regular surface. We’ll say that a curve α : I → M is a geodesic if Dα0 /dt = 0 or, equivalently, if α00 (t) is normal to Tα(t) M, for all t ∈ I. A basic fact about geodesics that must be immediately registered is the: Proposition 3.6.5. Let M ⊆ R3ν be a non-degenerate regular surface, and α : I → M be a geodesic. Then α has constant speed. Proof: It suffices to note that since α00 (t) is always normal to M and α0 (t) is always tangent, then d 0 hα (t), α0 (t)i = 2hα00 (t), α0 (t)i = 0. dt Remark. In particular, we have that geodesics have constant causal character. Moreover, we’ll say that a geodesic is normalized if it has unit speed. Example 3.6.6. It follows from item (1) in Example 3.6.3 above that the geodesics in any non-degenerate plane Π are precisely the lines contained in it. To wit, every geodesic α satisfies α00 (t) = ( Dα0 /dt)(t) = 0. Example 3.6.7. Consider the cylinder S1 (r ) × R ⊆ R3ν , with the usual parametrization x : ]0, 2π [ × R → S1 (r ) × R given by x(u, v) = (r cos u, r sin u, v). Let α : I → S1 (r ) × R be a curve given in coordinates as α(t) = x(u(t), v(t)) = (r cos u(t), r sin u(t), v(t)). Differentiating twice, we obtain   α00 (t) = − ru00 (t) sin u(t) − ru0 (t)2 cos u(t), ru00 (t) cos u(t) − ru0 (t)2 sin u(t), v00 (t)     = − ru00 (t) sin u(t), ru00 (t) cos u(t), v00 (t) − ru0 (t)2 cos u(t), sin u(t), 0 . Noting that a Gauss map for the cylinder, in both ambient spaces, is given by N ( x(u, v)) = (cos u, sin u, 0), we conclude that Dα0 (t) = (−ru00 (t) sin u(t), ru00 (t) cos u(t), v00 (t)). dt This way, we see that α is a geodesic if and only if u00 (t) = v00 (t) = 0, that is, if u(t) = at + b and v(t) = ct + d. So the geodesics of the cylinder are given by α(t) = (r cos( at + b), r sin( at + b), ct + d), for certain constants a, b, c, d ∈ R. In particular, note that the geodesics are the same, no matter the ambient space where the cylinder is. There are four possibilities:

Surfaces in Space  221

• if a = c = 0, the curve collapses to a point; • if a 6= 0 and c = 0, we have horizontal circles; • if a = 0 and c 6= 0, we have vertical lines; • if a 6= 0 and c 6= 0, we have helices.

Figure 3.31: Geodesics in the cylinder. We’ll leave more examples for later, when we have more powerful variational techniques at our disposal. Such techniques are also useful and interesting on their own. Moreover, we will see a much larger class of examples when we discuss abstract metrics. Occasionally, the following notion is also useful: Definition 3.6.8 (Pre-geodesics). Let M ⊆ R3ν be a non-degenerate regular surface. We’ll say that a curve α : I → M is a pre-geodesic if α admits a reparametrization which is a geodesic. In other words, a pre-geodesic is nothing more than a geodesic, up to reparametrization. Pre-geodesics are characterized in the following: Proposition 3.6.9. Let M ⊆ R3ν be a non-degenerate regular surface, and α : I → M be a regular curve. Then α is a pre-geodesic if and only if ( Dα0 /dt)(t) and α0 (t) are collinear, for all t ∈ I. Proof: Suppose initially that α is a pre-geodesic and let h : J → I be a diffeomorphism . such that the curve β = α ◦ h : J → M is a geodesic. Differentiating twice the relation β(s) = α(h(s)) with respect to s and setting t = h(s) we obtain β00 (s) = h0 (h−1 (t))2 α00 (t) + h00 (h−1 (t))α0 (t). Projecting the above in the tangent plane Tβ(s) M = Tα(t) M, we have that 0=

Dβ0 Dα0 (s) = h0 (h−1 (t))2 (t) + h00 (h−1 (t))α0 (t), ds dt

222  Introduction to Lorentz Geometry: Curves and Surfaces

whence

h00 (h−1 (t)) 0 Dα0 ( t ) = − 0 −1 α (t) dt h (h (t))2

are proportional for all t ∈ I, as wanted. Conversely, suppose that we have ( Dα0 /dt)(t) = f (t)α0 (t), for some smooth function f : I → R. Let’s see what condition the diffeomorphism h must satisfy for α ◦ h to be a geodesic, and use such condition to try and define it. Well, the computation done above shows that
h0 (s)2 f (h(s)) + h00 (s) α0 (t) = 0 must hold for all s in the domain of h, which by regularity of α implies that h00 (s) + f (h(s)) h0 (s)2 = 0

R for all s. We then look for a solution of this differential equation. Define F (t) = f (t) dt R and g(t) = eF(t) dt. Since g0 (t) = eF(t) > 0, g is a positive diffeomorphism of I onto . . J = g( I ), and so we may consider h = g−1 : J → I. Let’s show that h is a solution for the latter ODE (which ensures that α ◦ h is a geodesic). Set t = h(s) and s = g(t), as before. Differentiating h( g(t)) = t with respect to t twice gives us the relations h0 (s) =

1 0 g (t)

and h00 (s) = −

g00 (t) . g 0 ( t )3

Moreover, g00 (t) = f (t)eF(t) , and with this we have: g00 (t) f (t) + 0 2 g 0 ( t )3 g (t) 1 = 0 3 (− g00 (t) + f (t) g0 (t)) g (t) 1 = 0 3 (− f (t)eF(t) + f (t)eF(t) ) = 0, g (t)

h00 (s) + f (h(s))h0 (s)2 = −

as wanted. Remark. It also follows from this that pre-geodesics with non-zero constant speed are, in fact, geodesics. To wit, differentiating the relation hα0 (t), α0 (t)i = c 6= 0 gives us that hα00 (t), α0 (t)i = 0. Discarding the normal component of α00 (t) we obtain   Dα0 Dα0 0= (t), α0 (t) = c f (t) =⇒ f (t) = 0 =⇒ = 0. dt dt On the other hand, what we’ll do in what follows motivates a relatively simple proof that every lightlike curve is a pre-geodesic (see Exercise 3.6.10). 3.6.1

Darboux-Ribaucour frame

In Chapter 2, our main tool in the study of curves was the Frenet-Serret trihedron: a frame for R3ν adapted to the curve. To simultaneously study lines of curvature, asymptotic lines and geodesics, we’ll introduce a new trihedron, adapted not only to the curve itself, but also to the surface where the curve lies. Let M ⊆ R3ν be a non-degenerate regular surface, N be a Gauss map for M, and α : I → M be a unit speed admissible curve. Motivated by the Frenet-Serret trihedron, but this time requiring the unit normal field to M along α to play the role of the binormal

Surfaces in Space  223


field Bα , we look for a third element V α such that T α (s), V α (s), N (α(s)) is a positive orthonormal basis for R3ν , for all s ∈ I.

N (α(s)) T α (s)

V α (s) α(s)

M

Figure 3.32: Constructing a frame adapted to M. The orientability analysis done in Section 1.6 tells us that . V α (s) = (−1)ν+1 eα e M T α (s) × N (α(s)) is the vector we are looking for. Definition 3.6.10 (Darboux-Ribaucour Trihedron). Let M ⊆ R3ν be a non-degenerate regular surface, N be a Gauss map for M, and α : I → M be a unit speed admissible curve. The Darboux-Ribaucour frame of α at s ∈ I is the positive orthonormal basis
T α (s), V α (s), N (α(s)) constructed above. The next step, naturally, is to express the derivatives of those vectors as combinations of the vectors themselves, via orthonormal expansions. As before, some coefficients cannot be expressed in terms of objects already known, and so they will be baptized: Definition 3.6.11 (Geodesic curvature and torsion). Let M ⊆ R3ν be a non-degenerate regular surface, N be a Gauss map for M, and α : I → M be a unit speed admissible curve. (i) The geodesic curvature of α at s is the component of α00 (s) = T 0α (s) in the direction of V α (s), and it is denoted by κ g,α (s). (ii) The geodesic torsion of α at s is the component of V 0α (s) in the direction of N (α(s)), and it is denoted by τg,α (s). Remark.

• We may express κ g,α and τg,α in terms of projections: Dα0 (s) = κ g,α (s)V α (s) and proj N (α(s)) V 0α (s) = τg,α (s) N (α(s)), ds

224  Introduction to Lorentz Geometry: Curves and Surfaces

where in the first expression T 0α (s) = α00 (s) could be replaced by the covariant derivative of α0 , since V α (s) is always tangent to M. In particular, we’ll have that   Dα0 Dα0 2 ν ( s ), (s) . κ g,α (s) = (−1) eα e M ds ds

• Since α has constant speed, T 0α (s) has no component in the direction of T α (s), so that the geodesic curvature effectively measures how much T 0α (s) deviates from being normal to M and, hence, how much α deviates from being a geodesic. • The definition of geodesic torsion is meant to mimic what we have for torsion, when studying the Frenet-Serret trihedron in Chapter 2, with V α and N ◦ α playing the roles of N α and Bα , respectively. We are then ready to state and establish the: Theorem 3.6.12 (Darboux equations). Let M ⊆ R3ν be a non-degenerate regular surface, N be a Gauss map for M, and α : I → M be a unit speed admissible curve. We have that:      0 κ g,α (s) eα e M κ n ( s ) T α (s) T 0α (s)  V 0α (s)  = (−1)ν+1 e M κ g,α (s) 0 τg,α (s)   V α (s)  , 0 ν + 1 N (α(s)) ( N ◦ α) (s) −κ n ( s ) (−1) eα τg,α (s) 0 where we abbreviate the normal curvature of M at α(s) by κn (α0 (s)) ≡ κn (s). Proof: We’ll repeatedly use orthonormal expansion. Since all the vectors in the DarbouxRibaucour trihedron are unit vectors, it follows that the diagonal of the matrix of coefficients indeed consists of only zeros. The component of T 0α (s) in the direction of N (α(s)) is e M h T 0α (s), N (α(s))i = e M h T α (s), −( N ◦ α)0 (s)i

= e M h T α (s), −dN α(s) ( T α (s))i = eα e M κ n ( s ), and thus we have obtained the first equation. For the second equation, it only remains to compute the component of V 0α (s) in the direction of T α (s), observing that by definition of geodesic curvature, we have that κ g,α (s) = (−1)ν eα e M h T 0α (s), V α (s)i. We then have that eα hV 0α (s), T α (s)i = −eα hV α (s), T 0α (s)i = (−1)ν+1 e M κ g,α (s), and so we have obtained the second equation. For the last equation, note that from the definition of geodesic torsion we have τg,α (s) = e M hV 0α (s), N (α(s))i. So, the component of ( N ◦ α)0 (s) in the direction of T α (s) is eα h( N ◦ α)0 (s), T α (s)i = −eα h−dN α(s) ( T α (s)), T α (s)i = −κn (s), while the component in the direction of V α (s) is

(−1)ν eα e M h( N ◦ α)0 (s), V α (s)i = (−1)ν+1 eα e M h N (α(s)), V 0α (s)i = (−1)ν+1 eα τg,α (s).

Surfaces in Space  225

With these equations in place, we may decompose the curvature of α, seen as a curve in R3ν , into a component tangent to M and into a component normal to M. In other words, the geodesic curvature is the part of the curvature of α detected by an inhabitant of M. Or yet: in M, a geodesic α is “straight”, and so the curvature of α in R3ν is forced by the curvature of M in R3ν . These more qualitative interpretations are formalized in the: Corollary 3.6.13 (Pythagoras). Let M ⊆ R3ν be a non-degenerate regular surface, N be a Gauss map for M, and α : I → M be a unit speed admissible curve. Then ηα κα (s)2 = (−1)ν eα e M κ g,α (s)2 + e M κn (s)2 , for all s ∈ I. Proof: It suffices to apply h·, ·i in both sides of the first Darboux equation, bearing in mind that T 0α (s) = κα (s) N α (s) and that κn (s) = eα κα (s)h N (α(s)), N α (s)i. Remark. In particular note that, in R3 , the above formula boils down to κα2 = κ 2g,α + κn2 . The usefulness of this corollary is to provide a practical method for computing κ g,α , when necessary. And lastly, the geodesic torsion “completes the toolbox” provided by the DarbouxRibaucour trihedron: Theorem 3.6.14. Let M ⊆ R3ν be a non-degenerate regular surface, and α : I → M be a unit speed admissible curve. Then:   α is a geodesic ⇐⇒ κ g,α ≡ 0; α is an asymptotic line ⇐⇒ κn ≡ 0;   α is a line of curvature ⇐⇒ τg,α ≡ 0. Remark. It is not complicated to redo the construction of the Darboux-Ribaucour trihedron for curves not necessarily having unit speed. The Pythagorean relation still holds without modifications, but “geodesic” has to be replaced by “pre-geodesic” in the criterion above. The proof of this last theorem is direct and we ask you to carry it out in Exercise 3.6.5.

Exercises Exercise 3.6.1. Let M ⊆ R3ν be a non-degenerate regular surface, α : I → M be a regular curve, V , W : I → R3ν be vector fields along α, and f : I → R be a smooth function. Show that for all t ∈ I the following hold: (a)

D (V + W ) DV DW (t) = (t) + ( t ); dt dt dt

(b)

D( f V ) DV ( t ) = f 0 ( t )V ( t ) + f ( t ) ( t ). dt dt

226  Introduction to Lorentz Geometry: Curves and Surfaces

Exercise 3.6.2. Let M ⊆ R3ν be a non-degenerate regular surface, α : I → M be a regular curve, and V , W : I → R3ν be vector fields along α. Show that if V and W are both parallel vector fields along α, then hV (t), W (t)i is constant. Conclude that parallel fields have constant speed and causal character. Remark. In particular, this shows that if M is spacelike, parallel vector fields always make a constant angle along α. Exercise 3.6.3. Let M ⊆ R3ν be a non-degenerate regular surface, and α : I → M be a . (regular) geodesic. Show that if h : J → I is a diffeomorphism such that β = α ◦ h is also a geodesic, then h has the form h(s) = as + b, for certain a, b ∈ R, a 6= 0. Exercise† 3.6.4. Let M ⊆ R3ν be a non-degenerate regular surface, α : I → M be a unit speed admissible curve, and V α : I → R3ν defined by V α (s) = (−1)ν+1 eα e M T α (s) × N (α(s)).
Verify that T α (s), V α (s), N (α(s)) is a positive basis for R3ν for all possibilities of causal types for α and M. Exercise 3.6.5. Prove Theorem 3.6.14 (p. 225). Exercise 3.6.6. Let M ⊆ R3ν be a non-degenerate regular surface, N be a Gauss map . for M, α : I → M be a unit speed admissible curve, and p = α(s0 ) ∈ M. Show that if β : I → M given by . β(s) = α(s) − proj N ( p) (α(s) − p) is the orthogonal projection of α onto the affine tangent plane p + Tp M, then we have that |κ g,α (s0 )| = κ β (s0 ). Exercise 3.6.7. Let M ⊆ R3ν be a non-degenerate regular surface, N be a Gauss map for M, and α : I → M be a unit speed admissible curve. Show that α is an asymptotic line for M if and only if for each s ∈ I the osculating plane to α at s is precisely Tα(s) M. Exercise 3.6.8. Let M ⊆ R3ν be a non-degenerate regular surface, N be a Gauss map for M, and α : I → M be a unit speed admissible curve. Show that: (a) α is simultaneously a geodesic and a line of curvature if and only the image of α is contained in a plane normal to M along α; (b) α is simultaneously a line of curvature and an asymptotic line if and only if the image of α is contained in a plane tangent to M along α. Exercise 3.6.9. Let M ⊆ R3ν be a non-degenerate regular surface and α : I → M be a unit speed admissible curve. (a) Show that if α is a plane geodesic which is not a straight line, then α is a line of curvature. Hint. Fix a Gauss map N for M, consider the Darboux frame for α, and show that the normal v to the plane containing the image of α satisfies v = λ(s)V α (s) for all s ∈ I, where λ : I → R is a non-vanishing smooth function. (b) Give an example of a line of curvature which is a plane curve but not a geodesic.

Surfaces in Space  227

Exercise 3.6.10. Let M ⊆ L3 be a timelike surface and α : I → M be a lightlike curve. Show that α is a pre-geodesic. Hint. For all t ∈ I, the vector N (α(t)) is spacelike, and so you may write ( Dα0 /dt)(t) as a linear combination of α0 (t) and N (α(t)) × L α0 (t). Exercise 3.6.11. Let M ⊆ R3ν be a non-degenerate regular surface and (U, x) be a parametrization for M. Show that: (a) The coordinate curve x(u, v0 ), with v0 constant, is a geodesic if and only if Eu (u, v0 ) = 0 and Ev (u, v0 ) = 2Fu (u, v0 ) for all u in its domain. (b) The coordinate curve x(u0 , v), with u0 constant, is a geodesic if and only if Gv (u0 , v) = 0 and Gu (u0 , v) = 2Fv (u0 , v) for all v in its domain. Hint. Compute the products h xuu , xu i, etc., in a convenient way. Exercise 3.6.12. Let f : U ⊆ R2 → R be a smooth function with the symmetry f (u, v) = f (u, −v) for all (u, v) ∈ U (of course, we also assume that U is symmetric about the u-axis) and consider the usual Monge parametrization x : U → gr( f ) ⊆ R3ν , x(u, v) = (u, v, f (u, v)). Show that the coordinate curve v = 0 is a pre-geodesic. Exercise 3.6.13 (Horocycles, again). Let v ∈ L3 be a future-directed lightlike vector and c < 0. We have seen in Exercise 3.1.11 (p. 153) that the horocycle Hv,c ⊆ H2 admits a parametrization α : R → Hv,c for the form α(s) = −

s2 v + sw1 + w2 , 2c

where w1 and w2 are orthogonal unit vectors, with w1 spacelike, w2 timelike, w1 orthogonal to v and hw2 , vi L = c. (a) Show that the geodesic curvature of α is constant, κ g,α = 1. (b) The following converse holds: if α : I → H2 ⊆ L3 is a unit speed curve with constant geodesic curvature equal to 1, then the image of α is contained in some horocycle Hv,c . Hint. Show that in these conditions, α is semi-lightlike. By Theorem 2.3.32 (p. 123), the image of α is contained in an (affine) lightlike plane. Take an orthogonal basis {v, w1 } for this plane and proceed from there. Exercise 3.6.14. Compute the geodesic curvature of a circle of “latitude” u = u0 in S2 and S21 (use the usual parametrizations of revolution). Exercise 3.6.15. Find all the curves in S2 with constant geodesic curvature. Exercise 3.6.16. Let M ⊆ R3ν be a non-degenerate regular surface, N be a Gauss map for M, p ∈ M such that −dN p is diagonalizable, and α : I → M be a unit speed curve such that α(0) = p. Assume that the principal vectors u1 , u2 ∈ Tp M are such that u1 is spacelike and (u1 , u2 , N ( p)) is a positive basis for R3ν . (a) If M is spacelike, show that τg,α (0) =

(−1)ν (κ2 ( p) − κ1 ( p)) sin 2θ, 2

where θ is the (oriented) angle between the principal vector u1 and α0 (0).

228  Introduction to Lorentz Geometry: Curves and Surfaces

(b) If M is timelike, show that 1 τg,α (0) = ± (κ2 ( p) − κ1 ( p)) sinh 2ϕ, 2 no matter whether the velocity vector α0 (0) = ± cosh ϕ u1 + sinh ϕ u2 is spacelike or α0 (0) = sinh ϕ u1 ± cosh ϕ u2 is timelike. That is, the conclusion here is that the geodesic torsion of a curve in a given instant depends only on the position of the curve and on its velocity vector in this instant. Hint. Recall the proof of Euler’s Formula, and the diagrams for cross products (p. 56, Chapter 1) may be useful.

3.6.2

Christoffel symbols

We concluded the previous subsection by discussing the Darboux-Ribaucour trihedron, yet another moving frame associated to a curve. But, if M ⊆ R3ν is a non-degenerate regular surface, N is a Gauss map for M, and (U, x) is a parametrization for M, we have an “obvious” trihedron: the basis for Tx(u,v) M associated to the parametrization x gives us, for each (u, v) ∈ U, a basis { xu (u, v), xv (u, v), N ( x(u, v))} for R3ν . With this, we may express any vector as a combination of xu , xv and N ≡ N ◦ x, at adequate points. In particular, the second order derivatives of the parametrization x itself. Omitting points of application, and identifying u ↔ 1, v ↔ 2, we write: 2

xij =

∑ Γijk xk + e M hij N,

1 ≤ i, j ≤ 2,

k =1

where the hij are precisely the coefficients of the Second Fundamental Form of the parametrization x. Definition 3.6.15 (Christoffel Symbols). Keeping the above notation, the functions Γijk : U → R, 1 ≤ i, j, k ≤ 2, are called the Christoffel symbols of (U, x). A priori, the Christoffel symbols depend on the second order derivatives of the parametrization, which are not necessarily tangent to the surface. The next result allows us to express them only in terms of the First Fundamental Form: Proposition 3.6.16. If M ⊆ R3ν is a non-degenerate regular surface and (U, x) is a parametrization for M, the Christoffel symbols of x are given by: Γijk =

2

1 ∑ 2 gkr r =1



∂g jk ∂gij ∂gik + − ∂ur ∂ui ∂u j

 .

Proof: Omitting points of application, by definition of the Christoffel symbols Γijk we have that xij = ∑2k=1 Γijk xk + e M hij N. Applying h·, xr i to both sides of this relation, we have: 2

h xij , xr i =

∑ Γijk gkr .

k =1

Surfaces in Space  229

Multiplying both sides by gr` and summing over r, we get: 2

∑

gr` h xij , xr i =

r =1

2

∑

Γijk gkr gr` =

k,r =1

2

∑ Γijk δk` = Γij` ,

k =1

which after renaming indices reads as Γijk = ∑2r=1 gkr h xij , xr i. It remains to find out what this product is. Using that mixed partial derivatives commute and the product rule, we have: ∂ h x j , xr i − h x j , xri i ∂ui ∂g jr ∂ = − r h x j , xi i + h x jr , xi i i ∂u ∂u ∂g jr ∂gij ∂ = − r + j h xr , xi i − h xr , xij i, i ∂u ∂u ∂u

h xij , xr i =

whence: 2h xij , xr i =

∂g jk ∂gij ∂gik + − ∂ur ∂ui ∂u j

and finally: Γijk =

2

1 ∑ 2 gkr r =1



∂g jk ∂gij ∂gik + − ∂ur ∂ui ∂u j

 ,

as wanted. Remark. Note that the Christoffel symbols are symmetric in the lower indices (actually, we already knew that since xij = x ji ). Furthermore, in practice, many of the parametrizations we will encounter have F = g12 = g21 = 0. In this case, the expressions given for the Γijk boil down to   ∂g jk ∂gij 1 kk ∂gik k Γij = g + − k . 2 ∂ui ∂u ∂u j In particular, we see that:

• If i 6= j only, since Γijk = Γkji and gij = 0, we have Γiij =

1 ii ∂gii , g 2 ∂u j

that is, Γuuv = Γuvu =

Ev 2E

and Γvuv = Γvvu =

• If i = j 6= k, we have gik = g jk = 0 and so ∂g 1 Γiik = − gkk iik , 2 ∂u or, equivalently, Γvuu = −

Ev 2G

and Γuvv = −

Γiii =

1 ii ∂gii g , 2 ∂ui

Gu . 2E

• If i = j = k, we have: or, Γuuu =

Eu 2E

and Γvvv =

Gv . 2G

Gu . 2G

230  Introduction to Lorentz Geometry: Curves and Surfaces

In particular, Proposition 3.2.17 (p. 170) now yields the following: Corollary 3.6.17. Let M1 , M2 ⊆ R3ν be two non-degenerate regular surfaces, and φ : M1 → M2 be a local isometry. If (U, x) is a parametrization for M1 and (U, φ ◦ x) is the corresponding parametrization for M2 , we have that ek (u, v), Γijk (u, v) = Γ ij ek denote the Christoffel Symbols of the for all (u, v) ∈ U and 1 ≤ i, j, k ≤ 2, where the Γ ij parametrization φ ◦ x. In terms of the objects above, we may express geodesics in a surface locally as solutions to a second order system of ODEs: Proposition 3.6.18 (Geodesic Differential Equations). If M ⊆ R3ν is a non-degenerate regular surface, (U, x) is a parametrization for M, and α : I → x(U ) is given in coordinates by α(t) = x(u1 (t), u2 (t)), for smooth functions u1 , u2 : I → R, then α is a geodesic if and only if: u¨ k +

2

∑

Γijk u˙ i u˙ j = 0,

k = 1, 2,

i,j=1

where the u˙ i are evaluated in t and the Γijk are evaluated in (u1 (t), u2 (t)). Proof: Continuing to omit all the points of application, let’s obtain an expression for α00 in terms of the Christoffel symbols and of the Second Fundamental Form of x. The components of α00 in the tangent directions will be precisely the expressions in the statement of the proposition. We have that α0 = ∑2i=1 u˙ i xi , and differentiating again: ! α00 =

=

2

∑

u¨ i xi + u˙ i

j =1

2

2

2

j =1

k =1

∑ 2

u¨ i xi + u˙ i

∑

u¨ k xk +

k =1 2

=

∑ u˙ j xij

i =1

i =1

=

2

∑

k =1

∑ u˙ j ∑ Γijk xk + e M hij N

2

∑

2

∑

i,j=1

! Γijk u˙ i u˙ j

∑

e M u˙ i u˙ j hij N

i,j=1

i,j,k=1

u¨ k +

2

Γijk u˙ i u˙ j xk +

!!

2

xk +

∑

e M u˙ i u˙ j hij N,

i,j=1

and α00 is normal to M if and only if u¨ k + ∑2i,j=1 Γijk u˙ i u˙ j = 0, for k = 1, 2. Remark. In particular, the above proof shows that, in coordinates, the covariant derivative of α0 is given by ! 2 2 Dα0 = ∑ u¨ k + ∑ Γijk u˙ i u˙ j xk . dt i,j=1 k =1 See the analogous expression for general vector fields along α in Exercise 3.6.17. Combining the above with the previous Corollary 3.6.17, we have the: Corollary 3.6.19. Let M1 , M2 ⊆ R3ν be two non-degenerate regular surfaces, and φ : M1 → M2 be a local isometry. If α : I → M1 is a geodesic of M1 , then φ ◦ α : I → M2 is a geodesic of M2 .

Surfaces in Space  231

Proof: Suppose without loss of generality that α( I ) ⊆ x(U ) for some parametrization (U, x) for M1 . If α(t) = x(u1 (t), u2 (t)), then φ(α(t)) = (φ ◦ x)(u1 (t), u2 (t)). Keeping notation from Corollary 3.6.17, we see that the geodesic differential equations are satisfied for φ ◦ α: u¨ k +

2

∑

ek u˙ i u˙ j = u¨ k + Γ ij

i,j=1

2

∑

Γijk u˙ i u˙ j = 0,

(1 ≤ k ≤ 2)

i,j=1

as wanted. Furthermore, the geodesic differential equations also give us the: Theorem 3.6.20. Let M ⊆ R3ν be a non-degenerate regular surface, p ∈ M and v ∈ Tp M. Then there is an interval I containing 0 and a geodesic α : I → M with α(0) = p and α0 (0) = v. Proof: Take a parametrization (U, x) around the point p = x(u0 , v0 ) and, for certain a, b ∈ R, write v = axu (u0 , v0 ) + bxv (u0 , v0 ). The usual existence and uniqueness result from the theory of ODEs applied to the initial value problem  2   u¨ k + ∑ Γijk u˙ i u˙ j = 0, k = 1, 2 i,j=1

   u1 (0) = u , 0

u2 (0) = v0 ,

u˙ 1 (0) = a,

u˙ 2 (0) = b

provides an interval I containing 0 and unique functions u1 , u2 : I → R satisfying both equations above. We then define α : I → M by α(t) = x(u1 (t), u2 (t)). Clearly α satisfies all the requirements. Remark. In the above proof, the image of the curve lies in the image of the parametrization initially fixed, but the theorem may be stated so that the interval I is maximal, in the following sense: if I ⊆ J and β : J → M is another geodesic with β(0) = p and β0 (0) = v, then J = I and β = α. In particular, the geodesic α is now unique under these conditions. Corollary 3.6.21. Let M ⊆ R3ν be a non-degenerate regular surface, and φ : M → M an isometry. Let α : I → M be a regular curve whose image is precisely the set of points fixed by φ: φ(α(t)) = α(t) for all t ∈ I. Then α is a pre-geodesic. Proof: Suppose without loss of generality that 0 ∈ I; let’s show that α restricted to a neighborhood of 0 is a pre-geodesic. The previous theorem says that there is e > 0 and a unique geodesic β : ]−e, e[ → M such that β(0) = α(0) and β0 (0) = α0 (0). By Corollary 3.6.19 above, we know that φ ◦ β is also a geodesic. And, moreover, we have that (φ ◦ β)(0) = φ( β(0)) = φ(α(0)) = α(0) = β(0), as well as

(φ ◦ β)0 (0) = dφβ(0) ( β0 (0)) = dφα(0) (α0 (0)) = (φ ◦ α)0 (0) = α0 (0) = β0 (0). The remark following the previous theorem also ensures that a geodesic is determined by its initial conditions, whence we conclude that φ ◦ β = β. Since the image of β is fixed by φ, it follows that the image of β is contained in the image of α, and so β is a reparametrization of the restriction α ]−e,e[ .

232  Introduction to Lorentz Geometry: Curves and Surfaces

3.6.3

Critical points of the energy functional

To proceed in a more efficient way with our study of geodesics, it will be convenient to present a brief introduction to Variational Calculus: where the independent variables are functions instead of real numbers. We’ll study functionals, in general given by certain integrals, with the goal to find its extremizers. Variational Calculus also has several applications in areas other than Mathematics itself, such as Physics, Engineering, Economics, and Control Theory, among others. Our presentation here will barely touch the tip of the iceberg, and so we’ll also recommend [24] and [70] for more details. Previously, we have characterized geodesics in terms of the geometry of the surface on where they lie, but we can also provide a variational characterization. For this end, consider at first the problem of finding a function y : [ a, b] → R, of class C1 , with fixed endpoints y( a) = y0 and y(b) = y1 , which minimizes the value of the integral: . J [y] =

Z b a

L(t, y(t), y0 (t)) dt,

for some prescribed class C2 map L : [ a, b] × R2 → R. The above situation is known as a variational problem, the map J is called a functional (objective, action), and L is called the Lagrangian for the variational problem. We will say that y is a local minimum for J if J [y] ≤ J [y + sη ] for all η : [ a, b] → R of class C1 with η ( a) = η (b) = 0 and s ∈ R sufficiently small. The boundary condition η ( a) = η (b) = 0 is necessary since we want y + sη to also be a legitimate candidate for solution for the fixed endpoints variational problem. Moreover, we will say that such an η is an admissible variation. Similarly, one defined a local maximum for J, and we’ll say that y is a critical point of J if it is either a local minimum or local maximum. Just like what happens for smooth functions of a single real variable, where the derivative vanishes at critical points in the interior of its domain, we have the following “first derivative test” for functionals: Theorem 3.6.22 (Euler-Lagrange). If y : [ a, b] → R is a function of class C1 , which is a critical point for the variational problem  Z b   J [y] = L(t, y(t), y0 (t)) dt a

 

y ( a ) = y0 ,

y ( b ) = y1 ,

where L : [ a, b] × R2 → R is a given class C2 Lagrangian, then y satisfies the EulerLagrange equation:   ∂L d ∂L − = 0, ∂y dt ∂y0 where the partial derivatives of L are evaluated in (t, y(t), y0 (t)). Remark. To emphasize the similarity with functions of a single real variable, the quantity   d ∂L . ∂L δJ [y] = − ∂y dt ∂y0 is also known as the variational derivative of J, or the first variation of J.

Surfaces in Space  233

Proof: Fixed an arbitrary admissible variation η and introducing a real parameter s, we look for a critical point of a single variable function (one function for each η), s 7→ J [y + sη ], whence: Z b d L(t, y(t) + sη (t), y0 (t) + sη 0 (t)) dt = 0. ds s=0 a Differentiating under the integral sign, we have that: Z b a

η (t)

∂L ∂L (t, y(t), y0 (t)) + η 0 (t) 0 (t, y(t), y0 (t)) dt = 0. ∂y ∂y

To eliminate η 0 (t) from this expression, we’ll integrate the last term by parts, obtaining: Z b a

 η (t)

d ∂L (t, y(t), y0 (t)) − ∂y dt



∂L (t, y(t), y0 (t)) ∂y0



b ∂L 0 dt + η (t) 0 (t, y(t), y (t)) = 0 ∂y a

Using η ( a) = η (b) = 0, this boils down to:    Z b d ∂L ∂L 0 dt = 0 (t, y(t), y0 (t)) − ( t, y ( t ) , y ( t )) η (t) ∂y dt ∂y0 a Since η was arbitrary, it follows that: ∂L d (t, y(t), y0 (t)) − ∂y dt



∂L (t, y(t), y0 (t)) ∂y0



= 0.

As a curiosity, there are many generalizations of this result for more complicated variational problems. See a few examples below: Example 3.6.23 (More variational problems). (1) We seek y : [ a, b] → R of class Cn optimizing the integral: . J [y] =

Z b a

L(t, y(t), . . . , y(n) (t)) dt,

for a prescribed class Cn+1 Lagrangian, L : [ a, b] × Rn → R, imposing endpoint conditions on y and on its first n − 1 derivatives. Repeating the argument given above and integrating by parts more times, the Euler-Lagrange equation takes the form:     n  ∂L  d ∂L d2 ∂L ∂L n d − + − · · · + (− 1 ) = 0, ∂y dt ∂y0 dtn ∂y(n) dt2 ∂y00 or, more concisely: n

dk (− 1 ) ∑ dtk k =0 k



∂L ∂y(k)



= 0.

(2) We seek n functions y1 , . . . , yn : [ a, b] → R of class C1 which optimize the integral . J [ y1 , . . . , y n ] =

Z b a

L(t, y1 (t), y10 (t), . . . , yn (t), y0n (t)) dt,

234  Introduction to Lorentz Geometry: Curves and Surfaces

for a prescribed class C2 Lagrangian, L : [ a, b] × R2n → R, imposing endpoint conditions on all functions. Considering admissible variations in the direction of each yi , one may show that now we’ll have one Euler-Lagrange equation for each argument:   ∂L d ∂L − = 0, i = 1, 2, . . . , n. ∂yi dt ∂yi0 Each Euler-Lagrange equation should be seen as one component of a “variational gradient” vanishing. (3) With the two items above in mind, it is natural to consider a variational problem where we look for for n functions, y1 , . . . , yn : [ a, b] → R, where yi is of class Cmi , which optimize the integral . J [ y1 , . . . , y n ] =

Z b a

( m1 )

L(t, y1 (t), . . . , y1

(mn )

( t ), . . . , y n ( t ), . . . , y n

(t)) dt,

for a suitable prescribed Lagrangian, imposing endpoint conditions on each yi and on its mi − 1 first derivatives. We will have n Euler-Lagrange equations: ! mi k d ∂L = 0, i = 1, 2, . . . , n. ∑ (−1)k dtk (k) ∂y k =0 i

(4) One may also consider a variational problem for which the function we seek depends on more than one real variable. For instance, if Ω ⊆ R2 is a bounded open set with regular boundary, we may seek a function y : Ω → R of class C1 optimizing the functional Z . L(u, v, y(u, v), yu (u, v), yv (u, v)) du dv, J [y] = Ω

R4

where L : Ω × → R is a prescribed class C2 Lagrangian, and we impose the boundary condition y ∂Ω = 0. Using the Green-Stokes Theorem (which will play the role of integration by parts in this case), it is possible to show that if y is a critical point of J, then y satisfies the following Euler-Lagrange:     ∂L ∂L ∂L ∂ ∂ − − = 0. ∂y ∂u ∂yu ∂v ∂yv For interesting applications of this version (and a direct generalization in higher dimensions), see Exercises 3.6.27 and 3.6.28. Variational techniques such as those may be used to model several phenomena in Physics: Example 3.6.24 (Hamilton’s Principle). We may use what was discussed so far to understand a bit better the situation described in Exercise 2.1.15 (p. 76, in Chapter 2). Suppose that a particle with mass m > 0 moves in Rn under the action of a potential V : Rn → R, with trajectory described by α : I → Rn . Recall that the kinetic energy is the smooth map T : Rn → R defined by T (v) = mkvk2E /2. Consider the Lagrangian L : R2n → R defined by . L( p, v) = T (v) − V ( p). Lagrangians such as the one above, which do not explicitly depend on t, are called

Surfaces in Space  235

autonomous. The action of the trajectory between the instants t0 and t1 is defined by the integral Z t 1 . S[α] = L(α(t), α0 (t)) dt. t0

Then Hamilton’s Principle states that particles follow trajectories that minimize this action. If α(t) = ( x1 (t), . . . , xn (t)), the derivatives of this Lagrangian satisfy ∂L ∂V =− ∂xi ∂xi

and

∂L = m x˙ i , ∂ x˙ i

from where we see that the Euler-Lagrange equations (for each i) give us Newton’s Second Law mα00 (t) = −∇V (α(t)). For this reason, we will say that the critical points of the action functional are physical trajectories, and we’ll keep using this terminology even for Lagrangians which are not of the form T − V (called “natural”), such as the one here. See Exercise 3.6.25 for one more contextualization in Physics. Now, let’s analyze the Euler-Lagrange equations for the energy functional, given by E[α] =

1 2

Z I

hα0 (t), α0 (t)i dt,

to obtain the following characterization of geodesics: Theorem 3.6.25. Let M ⊆ R3ν be a non-degenerate regular surface, (U, x) be a parametrization for M, and α : I → M given in coordinates by α(t) = x(u1 (t), u2 (t)). Then the Euler-Lagrange equations for the energy of α are equivalent to the geodesic differential equations. Thus, all the critical points of the energy functional are geodesics. Proof: Again omitting the points of evaluation and bearing in mind that u˙ i = u˙ i (t) and gij = gij (u1 (t), u2 (t)), we write the (autonomous) Lagrangian for the energy of α as: L= We have that

1 2 gij u˙ i u˙ j . 2 i,j∑ =1

2 ∂L ∂ 1 2 1 ∂gij i j i j ˙ ˙ = g u u = u˙ u˙ , ij ∑ ∑ k k 2 k ∂u ∂u 2 i,j=1 i,j=1 ∂u

and also: 2 ∂L ∂ 1 2 1 j i j ˙ ˙ = g u u = gij (u˙ i δk + δki u˙ j ) ij ∑ ∑ k k 2 ∂u˙ ∂u˙ 2 i,j=1 i,j=1

=

2 2 2 1 1 1 1 i j j i i ˙ ˙ ˙ g u δ + g u δ = g u + ∑ 2 gjk u˙ j ∑ 2 ij k ∑ 2 ij k ∑ 2 ik i,j=1 i,j=1 i =1 j =1

=

∑ gik u˙ i .

2

2

i =1

With this, it follows that: !   2 2 2 2 2 ∂L d d d i i i j ∂gik ˙ ˙ ˙ ˙ = g u = ( g + u ) = u u ∑ dt ik ∑ gik u¨ i , ∑∑ ik j dt ∂u˙ k dt i∑ ∂u =1 i =1 i =1 j =1 i =1

236  Introduction to Lorentz Geometry: Curves and Surfaces

whence we obtain the Euler-Lagrange equations: ∂gik i j 1 2 ∂gij i j u˙ u˙ − ∑ u˙ u˙ = 0. j 2 i,j=1 ∂uk i,j=1 ∂u

2

2

∑ gik u¨ i + ∑

i =1

Observing that 2

∂gik i j 1 2 ∂gik i j 1 2 ∂g jk i j ˙ ˙ u u = u˙ u˙ + ∑ u˙ u˙ , ∑ j j 2 i,j∑ 2 i,j=1 ∂ui i,j=1 ∂u =1 ∂u we may reorganize the Euler-Lagrange equations as:   2 2 ∂g jk ∂gij 1 ∂gik i ¨ g u + + − ∑ ik ∑ 2 ∂u j ∂ui ∂uk u˙ i u˙ j = 0. i =1 i,j=1 Multiplying the whole equation by gkr , summing over k, and renaming r → k, we obtain precisely: u¨ k +

2

∑

Γijk u˙ i u˙ j = 0.

i,j=1

In general, the converse of Theorem 3.6.22 fails: functions that satisfy the EulerLagrange equation are not necessarily extremizers for the functional J under discussion. But, when dealing with geodesics, we have the: Proposition 3.6.26. Let M ⊆ R3ν be a non-degenerate regular surface. Then the geodesics of M are critical points of the energy functional. Proof: Suppose that α : [ a, b] → M is a geodesic. The initial idea would be to show that for every admissible variation η : [ a, b] → M with η( a) = η(b) = 0, we’d have that 0 is a critical point of Z 1 b 0 e 7→ hα (t) + eη0 (t), α0 (t) + eη0 (t)i dt, 2 a but this argument has a fatal flaw: α + eη may leave the surface M. To remedy this, we will consider, for r > 0, admissible variations f : [ a, b] × [−r, r ] → M satisfying f (t, 0) = α(t) for all t, and f ( a, s) = α( a) and f (b, s) = α(b) for all s.

∂f ( t0 , s0 ) ∂s

s f

f ( t0 , s0 )

r

α(b)

α

a

−r

b

t α( a) M

Figure 3.33: The variation of a curve in M.

Surfaces in Space  237

We then consider . 1 E(t) = 2

Z b ∂f a

∂f (t, s), (t, s) ∂t ∂t

 dt.

Let’s see that E0 (0) = 0. Observe that the conditions under f ensure that ∂f ∂f ( a, 0) = (b, 0) = 0. ∂s ∂s With this, differentiating under the integral sign, we have:  Z b 2 ∂ f 0 0 (t, 0), α (t) dt E (0) = ∂s∂t a     Z b (1) ∂ ∂f ∂f 0 00 = (t, 0), α (t) − (t, 0), α (t) dt ∂s ∂s a ∂t   b Z b   (2) ∂ f Dα0 ∂f 0 = (t, 0), α (t) − (t, 0), (t) dt ∂s ∂s dt a a  Z b ∂f Dα0 =− (t, 0), (t) dt = 0, ∂s dt a where in (1) we integrate by parts and in (2) that ∂ f /∂s is always tangent to M. Remark.

• The formula for the first variation of energy,  Z b ∂f Dα0 0 E (0) = − (t, 0), (t) dt, ∂s dt a also provides an alternative proof for Theorem 3.6.25. The advantage of the first proof presented is that it illustrates a simpler way to compute Christoffel symbols, avoiding a direct use of the expression given in terms of the gij ’s and its derivatives, given in Proposition 3.6.16. We will see examples soon.

• Despite the above result, we still cannot guarantee that geodesics are (global) extremizers for the energy functional. For instance, we may consider any point in a surface which admits a closed geodesic4 passing through that point: the energy is minimized by the degenerate constant curve that never leaves the point, and not by the chosen closed geodesic. If we require regular curves, it suffices to consider two non-antipodal points in a sphere and arcs of great circles with different lengths. But when M is spacelike, we have a local result, see Exercise 3.6.26. Besides all that, we have seen in Proposition 2.1.11 (p. 68, in Chapter 2) that there is a relation between the energy and the arclength of a curve, given by the Cauchy-Schwarz inequality. Namely, that if α : [ a, b] → M is a curve, then the inequality q Lba [α] ≤ 2eα (b − a) Eab [α] holds, with equality if and only if α has constant speed. In particular, this inequality says that a constant speed curve is a critical point of the energy functional if and only if it is a critical point of the arclength functions. It also follows from this that geodesics in spacelike surfaces locally minimize arclength. compact surface in R3 admits a closed geodesic (this is a particular instance of the so-called Lyusternik-Fet theorem – if you know Russian, see [45]). 4 Every

238  Introduction to Lorentz Geometry: Curves and Surfaces

Example 3.6.27. (1) Let p ∈ R3ν , and Π ⊆ R3ν be a non-degenerate plane passing through p, with orthonormal basis {w1 , w2 }. Consider the parametrization x : R2 → Π given by x(u, v) = p + uw1 + vw2 . For a curve α : I → Π given by α(t) = x(u(t), v(t)), we have that the energy is 1 E[α] = 2

Z I

ew1 u0 (t)2 + ew2 v0 (t)2 dt,

and the associated Lagrangian is L(u(t), u0 (t), v(t), v0 (t)) =


1 ew1 u 0 ( t )2 + ew2 v 0 ( t )2 . 2

Directly one sees that the Euler-Lagrange equations are just u00 = 0 and v00 = 0, so that all the Christoffel symbols vanish. In particular, we conclude yet again that the geodesics in non-degenerate planes are actual straight lines. (2) Let σ : I → R3ν be an injective and regular smooth plane curve, of the form σ (u) = ( f (u), 0, g(u)), with f (u) > 0 for all u ∈ I. Suppose that σ is not lightlike. Considering the usual parametrization of revolution, x : I × ]0, 2π [ → R3ν given by x(u, v) = ( f (u) cos v, f (u) sin v, g(u)), we have that its First Fundamental Form is given by ds2 = eσ du2 + f (u)2 dv2 . If α : I 0 → x( I × ]0, 2π [) is given in coordinates by α(t) = x(u(t), v(t)), the energy of α is: Z 1 E[α] = eσ u0 (t)2 + f (u(t))2 v0 (t)2 dt. 2 I0 Writing the Lagrangian L(u(t), u0 (t), v(t), v0 (t)) =

1 (eσ u0 (t)2 + f (u(t))2 v0 (t)2 ), 2

we have that the Euler-Lagrange equations are:   d ∂L ∂L − = 0 =⇒ u00 (t) − eσ f (u(t)) f 0 (u(t))v0 (t)2 = 0 ∂u dt ∂u0   ∂L d ∂L f 0 (u(t)) 0 00 − = 0 =⇒ v ( t ) + 2 u (t)v0 (t) = 0. ∂v dt ∂v0 f (u(t)) We then see that the non-vanishing Christoffel symbols are: Γuvv (u, v) = −eσ f (u) f 0 (u) and Γvuv (u, v) = Γvvu (u, v) =

f 0 (u) . f (u)

From such equations it is easy to see that the meridians, parametrized by α(t) = x( at + b, v0 ), are geodesics in both ambient spaces. Each parallel u = u0 , in turn, is a geodesic if and only if f 0 (u0 ) = 0. If all the parallels are geodesics, we have that f is a positive constant and, in this case, the surface is part of a straight circular cylinder.

Surfaces in Space  239

(3) Let’s reobtain the geodesics in the cylinder S1 (r ) × R ⊆ R3ν of radius r > 0, previously found in Example 3.6.7 (p. 220). Consider again the parametrizations x : ]0, 2π [ × R → S1 (r ) × R, given by x(u, v) = (r cos u, r sin u, v). We know that its First Fundamental Form is given by ds2 = r2 du2 + (−1)ν dv2 . The energy of a curve α : I → S1 × R given in coordinates by α(t) = x(u(t), v(t)) is just Z 1 E[α] = r2 u0 (t)2 + (−1)ν v0 (t)2 dt. 2 I The Lagrangian is: L(u(t), u0 (t), v(t), v0 (t)) =

1 2 0 2 (r u (t) + (−1)ν v0 (t)2 ), 2

and the Euler-Lagrange equations are:   ∂L d ∂L − = 0 =⇒ u00 (t) = 0 ∂u dt ∂u0   ∂L d ∂L − = 0 =⇒ v00 (t) = 0, ∂v dt ∂v0 whence all the Christoffel symbols vanish, and the geodesics are given by α(t) = (r cos( at + b), r sin( at + b), ct + d), where a, b, c, d ∈ R. At least when r = 1, this was to be expected: x itself is a local isometry, so not only all the Christoffel symbols vanish (see (1) above), but also the geodesics of the cylinder are precisely the images via x of geodesics in the plane, that is, straight lines. Example 3.6.28 (Geodesics of S2 (r ), S21 (r ) and H2± (r )). Suppose that M = S2 (r ), S2 (r ) or H2± (r ). Let’s see that the geodesics of M are precisely the intersections M ∩ Π, where Π is a plane passing through the origin such that M ∩ Π 6= ∅ (such intersection is empty when Π is spacelike or lightlike and M = H2± (r )). In view of the remark following Theorem 3.6.20 (p. 231), it suffices to show that all the intersections M ∩ Π may be parametrized by geodesics, and note that given p ∈ M and v ∈ Tp M, the plane Π = span{ p, v} has as tangent line at p precisely the line passing through p with direction v. We already know that in S21 (r ), every lightlike curve is a pre-geodesic by Exercise 3.6.10 (p. 227), so assume that Π has a non-lightlike fixed normal vector n ∈ R3ν . Let α : I → M ∩ Π be a unit speed curve. For each s ∈ I, write α00 (s) = a(s)α(s) + b(s)α0 (s) + c(s)n. The condition hα(s), α(s)i = cte. implies that hα0 (s), α(s)i = 0; now α having constant speed gives us that hα00 (s), α0 (s)i = 0 and, lastly, the condition hα(s), ni = 0 yields hα0 (s), ni = hα00 (s), ni = 0. This way, applying h·, α0 (s)i and h·, ni in the above gives us that b(s) = c(s) = 0, whence α00 (s) = a(s)α(s). But the position vector of an arbitrary point in M is always normal to M, and so we conclude that α00 (s) is always normal M as well. In other words, α is a geodesic, as wanted.

240  Introduction to Lorentz Geometry: Curves and Surfaces

2

2 0 -2

-2 0

0

2

-2

Figure 3.34: The geodesics in S21 . As a last application of the variational techniques presented in this section, let’s discuss a specific type of parametrization, convenient for the study of geodesics, and which generalizes the parametrizations of revolution we have encountered so far: Definition 3.6.29 (Clairaut Parametrizations). Let M ⊆ R3ν be a non-degenerate regular surface and (U, x) be a parametrization for M. If E, F and G denote the coefficients of the First Fundamental Form of M relative to x, we’ll say that x is a u-Clairaut parametrization if Ev = Gv = F = 0. Remark.

• That is, if x is u-Clairaut, its First Fundamental Form written in differential notation is ds2 = E(u) du2 + G (u) dv2 . Moreover, since M is non-degenerate, we always have EG 6= 0. • Clearly the parametrizations of revolution seen so far are u-Clairaut. To emphasize the comparison, we will refer to the coordinate curves u = u0 as parallels, and to the coordinate curves v = v0 as meridians. • Similarly, we’ll say that x is a v-Clairaut parametrization if Eu = Gu = F = 0. All the results proven for one case have obvious analogues for the other case. Lemma 3.6.30. Let M ⊆ R3ν be a non-degenerate regular surface, and (U, x) be a u-Clairaut parametrization. The geodesic equations for the parametrization x are: Eu 0 2 Gu 0 2 Gu 0 0 (u ) − (v ) = 0 and v00 + u v = 0. 2E 2E G In particular, the non-vanishing Christoffel symbols are u00 +

Γuuu (u, v) =

Eu (u) , 2E(u)

Γuvv (u, v) = −

Gu (u) 2E(u)

and

Γvuv (u, v) = Γvvu (u, v) =

Gu (u) . 2G (u)

Surfaces in Space  241

Proof: We’ll produce the geodesic equations via Euler-Lagrange equations, with the Larangian L(u(t), u0 (t), v(t), v0 (t)) =

1 ( E(u(t))u0 (t)2 + G (u(t))v0 (t)2 ). 2

Omitting points of evaluation, we have:  
d ∂L 1 d ∂L − = ( Eu (u0 )2 + Gu (v0 )2 ) − Eu0 0 ∂u dt ∂u 2 dt 1 1 = Eu (u0 )2 + Gu (v0 )2 − Eu (u0 )2 − Eu00 = 0. 2 2 Grouping similar terms and dividing the equation by − E it follows that: u00 +

Eu 0 2 Gu 0 2 (u ) − (v ) = 0. 2E 2E

For the second equation, we have:  
∂L d ∂L d 0 Gv − = − ∂v dt ∂v0 dt = − Gu u0 v0 − Gv00 = 0. Dividing the above by − G we obtain: v00 +

Gu 0 0 u v = 0. G

Remark. For the geodesic equations in the v-Clairaut case, see Exercise 3.6.33. The next proposition generalizes what we have discussed before about meridians and parallels in surfaces of revolution: Proposition 3.6.31. Let M ⊆ R3ν be a non-degenerate regular surface and (U, x) be a u-Clairaut parametrization. Then: (i) The meridians v = v0 are pre-geodesics. (ii) The parallels u = u0 are geodesics if and only if Gu (u0 ) = 0. Proof: (i) Let’s prove that if α : I → M given by α(u) = x(u, v0 ) parametrizes a meridian, then ( Dα0 /du)(u) is proportional to α0 (u), for all u ∈ I. Indeed, we have that α0 (u) = xu (u, v0 ) and α00 (u) = xuu (u, v0 ). By definition of covariant derivative we have: Dα0 Eu (u) 0 (u) = Γuuu (u, v0 ) xu (u, v0 ) + Γvuu (u, v0 ) xv (u, v0 ) = α ( u ). du 2E(u) By Proposition 3.6.9 (p. 221), α is a pre-geodesic, as wanted.

242  Introduction to Lorentz Geometry: Curves and Surfaces

(ii) Let α : I → M, α(v) = x(u0 , v), be a parametrization of a parallel. We have α0 (v) = xv (u0 , v) and α00 (v) = xvv (u0 , v). As before, we have that: Dα0 Gu (u0 ) (v) = Γuvv (u0 , v) xu (u0 , v) + Γvvv (u0 , v) xv (u0 , v) = − x u ( u0 , v ), dv 2E(u0 ) which vanishes if and only if Gu (u0 ) = 0.

Theorem 3.6.32 (Clairaut Relation). Let M ⊆ R3ν be a regular spacelike surface and (U, x) be a u-Clairaut parametrization for M. If α : I → x(U ) is a unit speed geodesic given in coordinates by α(s) = x(u(s), v(s)), there is c(α) ∈ R such that q G (u(s)) sin θ (s) = c(α), for all s ∈ I, where θ (s) is the angle formed between xu (u(s), v(s)) and α0 (s). In particular, the image of α in M does not leave the region in M where G ≥ c(α)2 . Proof: Since F = 0, we have that the angle between xv (u(s), v(s)) and α0 (s) is exactly π 2 − θ ( s ). Using that Gv = 0, we may integrate the second geodesic equation: v00 +

Gu 0 0 u v = 0 =⇒ Gv00 + Gu u0 v0 = 0 =⇒ ( Gv0 )0 = 0 =⇒ Gv0 = c(α), G

for some c(α) ∈ R. With this, on one hand (omitting points of application in xu and xv ):

hα0 (s), xv i = hu0 (s) xu + v0 (s) xv , xv i = G (u(s))v0 (s) = c(α), and, on the other hand,

hα0 (s), xv i = kα0 (s)kk xv k cos

π 2

 q − θ (s) = G (u(s)) sin θ (s).

p

Thus, we have G (u(s)) sin θ (s) = c(α) for all s ∈ I. Finally, we obtain the inequality c(α)2 = G (u(s)) sin2 θ (s) ≤ G (u(s)), for all s ∈ I. For timelike surfaces the situation is slightly more complicated, once we do not have a general notion of angle between vectors of arbitrary causal type. Imposing a few extra restrictions on x, we may obtain something similar to the above result, just with the tools we have so far: Proposition 3.6.33. Let M ⊆ L3 be a regular timelike surface, and (U, x) be a uClairaut parametrization for M such that xu is always spacelike. If α : I → M is a proper time parametrized timelike geodesic given in coordinates by α(t) = x(u(t), v(t)), there is a constant c(α) ∈ R, such that q | G (u(t))| cosh ϕ± (t) = ∓c(α), for all t ∈ I, where the sign ± indicates which among α0 (t) and −α0 (t) lies in the same timecone as xv (u(t), v(t)), and ϕ± (t) denotes the corresponding hyperbolic angle. In particular, the image of α does not leave the region in M where G ≥ −c(α)2 .

Surfaces in Space  243

Proof: As done before, we integrate the second geodesic equation to obtain G (u(t))v0 (t) = c(α) for some c(α) ∈ R and, omitting the point of evaluation for xv , we have that hα0 (t), xv i L = c(α). Let’s do the case where α0 (t) is in the same timecone as xv . If ϕ+ (t) is the hyperbolic angle between them, we have q hα0 (t), xv i L = −kα0 (t)k L k xv k L cosh ϕ+ (t) = − | G (u(t))| cosh ϕ+ (t), p and thus | G (u(t))| cosh ϕ+ (t) = −c(α) for all t ∈ I. When α0 (t) and xv are in opposite timecones, the calculation is similar. So c(α)2 = | G (u(t))| cosh2 ϕ± (t) ≥ | G (u(t))| = − G (u(t)) gives us that G (u(t)) ≥ −c(α)2 for all t ∈ I, as wanted. Remark. The constant c(α) in the previous two results is called the slant of α in M. One of the main advantages of working with Clairaut parametrizations is that the geodesic equations together boil down to a single ODE, as the next two results show: Theorem 3.6.34 (Clairaut). Let M ⊆ R3ν be a regular spacelike surface, (U, x) be a u-Clairaut parametrization for M, and α : I → x(U ) be a unit speed geodesic which never crosses the meridians of x orthogonally. Then α admits a reparametrization of the form β(u) = x(u, v(u)), where v is a solution for p c(α) E(u) dv p = ±p . du G ( u ) G ( u ) − c ( α )2 Remark. In the above statement, the signs ± only indicate reverse parametrizations. Proof: Start writing α(s) = x(u(s), v(s)). Omitting points of application, substituting Gv0 = c(α) into E(u0 )2 + G (v0 )2 = 1 yields

( u 0 )2 =

G − c ( α )2 . EG

The Clairaut relation seen in Theorem 3.6.32 now says that G − c(α)2 ≥ 0, but the condition on the meridians ensures strict inequality. So not only we may write p G − c ( α )2 0 √ u =± , EG we may also use that u0 6= 0, so that the Inverse Function Theorem allows us to use u . as parameter (explicitly, we’ll have β = α ◦ u−1 ) and conclude that √ dv v0 c(α)/G c(α) E √ = 0 = ±p = ±√ p . du u G − c(α)2 / EG G G − c ( α )2

Lastly, Proposition 3.6.33 allows us to use a similar argument to the above and obtain the:

244  Introduction to Lorentz Geometry: Curves and Surfaces

Theorem 3.6.35. Let M ⊆ L3 be a regular timelike surface, (U, x) be a u-Clairaut parametrization for M such that xu is always spacelike, and α : I → x(U ) be a proper time parametrized timelike geodesic which never crosses the meridians of x orthogonally. Then α admits a reparametrization of the form β(u) = x(u, v(u)), where v is a solution for p c(α) E(u) dv p = ±p . du | G (u)| G (u) + c(α)2 We ask you to carry out this last proof in Exercise 3.6.35. For more geometric interpretations of Clairaut parametrizations, see [55].

Exercises Exercise† 3.6.17. Let M ⊆ R3ν be a non-degenerate regular surface, (U, x) be a parametrization for M, α : I → x(U ) be a regular curve, and V : I → R3ν a vector field along α. Suppose that, in coordinates, we write α(t) = x(u1 (t), u2 (t)) and V (t) =

∑ ai (t)xi (u1 (t), u2 (t)).

i =1

Show that, omitting points of evaluation, we have  2 2  DV k k i j = ∑ a˙ + ∑ Γij a u˙ xk . dt i,j=1 k =1 Exercise† 3.6.18 (Parallel Translation). Let M ⊆ R3ν be a non-degenerate regular surface and α : [0, 1] → M be a regular curve joining p = α(0) to q = α(1). (a) Show that given v ∈ Tp M, there is a unique parallel vector field V : [0, 1] → R3ν along α such that V (0) = v. The vector V (1) ∈ Tq M is called the parallel translation of v from p to q via α. Hint. To make it easier, you may assume that the image of α lies in the image of a single parametrization for M. (b) How does parallel translation work when M is a non-degenerate plane? (c) If M ⊆ R3 is the surface described by z = y2 , consider α : [0, 1] → M given by α(t) = (0, t, t2 ), and compute the parallel translation of (1, 1, 0) ∈ T(0,0,0) M to T(0,1,1) M via α. Hint. Work with the obvious parametrization. (d) In the notation from item (a), show that Pα : Tp M → Tq M given by Pα (v) = V (1) is a linear isometry. What is ( Pα )−1 ? Remark. Given p ∈ M, the collection of all maps Pα , where α is a smooth curve in M that starts and ends at p, equipped with the operation of composition of functions, is a group (called the holonomy group of M at p). Exercise† 3.6.19. Let M1 , M2 ⊆ R3ν be two non-degenerate regular surfaces, α : [0, 1] → M1 be a regular curve joining p = α(0) to q = α(1), V a vector field along α, and φ : M1 → M2 smooth.

Surfaces in Space  245

(a) Suppose that φ is a local isometry. If V : I → R3ν is the vector field along φ ◦ α . defined by V (t) = dφα(t) (V (t)), show that  dφα(t)

DV (t) dt



=

DV ( t ), dt

for all t ∈ I. In particular, (local) isometries take parallel fields to parallel fields. (b) Suppose that φ is an isometry. Show that for all v ∈ Tp M we have that

Pφ◦α dφ p (v) = dφq Pα (v) . That is, isometries are compatible with parallel translations. Exercise 3.6.20 (Fermi-Walker Parallelism). Let M ⊆ R3ν be a non-degenerate regular surface and α : [0, 1] → M be a regular curve joining p = α(0) to q = α(1). In Exercise 3.6.18 above we have seen that parallel translation along α is a linear isometry between different tangent planes and, thus, it takes orthonormal bases of Tp M onto orthonormal bases of Tq M. However, if α0 (0) is an element of the initial basis, it is not necessarily true that α0 (1) is an element of the basis obtained, unless α itself is a geodesic (to wit, if α is a geodesic, the velocity field α0 is parallel along α and so α0 (1) would be the parallel translation of α0 (0) from p to q via α). Suppose that M and α are both timelike, with α parametrized with proper time. Aiming to fix the above deficiency, we define the FW-derivative of a smooth vector field V : [0, 1] → R3ν along α, by introducing a certain correction term: Dα0 0 ( t ) α ( t ) dt FV . DV  (t) = (t) +  0 . Dα dt dt 0 hα (t), V (t)i L dt (t), V (t) L We’ll say that V is FW-parallel if FV /dt = 0. (a) Show that the operator F/dt is R-linear and satisfies the Leibniz rule, as seen for the operator D/dt in Exercise 3.6.1 (p. 225). This way, the operator F/dt also deserves the name of “covariant derivative”. (b) Show that the velocity field α0 is always FW-parallel and, moreover, if α is a geodesic, then F/dt = D/dt. Thus, the new operator F/dt is indeed a type of generalization of D/dt, which makes all timelike curves “FW-geodesics”. (c) Show that if V is a vector field along α which is always orthogonal to α0 , then ( FV /dt)(t) is the projection of ( DV /dt)(t) onto the orthogonal complement of α 0 (t). (d) If V and W are smooth vector fields along α, then     FW d FV hV (t), W (t)i L = (t), W (t) + V (t), (t) dt dt dt L L holds. In particular, it follows that FW-parallel fields also have constant speed and causal type (compare with Exercise 3.6.2, p. 226).

246  Introduction to Lorentz Geometry: Curves and Surfaces

(e) It is possible to show (as done in Exercise 3.6.18) that in these conditions, given v ∈ Tp M, there is a unique FW-parallel vector field V along α such that V (0) = v, and the vector V (1) is called the FW-parallel translation of v from p to q via α. Show that the map Pα,FW : Tp M → Tq M so defined is also a linear isometry. (f) Show that local isometries preserve FW-derivatives and conclude that isometries are compatible with FW-parallel translations (in the sense of the previous exercise). Remark. For more details, see [48]. Exercise 3.6.21 (Transformation law for Γijk ). Let M ⊆ R3ν be a non-degenerate regular e e e ) 6 = ∅. surface, and (U, x) and (U, x) be two parametrizations such that x(U ) ∩ e x (U −1 In suitable domains, the change of parameters e x ◦ x defines inverse smooth relations i i 1 2 i i 1 2 ek the First Fundamental Form ue = ue (u , u ) and u = u (ue , ue ). Denote by geij and Γ ij and the Christoffel symbols of e x. (a) Show that the First Fundamental Form transforms as 2

geij =

∂uk ∂u` gkl ∂uei ∂uej k,`=1

∑

and

geij =

2

∂uei ∂uej kl g , k ` k,`=1 ∂u ∂u

∑

where all the functions above are evaluated in the correct points. (b) Show that ek = Γ ij

2 ∂uek ∂2 u` ∂ur ∂us ∂uek ` Γ + . ∑ rs ` ei ∂ u ∂uei ∂uej ∂u` ej `=1 ∂u ∂ u r,s,`=1 2

∑

Remark. The formula in item (b) says that the Christoffel symbols do not represent any type of linear transformation in R2 , no matter which is the fixed index i, j or k. The term involving second order derivatives accounts for the action of the Γijk ’s as a correction in the local expression for covariant derivatives. Exercise 3.6.22. Let M ⊆ R3ν be a connected and non-degenerate regular surface such that all the geodesics of M are plane curves. Show that M is contained in a plane, S2 (c, r ), S21 (c, r ) or H2± (c, r ), for certain c ∈ R3ν and r > 0. Hint. Show that M is totally umbilic. Exercise 3.6.23. Find extremizers for the following variational problems:  Z 2 0  y ( x )2  J [y] = dx 1 x3 (a) 1  y(1) = 0, y(2) = 15.  Z 1   J [y] = y( x )2 + y0 ( x )2 + 2y( x )ex dx 2 0 (b)  y(0) = 0, y(1) = e/2. It is more complicated, in general, to decide whether a given extremizer is a local maximum or minimum for a given functional. In some specific cases, the situation is more treatable. If y is the extremizer found in item (a), show that Z 2 0 η1 ( x )η20 ( x ) ∂2 . Hy [η1 , η2 ] = J [ y + e η + e η ] = 2 dx 2 2 1 1 1 ∂e1 ∂e2 e =e =0 x3 1 1

2

Surfaces in Space  247

for all admissible variations η1 and η2 . Note that given an admissible variation η, it holds that Hy [η, η ] ≥ 0 and Hy [η, η ] = 0 if and only if η = 0. Conclude that y is a strict local minimum for J1 . Remark. The map Hy is a sort of Hessian, useful as a tool to decide the nature of critical points of J1 . Exercise† 3.6.24. Repeat the idea given in the proof of Theorem 3.6.22 (p. 232) to show that if y : [ a, b] → R is a class C2 critical point for the variational problem  Z b   J [y] = L(t, y(t), y0 (t), y00 (t)) dt a

 

y ( a ) = y0 ,

y ( b ) = y1 ,

y0 ( a) = y0∗ ,

y0 (b) = y1∗ ,

where L : [ a, b] × R2 → R is a prescribed class C3 Lagrangian, then y satisfies the Euler-Lagrange equation:     d ∂L d2 ∂L ∂L − + 2 = 0, ∂y dt ∂y0 ∂y00 dt where the partial derivatives of L are evaluated in (t, y(t), y0 (t), y00 (t)). Exercise† 3.6.25 (Some conservation laws). Consider a class C2 Lagrangian L : I × R2n → R. The force of the Lagrangian in the direction xi is the quantity ∂L/∂xi , its momentum in the direction xi is the quantity ∂L/∂ x˙ i and, lastly, the quantity . H = ∑in=1 x˙ i (∂L/∂ x˙ i ) − L is called the associated Hamiltonian. (a) Show that if L does not explicitly depend on xi , that is, if the force of the Lagrangian vanishes in the direction of xi , the corresponding momentum is constant along physical trajectories. (b) Show that if L is autonomous (i.e., does not explicitly depend on t, or yet ∂L/∂t = 0), then the Hamiltonian is constant along physical trajectories. Remark. Many other conservations laws (such as Clairaut’s relation) follow from the celebrated Noether’s Theorem which, informally, says that if a Lagrangian L is invariant under a 1-parameter group of diffeomorphisms5 of Rn , ( ϕs )s∈R (i.e., if fixed s ∈ R, we have L(t, x(t), x˙ (t)) = L(t, y(t), y˙ (t)) for all t, where y = ϕs ◦ x), then the Noether charge J: I × R2n → R defined by ∂ϕis ( x(t)) . n ∂L J(t, x(t), x˙ (t)) = ∑ (t, x(t), x˙ (t)) , ∂ x˙ i ∂s s =0 i =1 where ϕis is the i-th coordinate of ϕs , is conserved along physical trajectories, that is, J = cte. In other words, in view of Noether’s Theorem, to find out conservation laws we must seek symmetries of the Lagrangian. Can you exhibit suitable 1-parameter groups of diffeomorphisms to conclude again the results from items (a) and (b) above using Noether’s Theorem? collection of diffeomorphisms ϕs : Rn → Rn satisfying ϕ0 = IdRn and ϕs1 +s2 = ϕs1 ◦ ϕs2 , for all s1 , s2 ∈ R. 5A

248  Introduction to Lorentz Geometry: Curves and Surfaces

Exercise† 3.6.26. Consider the energy functional in Rnν . Show that given α : I → Rnν and admissible variations η1 , η2 , we have that Z ∂2 . Hα [η1 , η2 ] = E [ α + e η + e η ] = hη10 (t), η20 (t)i dt. 2 2 1 1 ∂e1 ∂e2 e =e =0 I 2

1

Note that for any admissible variation η in Rn , we have Hα [η, η] ≥ 0 and Hα [η, η] = 0 if and only if η = 0. Conclude that in Rn , geodesics locally minimize the energy functional. Exercise 3.6.27. Let f : U ⊆ R2 → R be a smooth function. (a) Consider the functional A[ f ] =

Z q U

1 + f u (u, v)2 + f v (u, v)2 du dv.

Explicitly write the Euler-Lagrange equation in this case. Conclude (again) that the graph of f is a minimal surface in R3 if and only if f uu (1 + f v2 ) − 2 f u f v f uv + f vv (1 + f u2 ) = 0. (b) Considering another suitable functional, repeat the idea from the previous item, and conclude that the graph of f is a critical surface in L3 if and only if f uu (−1 + f v2 ) − 2 f u f v f uv + f vv (−1 + f u2 ) = 0. Exercise 3.6.28. Let φ : Rn → R be a smooth function with compact support, i.e., with { x ∈ Rn | φ( x) 6= 0} compact. (a) Define the Euclidean Dirichlet energy of φ as . 1 DE [ φ ] = 2

Z Rn

h∇φ( x), ∇φ( x)i E dx.

Show that the critical points of the Euclidean Dirichlet energy are harmonic functions, i.e., solutions of the equation 4φ = 0, where the Laplacian of φ is defined by . 4φ =

n

∂2 φ ∑ 2. i =1 ∂xi

(b) One may also consider a Lorentzian version of the above:: . 1 DL [φ] = 2

Z Rn

h∇φ( x), ∇φ( x)i L dx.

Show that the critical points of this Lorentzian Dirichlet energy are solutions to the wave equation φ = 0, where  is the (stationary) wave operator (d’Alembertian), defined by . n −1 ∂ 2 φ ∂ 2 φ φ = ∑ 2 − 2 . ∂xn i =1 ∂xi Remark. The assumption that φ has compact support is meant just to ensure that the integrals in question are not really improper. The operators 4 and  will play an important role when we study critical surfaces in more detail in Chapter 4.

Surfaces in Space  249

Exercise 3.6.29 (Surfaces of hyperbolic revolutions – III). Consider again an injective, regular, and smooth curve σ : I → L3 of the form σ (u) = ( f (u), 0, g(u)), with g(u) > 0 for all u ∈ I, as well as the surface of hyperbolic revolution it generated around the x-axis, with parametrization x : I × ]0, 2π [ → x( I × ]0, 2π [) ⊆ L3 given by: x(u, v) = ( f (u), g(u) sinh v, g(u) cosh v). Suppose that σ has unit speed. (a) Compute the geodesic equations for x, and conclude that the non-vanishing Christoffel symbols are Γuvv (u, v) = −eσ g(u) g0 (u) and Γvuv (u, v) = Γvvu (u, v) =

g0 (u) . g(u)

(b) Conclude that the meridians v = v0 are geodesics, and that each parallel u = u0 is a geodesic if and only if g0 (u0 ) = 0. What does the surface look like if all the parallels are geodesics? Exercise 3.6.30. (a) Recall that in polar coordinates, the metric in R2 \ {0} is given in differential notation by ds2 = dr2 + r2 dθ 2 . Show that the non-vanishing Christoffel symbols are Γrθθ (r, θ ) = −r

θ and Γrθ (r, θ ) = Γθθr (r, θ ) = 1/r.

(b) In Rindler coordinates, the metric in the Rindler wedge is given in differential notation by ds2 = dρ2 − ρ2 dϕ2 . Show that the non-vanishing Christoffel symbols are Γ ϕϕ (ρ, ϕ) = ρ and Γρϕ (ρ, ϕ) = Γ ϕρ (ρ, ϕ) = 1/ρ. ρ

ϕ

ϕ

(c) Solve the geodesic equations and conclude (again) that the geodesics in each case are straight lines. Exercise 3.6.31. Let h : R>0 → R be a smooth function and consider the injective and regular parametrized surface x : R>0 × ]0, 2π [ → R3ν given by x(u, v) = (u cos v, u sin v, h(u)). Show that the non-vanishing Christoffel symbols for x are Γuuu (u, v) = (−1)ν

h0 (u)h00 (u) , 1 + (−1)ν h0 (u)2

Γuvv (u, v) =

and Γvuv (u, v) = Γvvu (u, v) =

−u , 1 + (−1)ν h0 (u)2

1 . u

Exercise 3.6.32. Fix 0 < α0 < π/2 and consider x : R>0 × ]0, 2π [ → R3ν given by x(u, v) = (u cos v tan α0 , u sin v tan α0 , u). . We have seen in Exercise 3.2.2 (p. 164) that M = x(R>0 × ]0, 2π [) is a regular surface, and that for α0 6= π/4 in L3 , M is non-degenerate. Determine the Christoffel symbols for x, and solve the geodesic equations to find the geodesics of M.

250  Introduction to Lorentz Geometry: Curves and Surfaces

Exercise 3.6.33. Let M ⊆ R3ν be a non-degenerate regular surface and (U, x) be a v-Clairaut parametrization for M. Show that as geodesic equations in this case are: u00 +

Ev 0 0 Ev 0 2 Gv 0 2 u v = 0 and v00 − (u ) + (v ) = 0. E 2G 2G

Also read the Christoffel symbols from the above equations. Exercise 3.6.34. State and prove a result analogous to Proposition 3.6.31 (p. 241) for v-Clairaut parametrizations. Exercise† 3.6.35. Prove Theorem 3.6.35 (p. 244).

3.7

THE FUNDAMENTAL THEOREM OF SURFACES

In the beginning of this chapter, we promised a surface analogue to the Fundamental Theorem of Curves, seen in Chapter 2 (Theorem 2.3.20, p. 112) and, moreover, at this point we still owe you a proof of Gauss’ Theorema Egregium (Theorem 3.3.14, p. 186). In the brief discussion about geodesics we had in the previous section, one of the main concepts used to describe geodesics in coordinates were the Christoffel symbols — which now will be used to pay all our debts. 3.7.1

The compatibility equations

In the Fundamental Theorem of Curves, we have used the curvature and torsion of an admissible curve to “reconstruct it”, up to a rigid motion of the ambient space. More precisely, we solve the Frenet-Serret system of the curve, to then determine the curve completely from given initial conditions, by using the existence and uniqueness theorem for systems of ODEs. The natural attack strategy in this case would be to try and reconstruct a nondegenerate regular surface M from its Gaussian and mean curvatures. However, the same geometric information is already captured by the First and Second Fundamental Forms of M. In our study of curves, we dealt with parametrizations themselves, instead of looking at subsets of the space, as we have done for surfaces. So, aiming to repeat the same strategy, we will use a parametrization (U, x) to do a local analysis, and then determine the system that has to be satisfied by the coefficients ( gij )1≤i,j≤2 and (hij )1≤i,j≤2 to then, formally solve it. The first problem we encounter, though, is that this system will now consist of PDEs instead of ODEs. Thus, we need a more powerful existence and uniqueness theorem for solutions. This means that at this point, our progress depends on the following result, whose proof is out of our current reach: Theorem 3.7.1 (Frobenius). Let Fij : Rm+n → R be class C2 functions, for 1 ≤ i ≤ n and 1 ≤ j ≤ m, satisfying the compatibility equations n ∂F n ∂Fij ∂F ∂F ij +∑ Fk` = ik + ∑ ik Fj` , ∂xk `=1 ∂y` ∂x j ∂y` `=1

for all possibilities of i, j and k. If x = ( x1 , . . . , xm ) and y = (y1 , . . . , yn ), given a ∈ Rm

Surfaces in Space  251

and b ∈ Rn , the initial value problem   ∂yi = F ( x, y) ij ∂x j  y( a) = b has a unique solution in some neighborhood of a. Remark. We’ll say that a system of PDEs as above is integrable if the compatibility equations from the Frobenius Theorem are satisfied. For a proof see, for instance, [68]. Well, if N is a Gauss map for M, we know that the three identities

( xuu )v = ( xuv )u ,

( xvv )u = ( xuv )v ,

and ( N ◦ x)uv = ( N ◦ x)vu

must hold. We may write both sides of all relations as linear combinations of xu , xv and N ◦ x, and thus obtain nine relations which must necessarily be satisfied by the coefficients ( gij )1≤i,j≤2 and (hij )1≤i,j≤2 . Before highlighting some of these relations, note that this tells us explicitly that given smooth maps gij , hij : U → R, for 1 ≤ i, j ≤ 2,
which are symmetric in the indices i and j, such that det ( gij )1≤i,j≤2 6= 0, it is not necessarily true that they represent the Fundamental Forms of a surface. We will bring this up again soon. In any case, we move on: Proposition 3.7.2 (Compatibility). Let M ⊆ R3ν be a non-degenerate regular surface and (U, x) be a parametrization for M. Then the Gauss equations hold:  2 2 1 2 2 2 1 2 2 2    EK = (Γ11 )v − (Γ12 )u + Γ11 Γ12 + Γ11 Γ22 − Γ12 Γ11 − Γ12 Γ21   FK = (Γ1 ) − (Γ1 ) + Γ2 Γ1 − Γ2 Γ1 12 u 11 v 12 12 11 22 2 2 1 2  FK = (Γ12 )v − (Γ22 )u + Γ12 Γ12 − Γ122 Γ211     GK = (Γ1 ) − (Γ1 ) + Γ1 Γ1 + Γ2 Γ1 − Γ1 Γ1 − Γ2 Γ1 , 22 u 22 11 22 12 12 v 21 12 12 22 as well as the Codazzi-Mainardi equations: ( ev − f u = eΓ112 + f (Γ212 − Γ111 ) − gΓ211 f v − gu = eΓ122 + f (Γ222 − Γ112 ) − gΓ212 , where K ≡ K ◦ x denotes the Gaussian curvature of M. Remark. These equations are known as the compatibility equations, because they are precisely the conditions ensuring that the systems we will encounter in the proof of the Fundamental Theorem of Surfaces are integrable. Proof: By way of information, let’s study the identity xuuv = xuvu to obtain the first Gauss equation and the first Codazzi-Mainardi equation. Identifying u ↔ 1, v ↔ 2, N ≡ N ◦ x and omitting points of evaluation as usual, on one hand we have that ! 2

k xk + e M h11 N ∑ Γ11

xuuv =

k =1

=

2

k =1

k,r =1

k )v xk + ∑ ∑ (Γ11 2

=

v

2

∑

r =1

r (Γ11 )v +

2

k r Γ11 Γk2 xr + e M

2

r =1

k =1

k r Γk2 − e M h11 hr2 ∑ Γ11

k =1

2

k hk2 N + e M (h11 )v N − e M h11 ∑ hr2 xr ∑ Γ11

!

2

xr + e M

(h11 )v +

k hk2 ∑ Γ11

k =1

! N.

252  Introduction to Lorentz Geometry: Curves and Surfaces

On the other hand: 2

∑

xuvu =

! k Γ12 xk

+ e M h12 N

k =1

=

2

k =1

k,r =1

k )u xk + ∑ ∑ (Γ12 2

=

u

2

∑

r (Γ12 )u +

r =1

k r Γ12 Γk1 xr + e M

2

2

k hk1 N + e M (h12 )u N − e M h12 ∑ hr1 xr ∑ Γ12 r =1

k =1

2

k r Γk1 − e M h12 hr1 ∑ Γ12

!

2

xr + e M

(h12 )u +

k =1

k hk1 ∑ Γ12

! N.

k =1

Now recall that Lemma 3.3.7 (p. 180) gives us that: h22 =

2

∑ h2j g j2 =

j =1

− f F + gE EG − F2

and h21 =

2

∑ h1j g j2 =

j =1

−eF + f E . EG − F2

With this, equating the coefficients of xv on both expressions (i.e., setting r = 2), we obtain:     −e f F + f 2 E −e f F +egE 2 ) + Γ1 Γ2 + Γ2 Γ2 − e = ( Γ . (Γ211 )v + Γ111 Γ212 + Γ211 Γ222 − e M M 12 u 12 11 12 21 EG − F2 EG − F2 Reorganizing the above expression by moving all the Christoffel symbols to the left and all the terms with Fundamental Forms to the right, the Gauss equation

(Γ211 )v − (Γ212 )u + Γ111 Γ212 + Γ211 Γ222 − Γ112 Γ211 − Γ212 Γ221 = EK follows, as wanted. And equating the coefficients of N directly yields ev − f u = eΓ112 + f (Γ212 − Γ111 ) − gΓ211 .

As an immediate corollary of the above proposition, we have Gauss’ Theorema Egregium, since the First Fundamental Form, the Christoffel symbols (as well as their derivatives) are preserved by local isometries. Despite this, it is actually possible to prove the Theorema Egregium completely bypassing Christoffel symbols (see the steps to do this in Exercise 3.7.4). In particular, we may now formally present the explicit formula for K mentioned in Section 3.3, only in terms of the First Fundamental Form: Proposition 3.7.3. Let M ⊆ R3ν be a non-degenerate regular surface and (U, x) be an orthogonal parametrization for M, that is, satisfying F = 0. Then ! ! ! p p ( | G |)u ( | E|)v −1 p p K◦x= p + ev eu . | EG | | E| |G| v u Remark. This formula is also frequently written as ! Gu − eu ev p + K◦x= p 2 | EG | | EG | u For an example, see Exercise 3.7.3.

Ev p

| EG |

! ! . v

Surfaces in Space  253

Proof: Since F = 0, we directly have the expressions for the Christoffel symbols in terms of E, G, and their derivatives (see the remark following the proof of Proposition 3.6.16, p. 228). With this, solving for K ≡ K ◦ x in the first Gauss equation (given in the statement of Proposition 3.7.2 above) gives us that:     1 Ev Ev Gv Ev2 Gu2 1 Gu Eu Gu K=− + − . − + 2 − E 2G v E 2G u 4E G 4G2 4E2 G 4EG2 Computing explicitly the expression given in the statement of this result also gives the above after simplifications. Back to our main goal: the Fundamental Theorem of Surfaces. Not too long ago we saw that given smooth functions gij , hij : U → R, for 1 ≤ i, j ≤ 2, symmetric in the
indices i and j, such that det ( gij )1≤i,j≤2 6= 0, they will have to satisfy the compatibility equations where, of course, the Christoffel symbols associated to ( gij )1≤i,j≤2 are formally defined by the expression given in Proposition 3.6.16: Γijk

2

1 = ∑ gkr 2 r =1



∂g jk ∂gij ∂gik + − ∂ur ∂ui ∂u j

 .

We know that the compatibility equations are a necessary condition for the ( gij )1≤i,j≤2 and (hij )1≤i,j≤2 to locally determine a regular surface, but are they also sufficient? In other words, by differentiating the relations

( xuu )v = ( xuv )u ,

( xvv )u = ( xuv )v ,

and ( N ◦ x)uv = ( N ◦ x)vu

even further, can’t we obtain more compatibility equations? The negative answer to this question is the: Theorem 3.7.4 (Bonnet). Let U be an open subset of R2 and, for 1 ≤ i, j ≤ 2, gij , hij : U → R be smooth functions formally satisfying the Gauss and Codazzi-Mainardi equations, and symmetric: gij = g ji , hij = h ji for 1 ≤ i, j ≤ 2. Also, let (u0 , v0 ) ∈ U and p0 ∈ R3 be points, and ( xu,0 , xv,0 , N 0 ) be a basis for R3ν such that gij (u0 , v0 ) = h xi,0 , x j,0 i for 1 ≤ i, j ≤ 2, with N 0 unit and orthogonal to both
xu,0 and xv,0 . Suppose in addition that det ( gij (u, v))1≤i,j≤2 never vanishes on U and has a constant sign, equal to (−1)ν e N 0 . Then there are a neighborhood V of (u0 , v0 ) in R2 , a unique injective and regular . parametrized surface x : V → R3ν , and a Gauss map N ≡ N ◦ x for M = x(V ) satisfying:

• x ( u0 , v0 ) = p0 ; •

∂x (u0 , v0 ) = xi,0 , for 1 ≤ i ≤ 2; ∂ui

• N ( p0 ) = N 0 ; • the coefficients of the Fundamental Forms of M relative to x are precisely gij V and hij V . In particular, e M = e N 0 . Proof: Introduce new vector variables w1 =

∂x ∂u1

and w2 =

∂x . ∂u2

254  Introduction to Lorentz Geometry: Curves and Surfaces

Let’s first solve a new system of PDEs for the vector variables w1 , w2 and N, to then solve the initial system for x. Consider:  2  ∂wi   = Γijk wk + e N 0 hij N,  ∑ j  ∂u  k =1  2 ∂N  = − gik hkj wi , ∑  j  ∂u  i,k =1    (w1 (u0 , v0 ), w2 (u0 , v0 ), N (u0 , v0 )) = ( xu,0 , xv,0 , N 0 ). The compatibility equations for this system are precisely the Gauss and Codazzi-Mainardi equations, which are satisfied by assumption. Thus, the Frobenius Theorem provides a solution (w1 , w2 , N ) in some neighborhood of (u0 , v0 ). The next step is, naturally, to verify that

h N, N i = e N 0 ,

h N, wi i = 0, and hwi , w j i = gij ,

(1 ≤ i, j ≤ 2)

for all parameters (u, v), and not only at (u0 , v0 ). Indeed, using the first system we have that:  2  ∂   h N, N i = − 2 gik hkj h N, wi i  ∑  j  ∂u  i,k =1    ∂ 2 2 `k h N, w i = − g h h w , w i + i i ∑ ∑ Γijk h N, wk i + eN 0 hij h N, N i kj ` j  ∂u  `,k=1 k =1    2 2  ∂  `  h h N, w i + h w , w i = Γ h w , w i + e  N 0 ik j ∑ Γ`jk hw` , wi i + eN 0 h jk h N, wi i. ∑ ik ` j  ∂uk i j `=1

`=1

Note that since this last system was obtained from an integrable system, it is also integrable. Moreover, the functions e N 0 , 0 and gij are also solutions for this system, with the same initial conditions (we ask you to check this in Exercise 3.7.5). Hence the Frobenius Theorem ensures that

h N, N i = e N 0 ,

h N, wi i = 0, and hwi , w j i = gij ,

(1 ≤ i, j ≤ 2)

on the neighborhood where the solution (w1 , w2 , N ) is defined. In particular, this ensures that the solution remains a basis for R3ν in all points. Let’s finally go back to the initial system considered: w1 =

∂x ∂u1

and w2 =

∂x . ∂u2

Since Γijk = Γkji and hij = h ji for 1 ≤ i, j, k ≤ 2, this system is integrable and we obtain a neighborhood V of (u0 , v0 ) small enough where all the obtained conditions on w1 , w2 and N so far still hold, and a regular map x : V → R3ν that solves the system, with initial condition x(u0 , v0 ) = p0 . Such map may be assumed to be injective, reducing V further if necessary. Now, note that   ∂x ∂x , = hwi , w j i = gij , (1 ≤ i, j ≤ 2) ∂ui ∂u j . and so the gij are indeed the coefficients of the First Fundamental Form of M = x(V ) relative to x. Similarly, we have that   ∂x N, i = h N, wi i = 0, (1 ≤ i ≤ 2) ∂u

Surfaces in Space  255

and thus N is a Gauss map for M. This allows us to compute the coefficients of the Second Fundamental Form of M relative to x by using the previous systems:     ∂2 x ∂wi N, i j = N, ∂u ∂u ∂u j * + 2

N,

=

∑ Γijk wk + eN 0 hij N

k =1 2

=

∑ Γijk h N, wk i + eN 0 hij h N, N i = hij ,

(1 ≤ i, j ≤ 2)

k =1

as wanted. n Remark (Flat immersions). One noteworthy situation is when ( gij )i,j =1 is a matrix of constant functions, which forces all the Christoffel symbols to vanish. Then the Gauss equations should impose that K = 0 as well. This means that the Gauss equations are satisfied if and only if the matrix   e f f g

is always singular, while the Codazzi-Mainardi equations are satisfied if and only if we have ev = f u and f v = gu . If this is the case, in particular note that f = 0 if and only if e = 0 and g depends only on v or g = 0 and e depends only on u. Solving the compatibility system in general is very difficult. You can play around in Exercise 3.7.7 with the (reasonable) case where e, f and g are all constant as well to convince yourself of this. Bonnet’s Theorem ensures, under suitable conditions, the existence and uniqueness of the surface, once initial conditions are given. In the general case, we have uniqueness up to rigid motions of the ambient space: Proposition 3.7.5. Let x, e x : U → R3ν be two non-degenerate regular parametrized surfaces, with U connected, such that gij = geij and hij = e hij on U. Then there is a positive F ∈ Eν (3, R) such that e x = F ◦ x. e compatible with x and e Proof: Consider the Gauss maps N and N x, given by N (u, v) =

xu (u, v) × xv (u, v) , k xu (u, v) × xv (u, v)k

. e Since gij = geij , in particular we have that e N (u,v) = e e and similarly for N. N (u,v) = e and, for each (u, v) ∈ U, the linear map A(u, v) defined by A(u, v)( xu (u, v)) = e xu (u, v), A(u, v)( xv (u, v)) = e xv (u, v) and e (u, v) A(u, v)( N (u, v)) = N is in SOν (3, R). Let’s then check that A(u, v) is actually constant, by verifying that its partial derivatives are the zero map. We identify u ↔ u1 , v ↔ u2 , as always. Omitting

256  Introduction to Lorentz Geometry: Curves and Surfaces

points of evaluation everywhere except for A(u1 , u2 ), on one hand we have that e xij = A j (u1 , u2 )( xi ) + A(u1 , u2 )( xij ) 1

2

1

2

= A j (u , u )( xi ) + A(u , u )

2

∑

! Γijk xk

+ ehij N

k =1 2

= A j (u1 , u2 )( xi ) +

e ∑ Γijk exk + ehij N.

k =1

On the other hand:

2

e xij =

e ∑ Γeijk exk + eehij N.

k =1

ek . This, together with the fact that Note that since gij = geij , we also have that Γijk = Γ ij e hij = hij , allows us to compare both expressions and conclude that A j (u1 , u2 )( xi ) = 0

(1 ≤ i, j ≤ 2).

Moreover, we have that e j. A j (u1 , u2 )( N ) + A(u1 , u2 )( N j ) = N e j . Hence But, we also have that hi j = e hi j , and this gives us that A(u1 , u2 )( N j ) = N A j (u1 , u2 )( N ) = 0 and, since j was arbitrary and U is connected, we conclude that A j (u1 , u2 ) is the zero map. Thus, A(u1 , u2 ) ≡ A is constant. Now, since

(ex − A( x)) j = ex j − A( x j ) = 0 for all j, it follows that e x − A( x) = b ∈ R3ν is constant, so that F = Tb ◦ A ∈ Eν (3, R) is the positive rigid motion we’re looking for.

Exercises Exercise† 3.7.1. Let M ⊆ R3ν be a non-degenerate regular surface, N be a Gauss map for M, and (U, x) be a parametrization for M. Deduce again the Codazzi-Mainardi equations from the relation ( N ◦ x)uv = ( N ◦ x)vu . Exercise 3.7.2. Fill the details in the proof of Proposition 3.7.3 (p. 252). Exercise 3.7.3. Let M ⊆ R3ν be a non-degenerate regular surface and (U, x) be a parametrization for M. We’ll say that x is isothermal if F = 0 and | E| = | G | = λ2 , for some non-vanishing smooth function λ : U → R. Show that K◦x=−

4(log λ) λ2

if M is spacelike and that

(log λ) λ2 if M is timelike, where 4 and  denote the Laplacian and d’Alembertian operators, seen previously in Exercise 3.6.28 (p. 248). K ◦ x = − eu

Surfaces in Space  257

Remark. It is possible to define a single operator 4h·,·i (known as the Laplace-Beltrami operator) on functions over M which generalizes the operators 4 and . The name “isothermal” is motivated by the fact that such coordinates satisfy 4h·,·i x = 0, being stationary solutions to the heat equation in M. Exercise† 3.7.4. Let M ⊆ R3ν be a non-degenerate regular surface and p ∈ M. In this exercise, we will see an alternative proof for Gauss’ Theorema Egregium (slightly adapting the argument given by Sternberg in [67]), expressing the Gaussian curvature K ( p) in terms only of the coefficients of the First Fundamental Form of M relative to an inertial parametrization centered at p (whose existence was shown in Theorem 3.4.9, p. 195). We will treat the situation where M is spacelike, with the timelike case being similar. Recall that rigid motions of the ambient space (that is, elements of Eν (3, R)) preserve the Fundamental Forms of M, by Proposition 3.3.13 (p. 185). In particular, the Gaussian curvature is also preserved. And clearly the composition of an inertial parametrization with an isometry is again inertial. Thus, we may assume that we have an inertial parametrization (U, x) centered at p = 0, of the form x(u, v) = (u + a(u, v), v + b(u, v), c(u, v)), such that xu (0, 0) = (1, 0, 0) and xv (0, 0) = (0, 1, 0), for certain smooth functions a, b, c : U → R. When M is timelike, one must apply a rigid motion that takes Tp M in some coordinate timelike plane in L3 . (a) Use that x(0, 0) = (0, 0, 0) and ( xu (0, 0), xv (0, 0)) = (e1 , e2 ) to show that a, b, c and their first order partial derivatives all vanish at the origin. (b) Omitting points of evaluation, show that E = (1 + au )2 + bu2 + (−1)ν c2u , F = (1 + au ) av + (1 + bv )bu + (−1)ν cu cv G=

a2v

+ ( 1 + bv )

2

and

+ (−1)ν c2v .

(c) Use that the partial derivatives of E, F and G vanish at the origin (since x is inertial) to show that the second order partial derivatives of a and b also vanish at the origin. (d) Compute the coefficients hij (0, 0), for 1 ≤ i, j ≤ 2, and show that K ( p) = det Hess c(0,0) = cuu (0, 0)cvv (0, 0) − cuv (0, 0)2 . (e) Show that Fuv (0, 0) = auvv (0, 0) + buuv (0, 0) + (−1)ν (cuu (0, 0)cvv (0, 0) + cuv (0, 0)2 ), Evv (0, 0) = 2( auvv (0, 0) + (−1)ν cuv (0, 0)2 ) and Guu (0, 0) = 2(bvuu (0, 0) + (−1)ν cvu (0, 0)2 ). (f) Show that Evv (0, 0) Guu (0, 0) − 2 2 and conclude Gauss’ Theorema Egregium. K ( p) = Fuv (0, 0) −

Remark. This proof indeed could have been presented soon after Theorem 3.4.9, but we believe that it would be better appreciated only now. Do you agree?

258  Introduction to Lorentz Geometry: Curves and Surfaces

Exercise† 3.7.5. In the proof of Theorem 3.7.4 (p. 253), verify that the functions e N 0 , 0 and gij are solutions to the system used to show that the relations

h N, N i = e N 0 ,

h N, wi i = 0, and hwi , w j i = gij ,

(1 ≤ i, j ≤ 2)

were valid in the domain of definition of the solution (w1 , w2 , N ). Exercise 3.7.6. Show that 
1 (a) gij (u, v) 1≤i,j≤2 = 0 
1 (b) gij (u, v) 1≤i,j≤2 = 0

there is not, in R3ν , any regular parametrized surface with:   
0 v 0 and hij (u, v) 1≤i,j≤2 = . 1 0 1   2 
0 cos u 0 and hij (u, v) 1≤i,j≤2 = . cos2 u 0 1

Hint. Use the Codazzi-Mainardi equations. Exercise 3.7.7. Determine, when possible, a parametrized regular surface x : U → R3ν such that:    

1 0 −1 0 (a) gij (u, v) 1≤i,j≤2 = and hij (u, v) 1≤i,j≤2 = , 0 1 0 1 with x(0, 0) = (0, 0, 1), xu (0, 0) = (1, 0, 0) and xv (0, 1, 0).    

1 0 1 0 (b) gij (u, v) 1≤i,j≤2 = and hij (u, v) 1≤i,j≤2 = , 0 1 0 0 with x(0, 0) = (0, 0, 1), xu (0, 0) = (1, 0, 0) and xv (0, 1, 0).    

−1 0 0 0 (c) gij (u, v) 1≤i,j≤2 = and hij (u, v) 1≤i,j≤2 = , 0 1 0 −1 with x(0, 0) = (1, 0, 0), xu (0, 0) = (0, 0, 1) and xv (0, 0) = (0, 1, 0). Exercise 3.7.8. Let M ⊆ R3ν be a non-degenerate regular surface and (U, x) be a principal parametrization for M. Show that the Codazzi-Mainardi equations boil down to: g Gu  e g Ev  e + and gu = + , (a) ev = 2 E G 2 E G 1 ∂κ1 Ev 1 ∂κ2 Gu = and = , κ2 − κ1 ∂v 2E κ1 − κ2 ∂u 2G under the assumption that M is free of umbilic points.

(b) or

Remark. Can you see how H is hidden in the formulas in (a)?

CHAPTER

4

Abstract Surfaces and Further Topics

INTRODUCTION We approach the end of this pleasant excursion through Lorentz Geometry of curves and surfaces with this final chapter, which has the goal of pointing new directions of study for the reader. Here, we will slightly change the dynamics of the text: the Sections 4.2, 4.3, and 4.4 are independent of each other, and may be read in any order. However, here we’ll push a few limits, assuming for the first time some familiarity with point-set topology and complex analysis. In Section 4.1, we seek to free ourselves from the Codazzi-Mainardi equations, seen at the end of the previous chapter, presenting the definition of an abstract surface, which is easily generalized to give us the notion of a differentiable manifold, one of the most important in Differential Geometry. We keep adapting most of the concepts seen for regular surfaces in R3ν for this new setting, presenting examples. In particular, we have the concept of a pseudo-Riemannian metric, that is, a First Fundamental Form which does not come from any ambient space, but is instead prescribed. From this, we again define isometries, geodesics, and Gaussian curvature, which are all illustrated in models for Hyperbolic Geometry. In Section 4.2, we apply what has been discussed in Section 4.1 to obtain a local classification of the geometric surfaces with constant Gaussian curvature, employing the so-called Fermi parametrizations. We also illustrate some realizations of the obtained metrics in the Lorentzian spaces L3 and R32 . The long-awaited presentation of the set of split-complex numbers, denoted by C0 , is finally given at the beginning of Section 4.3. We briefly discuss the relation between complex and split-complex numbers with critical surfaces in R3ν , via two constructions: the Bonnet rotations, which are isometric deformations between conjugate or Lorentzconjugate critical surfaces; and the Enneper-Weierstrass representation formulas, which say exactly how critical surfaces are related to triples of holomorphic or split-holomorphic functions satisfying certain conditions. In particular, we’ll employ the so-called isothermal parametrizations, which are convenient for the study of critical surfaces, and also the complex Lorentzian space C31 . We’ll close the curtains with Section 4.4, where we invite you to proceed with your studies in Riemannian Geometry and General Relativity, presenting a few ideas about completeness in Riemannian manifolds and causality in Lorentzian manifolds. We introduce the notion of intrinsic distance in Riemannian manifolds, and state the famous Hopf-Rinow Theorem, which provides a bridge between the geometry and the topology of 259

260  Introduction to Lorentz Geometry: Curves and Surfaces

the manifold. Motivated to find an analogous result for Lorentzian manifolds, we present the formal definition of a spacetime, taking the first step towards General Relativity. The time separation in a spacetime plays the same role as the intrinsic distance in a Riemannian manifold, and it is closely related to the causality of the spacetime itself. With this, we present the several conditions defining the causal hierarchy of spacetimes, and we conclude the discussion with the statement of the Avez-Seifert Theorem: the analogue of the Hopf-Rinow Theorem for globally hyperbolic spacetimes. “The views of space and time which I wish to lay before you have sprung from the soil of experimental physics, and therein lies their strength. They are radical. Henceforth space by itself, and time by itself, are doomed to fade away into mere shadows, and only a kind of union of the two will preserve an independent reality.” – Hermann Minkowski (address to the 80th Assembly of German Natural Scientists and Physicians, September 21st , 1908).

4.1

PSEUDO-RIEMANNIAN METRICS

In the previous chapter we have seen that surfaces in R3ν must satisfy the compatibility equations given in Proposition 3.7.2 (p. 251). The Codazzi–Mainardi equations can be quite restrictive and, since they involve the coefficients of the Second Fundamental Form, are not intrinsic to the surfaces. In this section we will study a more general notion of surface, eliminating the dependence of an ambient space and, consequently, of those equations. Definition 4.1.1 (Abstract surface). An abstract surface is a set M equipped with a collection A = { xα : Uα ⊆ R2 → xα (Uα ) ⊆ M | α ∈ A} of abstract parametrizations (i.e., injective maps defined in open subsets of R2 ) satisfying the following conditions: (i)

S

α∈ A

xα (Uα ) = M;

1 2 (ii) given α, β ∈ A, x− β ◦ xα is a smooth map between open subsets of R (in the correct domain), whenever xα (Uα ) ∩ x β (Uβ ) 6= ∅;

(iii) A is maximal relative to (ii): if y : V → y(V ) ⊆ M is an abstract parametrization satisfying (ii), then y ∈ A. The collection A is called a (maximal) atlas for M. Remark.

• If U is an open subset of Rn instead of R2 , we get what is called a differentiable manifold of dimension n. • We say that W ⊆ M is open if x−1 (W ) is an open subset of R2 , for every x ∈ A. This indeed defines a topology on M, which turns all the abstract parametrizations into homeomorphisms. Note that it still does not make sense to ask whether such parametrizations are diffeomorphisms. • The topology on M induced by A is not necessarily Hausdorff or second-countable. One usually assumes from the beginning that all the manifolds in the discussion have those properties. Being Hausdorff is very natural, and second-countability essentially says that the manifold does not have “too many” open sets, avoiding pathological examples and ensuring the existence of the so-called partitions of unity.

Abstract Surfaces and Further Topics  261

Example 4.1.2. (1) Open subsets of R2 and regular surfaces M ⊆ R3ν are the first examples of abstract surfaces. (2) If x : U ⊆ R2 → Rn is injective and differentiable, such that Dx(u, v) has full rank, then x(U ) is an abstract surface, when equipped with the maximal atlas containing x itself. (3) If on the sphere S2 ⊆ R3 we define the equivalence relation ∼ by setting p ∼ q if and only if q ∈ { p, − p} (i.e., identifying antipodal points), we may define an atlas . in the quotient RP2 = S2 /∼ , which turns it into an abstract surface as well. For more details, see [16]. Remark. A rich class of examples of differentiable manifolds consists of matrix groups. For example, one may show that O(n, R) and SO(n, R) are differentiable manifolds of dimension n(n − 1)/2. To study geometry in M, we need some notion of Differential Calculus on M. We define what it means for a function f : M → R to be smooth by copying Definition 3.1.16 (p. 141) word by word, mimicking the situation regarding surfaces in R3ν . Similarly, we define what it means for a function between two abstract surfaces to be smooth and, in particular, we now know what is a diffeomorphism between abstract surfaces. The next step would be to define what is the tangent plane at a point p ∈ M. Before, however, we defined the tangent plane as a certain subspace of R3ν (according to Definition 3.1.12, p. 139), so we must be careful and take as a definition something which does not depend on the (now non-existent) ambient space. The characterization given in Proposition 3.1.14 (p. 140) sounds like a good idea, once we define the velocity vector of a curve in M. Differentiating component by component is no longer a valid option, as M is, a priori, an abstract set. Bearing in mind that when we have the ambient space R3ν , tangent vectors act in functions f ∈ C∞ ( M ) via directional derivatives, we have the: Definition 4.1.3. Let M be an abstract surface, p ∈ M and α : I → M be a smooth curve such that α(t0 ) = p. The tangent vector to α at t0 is the map α0 (t0 ) : C∞ ( M ) → R defined by . d 0 α (t0 )( f ) = f (α(t)). dt t=t 0

Remark. The tangent vector to α at t0 is also called the velocity vector of α at t0 , like before. With this, we finally define the tangent plane to M at p, Tp M, by Proposition 3.1.14. Similarly, we define the tangent space to a manifold M of any dimension n, in any point p, as the space of velocities of curves which start at p. Thus, Definition 3.1.23 (p. 145) of differential of a smooth function between regular surfaces now makes sense for abstract surfaces as well. So, if (U, x) is a parametrization for M, we write xu (u, v) = dx(u,v) (1, 0) and

xv (u, v) = dx(u,v) (0, 1),

and continue to identify u ↔ u1 and v ↔ u2 whenever convenient. However, we have no means to directly define the First Fundamental Form of M, as we no longer have the ambient scalar product available. This is the main reason we presented the definition of abstract surface: now the geometry of M will not come from R3ν , but will be prescribed:

262  Introduction to Lorentz Geometry: Curves and Surfaces

Definition 4.1.4 (Pseudo-Riemannian metric). Let M be an abstract surface. A pseudoRiemannian metric on M is a choice of non-degenerate scalar products h·, ·i p in each tangent plane Tp M, all with the same index, such that for every parametrization x of M, . gij (u, v) = h xi (u, v), x j (u, v)i x(u,v) is a smooth function of (u, v). We say that ( M, h·, ·i) is a geometric surface. Remark.

• We say that the metric is Riemannian if every scalar product is positive-definite, and that the metric is Lorentzian if the index of each scalar product is equal to 1. For geometric surfaces, those two types can be characterized by the sign of the determinant of the Gram matrix ( gij )1≤i,j≤2 . • It is usual to omit p in h·, ·i p , when there’s no risk of confusion. Moreover, if . p v ∈ Tp M, we write kvk = |hv, vi|. When the metric is Riemannian, we have norms in all the tangent planes. • When the metric is Lorentzian, we define the causal type and the indicator of v ∈ Tp M by the sign of hv, vi, like in Definition 1.1.3 (p. 3, Chapter 1). • Like before, we’ll also express a pseudo-Riemannian metric in terms of coordinates with the differential notation ds2 =

2

∑

gij dui du j = E(u, v) du2 + 2F (u, v) du dv + G (u, v) dv2 .

i,j=1

• Non-degeneracy of the metric ensures the existence of the inverse coefficients ( gij )1≤i,j≤2 , of fundamental importance for the development of the theory. • The above definition naturally extends to differentiable manifolds of any dimension. We say that ( M, h·, ·i) is a Riemannian manifold or a Lorentzian manifold, according to whether the index of all the scalar products is 0 or 1. The names “Riemann surface” and “Lorentz surface”, in turn, have precise meanings in other areas of geometry, and should be avoided in this setting. Example 4.1.5. (1) The First Fundamental Form of a non-degenerate regular surface M ⊆ R3ν is a pseudo-Riemannian metric on M (regarded as an abstract surface). Such metric is Riemannian if M is spacelike, and Lorentzian if timelike. (2) The above idea may be generalized by considering surfaces inside Rnν , and taking the restriction of the ambient scalar product in each tangent plane, whenever nondegenerate. The only difference is that while each tangent plane Tp M has dimension 2, the normal space Tp M⊥ is no longer a line for n > 3. It is possible to study how the geometry of M relates to the geometry of Rnν by using a vector-valued Second Fundamental Form II, but the theory becomes more sophisticated, and it is better approached with tools beyond the scope of this text. See, for example, [54]. Bearing this in mind, all the concepts previously studied which are intrinsic to the surface may be studied for abstract surfaces. For example, if α : I → M is given, we

Abstract Surfaces and Further Topics  263

define the arclength and the energy of α according to the pseudo-Riemannian metric via the usual formulas Z Z 1 L[α] = kα0 (t)kα(t) dt and E[α] = hα0 (t), α0 (t)iα(t) dt. 2 I I We also define the area of a region R ⊆ M according to Definition 3.2.12 (p. 161). Example 4.1.6 (Warm-up). In M = R × R>0 , take the abstract parametrization given by the identity. (1) Consider the Riemannian metric ds2 = dx2 + y dy2 , given in Cartesian coordinates. All the tangent planes to M are (naturally) isomorphic to R2 (as vector spaces). For instance, we have that

h(1, 3), (−2, 3)i(0,2) = 1 · (−2) + 2 · 3 · 3 = 16, but

h(1, 3), (−2, 3)i(0,1) = 1 · (−2) + 1 · 3 · 3 = 7. If α : ]0, 1[ → M is given by α(t) = (0, t), we have that Z 1p Z 1√ 2 2 1 L[α] = 0 + t · 1 dt = t dt = . 3 0 0 In particular, e α(s) = (0, (3s/2)2/3 ) is a unit speed reparametrization of α. Lastly, . let’s see what the area of the square R = ]0, 1[2 is according to this metric. We have that Z q Z 1Z 1 √ 2 A( R) = 1 · y − 02 dx dy = y dx dy = . 3 R 0 0 (2) Consider now the Lorentzian metric ds2 = −dx2 + y dy2 , given in Cartesian coordinates. We have that

h(1, 1), (1, 1)i(0,2) = −12 + 2 · 1 · 1 = 1 h(1, 1), (1, 1)i(1,1) = −12 + 1 · 1 · 1 = 0 1 1 h(1, 1), (1, 1)i(−2,1/2) = −12 + · 1 · 1 = − . 2 2 Thus, we see that the causal type of a tangent vector may depend on its base point.

Space

Time

Light

Figure 4.1: Our Euclidean eyes see the vector (1, 1) always in the same way, but its length depends on its location.

264  Introduction to Lorentz Geometry: Curves and Surfaces

For a Lorentzian metric, there might or might not exist a vector which is timelike at all points. In this example, we see that (1, 0) satisfies h(1, 0), (1, 0)i( x,y) = −1, for all ( x, y) ∈ M. We say that (1, 0) induces a time orientation on M. To understand how this happens, let’s determine the lightlike curves on M: let α : I → M, given by α(t) = ( x (t), y(t)), be a lightlike curve. Its coordinates then satisfy the differential equation − x 0 (t)2 + y(t)y0 (t)2 = 0. If y0 (t) = 0, then x 0 (t) = 0 and thus α degenerates to a point. Otherwise, the Inverse Function Theorem allows us to use y as a parameter, and assume that the curve has the form α(y) = ( x (y), y). We then have that

− x 0 (y)2 + y = 0, √ whence x 0 (y) = ± y, and so x (y) = ±2y3/2 /3 + c, for some constant c. The lightlike curves passing through a given point play the same role as the diagonals in the plane L2 , and the vector (1, 0) will determine which one of the “timecones” will be the future. For example, for (1, 1) ∈ M, if we say that a timelike vector v ∈ T(1,1) M is future-directed if hv, (1, 0)i(1,1) < 0, the future timecone of (1, 1) is indicated as follows:

Figure 4.2: The future of (1, 1) according to this metric. In other words, pseudo-Riemannian metrics will give us a new way of measuring lengths and areas. And with a “suitable” Lorentzian metric, one may even define notions of future and past in M. We will discuss this further still in this chapter. Let’s continue to translate the results already established for our new setting: Definition 4.1.7. Let ( M1 , h·, ·i1 ) and ( M2 , h·, ·i2 ) be geometric surfaces. We say that that a (local) diffeomorphism φ : M1 → M2 is a (local) isometry if given any p ∈ M1 , and v, w ∈ Tp M1 , it holds that

hv, wi1 = hdφ p (v), dφ p (w)i2 . Remark. Clearly there can be no local isometry if one of the metrics is Riemannian and the other is Lorentzian. Furthermore, Propositions 3.2.17, 3.2.18, 3.2.19, 3.2.20 and 3.2.21 (p. 170, 171 and 172) all remain valid, and so the verification that certain maps are isometries may be simplified by employing differential notation, just as before. Example 4.1.8 (Models of Hyperbolic Geometry). In the mid-19th century, Bolyai and Lobachevsky found out that Euclid’s fifth postulate (the parallel postulate) is in fact independent of the remaining ones — giving birth to non-Euclidean geometries, with hyperbolic geometry being one of the most important ones. With the tools we have developed so far, we can present a few models:

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. (1) The Poincaré disk is the disk D2 = {( x, y) ∈ R2 | x2 + y2 < 1}, equipped with the Riemannian metric . 4(dx2 + dy2 ) ds2P = . (1 − x 2 − y2 )2 Let’s see that D2 and H2 are isometric, mimicking the idea of the second map seen in Example 3.1.21 (p. 143). The idea is to project D2 onto H2 , now using the point (0, 0, −1) as the base of the projection. Seeing the disk D2 inside L3 , in the plane z = 0 and centered at the origin, consider the intersection of the ray starting at (0, 0, −1) and passing through ( x, y, 0) with the hyperbolic plane H2 .

H2 φ( p)

D2

p

(0, 0, −1) Figure 4.3: The isometry between D2 and H2 . This ray can be parametrized by r (t) = (tx, ty, −1 + t), and so we seek t > 0 such that hr (t), r (t)i L = −1. Directly, we obtain t = 2/(1 − x2 − y2 ), and so we have φ : D2 → H2 given by  φ( x, y) =

2x 2y 1 + x 2 + y2 , , 1 − x 2 − y2 1 − x 2 − y2 1 − x 2 − y2

 .

We claim that φ is an isometry. Clearly φ is a diffeomorphism, so let’s just check that it preserves the metrics. Consider the parametrizations x : ]0, 1[ × ]0, 2π [ → D2 and e x : R>0 × ]0, 2π [ → H2 given by . x(u, v) = (u cos v, u sin v) . e x(ue, ve) = (sinh ue cos ve, sinh ue sin ve, cosh ue). Thus, we have that the metrics on D2 and H2 are expressed, respectively, as 4(du2 + u2 dv2 ) (1 − u2 )2

and de u2 + sinh2 ue de v2 ,

while φ is expressed as  φ( x(u, v)) = e x arcsinh



2u 1 − u2



 ,v .

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This way, if ue = arcsinh(2u/(1 − u2 )) and ve = v, we have that de u=

2 du and de v = dv. 1 − u2

So: 2 2  2 2u dv2 de u + sinh ue de v = du + 1 − u2 1 − u2 4 du2 4u2 dv2 = + (1 − u2 )2 (1 − u2 )2 2

2



2

=

4(du2 + u2 dv2 ) , (1 − u2 )2

as wanted. (2) The Poincaré half-plane is the half-plane {( x, y) ∈ R2 | y > 0} equipped with the Riemannian metric dx2 + dy2 ds2H = , y2 and it is also denoted by H2 . The Poincaré half-plane is also isometric to the Poincaré disk and the hyperbolic plane in L3 . The most efficient way to exhibit the isometry is to explore the relation of R2 with the complex numbers C, writing H2 = {z ∈ C | Im(z) > 0} and D2 = {w ∈ C | |w| < 1}, with metrics given, respectively, by ds2H =

dz dz Im(z)2

and ds2P =

4 dw dw , (1 − | w |2 )2

where we use dz = dx + i dy, dz = dx − i dy, etc. Define the so-called Cayley transform ψ : H2 → D2 by z−i ψ(z) = . z+i Firstly, we have to check that if Im(z) > 0, then |ψ(z)| < 1, so that the mapping does in fact take values in D2 . Indeed, using the known expression z − z = 2i Im(z), we have that 2 z − i 2 2 = (z − i)(z + i) = 1 + |z| − 2 Im(z) < 1. |ψ(z)| = z+i (z + i)(z − i) 1 + |z|2 + 2 Im(z) In fact, since ψ is a Möbius transformation, it maps the asymptotic boundary of H2 (the real axis) into the asymptotic boundary of D2 (the unit circle): the above calculation also says that Im(z) = 0 if and only if |ψ(z)| = 1. One can show that ψ is a diffeomorphism by directly exhibiting its inverse, for instance. Let’s see now that ψ preserves the metrics. Set w = ψ(z). With this, we have w= whence dw =

z−i z+i

and w =

z+i , z−i

2i 2i dz and dw = − dz. 2 ( z + i) ( z − i)2

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Thus: 4 dw dw = (1 − | w |2 )2



4

1 + |z|2 − 2 Im(z) 1− 1 + |z|2 + 2 Im(z) dz dz 16 = 2 | z + i|4 4 Im(z)

2

2i dz ( z + i)2



2i − dz ( z − i)2



| z + i|2 dz dz = , Im(z)2 as wanted. For further models, see Exercises 4.1.9 and 4.1.10. The adaptations to be made go on: Definition 4.1.9. Let ( M, h·, ·i) be a geometric surface, and consider a parametrization (U, x) for M. For 1 ≤ i, j, k ≤ 2, the Christoffel symbols of x are defined by   2 ∂g jk ∂gij 1 kr ∂gik k Γij = ∑ g − r . + 2 ∂u ∂ui ∂u j r =1 Definition 4.1.10. Let ( M, h·, ·i) be a geometric surface and α : I → M be a curve on M. We say that α is a geodesic of M if α is a critical point of the energy functional associated to the metric h·, ·i. Those two definitions validate all the discussion done in Section 3.6, so that we may continue to use variational methods to compute Christoffel symbols and geodesics in a geometric surface. All the relevant results remain true, in particular, the characterization given in coordinates by the geodesic differential equations u¨ k +

2

∑

Γijk u˙ i u˙ j = 0,

i,j=1

and the results regarding Clairaut parametrizations. Example 4.1.11. (1) Let’s find the Christoffel symbols for the Riemannian metric given in coordinates by ds2 = y2 dx2 + x2 dy2 in R2 \ {( x, y) ∈ R2 | xy 6= 0}. Define the Lagrangian L(t, x (t), x 0 (t), y(t), y0 (t)) =

1 ( y ( t )2 x 0 ( t )2 + x ( t )2 y 0 ( t )2 ). 2

Omitting points of evaluation, we have:   ∂L d ∂L 2 2 0= − = xy0 − (y2 x 0 )0 = xy0 − x 00 y2 − 2yx 0 y0 = 0 0 ∂x dt ∂x   ∂L d ∂L 2 2 0= − = yx 0 − ( x2 y0 )0 = yx 0 − y00 x2 − 2xx 0 y0 = 0, ∂y dt ∂y0 so that the geodesic equations are 2 x 2 y 2 2 x 00 + x 0 y0 − 2 y0 = 0 and y00 − 2 x 0 + x 0 y0 = 0, y x y x

268  Introduction to Lorentz Geometry: Curves and Surfaces

and the non-zero Christoffel symbols are x 1 x , Γyy ( x, y) = − 2 y y 1 y y y y Γ xy ( x, y) = Γyx ( x, y) = and Γ xx ( x, y) = − 2 . x x x Γ xxy ( x, y) = Γyx ( x, y) =

We will not bother to solve such differential equations. Surprisingly, this metric is the usual Euclidean one after a convenient change of coordinates xe = xe( x, y), ye = ye( x, y). To deduce such change is non-trivial (convince yourself of this), but we’ll register it here: if √ √ √ 2 2 2 2 2 xe = xy and ye = y − x , 2 4 4 then de x2 + de y2 = y2 dx2 + x2 dy2 holds (see Exercise 4.1.11). (2) Let’s determine the geodesics in the Poincaré disk D2 , by exploring some of its isometries. Recall that the metric is given by ds2P =

4(dx2 + dy2 ) . (1 − x 2 − y2 )2

From such expression, it is easy to see that the reflection ( x, y) 7→ (− x, y) is an isometry of D2 . With this, Corollary 3.6.21 (p. 231) says that the vertical diameter is a geodesic. Moreover, rotations about the origin are also isometries of D2 (to wit, they are Euclidean isometries which preserve the quantity x2 + y2 ), from where we conclude that actually any diameter is a geodesic. In particular, we now know that every geodesic that passes through the origin is (part of) a diameter. In Exercise 4.1.7, we ask you to verify that, for each z0 ∈ D2 and θ ∈ R, the map F : D2 → D2 given by z − z0 F (z) = eiθ 1 − zz0 is an isometry of D2 . In particular, F and its inverse are Möbius transformations, that send lines and circles into lines or circles. Well, if α is a geodesic that passes through z0 , F ◦ α is a geodesic that passes through the origin, and thus is a diameter. Hence, α = F −1 ◦ ( F ◦ α) is necessarily a line segment (if z0 = 0) or an arc of circumference contained in D2 . Furthermore, such an arc crosses the boundary of D2 orthogonally: indeed, since it is an isometry, F preserves Euclidean angles (to wit, it preserves angles measured according to the Poincaré disk metric, which is conformal to the Euclidean one) — and diameters do cross the boundary orthogonally. Conclusion: the geodesics in D2 are its diameters, and arcs of circumferences contained in D2 which cross its boundary orthogonally.

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Figure 4.4: Geodesics in the Poincaré disk. Of course, to conclude that those are the geodesics of D2 , we could also geometrically see the images of geodesics in H2 ⊆ L3 under the inverse of the isometry φ seen in Example 4.1.8 above. Also, Clairaut parametrizations are efficient to find geodesics in other models, for example, in the Poincaré half-plane (see Exercise 4.1.12). Among the intrinsic concepts studied for surfaces in R3ν , we still have to see what happens with the Gaussian curvature in our new setting. The Gauss equations, seen in Proposition 3.7.2 (p. 251), which do not depend on the Second Fundamental Form, give us expressions for the Gaussian curvature K ◦ x, where (U, x) is any parametrization for M. It is true, and we should verify it, that the local expressions obtained for K don’t actually depend on the chosen parametrization and so can be used to define the Gaussian curvature of a geometric surface ( M, h·, ·i). We have seen that these expressions are much simpler when the parametrization is orthogonal. In practice, we will keep using the expression given in Proposition 3.7.3 (p. 252): ! ! ! p p ( | G |)u ( | E|)v −1 p p eu + ev , K ( p) = p | EG | | E| | G | u v where p = x(u0 , v0 ), and all of the coefficients in the right side are evaluated in (u0 , v0 ), since given any point in a geometric surface, there is a parametrization around it with F = 0. For a proof of this result, which does not depend on the index of h·, ·i, see [17]. With this definition, K is automatically invariant under isometries. Example 4.1.12. (1) Since the Gaussian curvature of the hyperbolic plane H2 ⊆ L3 is constant and equal to −1, we immediately know that the curvatures of the Poincaré disk D2 and of the Poincaré half-plane H2 are also equal to −1. This can be verified directly. For

270  Introduction to Lorentz Geometry: Curves and Surfaces

instance, for the half-plane we have:  p ! −1  ( 1/y2 ) x p K= p + 1/y2 1/y4 x   ! 1 = − y2 y y y y   1 1 = − y2 − = − y2 2 y y y

!  p ( 1/y2 )y  p 1/y2 y

= −1. See in Exercise 4.1.15 how to compute the Gaussian curvature of any metric proportional (conformal) to the usual metrics in R2 and L2 . . (2) Consider r1 , r2 > 0, and the product of circles M = S1 (r1 ) × S1 (r2 ) ⊆ R4 . When 2 2 r1 + r2 = 1, M is called a Clifford torus. The usual inner product in R4 restricted to the tangent planes to M defines a Riemannian metric on M. If α1 and α2 are the usual parametrizations for the given circles, we define a product parametrization x : ]0, 2π [2 → x(]0, 2π [2 ) ⊆ S1 (r1 ) × S1 (r2 ) by . x(u, v) ≡ (α1 ⊕ α2 )(u, v) = (r1 cos u, r1 sin u, r2 cos v, r2 sin v). Let’s see how to express the metric induced by R4 on these coordinates. We have xu (u, v) = (−r1 sin u, r1 cos u, 0, 0) and xv (u, v) = (0, 0, −r2 sin v, r2 cos v), whence dx(u,v) has full rank, for all (u, v) ∈ ]0, 2π [2 . Evaluating the coefficients E(u, v) = h xu (u, v), xu (u, v)i E , etc., we have that ds2 = r12 du2 + r22 dv2 along the image of x. As the image of x is dense in M, K is a continuous function, and all the coefficients above are all constants, we have that K ( p) = 0 for all p ∈ M. (3) In the previous example, we have considered a “direct product” of the circles. If α1 and α2 are given as above, we now consider their “tensor product”, i.e., the map x : ]0, 2π [2 → R4 given by x(u, v) ≡ (α1 ⊗ α2 )(u, v)

≡ α1 ( u ) ⊗ α2 ( v )  
. r1 cos u r2 cos v r2 sin v = r1 sin u   r1 r2 cos u cos v r1 r2 cos u sin v = r1 r2 sin u cos v r1 r2 sin u sin v ≡ (r1 r2 cos u cos v, r1 r2 cos u sin v, r1 r2 sin u cos v, r1 r2 sin u sin v), using the identification Mat(2, R) ∼ = R4 . Let’s see what the Riemannian metric induced by the usual inner product of R4 looks like. We have: xu (u, v) = r1 r2 (− sin u cos v, − sin u sin v, cos u cos v, cos u sin v) and xv (u, v) = r1 r2 (− cos u sin v, cos u cos v, − sin u sin v, sin u cos v),

Abstract Surfaces and Further Topics  271

and one can verify that dx(u,v) always has full rank, just as above. So, it follows that ds2 = r12 r22 (du2 + dv2 ), and the Gaussian curvature again vanishes. For more about tensor products in this setting, see [71]. A natural continuation of the ideas presented in this text so far will lead us to an area of Geometry called Pseudo-Riemannian Geometry, which consists of the study of the geometry of differentiable manifolds of arbitrary dimension, equipped with metrics of arbitrary index. A standard reference for this subject is [54], for example.

Exercises Exercise 4.1.1. Let 0 < q < 1 and consider in R>0 × R the Lorentzian metric given by ds2 = −dt2 + t2q dx2 . Determine the lightlike curves for this metric. Remark. In higher dimensions, metrics of the form

−dt2 + a(t)2 (dx2 + dy2 + dz2 ) model universes for which space, in each fixed instant of time, is a Euclidean space with curvature, but that expands as a function of t. Of particular interest in Physics are the scales of the form a(t) = tq , with 0 < q < 1. For more details, see [10]. Exercise 4.1.2. Let’s rotate the lightcones in L2 . Consider in R2 the metric ds2 = cos(2πx )(dx2 − dy2 ) − 2 sin(2πx ) dx dy. (a) Verify that this metric is Lorentzian. (b) For every non-zero v ∈ R2 there is ( x, y) ∈ R2 (actually, infinitely many of them) such that hv, vi( x,y) is negative, zero, or positive. In other words, according to this metric, every vector assumes all causal types for convenient base points, infinitely many times. Hint. Since v is non-zero, assume without loss of generality that v = (cos θ, sin θ ) for some θ ∈ [0, 2π [, and choose x in terms of θ. (c) Exhibit non-vanishing smooth vector fields V : R2 → R2 such that the product hV ( x, y), V ( x, y)i(x,y) is always negative, zero, or positive, for all ( x, y) ∈ R2 . That is, despite item (b), there are fields with constant causal type. Hint. Try to rotate the vectors (0, 1), (1, 1) and (1, 0) “together with the metric of L2 ”. Remark. This metric passes to the quotient, defining a time-orientable Lorentzian metric in the Möbius strip.

272  Introduction to Lorentz Geometry: Curves and Surfaces

Exercise 4.1.3. Consider in R2 the Riemannian metric given by ds2 =

dx2 + dy2 . 1 + ( x 2 + y2 )2

Show that, according to this metric, A(R2 ) < +∞. Exercise 4.1.4. Let 0 < r < 1 and consider the region . R = {( x, y) ∈ R2 | x2 + y2 < r2 } in the Poincaré disk D2 . Compute A( R). What happens when r → 1− ? Remark. Compare this with Exercise 3.2.14 (p. 168). Exercise 4.1.5 (Pseudo-sphere). The image of the regular parametrized surface x : R>0 × ]0, 2π [ → R3 given by x(u, v) = (sech u cos v, sech u sin v, u − tanh u) is known as the pseudo-sphere.

Figure 4.5: The pseudo-sphere. Consider the Poincaré half-plane H2 , with Riemannian metric ds2H =

dx2 + dy2 . y2

Show that F : x (]0, π/2[ × ]0, 2π [) → H2 given by F ( x(u, v)) = (v, cosh u) is an isometry onto its image. Exercise 4.1.6. Consider the Poincaré half-plane H2 with the Riemannian metric given in complex coordinates by dz dz ds2H = . Im(z)2 If we have numbers a, b, c, d ∈ R with ad − bc = 1, show that the Möbius transformation T : H2 → H2 given by az + b w = T (z) = cz + d is an isometry of H2 onto itself.

Abstract Surfaces and Further Topics  273

Hint. Don’t be afraid of dz and dz, and do the computation just like in Example 4.1.8 (p. 264). Exercise† 4.1.7. Let θ ∈ R and z0 ∈ D2 . Consider F : C → C given by z − z0 . F (z) = eiθ . 1 − zz0 (a) Compute | F (z)|2 in terms of |z| and conclude not only that F (S1 ) ⊆ S1 , but also that F (D2 ) ⊆ D2 . (b) Show that F is bijective, with inverse given by F −1 (w) = e−iθ

w + z0 eiθ . 1 + e−iθ z0 w

In particular, we have the equalities F (S1 ) = S1 and F (D2 ) = D2 . (c) Show that F D2 is an isometry of the Poincaré disk, recalling that its metric in complex coordinates is given by ds2 =

4 dz dz . (1 − | z |2 )2

Exercise† 4.1.8. Consider the hyperbolic plane H2 ⊆ L3 , and also the Poincaré halfplane, denoted only in this exercise by H2hp . Show that the map F : H2 → H2hp given by   x 1 F ( x, y, z) = , z−y z−y is an isometry. Hint. Consider, as usual, the revolution parametrization x : R≥0 × [0, 2π [ → H2 given by x(u, v) = (sinh u cos v, sinh u sin v, cosh u) and compute the products between the derivatives of F ( x(u, v)) using the half-plane metric. Exercise† 4.1.9 (Klein disk). We have one more model for hyperbolic geometry: the Klein disk is the set D2 equipped with the Riemannian metric ds2K =

dx2 + dy2 ( x dx + y dy)2 + . 1 − x 2 − y2 (1 − x 2 − y2 )2

(a) Show that this metric is indeed Riemannian (positive-definite). Hint. Sylvester’s Criterion. (b) Show that if x : ]0, 1[ × ]0, 2π [ → D2 \ {(0, 0)} is the usual polar parametrization x(u, v) = (u cos v, u sin v), then ds2K =

u2 1 2 du + dv2 . (1 − u2 )2 1 − u2

(c) Determine an isometry between the Klein disk and the hyperbolic plane H2 ⊆ L3 . Hint. One possible isometry already appeared in the text. Can you find which one it is?

274  Introduction to Lorentz Geometry: Curves and Surfaces

Remark. The Klein disk is less used than the Poincaré disk, since its metric is not conformal to the usual metric in R2 , and so we see distorted angles, while this does not occur in the Poincaré disk. Exercise† 4.1.10 (Hemisphere). Consider the Hemisphere . J2 = {( x, y, z) ∈ R3 | x2 + y2 + z2 = 1 and z > 0}, equipped with the Riemannian metric . dx2 + dy2 + dz2 ds2J = . z2 (a) Show that if x : ]0, π/2[ × ]0, 2π [ → J2 is the usual parametrization of revolution given by x(u, v) = (cos u cos v, cos u sin v, sin u), then ds2J =

du2 + cos2 u dv2 . sin2 u

(b) The Hemisphere equipped with this metric may also be used as a model for hyperbolic geometry: show that J2 is isometric to the hyperbolic plane H2 ⊆ L3 via the central projection based on (0, 0, −1), Π : H2 → J2 given by   x y 1 Π( x, y, z) = , , . z z z

H2 φ( p) J2

p

(0, 0, −1) Figure 4.6: The isometry φ = Π−1 between J2 and H2 . Hint. If e x is the usual parametrization for H2 , used in Example 4.1.8 (p. 264), show that Π(e x(ue, ve)) = x(arccos(tanh ue), ve). Exercise 4.1.11 (Parabolic coordinates in R2 ). Following Example 4.1.11 (p. 267), show that in R2 \ {( x, y) ∈ R2 | xy = 0}, if √ √ √ 2 2 2 2 2 xe = xy and ye = y − x , 2 4 4 then de x2 + de y2 = y2 dx2 + x2 dy2 .

Abstract Surfaces and Further Topics  275

Exercise 4.1.12. Consider the Poincaré half-plane H2 with the Riemannian metric ds2H =

dx2 + dy2 . y2

Observing that the identity is a y-Clairaut parametrization, use Theorem 3.6.34 (p. 243) and show that all the geodesics in H2 are vertical lines and arcs of circumferences which orthogonally cross the axis y = 0 (the so-called asymptotic boundary of H2 , denoted on literature by ∂∞ H2 ).

y

x Figure 4.7: Geodesics in the Poincaré half-plane. Hint. In the statement of Theorem 3.6.34, make u = y and v = x. Since in this case we have E = G, there’s no risk of switching them. Exercise 4.1.13. Let ( M, h·, ·i) be a geometric surface equipped with a Lorentzian metric and (U, x) be a parametrization of M for which both xu and xv are always lightlike. (a) Show that 1 K◦x=− F



Fu F

 v

1 =− F



Fv F

 . u

(b) Show that the geodesic differential equations simply boil down to u00 +

Fu 0 2 Fv 2 u = 0 and v00 + v0 = 0. F F

Exercise† 4.1.14. Show that switching the sign of a Lorentzian metric also switches the sign of its Gaussian curvature. Thus, Lorentzian metrics with K > 0 differ from those with K < 0 only in causal type. Exercise† 4.1.15. Let U ⊆ R2 be an open subset and h : U → R be a nowhere vanishing smooth function. (a) Show that the curvature of the Riemannian metric (dx2 + dy2 )/h( x, y)2 is given by K = h(h xx + hyy ) − (h2x + h2y ). (b) Show that the curvature of the Lorentzian metric (dx2 − dy2 )/h( x, y)2 is given by K = h(h xx − hyy ) − (h2x − h2y ).

276  Introduction to Lorentz Geometry: Curves and Surfaces

Exercise 4.1.16 (The Schwarzschild half-plane). Let M > 0 be a constant. Consider in . PI = {(t, r ) ∈ R2 | r > 2M} the Lorentzian metric given by ds2 = − h(r ) dt2 + h(r )−1 dr2 , where

2M . h(r ) = 1 − r is the Schwarzschild horizon function. Note that in PI , this horizon function is positive. (a) Show that the Gaussian curvature of PI is always positive, given by the expression K (t, r ) = 2M/r3 > 0. (b) Show that the only non-zero Christoffel symbols are: t Γttr (t, r ) = Γrt (t, r ) = −Γrrr (t, r ) =

M , 2 r − 2Mr

and Γrtt (t, r ) =

Mr − 2M2 . r3

(c) Show that α : R>0 → PI given by α(s) = (s + 2M log s, s + 2M) is a lightlike geodesic in PI . (d) Verify that horizontal translations and time reflection (i.e., the maps of the form PI 3 (t, r ) 7→ (±t + b, r ) ∈ PI ) are isometries of PI . Conclude that all lightlike geodesics in PI are images of the curve given in item (b) under transformations of this type.

t

r = 2M

Figure 4.8: Light rays in PI . Remark. The Schwarzschild half-plane has several applications in General Relativity. It is a bidimensional slice of a model of a 4-dimensional spacetime, which takes into account gravitational forces and, thus, has curvature. The constant M is interpreted as the mass of a massive particle. We have two important applications. The Schwarzschild space may model:

• the solar system in a more precise way than the models adopted in Newtonian mechanics, when such particle is regarded as the Sun; • the proximities of a black hole, situation on which M is its mass, the region r < 2M (where the sign of the metric is reversed) is the interior of the black hole, the region r > 2M is its exterior, and r = 2M is the event horizon (where h is singular).

Abstract Surfaces and Further Topics  277

Conveniently altering the function h, we obtain other models of spacetimes. For example, the metric in the Reissner-Nordström space uses the slightly more sophisticated horizon function 2M q2 h(r ) = 1 − + 2, r r taking into account some electric charge q that the black hole may have. For more details, see [10]. Exercise 4.1.17 (Lorentz-Poincaré half-plane). We may consider a Lorentzian version of the Poincaré half-plane. Take, in M = R × R>0 , the Lorentzian metric given by ds2 =

dx2 − dy2 . y2

(a) Verify that its Gaussian curvature is constant, K = 1; (b) Show that all non-zero Christoffel symbols are 1 y y x Γ xxy ( x, y) = Γyx ( x, y) = Γ xx ( x, y) = Γyy ( x, y) = − . y (c) Show that:

• α : R → M given by α(t) = (0, et ) is a timelike geodesic, parametrized with proper time; • the curve β : ]−π/2, π/2[ → M given by β(t) = (r sec t, r tan t), for each r > 0, is a timelike geodesic, parametrized with proper time; • the curve γ : ]−π/2, π/2[ → M given by γ(s) = (r tan s, r sec s), for each r > 0, is a spacelike geodesic, parametrized with unit speed; • the lightlike curves in M are precisely lines with slope π/4. (d) Verify that horizontal translations and horizontal reflections are isometries and conclude, in analogy with the Poincaré half-plane, that the geodesics in M are branches of hyperbolas and light rays, while the hyperbola branches are spacelike and crossing the axis y = 0 orthogonally.

y

x Figure 4.9: Geodesics in the Lorentz-Poincaré half-plane. (e) In the same fashion that the Poincaré half-plane is related to the hyperbolic plane H2 ⊆ L3 , M is related to de Sitter space S21 . Consider ζ : M → S21 given by   x 1 − x 2 + y2 1 + x 2 − y2 ζ ( x, y) = − , , . y 2y 2y

278  Introduction to Lorentz Geometry: Curves and Surfaces

Verify that ζ indeed takes values in S21 and that it is an isometry onto its image (determine it). Hint. In this case, it is much simpler to just compute the both derivatives dζ ( x,y) (1, 0) = (∂ζ/∂x )( x, y) and dζ ( x,y) (0, 1) = (∂ζ/∂y)( x, y), and verify that 1 , y2 1 hdζ (x,y) (0, 1), dζ (x,y) (0, 1)i L = − 2 and y

hdζ (x,y) (1, 0), dζ (x,y) (1, 0)i L =

hdζ (x,y) (1, 0), dζ (x,y) (0, 1)i L = 0. Remark. Another strategy to find the timelike geodesics is to mimic what was done in Exercise 4.1.12 above, using Theorem 3.6.35 (p. 244). For more details about this metric, see [53]. Exercise† 4.1.18 (Anti-de Sitter). Consider the space R32 = (R3 , h·, ·i2 ), with the product h·, ·i2 is defined by . hv, wi2 = v1 w1 − v2 w2 − v3 w3 , where v = (v1 , v2 , v3 ) and w = (w1 , w2 , w3 ). The anti-de Sitter space is defined by . H21 = {v ∈ R32 | hv, vi2 = −1}. Show that the restriction of h·, ·i2 to the tangent planes of H21 defines a Lorentzian metric in H21 , whose Gaussian curvature is constant and equal to −1. Hint. Try a parametrization of revolution around the x-axis. Remark. In the same way that S21 is a Lorentzian version of S2 , H21 is a Lorentzian version of H2 . As subsets of R3 , the spaces S21 and H21 differ only by a permutation of axes. Such phenomenon is particular to dimension 2, and the higher dimensional versions of the anti-de Sitter have applications in relativity, serving as models for spacetimes which take into account gravitational forces. Exercise 4.1.19 (What if?). Consider M = R × R>0 . Instead of searching a Lorentzian version of the Poincaré half-plane metric by dividing the metric of L2 by a timelike coordinate, like in Exercise 4.1.17, let’s see what happens if we divide by a spacelike coordinate instead. Take, in M, the Lorentzian metric given by ds2 =

−dx2 + dy2 . y2

By Exercise 4.1.14 above, we know that this metric has constant Gaussian curvature, equal to −1. Moreover, the geodesics are the same as the geodesics for the metric studied in Exercise 4.1.17, but with flipped causal types. The next question that occurs is if M, equipped with such a metric, is isometric to an open subset of any of our known surfaces. With a Lorentzian metric of constant and negative curvature, we have a natural candidate: the anti-de Sitter space H21 . Seems reasonable to expect, then, that the map we look for is similar to the isometry ζ given in Exercise 4.1.17, switching the axes.

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Show that ξ : M → H21 given by  ξ ( x, y) =

1 + x 2 − y2 1 − x 2 + y2 x , ,− 2y 2y y



indeed takes values in H21 , and it is an isometry onto its image. Remark. For more details about this metric, see [47]. Exercise 4.1.20. Let f : U ⊆ C → C be a holomorphic function. Writing z = x + iy and f (z) = φ( x, y) + iψ( x, y), consider the graph of f : . M = {( x, y, φ( x, y), ψ( x, y)) ∈ R4 | ( x, y) ∈ U }. Let’s compute the Gaussian curvature of the metric in M induced by h·, ·i E . Omit points of evaluation in what follows. (a) Show that the induced metric is given by ds2 = (1 + φx2 + φy2 )(dx2 + dy2 ). . (b) If h = (1 + φx2 + φy2 )−1/2 , show that h x = −h3 (φx φxx + φy φyx ),

hy = −h3 (φx φxy + φy φyy ),

and that 2 2 h2x + h2y = h6 (φx2 + φy2 )(φxx + φxy ).

(c) Verify that 3h2x 2 2 − h3 (φxx + φx φxxx + φyx + φy φyxx ), h 3h2y 2 2 = − h3 (φxy + φx φxyy + φyy + φy φyyy ), h

h xx = hyy and that

h xx + hyy =

3(h2x + h2y ) h

2 2 − 2h3 (φxx + φxy ).

(d) Put all the previous items together and finally conclude that K (( x, y, φ( x, y), ψ( x, y))) =

2 + φ2 ) −2(φxx xy . (1 + φx2 + φy2 )3

Hint. Use the Cauchy-Riemann equations and recall that if f is holomorphic, φ is harmonic. Remark.

• In the theory of submanifolds of Rn , we again have the notion of mean curvature vector. One may show that every graph of a holomorphic function, as above, has zero mean curvature vector (and thus is a minimal surface in R4 ). Compare this with the situation on R3 , where minimal surfaces have non-positive Gaussian curvature in all points.

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• There’s a Lorentzian analogue of the calculations done in this exercise, considering instead of graphs of holomorphic functions in R4 , graphs of “split-holomorphic” functions (defined in C0 ) in L4 . We will formally introduce C0 soon. Exercise 4.1.21. Consider the Artin space R42 = (R4 , h·, ·i2 ), with . hv, wi2 = v1 w1 + v2 w2 − v3 w3 − v4 w4 , where v = (v1 , v2 , v3 , v4 ) and w = (w1 , w2 , w3 , w4 ). The product h·, ·i2 is called the neutral product in R4 . Consider the tensor product of the usual parametrizations for the right branch of S11 , and S1 : x : R × ]0, 2π [ → R42 given by x(u, v) = (cosh u cos v, cosh u sin v, sinh u cos v, sinh u sin v). If M is the image of x, show that the restriction of the neutral product to the tangent planes to M defines a Lorentzian metric on M, which makes it isometric to R × ]0, 2π [, equipped with the usual metric from L2 . Remark.

• Even though the codomain of x is R42 instead of R3ν , the derivative Dx(u, v) still has full rank for all (u, v) ∈ R × ]0, 2π [. This is the reason why we still use the terminology of regular parametrized surfaces even in this case. • The Artin space also has applications in Physics, serving as the ambient space for spacetime models other than Lorentz-Minkowski space, for example, the tridimensional version of the anti-de Sitter space, . H31 = {v ∈ R42 | hv, vi2 = −1}. Exercise 4.1.22. Determine the Gaussian curvature of the metric induced in the surface M of the previous exercise, by the usual scalar product in R4 .

4.2

RIEMANN’S CLASSIFICATION THEOREM

Up until this moment, we have seen some surfaces with constant Gaussian curvature. Namely, we have met:

• the planes R2 and L2 , with K = 0; • the sphere S2 and the de Sitter space S21 , with K = 1; • the hyperbolic plane H2 and the anti-de Sitter H21 , with K = −1. Our goal here is to show that, locally, every surface with constant K “is” one of those surfaces described above. More precisely, we want to prove the: Theorem 4.2.1 (Riemann). Let ( M, h·, ·i) be a geometric surface with constant Gaussian curvature K ∈ {−1, 0, 1}. Then: (A) if the metric is Riemannian, every point in M has a neighborhood isometric to an open subset of

Abstract Surfaces and Further Topics  281

(i) R2 , if K = 0; (ii) S2 , if K = 1; (iii) H2 , if K = −1, (B) if the metric is Lorentzian, to an open subset of (i) L2 , if K = 0; (ii) S21 , if K = 1; (iii) H21 , if K = −1. The proof strategy consists of constructing parametrizations for which the metric assumes a simple form. To actually do this, we will use geodesics, which are known to be plentiful in any geometric surface. To avoid singularities, we won’t consider lightlike geodesics. Thus, we fix throughout this section a geometric surface ( M, h·, ·i), with metric tensor of index ν ∈ {0, 1}, and a unit speed geodesic γ : I → M. For each v ∈ I, consider a unit . speed geodesic γv : Jv → M, which crosses γ orthogonally at the point γv (0) = γ(v). Setting . U = {(u, v) ∈ R2 | v ∈ I and u ∈ Jv }, define x : U → x(U ) ⊆ M by x(u, v) = γv (u).

γ γ(v)

γv

Figure 4.10: Construction of a Fermi chart x. Definition 4.2.2. The above chart x is called a Fermi chart for M, centered at γ. We’ll also fix, until the end of the section, this Fermi chart (U, x) so constructed. Remark.

• When h·, ·i is Lorentzian, we’ll have two types of Fermi charts, according to the causal character of γ. Moreover, recalling that geodesics have automatically constant causal character (hence determined by a single velocity vector), it follows that if γ is spacelike (resp. timelike), then all the γv are timelike (resp. spacelike), since {γ0 (v), γ0v (0)} is an orthonormal basis of Tγ(v) M, for all v ∈ I. • When necessary, if γ is timelike, we might denote the coordinates by (τ, ϑ) instead of (u, v). Proposition 4.2.3. The Fermi chart x is indeed regular in a neighborhood of {0} × I (so that reducing U if necessary, we may assume that (U, x) itself is regular).

282  Introduction to Lorentz Geometry: Curves and Surfaces

Proof: We’ll show that for all v ∈ I, the vectors xu (0, v) and xv (0, v) are orthogonal. To wit, we have by construction that

h xu (0, v), xv (0, v)i = hγ0v (0), γ0 (v)i = 0. Since none of those vectors is lightlike, orthogonality implies linear independence. By continuity of x, the vectors xu (u, v) and xv (u, v) remain linearly independent for small enough values of u. Proposition 4.2.4. The coordinate expression of h·, ·i with respect to the Fermi chart (U, x) is ds2 = (−1)ν eγ du2 + G (u, v) dv2 . Proof: All the γv are unit speed curves with the same indicator eγv . We have that E(u, v) = h xu (u, v), xu (u, v)i = hγ0v (u), γ0v (u)i = eγv . Now, eγ eγv = (−1)ν for all v ∈ I, whence E(u, v) = (−1)ν eγ . Proceeding, we see that by construction, F (0, v) = 0 for all v ∈ I, so that it suffices to check that F does not depend on the variable u. Fixed v0 ∈ I, we have the expression x(u, v0 ) = γv0 (u), and so the second geodesic equation for γv0 yields Γ211 (u, v0 ) = 0. Since v0 was arbitrary, it follows that Γ211 = 0. On the other hand, by definition of Γ211 , we have (−1)ν eγ Γ211 (u, v) = Fu (u, v), (−1)ν eγ G (u, v) − F (u, v)2 so that Fu (u, v) = 0, and we conclude that F (u, v) = 0 for all (u, v) ∈ U, as desired. Remark. Since G (0, v) = eγ 6= 0, the continuity of G allows us to assume, by reducing U again if necessary, that G (u, v) has the same sign as eγ for all (u, v) ∈ U. Corollary 4.2.5. The Gaussian curvature of ( M, h·, ·i) is expressed in terms of the Fermi chart (U, x) by p ( | G |)uu ν +1 K ◦ x = (−1) eγ p . |G| Before starting the proof of Theorem 4.2.1 (p. 280), we only need to get one more technical lemma out of the way: Lemma 4.2.6 (Boundary conditions). The Fermi chart (U, x) satisfies Gu (0, v) = 0, for all v ∈ I. Proof: As γ(v) = x(0, v), the first geodesic equation for γ boils down to Γ122 (0, v) = 0, for all v ∈ I. Since F (0, v) = 0, it directly follows that Γ122 (0, v) = −

Gu (0, v) , 2eγ

whence Gu (0, v) = 0, as desired. Finally: Proof (of Theorem 4.2.1): In all possible cases, the coefficient G must satisfy the following differential equation: q q

| G |)uu + (−1)ν eγ K | G | = 0. p Now, we solve this equation (in each case) for | G |, and use the boundary conditions G (0, v) = eγ and Gu (0, v) = 0 to determine G explicitly. (

Abstract Surfaces and Further Topics  283

(A) Assume that h·, ·i is Riemannian. √ p (i) For K = 0, we have ( G )uu = 0, and so G (u, v) = A(v)u + B(v). The boundary conditions then give A(v) = 0 and B(v) = 1, so that G (u, v) = 1 for all (u, v) ∈ U, and ds2 = du2 + dv2 . p √ (ii) WhenpK = 1, we have ( G )uu + | G | = 0, whose solutions are of the form G (u, v) = A(v) cos u + B(v) sin u. Now, the boundary conditions give A(v) = 1 and B(v) = 0, and so G (u, v) = cos2 u, and it follows that ds2 = du2 + cos2 u dv2 : the metric in S2 . p √ (iii) If K p = −1, the equation to be solved is ( G )uu − | G | = 0. We have that G (u, v) = A(v)eu + B(v)e−u , and now the boundary conditions give A(v) = B(v) = 1/2, whence G (u, v) = cosh2 u and we obtain the local expression ds2 = du2 + cosh2 u dv2 . To recognize this in an easier way as the metric in H2 , we may let x = ev tanh u and y = ev sech u, so that ds2 =

dx2 + dy2 , y2

as desired. (B) Assume now that h·, ·i is Lorentzian. (i) For K = 0, just like above, we have ds2 = −du2 + dv2 = dτ 2 − dϑ2 . (ii) If K =√1, we now √ have two cases to discuss. If γ is spacelike, we again obtain that ( G )uu − G = 0, from where it follows that G (u, v) = cosh2 u and we get the S21 metric: ds2 = −du2 + cosh2 u dv2 (expressed in the usual revolution √ √ parametrization). If γ is timelike instead, we have ( − G )ττ + − G = 0, whose solution is G (τ, ϑ ) = − cos2 τ, and so ds2 = dτ 2 − cos2 τ dϑ2 . (iii) If K = −1, the situation is dual to the previous one, switching “spacelike” and “timelike”, and also the signs of the metric expressions. Omitting repeated calculations, we obtain ds2 = −du2 + cos2 u dv2 = dτ 2 − cosh2 τ dϑ2 , which is the metric of H21 in suitable coordinates.



We will conclude the section by presenting surfaces in the ambient spaces L3 and R32 whose metric’s coordinate expressions are the ones discovered in the proof above. For K = 0 the situation is completely uninteresting. But for K 6= 0 we have the following: Example 4.2.7. (1) K = 1:

• The metric ds2 = −du2 + cosh2 u dv2 may be realized by the usual revolution parametrization x : R2 → S21 ⊆ L3 given by x(u, v) = (cosh u cos v, cosh u sin v, sinh u), √
and also by y : cosh−1 ]1, 2[ × R → R32 given by   Z uq 2 y(u, v) = cosh u cosh v, cosh u sinh v, 2 − cosh t dt . 0

284  Introduction to Lorentz Geometry: Curves and Surfaces

• For ds2 = dτ 2 − cos2 τ dϑ2 , consider x : ]0, 2π [ × R → S21 ⊆ L3 given by x(τ, ϑ ) = (sin τ, cos τ cosh ϑ, cos τ sinh ϑ ), and also by y : ]−π/2, π/2[ × R → R32 , given by Z τ p  2 y(τ, ϑ ) = 1 + sin t dt, cos τ cos ϑ, cos τ sin ϑ . 0

Remark. The periodicity condition y(τ, ϑ ) = y(τ + π, ϑ ) in the last given parametrization along with the fact that translations are isometries in R32 allow us to restrict everything to the given domains, which is maximal for nondegenerability. To summarize, when K = 1 we have the following visualizations:

(a) In L3

(b) In R32

Figure 4.11: Constant Gaussian curvature K = 1. (2) K = −1:

• The metric ds2 = −du2 + cos2 u dv2 may be realized by the parametrization x : ]−π/2, π/2[ × R → L3 , given by   Z up 2 x(u, v) = cos u cos v, cos u sin v, 1 + sin t dt , 0

and also by y : ]0, 2π [ × R → H21 ⊆ R32 : y(u, v) = (cos u sinh v, cos u cosh v, sin u). In this case, the same remark made for y in the case K = 1 holds for x here.

• The metric ds2√=
dτ 2 − cosh2 τ dϑ2 may be realized by the parametrization x : cosh−1 ]1, 2[ × R → L3 given by Z τ q  2 x(τ, ϑ ) = 2 − cosh t dt, cosh τ cosh ϑ, cosh τ sinh ϑ 0

and by y : R × ]0, 2π [ → H21 ⊆ R32 , y(τ, ϑ ) = (sinh τ, cosh τ cos ϑ, cosh τ sin ϑ ).

Abstract Surfaces and Further Topics  285

So in this case, we have:

(a) In R32

(b) In L3

Figure 4.12: Constant Gaussian curvature K = −1. Lastly, we observe that the surfaces in Figures 4.11(a) and 4.12(a) are isometric when equipped with the metrics induced by R3 but, on the pseudo-Riemannian ambient spaces considered, they have rotational symmetry along axes of distinct causal characters. The same holds for the surfaces given in Figures 4.11(b) and 4.12(b). Furthermore, note that S21 and H21 “fit better” in L3 and R32 , respectively — switching the ambient spaces requires the use of parametrizations depending on certain elliptic integrals.

Exercises Exercise 4.2.1 (Riemann’s Formula). Let ( M, h·, ·i) be a geometric surface equipped with a Riemannian metric, and (U, x) be a Fermi chart for M (on which the metric is expressed by ds2 = du2 + G (u, v) dv2 ). In some adequate domain, consider the reparametrization x = u cos v and y = u sin v. Show that ds2 = dx2 + dy2 + H ( x, y)( x dy − y dx )2 , where H ( x, y) = ( G (u, v) − u2 )/u4 . Hint. Differentiating the three relations u2 = x2 + y2 , x = u cos v and y = u sin v, respectively, we obtain that u du = x dx + y dy, dx = cos v du − u sin v dv and finally dy = sin v du + u cos v dv. Proceed from there. Remark. For metrics of the above form, with H defined in some open set around the origin (it is not the case here), one can show by using the so-called Brioschi formula (see for example [56]), that the Gaussian curvature of the metric in the point of coordinates ( x, y) = (0, 0) is −3H (0, 0). That is, this metric is “Euclidean up to second order, near the origin”, and the function H measures this deviation. Exercise 4.2.2 (Revolution surfaces with constant K). Let α : I → R3ν be smooth, regular, non-degenerate, injective and of the form α(u) = ( f (u), 0, g(u)), for certain functions f and g with f (u) > 0 for all u ∈ I, and let M be the revolution surface spanned by α, around the z-axis. Assume that α has unit speed, M has constant Gaussian curvature K, and consider the parametrization x : I × ]0, 2π [ → I → x(U ) ⊆ M given by x(u, v) = ( f (u) cos v, f (u) sin v, g(u)).

286  Introduction to Lorentz Geometry: Curves and Surfaces

(a) Show that, in general, f and g satisfy 00

f (u) + eα K f (u) = 0 and

g(u) =

Z q

(−1)ν (eα − f 0 (u)2 ) du.

(b) Verify that  √ √  if eα K > 0  A cos( eα Ku) + B sin( eα Ku), f (u) = Au + B, if K = 0,  √ √  A cosh( −eα Ku) + B sinh( −eα Ku) if eα K < 0, where, in the case K = 0, we necessarily have | A| ≤ 1 if the ambient space is R3 , while | A| ≥ 1 if the curve is spacelike in L3 (for timelike curves there are no restrictions). (c) Identify all the revolution surfaces with constant Gaussian curvature K ∈ {−1, 0, 1}. Hint. Draw sketches of the generating curve on the plane y = 0. For the discussion in R3 , see [39]. Exercise 4.2.3 (Hyperbolic revolution surfaces with constant K). Investigate the hyperbolic revolution surfaces in L3 with constant Gaussian curvature K, following the idea of the previous exercise.

4.3

SPLIT-COMPLEX NUMBERS AND CRITICAL SURFACES

In this section we will briefly present the strong relation between complex variables and spacelike surfaces in R3ν . For timelike surfaces in L3 , we will need the set of splitcomplex numbers. Thus, we will start the discussion formalizing this concept and presenting basic facts about Calculus in a single split-complex variable. Next, we will apply this to study Bonnet rotations and the Enneper-Weierstrass representation formulas. 4.3.1

A brief introduction to split-complex numbers

Let’s recall one possible construction of the complex numbers: define in R2 the operations . ( a, b) + (c, d) = ( a + c, b + d) and . ( a, b)(c, d) = ( ac − bd, ad + bc). Such operations turn R2 into a field, which is then denoted by C. Since we have that ( a, b) = ( a, 0) + (b, 0)(0, 1) and (0, 1)2 = (−1, 0), we may identify R with the set . {( a, 0) ∈ R2 | a ∈ R}, write i = (0, 1), to finally recover the usual description C = { a + bi | a, b ∈ R and i2 = −1}. . . Given z = a + bi ∈ C, the projections Re(z) = a and Im(z) = b are called the real and . imaginary parts of z. The conjugate of z is defined as z = a − bi, and the absolute value . √ 2 of z as |z| = a + b2 = k( a, b)k E . Moreover, if z1 = a1 + b1 i and z2 = a2 + b2 i are two complex numbers, we have that Re(z1 z2 ) = h( a1 , b1 ), ( a2 , b2 )i E ,

Abstract Surfaces and Further Topics  287

which says that C captures the geometry of the usual inner product in R2 . With this in place, one proceeds to develop the theory of Calculus in a complex variable, with which we will assume some familiarity (we recommend [40] and [61], for good measure). Our goals, then, are to construct a Lorentzian version of C, based in the above brief review; and to understand the basics about how Calculus works in this new setting. Definition 4.3.1 (Split-complex numbers). The set of split-complex numbers, denoted by C0 , is the space L2 equipped with the operations . ( a, b) + (c, d) = ( a + c, b + d) and . ( a, b)(c, d) = ( ac + bd, ad + bc). Remark. The split-complex numbers are also known as hyperbolic numbers. For a reason for this terminology, see Exercise 4.3.6. The verification that such operations turn C0 into a commutative ring with unit is straightforward. Since we have that ( a, b) = ( a, 0) + (b, 0)(0, 1) and also (0, 1)2 = (1, 0), . we may again identify R with {( a, 0) ∈ L2 | a ∈ R} and write h = (0, 1) to obtain a description similar to the one previously given for C: C0 = { a + bh | a, b ∈ R and h2 = 1}. Definition 4.3.2. Let w = a + bh ∈ C0 . . (i) The split-conjugate of w is given by w = a − bh. . p (ii) The split-complex absolute value of w is given by |w| = | a2 − b2 | = k( a, b)k L . . . (iii) The real part of w is given by Re(w) = a, and its imaginary part by Im(w) = b. Remark. Rigorously, the symbol | · | appears in (ii) with two different meanings, but the context should prevent any risk of confusion. Let’s register a few basic properties of split-complex conjugation and absolute value: Proposition 4.3.3. Let w, w1 , w2 ∈ C0 . (i) w1 + w2 = w1 + w2 , w1 w2 = w1 w2 , w = w, and w = w if and only if w ∈ R. In other words, conjugation in C0 is still an involutive automorphism which preserves R; (ii) if the inverse 1/w exists, then 1/w = 1/w; (iii) |w| = |w|, |ww| = |w|2 ; (iv) |w1 w2 | = |w1 ||w2 | and, if the inverse 1/w exists, |1/w| = 1/|w|. In particular, if the inverse 1/w exists, we necessarily have |w| 6= 0. Remark. The proof is an instructive warm-up, and we ask you to do it in Exercise 4.3.1. The strength of C0 resides in the fact that if w1 = a1 + b1 h and w2 = a2 + b2 h are two split-complex numbers, it holds that Re(w1 w2 ) = h( a1 , b1 ), ( a2 , b2 )i L , so that C0 captures the geometry of L2 in the same fashion that C captures the geometry of R2 . This observation gives us a geometric intuition for the fact that C0 is not a field,

288  Introduction to Lorentz Geometry: Curves and Surfaces

while C is: zero divisors in C0 correspond precisely to lightlike directions in L2 (we ask that you verify this in Exercise 4.3.3). To proceed, we take in C0 the usual topology of C. That is, open subsets of C0 are the same ones as in C, and continuous functions remain the same. In particular, if U ⊆ C0 is open and f : U → C0 is written in the form f ( x + hy) = φ( x, y) + hψ( x, y) for certain φ, ψ : U → R, then f is continuous if and only if both φ and ψ are. In this aspect, there is nothing new. But to define holomorphicity in C0 , we’ll again mimic the definition adopted in C, being careful to avoid taking the limit along light rays in the definition of derivative: Definition 4.3.4. Let U ⊆ C0 be an open subset, w0 ∈ U, and f : U → C0 be a function. We’ll say that f is C0 -differentiable at w0 if the limit . f 0 ( w0 ) =

lim

∆w→0 ∆w6∈CL (0)

f (w0 + ∆w) − f (w0 ) ∆w

exists. This being the case, f 0 (w0 ) is called the derivative of f at w0 . Moreover, f is called split-holomorphic at w0 if it is C0 -differentiable at all points in some neighborhood of w0 . The usual Calculus rules hold, with the same proofs (which are then omitted): Proposition 4.3.5. Let U ⊆ C0 be an open subset, w0 ∈ U, and f , g : U → C0 be two C0 -differentiable functions at w0 . Then: (i) f + g is C0 -differentiable at w0 and ( f + g)0 (w0 ) = f 0 (w0 ) + g0 (w0 ); (ii) f g is C0 -differentiable at w0 and ( f g)0 (w0 ) = g(w0 ) f 0 (w0 ) + f (w0 ) g0 (w0 ); (iii) if g never takes values in the light rays, then f /g is C0 -differentiable at w0 and ( f /g)0 (w0 ) = ( f 0 (w0 ) g(w0 ) − f (w0 ) g0 (w0 ))/g(w0 )2 . Example 4.3.6. It is easy to see that the identity mapping idC0 : C0 → C0 and constant maps are split-holomorphic, with constant derivatives equal to 1 and 0, respectively. Hence, the above result says that all polynomials in the variable w are split-holomorphic, as well as all rational functions (quotients of polynomials, when the denominator does not take values in the light rays), with derivatives given by the usual formulas. Proposition 4.3.7 (Chain rule). Let U1 , U2 ⊆ C0 be open, and also f : U1 → C0 , g : U2 → C0 be functions with f (U1 ) ⊆ U2 . If f is C0 -differentiable at w0 and g is C0 -differentiable at f (w0 ), then g ◦ f is C0 -differentiable at w0 and we have the relation ( g ◦ f )0 (w0 ) = g0 ( f (w0 )) f 0 (w0 ). In the usual theory of Calculus in a single complex variable, we know that the real and imaginary parts of a holomorphic function must satisfy the Cauchy-Riemann equations. Let’s see what these equations become for split-holomorphic functions: Proposition 4.3.8 (Revised Cauchy-Riemann). Let U ⊆ C0 be an open subset and w0 ∈ U be a point. If f : U → C0 is C0 -differentiable at w0 , and we write the function as f ( x + hy) = φ( x, y) + hψ( x, y), then we have that ∂φ ∂ψ ( w0 ) = ( w0 ) ∂x ∂y

and

∂φ ∂ψ ( w0 ) = ( w0 ) . ∂y ∂x

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Proof: Writing ∆w = ∆w1 + h∆w2 , we have that ∆w → 0 if and only if ∆w1 → 0 and ∆w2 → 0. We know that the limit in the definition of the derivative of f at w0 exists, and so we may compute it by using any path away from the light rays in L2 . In particular, we may consider the paths ∆w1 = 0 and ∆w2 = 0. On one hand, we have that φ(w0 + ∆w1 ) − φ(w0 ) + h(ψ(w0 + ∆w1 ) − ψ(w0 )) ∆w1 ∂φ ∂ψ = ( w0 ) + h ( w0 ) , ∂x ∂x

f 0 (w0 ) = lim

∆w1 →0

and on the other hand that φ(w0 + ∆w2 ) − φ(w0 ) + h(ψ(w0 + ∆w2 ) − ψ(w0 )) h∆w2 ∂φ ∂ψ = h ( w0 ) + ( w0 ) . ∂y ∂y

f 0 (w0 ) = lim

∆w2 →0

Equating real and imaginary parts, the result follows. Remark. The revised Cauchy-Riemann equations may be expressed in a shorter way by introducing split-complex versions of the Wirtinger operators:     ∂ . 1 ∂ ∂ ∂ . 1 ∂ ∂ = +h and = −h . ∂w 2 ∂x ∂y ∂w 2 ∂x ∂y This way, in analogy with the complex case, the revised Cauchy-Riemann equations become just ∂f =0 ∂w and, being satisfied, imply that f 0 (w) =

∂f ( w ). ∂w

Proposition 4.3.9. Let U ⊆ C0 be an open subset and φ, ψ : U → R be two realdifferentiable functions, whose partial derivatives satisfy the revised Cauchy-Riemann . equations on U. Then the function f : C0 → C0 defined by f = φ + hψ is splitholomorphic. These ideas may be interpreted in a convenient way in terms of the total derivative of f , seen as a map on R2 , see Exercise 4.3.7. Example 4.3.10. Motivated by Euler’s formula ex+iy = ex (cos y + i sin y) in C, we define expC0 : C0 → C0 by expC0 (w) = ex (cosh y + h sinh y), where w = x + hy. It follows from the previous results that expC0 is split-holomorphic, with derivative given by (expC0 )0 = expC0 . When there’s no risk of confusion, we denote the split-complex exponential of w simply by ew . An important consequence of the revised Cauchy-Riemann equations is the:

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Corollary 4.3.11. Let U ⊆ C0 be an open subset and f : U → C0 be split-holomorphic. If f = φ + hψ, then φ and ψ are solutions of the wave equation: φ = ψ = 0. In these conditions, we say that φ and ψ are Lorentz-harmonic. Remark. In terms of the differential operators defined above, we may write the d’Alembertian as ∂ ∂ . =4 ∂w ∂w Here we have the first crucial difference between holomorphicity and splitholomorphicity: in contrast to what happens with the heat equation, we can explicitly solve the wave equation in suitable domains, by employing a convenient change of variables. However, we will adopt a slightly simpler strategy to classify split-holomorphic functions on convex domains: Definition 4.3.12. The Hadamard product in R2 is the coordinatewise multiplication ? : R2 × R2 → R2 defined by . ( u1 , v1 ) ? ( u2 , v2 ) = ( u1 u2 , v1 v2 ). Proposition 4.3.13. The map σ : R2 → C0 defined by     1+h 1−h u+v u−v . σ (u, v) = u +v = +h 2 2 2 2 is an isomorphism of R-algebras when R2 is equipped with the Hadamard product, that is, it satisfies: (i) σ is bijective; (ii) σ is R-linear; (iii) σ((u1 , v1 ) ? (u2 , v2 )) = σ(u1 , v1 )σ(u2 , v2 ). Moreover, its inverse is given by σ−1 ( x + hy) = ( x + y, x − y). . Proof: Let’s verify only item (iii). Put ` = (1 + h)/2, so that σ may be written as σ (u, v) = u` + v`. Noting that `` = 0 and `2 = `, we have: σ (u1 , v1 )σ (u2 , v2 ) = (u1 ` + v1 `)(u2 ` + v2 `)

= u1 u2 `2 + u1 v2 `` + v1 u2 `` + v1 v2 `

2

= u1 u2 ` + v1 v2 ` = σ ( u1 u2 , v1 v2 ) = σ((u1 , v1 ) ? (u2 , v2 )).

The idea is to “transfer” split-holomorphic functions from C0 to R2 via σ−1 , where their treatment becomes simpler. Let’s start with a “static” version of what we intend to do later:

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Lemma 4.3.14. Let A ∈ Mat(2, R) be any matrix. So, denoting by can both the standard basis of R2 and also the basis (1, h) of C0 , we have that A is of the form   a b , b a −1 A [ σ ] for certain a, b ∈ R, if and only if [σ ]can can is diagonal.

Proof: Suppose that A has the form given in the statement above. We have that:     1 1 a b 1/2 1/2 −1 [σ]can A[σ]can = 1 −1 b a 1/2 −1/2    a+b b+a 1/2 1/2 = a−b b−a 1/2 −1/2   a+b 0 = . 0 a−b Since the map ( a, b) 7→ ( a + b, a − b) is bijective, this directly implies that if −1 A [ σ ] [σ]can can = diag( c, d ), then   1 c+d c−d , A= 2 c−d c+d as wanted. Proposition 4.3.9 above and the characterization given in Exercise 4.3.7 then give us the: Proposition 4.3.15. Let U ⊆ C0 be an open subset and f : U → C0 be a function of real-differentiable real and imaginary parts. If . fe = σ−1 ◦ f ◦ σ : σ−1 (U ) → R2 , we have that f is split-holomorphic if and only if the Jacobian matrix of fe is diagonal in all points. As a consequence of this result, we have the: Theorem 4.3.16. Let U ⊆ C0 be a connected open set. Then, if f : U → C0 is splite and ψ, e such holomorphic, there are differentiable functions of a single real variable, φ that   e( x + y) + ψ e( x − y) e( x + y) − ψ e( x − y) φ φ f ( x + hy) = +h , 2 2 for all x + hy ∈ U. Proof: We’ll maintain the notation adopted in the statement of the previous proposition. As σ is linear and U is convex, we have that σ−1 (U ) is also convex. So, f being splitholomorphic implies that the Jacobian matrix of fe is diagonal, and σ−1 (U ) being convex e(u), ψ e(v)) for certain differentiable functions now implies that fe has the form fe(u, v) = (φ e e of a single real variable, φ and ψ. With this:
f ( x + hy) = σ fe(σ−1 ( x + hy))
= σ fe( x + y, x − y)
= σ φe( x + y), ψe( x − y)   e( x + y) + ψ e( x − y) e( x + y) − ψ e( x − y) φ φ = +h , 2 2 as wanted.

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e and ψ e to be differentiable, but not Remark. In particular, note that we may choose φ twice differentiable. This means that, in contrast to what happens with holomorphic functions in C, split-holomorphic functions are not necessarily of class C∞ (not even C2 ). To conclude our discussion about differentiation in C0 , we’ll introduce the notions of pole and split-meromorphic function, needed for what will be done in Subsection 4.3.3. Definition 4.3.17. Let U ⊆ C0 be open, w0 ∈ U and f : U \ {w0 } → C0 . We say that w0 is an order k ≥ 1 pole of f if k is the least integer for which (w − w0 )k f (w) is split-holomorphic. Definition 4.3.18. Let U ⊆ C0 be open and P ⊆ U discrete. We say that a splitholomorphic function f : U \ P → C0 is split-meromorphic on U if P is precisely the set of poles of f . Let’s proceed with integration: Definition 4.3.19. Let U ⊆ C0 be an open subset, f : U → C0 be a continuous function, and γ : I → U be a smooth curve. The integral of f along γ is defined as . f (w) dw =

Z γ

Z I

f (γ(t))γ0 (t) dt.

Remark.

• The expression f (γ(t))γ0 (t) is a product of two split-complex numbers. • Suppose that f = φ + hψ. Noting that w = x + hy implies that dw = dx + h dy, and that if γ(t) = x (t) + hy(t) then dx = x 0 (t) dt and dy = y0 (t) dt along γ, we have that Z Z Z . f (w) dw = φ( x, y) dx + ψ( x, y) dy + h ψ( x, y) dx + φ( x, y) dy, γ

γ

γ

where the integrals on the right side are usual real line integrals.

• When γ is closed, we denote the integral by

I

f (w) dw, as usual. γ

• This definition is naturally extended to the case where γ is piecewise smooth. It follows directly from the definition that this integral operator is R-linear over functions. And more importantly, we have the: Theorem 4.3.20 (Fundamental Theorem of Calculus). Let U ⊆ C0 be open, f : U → C0 be continuous, and γ : [ a, b] → U be a piecewise C1 curve. If F : U → C0 is an antiderivative for f (that is, F is split-holomorphic and satisfies F 0 = f ), then Z

f (w) dw = F (γ(b)) − F (γ( a)). γ

Proof: Suppose without loss of generality that γ is smooth. By the chain rule, we have that F ◦ γ is real-differentiable, and an anti-derivative for ( f ◦ γ)γ0 , so that the Fundamental Theorem of Calculus for real functions gives us that Z

f (w) dw = γ

as wanted.

Z b a

f (γ(t))γ0 (t) dt = F (γ(b)) − F (γ( a)),

Abstract Surfaces and Further Topics  293

For what’s next, we’ll need integrals of split-holomorphic functions along curves, under some conditions, to depend only on the endpoints of the curve, just like what we had in C. This follows from the: Theorem 4.3.21 (Revised Cauchy-Goursat). Let U ⊆ C0 be a simply connected open set, f : U → C0 be a split-holomorphic function with continuous derivative, and γ : [ a, b] → U be a piecewise C1 curve, closed and injective on ] a, b[. Then I

f (w) dw = 0. γ

Proof: The strategy is to apply the Green-Stokes Theorem twice, together with the revised Cauchy-Riemann equations. Let R be the interior of the region in the plane bounded by γ. Writing f = φ + hψ, we have: I

Z

γ

Z

f (w) dw = φ( x, y) dx + ψ( x, y) dy + h ψ( x, y) dx + φ( x, y) dy γ γ    Z Z  ∂φ ∂ψ ∂ψ ∂φ = ( x, y) − ( x, y) dx dy + h ( x, y) − ( x, y) dx dy ∂x ∂y ∂y R R ∂x

= 0 + h0 = 0.

Remark. The assumptions over γ and U allow us to apply Green-Stokes. If U is not simply connected, the region R bounded by γ may contain points outside U (think of a punctured disk), and we need f split-holomorphic on all of R to conclude the argument (this is not possible if f is not defined on all of R, to begin with). Corollary 4.3.22. The line integral of a split-holomorphic function, in the conditions of the previous theorem, depends only on the endpoints of the curve, and not on the curve itself. In this case, we write Z

f (ω ) dω =

Z w

γ

w0

f (ω ) dω,

where γ joins w0 to w. Corollary 4.3.23. Let U ⊆ C0 be a simply connected open set, w0 ∈ U and f : U → C0 be a continuous function. Then F : U → C0 given by F (w) =

Z w w0

f (ω ) dω

is split-holomorphic and satisfies F 0 = f . Similar to the complex case, we have the: Definition 4.3.24. Two functions φ, ψ : U ⊆ R2 → R (of class C2 ) are called Lorentzconjugates if φu = ψv and φv = ψu . Such condition implies that both φ and ψ are Lorentz-harmonic. Theorem 4.3.25. Let U ⊆ R2 ≡ C0 be a simply connected open set, and φ : U → R be a C2 Lorentz-harmonic function. Then there exists a split-holomorphic function f : U → C0 whose real part is φ. In particular, there is a function Lorentz-conjugate to φ.

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. Proof: Define g = φu + hφv . The condition φ = 0 ensures that g is split-holomorphic and, in particular, continuous, so that U being simply connected gives us the existence of an anti-derivative G = ψ + hζ for g. With this in place, using that G 0 = g with the revised Cauchy-Riemann equations for G yield that φu + hφv = ψu + hζ u = ψu + hψv . Equating real and imaginary parts, we obtain that ψ = φ + c for some c ∈ R. Thus, . f = G − c is the split-holomorphic function we seek. The next natural step is to look for, in C0 , an analogue for the Cauchy integral formula, which has as a consequence the fact that to know the value of a holomorphic function in some point, it suffices to know the value of this function in some circle centered at this point. Thinking in Euclidean terms, we have the: Example 4.3.26. Let a ∈ C0 , a 6= 0, and consider f : C0 → C0 given by f (w) = w2 + a. For γ : [0, 2π [ → C0 given by γ(t) = cos t + h sin t, one may check (by doing a long calculation with improper integrals, since w takes values in the intersection of γ with the two principal light rays four times) that 1 a = f (0) 6 = 2πh

I γ

f (w) dw = 0. w−0

For more details, see [35]. Thinking in Lorentzian terms in C0 , the “circles” are no longer compact and connected, so that the relevant integrals again become improper. In this case, we do not have a direct analogue for the Cauchy integral formula. The best we get, aiming towards the above conclusion, is the: Proposition 4.3.27. Let U ⊆ C0 be a connected open set and f : U → C0 be a splitholomorphic function. Put ` = (1 + h)/2. So, given s, t ∈ R, for every w ∈ U such that w + s`, w + t` ∈ U we have that f (w) = ` f (w + t`) + ` f (w + s`). Proof: Fix an arbitrary w ∈ U and consider F (s, t) = ` f (w + t`) + ` f (w + s`). We have that F (0, 0) = f (w), and since `` = 0, it follows that ∂F ∂F = = 0. ∂s ∂t Thus F is constant and equals f (w). For more general facts about split-complex numbers we recommend, for example, [4] and [11].

Exercises Exercise† 4.3.1. (a) Show Proposition 4.3.3 (p. 287);

Abstract Surfaces and Further Topics  295

(b) Show that the split-complex absolute value | · | : C0 → R does not satisfy the properties of a norm. Hint. Think about how | · | is related to the “norm” k · k L . Exercise 4.3.2. If w = x + hy ∈ C0 and we assume that all the necessary inverses below exist, express in terms of x and y the real and imaginary parts of: (a) w2 ; (b)

w−1 ; w+1

(c) 1/w2 . Exercise† 4.3.3. Show that the zero divisors in C0 are precisely the real multiples of 1 + h and 1 − h. That is, zero divisors in C0 correspond to lightlike directions in L2 . Exercise 4.3.4 (More avatars of C0 ). (a) Show that the map Φ : C0 → Mat(2, R) given by   x y Φ( x + hy) = y x is a ring monomorphism. Moreover, note that given any w ∈ C0 , we have that ww = det Φ(w). This way, C0 may be seen as a collection of matrices. (b) Consider the evaluation at h, evalh : R[ x ] → C0 , given by . evalh ( p( x )) = p(h). Show that evalh is a ring epimorphism, with ker evalh = ( x2 − 1). Conclude that C0 ∼ = R[ x ] / ( x 2 − 1). Hint. Use the division algorithm in R[ x ]. Exercise 4.3.5. Let T : R2 → C0 be a R-linear map, written as T ( x, y) = ( ax + by) + h(cx + dy), for certain a, b, c, d ∈ R. Identifying ( x, y) ≡ w = x + hy, show that there are α, β ∈ C0 such that T (w) = αw + βw and αα − ββ = ad − bc, so that T is an isomorphism if and only if αα 6= ββ. Exercise 4.3.6 (Generalized complex numbers). Let u be any symbol, and consider the real commutative algebra generated by {1, u}, subject to the relation u2 = α + βu, for certain structure constants α, β ∈ R. Denote such algebra by Cα,β , that is: Cα,β = { a + ub | a, b ∈ R and u2 = α + βu}. In particular, note that C−1,0 = C and C1,0 = C0 . (a) Show that an element a + ub ∈ Cα,β has a multiplicative inverse if and only if . D = a2 + βab − αb2 6= 0.

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(b) Let a + ub ∈ Cα,β be nonzero. If b 6= 0, then a + ub has an inverse. In this case, note that D/b2 = 0 may be seen as a quadratic equation in the variable a/b. Verify that . the discriminant of such equation is ∆ = β2 + 4α. (c) The position of a point (α, β) in the parameter plane, relative to the parabola ∆ = 0, determines the possibility or not of performing divisions in Cα,β . More precisely, show that:

• if ∆ < 0, every nonzero element of Cα,β has an inverse; • if ∆ = 0, the zero divisors are the elements a + ub with a + βb/2 = 0, and all the remaining ones have inverses; √ • if ∆ > 0, the√zero divisors are the elements a + ub with a + ( β + ∆)b/2 = 0 or a + ( β − ∆)b/2 = 0, and all the remaining ones have inverses. (d) According to the three cases listed in the item above, we’ll say that Cα,β is, respectively, a system of elliptic, parabolic, or hyperbolic numbers. Justify this terminology by describing, in terms of ∆, the conic x2 + βxy − αy2 = 0 in the plane. Remark. . • Item (a) motivates us to define a + ub = a + βb − ub, thus making D a “squared norm” of a + ub. The behavior of this map D : Cα,β → R ends up being controlled by the discriminant ∆ given in item (b). It follows from Sylvester’s Criterion that D is positive-definite when ∆ < 0, degenerate for ∆ = 0, and indefinite for ∆ > 0. Polarizing D, we have that Cα,β is an algebraic model for the geometry of the symmetric bilinear form β β . h( a, b), (c, d)iα,β = ac + ad + bc − αbd 2 2 in R2 .

• Note that item (c) generalizes Exercise 4.3.3 above. • For α = β = 0, the set C0,0 = { a + bε | a, b ∈ R and ε2 = 0} is called the set of dual numbers, and it also has applications in several areas of Mathematics and Physics. One may show that according to the discriminant ∆, Cα,β is isomorphic to either C, C0 or C0,0 . Exercise† 4.3.7. (a) Let U ⊆ C0 be an open subset, w0 = x0 + hy0 ∈ U and f : U → C0 be any function. Show that f is C0 -differentiable at w0 if and only if it is differentiable at ( x0 , y0 ) when seen as a map from L2 to L2 and the total derivative D f ( x0 , y0 ) is C0 -linear. Hint. Saying that D f ( x0 , y0 ) is C0 -linear is equivalent (why?) to saying that     0 1 0 1 D f ( x0 , y0 ) = D f ( x0 , y0 ). 1 0 1 0 (b) Show Proposition 4.3.7 (p. 288). Exercise† 4.3.8. Let U ⊆ C0 be a connected open set and f : U → C0 be splitholomorphic. Show that if f 0 (w) = 0 for all w ∈ U, then f is constant.

Abstract Surfaces and Further Topics  297

Exercise 4.3.9. Let U ⊆ C0 be an open and connected set and f : U → C0 be splitholomorphic. (a) Show that if f only assumes real values or only purely imaginary values, then f is constant. (b) Show that if the function g : U → C0 given by g(w) = f (w) f (w) is a non-zero constant, then f is constant. Give a counter-example where g = 0 and f is nonconstant. Exercise 4.3.10. Let U ⊆ C0 be an open and connected set, and f : U → C0 be splitholomorphic, written as f = φ + hψ. If ξ : U → R, show that φ + hξ is split-holomorphic if and only if ψ − ξ is constant. Exercise 4.3.11. Let f : C0 → C0 be split-holomorphic. Define a function g : C0 → C0 by g(w) = f (w). Show that g is also split-holomorphic, and that its derivative is given by g0 (w) = f 0 (w). Exercise† 4.3.12. Show that: (a) given an open set U ⊆ C0 and a function f : U → C0 , f satisfies the revised Cauchy∂f = 0; Riemann equations if and only if ∂w (b) if f is split-holomorphic, then f 0 (w) = (c)  = 4

∂f (w), for each w ∈ U; ∂w

∂ ∂ . ∂w ∂w

Exercise 4.3.13. Liouville’s Theorem claims that every bounded holomorphic function defined on the whole complex plane C is necessarily constant. In C0 , this theorem is lost. Show that f : C0 → C0 given by f ( x + hy) =

1+h 1 + e− x e− y

is a counter-example. Exercise 4.3.14. Let U ⊆ C0 be an open set, f : U → C0 be a continuous map, and γ : [ a, b] → U be a C1 curve. (a) Suppose that the image of f is contained in a single light ray passing through the origin. Show that Z f (w) dw = 0. γ (b) Assume that γ parametrizes a portion of a light ray (possibly affine). Show that Z f (w) dw = 0. γ

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4.3.2

Bonnet rotations

We have previously seen that some parametrizations of a non-degenerate regular surface may emphasize some of its geometric aspects better than others. For critical surfaces, it is convenient to work with isothermal parametrizations: Definition 4.3.28. Let M ⊆ R3ν be a non-degenerate regular surface and (U, x) be a parametrization for M. We say that x is isothermal if F = 0 and | E| = | G | = λ2 , for some smooth function λ : U → R. Remark.

• We are free to assume, when necessary, that xu is always spacelike. No generality is lost, since if M ⊆ L3 and xu are timelike, we consider the reflected open set U 0 = (u, v) ∈ U | (v, u) ∈ U and the reparametrization (U 0 , y) given by y(u, v) = x(v, u), whose image is the same as the image of x, now with yu spacelike. • Under these conditions, the First Fundamental Form of M is expressed in terms of x by ds2 = λ(u, v)2 (du2 + (−1)ν dv2 ). The existence of such parametrizations is guaranteed by the: Theorem 4.3.29 (Korn-Lichtenstein). Let M ⊆ R3ν be a non-degenerate regular surface. Then, for each point p ∈ M there is an open set U ⊆ R2 and an isothermal parametrization (U, x) for M around p. The proof of this result is outside the scope of this text, but it may be found in [3]. Naturally, the first step in what follows is to find expressions for the mean curvature in terms of such parametrizations: Proposition 4.3.30. If M ⊆ R3ν is a non-degenerate regular surface and (U, x) is an isothermal parametrization for M, then H ◦ x = (−1)ν

ev e + eu g , 2λ2

where e and g denote the coefficients of the Second Fundamental Form of x (computed relative to a Gauss map compatible with x), and eu and ev are its partial indicators (recall Definition 3.2.2, p. 156). Proof: Using the expression for H ◦ x given in Proposition 3.3.10 (p. 182), we have that H◦x=

e M eG − 2 f F + Eg e M eev λ2 + geu λ2 ev e + eu g = = (−1)ν . 2 4 2 2 EG − F 2λ2 eu ev λ

Remark. The expressions for the Gaussian curvature have already been registered in Exercise 3.7.3 (p. 256). The operators 4 and  also act on vector-valued functions in a natural way: if . Φ : U ⊆ R2 → R3 is differentiable, the Laplacian of Φ is defined by 4Φ = Φuu + Φvv , . and the d’Alembertian of Φ is defined by Φ = Φuu − Φvv . We say that Φ is harmonic if 4Φ = 0 and Lorentz-harmonic if Φ = 0. Moreover, if Ψ : U ⊆ R2 → R3 is another differentiable function, we say that Ψ is conjugate or Lorentz-conjugate to Φ if its components are (see Definition 4.3.24, p. 293). These operators have a close relation with the mean curvature, given in the:

Abstract Surfaces and Further Topics  299

Proposition 4.3.31. Let M ⊆ R3ν be a non-degenerate regular surface and (U, x) be an isothermal parametrization for M. We have that: (i) if M is spacelike, then 4 x = 2λ2 H; (ii) if M is timelike, then x = 2λ2 eu H, where we abbreviate the mean curvature vector by H ≡ H ◦ x. Proof: Suppose initially that M is spacelike and fix a Gauss map N compatible with x. We have that

h4 x, xu i = h xuu , xu i + h xvv , xu i = h xuu , xu i − h xv , xuv i 1 ∂ = (h xu , xu i − h xv , xv i) = 0 2 ∂u and, similarly, h4 x, xv i = 0. This way, we have that 4 x is normal to M and, also identifying N ≡ N ◦ x, we have that 4 x = (−1)ν h4 x, N i N. But:

h4 x, N i = h xuu , N i + h xvv , N i = e + g = 2(−1)ν λ2 H, and the result follows. Suppose now that M is timelike. Similarly as done above, one may show that hx, xu i L = hx, xv i L = 0. Since N is spacelike, we have that x = hx, N i L N and, moreover, ev = −eu . Thus

hx, N i L = h xuu , N i L − h xvv , N i L = e − g = 2λ2 eu H.

Corollary 4.3.32. Let M ⊆ R3ν be a non-degenerate regular surface and (U, x) be an isothermal parametrization for M. Then x(U ) has zero mean curvature if and only if x is harmonic or Lorentz-harmonic, depending on whether M is spacelike or timelike. Definition 4.3.33. Given parametrized regular surfaces x, y : U → R3ν , we define the 1-parameter families of maps zt , wt : U → R3ν , with t ∈ R, by: . (i) zt = (cos t) x + (sin t)y, if x and y are spacelike and conjugate. . (ii) wt = (cosh t) x + (sinh t)y, if x and y are timelike and Lorentz-conjugate. We say that zt and wt are the Euclidean and Lorentzian Bonnet rotations, respectively, associated to x and y. Remark. Since the functions cosh and sinh have no real period and cosh never vanishes, the maps x and y play different roles in the definition of the Lorentzian Bonnet rotation wt , while for zt , their order is irrelevant. Let’s start looking for properties shared by both situations, despite this loss of periodicity in the timelike case. In the following proofs, until the end of this subsection, we will work under the assumption that xu is always spacelike. Theorem 4.3.34. Let x, y : U → R3ν be two parametrized regular critical surfaces, conjugate and isothermal. The following objects are invariant under a Bonnet rotation associated to x and y:

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(i) the coefficients of the First Fundamental Form (and, thus, all the obtained surfaces are also isothermal and pairwise isometric); (ii) the direction of the Gauss maps N t (and, thus, all the tangent planes for fixed (u, v) ∈ U are parallel). Furthermore, all the obtained surfaces are critical. Proof: (i) Let’s verify the result for Euclidean Bonnet rotations: both x and y have their First Fundamental Form coefficients given by E = G = λ2 and F = 0 (i.e., the same parameter λ for both). With this: ztu = (cos t) xu + (sin t)yu

and ztv = (cos t) xv + (sin t)yv .

Rewriting this only in terms of derivatives of x, we get that: ztu = (cos t) xu − (sin t) xv

and ztv = (sin t) xu + (cos t) xv .

It follows from this that if Et , F t and G t denote the coefficients of the First Fundamental Form of zt , then E t = G t = λ2

and

F t = 0,

as wanted, by using that F = 0 and cos2 t + sin2 t = 1. The Lorentzian case is analogous, using cosh2 t − sinh2 t = 1 instead. (ii) It suffices to show that ztu × ztv is always proportional to xu × xv , for every t. To wit, they’re actually equal. In the Euclidean case, we have: ztu × ztv = ((cos t) xu − (sin t) xv ) × ((sin t) xu + (cos t) xv )

= (cos2 t) xu × xv − (sin2 t) xv × xu = xu × xv . The other case, again, is similar. Lastly, we have that: ztuu = (cos t) xuu − (sin t) xvu ,

ztvv = (sin t) xuv + (cos t) xvv ,

and adding we obtain that 4zt = (cos t)4 x = 0, since x is critical. In the Lorentzian case we’ll have wt = (cosh t)x = 0 instead. Remark. In the Euclidean case, in the same way that x and y are conjugate, so are zt and zt+π/2 , for all t ∈ R. Corollary 4.3.35. In the above conditions, the Weingarten maps of all the wt will be diagonalizable if the Weingarten map of x is.

Abstract Surfaces and Further Topics  301

Example 4.3.36. (1) Isometric deformation from the catenoid to the helicoid: consider the catenoid parametrized by x : R × ]0, 2π [ → R3 , given by x(u, v) = (cosh u cos v, cosh u sin v, u). We have that xu (u, v) = (sinh u cos v, sinh u sin v, 1) and xv (u, v) = (− cosh u sin v, cosh u cos v, 0), whence ds2 = (cosh2 u)(du2 + dv2 ). Moreover, 4 x = 0. Let’s find a surface y conjugate to x. Integrating xu with respect to v, we have y(u, v) = (sinh u sin v, − sinh u cos v, v) + c(u), and differentiating with respect to u, we must have c0 (u) = 0. This shows that c is a constant, which may take to be 0. Therefore a conjugate surface to x is given by y(u, v) = (sinh u sin v, − sinh u cos v, v), which also satisfies 4y = 0 and has the First Fundamental Form expressed by ds2 = (cosh2 u)(du2 + dv2 ). We have a family of minimal and pairwise isometric surfaces zt , t ∈ R. Let’s illustrate this Bonnet rotation:

(a) t = 0

(b) t = π/6

(c) t = π/3

(d) t = π/2

Figure 4.13: Minimal isometric deformation from the catenoid to the helicoid. (2) Bonnet rotation for the Lorentzian (hyperbolic catenoid): consider the parametrization x : R2 → L3 , given by x(u, v) = (u, cosh u cosh v, cosh u sinh v). We have that xu (u, v) = (1, sinh u cosh v, sinh u sinh v) and xv (u, v) = (0, cosh u sinh v, cosh u cosh v),

302  Introduction to Lorentz Geometry: Curves and Surfaces

whence ds2 = (cosh2 u)(du2 − dv2 ). Moreover, we have x = 0. Let’s find a surface y Lorentz-conjugate to x. Integrating xu with respect to v, we have y(u, v) = (v, sinh u sinh v, sinh u cosh v) + c(u), and differentiating this with respect to u we get that c0 (u) = 0, whence c is a constant, which we’ll take to be zero. Thus, we obtain y(u, v) = (v, sinh u sinh v, sinh u cosh v). We have that y also satisfies y = 0 and has its Minkowski First Fundamental Form expressed by ds2 = (cosh2 u)(du2 − dv2 ). So we have a family wt of critical and pairwise isometric surfaces, illustrated as follows:

(a) t = 0

(b) t = 1/4

(c) t = 1/2

(d) t = 3/4

Figure 4.14: Minimal isometric deformation of a Lorentzian catenoid. Now, let’s see the relation between the Second Fundamental Forms of the surfaces obtained via a Bonnet rotation and the Second Fundamental Form of the initial surfaces: Lemma 4.3.37. Let x, y : U → R3ν be two parametrized regular surfaces, conjugate and isothermal. Denote by e and f the coefficients of the Second Fundamental Forms of x and y, and by et and f t the coefficients of the Second Fundamental Form of the surfaces obtained via the Bonnet rotation associated to x and y. (i) In the Euclidean case, we have that et = e cos t − f sin t

and

f t = e sin t + f cos t.

In particular, et+π/2 = − f t and f t+π/2 = et . (ii) In the Lorentzian case, we have that et = e cosh t + f sinh t

and

f t = e sinh t + f cosh t.

Remark. Do not confuse et with et (an exponential). Proof: Let’s verify the Lorentzian case this time. Denoting the common normal direction to all the wt by N ≡ N ◦ wt , we have et = hwtuu , N i L

= h(cosh t) xuu + (sinh t) xuv , N i L = cosh th xuu , N i L + sinh th xuv , N i L = e cosh t + f sinh t,

Abstract Surfaces and Further Topics  303

and also f t = hwtuv , N i L

= h(cosh t) xuv + (sinh t) xvv , N i L = cosh th xuv , N i L + sinh th xvv , N i L = f cosh t + g sinh t = e sinh t + f cosh t, by using that x is critical and thus g = e. The periodicity stated above for the Euclidean case has the following result as a consequence: Theorem 4.3.38 (Special curves). Let x, y : U → R3ν be two parametrized regular spacelike surfaces, conjugate and isothermal, and zt be the Euclidean Bonnet rotation associated to x and y. For each t ∈ R, consider the curve αt : I → zt (U ) ⊆ R3ν given in coordinates by αt (s) = zt (u(s), v(s)). Then: (i) αt is an asymptotic line for zt if and only if αt+π/2 is a curvature line for zt+π/2 . (ii) αt is a curvature line for zt if and only if αt+π/2 is an asymptotic line for zt+π/2 . Proof: We will use Propositions 3.5.6 (p. 210) and 3.5.9 (p. 212): (i) It suffices to note that, avoiding points of evaluation, we have: 02 v λ2 et+π/2

2 −u0 v0 u0 2 = λ2 ( e t u 0 2 + 2 f t u 0 v 0 − e t v 0 2 ). 0 λ f t+π/2 −et+π/2

(ii) As in the previous item, the conclusion follows from the following identity: 2 v0 −u0 v0 u0 2 2 = λ2 (− et+π/2 u0 2 − 2 f t+π/2 u0 v0 + et+π/2 v0 2 ). λ2 0 λ et ft −et

The corresponding result for geodesics follows from the fact that all the surfaces obtained via a Bonnet rotation have the same coefficients for their First Fundamental Forms. We register that: Proposition 4.3.39. Let x, y : U → R3ν be parametrized regular surfaces, conjugate or Lorentz-conjugate, and isothermal. For each stage t, the Bonnet rotations map the geodesics in x into geodesics of zt and wt .

304  Introduction to Lorentz Geometry: Curves and Surfaces

4.3.3

Enneper-Weierstrass representation formulas

Given a regular spacelike surface M ⊆ R3ν and a parametrization (U, x), the idea here is to make the usual identification of R2 with C and use z = u + iv as the surface parameter. Recall that z+z z−z u= and v = . 2 2i We also have the Wirtinger operators     ∂ 1 ∂ ∂ ∂ 1 ∂ ∂ = −i and = +i . ∂z 2 ∂u ∂v ∂z 2 ∂u ∂v With this, the Laplacian operator can be written as    ∂2 ∂2 ∂ ∂ ∂ ∂ ∂ ∂ 4= 2+ 2 = +i −i =4 . ∂u ∂v ∂u ∂v ∂z ∂z ∂u ∂v Abusing notation, we may also write x(z, z) = ( x1 (z, z), x2 (z, z), x3 (z, z)). It will also be convenient to consider the above parametrizations as the real part of curves in C3 . For this end, we consider an extension of the inner product of R3 to C3 , to be also denoted by h·, ·i E , defined by

h(z1 , z2 , z3 ), (w1 , w2 , w3 )i E = z1 w1 + z2 w2 + z3 w3 . In a similar fashion, to study timelike surfaces in L3 , we will use a split-complex parameter, and then extend the scalar product in L3 to the complex Lorentzian space C31 , i.e., the vector space C3 equipped with the product h·, ·i L , defined by

h(z1 , z2 , z3 ), (w1 , w2 , w3 )i L = z1 w1 + z2 w2 − z3 w3 . We will keep the usual terminology about causal types in this new setting. Definition 4.3.40. Let U be an open subset of C or C0 , and x : U → R3ν be a nondegenerate parametrized regular surface. (i) The complex derivative of x is φ≡

∂x . 1 ≡ x z = ( x u − i x v ). ∂z 2

(ii) The split-complex derivative of x is ψ≡

∂x . 1 ≡ x w = ( x u + h x v ). ∂w 2

Remark. Note that hφ, φi E = 0 does not imply that φ = 0, since φ(z, z) ∈ C3 for all z, and not necessarily in R3 . This holds a fortiori for ψ. Proposition 4.3.41. If M ⊆ R3ν is a non-degenerate regular surface and (U, x) is a parametrization for M, then x is isothermal if and only if: (i) hφ, φi E = 0, for M ⊆ R3 ;

Abstract Surfaces and Further Topics  305

(ii) hφ, φi L = 0, for spacelike M ⊆ L3 ; (iii) hψ, ψi L = 0, for timelike M ⊆ L3 . Proof: We’ll do here the cases where M is spacelike, and just state the equivalent expression in the remaining case. If E, F and G are the coefficients of the First Fundamental Form of M relative to the parametrization x, we have that j ( x z )2



=

1 j j ( xu − i xv ) 2

2

1 j j j j = (( xu )2 − ( xv )2 − 2ixu xv ), 4

and summing over j we obtain 1 hφ, φi = ( E − G − 2iF ), 4 so that the conclusion follows from the fact that a complex number is zero if and only if both its real and imaginary parts are also zero. When M is timelike, we’ll have 1 hψ, ψi L = ( E + G + 2hF ) 4 instead. Lemma 4.3.42. Let M ⊆ R3ν be a non-degenerate regular surface and (U, x) be an isothermal parametrization for M. Then: (i) hφ, φi E = λ2 /2 6= 0, for M ⊆ R3 ; (ii) hφ, φi L = λ2 /2 6= 0, for spacelike M ⊆ L3 ; (iii) hψ, ψi L = eu λ2 /2 6= 0, for timelike M ⊆ L3 . Proof: Again we’ll do the proof only in the case where M is spacelike: j j

xz xz =

1 j 1 j j j j j j j ( xu − i xv )( xu + i xv ) = (( xu )2 + ( xv )2 − 2ixu xv ) 4 4

and, summing over j, it follows that 1 λ2 hφ, φi = (λ2 + λ2 − 2i · 0) = . 4 2 The verification of this result for timelike surfaces is Exercise 4.3.18. Proposition 4.3.43. Let M ⊆ R3ν be a non-degenerate regular surface and (U, x) be an isothermal parametrization for M. Then x is critical if and only if φ is holomorphic or ψ is split-holomorphic, according to whether M is spacelike or timelike, respectively. Proof: This follows directly from the expressions ∂φ ∂2 x 1 = = 4 x and ∂z ∂z∂z 4

∂ψ ∂2 x 1 = = x. ∂w ∂w∂w 4

In view of this result we may conclude that every non-degenerate critical surface may be, at least locally, represented by a triple:

306  Introduction to Lorentz Geometry: Curves and Surfaces

• φ = (φ1 , φ2 , φ3 ) of holomorphic functions satisfying (φ1 )2 + (φ2 )2 + (φ3 )2 = 0, if M ⊆ R3 ;

• φ = (φ1 , φ2 , φ3 ) of holomorphic functions satisfying (φ1 )2 + (φ2 )2 − (φ3 )2 = 0, if M ⊆ L3 is spacelike;

• ψ = (ψ1 , ψ2 , ψ3 ) of split-holomorphic functions satisfying (ψ1 )2 + (ψ2 )2 − (ψ3 )2 = 0, if M ⊆ L3 if timelike. These motivate the following: Definition 4.3.44. A spacelike (resp. timelike) critical curve is a map ζ : U → C3ν with holomorphic (resp. split-holomorphic) components satisfying hζ 0 , ζ 0 i = 0. The curve is said to be regular if hζ 0 , ζ 0 i 6= 0. The Bonnet rotations seen previously may be translated in terms of critical curves by using complex and split-complex numbers: regular critical curves give rise to families of associated critical surfaces via zt (u, v) = Re(e−it ζ (u + iv)) and wt (u, v) = Re(e−ht ζ (u + hv)), which are spacelike and timelike for all t, according to whether ζ takes values in C3 or C31 . We emphasize that for spacelike surfaces, the regularity condition on ζ is relative to the Euclidean or Lorentzian product, according to the ambient where the surface lies. With this in place, we may define the Gaussian curvature and the normal direction of a critical curve, from zt and wt . For more details in the Euclidean case, see [27]. Proposition 4.3.45. Let M ⊆ R3ν be a non-degenerate regular and critical surface, U be a simply connected domain, and (U, x) be an isothermal parametrization for M. Then the components of x satisfy: (i) x j (z, z) = c j + 2 Re j

Z z

(ii) x (w, w) = c j + 2 Re

z0

φ j (ξ ) dξ, for some z0 ∈ U, if M is spacelike, and

Z w w0

ψ j (ω ) dω, for some w0 ∈ U, if M is timelike,

where the c j ∈ R are suitable constants. Proof: For a change, let’s do the proof in the case where M is timelike. First observe that since U is simply connected, x is isothermal and M is critical, then ψ is splitholomorphic, so that the integrals in the statement of the result are all path-independent. With differentials, we have that: 1 j 1 j j j j j ( xu + h xv )(du + h dv) = ( xu du + xv dv + h( xv du + xu dv)) 2 2 1 j 1 j j j j j ψ j dw = ( xu − h xv )(du − h dv) = ( xu du + xv dv − h( xv du + xu dv)) 2 2

ψ j dw =

Abstract Surfaces and Further Topics  307

Adding, we get that j

j

dx j = xu du + xv dv = ψ j dw + ψ j dw = 2 Re ψ j dw, whence j

x (w, w) = c j + 2 Re

Z w w0

ψ j (ω ) dω,

for some c j ∈ R and w0 ∈ U. Theorem 4.3.46 (Enneper-Weierstrass I). Let U ⊆ C be simply connected, z0 ∈ U, and f , g : U → C be functions such that f is holomorphic, g is meromorphic, but f g2 is holomorphic. Then the map x : U → R3ν defined by x(z, z) = ( x1 (z, z), x2 (z, z), x3 (z, z)), where (i) x1 (z, z) = Re 2

x (z, z) = Re

Z z Zz0z

x3 (z, z) = 2 Re (ii) x1 (z, z) = Re x2 (z, z) = Re

zZ0

i f (ξ )(1 + g(ξ )2 ) dξ and, z

z0

Z z Zz0z

x3 (z, z) = 2 Re

f (ξ )(1 − g(ξ )2 ) dξ,

z0 Z

f (ξ ) g(ξ ) dξ, for x in R3 or;

f (ξ )(1 + g(ξ )2 ) dξ, i f (ξ )(1 − g(ξ )2 ) dξ and, z

z0

− f (ξ ) g(ξ ) dξ, for x in L3

is a parametrized surface, regular on the points where the zeros of f have precisely twice the order of the poles of g, and | g| 6= 1 (this last condition only in L3 ). Moreover, its image is a critical spacelike surface. Proof: The conditions on U, f and g ensure that all the integrals are path-independent. Also, in R3 , the complex derivative of x is precisely   1 2 i 2 φ= f (1 − g ), f (1 + g ), f g , 2 2 which satisfies 

hφ, φi E =

1 f (1 − g2 ) 2

2



+

i f (1 + g2 ) 2

2

+ ( f g)2 = 0,

so that the expression for hφ, φi E given in Proposition 4.3.41 ensures that E = G and F = 0. A similar computation gives the same conclusion in L3 . So, x is regular precisely when E = G 6= 0, which is equivalent to the condition given in the statement regarding the orders of the zeros of f and poles of g in R3 . The condition | g| ≤ 1 in L3 follows from the fact that if   1 2 i 2 f (1 + g ), f (1 − g ), − f g , φ= 2 2 then 2 2 1 i λ2 | f |2 2 2 = hφ, φi L = f (1 + g ) + f (1 − g ) − |− f g|2 = (1 − | g |2 )2 . 2 2 2 2 This being the case, x is isothermal and spacelike. Furthermore, φ is holomorphic, and thus the image of x is critical.

308  Introduction to Lorentz Geometry: Curves and Surfaces

The pair ( f , g) is known as the Weierstrass data for the surface. It is possible to write expressions for the coefficients of the First Fundamental Form and the Gaussian curvature in terms of the Weierstrass data ( f , g). When M ⊆ R3 , the function g is closely related to the Gauss map of the surface (which takes values in the sphere S2 ), via the stereographic projection (mentioned in Exercise 3.1.2, p. 152). For more details, see [57]. In the expression for the spacelike parametrization in L3 , the sign in the third component represents a reflection about the plane z = 0, seemingly irrelevant, but which allows us to look for a relation with the Gauss map also in this case. For details, see [36]. When the function g is holomorphic and invertible, we may use it as a parameter to obtain an alternative representation: Theorem 4.3.47 (Enneper-Weierstrass II). Let U ⊆ C be open and simply connected, z0 ∈ U, and F : U → C be a holomorphic function. Then the map x : U → R3ν given by x(z, z) = ( x1 (z, z), x2 (z, z), x3 (z, z)), where (i) x1 (z, z) = Re x2 (z, z) = Re

Z z Zz0z

x3 (z, z) = 2 Re (ii) x1 (z, z) = Re x2 (z, z) = Re 3

(1 − ξ 2 ) F (ξ ) dξ, i(1 + ξ 2 ) F (ξ ) dξ and,

zZ0

z z0

Z z

(1 + ξ 2 ) F (ξ ) dξ,

Zz0z

i(1 − ξ 2 ) F (ξ ) dξ, and

z0Z

x (z, z) = −2Re

ξ F (ξ ) dξ, for x in R3 , or;

z z0

ξ F (ξ ) dξ, for x in L3 ,

is a parametrized surface, which is regular on the points where F (z) 6= 0 (in R3 ) or F (z) 6= 0 and |z| 6= 1 (in L3 ). Moreover, its image is a critical spacelike surface. Example 4.3.48 (Critical spacelike surfaces in R3ν ). (1) Enneper surface in R3 : consider the Weierstrass data f (z) = 1 and g(z) = z. We obtain the parametrization x : R2 → R3 given by   v3 u3 2 2 2 2 + uv , −v + − u v, u − v . x(u, v) = u − 3 3

Figure 4.15: Enneper surface in R3 . The same surface could be obtained via the second representation with F (z) = 1.

Abstract Surfaces and Further Topics  309

(2) Spacelike Enneper surface in L3 : consider the same Weierstrass data from the previous example, now in the Lorentzian setting. This time we obtain the parametrization x : R2 → R3 given by   u3 v3 x(u, v) = u + − uv2 , −v − + u2 v, v2 − u2 . 3 3

Figure 4.16: Spacelike Enneper surface in L3 . (3) Catalan surface in

R3 :

 using the single type II data F (z) = i

 1 1 − , we produce z z3

the parametrization  u  v  x(u, v) = u − sin u cosh v, 1 − cos u cosh v, −4 sin sinh 2 2

Figure 4.17: Catalan surface in R3 . (4) Spacelike catenoid in L3 : this time we take the type II Weierstrass data F (z) = 1/z2 , which gives the parametrization   u v 2 2 x(u, v) = u − 2 ,v− 2 , −2 log(u + v ) . u + v2 u + v2

310  Introduction to Lorentz Geometry: Curves and Surfaces

Figure 4.18: Spacelike Lorentzian catenoid. (5) Henneberg surface in R3 : it is given by F (z) = 1 −

1 , with the parametrization z4

x(u, v) = (2 sinh u cos v − (2/3) sinh(3u) cos(3v), 2 sinh(u) sin(v) + (2/3) sinh(3u) sin(3v), 2 cosh(2u) cos(2v)) .

Figure 4.19: Henneberg surface. Proceeding, we may express the Gaussian curvature in terms of the type II Weierstrass data. We register this in the: Proposition 4.3.49. Let M ⊆ R3ν be a spacelike critical surface and (U, x) a type II Weierstrass parametrization induced by a holomorphic function F. Then its Gaussian curvature is given by K ( x(u, v)) = (−1)ν+1

4

| F (u + iv)|2 ((−1)ν

+ u2 + v2 )4

.

Proof: To compute K, it suffices to know the value of λ2 = 2hφ, φi and then apply the result from Exercise 3.7.3 (p. 256). In this case, we may study the situation in both

Abstract Surfaces and Further Topics  311

ambient spaces at the same time. If M ⊆ R3 , the complex derivative of x is   i 1 2 2 (1 − z ) F (z), (1 + z ) F (z), zF (z) , φ(z) = 2 2 while if M ⊆ L3 we have  φ(z) =

 1 i 2 2 (1 + z ) F (z), (1 − z ) F (z), −zF (z) , 2 2

but in any case we get that: ! 2 2 1 i λ(z)2 = 2 (1 − z2 ) F (z) + (1 + z2 ) F (z) + (−1)ν |zF (z)|2 2 2   1 = | F (z)|2 |1 − z2 |2 + |1 + z2 |2 + 4(−1)ν |z|2 . 2 Writing z = u + iv, we have  2 2 2 2 2 2 2  |1 − z | = (u − v − 1) + 4u v |1 + z2 |2 = (u2 − v2 + 1)2 + 4u2 v2   4| z |2 = 4( u2 + v2 ). Substituting that in the expression for λ2 , canceling and factoring terms, it follows that λ(u, v)2 = | F (u + iv)|2 ((−1)ν + u2 + v2 )2 . With this, we have log λ(u, v)2 = log F (z) F (z) + 2 log((−1)ν + u2 + v2 ). On one hand, a direct calculation shows that

4 log((−1)ν + u2 + v2 ) = (−1)ν

((−1)ν

4 . + u2 + v2 )2

On the other hand, we claim that 4 log F (z) F (z) = 0. To wit, this is trivial when F is constant. Else, F being holomorphic implies that ∂F/∂z = 0. And moreover, if F is holomorphic, F is not, so that ∂F/∂z = 0. Thus:

4 log F (z) F (z) = 4(log F (z) + log F (z)) = 4 log F (z) + 4 log F (z)     ∂ ∂ ∂ ∂ log F (z) + 4 log F (z) =4 ∂z ∂z ∂z ∂z !   1 ∂F ∂ 1 ∂F ∂ =4 (z) + 4 (z) ∂z F (z) ∂z ∂z F (z) ∂z = 0. Substituting everything, we obtain K ( x(u, v)) = − as wanted.

4 log λ(u, v)2 4 = (−1)ν+1 , 2λ(u, v)2 | F (u + iv)|2 ((−1)ν + u2 + v2 )4

312  Introduction to Lorentz Geometry: Curves and Surfaces

Now, we may establish the corresponding results for timelike surfaces in L3 : Theorem 4.3.50 (Enneper-Weierstrass I). Let U ⊆ C0 be open and simply connected, w0 ∈ U, and f , g : U → C be two functions such that f is split-holomorphic, g is splitmeromorphic, and f g2 is split-holomorphic. Then the parametrized surface x : U → L3 given by x(w, w) = ( x1 (w, w), x2 (w, w), x3 (w, w)), where x1 (w, w) = Re

Z w

Z w

x2 (w, w) = 2 Re x3 (w, w) = Re

f (ω )(1 − g(ω )2 ) dω

w0

w Z w0 w0

f (ω ) g(ω ) dω

f (ω )(1 + g(ω )2 ) dω,

is regular on the points where the zeros of f have precisely twice the order of the poles of g, f (w) is not a zero divisor and g(w) is not real. Moreover, its image is a timelike critical surface. Proof: The conditions on U, f and g again ensure that all the integrals are pathindependent. In this case the split-complex derivative of x is   1 1 2 2 ψ= f (1 − g ), f g, f (1 + g ) . 2 2 We have that 2  2 1 1 2 2 2 f (1 − g ) + ( f g ) − f (1 + g ) = 0 and hψ, ψi L = 2 2  ff  ff hψ, ψi L = (1 − g2 )(1 − g2 ) + 4gg − (1 + g2 )(1 + g2 ) = − ( g − g)2 , 4 2 

from where all conclusions follow. Theorem 4.3.51 (Enneper-Weierstrass II). Let U ⊆ C0 be open and simply connected with U ∩ R = ∅, w0 ∈ U, and F : U → C0 be a split-holomorphic function. Then the parametrized surface x : U → L3 given by x(w, w) = ( x1 (w, w), x2 (w, w), x3 (w, w)), where x1 (w, w) = Re

Z w w0

2

x (w, w) = 2 Re x3 (w, w) = Re

(1 − ω 2 ) F (ω ) dω

Z w w0

Z w w0

ωF (ω ) dω

(1 + ω 2 ) F (ω ) dω,

is regular in the points where F (w) is not a zero divisor. Moreover, its image is a timelike critical surface. Example 4.3.52 (Timelike critical surfaces in L3 ). (1) Timelike Enneper surface: for the data F (w) = 1 we obtain the parametrization  x(u, v) =

v3 v3 v − u v − , 2uv, v + u2 v + 3 3 2

 .

Abstract Surfaces and Further Topics  313

Figure 4.20: Timelike Enneper surface. (2) Timelike catenoid: for the data F (w) = 1/w2 we obtain the parametrization     u u 2 2 2 x(u, v) = − 2 − u, log (u − v ) , − 2 +u , u − v2 u − v2 defined everywhere, except on the light rays (u2 = v2 ), and regular on its domain, except on the real axis (v = 0).

Figure 4.21: Timelike catenoid.

For more details about such representation formulas, you may consult [37] and [46]. Such techniques also have applications in the study of the so-called Björling problems — see, for example, [2], [12], [18], and [19].

314  Introduction to Lorentz Geometry: Curves and Surfaces

Exercises Exercise 4.3.15 (Mercator’s Projection). We have seen that every surface locally admits isothermal parametrizations. However, it is not always easy to find them. Show that Mercator’s projection x : R × ]0, 2π [ → R3 given by   cos v sin v x(u, v) = , , tanh u cosh u cosh u is a regular and isothermal parametrization of the sphere S2 . Also verify that the meridians and parallels in S2 correspond via x to straight lines in the plane. Exercise 4.3.16. Let x, y : U → R3ν be two regular and conjugate (resp., Lorentzconjugate) parametrized surfaces. Show that if x is isothermal, so is y. Exercise 4.3.17. Let θ ∈ R be any angle. Show that the parametrized surface x : ]0, 2π [ × R → R3 given by x(u, v) = (u cos θ ± sin u cosh v, v ± cos θ cos u sinh v, ± sin θ cos u cosh v) is isothermal and minimal. Exercise 4.3.18. Prove Lemma 4.3.42 (p. 305) for spacelike surfaces. Exercise 4.3.19. Let (U, x) be a regular, injective, isothermal and minimal parametrized . surface, and N be a Gauss map for M = x(U ), compatible with x. Show that if M is not a plane, then N is a locally conformal map (in the sense of Exercise 3.2.28, p. 177). Hint. Also compute the Gaussian curvature via the expression given in Proposition 3.3.10 (p. 182). Exercise 4.3.20. There is a surprising “converse” to the result given in the previous exercise: let M ⊆ R3ν be a non-degenerate and connected regular surface, and N be a Gauss map for M. Suppose that N is locally conformal. Show that M is a critical surface, or is a piece of a sphere, de Sitter space, or hyperbolic plane. Hint. Consider an orthogonal parametrization x for M, compatible with N. Use the conformality of N to show that f (eG + Eg) = 0. If eG + Eg = 0, then x is critical. If f = 0, use again the conformality of N to show that e/E = ± g/G. If the equality holds with the negative sign, then x is critical. Else, x is totally umbilical.

4.4

DIGRESSION: COMPLETENESS AND CAUSALITY

In Section 1.3 of Chapter 1, we discussed some interpretations of the definitions and results proven until then, in the context of Special Relativity, giving emphasis to LorentzMinkowski space Ln . But now, with the language of manifolds, we can formalize such ideas. To motivate what we are going to do in this section, we first present the notions of intrinsic distance (in the Riemannian case) and geodesic completeness. We begin recalling a few definitions:

Abstract Surfaces and Further Topics  315

Definition 4.4.1. Let M be a smooth manifold. A vector field along M is a smooth S mapping V : M → p∈ M Tp M such that V ( p) ∈ Tp M for all p ∈ M. . S Remark. The set TM = p∈ M Tp M is called the tangent bundle of M, and it possesses a natural smooth manifold structure. This way, it makes sense to consider the smoothness of a map M → TM. Definition 4.4.2. Let M be a smooth manifold. A pseudo-Riemannnian metric in M is a choice of non-degenerate scalar products h·, ·i p in each tangent space Tp M such that (i) the index of h·, ·i p is the same, for all p ∈ M; (ii) the choice depends smoothly on p, that is: given any two smooth vector fields V and W along M, the map taking p to the number hV ( p), W ( p)i p is smooth. We will say that the pair ( M, h·, ·i) is a pseudo-Riemannian manifold. Remark.

• If M is connected, condition (i) is automatically satisfied. We will assume from now on that this is the case. • If the index of the pseudo-Riemannian metric is 0, the pair ( M, h·, ·i) is called a Riemannian manifold. If the index is 1, a Lorentz manifold. • The causal type of a vector v ∈ Tp M is defined through the sign of hv, vi p and the causal type of a curve is defined by the causal type of its velocity vectors, like always. Definition 4.4.3. Let ( M, h·, ·i) be a pseudo-Riemannian manifold. The energy functional and the arclength functional associated to h·, ·i are the functionals E and L defined by Z Z r . 1 . 0 0 0 E[α] = hα (t), α (t)iα(t) dt and L[α] = hα (t), α0 (t)iα(t) dt, 2 I I for all smooth curves α : I → M. The critical points of E are called geodesics. Remark.

• If α : I → M is timelike, we denote its proper time by Z q t[ α ] = −hα0 (t), α0 (t)iα(t) dt. I

• The variational characterizations given in Section 3.6 of Chapter 3 still hold in this context. One can prove that a constant-speed curve is a critical point of one functional if and only if it is a critical point of the other one. With these definitions in hand, we are ready to start the discussion. A fundamental concept in geometry is that of distance between two points. When ( M, h·, ·i) is a Riemannian manifold, for each p ∈ M we have that h·, ·i p is a positivedefinite inner product. Since we are assuming that M is connected, we have that if . Ω( p, q) = {α : [ a, b] → M | α is piecewise smooth, α( a) = p and α(b) = q}, then d : M × M → R defined by d( p, q) = inf{ L[α] | α ∈ Ω( p, q)} is a distance function on M, that is, it satisfies the following properties:

316  Introduction to Lorentz Geometry: Curves and Surfaces

(i) d( p, q) ≥ 0, and d( p, q) = 0 if and only if p = q; (ii) d( p, q) = d(q, p) and (iii) d( p, r ) ≤ d( p, q) + d(q, p). In this context, d is called the intrinsic distance on M (see a concrete example of this in Exercise 4.4.1). Of the above properties, the only one whose verification poses some difficulty is the first one, when it should be proven that if p 6= q, then d( p, q) > 0. Then, the pair ( M, d) is a metric space. Bearing in mind the construction of the distance function d, it is reasonable to expect its properties to have a strong relation with the geometry and topology of M itself. This begins when we prove that d is, in fact, compatible with the original topology in M: every open ball according to d is an open set in M, and on the other hand, every open set in M is a union of open balls according to d. The proof of these facts is made easier by the notion of exponential map (valid in any pseudo-Riemannian manifold): it can be shown that given p ∈ M, there is an open set D p ⊆ Tp M on which the map exp p : D p → M given by exp p (v) = α(1), where α is the unique (maximal) geodesic such that α(0) = p and α0 (0) = v, is well-defined. That is, the exponential map at a point collects all the geodesics starting at p. It would be natural to wonder, now, how big can this domain D p be. This leads us to the following general definition: Definition 4.4.4. Let ( M, h·, ·i) be a pseudo-Riemannian manifold. We will call a geodesic α : I → M (with I maximal) complete if I = R. And we will say that M is geodesically complete if every geodesic in M is complete. Besides the questioning about the size of D p , we now have two notions of completeness for M (metric completeness and geodesic completeness), and it is natural to wonder if there is any relation between them. The following theorem answers all of our questions: Theorem 4.4.5 (Hopf-Rinow). Let ( M, h·, ·i) be a Riemannian manifold and p0 ∈ M. The following are equivalent: (i) M is geodesically complete; (ii) ( M, d) is a complete metric space; (iii) exp p0 is defined on all of Tp0 M; (iv) exp p is defined on all of Tp M, for all p ∈ M; (v) the closed and bounded subsets of M are compact; (vi) there is a sequence of compact subsets (Kn )n≥0 of M such that Kn ⊆ int(Kn+1 ) S for all n ≥ 0, M = n≥0 Kn , and (Kn )n≥0 has the following property: for every sequence (qn )n≥0 of points in M such that qn 6∈ Kn for all n ≥ 0, we have d( p, qn ) → +∞ for all p ∈ M. Besides, each one of these properties implies that (vii) given p, q ∈ M, there is a smooth geodesic α ∈ Ω( p, q) such that L[α] = d( p, q). For a proof, we suggest [16]. We have two important consequences:

Abstract Surfaces and Further Topics  317

Corollary 4.4.6. Let ( M, h·, ·i) be a compact Riemannian manifold. Then M is geodesically complete. Definition 4.4.7. Let ( M, h·, ·i) be a Riemannian manifold. A (piecewise) smooth curve α : R≥0 → M is called divergent if for every K ⊆ M compact there is t0 > 0 such that α(t) 6∈ K for all t > t0 . Remark. That is, α escapes from every compact set in M, without returning (so that its trace is, in a certain way, “divergent”). Corollary 4.4.8. Let ( M, h·, ·i) be a Riemannian manifold. Then M is complete if and only if the arclength of every divergent curve in M is infinite. h i Remark. The arclength of a divergent curve α : R≥0 → M is L[α] = lim L α [0,b] . b→+∞

In general, this theorem allows us to use tools from analysis and topology to study the geometry of a Riemannian manifold ( M, h·, ·i). We would like very much to have an analogous result for pseudo-Riemannian manifolds. This is, in general, not possible: everything we have discussed so far fails, since we do not have a distance function anymore. Another difficulty that appears is that, for pseudo-Riemannian manifolds, we don’t have only one notion of geodesic completeness, but one for each causal type, and it is not possible in general to control all three of them simultaneously. Surprisingly, for certain Lorentz manifolds there is an analogous result, and the adequate tools for healing this deficiency in this case are the notions of chronological precedence () and causal precedence (4), discussed briefly in Chapter 1. Let us see a few ideas: Definition 4.4.9 (Spacetime). Let ( M, h·, ·i) be a Lorentz manifold. We will say that M is time-orientable if there is a timelike vector field V defined globally on all of M (that is, hV ( p), V ( p)i p < 0 for all p ∈ M). When V is fixed, we will say that M is time-oriented by V . A spacetime is a Lorentz manifold of dimension greater than or equal to 2, which is time-oriented. In this case, the points of M are called events. To know if it is possible to time-orient a given manifold, we appeal to a result from algebraic topology (see [34] for a proof): Theorem 4.4.10. Let M be a smooth manifold. The following are equivalent: (i) There is a Lorentz metric on M. (ii) There is a time-orientable Lorentz metric on M. (iii) There is a non-vanishing vector field defined on all of M. (iv) M is non-compact, or its Euler-Poincaré characteristic χ( M) is zero. In particular, it follows that connected timelike surfaces in L3 are always timeorientable, perhaps seeing M as an abstract surface and changing its metric. The importance of all of this resides in the fact that a time orientation allows us to tell the direction of time in M, according to the: Definition 4.4.11. Let ( M, h·, ·i) be a spacetime oriented by a timelike vector field V . A timelike or lightlike vector v ∈ Tp M is called future-directed if hv, V ( p)i p ≤ 0, and past-directed if hv, V ( p)i p ≥ 0.

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Remark.

• In Section 1.3, the standard time orientation for Ln is given by the field that associates to each point in Ln the vector en . • Note that in the above definition, if v is timelike, the inequalities must be strict (why?). • As in Chapter 2, a timelike or lightlike curve is future-directed or past-directed according to whether one (hence all) of its velocity vectors is. It is time to formalize Definition 1.3.3 (p. 20), given in Chapter 1. If α : I → M is a piecewise smooth timelike curve, and t0 ∈ I, we will say that the causal type of α in t0 is the causal type of the one-sided derivatives of α in t0 , when they coincide. This way, let . Ωc ( p, q) = {α ∈ Ω( p, q) | α is timelike or lightlike, and future-directed}. We have: Definition 4.4.12 (Future and Past). Let ( M, h·, ·i) be a spacetime and p ∈ M. We define the chronological future and the causal future of p, respectively, as . I + ( p) = {q ∈ M | there exists a timelike α ∈ Ωc ( p, q)} . J + ( p) = {q ∈ M | there exists α ∈ Ωc ( p, q)}. The definitions of chronological past and causal past, denoted respectively by I − ( p) and J − ( p), are dual to the given above, with “past-directed” instead of “future-directed”. If S ⊆ M is any subset, the chronological future of S is defined by . [ + I + (S) = I ( p ), p∈S

and similarly for the other notions of future and past (I − (S), J + (S) and J − (S)). Remark. For some basic properties of I + , see Exercise 4.4.7. This allows us to define  and 4 in M: Definition 4.4.13 (Causality). Let ( M, h·, ·i) be a spacetime, and p, q ∈ M. We will say that p chronologically (resp. causally) precedes q if q ∈ I + ( p) (resp. q ∈ J + ( p) or q = p). Such relations are denoted by p  q and p 4 q. The futures and pasts of points in M end up being closely related to the topology in M. It is possible to prove, using the exponential map, the following result: Proposition 4.4.14. If ( M, h·, ·i) is a spacetime and we’re given p, q ∈ M such that p  q, then there are neighborhoods U and V of p and q in M such that p0  q0 , for all p0 ∈ U and q0 ∈ V. In other words, the binary relation  is open in M × M. In particular, for all p ∈ M, I + ( p) and I − ( p) are open sets in M. Assuming this result, we give an example of how the topology in M can influence its causality: Proposition 4.4.15. Let ( M, h·, ·i) be a compact spacetime. Then there is a (piecewise smooth) closed timelike curve in M.

Abstract Surfaces and Further Topics  319

Proof: Consider the open cover ( I + ( p)) p∈ M of M, consisting of the chronological futures of all points in M. By compactness of M, there are events p1 , . . . , pk ∈ M such that S M = ik=1 I + ( pi ). If 1 ≤ i ≤ k is such that there is j distinct from i with pi ∈ I + ( p j ), we would have I + ( pi ) ⊆ I + ( p j ), and so we could remove I + ( pi ) from the open cover. With this in mind, we can assume that pi 6∈ I + ( p j ), for all distinct 1 ≤ i, j ≤ k. So, p1 ∈ M S and p1 6∈ ik=2 I + ( pi ) means that p1 ∈ I + ( p1 ), and the definition of chronological future gives us the desired curve. That is, in such a spacetime, causality breaks, since every point in the closed timelike curve is simultaneously in its own future and past. In other words, in this model, some observers could travel back in time. This indicates that this notion of “spacetime” is too broad. Bearing this in mind, we can impose several conditions on the causality of the spacetime to avoid pathological situations such as the one given by the above proposition. Let us register them: Definition 4.4.16 (Causal hierarchy). Let ( M, h·, ·i) be a spacetime. We will say that M is: (i) chronological, if there is no closed timelike curve in M; (ii) causal, if there is no closed timelike and no lightlike curve in M; (iii) distinguishing, if given p, q ∈ M with I + ( p) = I + (q) or I − ( p) = I − (q), then p = q; (iv) strongly causal, if the collection of chronological diamonds . B♦, = { I + ( p) ∩ I − (q) | p, q ∈ M } form a basis for the original topology in M; (v) stably causal, if M admits a time function, that is, a continuous real-valued function on M that is strictly increasing along every timelike or lightlike future-directed curve; (vi) causally simple if it is causal, and the sets J + ( p) and J − ( p) are closed in M, for all p ∈ M; (vii) globally hyperbolic, if it is causal, and each causal diamond J + ( p) ∩ J − (q) is compact, for all p, q ∈ M. Remark.

• This set of definitions receives the name of causal hierarchy (or causal ladder) because the following implications hold: (vii ) =⇒ (vi ) =⇒ (v) =⇒ (iv) =⇒ (iii ) =⇒ (ii ) =⇒ (i ). However, all of the converses are false. We recommend [7] and [29] for more details.

• The topology generated by B♦, is called the Alexandrov topology in M. In other words, a spacetime is strongly causal if the Alexandrov topology coincides with the original one.

320  Introduction to Lorentz Geometry: Curves and Surfaces

• With this terminology, Proposition 4.4.15 above tells us that no compact spacetime is chronological. Despite that, there are still interesting examples of compact spacetimes, such as Gödel’s spacetime: the first model of the universe where time travel is possible. This space has closed timelike curves, but no closed timelike geodesics. It’s metric is a solution of the famous Einstein’s field equations, and it was given to Einstein by Gödel as a birthday present in 1949. For more details, see [29]. These notions will allow us to state a result analogous to the Hopf-Rinow theorem for timelike surfaces. We start adapting the concept of distance: Definition 4.4.17. Let ( M, h·, ·i) be a spacetime. The time separation (or Lorentz distance) in M is the map t : M × M → [0, +∞] defined by . t( p, q) = sup{t[α] | α ∈ Ωc ( p, q)} if Ωc ( p, q) 6= ∅, and t( p, q) = 0 otherwise. The wrong-way triangle inequality gives us some properties of t analogous to some of the intrinsic distance d: Proposition 4.4.18. Let ( M, h·, ·i) be a spacetime, and p, q, r ∈ M. We have: (i) t( p, q) > 0 if and only if p ∈ I − (q), if and only if q ∈ I + ( p). (ii) t( p, p) = +∞ if there is a timelike α ∈ Ωc ( p, p), and it is zero otherwise. (iii) If 0 ≤ t( p, q) < +∞, then t(q, p) = 0. In particular, t is not, in general, symmetric. (iv) If p 4 q and q 4 r, then t( p, q) + t(q, r ) ≤ t( p, r ). Remark.

• There are spacetimes for which the time separation is constant and equal to +∞. Such spaces are said to be totally vicious, and their Alexandrov topology is the chaotic one. • The inequality in (iv) above still holds if p = q, q = r or p = r. Another important property is given by the: Proposition 4.4.19. Let ( M, h·, ·i) be a spacetime. Then t is lower semi-continuous, that is, if p, q ∈ M and ( pn )n≥0 and (qn )n≥0 are two sequences in M such that pn → p and qn → q, then lim inf t( pn , qn ) ≥ t( p, q). n→+∞

In general, the time separation is not well-behaved, perhaps being discontinuous and assuming the value +∞. However, t has better properties according to how far we go in the causal ladder. This climb leads to the: Theorem 4.4.20 (Avez-Seifert). Let ( M, h·, ·i) be a globally hyperbolic spacetime. Then: (i) if p, q ∈ M are such that p 4 q, then there is α ∈ Ωc ( p, q) such that t[α] = t( p, q); (ii) the time separation t is finite and continuous. For more technical details and deeper results about the topics discussed here, we refer the reader to [16], [54], [41] (for Riemannian Geometry) and [7], [29], [58], [34], [51] and [54] (for the interplay between Differential Geometry and General Relativity).

Abstract Surfaces and Further Topics  321

Exercises Exercise† 4.4.1 (Introduction to Hyperbolic Geometry). We have seen that the hyperbolic plane H2 ⊆ L3 may be used as a model for hyperbolic geometry. Let’s see a few basic facts about the lines in such geometry: (a) Let p, q ∈ H2 . Show that there is a unique geodesic in H2 joining p to q. Hint. Every geodesic in H2 is related to some spacelike direction in L3 . (b) Verify that the Parallel Postulate does not hold here: if α is a geodesic in H2 and p ∈ H2 is not in the trace of α, there are infinitely many geodesics containing p which do not intersect α. Hint. Every geodesic is determined by a point and a tangent vector. (c) Let p, q ∈ H2 . Show that the length of the geodesic in H2 joining p to q is cosh−1 (−h p, qi L ). Hint. Parametrize the geodesic by α(s) = (cosh s)v + (sinh s)w, with v and w unit and orthogonal, v timelike, and w spacelike. Compute hα(s0 ), α(s1 )i L , for any s0 , s1 ∈ R. (d) Define1 d : H2 × H2 → R by d( p, q) = cosh−1 (−h p, qi L ). Show that d turns H2 into a metric space, that is, given any p, q, r ∈ H2 , the following properties hold: (i) d( p, q) ≥ 0, and d( p, q) = 0 ⇐⇒ p = q. (ii) d( p, q) = d(q, p). (iii) d( p, q) ≤ d( p, r ) + d(r, q). (e) A hyperbolic isometry is a function Λ0 : H2 → H2 such that d(Λ0 ( p), Λ0 (q)) = d( p, q), for any p, q ∈ H2 . Prove that if Λ0 is a hyperbolic isometry, then Λ0 = Λ H2 for some orthochronous Lorentz transformation Λ in L3 . Remark. For more details about this model, see [62]. Exercise† 4.4.2. Show Corollary 4.4.8 (p. 317). Exercise 4.4.3. Consider in M = R × R>0 the Riemannian metric ds2 = dx2 +

dy2 . y

Show that the vertical segment {0} × ]0, 1[ is divergent and that its length equals 2. Conclude that this metric is not complete. 1 Recall

that if p and q are timelike vectors in the same timecone, there is a unique positive ϕ such that h p, qi L = −k pk L kqk L cosh ϕ.

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Exercise 4.4.4. Let ( M, h·, ·i) be a connected Riemannian manifold. Show that any two points may be joined by a geodesic polygonal, that is, a piecewise smooth curve for which each smooth arc is a geodesic. Hint. Note that if p ∈ M, each point in the image exp p ( D p ) may be joined to p by a geodesic there contained. Use this to show that a certain subset of M is both open and closed. Exercise 4.4.5. Let ( M, h·, ·i) be a Riemannian manifold. Suppose that there is a function f : M → R which is both:

• Lipschitz-continuous: there is C > 0 such that | f ( p) − f (q)| < Cd( p, q), for all p, q ∈ M, and • proper: for all compact K ⊆ R, f −1 (K ) ⊆ M is compact. Show that ( M, h·, ·i) is geodesically complete. Exercise 4.4.6. Show that the relation  is transitive. Remark. We have already seen this for 4 in Exercise 1.3.4 (p. 28, Chapter 1), when M = Ln , but it is also true in this more general setting. Exercise† 4.4.7. Let ( M, h·, ·i) be a spacetime. (a) Show that p ∈ I + (q) if and only if q ∈ I − ( p). (b) Show that for all S ⊆ M, we have I + (S) = I + (S). (c) Show that if (Si )i∈ I is any collection of subsets of M, then I + (

S

(d) Show that if (Si )i∈ I is any collection of subsets of M, then I + ( and look for an example showing that this inclusion is strict.

T

i∈ I

Si ) =

S

i∈ I

Si ) ⊆

T

i∈ I

I + ( Si ) .

i∈ I

I + ( Si ) ,

(e) Investigate if the relations given in items (c) and (d) also hold for causal futures instead of chronological futures. Exercise 4.4.8. Consider the punctured plane M = L2 \ {(1, 1)} with the usual flat metric given by ds2 = dx2 − dy2 . Show that I + (0) 6= J + (0) and that J + (0) is not closed. Deleting even a single event from a spacetime may have disastrous consequences for its causality. Exercise 4.4.9. Let ( M, h·, ·i) be a spacetime and F ⊆ M be any subset. We’ll say that F is a future-set if I + ( F ) ⊆ F. (a) Show that for every S ⊆ M, I + (S) is a future-set (surprised?). (b) Suppose that F ⊆ M is open. Show that F is a future-set if and only if F = I + (S) for some S ⊆ M. Give a counter-example when F is not open. Hint. If F is open, every x ∈ F has an open neighborhood W contained in F such that every geodesic starting at x is contained in W, for small instants of time. In particular, past-directed timelike geodesics. Exercise 4.4.10. Let ( M, h·, ·i) be a spacetime and A ⊆ M be any subset. We’ll say that A is achronal if any two events in A are not chronologically related or, equivalently, if I + ( A) ∩ A = ∅. Show that if F ⊆ M is a future-set, then its boundary ∂F is achronal.

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Hint. Show the contrapositive statement by using the definition of ∂F twice, in a “good order”:

M\F

F ∂F

Figure 4.22: The indicated points separated by ∂F show that I + ( F ) 6⊆ F. Exercise 4.4.11. Let ( M, h·, ·i) be a spacetime. Show that the chronological diamonds { I + ( p) ∩ I − (q) | p, q ∈ M} indeed form a basis for a topology in M. Hint. Since chronological futures and pasts are open in M, every event x in the intersection of two chronological diamonds has a neighborhood W still contained in such neighborhood and such that every event in W may be reached by a geodesic starting at x, completely inside W. Consider timelike geodesics, one going to the future and another one going to the past. Exercise 4.4.12. Let ( M1 , h·, ·i1 ) and ( M2 , h·, ·i2 ) be two spacetimes. A bijection φ : M1 → M2 is called a chronological isomorphism if given p, q ∈ M1 , it holds that p  q if and only if φ( p)  φ(q). Show that φ is a homeomorphism, when M1 and M2 are equipped with their Alexandrov topologies. Hint. Check that φ−1 ( I + (φ( p)) ∩ I − (φ(q))) = I + ( p) ∩ I − (q). Exercise 4.4.13. Suppose that ( M, h·, ·i) is a Riemannian manifold, and that there is a non-vanishing smooth vector field V on all of M. Show that hh·, ·ii defined by

hv, V ( p)i p hw, V ( p)i p . hhv, wii p = hv, wi p − 2 hV ( p), V ( p)i p is a Lorentz metric for which V ( p) is timelike, for each p ∈ M. Remark. Applying this construction to R3 equipped with the usual metric and taking the field V ( x, y, z) = (0, 0, 1) produces L3 , as hV , ·i = dz. Exercise 4.4.14. Give a counter-example for the Avez-Seifert Theorem when the spacetime is not globally hyperbolic. Exercise 4.4.15 (Zeeman Topology). The usual topology in Lorentz-Minkowski space Ln does not reflect in an adequate way its physical aspects, as it does not make any reference to causality. In this exercise we will see a more adequate topology that heals this deficiency. We’ll say that a subset Z ⊆ Ln is Zeeman-open if for every affine spacelike hyperplane or timelike line Σ in Ln , we have that Z ∩ Σ is open in Σ. (a) Show that the collection of Zeeman-open sets indeed forms a topology in Ln , called the Zeeman topology if Ln .

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(b) Show that every open set is Zeeman-open, but that the converse is not in general true. In particular, the Zeeman topology is Hausdorff. Hint. Given p ∈ Ln and r > 0, consider the natural candidate to “open ball”: . Zr ( p) = ( Br ( p) \ CL ( p)) ∪ { p}.

p

Figure 4.23: Illustrating Zr ( p). (c) Show that the Zeeman topology induced in lightlike rays is the discrete topology. In particular, conclude that every sequence of points in Ln which converges along a lightlike ray is eventually constant. This is motivated by the experimental evidence that photons may only be observed during discrete events of emission and absorption. (d) Show that ( Zr ( p))r>0 is not a local basis at p of Zeeman-open sets. Remark.

• Besides those basic facts, it is possible to show that the Zeeman topology is connected and locally connected, while it is not normal, compact, or locally compact. Non-normality, for instance, is an application of the Baire Category Theorem. In particular, the Zeeman topology is not metrizable. See [51] and also Zeeman’s original paper, [72]. • The Alexandrov-Zeeman Theorem stated in Chapter 1 (Theorem 1.3.16, p. 26) may also be stated in this setting: all the homeomorphisms of Ln with the Zeeman topology are compositions of translations, positive homotheties and orthochronous Lorentz transformations.

APPENDIX

Some Results from Differential Calculus

In this appendix, for completeness purposes, we will list a few results from Differential Calculus, convenient for a better understanding of the text. For proof of the results and more details, we recommend, for example, [43], [50] and [66]. Here, A will always denote an open subset of Rn , cann = (e1 , . . . , en ) the standard basis of Rn , and Lin(Rn , Rk ) the set of linear maps from Rn to Rk . Definition A.1. Let f : A ⊆ Rn → Rk and p ∈ A. The function f is said to be differentiable at p if there is T ∈ Lin(Rn , Rk ) such that lim

h→0

f ( p + h) − f ( p) − Th = 0. khk

If such transformation T exists, it is unique. In this case, T is called the total derivative of f at p, and it is denoted by D f ( p). We will just say that f is differentiable, if it is differentiable at all points. And if f is bijective and differentiable, with differentiable inverse, f is called a diffeomorphism. Remark. It immediately follows from the definition that D f ( p)(ei ) =

∂f ( p ). ∂xi

As particular cases of this, if I denotes an open interval in the real line, we have that:

• if α : I ⊆ R → Rn is differentiable, then Dα(t)(1) = α0 (t); • if f : I → R is differentiable, D f ( x ) is a linear map from R to R, and so it is a scalar multiple of the identity. This scalar is precisely the number f 0 ( x ), seen in a first Calculus course; • if f : A ⊆ Rn → R is differentiable, D f ( p) is a linear map from Rn to R and D f ( p)(v) = h∇ f ( p), vi, where   ∂f ∂f · ( p ), . . . , ( p) ∇ f ( p) = ∂x1 ∂xn is the gradient vector of f at p, seen in a second Calculus course.

325

326  Introduction to Lorentz Geometry: Curves and Surfaces

Definition A.2. Let f : A ⊆ Rn → Rk be differentiable at p ∈ A. The matrix ! ∂ fi · ( p) J f ( p) = [ D f ( p)]cann ,cank = ∂x j 1≤ i ≤ k 1≤ j ≤ n

is called the Jacobian matrix of f at p, where f = ( f 1 , . . . , f k ). Remark. In general, we’ll identify J f ( p) with D f ( p). In the case n = k = 1, we have J f ( p) = ( f 0 ( p)). Theorem A.3. If f : A ⊆ Rn → Rk is differentiable at a point p ∈ A, then f is continuous at p. Proposition A.4. Let f : A ⊆ Rn → Rk and p ∈ A. Then, writing f = ( f 1 , . . . , f k ), it holds that f is differentiable at p if and only if each component function f i is. In this case, we have that D f ( p) = ( D f 1 ( p), . . . , D f k ( p)) or, in matrix terms,   ∇ f 1 ( p)   J f ( p) =  ...  . ∇ f k ( p) Theorem A.5 (Chain Rule). Let f : A1 ⊆ Rn → Rk and g : A2 ⊆ Rk → R p be two functions, differentiable at p ∈ A1 and at f ( p) ∈ A2 , respectively, where A1 and A2 are open subsets with A2 ⊆ f ( A1 ), then g ◦ f is differentiable at p and, moreover, the formula D ( g ◦ f )( p) = Dg( f ( p)) ◦ D f ( p) holds. Example A.6. (1) If f : Rn → Rk is constant, then f is differentiable and D f ( p) is the zero map, for every p ∈ Rn . (2) Translations. Given a ∈ Rn , Ta : Rn → Rn given by Ta ( x) = x + a is differentiable and DTa ( p) = idRn , for every p ∈ Rn . (3) If T : Rn → Rk is linear, then T is differentiable and DT ( p) = T for every p ∈ Rn . In particular, the sum s : Rn × Rn → Rn is linear when seen as a map s : R2n → Rn , and thus it is differentiable. (4) If f , g : A ⊆ Rn → Rk are differentiable at p ∈ A, then f + g is also differentiable, and D ( f + g)( p) = D f ( p) + Dg( p). To wit, f + g = s ◦ ( f , g) is the composition of differentiable maps. (5) If B : Rn × Rk → R p is bilinear, then B is differentiable and DB( x, y)(h, k) = B( x, k) + B(h, y). In particular, the multiplication m : R × R → R is bilinear, due to the distributive property in R.

Some Results from Differential Calculus  327

(6) If f , g : A ⊆ Rn → R are differentiable at p ∈ A, then f g is also differentiable, and D ( f g)( p) = g( p) D f ( p) + f ( p) Dg( p). Indeed, f g = m ◦ ( f , g) is the composition of differentiable maps. In particular, we have that if λ ∈ R, then D (λ f )( p) = λD f ( p). Observe that if one of the functions takes values in Rk , the result holds for each component function, and so the product is also differentiable. (7) If f : A ⊆ Rn → R is differentiable at p ∈ A and f ( p) 6= 0, then 1/ f is differentiable at p and we have that   1 1 D ( p) = − D f ( p ). f f ( p )2 If f : A ⊆ Rn → Rk is differentiable, then D f : A → Lin(Rn , Rk ), and eventually identifying Lin(Rn , Rk ) with Rnk , it makes sense to wonder whether D f itself is continuous or differentiable. The latter being the case, we write ·

D2 f ( p) = D ( D f )( p) : Rn → Lin(Rn , Rk ). If Linr (Rn , Rk ) denotes the space of r-linear maps defined in (Rn )r taking values in Rk , we identify Lin2 (Rn , Rk ) ∼ = Lin(Rn , Lin(Rn , Rk )) and see D2 f ( p) as a bilinear map. A similar reasoning applies for derivaties of higher order, which are then identified with multilinear maps. For k = 1, this is used to study maxima and minima of functions. Definition A.7. We’ll say that a differentiable function f : A ⊆ Rn → Rk is of class C1 if D f : Rn → Lin(Rn , Rk ) is continuous. In general, for r > 1, we’ll say that f is of class Cr if D f is of class Cr−1 . Theorem A.8 (Clairaut-Schwarz). Let f : A ⊆ Rn → R be a function of class C2 . Then ∂2 f ∂2 f = . ∂xi ∂x j ∂x j ∂xi Remark. In the above result, it actually suffices for one of the mixed partial derivatives to be continuous and the other to exist, so they commute. Theorem A.9. Let f : A ⊆ Rn → R be a differentiable function. If p ∈ A is a local maximum or minimum of f , then D f ( p) = 0. Theorem A.10. Let f : A ⊆ Rn → R be a function of class C2 . If p ∈ A is a critical point of f , that is, with D f ( p) = 0, then we have that: (i) if D2 f ( p) is positive-definite, p is a local minimum for f ; (ii) if D2 f ( p) is negative-definite, p is a local maximum for f . Remark. If f is of class C2 , each D2 f ( p) is a symmetric bilinear form, so that results such as Sylvester’s Criterion may be used to decide whether D2 f ( p) is positive-definite or not. The above result is a consequence of the following one:

328  Introduction to Lorentz Geometry: Curves and Surfaces

Theorem A.11 (Taylor Formula). Let f : A ⊆ Rn → Rm be a function r times differentiable, and p ∈ A such that D (r+1) f ( p) exists. So, if h ∈ Rn is such that p + h ∈ A, we have that r 1 f ( p + h) = ∑ D (k) f ( p)(h(k) ) + r (h), k! k =0 where the remainder satisfies r (h)/k hkr+1 → 0 if h → 0 and, for each k, h(k) denotes the vector (h, . . . , h) ∈ (Rn )k . Theorem A.12 (Inverse Function Theorem). Let f : A ⊆ Rn → Rn be of class Ck , with k ≥ 1. If p ∈ A is such that D f ( p) is non-singular, there are open subsets U ⊆ A and V ⊆ Rn containing p and f ( p), respectively, such that the restriction f : U → V is bijective, and the inverse f −1 : V → U is also of class Ck . In this case, we have that D ( f −1 )( f ( x)) = D f ( x)−1 for each x ∈ U. Every partition of the standard basis in Rm determines a decomposition of the form Rn1 × · · · × Rnr , with m = n1 + · · · + nr , and each such decomposition induces “fat partial derivatives”. For example, if f : Rn × Rk → R p is differentiable at a point ( x0 , y0 ), we have two linear maps Dx f ( x0 , y0 ) : Rn → R p and Dy f ( x0 , y0 ) : Rk → R p given by ·

Dx f ( x0 , y0 )(h) = D f ( x0 , y0 )(h, 0) and

·

Dy f ( x0 , y0 )(k) = D f ( x0 , y0 )(0, k).

Note that D f ( x0 , y0 )(h, k) = Dx f ( x0 , y0 )(h) + Dy f ( x0 , y0 )(k). With this, we may state the: Theorem A.13 (Implicit Function Theorem). Consider a function f : A × B ⊆ Rn × Rk → Rk of class Cr , with r ≥ 1. Suppose that ( x0 , y0 ) ∈ A × B is such that f ( x0 , y0 ) = c. If Dy f ( x0 , y0 ) is non-singular, there are open subsets U ⊆ A and V ⊆ B containing x0 and y0 , and a function ϕ : U → V of class Cr such that f ( x, ϕ( x)) = c and f ( x, y) = c implies y = ϕ( x), for each x ∈ U. Remark. The equation f ( x, ϕ( x)) = c allows us to effectively compute the derivative of the implicit function ϕ. We have that Dx f ( x, ϕ( x)) + Dy f ( x, ϕ( x)) ◦ Dϕ( x) = 0, and so Dϕ( x) = − Dy f ( x, ϕ( x))−1 ◦ Dx f ( x, ϕ( x)).

Some Results from Differential Calculus  329

Exercises Exercise A.1. Consider f : R2 → R2 given by f ( x, y) = (ex cos y, ex sin y), and the · linear map T = D f (3, π/6). Find the angle between T 2017 (1, 0) and T 2018 (1, 1). Exercise A.2. Compute the total derivative of F in the following cases: (a) F ( x, y) = ( x, f (y)), where f is differentiable; (b) F ( x) = h x, x0 i, where x0 is fixed; (c) F ( x) = h x, Txi, where T is a fixed linear operator; (d) F ( x, y) = f ( x) + g(y), where f and g are differentiable; (e) F ( x, y) = A( x)y, where A : Rn → Lin(Rn , Rk ) is differentiable and y ∈ Rk . Exercise A.3. Let f : A ⊆ Rn → Rk be a differentiable function and define two maps ϕ : A → Rn × Rk and F : A × Rk → Rk , respectively, by ϕ( x) = ( x, f ( x)) and F ( x, y) = y − f ( x). Show that ϕ and F are both differentiable, and that for all x0 ∈ A and y0 ∈ Rk we have that ker DF ( x0 , y0 ) = Im Dϕ( x0 ). Exercise A.4. Let B : Rn × Rk → R p be a bilinear map. Show that there is C > 0 such that k B( x, y)k ≤ C k xkkyk for all x ∈ Rn and y ∈ Rk . Use this to show that B is differentiable at all points, and the formula DB( x, y)(h, k) = B( x, k) + B(h, y) holds. State analogous results for multilinear maps T : Rn1 × · · · × Rnk → R p . Exercise A.5. Show that Φ : Lin2 (Rn , Rk ) → Lin(Rn , Lin(Rn , Rk )) given by Φ( B)( x)(y) = B( x, y) is an isomorphism of vector spaces. Exercise A.6. Let p0 ( x ) = a0 x3 − b0 x2 + c0 x − d0 , with a0 6= 0, be a polynomial with real coefficients and three distinct real roots. Show that every polynomial of the form p( x ) = ax3 − bx2 + cx − d with coefficients ( a, b, c, d) sufficiently close to ( a0 , b0 , c0 , d0 ) also has three distinct real roots, which depend smoothly on the coefficients of p( x ). Hint. Vieta’s Relations and Inverse Function Theorem. Exercise A.7. Let f : Rn → Rn be differentiable, with f (0) = 0. If D f (0) does not have 1 as an eigenvalue, then f ( x) 6= x for x sufficiently close to 0, but not equal to 0. Exercise A.8 (Local form of Immersions). We’ll say that a function f : A ⊆ Rn → Rn+k is an immersion if D f ( p) is injective for each p ∈ A. Show that if f is an immersion of class C1 , there are open subsets U × V and W of Rn+k containing ( p, 0) and f ( p), and a diffeomorphism ψ : W → U × V such that ψ ◦ f ( x) = ( x, 0), for each x ∈ U. Remark. That is, up to a diffeomorphism, every immersion is an inclusion.

Bibliography [1] Alexandrov, D., A contribution to chronogeometry, Canadian J. Math. 19, pp. 1119–1128, 1967. [2] Alias, L.; Chaves, R. M. B.; Mira, P.; Bjorling problem for maximal surfaces in Lorentz-Minkowski space, Math. Proc. Camb. Phil. Soc. (no. 134, pp. 289–316), 2003. [3] Anciaux, H., Minimal Submanifolds in Pseudo-Riemannian Geometry, World Scientific, 2011. [4] Antonuccio, F., Semi-Complex Analysis & Mathematical Physics (Corrected Version), eprint arXiv:gr-qc/9311032, 1993, https://arxiv.org/pdf/gr-qc/ 9311032.pdf. [5] Araújo, P. V., Geometria Diferencial, IMPA (Coleção Matemática Universitária), 2012. [6] Bär, C., Elementary Differential Geometry, Cambridge University Press, 2010. [7] Beem, J. K.; Ehrlich, P. E.; Easley, K. L., Global Lorentzian Geometry, CRC Press, 1996. [8] Calioli, C. A.; Domingues, H. H.; Costa, R. C. F., Álgebra Linear e Aplicações, Atual, 1995. [9] Cannon, J. W.; Floyd W. J.; Kenyon, R.; Parry, W. R., Hyperbolic Geometry, MSRI Publications, Flavors of Geometry 31, 1997. [10] Carroll, S., Spacetime and Geometry, An Introduction to General Relativity, Pearson Education, 2004. [11] Catoni et al., Geometry of Minkowski Spacetime, Springer-Verlag (Springer Briefs in Physics), 2011. [12] Chaves, R. M. B.; Dussan, M. P.; Magid, M.; Bjorling Problem for timelike surfaces in the Lorentz-Minkowski space, Journal of Mathematical Analysis and Applciations (377, no. 2, pp; 481–494), 2011. [13] Chen, B. Y.; Pseudo-Riemannian Geometry, δ-invariants and Applications, World Scientific, 2011. [14] Chern, S. S., Curves and Surfaces in Euclidean Spaces, Studies in Global Analysis, MAA Studies in Mathematics, The Mathematical Association of America, 1967. [15] Dieudonné, J., La Géométrie des Groupes Classiques, Springer-Verlag, 1971. [16] do Carmo, M. P., Geometria Riemanniana, IMPA (Coleção Projeto Euclides), 2005. 331

332  Bibliography

[17] do Carmo, M. P., Geometria Diferencial de Curvas e Superfícies, SBM (Coleção Textos Universitários), 2014. [18] Dussan, M. P.; Franco Filho, A. P.; Magid, M.; The Bjorling problem for timelike minimal surfaces in R41 , Annali di Matematica Pura ed Applicata (pp. 1–19), 2016. [19] Dussan, M. P.; Magid, M.; Bjorling problem for timelike surfaces in R42 , Journal of Geometry and Physics (no. 73, pp. 187–199), 2013. [20] Faber, R., Differential Geometry and Relativity Theory, CRC Press, 1983. [21] Fang, Y., Lectures on Minimal Surfaces in R3 , Proceedings of the Centre for Mathematics and Its Applications, Australian National University 35, 1996. [22] Foldes, S., Hyperboloid preservation implies the Lorentz and Poincaré groups without dilations, eprint arXiv:1009:3910, https://arxiv.org/pdf/1009.3910.pdf, 2010. [23] Fraleigh, J. B., A First Course in Abstract Algebra, Addison-Wesley, 1994. [24] Gelfand, I. M.; Fomin, S. V., Calculus of Variations, Prentice-Hall, 1963. [25] Goldman, R., Curvature formulas for implicit curves and surfaces, Computer Aided Geometric Design - Special issue: Geometric modelling and differential geometry 22, pp. 632–658, 2005. [26] Goldman, W. M.; Margulis, G. A., Flat Lorentz 3-manifolds and cocompact Fuchsian groups. Crystallographic groups and their generalizations (Kortrijk, 1999) Contemporary Mathematics 262, pp. 135–145, 2000. [27] Gray, A.; Abbena, E.; Salamon, S., Modern Differential Geometry of Curves and Surfaces with Mathematica, Chapman and Hall, 2016. [28] Hall, B. C., An Elementary Introduction to Groups and Representations, eprint arXiv:math-ph/0005032, https://arxiv.org/pdf/math-ph/0005032, 2000. [29] Hawking, S.; Ellis G., The Large Scale Structure of Spacetime, Cambridge Monographs on Mathematical Physics, 1973. [30] Hitzer, E., Non-constant bounded holomorphic functions of hyperbolic numbers — Candidates for hyperbolic activation functions, Proceedings of the First SICE Symposium on Computational Intelligence, pp. 23–28, 2011. [31] Hoffman, K.; Kunze, R., Linear Algebra, Prentice-Hall, 1971. [32] Horn, R. A.; Johnson, C. R., Matrix Analysis, Cambridge University Press, 2013. [33] Jaffe, A., Lorentz Transformations, Rotations and Boosts, Lecture Notes, 2013. [34] Javaloyes, M. A.; Sánchez, M., An Introduction to Lorentzian Geometry and its Applications, XVI Escola de Geometria Diferencial, São Paulo, 2010. [35] Khrennikov A.; Segre G., An Introduction to Hyperbolic Analysis, eprint arXiv:math-ph/0507053, https://arxiv.org/pdf/math-ph/0507053.pdf, 2005. [36] Kobayashi, O., Maximal Surfaces in the 3-Dimensional Minkowski Space L3 , Tokyo J. of Math. Volume 06, pp. 297–309, 1983.

Bibliography  333

[37] Konderak, J.; A Weierstrass Representation Theorem for Lorentz Surfaces, Complex Variables 50 (no. 5, pp. 319–332), 2005. [38] Kosheleva, O.; Kreinovich, V., Observable Causality Implies Lorentz Group: Alexandrov-Zeeman-Type Theorem for Space-Time Regions, Mathematical Structures and Modeling 30, pp. 4–14, 2014. [39] Kühnel, W., Differential Geometry: Curves – Surfaces – Manifolds, AMS, 2006. [40] Lang, S., Complex Analysis, 4th. ed., Springer-Verlag (Graduate Texts in Mathematics 103), 1999. [41] Lee, J. M., Riemannian Manifolds: An Introduction to Curvature, Springer-Verlag (Graduate Texts in Mathematics 176), 1997. [42] Lima, E. L., Grupo Fundamental e Espaços de Recobrimento, IMPA (Coleção Projeto Euclides), 2012. [43] Lima, E. L., Análise no Espaço Rn , IMPA (Coleção Matemática Universitária), 2013. [44] López, R., Differential Geometry of Curves and Surfaces in Lorentz-Minkowski space, eprint arXiv:0810.3351, https://arxiv.org/pdf/0810.3351, 2008. [45] Lyusternik, L. A.; Fet, A. I., Variational problems on closed manifolds, Dokl. Akad. Nauk. SSSR 81 (pp. 17–18), 1951. [46] Magid, M.; Minimal Timelike Surfaces via the Split-Complex Numbers, Proceedings of PADGE 2012, Shaker-Verlag, Aachen, 2013. [47] Matsuda, H., A note on an isometric imbedding of upper half-space into the anti de Sitter space, Hokkaido Math. J. 13, pp. 123–132, 1984. [48] Mian, J., Fermi-Walker holonomy, second-order vector bundles, and EPR correlations. Honours Thesis, National University of Singapore, 2015. [49] Müller, O.; Sánchez, M., Lorentzian Manifolds Isometrically Embeddable in L N , Trans. Amer. Math. Soc. 363, pp. 5367–5379, 2011. [50] Munkres, J. R., Analysis on Manifolds, Addison-Wesley, 1991. [51] Naber G. L., Spacetime and Singularities, An Introduction, Cambridge University Press, 1988. [52] Naber, G. L., The Geometry of Minkowski Spacetime: An Introduction to the Mathematics of the Special Theory of Relativity, Springer-Verlag (Applied Mathematical Sciences 92), 1992. [53] Nomizu, K., The Lorentz–Poincaré metric on upper half-space and its extension, Hokkaido Math. J. 11, pp. 253–261, 1982. [54] O’Neill, B., Semi–Riemannian Geometry with Applications to Relativity, Academic Press, 1983. [55] O’Neill, B., Elementary Differential Geometry, Academic Press, 2006. [56] Oprea, J., Differential Geometry and Its Applications, Prentice-Hall, 1997.

334  Bibliography

[57] Osserman, R., A Survey of Minimal Surfaces, Dover, 1986. [58] Penrose, R., Techniques of Differential Topology in Relativity, AMS Colloquium Publications (SIAM, Philadelphia), 1972. [59] Petersen, P., Classical Differential Geometry, eprint https: // www. math. ucla. edu/ ~petersen/ DGnotes. pdf , 2012. [60] Ratcliffe, J., Foundations of Hyperbolic Manifolds, Springer-Verlag (Graduate Texts in Mathematics 149), 1994. [61] Remmert, R., Theory of Complex Functions, Springer-Verlag (Readings in Mathematics 122), 1991. [62] Ryan, P. J., Euclidean and Non-Euclidean Geometry: An Analytic Approach, Cambridge University Press, 1986. [63] Saloom, A., Curves in the Minkowski plane and Lorentzian surface, Durham theses, Durham University, 2012. [64] Shilov, G. E., Linear Algebra, Dover, 1977. [65] Simmonds, J. G., A Brief on Tensor Analysis, Springer-Verlag (Undergraduate Texts in Mathematics), 1982. [66] Spivak, M., Calculus on Manifolds, Addison-Wesley, 1965. [67] Sternberg, S., Curvature in Mathematics and Physics, Dover Publications, Inc., 2012. [68] Stoker, J. J., Differential Geometry, John Wiley & Sons, Inc., 1969. [69] Tenenblat, K. Introdução à Geometria Diferencial, Edgard Blücher, 2008. [70] Van Brunt, B., The Calculus of Variations, Springer-Verlag (Universitext), 2004. [71] Walrave, J., Curves and Surfaces in Minkowski Space, Doctoral Thesis, K.U. Leuven, Fac. of Science, Leuven, 1995. [72] Zeeman, E. C.; The Topology of Minkowski Space, Topology (vol 6., pp. 161–170), Pergamon Press, 1967.

Index

Catenoid, 173, 301 spacelike in L3 , 309 timelike in L3 , 313 Cauchy-Riemann equations (revised), 177, 288 equations, 177, 279 Causal automorphism, 25 diamond, 319 future, 20, 318 hierarchy of a spacetime, 319 type, 3, 262, 315 Cayley transform, 266 Center of curvature, 84 Central projection, 153 Chain rule, 146, 288, 326 Change of parameters, see also reparametrization Christoffel Symbols, 228 Christoffel symbols, 267 Chronological diamond, 319 future, 20, 264, 318 isomorphism, 323 Clairaut parametrization, 240 relation, 242 Clifford torus, 270 Codazzi-Mainardi equations, 251 Coindicator (of a curve), 98 Compatibility equations, see also Codazzi-Mainardi equations Complex derivative, 304 Lorentzian space, 304 Cone light, 4, 20 time, 20 Conformal map (between surfaces), 177 Congruence of surfaces, 170 Coordinate curves, 132

Achronal set, 322 Admissible variation (of a parametrized surface), 200 Alexandrov topology, 319 Anti-de Sitter space, 278 Antipodal map, 155 Arc-photon, 71 Archimedes Spiral, 95 Arclength, 67, 160, 263, 315 Area functional, 200 of a geometric surface, 263 of a surface, 161 Artin space, 280 Asymptotic boundary, 275 line, 211 parametrization, 212 vector, 207 Atlas, 260 Bernoulli equation, 203 Bilinear form indefinite, 5 negative-definite, 5 non-degenerate, 5 positive-definite, 5 symmetric, 5 trace, 57 Binormal indicatrix, 104 vector, 98, 118 Bonnet rotation, 299 Brioschi formula, 285 B-scroll (associated to a lightlike curve), 193 Cardioid, 93 Cartan curvature, see also Pseudo-torsion frame, 119 Catalan surface in R3 , 309 335

336  INDEX

Cotangent plane, 160 Covariant derivative, 216, 219 Critical curve, 306 surface, 201 value, 135 Cross product, 53 Curvature (of a space curve), 98 Gaussian, 181, 269 Mean, 181 Mean (vector), 181 principal, 179 Curve admissible, 98 biregular, 98 closed, 67 congruence, 72 divergent, 317 energy of a, 235 lightlike, 64 Logarithmic spiral, 74 Loxodromic, 74 parametrized, 64 regular, 66 semi-lightlike, 118 spacelike, 64 timelike, 64 d’Alembertian, 248, 298 Darboux vector, 104 Darboux-Ribaucour frame, 222 de Sitter space, 80, 108 Diffeomorphism between open sets in Rn , 325 between surfaces, 142 group (of a surface), 152 Differentiable manifold, 260 Differential, 145, 154 Dirichlet energy, 248 Distance function, 315 Dual numbers, 296 Dupin Indicatrix, 218 Elliptic helix, 110 linear map, 44 point, 197 Energy kinetic, 76, 234

of a curve, 67, 160, 263 potential, 76 Energy of a curve, 315 Energy-momentum map, 42 Enneper surface spacelike in L3 , 309 in R3 , 308 timelike in L3 , 312 Enneper-Weierstrass representation formula, 307, 308, 312 Euler formulas, 208 Euler-Lagrange Equations, 232 Evolute of a curve, 85 Fermi chart, 281 Fermi-Walker Parallelism, 245 First Fundamental Form, 158 First variation (of a functional), 232 Flip operator, 92, 96, 166, 174, 177 Folium of Descartes, 92 Frenet-Serret frame, 78, 98 Future-directed, 20, 264, 317 Future-set, 322 Gauss equations, 251 normal map, 178 Geodesic, 216, 220, 267, 281, 315 completeness, 316 curvature, 223 polygonal, 322 torsion, 223 Geometric surface, 262 Girard’s Formula, 167 Gradient of a smooth function defined in a surface, 165 Gram matrix, 6 Gram-Schmidt process, 13 Group Euclidean, 34 Holonomy, 244 Klein, 42 Lorentz, 33 Lorentz special, 35 orthogonal, 33 Poincaré, 34 pseudo-orthogonal, 33 pseudo-orthogonal special, 35 special orthochronous Lorentz, 38 Gödel’s spacetime, 320

INDEX  337

Hadamard product, 290 Hamilton’s Principle, 235 Hamiltonian, 247 Harmonic function, 279 Helicoid, 133, 164, 301 hyperbolic, 164 Hemisphere, 274 Henneberg surface (in R3 ), 310 Hessian, 92 Holomorphic function, 279 Horocycle, 153 Householder reflection, 41 Hyperbolic angle, 23, 160 helix, 110 isometry, 321 linear map, 44 numbers, see also Split-complex numbers plane, 108 point, 197 Identity Jacobi, 59 Lagrange, 56 polarization, 34 Immersion, 329 Index (of an inner product), 20 Indicator of a curve, 64 of a vector, 3, 262 Inequality reverse Cauchy-Schwarz, 23 reverse triangular, 24 Inertial parametrization, 195, 257 Integral along a curve, 292 Intrinsic distance, 316 Isometry between surfaces, 169, 264 Euclidean, 29 group (of a surface), 174 pseudo-Euclidean, 30 Isothermal parametrization, 256, 298 Klein bottle, 151 disk, 273 Kronecker’s Delta, 2 Lagrangian, 232 Laguerre transformation, 104

Lambert cylindrical projection, 143, 172 Lambert’s Formula, 168 Laplace-Beltrami operator, 257 Laplacian, 248, 298 Lightlike subspace, 5 vector, 3 Linear and affine parts, 34 Lines of curvature, 209 Lipschitz-continuous function, 322 Local canonical form of a curve in space, 107 of a lightlike curve, 124 of a plane curve, 84 Local coordinates, see also regular parametrization Local form of immersions, 329 Lorentz boost, 48 inner product, 2 factor, 28 manifold, 315 surface, 262 Lorentz-harmonic function, 290 Lorentz-Minkowski space, 2 Lorentz-Poincaré half-plane, 277 Lorentzian manifold, 262 metric, 158, 262 Margulis Invariant, 52 Maximal surface, 201 Mercator’s projection, 314 Metric determinant (of a bilinear map), 57 Minimal surface, 201 Minkowski metric, see also Lorentz inner product Monge parametrizations, 133 Möbius strip, 150 transformation, 272 Neil’s parabola, 76 Newton’s Second Law, 235 Noether charge, 247 Normal curvature, 207 plane, 107 section of a surface, 207

338  INDEX

vector, 82, 98, 118 Orientable Surface, 148 Oriented curvature (of a plane curve), 79 Orthochronous, 25, 35, 38 Orthogonal projection, 18 Osculating plane, 107 Outer semi-direct product for groups, 40 Parabolic helix, 110 linear map, 44 point, 197 Parallel field (along a curve), 219 surfaces, 189 translation, 244 Parametrization compatible with a given orientation, 148 abstract, 260 Partial differential, 155 indicator (of a parametrization), 156, 298 Past-directed, see also future-directed Pauli Matrix, 51 Penrose basis, 17 Planar point, 197 Poincaré disk, 265 half-plane, 266 Polar coordinates, 92 parametrization, 218 Pole, 292 Positive basis (for a lightlike plane), 118 Pre-geodesic, 221 Principal parametrization, 211 vector, 179 Proper function, 322 time, 67, 315 PseudoEuclidean norm, 3 Euclidean space of index ν, 2 orthogonal, 5 orthonormal, 5

Riemannian manifold, 315 Riemannian metric, 158, 262, 315 sphere, 272 torsion, 120 Radius of curvature, 84 Rectifying plane, 107 Regular parametrization, 131 parametrized surface, 131 point, 135 surfaces, 130 value, 135 Reissner-Nordström space, 277 Relativistic addition of speeds, 52 Reparametrization, 69 Riemann formula, 285 surface, 262 Riemannian manifold, 262, 315 metric, 158, 262 Rigid motion, see also Isometry, Euclidean Rindler coordinates, 94 parametrization, 218 Rotation (pure), 47 Schwarzschild half-plane, 276 horizon function, 276 space, 276 Second Fundamental Form, 180 Shape operator, see also Weingarten map Sherlock hat, 205 Singular point, 66 Slant of a curve (in a surface), 243 Smooth function (defined on a surface), 141 Spacelike subspace, 5 vector, 3 Spacetime, 317 Spatial and temporal parts, 36 Split-complex derivative, 304 numbers, 95, 177, 287 Split-holomorphic function, 280, 288

INDEX  339

Split-meromorphic function, 292 Stationary heat operator, see also Laplacian Stationary wave operator, see also d’Alembertian Stereographic Projection, 152 Structure constants, 295 Surface abstract, 260 lightlike, 155 of hyperbolic revolution, 166, 188, 249 of revolution, 134 spacelike, 155 timelike, 155 Sylvester’s Criterion, 8 Tangent bundle, 315 indicatrix, 104, 112 plane (to a surface), 261 plane (to a parametrization), 140 plane (to a surface), 139 space, 261 surface, 164, 176 vector, 82, 98, 118 vector (to a surface), 139 Tensor product (of curves), 270 Theorem Alexandrov-Zeeman, 26, 324 Beltrami-Enneper, 218 Bonnet, 253 Calabi-Bernstein, 202 Cartan–Dieudonné, 41 Cauchy-Goursat (revised), 293 Clairaut-Schwarz, 327 Egregium, 186, 252 Four Singularities, 90 Four Vertex, 91 Frobenius, 250 Fundamental of Calculus (split-complex version), 292 Fundamental of curves (lightlike/semi-lightlike case), 125 Fundamental of Plane Curves, 88 Fundamental para Curvas Admissíveis, 112 Gauss-Bonnet, 168 Green-Stokes, 201, 234

Hopf-Rinow, 316 Implicit Function (for products of surfaces), 155 Implicit Function (in Rn ), 328 Inverse Function (for products of surfaces), 154 Inverse Function (for surfaces), 146 Inverse Function (in Rn ), 328 Korn-Lichtenstein, 298 Lancret, 110 Lancret (lightlike version), 125 Liouville, 297 Lyusternik-Fet, 237 Meusnier, 208 Noether, 247 Riemann’s classification (for surfaces with constant Gaussian curvature), 280 Third Fundamental Form, 187 Time orientability, 317 orientation, 264 separation, 320 Timelike subspace, 5 vector, 3 Torsion, 99 Total derivative, 325 Totally umbilic surface, 190 Transformation isogonal, 42 Lorentz, 25, 30 Poincaré, see also Isometry, Lorentzian pseudo-orthogonal, 30 Translation surfaces, 205 Twins Paradox, 24 Umbilic point, 190 Unit normal field, 149 Variational derivative, see also First variation (of a functional) Vector indicator, see also indicator of a vector field, 315 field (along a curve), 219 spatial velocity, 19 velocity (of a curve), 64, 261

340  INDEX

Viviani’s window, 101 Weingarten Map, 179 Wirtinger operators

split-complex version, 289 complex version, 304 Zeeman Topology, 323