The content is wellpresented and instructive. My main gripe with this book is the very low quality paperback edition. H
291 109 22MB
English Pages 511 Year 1976
pifferen tial Geometry of Curves and Surfaces
Mantredo P. do $arm0 Institute de Mateinurica Pura e Aplicada (IMPA) R i o de Janeiro. Brazil
.
% 1 UP
1 COlLEGE OF SCIENCE I
PrenticeHah lnc., Englewood Cliffs, New Jersey
Carmo, Manfredo Perdigao do. Differential geometry of curves and surfaces "A free translation, with additional material, of a book and a set of notes, both published originally in Portuguese." Bibliography: p. Includes index. 1. Geometry, Differential. 2. Curves. 3. Surfaces. I. Title
To Leny @ 1976 k y PrenticeHall, Inc.
Englewood Cliffs, N e w Jersey
All rights book may o r by any in writing
reserved. N o part of this be reproduced in any form, means, without permission f r o m the publisher
Current printing:
1 0 9 8 7 6 5 4 3 2 1
Printed in the United States of America
PrenticeHall PrenticeHall PrenticeHall PrenticeHall PrenticeHall PrenticeHall
International, Inc., London of Australia Pty., Limited, Sydney of Canada, Ltd., Toronto of India Private Ltd., New Delhi of Japan, Inc., Tokyo of Southeast Asia Private Limited, Singapore
Conten ts
Preface v Some Remarks on Using this Book vii 1. Curves I 11 12 13 14 15 16 17
Introduction I Parametrized Curves 2 Regular Curves; Arc Length 5 The Vector Product in R3 1I The Local Theory of Curves Parametrized by Arc Length 16 The Local Canonical Form 27 Global Properties of Plane Curves 30
2. Regular Surfaces 51 Introduction 51 Regular Surfaces; Inverse Images of Regular Values 52 Change of Parameters; Differentia1 Functions on Surfaces 69 The Tangent Plane; the Differential of a Map 83 The First Fundamental Form; Area 92 Orientation of Surfaces 102 A Characterization of Compact Orientable Surfaces 109 A Geometric Definition of Area 114 Appendix: A Brief Review on Continuity and ~ i f f e r e n b a b i l i tI1 ~8
Contents
The Geometry of the Gauss Map 134 Introduction I34 The Definition of the Gauss Map and Its Fundamental Properties 135 The Gauss Map in Local Coordinates I53 Vector Fields 175 Ruled Surfaces and Minimal Surfaces 188 Appendix: SelfAdjoint Linear Maps and Quadratic Forms 214
The Intrinsic Geometry of Surfaces 217 Introduction 21 7 Isometrics; Conformal Maps 21 8 The Gauss Theorem and the Equations of Compatibility 231 Parallel Transport; Geodesics 238 The GaussBonnet Theorem and its Applications 264 The Exponential Map. Geodesic Polar Coordinates 283 Further Properties of Geodesics. Convex Neighborhoods 298 Appendix: Proofs of the Fundamental Theorems of The Local Theory of Curves and Surfaces 309
Global Differential Geometry 315 Introduction 315 The Rigidity of the Sphere 317 Complete Surfaces. Theorem of HopfRinow 325 First and Second Variations of the Arc Length; Bonnet's Theorem 339 Jacobi Fields and Conjugate Points 357 Covering Spaces; the Theorems of Hadamard 371 Global Theorems for Curves; the FaryMilnor Theorem 380 Surfaces of Zero Gaussian Curvature 408 Jacobi's Theorems 415 Abstract Surfaces; Further Generalizations 425 Hilbert's Theorem 446 Appendix: PointSet Topology of Euclidean Spaces 456
Bibliography and Comments 471 Hints and Answers to Some Exercises 475
Preface
This book is an introduction to the differential geometry of curves and surfaces, both in its local and global aspects. The presentation differs from the traditional ones by a more extensive use of elementary linear algebra and by a certain emph'asis plaCed on basicgeometrical facts, rather than on mjchinery or random details. We have tried to build each chapter of the book around some simple and fundamental idea. Thus, Chapter 2 develops around the concept of a regular surface in R3; when this concept is properly developed, it is probably the best model for differentiable manifolds. Chapter 3 is built on the Gauss normal map and contains a large amount of the local geometry of surfaces in R3. Chapter 4 unifies the intrinsic geometry of surfaces around the concept of covariant derivative; again, our purpose was to prepare the reader for the basic notion of connection in Riemannian geometry. Finally, in Chapter 5, we use the first and second variations of arc length to derive some global properties of surfaces. Near the end of chap& 5 (Sec. 510), we show how questions on surface theory, and the experience of Chapters 2 and 4, lead naturally to the consideration of differentiable manifolds and Riemannian metrics. To maintain the proper balance between ideas and facts, we have presented a large number of examples that are computed in detail. Furthermore, a reasonable supply of exercises is provided. Some factual material of classical differential geometry found its place in these exercises. Hints or answers are given for the exercises that are starred. The prerequisites for reading this book are linear algebra and calculus. From linear algebra, only the most basic concepts are needed, and a
vi
Preface
standard undergraduate course on the subject should suffice. From calculus, a certain familiarity with calculus of several variables (including the statement of the implicit function theorem) is expected. For the reader's convenience, we have tried to restrict our references to R. C. Buck, Advancd Calclil~~ New ~ , York: McGrawHill, 1965 (quoted as Buck, Advanced Calculus). A certain knowledge of differential equations will be useful but it is not required. This book is a free translation, with additional material, of a book and a set of notes, both published originally in Portuguese. Were it not for the enthusiasm and enormous help of Blaine Lawson, this book would not have come into English. A large part of the translation was done by Lcny Cavalcante. I am also indebted to my colleagues and students at IMPA for their comments and support. In particular, Elon Lima read part of the Portuguese version and made valuable comments. Robert Gardner, Jiirgea Kern, Blaine Lawson, and Nolan Wallach read critically the English manuscript and helped me to avoid several mistakes, both in English and Mathematics. Roy Ogawa prepared the computer programs for some beautiful drawings that appear in the book (Figs. 13, 18, 19, 110, 111, 345 and 44). Jerry Kazdan devoted his time generously and literally offered hundreds of suggestions for the improvement of the manuscript. This final form of the book has benefited greatly from his advice. To all these peopleand to Arthur Wester, Editor of Mathematics at PrenticeHall, and Wilson G6es at IMPAI extend my sincere thanks. Rio de Janeiro
Manfredo P. do Carmo
Some Remarks on Using This Book
We tried to prepare this book so it could be used in more than one type of differential geometry course. Each chapter starts with an introduction that describes the material in the chapter and explains how this material will be used later in the book. For the reader's convenience, we have used footnotes to point out the sections (or parts thereof) that can be omitted on a first reading. Although there is enough material in the book for a fullyear course (or a topics course), we tried to make the book suitable for a first course on differential geometry for students with some background in linear algebra and advanced calculus. For a short onequarter course (10 weeks), we suggest the use of the following material: Chapter 1: Secs. 12, 13, 14, 15 and one topic of Sec. 172 weeks. Chapter 2: Secs. 22 and 23 (omit the proofs), Secs. 24 and 253 weeks. Chapter 3: Secs. 32 and 332 weeks. Chapter 4: Secs. 42 (omit conformal maps and Exercises 4, 1318, 20), 43 (up to Gauss theorema egregium) , 44 (up to Prop. 4; omit Exercises 12, 13, 16, 1821 ), 45 (up to the local GaussBonnet theorem; include applications (b) and (f) ) 3 weeks. The 10week program above is on a pretty tight schedule. A more relaxed alternative is to allow more time for the first three chapters and to present survey lectures, on the last week of the course, on geodesics, the Gauss theorema egregium, and the GaussBonnet theorem (geodesics can then be defined as curves whose osculating planes contain the normals to the surface). In a onesemester course, the first alternative could be taught more vii
viii
Some Remarks on Using this Book
leisurely and the instructor could probably include additional material (for instance, Secs. 52 and 510 (partially), or Secs. 46, 53 and 54). Please also note that an asterisk attached to an exercise does not mean the exercise is either easy or hard. It only means that a solution or hint is provided at the end of the book. Second, we have used for parametrization a boldfaced x and that might become clumsy when writing on the blackboard. Thus we have reserved the capital X as a suggested replacement. Where letter symbols that would normally be italic appear in italic context, the letter symbols are set in roman. This has been done to distinguish these symbols from the surrounding text.
1
Curves
The differential geometry of curves and surfaces has two aspects. One, which may be called classical differential geometry, started with the beginnings of calculus. Roughly speaking, classical differential geometry is the study of local properties of curves and surfaces. By local properties we mean those properties which depend only on the behavior of the curve or surface in the neighborhood of a point. The methods which have shown themselves to be adequate in the study of such properties are the methods of differential calculus. Because of this, the curves and surfaces considered in differential geometry will be defined by functions which can be differentiated a certain number of times. The other aspect is the socalled global differential geometry. Here one studies the influence of the local properties on the behavior of the entire curve or surface. We shall come back to this aspect of differential geometry later in the book. Perhaps the most interesting and representative part of classical differential geometry is the study of surfaces. However, some local properties of curves appear naturally while studying surfaces. We shall therefore use this first chapter for a brief treatment of curves. The chapter has been organized in such a way that a reader interested mostly in surfaces can read only Secs. 12 through 15. Sections 12 through 14 contain essentially introductory material (parametrized curves, arc length, vector product), which will probably be known from other courses and is included here for completeness. Section 15 is the heart of the chapter
Curves
2
and contains the material of curves needed for the study of surfaces. For those wishing to go a bit further on the subject of curves, we have included Secs. 16 and 17.
72. Parametrized Curves We denote by R3 the set of triples (x, y, z) of real numbers. Our goal is to characterize certain subsets of R3 (to be called curves) that are, in a certain sense, onedimensional and to which the methods of differential calculus can be applied. A natural way of defining such subsets is through differentiable functions. We say that a' real function of a real variable is dzflerentiable (or smooth) if it has, at all points, derivatives of all orders (which are automatically continuous). A first definition of curve, not entirely satisfactory but sufficient for the purposes of this chapter, is the following.
DEFINITION. A parametrized differentiable curve is a dzflerentiable map a: I + R3 of an open interval I = (a, b) of the real line R into R3.t The word dtferentiable in this definition means that a is a correspondence which maps each t E I into a point a(t) = (x(t), y(t), z(t)) E R3 in such a way that the functions x(t), y(t), z(t) are differentiable. The variable t is called the parameter of the curve. The word interval is taken in a generalized sense, so that we do not exclude the cases a = oo, b = $ m. If we denote by x'(t) the first derivative of x at the point t and use similar notations for the functions y and z, the vector (xl(t), y'(t), zl(t)) = a'(t) E R3 is called the tangent vector (or velocity vector) of the curve a at t. The image set a(I) c R3 is called the trace of a . As illustrated by Example 5 below, one should carefully distinguish a parametrized curve, which is a map, from its trace, which is a subset of R3. A warning about terminology. Many people use the term "infinitely differentiable" for functions which have derivatives of a11 orders and reserve the word "differentiable" to mean that only the existence of the first derivative is required. We shall not follow this usage. ExampIe 1. The parametrized differentiable curve given by
a(?) = (a cos t, a sin t, bt),
t
E
R,
+
has as its trace in R3 a helix of pitch 2nb on the cylinder x2 y 2 = a2. The parameter t here measures the angle which the x axis makes with the line joining the origin 0 to the projection of the point a(t) over the xy plane (see Fig. 11). fin italic context, letter symbols will not be italicized so they will be clearly distinguished from the surrounding text.
Parametrized Curves
" 1C
Figure 11
Figure 12
Example 2. The map a: R 3 R2 given by a(t) = (t3, t2), t E R, is a parametrized differentiable curve which has Fig. 12 as its trace. Notice that a'(0) = (0,O); that is, the velocity vector is zero for t = 0. Example 3. The map a: R + R2 given by a(t) = (t3  4t, t2  4), t E R, is a parametrized differentiable curve (see Fig. 13). Notice that a(2) = a(2) = (0,O); that is, the map a is not onetoone.
Figure 13
Figure 14
Example 4, The map a: R +R2 given by u(t) = (t, 1 t I), t E R, is not a parametrized differentiable curve, since ] tl is not differentiable at t = 0 (Fig. 14). Example 5. The two distinct parametrized curves a(t)
= (cos
p(t)
=
t, sin t),
(cos 2t, sin 2t),
4
Curves
+
(0  E, 2n E), E > 0, have the same trace, namely, the circle x2 y Z = I. Notice that the velocity vector of the second curve is the double of the first one (Fig. 15).
where t
+
E
We shall now recall briefly some properties of the inner (or dot) product of vectors in R 3 . Let u = (u,, u,, 24,) E R3 and define its norm (or length) by
Geometrically, lul is the distance from the point (u,, u,, u,) to the origin 0 = (0,0, 0 ) . NOW,let u = (u,, u,, 24,) and v = (v,, v,, v , ) belong to R3, and let 8,O 8 n, be the angle formed by the segments Ou and Ov. The inner product u v is defined by (Fig. 16)
<
0, b < 0,be a
Regular Curves; Arc Length
Figure 110. Folium of Descartes.
a. Show that as t 3 +a,a(t) approaches the origin 0, spiraling around it (because of this, the trace of a is called the Iogarithmic spiral; see Fig. 111). b. Show that a'(&)+ (0, 0) as t + +CO and that
is finite; that is, a has finite arc length in [ t o ,co).
Figure 111. Logarithmic spiral.
Curves
70
7. A map a : I + R3 is called a curve of class C k if each of the coordinate functions in the expression a ( t ) = (x(t),y(t), z(t)) has continuous derivatives up to order k. If a is merely continuous, we say that a is of class C D .A curve a is calIed simple if the map a is onetoone. Thus, the curve in Example 3 of Sec, 12 is not simple. R3 be a simpIe curve of class C O .We say that a has a weak Let a : I has a limit tangent at t = to E I i f the line determined by @(to h ) and @(to) position when h 3 0. We say that a has a strong tangent at t = to if the line determined by a ( t , + h) and a(to + k) has a limit position when h, k + 0. Show that


a. a ( t ) = (t3, t2),t
+
E
R, has a weak tangent but not a strong tangent at t
*b. If a : 1 R3 is of class C 1and regular at t at t = t o .
= t o ,then
= 0.
it has a strong tangent
c. The curve given by
is of class C 1but not of class C2. Draw a sketch of the curve and its tangent vectors.

*8. Let a : 1 R3 be a differentiable curve and let [a, b] c I be a closed interval. For every partition
of [a, b], consider the sum CI= I &(ti) a(til) 1 = [(a,P ) , where P stands for the given partition. The norm I P ( of a partition P is defined as
Geometrically, l(a, P ) is the length of a polygon inscribed in a([a,b]) with vertices in a(ti) (see Fig. 112). The point of the exercise is to show that the arc length of @([a,b])is, in some sense, a limit of lengths of inscribed poIygons.
Figure 112
The Vector Product i n R3
77
Prove that given E > 0 there exists 8 > 0 such that if I PI < 8 then
9. a. Let a: 1 3 R3 be a curve of class C0(cf. Exercise 7). Use the approximation by polygons described in Exercise 8 to give a reasonable definition of arc
length of a. b. ( A NonrectrlfiableCurve.) The following example shows that, with any reasonable definition, the arc length of a C0 curve in a closed interval may be unbounded. Let a : [O, 13 R2 be given as a(t) = (t, t sin(n/t)) if t # 0, and a(0) = (0,O). Show, geometricaIly, that the arc length of the portion of the curve corresponding to l/(n f 1) _< t < l/n is at least 2/(n l).Use this to show that the length of the curve in the interval l/AT_< t < 1 is greater than 2 Cr= I/(n I), and thus it tends to infinity as N + m.

+
+
10. (Straight Lines as Shorfesr.) Let a:1 . R3 be a parametrized curve.Let fa, b'j c I and set &(a) = p, a(b) = q. a. Show that, for any constant vector v, I v 1 = 1,
b. Set
v
=
4P 14PI
and show that
that is, the curve of shortest length from &(a)to a(b) is the straight line joining these points.
7 4. The Vector Product,in R3 In this section, we shall present some properties of the vector product in R3. They will be found useful in our later study of curves and surfaces. It is convenient to begin by reviewing the notion of orientation of a vector 1, . . . , n, of a n ndimenspace. Two ordered bases e = (e,) and f = (A), i sional vector space V have the same orientatiovl if the matrix of change of basis has positive determinant. We denote this relation by e f. From elementary properties of determinants, it follows that e  f is an equivalence relation; i.e., it satisfies


Curves
72
The set of all ordered bases of V is thus decomposed into equivalence classes (the elements of a given class are related by ) which by property.3 are disjoint. Since the determinant of a change of basis is either positive or negative, there are only two such classes. Each of the equivalence classes determined by the above relation is called an orientation of V. Therefore, V has two orientations, and if we fix one of them arbitrarily, the other one is called the opposite orientation. In the case V = R3, there exists a natural ordered basis el = ( I , 0, O), e, = (0, 1, 0), e3 = (0,0, I), and we shall call the orientation corresponding to this basis the positive orientation of R3, the other one being the negative orientatiorz (of course, this applies equally well to any Rn). We also say that a given ordered basis of R3 is positive (or negative) if it belongs to the positive (or negative) orientation of R3. Thus, the ordered basis e,, e,, e2 is a negative basis, since the matrix which changes this basis into e,, e,, e3 has determinant equal to  1. We now come to the vector product. Let u, v E R3. The vector product of u and v (in that order) is the unique vector u A v E R3 characterized by
Here det(ti, v, w) means that if we express u, v, and w in the natural basis {ei),
then
where I a,, I denotes the.determinant of the matrix (a,,). It is immediate from the definition that
Remark. It is also very frequent to write u A v as 11 the cross product.
x v and refer to it .as
The following properties can easily be checked (actually they just express the usual properties of determinants):

1. ti A v v A u (anticommutativity). 2. u A v depends linearly on u and v ; i.e., for any real numbers a, 6, we have
The Vector Product in R3
(au f bw) A v
= au
A v
+ bw
v.
3. u A v = 0 if and only if u and v are linearly dependent. 4, (u V)*U= 0 , (u A V)*V= O . It follows from property 4 that the vector product u v # 0 is normal to a plane generated by u and v. To give a geometric interpretation of its norm and its direction, we proceed as follows. First, we observe that (u A v)*(u A v) = ( u A vlZ> 0. This means that the determinant of the vectors u, v, u A v is positive; that is, {u, v, u v] is a positive basis. Next, we prove the relation
where 11, v, x, y are arbitrary vectors. This can easily be done by observing that both sides are linear in u, v, x, y. Thus, it suffices to check that
for all i, j, k, I = 1, 2, 3. This is a straightforward verification. It follows that
where 0 is the angle of u and v, and A is the area of a parallelogram generated by u and v. In short, the vector product of u and v is a vector u A v perpendicular to a plane generated by u and v, with a norm equal to the area of a parallelogram generated by u and v and a direction such that (u, v, u A v) is a positive basis (Fig. 113).
14
Curves
The vector product is not associative. In fact, we have the following identity : v) A
(U
W = (u*w)v
 (v.w)u,
(2)
which can be proved as follows. First we observe that both sides are linear in zr, v, n7;thus, the identity will be true if it holds for all basis vectors. This last verification is, however, straightforward; for instance,
Finally, let u(t) = (u,(t), u2(t), u3(t)) and v(t) = (v,(t), vz(t), v,(t)) be differentiable maps from the interval (a, b) to R3, t E (a, b). It follows immediately from Eq. (1) that u(t) A v(t) is also differentiable and that
Vector products appear naturally in many geometrical constructions. Actually, most of the geometry of planes and lines in R 3 can be neatly expressed in terms of vector products and determinants. We shall review some of this material in the following exercises.
EXERCISES 1. Check whether the following bases are positive: a. The basis {(I, 3), (4, 2)] in R2.
b. The basis ((1, 3, 5), (2, 3, 7), (4, 8, 3)) in R3. "2. A plane P contained in R3 is given by the equation ax
+ by + cz + d = 0.
Show that the vector v = (a, b, c) is perpendicular to the plane and that 1 d l / d a 2 + b2 cz measures the distance from the plane to the origin (0, 0,O). *3. Determine the angle of intersection of the two planes 5x + 3y + 22  4 = 0 and 3x + 4y  72 = 0. *4. Given two planes aix biy 3 riz di = 0, i = 1, 2, prove that a necessary and sufficient condition for them to be parallel is
+
+
+
where the convention is made that if a denominator is zero, the corresponding numerator is also zero (we say that two planes are parallel if they either coincide or do not intersect). 5. Show that the equation of a plane passing through three noncolinear points PI = (XI,Y r , ZI),~2 = ( ~ 2YZ, , z2), ~3 = ( ~ 3~, 3 ~, 3is) given by
The Vector Product in R3
75
where p = (x, y, z) is an arbitrary point of the plane and p means thevector ( x  X I , y  y , , z  z l ) .
+
+ + d,
"6. Given two nonparallel planes aix biy ciz their line of intersection may be parametrized as
where (x,, yo, zo) belongs to the intersection and u product u = v l A v2, vi = (ai, bi, ci), i = 1 , 2.
=
=
0, i

p l , for instance,
=
1,2, show that
( u , , u2,u3)is the vector
*7. Prove that a necessary and sufficient condition for the plane
and the line x
 xo = u l t , y
 yo = u2t,z
 zo = u3t
to be parallel is
*8. Prove that the distance p between the nonparaIleI lines
is given by
where u
u2, ~ 3 1 v, = @ I , 212, ~ 3 1 r, = ( X O  X I , Y O  Y I ,zo  ZI). 9. Determine the angle of intersection of the plane 3x 4y 72 8 = 0 and the line x  2 = 3t, y  3 = 5t, z  5 = 9t. =
+ + +
10. The natural orientation of R2 makes it possible to associate a sign to the area A of a parallelogram generated by two linearly independent vectors u, v E R2. TO do this, let ( e i ] , i = 1 , 2, be the natural ordered basis of Rz, and write u = u l e l u2e2,v = vie, v2e2.Observe the matrix relation
+
+
U'U

v.3 C: ::)6:1:)
Uv
=
and conclude that
Since the last determinant has the same sign as the basis {u, v), we can say that A is positive or negative according to whether the orientation of (u, v ) is positive or negative. This is calIed the oriented area in R2.
Curves
16
11. a. Show that the volume V of a parallelepiped generated by three linearly independent vectors u, v, w E R3 is given by V = l(u A v) w 1, and introduce an orietzted volume in R3. b. Prove that
v 2=
U'U
uv
U'W
V'U
v*v
v*w
WU
Wv
WW
.
12. Given the vectors v # 0 and w, show that there exists a vector u such that u A v = w if and only if v is perpendicular to w. Is this vector u uniquely determined? If not, what is the most general solution? = (ul(t), u2(t), u3(t)) and v(t) = (vl(t), v2(t), v3(t)) be differentiable maps from the interval (a, b) into R3. If the derivatives u'(t ) and v'(t) satisfy the conditions
13. Let u(t)
where a, b, and c are constants, show that u(t) A v(t) is a constant vector. 14. Find all unit vectors which are perpendicular to the vector (2,2,1) and parallel to the plane determined by the points (O,0, O), (1, 2, I), (1, 1, 1).
75. The Local Theory of Curves Parametrized by Arc Length This section contains the main results of curves which will be used in the later parts of the book. R 3 be a curve parametrized by arc length s. Since Let a: 1 = (a, 6) the tangent vector af(s) has unit length, the norm I a"(s) I of the second derivative measures the rate of change of the angle which neighboring tangents make with the tangent at s. (aU(s)I gives, therefore, a measure of how rapidly the curve pulls away from the tangent line at s, in a neighborhood of s (see Fig. 1 14). This suggests the following definition.


DEFINITION. Let a: I R3 be a curve parametrized by arc length s E I. The number j aU(s)1 = k(s) is called the curvature of a at s.
+
If a is a straight line, a(s) = us v, where u and v are constant vectors (1 u I = I), then k = 0. Conversely, if k = Ia"(s)] = 0, then by integration a(s) = us v, and the curve is a straight line. Notice that by a change of orientation, the tangent vector changes its direction; that is, if P(S) = a(s), then
+
Curves Parametrized b y Arc Length
Figure 114
Therefore, a"(s) and the curvature remain invariant under a change of orientation. At points where k(s) # 0, a unit vector n(s) in the direction a"(s) is well defined by the equation a"(s) = k(s)n(s). Moreover, a"(s) is normal to mr(s), because by differentiating a'(s)a'@) = 1 we obtain a"(s)a'(s) = 0. Thus, n(s) is normal to a'(s) and is called the normal vector at s. The plane determined by the unit tangent and normal vectors, ar(s) and n(s), is called the osculating plane at s. (See Fig. 115.) At points where k(s) = 0, the normal vector (and therefore the osculating plane) is not defined (cf. Exercise 10). To proceed with the local analysis of curves, we need, in an essential way, the osculating plane. It is therefore
Figure 115
18
Curves
convenient to say that s E I is a singular point of order 1 if a"(s) = 0 (in this context, the points where al(s) = 0 are called singular points of order 0). In what follows, we shall restrict ourselves to curves parametrized by arc length without singular points of order 1. We shall denote by t(s) = al(s) the unit tangent vector of a at s. Thus, t'(s) = k(s)n(s). The unit vector b(s) = t(s) A n(s) is normal to the osculating plane and will be called the binormal vector at s. Since b(s) is a unit vector, the length I br(s) 1 measures the rate of change of the neighboring osculating planes with the osculating plane at s ; that is, bl(s) measures how rapidly the curve pulls away from the osculating plane at s, in a neighborhood of s (see Fig. 115). To compute br(s) we observe that, on the one hand, bl(s) is normal to b(s) and that, on the other hand,
that is, bl(s) is normal to t(s). It follows that bl(s) is parallel to n(s), and we may write
for some function z(s). (Warning: Many authors write z(s) instead of our z(s>.)
DEFINITION. Let a: I , R3 be a curve parametrized by arc length s such that aU(s)# 0, s E I. The number z(s) dejned by bl(s) = z(s)n(s) is called the torsion of a a t s.

If ac is a plane curve (that is, a(1) is contained in a plane), then the plane of the curve agrees with the osculating plane; hence, z = 0. Conversely, if z 0 (and k # 0), we have that b(s) = b, = constant, and therefore
It follows that a(s). b, = constant; hence, a(s) is contained in a plane normal to b,. The condition that k st 0 everywhere is essential here. In Exercise 10 we shall give an example where z can be defined to be identically zero and yet the curve is not a plane curve. In contrast to the curvature, the torsion may be either positive or negative. The sign of the torsion has a geometric interpretation, to be given later (Sec. 16). Notice that by changing orientation the binormal vector changes sign, since b = t /\ n. It follows that b'(s), and, therefore, the torsion, remains invariant under a change of orientation.
Curves Parametrized by Arc Length
79
Let us summarize our position. To each value of the parameter s, we have associated three orthogonal unit vectors t(s), n(s), b(s). The trihedron thus formed is referred to as the Frenet trihedron at s. The derivatives t'(s) = kn, bt(s) = zn of the vectors tts) and b(s), when expressed in the basis ( t , n, b), yield geometrical entities (curvature k and torsion z) which give us information about the behavior of a in a neighborhood of s. The search for other local geometrical entities would lead us to compute n'(s). However, since n = b /\ t, we have
and we obtain again the curvature and the torsion. For later use, we shall call the equations t'
= kn,
n'
= kt
b'
= zn

zb,
the Frenet.formulas (we have ommited the s, for convenience). In this context, the following terminology is usual. The tb plane is called the rectfyingplane, and the nb plane the normalplane. The lines which contain n(s) and b(s) and pass through a(s) are called the principal normal and the binormal, respectively. The inverse R = llk of the curvature is called the radius of curvature at s. Ofcourse, a circle of radius r has radius ofcurvature equal to r, as one can easily verify. Physically, we can think of a curve in R3 as being obtained from a straight line by bending (curvature) and twisting (torsion). After reflecting on this construction, we are led to conjecture the following statement, which, roughly speaking, shows that k and z describe completely the local behavior of the curve.
FUNDAMENTAL THEOREM OF THE LOCAL THEORY OF CURVES. Given dzfferentiable functions k(s) > 0 and z(s), s E I, there exists a regular parametrized curve a : I + R3such that s is the arc length, k(s) is the curvature, and z(s) is the torsion of a . Moreover, any other curve d, satisfying the same conditions, differs from a by a rigid motion; that is, there exists an orthogonal linear map p of R3,with positive determinant, and a vector c such that d = poa c.
+
The above statement is true. A complete proof involves the theorem of existence and uniqueness of solutions of ordinary differential equations and will. be given in the appendix to Chap. 4. A proof of the uniqueness, up to
20
Curves
rigid motions, of curves having the same s, k(s), and r(s) is, however, simple and can be given here. Proof of the Uniqt~enessPart of the Fundamental Theorem. We first remark that arc length, curvature, and torsion are invariant under rigid motions; that means, for instance, that if M: R3 ,R3 is a rigid motion and a = a(t) is a parametrized curve, then
That is plausible, since these concepts are defined by using inner or vector products of certain derivatives (the derivatives are invariant under translations, and the inner and vector products are expressed by means of lengths and angles of vectors, and thus also invariant under rigid motions). A careful checking can be left as an exercise (see Exercise 6). Now, assume that two curves a = a(s) and 6 = 6(s) satisfy the conditions k(s) = K(s) and z(s) = T(s), s E I. Let t o ,no, b, and fo,A,, 6, be the Frenet trihedrons at s = s o E I of a and 6, respectively. Clearly, there is a rigid motion which takes &(so)into a(s,) and f,, ii,, 6, into to, no, b,. Thus, after performing this rigid motion on a, we have that &(so)= a(s,) and that the Frenet trihedrons t(s), n(s), b(s) and ys), A(s), 6(s) of a and 6, respectively, satisfy the Frenet equations:
dn  = ktrb ds
dn' = kfrZ ds
with t(s,) = i(s,), n(s0) = A(so), b(s,) = &so). We now observe, by using the Frenet equations, that
for all s
E
I. Thus, the above expression is constant, and, since it is zero for
Curves Parametrized by Arc Length
27
is identically zero. It follows that t(s) = T(s), n(s) for all s E I. Since
s
= so,it
= ri(s),
h(s) = 6(s)
+
we obtain (dlds) (a  &) = 0. Thus, a(s) = d(s) a, where a is a constant vector. Since a(so) = &(so),we have a = 0; hence, a(s) = d(s) for all s E I. Q.E.D. Remark I. In the particular case of a plane curve a : I RZ,it is possible to give the curvature k a sign. For that, let (el, e,] be the natural basis (see Sec. 14) of R2 and define the normal vector n(s), s E I, by requiring the basis (t(s), n(s)) to have the same orientation as the basis (el, e2]. The curvature k is then defined by +
and might be either positive or negative. It is clear that jkl agrees with the previous definition and that k changes sign when we change either the orientation of a or the orientation of R2 (Fig. 116).
Figure 116

It should also be remarked that, in the case of plane curves (z O), the proof of the fundamental theorem, refered to above, is actually very simple (see Exercise 9). Remark 2. Given a regular parametrized curve a : I + R3 (not neces. sarily parametrized by arc length), it is possible to obtain a curve J 4 R3 parametrized by arc length which has the same trace as a . In fact, let
P:
Curves

Since ds/dt 1 a'@)( # 0, the function s = s(t) has a differentiable inverse t = t(s), s E ~ ( 1= ) J, where, by an abuse of notation, t also denotes the inverse function 8  of s. Now set p = aot :J R3. Clearly, P(J) = a ( I ) and I pl(s) I = I a l ( t ) (dtlds)] = 1. This shows that p has the same trace as a and is parametrized by arc length. I t is usual to say that P is a repnrametrizntion of a ( l ) by arc length. This fact allows us to extend all local concepts previously defined to regular curves with an arbitrary parameter. Thus, we say that the curvature k ( t ) of a : I R3 at t E I i s the curvature of a reparametrization p: J + R3 of a ( I ) by arc length a t the corresponding point s = s(t). This is clearly independent of the choice of p and shows that the restriction, made at the end of Sec. 13, of considering only curves parametrized by arc length is not essential. In applications, it is often convenient to have explicit formulas for the geometrical entities in terms of an arbitrary parameter; we shall present some of them in Exercise 12.



Unless explicity stated, a : I 4 R3 is a curve parametrized by arc length s, with curvature k(s) f 0,for all s E I .
1. Given the parametrized curve (helix) S a(s) = ( a cos T ,a sin
where c2 = a2 + 6 2 , a. Show that the parameter s is the arc length. b. Determine the curvature and the torsion of a. c. Determine the osculating plane of a. d. Show that the lines containing n(s) and passing through a($ meet the z axis under a constant angle equal to n/2. e. Show that the tangent lines to a make a constant angle with the z axis. *2. Show that the torsiop z of a is given by
3. Assume that a ( ~c) R z (i.e., a is a plane curve) and give k a sign as in the text. Transport the vectors t(s)parallel to themselves in such a way that the origins of
23
Curves Parametrized b y Arc Length
t(s) agree with the origin of R 2 ; the end points of t(s) then describe a parametrized curve s + f(s) called the indicatrix of tangents of a. Let $(s) be the angle from e l to t(s) in the orientation of R2. Prove (a) and (b) (notice that we are assuming that k # 0). a. The indicatrix of tangents is a regular parametrized curve. b. dilds = (d$/ds)n, that is, k = d6'lds.
"4. Assume that all normals of a parametrized curve pass through a fixed point. Prove that the trace of the curve is contained in a circle.
5. A regular parametrized curve a has the property that all its tangent lines pass through a fixed point. a. Prove that the trace of a is a (segment of a) straight line. b. Does the conclusion in part a still hold if
a is not regular?
6. A translation by a vector v in R3 is the map A : R3 + R3 that is given by A(p) = p v, p E R3. A linear map p : R3 + R3 is an orthogonal transformation when pu. pv = u v for all vectors u, v E R3. A rigid motion in R3 is the result of composing a translation with an orthogonal transformation with positive determinant (this last condition is included because we expect rigid motions to preserve orientation). a. Demonstrate that the norm of a vector and the angle 8 between two vectors, 0 5 6' < n, are invariant under orthogonal transformations with positive determinant. b. Show that the vector product of two vectors is invariant under orthogonal transformations with positive determinant. Is the assertion still true if we drop the cdndition on the determinant ? c. Show that the arc length, the curvature, and the torsion of a parametrized curve are (whenever defined) invariant under rigid motions.
+

*7. Let a : I Rz be a regular parametrized plane curve (arbitrary parameter), and define n = n(t) and k = k(t) as in Remark 1. Assume that k(t) # 0, t E I. In this situation, the curve
is called the evolufe of a (Fig. 117).
a. Show that the tangent at t of the evolute of a is the normal to a at t. b. Consider the normal lines of a at two neighboring points t 1, tZ,f # t2. Let t, approach t, and show that the intersection points of the normals converge to a point on the trace of the evolute of a. 8. The trace of the parametrized curve (arbitrary parameter) a(t)
is called the catenary.
=
(t, C O S t), ~
t
E
R,
Figure 117
a. Show that the signed curvature (cf. Kemark 1) of the catenary is
b. Show that the evolute (cf. Exercise 7) of the catenary is P(t)
= (t

sinh t cosh t, 2 cosh t ) .
9. Given a differentiable function k(s), s E I, show that the parametrized plane curve having k(s) = k as curvature is given by
a($)=
(1
cos ~ ( sds)
+ a, J sin ~ ( sds) + b),
where
and that the curve is determined up to a translation of the vector (a, 6 ) and a rotation of the angle p.
10. Consider the map (t, 0, e  l / t z ) ,
t
>0
(t,el~'~"Oo>, t < O (O,O,O),
t=O
a. Prove that a is a differentiable curve. b. Prove that
t# f
a is regular for all t and that the curvature k ( t ) # 0, for t # 0, and k(0) = 0.
Curves Parametrized by Arc Length
25
c. Show that the limit of the osculating planes as t + 0, t > 0, is the plane y = 0 but that the limit of the osculating planes as t  0, t < 0, is the plane z = 0 (this implies that the normal vector is discontinuous at t = 0 and shows why we excluded points where k = 0). d. Show that z can be defined so that
z

0, even though a is not a plane curve.
11. One often gives a plane curve in polar coordinates by p
= p(O),
a (6' (6.
a. Show that the arc length is
where the prime denotes the derivative relative to 0. b. Show that the curvature is

12. Let a : I R3 be a regular parametrized curve (not necessarily by arc length) and let p: J ,R3 be a reparametrization of U(I) by the arc length s = s(t), measured from t o c I(see Remark 2). Let t = t(s) be the inverse function of s and set dajdt = a', d2aldt2 = a", etc. Prove that a. dtjds = 111 a' I, d2t/ds2 = (ara"jla' 14). b. The curvature of
c. The torsion of
a at t
a at t
E
E
I is
1 is
z(t) =

(a' A ,'').a1// 1 U' A a'' l2
d. If a : I, R 2 is a plane curve ~ ( t=) (x(t),y(t)), the signed curvature (see Remark f ) of a at t is
C13.Assume that z(s) # 0 and k'(s) # 0 for all s E I. Show that a necessary and sufficient condition for ~ ( 1to)lie on a sphere is that R2 where R
=
llk, T
=
+ (R')3T2= const.,
117, and R' is the derivative of R relative to s.
14. Let a: (a, 6 ) + R2 be a regular parametrized plane curve. Assume that there exists t o , a < to < b, such that the distance I a(t)1 from the origin to the trace of
26
Curves
a will be a maximum at I k(t0) I 2 111goo) I.
t o . Prove
that the curvature k of a at to satisfies
*15. Show that the knowledge of the vector function b = b(s) (binormal vector) of a curve a, with nonzero torsion everywhere, determines the curvature k(s) and the absolute value of the torsion z(s) of a. *16. Show that the knowledge of the vector function n = n(s) (normal vector) of a curve a , with nonzero torsion everywhere, determines the curvature k(s) and the torsion z(s) of a.
17. In general, a curve a is called a helix if the tangent lines of a make a constant angle with a fixed direction. Assume that z(s) # 0, s E I, and prove that: *a. a is a helix if and only if k/z = const. *b. a is a helix if and only if the lines containing n(s) and passing through ~ ( s ) are parallel to a fixed plane. *c. a is a helix if and only if the lines containing b(s) and passing through a(s) make a constant angle with a fixed direction. d. The curve a($) =
:( J'
sin 6%) dr, T a
S
cos B(s) ds,
where a2 = b2 f c2, is a helix, and that klz
= bja.
*IS. Let a: 1 3 R3 be a parametrized regular curve (not necessarily by arc length) with k(t) # 0, z(t) # 0, t E I. The curve a is called a Bertrand curve if there exists a curve 8 : I + R3 such that the normal lines of a and 8 at t E I are equal. In this case, 8 is calIed a Bertrand mate of a, and we can write
Prove that
a. r is constant. b. a is a Bertrand curve if and only if there exists a linear relation
where A , B are nonzero constants and k and z are the curvature and torsion of a , respectively. c. If a has more than one Bertrand mate, it has infinitely many Bertrand mates. This case occurs if and only if a is a circular helix.
The Local Canonical form
16. The Local Canonical Form? One of the most effective methods of solving problems in geometry consists of finding a coordinate system which is adapted to the problem. In the study of local properties of a curve, in the neighborhood of the point s, we have a natural coordinate system, namely the Frenet trihedron at s. It is therefore convenient to refer the curve to this trihedron. Let a : 1 R3 be a curve parametrized by arc length without singular points of order 1. We shall write the equations of the curve, in a neighborhood of so, using the trihedron t(s,), n(so), b(so) as a basis for R3. We may assume, without loss of generality, that so = 0, and we shall consider the (finite) Taylor expansion 3
where lim R/s3 = 0, Since ai(0) = t, a"(0) aro
a"'(0)
= (kn)' =
k'n

kur, and
+ kn' = k'n  k2t

kzb,
we obtain
where all terms are computed at s = 0. Let us now take the system Oxyz in such a way that the origin 0 agrees with a(0) and that t = (1,0, O), n = (0, l,O), b = (0, 0, 1). Under these conditions, a(s) = (x(s), y(s), z(s)) is given by
kz Z(S)= s3 6
+ R,,
where R = (R,, R,, R,). The representation ( 2 ) is called the local canonical form of a, in a neighborhood of s = 0. In Fig. 118 is a rough sketch of the projections of the trace of a, for s small, in the tn, tb, and nb planes. ?This section may be omitted on a first reading.
Curves
Projection over the plane t n
Projection over the plane r b
Projection over the
rl b
Figure 118
Below we shall describe some geometrical applications of the local canonical form. Further applications will be found in the Exercises.
A first application is the following interpretation of the sign of the torsion. From the third equation of (1) it follows that if z < 0 and s is sufficiently small, then z(s) increases with s. Let us make the convention of calling the "positive side" of the osculating plane that side toward which b is pointing. Then, since z(0) = 0, when we describe the curve in the direction of increasing arc length, the curve will cross the osculating plane at s = 0, pointing toward the positive side (see Fig. 119). If, on the contrary, z > 0, the curve (described in the direction of increasing arc length) will cross the osculating plane pointing to the side opposite the positive side.
Positive torsion
Negative torsion
Figure 119
The Local Canonical Form
29
The helix of Exercise 1 of Sec. 15 has negative torsion. An example of a curve with positive torsion is the heIix S
S
a sin , C
b
C
obtained from the first one by a reflection in the xz plane (see Fig. 119). Remark. It is also usual to define torsion by b' = zn. With such a definition, the torsion of the helix of Exercise 1 becomes positive.
Another consequence of the canonical form is the existence of a neighborhood J c I of s = 0 such that a(J)is entirely contained in the one side of the rectifying plane toward which the vector n is pointing (see Fig. 118). In fact, since k > 0, we obtain, for s sufficiently small, y(s) 0, and y(s) = 0 if and only if s = 0. This proves our claim. As a last application of the canonical form, we mention the following property of the osculating plane. The osculating plane at s is the limit position of the plane determined by the tangent line at s and the point a(s h ) when h 0. To prove this, let us assume that s = 0. Thus, every plane containing the tangent at s = 0 is of the form z = ey or y = 0. The plane y = 0 is the rectifying plane that, as seen above, contains no points near a(0) (except a(0) itself) and that may therefore be discarded from our considerations. The condition for the plane z = cy to pass through s h is ( s = 0)
>
+
4
+
0, we see that c + 0. Therefore, the limit position of the plane Letting h z(s) = c(h)y(s)is the plane z = 0, that is, the osculating plane, as we wished. +
"1. Let a : I, R3 be a curve parametrized by arc length with curvature k(s) ff 0, s E I. Let P be a plane satisfying both of the following conditions:
1. P contains the tangent line at s. 2. Given any neighborhood J c 1 of s, there exist points of a(J) in both sides of P.
Prove that P is the osculating plane of a at s. 2. Let a : I + R3 be a curve parametrized by arc length, with curvature k(s) # 0, s E I. Show that
30
Curves
*a. The osculating plane at s is the limit position of the plane passing through a(s), a(s h,), a(s h,) when hl, h2 + 0. 6. The limit position of the circle passing through a($, a(s h,), U(s h2) when hl, hZ + 0 is a circle in the osculating plane at s, the center of which is on the line that contains n(s) and the radius of which is the radius of curvature l/k(s); this circle is called the osculating circle at s.
+
+
+
+
3. Show that the curvature k(t) # 0 of a regular parametrized curve a: I + R3 is the curvature at t of the plane curve noa, where n is the normal projection of a
over the osculating plane at
t.
17. Global Properties of Plane Curvesf I n this section we want to describe some results that belong to the global differential geometry of curves. Even in the simple case of plane curves, the subject already offers examples of nontrivial theorems and interesting questions. To develop this material here, we must assume some plausible facts without proofs; we shall try to be careful by stating these facts precisely. Although we want to come back later, in a more systematic way, to global differential geometry (Chap. 5), we believe that this early presentation of the subject is both stimulating and instructive. This section contains three topics in order of increasing difficulty: (A) the isoperimetric inequality, (B) the fourvertex theorem, and (C) the CauchyCrofton formula. The topics are entirely independent, and some or all of them can be omitted on a first reading. A diferentiable function on a closed interval [a, b] is the restriction of a differentiable function defined on an open interval containing [a, b]. A closedplane curve is a regular parametrized curve a : [a, b] + R2 such that a and all its derivatives agree at a and b; that is,
The curve a is simple if it has no further selfintersections; that is, if t,, t2 E [a, b), t, # t,, then a(t,) # a(r2) (Fig. 120). We usually consider the curve a : [0, I ] 4 R2 parametrized by arc length s; hence, I is the length of a. Sometimes we refer to a simple closed curve C, meaning the trace of such an object. The curvature of a will be taken with a sign, as in Remark I of Sec. 15 (see Fig. 120). We assume that a simple closed curve C in the plane bounds a region of this plane that is called the interior of C. This is part of the socalled Jordan tThis section may be omitted on a first reading.
Global Properties of Plane Curves A
(a) A simple closed curve
(b) A (nonsimple) closed curve
Figure 120 Interior of C
(a) A simple closed curve C on a torus T; C bounds no region on T
(b) C is positively oriented
Figure 321
curve theorem (a proof will be given in Sec. 56, Theorem I), which does not hold, for instance, for simple curves on a torus (the surface of a doughnut; see Fig. 121(a)). Whenever we speak of the area bounded by a simple closed curve C, we mean the area of the interior of C. We assume further that the parameter of a simple closed curve can be so chosen that if one is going along the curve in the direction of increasing parameters, then the interior of the curve remains to the left (Fig. 121(b)). Such a curve will be calledpositively oriented.
A. The Isoperimetric fnequafity This is perhaps the oldest global theorem in differential geometry and is related to the following (isoperimetric) problem, Of all simple closed curves in the plane with a given length I, which one bounds the largest area? In this form, the problem was known to the Greeks, who also knew the solution, namely, the circle. A satisfactory proof of the fact that the circle is a solution to the isoperimetric problem took, however, a long time to appear. The main
Curves
32
reason seems to be that the earliest proofs assumed that a solution should exist. It was only in 1870 that K. Weierstrass pointed out that many similar questions did not have solutions and gave a complete proof of the existence of a solution to the isoperimetric problem. Weierstrass' proof was somewhat hard, in the sense that it was a corollary of a theory developed by him to handle problems of maximizing (or minimizing) certain integrals (this theory is called calculus of variations and the isoperimetric problem is a typical example of the problems it deals with). Later, more direct proofs were found. The simple proof we shall present is due to E. Schmidt (1939). For another direct proof and further bibliography on the subject, one m'ay consult Reference [lo] in the Bibliography. We shall make use of the following formula for the area A bounded by a , where t E [a, b] positively oriented simple closed curve a(t) ( ~ ( t )y(t)), is an arbitrary parameter:

y(t)x1(t)dt
=
S:
x(t)y'(t) dt =
1:
(xy'  yx') dt
(1)
Notice that the second formula is obtained from the first one by observing that
since the curve is closed. The third formula is immediate from the first two. To prove the first formula in Eq. (I), we consider initialIy the case of Fig. 122 where the curve is made up of two straightline segments parallel
I I
0
x0
I XI
C X
Figure 122
to the J. axis and two arcs that can be written in the form
Global Properties of Plane Curves
Clearly, the area bounded by the curve is
Since the curve is positively oriented, we obtain, with the notation of Fig. 122,
since x'(t) = 0 along the segments parallel to the y axis. This proves Eq. (1) for this case. To prove the general case, it must be shown that it is possible to divide the region bounded by the curve into a finite number of regions of the above type. This is clearly possible (Fig. 123) if there exists a straight line E in the
Figure 123
plane such that the distance p(t) of a(t) to this line is a fimction with finitely n ~ a n ycritical points (a critical point is a point where pl(t) = 0). The last assertion is,true, but we shall not go into its proof. We shall mention, however, that Eq. (1) can also be obtained by using Stokes' (Green's) theorem in the plane (see Exercise 15).
THEOREM 1 (The Isoperimetric Inequality). Let C be a simple closed plane curve with length I, and let A be the area of the region bounded by C. Then
and equality holds
if and only Y C is a circle.
Curves
34
Proof Let E and E' be two parallel lines which do not meet the closed curve C, and move them together until they first meet C . We thus obtain two parallel tangent lines to C, L and L', so that the curve is entirely contained in
the strip bounded by L and L'. Consider a circle S1 which is tangent to both L and L' and does not meet C. Let O be the Center of ~1 and take a coordinate system with origin at O and the x axis perpendicular to L and L' (Fig, 124).
Figure 124
Parametrize C by arc length, a($) = (xb), ~(s)),so that it is positively oriented and the tangency points of L and L' are s =:= 0 and s = s,, respectively. We can assume that the equation of S1is &(s)= ( W , B(s))
= (~($1, j(s)),
s
E
[O, 11
where 3r is the distance between L and L'. by using Eq. (1) and denoting by 2 the area bounded by S1, we have A
Thus,
=
J
I 0
xy' cis,

A
=
nrz , 
j
I jjxt 0
ds.
Global Properties o f Plane Curves
35
We now notice the fact that the geometric mean of two positive numbers is smaller than or equal to their arithmetic mean, and equality holds if and only if they are equal. It follows that
I* ( A + nr2) < Ilr.
(4)
Therefore, 4nAr2 (12r2,and this gives Eq. (2). Now, assume that equality holds in Eq. (2). Then equality must hold everywhere in Eqs. (3) and (4). From the equality in Eq. (4) it follows that A = nr 2. Thus, I = 2nr and r does not depend on the choice of the direction of L. Furthermore, equality in Eq. (3) implies that (xyf  Yx')~= (xZ
+ Y 2)((x')2 4 (Y')~)
or
(xx'
+fi32= 0;
that is,
Thus, x = kry'. Since r does not depend on, the choice of the direction of L, we can interchange x and y in the last relation and obtain y = &rxf. Thus,
and C is a circle, as we wished.
Q.E.D.
Remark I. It is easily checked that the above proof can be applied to C' curves, that is, curves a(t) = (x(t), ~ ( t ) )t, E:[a, b], for which we require only that the functions x(t), y ( t ) have continuous first derivatives (which, of course, agree at a and b if the curve is closed). Remark 2. The isoperimetric inequality holds true for a wide class of curves. Direct proofs have been found that work as long as we can define arc length and area for the curves under consideration. For the applications, it is convenient to remark that the theorem holds far piecewise Cl curves, that is, continuous curves that are made up by a finite number of C1 arcs. These curves can have a finite number of corners, where the tangent is discontinuous (Fig. 125).
A piecewise C1 curve
Figure 125
Curves
B. The Four Vertex Theorem

We shall need further general facts on plane closed curves. R2 be a plane closed curve given by a(s) = (x(s),y(s)). Let a : [O, I] Since s is the arc length, the tangent vector t(s) = ( x f ( s ) ,y'(s)) has unit length. It is convenient to introduce the tangent indicatrix t : [0,I ] + R2 that is given by t(s) = (xl(s),yf(s));this is a differentiable curve, the trace of which is contained in a circle of radius 1 (Fig. 126). Observe that the velocity vector
Figure 126
of the tangent indicatrix is
d '= ( x"(s),y "(s)) ds = a"(s) =
kn,
where n is the normal vector, oriented as in Remark 2 of Sec. 15, and k is the curvature of a. Let O(s),0 < 8(s) < 2n, be the angle that t(s) makes with the x axis; that is, x'(s) = cos 8(s),yl(s) = sin O(s). Since
'( O(s) = arc tan ~ xl(s)
4,
8 = 8(s) is locally well defined (that is, it is well defined in a small interval about each s) as a differentiable function and dt = (cos d 8, sin 8 )
ds
ds
= 8'(
sin 6,cos 8 ) = O'n.
Global Properties o f Plane Curves

37
This means that Of(s)= k(s) and suggests defining a global differentiable function 8 : [O, I ] R by 8(s) =
Is
k ( s ) ds.
0
Since
this global function agrees, up to constants, with the previous locally defined 8. Intuitively, 8(s) measures the total rotation of the tangent vector, that is, the total angle described by the point t(s) on the tangent indicatrix, as we run the curve a from 0 to s. Since a is closed, this angle is an integer multiple I of 2n; that is,
The integer I is called the rotation index of the curve a. In Fig. 127 are some examples of curves with their rotation indices. Observe that the rotation index changes sign when we change the orientation of the curve. Furthermore, the definition is so set that the rotation index of a positively oriented simple closed curve is positive. An important global fact about the rotation index is given in the following theorem, which will be proved later in the book (Sec. 56, Theorem 2). T H E T H E O R E M OF T U R N I N G T A N G E N T S . The rotation index of a simple closed curve is & 1, where the sign depends on the orientation of the curve.
A regular, plane (not necessarily closed) curve a: [a, b] R2 is convex if, for all t E [a, b],the trace u([a,b ] )of a lies entirely on one side of the closed halfplane determined by the tangent line at t (Fig. 128). A vertex of a regular plane curve a : [a, b] + R2 is a point t E [a, b] where k'(t) = 0. For instance, an ellipse with unequal axes has exactly four vertices, namely the points where the axes meet the ellipse (see Exercise 3). It is an interesting global fact that this is the least number of vertices for all closed convex curves. +
T H E O R E M 2 (The FourVertex Theorem). A simple closed convex curve has at Ienst fozrr vertices.
Before starting the proof, we need a lemma. L E M M A . Let a : [0, 11 + R2 be a plane closed curve parametrized by arc length and let A, B, and C be arbitrary real numbers. Then
38
Curves
Tangent indicatrix
Figure 127
G/oba/ Properties o f P/ane Curves
39
Convex curves
Nonconvex curves
Figure 128
where thefunctions x the curvature of a .
= x(s),
y
= y(s)
are given by a(s)
= (x(s),
y(s)), and k is
Proof of the Lemma. Recall that there exists a differentiable function 0 : [0, I] + R such that x'(s) = cos 8, y'(s) = sin 0. Thus, k(s) = 8'(s) and
Therefore, since the functions involved agree at 0 and I,
Q.E.D. Proof of the Theorem. Parametrize the curve by arc length, a: LO, I] + RZ. Since k = k(s) is a continuous function on the closed interval [0,1], it reaches a maximum and a minimum on [0, I] (this is a basic fact in real functions; a proof can be found, for instance, in the appendix to Chap. 5, Prop. 10). Thus, a has at least two vertices, a(sl) = p and a(s2) = q. Let L be the straight line passing through p and q, and let P and y be the two arcs of C which are determined by the points p and q.
40
Curves
We claim that each of these arcs lies on a definite side of L. Otherwise, it meets L in a point r distinct from p and q (Fig. 129(a)). By convexity, and since p, q, r are distinct points on C, the tangent line at the intermediate point, say p, has to agree with L. Again, by convexity, this implies that L is tangent to C at the three points p, q, and r. But then the tangent to a point
Figure 129
near p (the intermediate point) will have q and r on distinct sides, unless the whole segment rq of L belongs to C (Fig. 129(b)). This implies that k = 0 a t p and q. Since these are points of maximum and minimum for k, k = 0 on C, a contradiction. C = 0 be the equation of L. If there are no further Let Ax + By vertices, kl(s) keeps a constant sign on each of the arcs P and y. We can then arrange the sign of all the coefficients A, B, C SO that the integral in Eq. (5) is positive. This contradiction shows that there is a third vertex and that kl(s) changes sign on P or y, say, on P. Since p and q are points of maximum and minimum, kl(s) changes sign twice on P. Thus, there is a fourth vertex. Q.E.D.
+
The fourvertex theorem has been the subject of many investigations. The theorem also holds for simple, closed (not necessarily convex) curves, but the proof is harder. For further literature on the subject, see Reference [lo]. Later (Sec. 56, Prop. 1) we shall prove that aplalze closed curve is convex ifand only if it is simple and can be oriented so that its curvature is positive or zero. From that, and the proof given above, we see that we can reformulate the statement of the fourvertex theorem as follows. The curvature function of a closed convex curve is (nonnegative and) either constant or else has at Zenst two maxima and two minima. It is then natural to ask whether such curvature functions do characterize the convex curves. More precisely, we can ask the following question. Let k: [a, b] + R be a diferentiable nonnegative function such that k agrees, with all its derivatives, at a and b. Assume that k is either
G/oba/ Properties of P/ane Curves
47
constant or else has at least two maxima and two minima. Is there a simple closed curve a : [a, b] + R2 such that the curvature of a at t is k ( t ) ? For the case where k ( t ) is strictly positive, H. Gluck answered the above question affirmatively (see H. Gluck, "The Converse to the Four Vertex Theorem," L'Ens~ignementMnthkmatigue T. XVII, fasc. 34 (1 971), 295309). His methods, however, do not apply to the case k 2 0.
C. The CauchyCrofton Formula Our last topic in this section will be dedicated to finding a theorem which, roughly speaking, describes the following situation. Let C be a regular curve in the plane. We look at all straight lines in the plane that meet C and assign which is the number of its interesection points to each such line a rnz~lfiplicit~ with C (Fig. 130).
Figure 130. n is the multiplicity of the corresponding straight line.
Figure 131 L is determined by g and 8.
We first want to find a way of assigning a measure t o a given subset of straight lines in the plane. It should not be too surprising that this is possible. After all, we assign a measure (area) to point subsets of the plane. Once we realize that a straight line can be determined by two parameters (for instance, p and 0 in Fig. 131), we can think of the straight lines in the plane as points in a region of a certain plane. Thus, what we want is to find a "reasonable" way of measuring "areas" in such a plane. Raving chosen this measure, we want to apply it and find the measure of the set of straight lines (counted with multiplicities) which meet C. The result is quite interesting and can be stated as follows. THEOREM 3 (The CauchyCrofton Formula). Let C be a regular plane curve with length I. The measure of the set of straight lines (counted with mu,ltiplicities) which meet C is equal to 21.
Curves
42
Before going into the proof we must define what we mean by a reasonable measure in the set of straight lines in the plane. First, let us choose a convenient system of coordinates for such a set. A straight line L in the plane is determined by the distance p 0 from L to the origin 0 of the coordinates and by the angle 0, 0 < 8 < 27r7which a halfline starting at 0 and normal to L makes with the x axis (Fig. 13 1). The equation of L in terms of these parameters is easily seen to be
>
x cos 8
+ y sin 0 = p .
Thus we can replace the set of all straight lines in the plane by the set
We will show that, up to a choice of units, there is only one reasonable measure in this set, To decide what we mean by reasonable, let us look more closely at the usual measure of areas in R 2 . We need a definition. A rigid motion in R2 is a map F: R2 R2 given by (2, y) + (x, y), where (Fig. 132) +
Figure 132
x = a +xcosp 7sinp
y
=b $
2 sin q
+ j cos p.
Now, to define the area of a set S c R2 we consider the double integral
that is, we integrate the "element of area" dx dy over S. When this integral exists in some sense, we say that S is measurable and define the area of S as the vaIue of the above integral. From now on, we shall assume that all the integrals involved in our discussions do exist. Notice that we could have chosen some other element of area, say, xy2 dx dy. The reason for the choice of dx dy is that, up to a factor, this is
43
Global Properties o f Plane Curves
the only element of area that is invariant under rigid motions. More precisely, we have the following proposition.
PROPOSITION 1. Let f(x, y) be a continuousftinction dqfined in R2. For any set S c R2, dejne the area A of S by
(of course, we are considering only those sets for which the above integral exists). Assume that A is invariant under rigid motions; that is, if S is any set and = F  ' ( S ) , where F is the rigid motion (6), we have
Then f(x, y)
= const.
Proof. We recall the fcrmula for change of variables in multiple integrals (Buck, Advanced Calculus, p. 301, or Exercise 15 of this section):
Here, x = x(2, j ) , y = y(2, j ) are functions with continuous partial  derivatives which define the transformation of variables T : R2 , R2, S = T  I ( S ) , and ldx
dxl
is the Jacobian of the transformation T. In our particular case, the transformation is the rigid motion (6) and the Jacobian is cos sin p,
sin 9 cos y,
By using this fact and Eq. (7), we obtain
Since this is true for all S, we have
44
Curves
We now use the fact that for any pair of points (x, y), (2, j ) in R2 there exists a rigid motion F such that F(2, j ) = (x, y ) . Thus,
and f (x, y )
= const.,
Q.E.D.
as we wished.
Remark 3. The above proof rests upon two facts: first, that the Jacobian of a rigid motion is 1, and, second, that the rigid motions are transitive on points of the plane; that is, given two points in the plane there exists a rigid motion taking one point into the other.
With these preparations, we can finally define a measwre in the set 2. We first observe that the rigid motion (6) induces a transformation on 2. In fact, Eq. (6) maps the line x cos 0 y sin 8 = p into the line
+
i cos(0  9)
+ j sin(0  9)
=p 
a cos 0  b sin 0.
This means that the transformation induced by Eq. (6) on 6: is p'
=p 
a cos 0
8=8~.

b sin 0,
It is easily checked that'the Jacobian of the above transformation is 1 and that such transformations are also transitive on the set of lines in the plane. We then define the measure of a set S c G as
In the same way as in Prop. 1, we can then prove that this is, up to a constant factor, the only measure on 6: that is invariant wnder rigid motions. This measwre is, therefore, as reasonable as it can be. We can now sketch a proof of Theorem 3. Sketch of Proof of Theorem 3. First assume that the curve C i s a segment of a straight line with length I. Since our measwre is invariant wnder rigid motions, we can assume that the coordinate system has its origin 0 in the middle point of C and that the x axis is in the direction of C. Then the measwre of the set of straight lines that meet C i s (Fig. 133) I cos 0 1 (1,'21
/~dpdO=~:*([o
dp)dO
1
2R
0
1 Zcos0d~21.
Global Properties o f Plane Curves
Figure 133
Next, let C be a polygonal line composed of a finite number of segments Ciwith length li (C li = I). Let n = n(p, 8) be the number of intersection points of the straight line ( p , 8) with C. Then, by summing up the results for each segment Ci, we obtain
which is the CauchyCrofton formula for a polygonal line. Finally, by a limiting process, it is possible to extend the above formula to any regular curve, and this will prove Theorem 3. Q.E.D. It should be remarked that the general ideas of this topic belong to a branch of geometry known under the name of integral geometry. A survey of the subject can be found in L. A. Santal6, "Integral Geometry," in Studies in Global Geometry and Analysis, edited by S. S. Chern, The Mathematical Association of America, 1967, 1471 93. The CauchyCrofton formula can be used in many ways. For instance, if a curve is not rectifiable (see Exercise 9, Sec. 13) but the lefthand side of Eq. (8) has a meaning, this can be used to define the "length" of such a curve. Equation (8) can also be used to obtain an efficient way of estimating lengths of curves, Indeed, a good approximation for the integral in Eq. (8) is given as follows.? Consider a family of parallel straight lines such that two consecutive lines are at a distance r. Rotate this family by angles of n/4,2n/4, 37114 in order to obtain four families of straight lines. Let n be the number of intersection points of a curve C with all these lines. Then
TI want to thank Robert Gardner for suggesting this application and the example that follows.
Curves
is an approximation to the integral 2
1 1 ndpdt9
= length
of C
and therefore gives an estimate for the length of C. To have an idea of how good this estimate can be, let us work out an example. Example. Figure 134 is a drawing of an electron micrograph of a circular D N A molecule and we want to estimate its length. The four families
Figure 134. Reproduced from H. Ris and B. C. Chandler, Cold Spring Harbor Symp. Quarzt. Biol. 28, 2 (1963), with permission.
of straight lines at a distance of 7 millimeters and angles of n/4 are drawn over the picture (a more practical way would be to have this family drawn once and for all on transparent paper). The number of interesection points is found to be 153. Thus,
Since the reference line in the picture represents 1 micrometer (= meter) and measures, in our scale, 25 millimeters, r = and thus the length of this D N A molecule, from our values, is approximately
y,
(275) 
60 
16.6 micrometers.
The actual value is 16.3 micrometers.
Global Properties of Plane Curves
*l. Is there a simple closed curve in the plane with length equal to 6 feet and bounding an area of 3 square feet? *2. Let ~ z b aesegment of straight line and let I > length of AB. Show that the B bounds curve C joining A and B, with length I, and such that together with A the largest possible area is an arc of a circle passing through A and B (Fig. 135),
Figure 135
Figure 136
3. Compute the curvature of the ellipse
x
=a
cos t,
y
=
b sin t,
t
E
[O, 2n], a + 6,
and show that it has exactly four vertices, namely, the points (a, 0), (a, 0), (0Yb)Y (0, 6). *4. Let C be a plane curve and let T be the tangent line at a point p E C. Draw a lineL parallel to the normal line a t p and at a distance d ofp (Fig. 136). Let h be the length of the segment determined on L by C and T (thus, h is the "height" of C relative to T). Prove that 2h 1 k(p) 1 = lim z2y d0
where k(p) is the curvature of C at p. *5. If a closed plane curve C is contained inside a disk of radius ryprove that there exists a point p E C such that the curvature k of C at p satisfies [ k ] 2 l/r.
6. Let ~ ( s )s,
E
[O, I ] be a closed convex plane curve positively oriented. The curve
where r is a positive constant and n is the normal vector, is called a parallel curve to a (Fig. 137). Show that a. Length of P = length of a f 2zr. b. A@) = A(U) + rZ nv2.
+
+
c* kB(s) = ka(~)j(l r).
Curves
Figure 137
For (a)(c), A( ) denotes the area bounded by the corresponding curve, and k,, k p are the curvatures of a and p, respectively. + R2 be a plane curve defined in the entire real line R. Assume that a does not pass through the origin 0 = (0,O) and that both
7. Let a: R
lim I a(t) [
= oo
and
t+m
lim I ~ ( t1 )= oo. t+ m
a. Prove that there exists a point to E R such that l&(to)l5 la(t)l for all t
E
R.
b. Show, by an example, that the assertion in part a is false if one daes not asI a(t) 1 = co and lim,, I a(t) I = a. sume that both Iim,,,,
8. *a. Let ~ ( s )s, E [O, I ] , be a plane simple closed curve. Assume that the curvature k(s) satisfies 0 < k(s) _( c, where c is a constant (thus, a is less curved than a circle of radius Ilc). Prove that length of
2n a 27 
b. In part a replace the assumption of being simple by N." Prove that
length of a
"a has rotation index
2ziV >. C
*9. A set K c Rz is convex if given any two points p, q E K the segment of straight line pqis contained in K (Fig. 138). Prove that a simple closed convex curve bounds a convex set. 10. Let C be a convex plane curve. Prove geometrically that C has no selfintersections.
*lI. Given a nonconvex simple closed plane curve C, we can consider its convex hull N (Fig. 139), that is, the boundary of the smallest convex set containing the interior of C. The curve H i s formed by arcs of C and by the segments of the tangents to C that bridge "the nonconvex gaps" (Fig. 139). It can be proved that H i s a C1 closed convex curve. Use this to show that, in the isoperimetric problem, we can restrict ourselves to convex curves.
Global Properties o f Plane Curves
Figure 138
Figure 139
*12. Consider a unit circle S 1 in the plane. Show that the ratio M l [ M 2= 3, where
M , is the measure of the set of straight lines in the plane that meet Sl and M 1 is the measure of all such lines that determine in S 1 a chord of length > Intuitively, this ratio is the probability that a straight line that meets Sl determines in S 1 a chord longer than the side of an equilateral triangle inscribed in S1 (Fig. 140).
n.
Figure 141
Figure 140
13. Let C be an oriented plane closed curve with curvature k > 0. Assume that C has at least one point p of selfintersection. Prove that
a. There is a point p' tangent at p.
E
C such that the tangent line T' at p' is paralIeI to some
b. The rotation angle of the tangent in the positive arc of C made up by pp'p is > n (Fig. 141).
c. The rotation index of C is
2 2.
50
Curves
14. a. Show that if a straight line L meets a closed convex curve C, then either L is tangent to C or L intersects C in exactly two points. b. Use part a to show that the measure of the set of lines that meet C (without multiplicities) is equal to the length of C. 15. Green's theorem in the plane is a basic fact of calculus and can be stated as follows. Let a simple closed pIane curve be given by a(t) = (x(t), y(t)), t E [a, b]. Assume that a is positively oriented, let C be its trace, and let R be the interior of C. Let p = p(x, y), q = q(x, y) be real functions with continuous partial derivatives P,, P,, q,, qy. Then
where in the second integral it is understood that the functions p and q are restricted to a and the integral is taken between the limits t = a and t = b. In parts a and b below we propose to derive, from Green's iheorem, a formula for the area of R and the formula for the change of variables in double integrals (cf. Eqs. (1) and (7) in the text).
a. Set q
=x
and p
A(R)
=
y in Eq. (9) and conclude that
=j/
dx dy
R
1
=
2
j: 2 (x(t)
dx
 y(tjJi)
dt,
b. Let f ( x , y ) be a real function with continuous partiaI derivatives and T: R 2 + R 2 be a transformation of coordinates given by the functions x = x(u, v), y = y(u, v), which also admit continuous partial derivatives. Choose in Eq. (9) p = 0 and q SO that q, =f.Apply successively Green's theorem, the map T, and Green's theorem again to obtain
Show that
Put that together with the above and obtain the transformation formula for double integrals :
2
Regular
In this chapter, we shall begin the study of surfaces. Whereas in the first chapter we used mainly elementary calculus of one variable, we shall now need some knowledge of calculus of several variables. Specifically, we need to know some facts about continuity and differentiability of functions and maps in R2 and R 3 . What we need can be found in any standard text of advanced calculus, for instance, Buck Advanced Calculus; we have included a brief review of some of this material in an appendix to Chap. 2. In Sec. 22 we shall introduce the basic concept of a regular surface in R3. In contrast to the treatment of curves in Chap. 1, regular surfaces are defined as sets rather than maps. The goal of Sec. 22 is to describe some criteria that are helpful in trying to decide whether a given subset of R3 is a regular surface. In Sec. 23 we shall show that it is possible to define what it means for a function on a regular surface to be differentiable, and in Sec. 24 we shall show that the usual notion of differential in R 2 can be extended to such functions. Thus, regular surfaces in R3 provide a natural setting for twodimensional calculus. Of course, curves can also be treated from the same point of view, that is, as subsets of R3 which provide a natural setting for onedimensional calculus. 11e shall mention them briefly in Sec. 23. Sections 22 and 23 are crucial to the rest of the book. A beginner may find the proofs in these sections somewhat difficult. If so, the proofs can be omitted on a first reading.
52
Regular Surfaces
In Sec. 25 I\e shall introduce the first fundamental form, a natural instrument to treat metric questions (lengths of curves, areas of regions, etc.) on a regular surface. This will become a very important issue when we reach Chap. 4. Sections 26 through 28 are optional on a first reading. In Sec. 26, we shall treat the idea of orientation on regular surfaces. This will be needed in Chaps. 3 and 4. For the benefit of those who omit this section, we shall review the notion of orientation at the beginning of Chap. 3.
22. Regular Surfaces; /nverse /mages of Regu/ar Va/uest In this section we shall introduce the notion of a regular surface in R 3 .Roughly speaking, a regular surface in R3 is obtained by taking pieces of a plane, deforming them, and arranging them in such a way that the resulting figure has no sharp points, edges, or selfintersections and so that it makes sense to speak of a tangent plane at points of the figure. The idea is to define a set that is, in a certain sense, twodimensional and that also is smooth enough so that the usual notions of calculus can be extended to it. By the end of Sec. 24, it should be completely clear that the following definition is the right one. DEFINITION 1. A subset S c R3 is a regular surface if, for each p E S, there exists a neighborhood V in R3 and a map x: U + V n S of an open set U c R2 onto V. n S c R 3 such that (Fig. 21)
1. x is differentiable. This means that if we write
thefunctions x(u, v), y(u, v), z(u, v) have continuouspartial derivatives of all orders in U.
2. x is a homeomorphism. Since x is continuous by condition 1 , this means that x has an inverse x l: V n S + U which is continuous; that is, x is the restriction of a continuous map F : W c R3 R2 4
de$ned on an open set W containing V n S.
3. (The regularity condition.) For each q dx,: R2 + R3 is onetoone.$
E
U , the differerztial
We shall explain condition 3 in a short while. ?Proofs in this section may be omitted on a first reading. $In italic context, letter symbols are roman so they can be distinguished from the surrounding text.
Figure 21
The mapping x is called aparametrization or a system of(loca1) coordinates in (a neighborhood of) p. The neighborhood V n S of p in S is called a coordinate neighborhood. To give condition 3 a more familiar form, let us compute the matrix of the linear map dx, in the canonical bases el = (1, 0), e, = (0, 1) of R2 with coordinates (u, v) and f, = ( l , O , O ) , f . = (0, 1, 0), f, = (0,0, 1) of R3, with coordinates (x, y, z). Let q = (u,, v,). The vector el is tangent to the curve u (u, v,) whose image under x is the curve t
This image curve (called the coordinate curve v = v,) lies on S and has at x(q) the tangent vector (Fig. 22)
(d", d y , dl) = du du du
g,
where the derivatives are computed at (u,, 0,) and a vector is indicated by its components in the basis { f ,,f ,, f ,). By the definition of differential (appendix to Chap. 2, Def. I),
v

Similarly, using the coordinate curve u = u , (image by x of the curve (u,, v)), we obtain
Thus, the matrix of the linear map dx, in the referred basis is
Condition 3 of Def. 1 may now be expressed by requiring the two column vectors of this matrix to be linearly independent; or, equivalently, that the vector product dxldu A dxldv # 0; or, in still another way, that one of the minors of order 2 of the matrix of dx,, that is, one of the Jacobian determinants
be different from zero a t q.
Remark 1. Definition 1 deserves a few comments. First, in contrast to our treatment of curves in Chap. 1, we have defined a surface as a subset S of R3, and not as a map. This is achieved by covering S with the traces of parametrizations which satisfy conditions 1, 2, and 3. Condition 1 is very natural if we expect to do some differential geometry on S. The onetooneness in condition 2 has the purpose of preventing self
55
Regular Surfaces
intersections in regular surfaces. This is clearly necessary if we are to speak about, say, the tangent plane at a point p E S (see Fig. 23(a)). The continuity of the inverse in condition 2 has a more subtle purpose which can be fully understood only in the next section. For the time being, we shall mention that this condition is essential to proving that certain objects defined in terms of a parametrization do not depend on this parametrization but only on the set S itself. Finally, as we shall show in Sec. 2.4, condition 3 will guarantee the existence of a "tangent plane" at all points of S (see Fig. 23(b)).
Figure 23. Some situations to be avoided in the definition of a regular surface.
Example 1. Let us show that the unit sphere
is a regular surface. We first verify that the map x, : U c R2 , R3 given by
+
[!.here R2 = ((x, y, z) E R3; z = 03 and U = {(x, y) E R 2 ;x2 y 2 < 11, is 3. parametrization of S 2 . Observe that x, (U) is the (open) part of S 2 above the . ~ plane. y Since x2 y2 < I , the function +dl  (xZ $ y2) has continuous partial derivatives of all orders. Thus, x, is differentiable and condition 1 holds. Condition 3 is easily verified, since
+
To check condition 2, we observe that x, is onetoone and that xy1 is the restriction of the (continuous) projection n(x, y, z) = (x, y) to the set x,(U). Thus, x;' is continuous in x,(U).
56
Regular Surfaces
We shall now cover the whole sphere with similar parametrizations as follows. We define x,: U c R2 R3 by +
check that x2 is a parametrization, and observe that x , ( U ) u x 2 ( U ) covers S 2 minus the equator
Then, using the xz and zy planes, we define the pmametrizations x,(x,
2) =
x ~ ( xz) , = ( x , dl
XAY,z) = (+Ji x ~ ( YZ) ,
+ z2),z), ( x 2 + z2),z), ( y 2 + z2), y, z), ( y 2 + z2),y, z),
( x , +dl  ( x 2
(dl


which, together with x , and x,, cover S completely (Fig. 24) and show that S 2 is a regular surface.
Figure 24
For most applications, it is convenient to relate parametrizations to the geographical coordinates on S 2 . Let V = {(B, p) ; 0 < 8 < n, 0 < q < 2x1 and let x : V R3 be given by x(8, p)
=
(sin 8 cos p, sin 8 sin q, cos 8).
Clearly, x ( V ) c S2. We shall prove that x is a parametrization of S 2 . 8 is
Regu/ar Surfaces
usually called the colatitude (the complement of the latitude) and p, the longitude (Fig. 25). It is clear that the functions sin 8 cos p, sin 8 sin p, cos 8 have continuous partial derivatives of all orders; hence, x is differentiable. Moreover, in order that the Jacobian determinants
w , p,) '(~7 z, w , p)
= cos B sin 8,
=
sin2 e cos p,
d(x9 1'  sin2 8 sin p, w ,
vanish simultaneously, it is necessary that cos2 8 sin2 8
+ sin48 cos2p + sin4 8 sin2 p = sin28 = 0.
This does not happen in V, and so conditions 1 and 3 of Def. 1 are satisfied. Next, we observe that given (x, y, z) t S2 C, where C is the semicircle
0 is uniquely determined by 8 = cosI
z, since 0
< 8 < n. By knowing 8,

we find sin p and cos p from x = sin 0 cos p, y sin 8 sin p, and this determines p uniquely (0 < p < 2n). It follows that x has an inverse x'. To complete the verification of condition 2, we should prove that xI is continuous. However, since we shall soon prove (Prop. 4) that this verification is not necessary provided we already know that the set.S is a regular surface, we shall not do that here. We remark that x(V) only omits a semicircle of S2 (including the two poles) and that S2can be covered with the coordinate neighborhoods of two parametrizations of this type.
58
Regular Surfaces
In Exercise 16 we shall indicate how to cover S2 with another useful set of coordinate neighborhoods. Example 1 shows that deciding whether a given subset of R3 is a regular surface directly from the definition may be quite tiresome. Before going into further examples, we shall present two propositions which will simplify this task. Proposition 1 shows the relation which exists between the definition of a regular surface and the graph of a function z = f (x, y). Proposition 2 uses the inverse function theorem and relates the definition of a regular surface with the subsets of the form f(x,y, z) = constant.
PROPOSITION 1. If f: U R is a differentiable function in an open set U of RZ,then the graph off, that is, the subset of R3given by (x, y, f(x, y)) for (x, y) E U, is a regular surface. 4
Proof: It suffices to show that the map x : U
R3 given by
is a parametrization of the graph whose coordinate neighborhood covers every point of the graph. Condition 1 is clearly satisfied, and condition 3 also offers no difficulty since d(x, y)/d(u, v) . I. Finally, each point (x, y, Z) of the graph is the image under x of the unique point (u, v) = (x, y) E U. x is therefore onetoone, and since x  l is the restriction to the graph o f f of the (continuous) projection of R3 onto the xy plane, xI is continuous. Q.E.D. Before stating Prop. 2, we shall need a definition.

DEFINITION 2. Given a diflerentiable map F : U c Rn RmdeJined in an open set U ofRfl we say that p E Uis a critical point ofF fthe direrential dF,: Rn + Rm is not a surjective (or onto) mapping. The image F(p) E Rm of a critical point is called a critical value of F. A point of Am which is not a critical value is called a regular value of F. The terminology is evidently motivated by the particular case in which f :U cR R is a realvalued function of a real variable. A point x, E U is critical iff '(x,) = 0, that is, if the differential df,, carries all the vectors in R to the zero vector (Fig. 26). Notice that any point a 6 f (U) is trivially a regular value off. If f:U c R3 R is a differentiable function, then dS'p applied to the vector (1, 0,O) is obtained by calculating the tangent vector at f(p) to the curve +
4
X
f CX,
+
Yo, zo).
Regu/ar Surfaces
Figure 26
It follows that
and analogously that
We conclude that the matrix of given by
df,
df, in the basis (I, 0, 0),
(0, 1, 0), ((40, 1) is
= (fX9fY9fZ).
Note, in this case, that to say that df, is not surjective is equivalent to saying that f, =f y =f, = 0 a t p. Hence, a E f(U) is a regular value of f : U c R3 ,R if and only iff,, fy, and f, do not vanish simultaneously at any point in the inverse image f '(a) = {(x, y, z ) E U: f(x,y, z )
=
a).
PROPOSITION 2. If f: U c R3 ' R is a diferentiable function and a E f(U) is a regular value off, then f  (a) is a reghlar surface in R3. Proof. Let p = (x,, y o , z,) be a point o f f '(a). Since a is a regular value off, it is possible to assume, by renaming the axis if necessary, that f, # 0 at p. We define a mapping F: U c R3 + R3 by
and we indicate by (u, v, t) the coordinates of a point in R3 where F takes its \alues. The differential of F a t p is given by
Regular Surfaces
whence
We can therefore apply the inverse function theorem (cf. the appendix to Chap. 2), which guarantees the existence of neighborhoods V of p and W of W is invertible and the inverse Fl : W V is differF(p) such that F: V entiable (Fig. 27). It follows that the coordinate functions of F  ' , i.e., the functions


Figure 27
are differentiable. In particuIar, z = g(u, v, a) = h(x, y) is a differentiable function defined in the projection of V onto the xy plane. Since
we conclude that the graph of h is f l(a) n V. By Prop. 1, f l(a) n V is a coordinate neighborhood of p. Therefore, every p E f '(a) can be covered by a coordinate neighborhood, and so f '(a) is a regular surface. Q.E.D.
Remark 2. The proof consists essentially of using the inverse function theorem "to solve for z" in the equation f (x, y, z ) = a, which can be done in a neighborhood o f p if f,(p) # 0. This fact is a special case of the general implicit function theorem, which follows from the inverse function theorem and is, in fact, equivalent to it.
Regular Surfaces
Example 2. The ellipsoid
is a regular surface. In fact, it is the set f '(O) where
is a differentiable function and 0 is a regular value off. This follows from the fact that the partial derivatives f, = 2x/a2, f, = 2y/b2, f, = 2z/c2 vanish simultaneously only at the point (0, 0, O), which does not belong to f l(O). This example includes the sphere as a particular case (a = b = c = I). The examples of regular surfaces presented so far have been connected subsets of R3. A surface S c R3 is said to be connected if any two of its points can be joined by a continuous curve in S. In the definition of a regular surface we made no restrictions on the connectedness of the surfaces, and the following example shows that the regular surfaces given by Prop. 2 may not be connected.
+
ExampIe 3. The hyperboloid of two sheets x2  y 2 z2 = 1 is a regular surface, since it is given by S = f (0), where 0 is a regular value of f (x, y, z ) = x2  y2 z2  1 (Fig. 28). Note that the surface S is not connected; that is, given two points in two distinct sheets (z > 0 and z < 0) it is not possible to join them by a continuous curve a(t) = (x(t), y(t), z(t)) contained in the surface; otherwise, z changes sign and, for some to, we have :(to) = 0, which means that a(t,) $ S.
+
Incidentally, the argument of Example 3 may be used to prove a property of connected surfaces that we shall use repeatedly. I f f : S c R3 R is a nonzero continuous function deJined on a connected surface S, then f does not change sign on S. To prove this, we use the intermediate value theorem (appendix to Chap. 2, Prop. 4). Assume, by contradiction, that f (p) > 0 and f (q) < 0 for some points p, q E S. Since S is connected, there exists a continuous curve x : [a, b] +S with %(a) = p, a(b) = q. By applying the intermediate value theorem to the continuous function f 0 cc : [a, b] +R, we find that there exists c E (a, b) with f 0 a(c) = 0; that is, f is zero at ~ ( c ) a, contradiction. 4
Example 4. The torus T is a "surface" generated by rotating a circle S1 of radius r about a straight line belonging to the plane of the circle and a t a distance a > r away from the center of the circle (Fig. 29).
Regular Surfaces
Let S1be the circle in the yz plane with its center in the point (0, a, 0). z2 = r 2 , and the points of the figure T obThen S1 is given by (y tained by rotating this circle about the z axis satisfy the equation
+
Therefore, T is the inverse image of r2 by the function
This function is differentiable for (x,y) # (0, 0), and since
df = 2r, dz
df  ~
6
a f  2 x ( J x 2 dx
Y ( N @ T~
dx2
+ y2
a),
++y y 2 a ) ,
Jx2
r2 is a regular value off. It follows that the torus T is a regular surface. Proposition 1 says that the graph of a differentiable function is a regular surface. The following proposition provides a local converse of this; that is, any regular surface is locally the graph of a differentiable function.
Figure 29
PROPOSITION 3. Let S c R3be a regular surface and p E S. Then there exists a neighborhood V of p in S such that V is the graph of a diferentiable function which has one of the following three forms: z = f(x, y), y = g(x, z), x = h(y, z). Proof: Let x : U c R2 t S be a parametrization of S in p, and write ~ ( uv ,) = ( ~ ( uv), , y(u, v), Z ( U , v)), (u, V ) E U. By condition 3 of Def. 1, one of the Jacobian determinants
is not zero at x ' ( p ) = q. Suppose first that (d(x,y)/d(u, v))(q)# 0, and consider the map n 0 x : U R 2 , where n is the projection n(x, y, z ) = ( x , y). Then n 0 x(u, v ) = ( ~ ( 2 1 ,v), y(u, v)), and, since (d(x,y)/d(u, v))(q)# 0 , we can apply the inverse function theorem to guarantee the existence of neighborhoods V, of q, V, of n 0 ~ ( qsuch ) that n 0 x maps V, diffeomorphically onto V, (Fig. 210). It follows that n restricted to x(V,) = Vis onetoone and that there is a differentiV 1 .Observe that, since x is a homeomorphism, able inverse (n 0 x)I : V 2 V is a neighborhood of p in S. Now, if we compose the map (n 0 x) : (x,y ) , (u(x,y), V ( X , y ) ) with the function (u, v ) + z(u, v), we find that V is the graph of the differentiable function z = z(u(x, y), v(x, y)) = f ( x , y), and this settles the first case. The remaining cases can be treated in the same way, yielding x = h(y, z) Q.E.D. and y = g(x, z).


The next proposition says that if we already know that S is a regular surface and we have a candidate x for a parametrization, we do not have to
Regular Surfaces
Figure 210
check that x1 is continuous, provided that the other conditions hold. This remark was used in Example 1.
S be a point of a regular suyface S and let E x(U) such that conditions 1 and 3 of DeJ 1 hold. Assume that x is onetoone. Then X' is continuous. PROPOSITION 4. Let p
E
x: U c R2 + R3 be a map with p
Proof. The first part of the proof is similar to the proof of Prop. 3. Write X(U,V) = (x(u, v), y(u, v), z(u, v)), (u, v) E U, and let q E U. By conditions 1 and 3 we may assume, interchanging the coordinate axis of R3 if necessary, that (d(x, y)/d(u, v))(q) + 0.Let n : R3 R2 be the projection n(x, y, 2). = (x, y). From the inverse function theorem, we obtain neighborhoods V, of q in U and V2 of n 0 x(q) in RZsuch that n 0 x maps V , diffeomorphically onto v2.
Assume now that x is onetoone. Then, restricted to x(V,),
(see Fig. 210). Thus, xL1is continuous as a composition of continuous maps. Since q is arbitrary, xI is continuous in x(U). Q.E.D.
Example 5. The onesheeted cone C, given by
is not a regular surface. Observe that we cannot conclude this from the fact alone that the "natural" parametrization
is not differentiable; there could be other pararnetrizations satisfying Def. 1. To show that this is not the case, we use Prop. 3. If C were a regular surface, it nould be, in a neighborhood of (0, 0, 0) E C, the graph of a
65
Regular Surfaces
differentiable function having one of three forms: y = h(x, z), x = g(y, z), 1 = f (x, y ) . The two first forms can be discarded by the simple fact that the projections of C over the xz and yz planes are not onetoone. The last form would have to agree, in a neighborhood of (O,0, 0), with z = +,J'x2 f y2. Since z = + d m is not differentiable at (0, 0), this is impossible. Example 6. A parametrization for the torus T of Example 4 can be given by (Fig. 29) X(U,v)
=
((r cos u
+ a ) cos v, (r cos u + a ) sin v, r sin u),
where 0 < u < 2n, 0 < v < 2n. Condition 1 of Def. I is easily checked, and condition 3 reduces to a straightforward computation, which is left as an exercise. Since we know that T is a regular surface, condition 2 is equivaIent, by Prop. 4, to the fact that x is onetoone. To prove that x is onetoone, we first observe that sin u = zlr; also, if < a, then n/2 (u ( 3n/2, and if ,J'x2 y 2 2 a, then either Jx2+y 0 v). Clearly, x( U ) c P. Is x a parametrization of
P?
?Those who have omitted the proofs in this section should also omit Exercises 1719.
Regular Surfaces
66
6. Give another proof of Prop. 1 by applying Prop. 2 to h(x, y, z) = f(x, y)  z.
+
7. Let f(x, y, z) = (x + y z a. Locate the critical points and critical values off. b. For what values of c is the set f (x,y,z) = c a regular surface?

c. Answer the questions of parts a and b for the function f(x,y, z) = xyz2.
8. Let x(u, v) be as in Def. 1.Verify that dx,: R2
R3 is onetoone if and only if
9. Let V be an open set in the xy plane. Show that the set
{(x,y, z) E R3; z = 0 and (x,y)
E
V)
is a regular surface.
10. Let C be a figure "8" in the xy plane and let S be the cylindrical surface over C (Fig. 211);that is,
Is the set S a regular surface ?
S
Figure 211
11. Show that the set S = {(x,y, z) E R3;z = x2  y2) is a regular surface and check that parts a and b are parametrizations for S : a. x(u, v) = (u V , u  V, ~ u v )(u, , v) E R2. *b. ~ ( uv), = ( ~ cosh i v,u sinh v,u2),(u,v) E R2,zi # 0. Which parts of S do these parametrizations cover?
+
12. Show that x: U
c
R2 + R3given by
x(u, v) = (asin tr cos v,b sin u sin v,c cos tr),
where 0 < u
a,6, c # 0,
< n, 0 < v < 2n, is a parametrization for the ellipsoid
Describe geometrically the curves u
=
const. on the elIipsoid.
Regular Surfaces
67
*13. Find a parametrization for the hyperboloid of two sheets {(x, y, z) x2  y2 z 2 = 1).
+
E
R3;
14. A halfline [O, co) is perpendicular to a line E and rotates about E from a given initial position while its origin 0 moves along E. The movement is such that when [0, oo) has rotated through an angle 6, the origin is at a distance d = sin2(6/2) from its initial position on E. Verify that by removing the line E from the image of the rotating line we obtain a regular surface. If the movement were such that d = sin(6/2), what else would need to be excluded to have a regular surface ? *15. Let two points p(t) and q(t) move with the same speed, p starting from (O,0,0) and moving along the z axis and q starting at (a, 0, 0), a # 0, and moving parallel to the y axis. Show that the line joining p!t) to q(t) describes a set in R3 given by y(x  a) f zx = 0. Is this a regular surface?
16. One way to define a system of coordinates for the sphere S 2 , given by x 2 f y2 (Z  1)2 = 1, is to consider the socalled stereographic projection n: S2 EN] 3 R2 which carries a point p = (x, y, z) of the sphere S2minus the north pole N = (0, 0, 2) onto the intersection of the xy plane with the straight line which connects N to p (Fig. 212). Let (u, v) = n(x, y, z), where (x, y, z) E S 2 ( N ] and (u, v) E xy plane.
+
a. Show that nll: R2 + S is given by
Figure 212. The stereographic projection.
Regular Surfaces
68
b. Show that it is possible, using stereographic projection, to cover the sphere with two coordinate neighborhoods.
17. Define a regular curve in analogy with a regular surface. Prove that a, The inverse image of a regular value of a differentiable function
is a regujar plane curve. Give an example of such a curve which is not connected. b. The inverse image of a regular value of a differentiable map
is a regular curve in R3. Show the relationship between this proposition and the classical way of defining a curve in R3 as the intersection of two surfaces. *c. The set C
=
{(x, y)
E
"18. Suppose that f (x, y, z)
R2; x2 = y3) is not a regular curve.
= u = const.,
h(x, y, z )
=
g(x, y, z ) w
=
=
v = const.,
const.,
describe three families of regular surfaces and assume that at (x,, yo, z,) the Jaco bian
Prove that in a neighborhood of (x,, yo, z,) the three families will be described by a mapping F(M,v, W) = (x, y, Z ) of an open set of R 3 into R3, where a local parametrization for the surface of the family f(x, y, z ) = u, for example, is obtained by setting u = const. in this mapping. Determine F for the case where the three families of surfaces are
*19.
f(x, y, z)
=
g(s, y, z)
=
h(x7Y , 2 )
=
x2 X
+ y2 + z 2 = u = const.; =v =
const.,
Iy Z = w = const.,
z
(spheres with center (0, 0, 0)); (planes through the z axis) ; (cones with vertex at (0, 0, 0)).
Let a : (  3, 0) , RZ be defined by (Fig. 213)
+
if t E (3, I), (t 2)), = regular parametrized curve joining p
I
= (0,
=
(0,
 1)
to q
=
Ce
of Parameters
Horizontal scale distinct from \iertical scale
Figure 213
It is possible to define the curve joining p to q so that all the derivatives of a are continuous at the corresponding points and a has no selfintersections. Let C be the trace of a. a. Is C a regular curve? b. Let a normal line to the plane R2 run through C so that it describes a "cylinder" S. Is S a regular surface?
23. Change of Parameters; Differentiable Functions on Surfaces? Differentia1 geometry is concerned with those properties of surfaces which depend on their behavior in a neighborhood of a point. The definition of a regular surface given in Sec. 22 is adequate for this purpose. According to this definition, each point p of a regular surface belongs to a coordinate neighborhood. The points of such a neighborhood are characterized by their coordinates, and we should be able, therefore, to define the local properties which interest us in terms of these coordinates. For example, it is important that we be able to define what it means for a function f :S 4 R to be differentiable at a point p of a regular surface S. A natural way to proceed is to choose a coordinate neighborhood of p, with coordinates u, v, and say that f is differentiable at p if its expression in the coordinates u and v admits continuous partial derivatives of all orders. The same point of S can, however, belong to various coordinate neighborhoods (in the sphere of Example 1 of Sec. 22 any point of the interior of the first octant belongs to three of the given coordinate neighborhoods). Moreover, other coordinate systems could be chosen in a neighborhood of p (the points referred to on the sphere could also be parametrized by geotProofs in this section may be omitted on a first reading.
70
Regular Surfaces
graphical coordinates or by stereographic projection; cf. Exercise 16, Sec. 22). For the above definition to make sense, it is necessary that it does not depend on the chosen system of coordinates. In other words, it must be shown that when p belongs to two coordinate neighborhoods, with parameters (u, v) and (c, q), it is possible to pass from one of these pairs of coordinates to the other by means of a differentiable transformation. The following proposition shows that this is true. PROPOSITION 1 (Change of Parameters). Let p be a point of a regular surface S, and let x: U c R2 S, y: V c R2 4S be two parametrizations of S such that p E x(U) n y(V) = W. Then the "change of coordinates" h = xl 0 y : yl(W) . x'(W) (Fig. 214) is a difleomorphism; that is, h is differentiable and has a differentiable inverse hl. +
Figure 214
In other words, if x and y are given by
then the change of coordinates h, given by
71
Change of Parameters
has the property that the functions u and v have continuous partial derivatives of all orders, and the map h can be inverted, yielding
where the functions
5 and q also have partial derivatives of all orders.
Since
this implies that the Jacobian determinants of both h and hl are nonzero everywhere. Proof of Prop. 1. h = xloy is a homeomorphism, since it is composed of homeomorphisms (cf. the appendix to Chap. 2, Prop. 3). It is not possible to conclude, by an analogous argument, that h is differentiabIe, since xI is defi~edin an open subset of S, and we do not yet know what is meant by a differentiable function on S. We proceed in the following way. Let r E y'(W) and set q = h(r). Since x(u, v) = (x(u, v), y(u, v), z(u, v)) is a parametrization, we can assume, by renaming the axis if necessary, that
We extend x to a map F: U x R
+ R3
F(u, v, t ) = (x(u, v), y(u, v), z ( ~V),
defined by
+ t),
(u, V) E U, t
E
R.
Geometrically, F maps a vertical cylinder C over U into a "vertical cylinder" over x(U) by mapping each section of C with height t into the surface x(u, v) te,, where e, is the unit vector of the z axis (Fig. 214). It is clear that F is differentiable and that the restriction F [ U x (01 = x. Calculating the determinant of the differential dF,, we obtain
+
It is possible therefore to apply the inverse function theorem, which guarantees the existence of a neighborhood M of x(q) in R3 such that FI exists and is differentiable in M.
72
Regular Surfaces
By the continuity of y, there exists a neighborhood N of r in V such that y ( N ) c M (appendix to Chap. 2, Prop. 2). Notice that, restricted to N , h I N = F 1 0 y I N is a composition of differentiable maps. Thus, we can apply the chain rule for maps (appendix to Chap. 2, Prop. 8) and conclude that h is differentiable at r. Since r is arbitrary, h is differentiable on yl(W). Exactly the same argument can be applied to show that the map hI is differentiable, and so h is a diffeomorphisrn. Q.E.D. We shall now give an explicit definition of what is meant by a differentiable function on a regular surface. DEFINITION 1. Let f: V c S + R be a function d e f i e d in an open subset V of a regular surface S. Then f is said to be differentiable at p E V if, for some parametrization x : U c R2 S with.p E x ( U ) c V , the composition f x : U c R2 4R is differentiable at x I(p). f is differentiable in V ifit is diflerentiable at all points of V .

0
It follows immediately from the last proposition that the definition given does not depend on the choice of the parametrization x. In fact, if y: V c R 2 S is another parametrization withp E x ( V ) , and if h = X  ~ O Y , then f 0 y =f 0 x h is also differentiable, whence the asserted independence.

0
Remark 1. We shaIl frequently make the notational abuse of indicating f and f 0 x by the same symbol f (u, v), and say that f (u, v ) is the expression off in the system of coordinates x. This is equivalent to identifying x ( U ) with U and thinking of (u, v ) , indifferently, as a point of Il' and as a point of x ( U ) with coordinates (u, v). From now on, abuses of language of this type will be used without further comment.
Example 1. Let S be a regular surface and V c R 3 be an open set such that S c V. Let f : V c R3+ R be a differentiable function. Then the restriction off to S is a differentiable function on S. In fact, for any p E S and any parametrization x : U c R 2 + S in ~p,the function f 0 x : U  R is differentiable. In particular, the following are differentiable functions:
1. The height function relative to a unit vector v E R 3 , h : S + R, given by h(p) = p v,p E S, where the dot denotes the usual inner product in R 3 . h ( p ) is the height ofp E S relative to a plane normal to v and passing through the origin of R3 (Fig. 215). 2. The square of the distance from a fixed pointp, E R3, f ( p ) = Ip  p, 12, p E S. The need for taking the square comes from the fact that the distance Ip  p, I is not differentiable at p = p,. Remark 2. The proof of Prop. 1 makes essential use of the fact that the inverse of a parametrization is continuous. Since we need Prop. 1 to be able
Change o f Peremeters
Figure 215
to define differentiable functions on surfaces (a vital concept), we cannot dispose of this condition in the definition of a regular surface (cf. Remark 1 of Sec. 22). The definition of differentiability can be easily extended to mappings between surfaces. A continuous map 9 :V , c S , ,S2 of an open set V , of a regular surface S , to a regular surface S, is said to be diflerentiable at p E V , if, given parametrizations
with p
E
x,(U) and q(x,(Ul)) c x2(U2),the map x,'
is differentiable at q
= x;
0
p 0 x,: U,
+
'(p) (Fig. 216).
Figure 216
U,
74
Regular Surfaces
In other words, p is differentiable if when expressed in local coordinates as p(ul, v , ) = (yl,(u,, vl), p2(u,, vl)) the functions p, and yl, have continuous partial derivatives of all orders. The proof that this definitions does not depend on the choice of parametrization~is left as an exercise. We should mention that the natural notion of equivalence associated with differentiability is the notion of diffeomorphism. Two regular surfaces S, and S, are difleomorphic if there exists a differentiable map p : S, S, with a differentiable inverse pI: S, + S ,. Such a p is called a difeomorphism from S, to S,. The notion of diffeomorphism plays the same role in the study of regular surfaces that the notion of isomorphism plays in the study of vector spaces or the notion of congruence plays in Euclidean geometry. In other words, from the point of view of differentiability, two diffeomorphic surfaces are indistinguishable.

Example 2. If x: U c R2 S is a parametrization, xI : x(U) R2 is differentiable. In fact, for any p r x(U) and any parametrization y: V c R2 + S i n p , we have that xI 0 y: y'(W) x'(W), where 4

is differentiable. This shows that U and x(U) are diffeomorphic (i.e., every regular surface is locally diffeomorphic to a plane) and justifies the identification made in Remark 1. Example 3. Let S, and S, be regular surfaces. Assume that S, c V c R3, where V is an open set of R3,and that yl : V + R3 is a differentiable map such that p(S,) c S,. Then the restriction yl ) S, : S, S , is a differentiable map. In fact, given p t S, and parametrizations x, : U, + S,, x,: U, , S,, with p E x,(U,) and p(x,(U,)) c x,(U,), we have that the map

x ; ' o p o x , : U1+
U2
is differentiable. The following are particular cases of this general example:

1. Let S be symmetric relative to the xy plane; that is, if (x, y, z) E S, then also (x, y,  2 ) E S. Then the map a : S S, which takes p E S into its symmetrical point, is differentiable, since it is the restriction to S of a: R3 4 R3, a(x, y, z) = (x, y, 2). This, of course, generalizes to surfaces symmetric relative to any plane of R3. 2. Let R,,,: R3 " R3 be the rotation of angle 13about the z axis, and let S c R3 be a regular surface invariant by this rotation; i.e., if p E S, R,,,(p) t S. Then the restriction R,.,: S S is a differentiabie map.

Regular Surfaces
Figure 217. A regular curve.
Example 4 (Surfaces of Revolution). Let S c R3 be the set obtained by rotating a regular plane curve C about an axis in the plane which does not meet the curve; we shall take the xz plane as the plane of the curve and the z axis as the rotation axis. Let
be a parametrization for C and denote by u the rotation angle about the z axis. Thus, we obtain a map
from the open set U = ((u, v) E R2; 0 < u i 2n, a < v < bj into S (Fig. 218). We shall soon see that x satisfies the conditions for a parametrization in the definition of a regular surface. Since S can be entirely covered by similar parametrizations, it follows that S is a regular surface which is called a surface of revolution. The curve C is called the generating curve of S, and the z axis is the rotation axis of S. The circles described by the points of C are called the parallels of S, and the various positions of C on S are called the meridians of S. To show that x is a parametrization of S we must check conditions I, 2, and 3 of Def. 1, Sec. 22. Conditions 1 and 3 are straightforward, and we leave them to the reader. To show that x is a homeomorphism, we first show that x is onetoone. In fact, since (f (v), g(u)) is a parametrization of C, given z and x2 y2 = (f ( v ) ) ~we , can determine v uniquely. Thus, x is onetoone.
+
Change of Parameters
Figure 218. A surface of revolution.
We remark that, again because (f(v), g(v)) is a parametrization of C, v is a continuous function of z and of ,,/x2 y2 and thus a continuous function of (x, y, 2 ) . To prove that xI is continuous, it remains to show that u is a continuous function of (x, y, 2). To see this, we first observe that if u # n, we obtain, since f (v) # 0,
+
tan, u 2
=
U sin 2 u cos 2
. 2 sin
U
cos

u
24
sin u

2 cos2 2
1
+ cos u
hence, u = 2 tan'
x
+ J xY2 + y 2
Thus, if u # n, u is a continuous function of (x, y, 2 ) . By the same token, if u is in a small interval about n, we obtain
Thus, u is a continuous function of (x, y, 2 ) . This shows that xI is continuous and completes the verification. Remark 4. There is a slight problem with our definition of surface of revolution. If C c R2 is a closed regular plane curve which is symmetric relative to an axis r of R3, then, by rotating C about r, we obtain a surface
Regular Surfaces
78
which can be proved to be regular and should also be called a surface of revolution (when C i s a circle and r contains a diameter of C, the surface is a sphere). T o fit it in our definition, we would have to exclude two of its points, namely, the points where r meets C. For technical reasons, we want to rnaintain the previous terminology and shall call the latter surfaces extended surfaces of revolution.
A final comment should now be made on our definition of surface. We have chosen to define a (regular) surface as a subset of R3. If we want to consider gIobal, as well as local, properties of surfaces, this is the correct setting. The reader might have wondered, however, why we have not defined surface simply as a parametrized surface, as in the case of curves. This can be done, and in fact a certain amount of the classical literature in differential geometry was presented that way. No serious harm is done as long as only local properties are considered. However, basic global concepts, like orientation (to be treated in Secs. 26 and 3l), have to be omitted, or treated inadequately, with such an approach. In any case, the notion of parametrized surface is sometimes useful and should be included here.
DEFINITION 2. A parametrized surface x:. U c R2 + R3 is a dzferentiable map x from an open set U c R2 into R 3 .The set x(U) c R3 is called the trace of x. x is regular fi the diferential dx,: R2 + R3 is onetoone for all q E U (i.e., the vectors dx/du, dx/dv are linearly independent for all q E U). A point p E U where dx, is not onetoone is called a singular point of x. Observe that a parametrized surface, even when regular, may have selfintersections in its trace. Example 5. Let a : I+ R3 be a regular parametrized curve. Define x(t, v)
= a(t)
+ vat(?),
(t, vj
E
I x R.
x is a parametrized surface called the tangent surface of a (Fig. 219). Assume now that the curvature k(t), t E I, of a is nonzero for all t and restrict the domain of x to U = {(t, V) E I x R ; v + 01. Then
and
E
I,
Change of Parameters
Figure 219. The tangent surface.
since, for all t, the curvature (cf. Exercise 12 of Sec. 15)
is nonzero. It follows that the restriction x: U 4 R3 is a regular parametrized surface, the trace of which consists of two connected pieces whose common boundary is the set a(I). The following proposition shows that we can extend the local concepts and properties of differential geometry to regular parametrized surfaces.
PROPOSITION 2. Let x : U c R2+ R3 be a regular parametrized stliface and let q E U. Then there exists a neighborhood V of q in R2 such [hat x(V) c R3 is a regular surface. Proof. This is again a consequence of the inverse function theorem. Write

By regularity, we can assume that (d(x, y)/dfu, v))(q) # 0. Define a map F: U x R R3 by F(u, 2 1 3 t ) = ( ~ ( uv), , y(u, v)?Z(U,v) 1 t),
(u, v)
E
U, t
E
R.
80
Regular Surfaces
Then
By the inverse function theorem, there exist neighborhoods W, of q and W2 of F(q) such that F: W, + W2 is a diffeomorphism. Set V = W , n U and observe that the restriction FI V = xl V. Thus, x(V) is diffeomorphic t o V, and hence a regular surface. Q.E.D.
EXER CISESt "1. Let S2 = {(x, y, z) E R 3 ; x 2 f y 2 $ z2 = 1) be the unit sphere and let A : S2+ S2be the (antipodal) map A(x, y, z) = (x, y,  2 ) . Prove that A is a diffeomorphism.

2. Let S c R3 be a regular surface and x : S R2 be the map which takes each p E S into its orthogonal projection over R2 = ((x, y, z) E R3; z = 0). IS n differentiable ? 3. Show that the paraboloid z
=
x2
+ y2 is diffeomorphic to a plane.
4. Construct a diffeomorphism between the ellipsoid
and the sphere xZ f y 2
+ z2 = 1.
"5. Let S c R3 be a regular surface, and let d : S ,R be given by d(p) = / p  poI, where p r S, p, E R3, po 6 S ; that is, d is the distance from p to a fixed point p, not in S. Prove that d is differentiable.
6. Prove that the definition of a differentiable map between surfaces does not depend on the parametrizations chosen.
7. Prove that the reIation "S1 is diffeomorphic to S," is an equivalence relation in the set of regular surfaces. "8. Let
+ +
S2= {(x, y, Z) E R 3 ; x 2 y2 z2 = l} and H = {(x,y, 2) E R 3 ; x 2 y 2  z2 = 1). Denote by N = (0,0, 1) and S = (0,0, 1) the north and south poles of S2, respectively, and let F: S 2  (N) u {S) + H be defined as follows: For each p E S2 ( N j U ( S ) let the perpendicular from p to the z axis meet Oz at q. Consider the halfline I starting at q and containingp. Then F(p) = 1 n H (Fig. 220). Prove that F is differentiable.
+
9. a. Define the notion of differentiable function on a regular curve. What does one need to prove for the definition to make sense? Do not prove it now. If tThose who have omitted the proofs of this section should also omit Exercises 1316.
Change o f Parameters
Figure 220
you have not omitted the proofs in this section, you will be asked to do it in Exercise 15. b. Show that the map E: R
E(t)
+S 1 =
=
{(x,y)
(cos t, sin t),
E
t
R 2 ;x 2 E
+ y 2 = 1) given by
R,
is differentiable (geometrically, E "wraps" R around S I). 10. Let C be a plane regular curve which Iies in one side of a straight Iine r of the plane and meets r at the points p, q (Fig. 221). What conditions shouId C satisfy to ensure that the rotation of C about r generates an extended (regular)
surface of revolution?
1
Figure 221
11, Prove that the rotations of a surface of revolution S about its axis are diffeo
rnorphisms of S. 12. Parametrized surfaces are often usefuj to describe sets E which are regular sarfaces except for a finite number of points and a finite number of lines. For inR3 which stance, let C be the trace of a regular parametrized curve a : (a, 6) does not pass through the origin 0 = (0,0,O). Let Z be the set generated by the

Regular Surfaces
Figure 222
displacement of a straight line I passing through a moving point p fixed point 0 (a cone with vertex 0; see Fig. 222).
E
C and the
a. Find a parametrized surface x whose trace is E. b. Find the points where x is not regular.
c. What should be removed from E so that the remaining set is a regular surface ?
"13. Show that the definition of differentiability of a functionf: V c S ,R given in the text (Def. 1) is equivalent to the following: f is differentiable in p E V if it is the restriction to V of a differentiable function defined in an open set of R3 containing p. (Had we started with this definition of differentiability, we could have defined a surface as a set which is locally diffeomorphic to R2; see Remark 31 14, Let A c S be a subset of a regular surface S. Prove that A is itself a regular surface if and only if A is open in S; that is, A = U n S, where Uis an open set in R3.

15. Let C be a regular curve and let a : I c R 4 C, P: J c R C be two parametrization~of C in a neighborhood of p E a(1) n P(1) = W. Let
be the change of parameters. Prove that
a. lz is a diffeomorphism. b. The absolute value of the arc length of C in W does not depend on which parametrization is chosen to define it, that is,
"16. Let R2 = {(x,y, Z) E R3; z setting (x, y,  1) = x iy mial
+
=
c
1) be identified with the complex plane by Let P: + be the complex polyno
=[ E
c.
a=
The Tangent P/ane
83
Denote by n, xZ
the stereographic projection of S2
= {(x,y ,
Z)E R 3 ;
+ y 2 + z 2 = 1) from the north pole N = (0,0, 1) onto R2.Prove that the
map F: S2
+
S2 given by
is differentiable.
24. The Tangent Plane; The Differential of a Map In this section we shall show that condition 3 in the definition of a regular surface S guarantees that for every p E S the set of tangent vectors to the parametrized curves of S, passing through p, constitutes a plane. By a tangent vector to S, at a point p E S, we mean the tangent vector a'(0) of a differentiable parametrized curve a : (  6 , e ) . S with a(0) = p. PROPOSITION I. Let x : U c R2 + S be a parametrization of a regular surface S and let q E U . The vector subspace of dimension 2,
coincides with the set of tangent vectors to S at x(q). Proof. Let w be a tangent vector at x(q), that is, let w = a'(O), where a : (6, 6 ) x ( U ) c S is differentiable and a(0) = x(q). By Example 2 of Sec. 23, the curve P = xI 0 a : (  6 , 6 ) , U is differentiable. By definition of the differential (appendix to Chap. 2, Def. I), we have dxq(Pf(0))= w, Hence, w E dx,(R2) (Fig. 223). On the other hand, let w = dxq(v),where v E R2. It is clear that v is the velocity vector of the curve y : (  6 , 6 ) U given by +

By the definition of the differential, w = a'(O), where a that w is a tangent vector.
=x
0
y. This shows
Q.E.D.
By the above proposition, the plane dx,(R2), which passes through x(q) = p, does not depend on the parametrization x. This plane will be called the
Regular Surfaces
Figure 223
tangent plane to S at p and will be denoted by T,(S). The choice of the parametrization x determines a basis ((dx/du(q), (dx/dv)(q)] of T,(S), called the basis associated to x. Sometimes it is convenient to write dxldu = xu and dxldv = x,. The coordinates of a vector w E Tp(S) in the basis associated to a parametrization x are determined as follows. w is the velocity vector af(0) of a curve a = x 0 p, where p: (E, E) + U is given by P(t) = ( ~ ( t )v(t)), , with P(0) = q = x ' ( p ) . Thus,
Thus, in the basis {x,(q), x,(q)], w has coordinates (uf(0), vf(0)), where (u(t), v(t)) is the expression, in the parametrization x, of a curve whose velocity vector at t = 0 is w. With the notion of a tangent plane, we can talk about the differential of a (differentiable) map between surfaces. Let S , and S, be two regular surfaces and let p: V c S, + S, be a differentiabIe mapping of an open set V of S1 into S2.I f p E V, we know that every tangent vector w E Tp(S,) is the velocV with ity vector al(0) of a differentiable parametrized curve a : (6, 6) a(0) = p . The curve P = p o a is such that P(0) = yl(p), and therefore Pf(0)is a vector of T,, (S,) (Fig. 224).

PROPOSITION 2. In the discussion above, given w, the zjector Pf(0) does not depend on the choice of a. The map dp,: T,(S,) + T,(,,(S,) defined by dp,(w) = Pf(0) is linear.
The Tangent Plane
Figure 224
Proof. The proof is similar to the one given in Euclidean spaces (cf. Prop. 4, appendix to Chap. 2). Let x(u, v), %(G, ,3 be parametrizations in neighborhoods of p and p,(p), respectively. Suppose that p is expressed in these coordinates by
and that a is expressed by
, the expression of Pt(0) in the Then P(t) = (pl(u(t),v(t)), p,(u(t), ~ ( t ) ) )and basis (%a, is
The relation above shows that Pf(0) depends only o n the map p and the coordinates (uf(0),~ ' ( 0 )of) w in the basis {xu,x,]. Pr(0)is therefore independent of a. Moreover, the same relation shows that
that is, dp, is a linear mapping of T,(S,) into T,,,(S,) whose matrix in the bases (x,, xu)of T,(S,) and (x,, 52,] of T (S,) is just the matrix given above. Q.E.D.
,,
The linear map dp,, defined by Prop. 2 is called the dflereniial of p, at p E S 1 . In a similar way we define the differential of a (differentiable) function f : U c S 4 R at p t U as a linear map df, : T,(S) 3 R. We leave the details to the reader.
86
Regular Surfaces
Example 1. Let v E R3 be a unit vector and let h : S + R, h(p) = v w p , p E S, be the height function defined in Example I of Sec. 23. To compute dh,(w), w E T,(S), choose a differentiable curve u : (  6 , r ) S with u(0) = p, uf(0)= w. Since h(u(t))= a(t)v,we obtain

Example 2. Let S2 c R2 be the unit sphere
and let Rz,,: R3 + R3 be the rotation of angle 8 about the z axis. Then R,,, restricted to S 2 is a differentiable map of S 2 (cf. Example 3 of Sec. 23). We shall compute (dR,&,(w),p E S 2 , w E Tp(S2).Let a : (  6 , r ) SZ be a differentiable curve with a(0) = p , ~ ' ( 0=) w. Then, since R,,, is linear,

Observe that R,,,leaves the north pole N = (0,0, 1) fixed, and that (dRz,e)N: T N ( S ) T N ( S )is just a rotation of angle 8 in the plane TN(S). In retrospect, what we have been doing up to now is extending the notions of differential calculus in R2 to regular surfaces. Since calculus is essentially a local theory, we defined an entity (the regular surface) which locally was a plane, up to diffeomorphisms, and this extension then became natural. It might be expected therefore that the basic inverse function theorem extends to differentiable mappings between surfaces. We shall say that a mapping 9 : U c S , + S, is a local difeomorphism atp E U if there exists a neighborhood V c U of p such that restricted to V is a diffeomorphism onto a n open set q ( V ) c S,. In these terms, the version of the inverse of function theorem for surfaces is expressed as follows.
PROPOSIlION 3. vSl and S , are regular surfaces and 9 : U c S , + S, is a diferentiable mapping of an open set U c S, such that the di'erential dp, of at p E U is an isomorphism, then 9 is a local difleomorphism at p. The proof is an immediate application of the inverse function theorem in R2 and will be left as a n exercise. Of course, all other concepts of calculus, like critical points, regular values, etc., do extend naturally to functions and maps defined on regular surfaces.
The Tangent Plane
87
The tangent plane also allows us to speak of the angle of two intersecting surfaces at a point of intersection. Given a point p on a regular surface S, there are two unit vectors of R3 that are normal to the tangent plane T,(S); each of them is called a unit normal vector at p. The straight line that passes through p and contains a unit norma1 vector at p is called the normal line at p. The angle of two intersecting surfaces at an intersection point p is the angle of their tangent planes (or their normal lines) at p (Fig. 225).
Figure 225
By fixing a parametrization x: U c R2 + S at p E S, we can make a definite choice of a unit normal vector at each point q E x(U) by the rule A Xv (q). N ( q ) = I xu A x, 1

Thus, we obtain a differentiable map N : x(U) R3.We shall see later (Secs. 26 and 31) that it is not always possible to extend this map differentiably to the whole surface S. Before leaving this section, we shall make some observations on questions of differentiability. The definition given for a regular surface requires that the pararnetrizations be of class C", that is, that they possess continuous partial derivatives af all orders. For questions in differential geometry we need in general the existence and continuity of the partial derivatives only up to a certain order, n.hich varies with the nature of the problem (very rarely do we need more than four derivatives). For example, the existence and continuity of the tangent plane depends only on the existence and continuity of the first partial derivatives. It could happen, therefore, that the graph of a function z = f (x, y ) admits a tangent
88
Regular Surfaces
plane at every point but is not sufficiently differentiable to satisfy the definition of a regular surface. This occurs in the following example.
+
ExampIe 3. Consider the graph of the function z = ,$'(x2 y2)2, generated by rotating the curve z = x4I3about the z axis. Since the curve is symmetric relative to the z axis and has a continuous derivative which vany2)2 admits the ishes at the origin, it is clear that the graph of z = ,$'(x2 xy plane as a tangent plane at the origin. However, the partial derivative z,, does not exist at the origin, and the graph considered is not a regular surface as defined above (see Prop. 3 of Sec. 22).
+
We do not intend to get involved with this type of question. The hypothesis C" in the definition was adopted precisely to avoid the study of the minimal conditions of differentiability required in each particular case. These nuances have their place, but they can eventually obscure the geometric nature of the problems treated here.
EXERCISES *I. Show that the equation of the tangent plane at ( x o ,yo, 2,) of a regular surface given by f (x,y, z) = 0, where 0 is a regular vaIue off; is fx(x0, Yo, zo)(x  xo)
+ fy(x0, Yo,
zO)(Y
2. Determine the tangent planes of x2
YO)
+ y2
+ fr(x0,Yo,
ZO)(Z

zo)
= 0.
 z2 = 1 at the points ( x ,y, 0) and
show that they are all parallel to the z axis. 3. Show that the equation of the tangent plane of a surface which is the graph of a differentiable function z = f (x, y), at the point po = (xo,yo),is given by
Recall the definition of the differential df of a function f: R2 + R and show that the tangent plane is the graph of the differential df,. *4. Show that the tangent planes of a surface given by z = xf (y/x),x f 0, where f is a differentiable function, all pass through the origin (0, 0, 0). 5. If a coordinate neighborhood of a regular surface can be parametrized in the form
where al and a2 are regular parametrized curves, show that the tangent planes along a fixed coordinate curve of this neighborhood are all parallel to a line. 6 . Let a : I + R3 be a regular parametrized curve with everywhere nonzero curvature. Consider the tangent surface of a (Example 5 of Sec. 23)
The Tangent Plane
Show that the tangent planes along the curve x(l, const.) are all equal. 7. L e t t S + R be given by f (p) = 1 p  po 12, w h e w E S a n d PO is a fixed point of R3 (see Example 1 of Sec. 23). Show that dfp(w) = 2w (P PO), w E T,(S).

8. Prove that if L : R3 4 R3 is a Iinear map and S c R3 is a regular surface invariant under L, i.e., L(S) c S, then the restriction L I S is a differentiable map and
9. Show that the parametrized surface
~ ( uv),
= (v
cos u, v sin u, au),
a f 0,
is regular. Compute its normal vector N(u, v) and show that along the coordinate line u = uo the tangent plane of x rotates about this line in such a way that the tangent of its angle with the z axis is proportional to the distance c (= d x 2 y2) of the point x(uo, V)to the z axis.
+
10. (Tubular Surfaces.) Let a: I +R3 be a regular parametrized curve with nonzero curvature everywhere and arc length as parameter. Let
X(S,v)
= a(s)
+ r(n(s) cos v + b(s) sin v),
r
=
const. # 0, s E I,
be a parametrized surface (the tube of radius r around a), where n is the normal vector and b is the binormal vector of a . Show that, when x is regular, its unit normal vector is
N(s, v)
=
(n(s) cos 7) 4 b(.~)sin v).
11. Show that the normals to a parametrized surface given by
X(U,v)
=
(f ( 4 cos v,
f (u) + 0, gf # 0,
f (u) sin v, g(u)),
a11 pass through the z axis. '12. Show that each of the equations (a, b, c # 0)
+ y2 + z2 = ax, x2 + y2 + z2 by, x2 + yz + zz cz x2
= =
define a regular surface and that they all intersect orthogonally. 13. A criticalpoint of a differentiable functionf: S , R defined on a regular surface S is a point p E S such that df, = 0.
*a. Let f: S + R be given by f(p)
=
p

po1, P
E
S,PO 6 S (cf Exercise 5,
Regular Surfaces
90
Sec. 23). Show that p E S is a critical point off if and only if the line joining p to po is normal to S at p. b. Let h: S + R be given by h(p) = p  v , where v E R3 is a unit vector (cf. Example 1, Sec. 23). Show thatp E S is a critical point off if and only if v is a normal vector of S at p.
"14. Let Q be the union of the three coordinate planes x p = (x, y, z) E R3  Q. a. Show that the equation in t, x2 y2 +z2 a  t + ~ r ct
= f ( f ) = 1,
= 0,
y
=
0, z
=
0. Let
a > b > c > 0,
has three distinct real roots: t , , t,, t 3 . b. Show that for each p E R3  Q, the sets given by f (t,) 1 = 0,f (t,)  1 = 0, f ( t 3 )  1 = 0 are regular surfaces passing throughp which are painvise
orthogonal.
15. Show that if all normals to a connected surface pass through a fixed point, the surface is contained in a sphere. 16. Let w be a tangent vector to a regular surface S at a point p E S and Iet x(u, v) and %(i,JC)be two parametrizations at p. Suppose that the expressions of w in the bases associated to x(u, v) and %(G, G) are
and W =
+ P2%+.
Show that the coordinates of w are related by
where ii nates.
= i(u,v)
and C
= 6(u, v )
are the expressions of the change of coordi
*17. Two reguIar surfaces S1and S, intersect transversally if whenever p E S1 n S , then T,(S1) # T,(S2). Prove that if S1intersects S , transversally, then S 1 n S 2 is a regular curve. 18. Prove that if a regular surface S meets a plane P in a single point p, then this plane coincides with the tangent plane of S at p.
19. Let S c R3 be a regular surface and P c R3 be a plane. If all points of S are on the same side of P,prove that P is tangent to S at all points of P n S. *20. Show that the perpendicular projections of the center (O,0, 0) 01 the ellipsoid
The Tangent Plane
onto its tangent planes constitute a regular surface given by {(x, y, z) E R3; (x2 4 y2 '21. Let
+ z2)2 = a2x2i.b2Y
c2z23 ((0, 0, 0)).
f: S + R be a differentiable function on a connected regular surface S.
Assume that df,
=0
for all p
E
S. Prove that f is constant on S.
'12. Prove that if all normal lines to a connected regular surface S meet a fixed straight line, then S is a surface of revolution.
23. Prove that the map F: S2+ S2defined in Exercise 16 of Sec, 23 has only a finite number of critical points (see Exercise 13). 51. (Clzain Rule.) Show that if q : S, + S2and tp : S2+ S g are differentiable maps and p E SI, then
25. Prove that if two regular curves C1 and C2 of a regular surface S are tangent a t a point p E S, and if q : S + S is a diffeomorphism, then q(C,) and q(C2) are
regular curves which are tangent at q(p). 26. Show that if p is a point of a regular surface S, it is possible, by a convenient choice of the (x, y, z) coordinates, to represent a neighborhood o f p in S in the form z = f (x, y ) so that f (0,O) = 0, f,(O, 0) = 0, f,(O, 0) = 0. (This is equivalent to taking the tangent plane to S at p as the xy plane.) 27. (Theory of Contact.) Two regular surfaces, S and 3 , in R3, which have a point p in common, are said to have contact oforder 2 1 at p if there exist parametrizations with the same domain x(u, v), 5(u, v) at p of S and 3 , respectively, such that xu = 5, and x, = 2, at p. If, moreover, some of the second partial deriva
tives are different at p, the contact is said to be of order exactly equal to 1. Prove that
a. The tangent plane T,(S) of a regular surface S at the point p has contact of order > 1 with the surface at p. b. If a plane has contact of order > 1 with a surface S at p, then this plane coincides with the tangent plane to S at p. c. Two regular surfaces have contact of order 2 1 if and only if they have a common tangent plane at p, i.e., they are tangent at p. d. If two regular surfaces S and 3 of R3 have contact of order 2 1 at p and if F: R 3 + R 3 is a diffeomorphism of R3, then the images F(S) and F($ are regular surfaces which have contact of order 2 1 at f (p) (that is, the notion of contact of order 2 1 is invariant under diffeomorphisms). e. If two surfaces have contact of order 2 1 at p, then lim,,o (dlr) = 0, where d is the length of the segment which is determined by the intersections with the surfaces of some parallel to the common normal, at a distance r from this normal.
92
Regular Surfaces
28. a. Define regular value for a differentiable functionf: S + R on a regular surface S, b. Show that the inverse image of a regular value of a differentiable function on a regular surface S is a regular curve on S.
25. The First Fundamental Form; Area So far we have looked at surfaces from the point of view of differentiability. In this section we shall begin the study of further geometric structures carried by the surface. The most important of these is perhaps the first fundamental form, which we shall now describe. The natural inner product of R3 3 S induces on each tangent plane T,(S) of a regular surface S an inntr product, to be denoted by ( , )p : If w l , w2 E T,(S) c R3, then (w,, w,), is equal to the inner product of w, and w2 as vectors in R3. To this inner product, which is a symmetric bilinear form (i.e., (w,, w,) = (w,, w,) and (w,, w,) is linear in both w , and w,), there corresponds a quadratic form I, : T,(S) + R given by I,(w)
=
(w, w),
=
1 w l2 2 0.
(1)
DEFINITION 1. The quadratic form I, on T,(S), defined by Eq. ( I ) , is called the first fundamental form of the regular surface S c R3 at p E S. Therefore, the first fundamental form is merely the expression of how the surface S inherits the natural inner product of R3. Geometrically, as we shall see in a while, the first fundamental form allows us to make measurements on the surface (lengths of curves, angles of tangent vectors, areas of regions) without referring back to the ambient space R3 where the surface lies. We shall now express the first fundamental form in the basis {xu,xu] associated to a parametrization x(u, v) atp. Since a tangent vector w E Tp(S) is the tangent vector to a parametrized curve a(t) = x(u(t); v(t)), t E (c, f), with p = a(0) = x(uo, v,), we obtain
where the values of the functions involved are computed for t
= 0,
and
93
The Firsf Fundemenfal Form
are the coefficients of the first fundamental form in the basis {xu, x,) of T,(S). By letting p run in the coordinate neighborhood corresponding to X ( U , v ) we obtain functions E(u, v ) , F(u, v), G(u, v ) which are differentiable in that neighborhood. From now on we shall drop the subscript p in the indication of the inner product ( , ), or the quadratic form I, when it is clear from the context which point we are referring to. It will also be convenient to denote the natural inner product of R3 by the same symbol ( , ) rather than the previous dot. Example 1. A coordinate system for a plane P c R3 passing through po = (x,, y o , z,) and containing the orthonormal vectors w , = ( a , , a,, a,), w , = ( b , , b , , 6 , ) is given as follows: X(U,
V ) = PO
+
UW,
f vw,,
(u, v )
g R2.
To compute the first fundamental form for an arbitrary point of P we observe that xu = w , , x, = w,; since w , and w , are unit orthogonal vectors, the functions E, F, G are constant and given by
In this trivial case, the first fundamental form is essentially the Pythagorean theorem in P; i.e., the square of the length of a vector w which has coordinates a, b in the basis (xu,x,] is equal to a2 b2.
+

ExampIe 2. The right cylinder over the circle x2 parametrization x : U R 3 , where (Fig. 226) X(U,v ) = (COS U , sin u, v ) , 0 O,O < 8 < 2n3.
Compute the coefficients of the first fundamental form of P in this parametrizatian. 11. Let S be a surface of revolution and C its generating curve (cf. Example 4, Sec. 23). Lets be the arc length of C and denote by p = p(s) the distance to the rotation axis of the point of C corresponding to s,
a. (Pappus' Theorem.) Show that the area of S is
where 1 is the length of C. b. Apply part a to compute the area of a torus of revolution. 12. Show that the area of a regular tube of radius r around a curve u (cf. Exercise 10, Sec. 24) is 2nr times the length of a. 13. (Generalized Helicoids.) A natural generalization of both surfaces of revolution
and helicoids is obtained as follows. Let a regular plane curve C, which does not meet an axis E in the plane, be displaced in a rigid screw motion about E, that is, so that each point of C describes a helix (or circle) with E as axis. The set S generated by the displacement of C is called a generalized helicoid with axis E and generator C . If the screw motion is a pure rotation about E, S is a surface of revolution; if C is a straight line perpendicular to E, S is (a piece of) the standard helicoid (cf. Example 3). Choose the coordinate axes so that E is the z axis and C lies in the yz pIane. Prove that a. If (f (s), g(s)) is a parametrization of C by arc length s, a then x: U + S, where
< s < 6 , f (s) > 0,
and X(S,u)
= (f
(s) cos U,f (s) sin u, g(s)
+ cu),
c
= const.,
is a parametrization of S. Conclude that S is a regular surface. b. The coordinate lines of the above parametrization are orthogonal (i.e., F = 0) if and only if x(U) is either a surface of revolution or (a piece of) the standard helicoid. 14. (Gradient on Surfaces.) The gradient of a differentiable function f: S 4 R is a differentiable map grad f: S + R3 which assigns to each point p E S a vector grad f ( p ) E Tp(S)c R3 such that
(grad f(p),v),=df,(v)
f o r a l l v ~T,(S).
7 02
Regular Surfaces
Show that a. If E, F, G are the coefficients of the first fundamental form in a parametrization x: U c R2 + S, then grad f on x ( U ) is given by grad f
=
f.G  f , F x u EG  F2
+
fUE  fUF EG  ~2
In particular, if S = R2 with coordinates x, y,
where {el, e2) is the canonical basis of R, (thus, the definition agrees with the usual dejinition of gradient in the plane). b. If you let p E S be fixed and v vary in the unit circle I v I = 1 in T,(s), then df,(v) is maximum if and only if v = gradfll grad f 1 (thus, grad f (p) gives the direction of maximum variation o f f atp). c. If grad f # 0 at all points of the level curve C = {q E S ; f ( q ) = const.), then C is a regular curve on S and grad f is normal to C at all points of C.
15. (Orthogonal Families of Curves.) a. Let E, F, G be the coefficients of the first fundamental form of a reguIar surface S in the parametrization x: U c R2 + S. Let ~ ( uv), = const. and ~ ( uV) , = const. be two families of regular curves on x(U) c S (cf. Exercise 28, Sec. 24). Prove that these two families are orthogonal (i.e., whenever two curves of distinct families meet, their tangent lines are orthogonal) if and only if
b. Apply part a to show that on the coordinate neighborhood x(U) of the helicoid of Example 3 the two families of regular curves
are orthogonal.
2 6. Orientation o f Surfaces? In this section we shall discuss in what sense, and when, it is possible to orient a surface. Intuitively, since every point p of a regular surface S has a tangent plane T,(S), the choice of an orientation of T,(S) induces an orientation in a neighborhood of p, that is, a notion of positive movement along sufficiently V h i s section may be omitted on a first reading.
Orientation o f Surfaces
103
small closed curves about each point of the neighborhood (Fig. 230). If it is pssible to make this choice for eachp E S so that in the intersection of any rn.0 neighborhoods the orientations coincide, then S is said to be orientable. If this is not possible, S is called nonorientable.
Figure 230
We shall now make these ideas precise. By fixing a parametrization x(u, v) af a neighborhood of a point p of a regular surface S, we determine an orientation of the tangent plane T,(S), namely, the orientation of the associated ordered basis (xu, x,). I f p belongs to the coordinate neighborhood of another parametrization ji(Li, G), the new basis jia] is expressed in terms of the first one by
{x,,
where u = u(G, v) and v = v(t7, C) are the expressions of the change of coordinates. The bases {xu,x,) and (ji,, 2,) determine, therefore, the same orientation of T,(S)if and only if the Jacobian
of the coordinate change is positive. DEFINITION 1. A regular surface S is called orientable if it is possible to cover it with a family of coordinate neighborhoods in such a way that &!a
104
Regular Surfaces
point p E S belongs to two neighborhoods of this family, then the change of coordinates has positive Jacobian at p. The choice of such a family is called an orientation of S, and S, in this case, is called oriented. v s u c h a choice is not possible, the surface is called nonorientable. Example 1. A surface which is the graph of a differentiable function (cf. Sec. 22, Prop. I) is an orientable surface. In fact, all surfaces which can be covered by one coordinate neighborhood are trivially orientable. Example 2. The sphere is an orientable surface. Instead of proceeding to a direct calculation, let us resort to a general argument. The sphere can be covered by two coordinate neighborhoods (using stereographic projection; see Exercise 16 of Sec. 22), with parameters (u, v) and (ii, fi), in such a way that the intersection W of these neighborhoods (the sphere minus two points) is a connected set. Fix a point p in W. If the Jacobian of the coordinate change a t p is negative, we interchange u and v in the first system, and the Jacobian becomes positive. Since the Jacobian is different from zero in Wand positive at p E W, it follows from the connectedness of W that the Jacobian is everywhere positive. There exists, therefore, a family of coordinate neighborhoods satisfying Def. 1, and so the sphere is orientabIe. By the argument just used, it is clear that $ a regular surface can be covered by two coordinate neighborhoods whose intersection is connected, then the surface is orientable.
Before presenting an example of a nonorientable surface, we shall give a geometric interpretation of the idea of orientability of a regular surface in R3. As we have seen in Sec. 24, given a system of coordinates x(u, v) a t p , we have a definite choice of a unit normal vector N at p by the rule
Taking another system of local coordinates %(S, 6)a t p , we see that
where d(u, v)/d(ii, fi) is the Jacobian of the coordinate change. Hence, N will preserve its sign or change it, depending on whether d(u, v)/d(zi, C) is positive or negative, respectively. By a differentiableJield of unit normal vectors on an open set U c S, we shall mean a differentiable map N: U + R3 which associates to each q E U a unit normal vector N(q) E R3 to S at q,
105
Orientation o f Surfaces
PROPOSITION 1. A regular surface S c R3 is orientable if and only if there exists ca diflerentiable jeld of unit normal vectors N : S + R3 on S .
Proof. If S is orientable, it is possible to cover it with a family of coordinate neighborhoods so that, in the intersection of any two of them, the change of coordinates has a positive Jacobian. At the points p = x(u, v) of each neighborhood, we define N(p) = N(u, v) by Eq. (1). N(p) is well defined, since i f p belongs to two coordinate neighborhoods, with parameters (u, v) and ( i , G), the normal vector N(u, v) and N(i, 6) coincide by Eq. (2). Moreover, by Eq. (I), the coordinates of N(u, v) in R3 are differentiable functions of (u, vj, and thus the mapping N: S R3 is differentiable, as desired. On the other hand, let N: S + R3 be a differentiable field of unit normal vectors, and consider a family of connected coordinate neighborhoods covering S. For the points p = x(u, v) of each coordinate neighborhood x(U), U c R2, it is possible, by the continuity of N and, if necessary, by interchanging u and v, to arrange that +
In fact, the inner product
is a continuous function on x(U). Since x(U) is connected, the sign off is constant. Iff = 1, we interchange u and v in the parametrization, and the assertion follows. Proceeding in this manner with all the coordinate neighborhoods, we have that in the intersection of any two of them, say, x(u, v) and ji(ii, C), the Jacobian
is certainly positive; otherwise, we would have
which is a contradiction. Hence, the given family of coordinate neighborhoods after undergoing certain interchanges of u and v satisfies the conditions of Def. 1, and S is, therefore, orientable. Q.E.D.
la6
Regular Surfaces
Remark. As the proof shows, we need only to require the existence of a continuous unit vector field on S for S to be orientable. Such a vector field will be automatically differentiable. Example 3. We shall now describe an example of a nonorientable surface, the socalled Mobius strip. This surface is obtained (see Fig. 231) by considering the circle S1 given by x2 y = 4 and the open segment AB given in the yz plane by y = 2, j z 1 < 1. We move the center c of AB along S1 and turn AB about c in the cz plane in such a manner that when c has passed through an angle u, AB has rotated by an angle 4 2 . When c completes one trip around the circle, AB returns to its initial position, with its end points inverted.
+
Figure 231
From the point of view of differentiability, it is as if we had identified the opposite (vertical) sides of a rectangle giving a twist to the rectangle so that each point of the side AB was identified with its symmetric point (Fig. 231). It is geometrically evident that the Mobius strip M is a regular, nonorientable surface. In fact, if M were orientable, there would exist a differentiable field N : M + R3 of unit normal vectors. Taking these vectors along the circle x 2 y 2 = 4 we see that after making one trip the vector N returns to its original position as N, which is a contradiction. We shall now give an analytic proof of the facts mentioned above. A system of coordinates x : U M for the Mobius strip is given by
+
+
x(u, v) =
((2

(
v sin  sm u, 2 2"
>.

>
>
v sin  cos u, v cos  ,
"
2
2
where 0 < u < 2n and 1 < v < 1. The corresponding coordinate neighborhood omits the points of the open interval u = 0. Then by taking the
Orientation o f Surfaces
707
origin of the u's at the x axis, we obtain another parametrization given by
ji(ii,
G)
whose coordinate neighborhood omits the interval u = n/2. These two coordinate neighborhoods cover the Mobius strip and can be used to show that it is a regular surface. Observe that the intersection of the two coordinate neighborhoods is not connected but consists of two connected components:
W , = x(u, v): 0 < u
{
1
0 satisfying the conditions in the statement; that is, given I/n there exist points p, and q, such that d(p,, q,) < l/n butp, and q, do not belong to the same open set of the family (U,). Setting n = 1,2, . . . , we obtain two infinite sets of points {p,) and {q,) which, by Property 1, have limit points p and q, respectively. Since d(p,, q,) < lln, we may choose these limit points in such a way that p = q. But p E U, for some a, because p E A = U, U,, and since U, is an open set, there is an open ball B,(p), with center in p, such that B,(p) c U,. Since p is a limit point of {p,] and {q,), there exist, for n sufficiently large, in B,(p) c U,; that is, p, and q, belong to the same U,, points p, and which is a contradiction. Q.E.D. Using Properties 2 and 3, we shall now prove the existence of a tubular neighborhood of an orientable compact surface.
PROPOSITION 3. Let S c R3 be a regular, compact, orientable surface. Then there exists a number E > 0 such that whenever p, q E S the segments of the normal lines of lelzgth 2 ~ centered , in p and q, are disjoint (that is, S has a tubular neighborhood). Proof. By Prop. 1, for eachp E S there exists a neighborhood W, and a number E, > 0 such that the proposition holds for points of W, with E = E,, Letting p run through S, we obtain a family {W,) with Up,, W, = S. By compactness (Property 2), it is possible to choose a finite number of the Wi S, W,'s, say, W,, . . . , Wk (corresponding to E , , . . . , E,) such that i = 1, . . . ,k. We shall show that the required E is given by

where S is the Lebesgue number of the family {Wi) (Property 3). In fact, let two pointsp, q E S. If both belong to some W,, i = 1, . . . ,k, the segments of the normal lines with centers i n p and q and length 26 do not meet, since E < ci. I f p and q do not belong to the same Wi, then d(p, q) 6; were the segments of the normal lines, centered in p and q and of length 2 ~ , to meet at point Q E R3, we would have
>
114
which contradicts the definition of
Regular Surfacas
Q.E.D.
E.
Putting together Props. 1,2, and 3, we obtain the following theorem, which is the main goal of this section.
THEOREM. Let S c R3 be a regular compact orientable surface. Then there exists a digerentiablefunction g : V + R, dejined in an open set V c R3, with V 2 S (precisely a tubular neighborhood of S), which has zero as a regular value and is such that S = gl(O). Remark 1, It is possible to prove the existence of a tubular neighborhood of an orientable surface, even if the surface is not compact; the theorem is true, therefore, without the restriction of compactness. The proof is, however, more technical. In this general case, the ~ ( p > ) 0 is not constant as in the compact case but may vary with p. Remark 2. It is possible to prove that a regular compact surface in R3 is orientable; the hypothesis of orientability in the theorem (the compact case) is therefore unnecessary. A proof of this fact can be found in H. Samelson, "Orientability of Hypersurfaces in Rn," Proc. A.M.S. 22 (1969), 301302.
28. A Geometric Definition of Areat In this section we shall present a geometric justification for the definition of area given in Sec. 25. More precisely, we shall give a geometric definition of area and shall prove that in the case of a bounded region of a regular surface such a definition leads to the formula given for the area in Sec. 25. To define the area of a region R c S we shall start with a partition 6 of R into a finite number of regions R,, that is, we write R = Ui R,, where the intersection of two such regions R, is either empty or made up of boundary points of both regions (Fig. 233). The diameter of Ri is the supremum ofthe distances (in R3) of any two points in R,; the Iargest diameter of the R,'s of a given partition 6 is called the norm ,u of 6. If we now take a partition of each R,, we obtain a second partition of R, which is said to rejine 6. Given a partition
of R, we choose arbitrarily points pi
E
Ri and project Ri onto the tangent
?This section may be omitted on a first reading.
Geometric Definition of Area
Ri
Figure 233
plane at pi in the direction of the normal line at pi;this projection is denoted by Ri and its area by A@,). The sum C i A@,) is an approximation of what we understand intuitively by the area of R. If, by choosing partitions 6 , , . . . , 6,, . . . more and more refined and such that the norm p, of 6. converges to zero, there exists a limit of X iA@,) and this limit is independent of all choices, then we say that R has an area A(R) defined by
An instructive discussion of this definition can be found in R. Courant, Dlflerential and Integral Calculus, Vol. 11, WileyInterscience, New York, 1936, p. 311. We shall show that a bounded region of a regular surface does have an area. We shall restrict ourselves to bounded regions contained in a coordinate neighborhood and shall obtain an expression for the area in terms of the coefficients of the first fundamental form in the corresponding coordinate system. PROPOSITION. Let x: U S be a coordinate system in a regular surface S and let R = x(Q) be a bounded region of S contained in x(U). Then R has an area given by 4
Ui
Ri,of R. Since R is bounded and Proox Consider a partition, R = closed (hence, compact), we can assume that the partition is sufficiently refined so that any two normal lines of Ri are never orthogonal. In fact, because the normal lines vary continuously in S, there exists for each p E R a neighborhood of p in S where any two normals are never orthogonal; these neighborhoods constitute a family of open sets covering R, and consid
176
Regular Surfaces
ering a partition of R the norm of which is smaller than the Lebesgue number of the covering (Sec. 27, Property 3 of compact sets), we shall satisfy the required condition. Fix a region Ri of the partition and choose a point pi E R,= x(Qi). We want to compute the area of the normal projection 8, of Ri onto the tangent plane zit pi.To do this, consider a new system of axes pi2j2 in R3, obtained from Oxyz by a translation Op,, followed by a rotation which takes the z axis into the normal line at pi in such a way that both systems have the same orientation (Fig. 234). In the new axes, the parametrization can be written q u , v)
= (2(u,
v), j(u, v), Z(u, v)),
where the explicit form of %(u,v) does not interest us; it is enough to know that the vector 5(u, v) is obtained from the vector x(u, v) by a translation followed by an orthogona1 linear map. We observe that d(2, j)/d(u, v) # 0 in Q,; otherwise, the ZI component of some normal vector in R, is zero and there are two orthogonal normal lines in R,, a contradiction of our assumptions.
Figure 234
The expression of A(&) is given by
Since d(Z,y)/d(u, v) # 0, we can consider the change of coordinates X = X(U,v), y = j(u, v) and transform the above expression into
777
Geometric Definition o f Area
We remark now that, at pi, the vectors 2, and Z, belong to the Zj plane; therefore,
hence,
It follows that
where ei(tl, v) is a continuous function in Q,with e,(xl(pi))= 0. Since the length of a vector is preserved by translations and orthogonal linear maps, we obtain
Now let Mi and mi be the maximum and the minimum of the continuous function ci(u, v) in the compact region Q,; thus,
hence, mi
JI
du dv 5 A(R) 
Qt
/IQ,1$
A
1
dudv 5 M j
!IQ.
Doing the same for all R,, we obtain

Now, refine more and more the given partition in such a way that the norm p + 0. Then Mi mi. Therefore, there exists the limit of C,A(&), given by
which is clearly independent of the choice of the partitions and of the point pi in each partition. Q.E.D.
Appendix
A Brief Review
of Continuity and Bifferen fiabill'ty
R" will denote the set of ntuples ( x , , . . . ,x,) of real numbers. Although we use only the cases R1 = R,R2,and R3,the more general notion of R" unifies the definitions and brings in no additional difficulties; the reader may think in R2or R3,if he wishes so. In these particular cases, we shall use the following more traditiona1 notation: x or t for R,( x , y ) or (u, v) for R2, and (x,y, z) for R3.
A. Continuity in R" We start by making precise the notion of a point being €close to a given point p , E R". A ball (or open ball) in Rnwith center p, = (xy, . . . , x,O) and radius E > 0 is the set B,(p,)
= ((x,,
. . . , x,)
R n ; (XI

x?l2 k
. k (xn
x:) < f 2 ) .
Thus, in R, B,(p,) is an open interval with center p , and length 26; in R2, B,(p,) is the interior of a disk with center p, and radius E ; in R3,Bt(pO)is the interior of a region bounded by a sphere of center p, and radius E (see Fig. A2 I). A set U c Rn is an open set if for each p E U there is a ball B,(p) c U; intuitively this means that points in U are entirely surrounded by points of U, or that points sufficiently close to points of U still belong to U. For instance, the set
Appendix :Continuity and Differentiabiiit y
Figure A21
is easily seen to be open in R2. However, if one of the strict inequalities, say x c b, is replaced by x < b, the set is no longer open; no ball with center at the point (b, ( d c)/2), which belongs to the set, can be contained in the set (Fig. A22). I t is convenient to say that an open set in Rncontaining a pointp E R" is a neighborhood of p. From now on, U c Rnwill denote an open set in R".
+
We recall that a real function f : U c R R of a real variable is continuous at x, E U if given an E > 0 there exists a S > 0 such that if Ixx,] < S , then +
I f(x>  f(xJ I < €.
Figure A22
Regular Surfaces
120
Similarly, a real function f : U c R2+ R of two real variables is continuous at (x,, y o ) E U if given an 6 > 0 there exists a S > 0 such that if
+
(X  x ~ )( y~ y J 2 < S2, then
l 6. I f @ , Y )  f i ~ o , ~ o >< The notion of ball unifies these definitions as particular cases of the following general concept : A map F: U c R" Rm is continlcous at p E U if given E > 0, there exists a 6 > 0 such that +
In other words, F is continuous at p if points arbitrarily close to F(p) are images of points sufficiently close to p. It is easily seen that in the particular cases of n = 1 , 2 and m = 1, this agrees with the previous definitions. We say that F is continuous in U if F is cantinuous for all p E U (Fig. A23).
Figure A23
Given a map F: U c R" Rm,we can determine m functions of n variables as follows. Let p = (x,, . . . , x,) E U and f ( p ) = ( y ,, . . . ,y,). Then we can write +
The functions fi:U + R, i
=
I, . . . , r??, are the component functions of F.
Appendix: Continuity and Differentiabitit y
121
Example 1 (Symmetry). Let F : R3 + R3 be the map which assigns to each p E R3 the point which is symmetric to p with respect to the origin 0 t R3. Then F(p) = p, or
and the component functions of F a r e

Example 2 (Inversion). Let F: R2  ((0, 0)) R2 be defined as folIows. Denote by lp ( the distance to the origin (0,O) 0 of a point p E R2. By definition, F(p), p + 0, belongs to the halfline Op and is such that I F(p) 1 . 1 ~I = 1. Thus, F(p) pll p 12, or

and the component functions of F are
Example 3 (Projection). Let n : R3 R2 be the projection n(x, y, z) (x, 2') Thenf,(x, Y, z) = x,f2lx, Y, z) = Y* 4
=
The following proposition shows that the continuity of the map F is equivalent to the continuity of its component functions.
PROPOSITION 1. F: U c R" , Rm is continuous if and only component function fi: U c Rn + R, i = 1, . . . , m, is contirmous.
if each
Proof. Assume that F is continuous at p E U. Then given E > 0, there exists 6 > 0 such that F(B,(p)) c B,(F(p)). Thus, if q E B,(p), then F(q) t BE(F(P)), that is,
which implies that, for each i = I , . . . , m, I J.(q) fi(p)] < E . Therefore, given E > 0 there exists 6 > 0 such that if q E Ss(p), then Ii.(q) fi(p) I < EHence, each L. is continuous at p. Conversely, let J1., i = 1, . . . , m, be continuous at p. Then given E > 0 there exists 6, > 0 such that if q t S,,(p), then \f,(q)  J . ( p ) I < E I J ~ .Set
Regular Surfaces
S
< min Si and let q
E
Ss(p). Then
and hence, the continuity of F at p.
Q.E.D.
It follows that the maps in Examples 1,2, and 3 are continuous. Example 4. Let F: U c R

Rm.Then
This is usually called a vectorvaluedfunction, and the component functions of F are the components of the vector F(t) E Rm.When F is continuous, or, equivalently, the functions xi(t), i 1, . . . , m , are continuous, we say that F is a continuous curve in R".

In most applications, it is convenient to express the continuity in terms of neighborhoods instead of balls.
PROPOSITION 2. A map F : U c Rn + Rm is continuous at p +E U if and only $ given a neighborhood V of F(p) in Rm there exists a neighborhood W .of p in Rn such that F(W) c V . Proof. Assume that F i s continuous at p. Since Vis an open set containing F(p), it contains a ball B,(F(p)) for some E > 0. By continuity, there exists a ball B,(p) = W such that
and this proves that the condition is necessary. Conversely, assume that the condition holds. Let E > 0 be given and set V = B,(F(p)). By hypothesis, there exists a neighborhood W of p in Rnsuch that F( W) c V. Since W is open, there exists a ball B, ( p ) c W. Thus,
and hence the continuity of F at p.
Q.E.D.
The composition of continuous maps yields a continuous map. More precisely, we have the following proposition.
PROPOSITION 3. Let F: U c Rn + Rm and G: V c Rm. Rk be continuous maps, where IJ and V are open sets such that F(U) c V . Then G 0 F : U c Rn Rk is a continuous map. )
123
Appendix: Continuity and Differentiebility
Proof: Let p E U and let V be a neighborhood of G F ( p ) in Rk. By continuity of G , there is a neighborhood Q of F ( p ) in R m with G(Q) c V. By continuity of F, there is a neighborhood W of p in Rnwith F ( W ) c Q. Thus, G F ( W ) c G(Q) c V, 0
0
and hence the continuity of G 0 F.
Q.E.D.
It is often necessary to deal with maps defined on arbitrary (not necessarily open) sets of R". To extend the previous ideas to this situation, we shall proceed as follows. Let F: A c R" + Rm be a map, where A is an arbitrary set in R". We say that F is continuous in A if there exists an open set U c R", U =I A, and a continuous map F : U Rm such that the restriction I'I A = F. In other words, F is continuous in A if it is the restriction of a continuous map defined in an open set containing A. It is clear that if F: A c R" + Rm is continuous, given a neighborhood V of F ( p ) in Rm,p E A, there exists a neighborhood W of p in Rnsuch that F ( W n A) c V. For this reason, it is convenient to call the set W n A a neighborhood of p in A (Fig. A24).

Figure A24
Example 5. Let
be an ellipsoid, and let x : R3+ R2be the projection of Example 3. Then the restriction of n to E is a continuous map from E to R2.

We say that a continuous map F: A c RK Rn is a homeomorphism anto F ( A ) if F is onetoone and the inverse FI : F ( A ) c Rn3 Rnis continuaus. In this case A and F(A) are homeomorphic sefs.
Reguler Surfaces
Example 6. Let F: R3 + R3 be given by F(x, y, z ) = ( x a , yb, zc). Fis clearly continuous, and the restriction of F to the sphere

is a continuous map f : S2 R3. Observe that ~ ( S V= E, where E is the ellipsoid of Example 5. It is also clear that F is onetoone and that
Thus, P' = F  I 1 E is continuous. Therefore, sphere S2 onto the ellipsoid E.
is a homeomorphism of the
Finally, we want to describe two properties of real continuous functions defined on a closed interval [a, b],
(Props. 4 and 5 below), and an important property of the closed interval [a, b] itself. They will be used repeatedly in this book. PROPOSITION 4 (The Intermediate Value Theorem). Let f: [a, b] R be a continuous function defined on the closed interval [a, b]. Assume that f(a) and f(b) have opposite signs; that is, f(a)f(b) < 0. Then there exists a point c E (a, b) such that f(c) = 0. +
PROPOSITION 5. Let f : [a, b] be a continuous function defined in the closed interval [a, b]. Then f reaches its maximum and its minimum in [a, b]; that is, there exist points x,, x, E [a, b] such that f(x,) < f(x) < f(x2)for all x E [a, b]. PROPOSITION 6 (HeineBore]). Let [a, b] be a closed interval and let I,, a E A, be a collection of open intervals in [a, b] such that U, 1, = [a, b]. Then it is possible to choose a finite number I,,, I,,, . . . , I,, of I, such that U I k i= I , i = 1,..., n. These propositions are standard theorems in courses on advanced calculus, and we shall not prove them here. However, proofs are provided in the appendix to Chap. 5 (Props. 6, 13, and 11, respectively).
Appendix .Continuity and Differentiabitit y
B. Differentiability in R" Let f : U c R it exists)

R. The derivative f '(x,) off at x,
E
U is the limit (when
When f has derivatives at all points of a neighborhood V of x,, we can consider the derivative o f f ' : V R at x,, which is called the second derivative f "(x,) off at x,, and so forth. f is diferentiable at x, if it has continuous derivatives of all orders at x,. f is diferentiable in U if it is differentiable at all points in U. 4
Remark. We use the word differentiable for what is sometimes called infinitely differentiable (or of class C"). Our usage should not be confused with the usage of elementary calculus, where a function is called differentiable if its first derivative exists.

Let F: U c R2 R. The partial derivative of f with respect to x at (x,, yo) E U, denoted by (df/dx)(x,, yo), is (when it exists) the derivative at M, of the function of one variable: x f (x, yo). Similarly, the partial derivative with respect to y at (x,, yo), (df/dy)(xo, yo),is defined as the derivative at y o of y f (x,, y). When f has partial derivatives at a11 points of a neighborhood V of (x,, yo), we can consider the secondpartial derivatives at (x,, yo):


and so forth.f is differentiable at (x,, yo)if it has continuous partial derivatives of all orders at (x,, yo).f is differentiable in Uif it is differentiable at all points of U. We sometimes denote partial derivatives by
It is an important fact that when f is differentiable the partial derivatives off are independent of the order in which they are performed; that is, d2f  dzf mdydYdx'
d3f
d2xdy

d 3f dxdydx'
etc.
Regular Surfaces
326
The definitions of partial derivatives and differentiability are easily extended to functions f : U c Rn , R. For instance, (df/dx3)(xy, xi, . . . , x,") is the derivative of the function of one variable
A further important fact is that partial derivatives obey the socalled chain rule. For instance, if x = x(u, v), y = y(u, v), z = z(u, v) are real differentiable functions in U c R2 and f (x, y, z) is a real differentiable function in R3,then the composition f (x(u, v), y(u, v), z(u, v)) is a differentiable function in U, and the partial derivative off with respect to, say, ~ris given by

We are now interested in extending the notion of differentiabihty to maps F: U c R" Rm.We say that F is diflerentiable at p E U if its component functions are differentiable at p; that is, by writing
the functions f;., i = 1, . . . , m, have continuous partial derivatives of a11 orders at p. F is diflerentiable in U if it is differentiable at all points in U, For the case rn = 1, this repeats the previous definition. For the case n = 1, we obtain the notion of a (parametrized) diflerentiable curve in Rm, In Chap. 1, we have already seen such an object in R3. For our purposes, we need to extend the definition of tangent vector of Chap. 1 to the present Rmat to E U is the vector situation. A tangent vector to a map a : U c R in Rm

a'(t0)
= (xf1(f0),
. . . , xL(t0)).
Example 7. Let F: U c R2 , R3 be given by F(U, V)
=
(COSu cos v, cos u sin v, cos2 v),
(u, v)
U
The component functions of F, namely,
have continuous partial derivatives of all orders in U. Thus, F i s differentiable in U. Example 8. Let a: U c R a(t)

R4 be given by
= (t4, t3, t2, t),
t
E
U.
127
Appendix: Continuity and Differentiability
Then a is a differentiable curve in R4, and the tangent vector to a at t js af(t)= (4t3,3t2, 2t, I ) . Example 9. Given a vector w E Rm and a point p, E U c Rm,we can always find a differentiable curve a : ( E , 6 ) + U with a(0) = p, and a'(0) = w. Simply define a(t) = p, tw, t E (  E , E ) . By writing p, = (xy, . . . , x;) and w = (w,, . . . , w,), the component functions of a are xi(t) = xp tw,, i = 1, . . . , m . Thus, a is differentiable, a(0) = p, and
+
+
We shall now introduce the concept of differential of a differentiable map. It will play an important role in this book.

DEFLWITION I. Let F: U c Rn 3 Rmbe a diferentiable map. T o each p E U we associate a linear map dF, : R n Rmwhich is called the differential of F at p and is defined as follows. Let w E Rn and let a : (  E , E ) U be a diferentiable curve such that a(0) = p, a'(0) = w. By the chain rule, the curve = F 0 a : ( E , IF) Rmis also diferentiable. Then (Fig. A25)


Figure A25
PROPOSITION 7. The above definition of dF, does not depend on the choice of the curve which passes through p with tangent vector w, and dF, is, in fact, a linear map. Proof. T o simplify notation, we work with the case F: U c R2 4R3. Let (u, v) be coordinates in R2 and (x, y, z) be coordinates in R 3 . Let
Regular Surfaces
728
0), e2 = (0, I) be the canonical basis in R2 and fl = (I, 0, O), fi = (0,1, 0), f, = (0,0, 1) be the canonical basis in R3. Then we can write a(t) = (u(t), v(t)), t E (  E , €1, el
= (I,
F(u, v)
= Mu,
v), y(u, v), z(u, v)), and
Thus, using the chain rule and taking the derivatives at t
dzdu
= 0,
we obtain
dzdv
+(z5+zz)f3
This shows that dFp is represented, in the canonical bases of R 2 and R3, by a matrix which depends only on the partial derivatives at p of the component functions x, y, z of F. Thus, dF, is a linear map, and clearly dFp(w) does not depend on the choice of a. The reader will have no trouble in extending this argument to the more general situation. Q.E.D.

The matrix of dFp:Rn Rm in the canonical bases of Rn and Rm,that is, the matrix (d'ldx,), i = 1, . . . , m, j = I , . . . , n, is called the Jucobian matrix of F at p. When n = m, this is a square matrix and its determinant is called the Jacobian determinant; it is usual to denote it by
Remark. There is no agreement in the literature regarding the notation for the differential. It is also of common usage to call dFp the derivative of F a t p and to denote it by F'(p).
Example 10. Let F: R2 +R2 be given by
129
Appendix :Continuity and Differentiability
F is easily seen to be differentiable, and its differential dF, at p
For instance, dF,, , ,,(2, 3)
= (2,
= (x, y)
is
10).
One of the advantages of the notion of differential of a map is that it allows us to express many facts of calculus in a geometric language. Consider, for instance, the following situation : Let F: U c R2 + R3,G: V c R3 4 R2 be differentiable maps, where U and V are open sets such that F(U) c V. Let us agree on the following set of coordinates, ~c R2  VF C
R3
( 4 0)
( x , Y , z)
(t, G
R2
v)
and let us write , y(u, v), z(u, v)), F(u, v) = ( ~ ( uv), G(x, y, Z) = ( = 0. Hence,
Therefore, rl,(af(0>>= (dNp(a'(O)), ~ ' ( 0 ) ) = (N'(O), ar(o>) = (NO), ~ " ( 0 ) ) =
DEFINITION 4. The maximum normal curvature k , and the minimum normal curvature k, are called the principal curvatures at p; the corresponding directions, that is, the directions given by the eigenvectors e , , e,, are called principal directions at p. For instance, in the plane all directions at all points are principal direc
Definition o f the Gauss M a p
145
tions. The same happens with a sphere. In both cases, this comes from the fact that the second fundamental form at each point is constant (cf. Example 7); thus, all directions are extremals for the normal curvature. In the cylinder of Example 3, the vectors v and w give the principal directions at p, corresponding to the principal curvatures 0 and 1, respectively. In the hyperbolic paraboloid of Example 4, the x and y axes are along the principal directions with principal curvatures 2 and 2, respectively. DEFINITION 5. If a regular connected curve C on S is such that for all p E C the tangent line of C is a principal direction at p, then C is said to be a line of curvature of C. PROPOSITION 3 (Olinde Rodrigues). A necessary and sufficient condition for a connected regular curve C on S to be a line of curvature of S is that
for any parametrization a(t) of C , where N(t) = Noa(t) and R(t) is a dzflerentiable function o f t . In this case, R(t) is the (principal) curvature along al(t). Proof. It suffices to observe that if af(t) is contained in a principal direction, then af(t) is an eigenvector of dN and
The converse is immediate.
Q.E.D.
The knowledge of the principal curvatures at p allows us to compute easily the normal curvature along a given direction of Tp(S). In fact, let v E Tp(S) with I v I = 1. Since e , and e, form an orthonormal basis of T,(S), we have v
= el
cos 0
+ e, sin 0,
where 0 is the angle from e, to v in the orientation of Tp(S). The normal curvature k , along v is given by
k,
= IIP(v) =
(dNP(v), V) = (dNp(e, cos 0 e2 sine), el cose e, sine) = ( e l k l cos 9 e2k2sin 8, e, cos 0 e, sin 9) = k 1 cos2 0 k , sin2 0. 
+
+
+
+
+
The last expression is known cIassically as the Euler formula; actually, it is just the expression of the second fundamental form in the basis { e l ,e,].
146
The Geometry of the Gauss Map
Given a linear map A : V + V of a vector space of dimension 2 and given a basis (v,, v,) of V, we recall that determinant of A
= a, ,a2, 
a,,a,, ,
trace of A
= a,,
+ a,,,
where (a,,) is the matrix .of A in the basis (v,, v,). It is known that these numbers do not depend on the choice of the basis (v,, vz) and are, therefore, attached to the linear map A. In our case, the determinant of dN is the product (k,)(k,) = k,k2 of k,) of the principal curvatures, and the trace of dN is the negative (k, the sum of principal curvatures. If we change the orientation of the surface, the determinant does not change (the fact that the dimension is even is essential here); the trace, however, changes sign.
+
DEFINITION 6. Let p E S and let dN,: T,(S) .Tp(S) be the diflerential of the Gauss map. The determinant of dNp is the Gaussian curvature K of S at p. The negative of halfof the trace of dN, is called the mean curvature H of S at p. In terms of the principal curvatures we can write
DEFINITION 7. A point of a surface S is called 1. Elliptic ifdet(dN,) > 0. 2. Hyperbolic ifdet(dN,) < 0. 3. Parabolic ifdet(dN,) = 0, with dN, # 0. 4. Planar if dN, = 0. It is clear that this classification does not depend on the choice of the orientation. At an elliptic point the Gaussian curvature is positive. Both principal curvatures have the same sign, and therefore all curves passing through this point have their normal vectors pointing toward the same side of the tangent plane. The points of a sphere are elliptic points. The point (0, 0, 0) of the paraboloid z = x2 ky2, k > 0 (cf. Example 5), is also an elliptic point. At a hyperboIic point, the Gaussian curvature is negative. The principal curvatures have 03osite signs, and therefore there are curves through p whose normal vectors at p point toward any of the sides of the tangent plane at p. The point (0, 0,0) of the hyperbolic paraboloid z = y 2  x2 (cf. Example 4) is a hyperbolic point. At a parabolic point, the Gaussian curvature is zero, but one of the prin
+
Definition o f the Gauss Map
747
cipal curvatures is not zero. The points of a cylinder (cf. Example 3) are parabolic points, Finally, at a planar point, all principal curvatures are zero. The points of a plane trivially satisfy this condition. A nontrivial example of a planar point was given in Example 6.
DEFINITION 8. f l a t p E S, k, = k,, then p is called an umbilical point of S; in particular, the planar points (k, = k2 = 0) are umbilical points. All the points of a sphere and a plane are umbilical points. Using the method of Example 6, we can verify that the point (0, 0, 0) of the paraboloid 1 = x2 y 2 is a (nonplanar) umbilical point. We shall now prove the interesting fact that the only surfaces made up entirely of umbilical points are essentially spheres and planes.
+
PROPOSITION 4, I f all points of a connected surface S are umbilical points, then S is either contained in a sphere or in aplane. Proof. Let p E S and let x(u, v) be a parametrization of S at p such that the coordinate neighborhood V is connected. Since each q E V is an umbilical point, we have, for any vector w = a,x, a,% in T*(S),
+
dN(w)
= l(q)w,
where ;l = A(q) is a real differentiable function in V. We first show that a(q) is constant in V. For that, we write the above equation as
+
NU@,
= l(x,a,
+ xva2);
hence, since w is arbitrary,
Nu = ax,, Nu = axv. Differentiating the first equation in u and the second one in v and subtracting the resulting equations, we obtain
Since x, and x, are linearly independent, we conclude that
for all q
E
V. Since V is connected, 2 is constant in V, as we claimed.
The Geometry o f the Gauss Map
748
If a = 0, Nu = N, = 0 and therefore N ( ~ ( u v), , No), = (x(u, v), No), = 0 ; hence, (x(u, v), No)
= No = constant
= const.,
and all points x(u, v) of V belong to a plane. If A # 0, then the point x(u, v)  (l/a)N(u, v) (x(u, v)

1
a
N(u, v)),

in V. Thus,
= (x(u,
v)

= y(u,
v) is fixed, because
1 XN(u, v))v
=
0
Since
all points of V are contained in a sphere of center y and radius 11111. This proves the proposition locally, that is, for a neighborhood of a point p E S. To complete the proof we observe that, since S is connected, given any other point r E S, there exists a continuous curve a : [0, 11 3 S with a(0) = p , a(1) = r . For each point a(t) E S of this curve there exists a neighborhood V, in S contained in a sphere or in a plane and such that a'(Vt) is an open interval of [0, 11. The union U al(K), t E [O, 11, covers [O, 11 and since [0, I] is a closed interval, it is covered by finitely many elements of the family {a'(V,)) (cf. the HeineBore1 theorem, Prop. 6 of the appendix to Chap. 2). Thus, c([O, 11) is covered by a finite number of the neighborhoods V,. If the points of one of these neighborhoods are on a plane, a11 the others will be on the same plane. Since r is arbitrary, all the points of S belong to this plane. If the points of one of these neighborhoods are on a sphere, the same argument shows that all points on S belong to a sphere, and this completes Q.E.D. the proof.
DEFINITION 9. Let p be apoint in S. An asymptotic direction of S at p is a direction of Tp(S)for which the normal curvature is zero. An asymptotic curve of S is a regular connected curve C c S such that for each p E C the tangent line of C at p is an asymptotic direction. It follows at once from the definition that at an elliptic point there are no asymptotic directions. A useful geometric interpretation of the asymptotic directions is given by means of the Dupin indicatrix, which we shall now describe. Let p be a point in S. The Dupin indicatrix at p is the set of vectors w of Tp(S) such that IIp(w) = & 1.
Definition o f the Gauss Map
749
To write the equations of the Dupin indicatrix in a more convenient form, let (t,q ) be the cartesian coordinates of T p ( S )in the orthonormal basis ( e l ,e,), where e , and e2 are eigenvectors of dNp. Given w E Tp(S),let 1 and v = p and 8 be "polar coordinates" defined by w = pv, with I v [ e, cos 8 e, sin 8, if p # 0. By Euler's formula,

+
+
where w = t e , qe2. Thus, the coordinates indicatrix satisfy the equation
(c, q ) of a point of the Dupin
hence, the Dupin indicatrix is a union of conics in Tp(S).We notice that the normal curvature along the direction determined by w is k,(v) = Il,(v) =i(l/p2). For an elliptic point, the Dupin indicatrix is an ellipse ( k , and k2 have the same sign); this ellipse degenerates into a circle if the point is an umbilical nonplanar point (k, = k2 # 0). For a hyperbolic point, k 1 and k2 have opposite signs. The Dupin indicatrix is therefore made up of two hyperbolas with a common pair of asymptotic lines (Fig. 313). Along the directions of the asymptotes, the normal curvature is zero; they are therefore asymptotic directions. This justifies the terminology and shows that a hyperbolic point has exactly two asymptotic directions.
Elliptic point
Hyperbolic point
Figure 313. The Dupin indicatrix.
For a parabolic point, one of the principal curvatures is zero, and the Dupin indicatrix degenerates into a pair of parallel lines. The common direction of these lines is the only asymptotic direction at the given point. In Example 5 of Sec. 34 we shall show an interesting property of the Dupin indicatrix. Closely related with the concept of asymptotic direction'is the concept of conjugate directions, which we shall now define.
750
The Geometry o f the Gauss Map
DEFINITION 10, Let p be a point on a surface S. Two nonzero vectors w,,w, E T p ( S )are conjugate if (dN,(wl), w,) = (w,,dNp(w,)> = 0. Two directions r , , r, at p are conjugate if a pair of nonzero vectors w,, w, paralEel to r, and r,, respectively, are conjugate. It is immediate to check that the definition of conjugate directions does not depend on the choice of the vectors w , and w, on r, and r,. It follows from the definition that the principal directions are conjugate and that an asymptotic direction is conjugate to itself. Furthermore, at a nonplanar umbilic, every orthogonal pair of directions is a pair of conjugate directions, and at a planar umbilic each direction is conjugate to any other direction. Let us assume that p E S is not an umbilical point, and let { e l ,e,) be the orthonormal basis of T p ( S ) determined by dN,(e,) = k,e,, dN,(e,) = k,e,, Let 8 and p be the angles that a pair of directions r , and r2 make with e , . We claim that r , and r, are conjugate if and only if k , cos 8 cos p
=
k2 sin 8 sin p.
(2)
In fact, r , and r, are conjugate if and only if the vectors W,
= el cos 8
+ e, sin 8,
w2 = el cos 9
+ e2 sin p
are conjugate. Thus, 0
= 0 be a small number such that
is a regular curve (we may have to change the orientation of the surface to achieve E > 0). We want to show that i f p is not a planar point, the curve Cis "approximately" similar to the Dupin indicatrix of S at p (Fig. 320). To see this, let us assume further that the x and y axes are directed along the principal directions, with the x axis along the direction of maximum principal curvature. Thus, f = hxy(O,0) = 0 and
By developing h(x, y) into a Taylor's expansion about (0, O), and taking into account that h,(O, 0) = 0 = hy(O, 0), we obtain
+
h(x, 4') = +(h,,(O, O)x2 h,,(O, = g k l x 2 $ k2y2) R? where
Thus, the curve C is given by
+
t hn(O, Oly
+R
The Gauss Map in Loca/ Coordinates
A plane parailel to T p ( S )
A77 +
Now, if p is not a planar point, we can consider k , x 2 k 2 y 2 = 26 as a firstorder approximation of C. By using the similarity transformation,
we have that k , x 2
+ k 2 y 2 = 26 is transformed into the curve
which is the Dupin indicatrix at p. This means that i f p is a nonplanarpoint, the intersection with S of a plane parallel to T,(S) and close to p is, in afirslorder approximation, a curve similar to the Dupin indicatrix at p. If p is a planar point, this interpretation is no longer vaIid (cf. Exercise 11). T o conclude this section we shall give a geometrical interpretation of the Gaussian curvature in terms of the Gauss map N: S ,S 2 . Actually this was how Gauss himself introduced this curvature. To do this, we first need a definition. Let S and be two oriented regular surfaces. Let 4:S  3 be a differentiable map and assume that for some p E S, dip, is nonsingular. We say that q is orientationpreserving at p if given a positive basis {w,, w23in T,(S),
7 66
The Geometry of the Gauss Map
then {dp,(w,), dg,(w2)) is a positive basis in T,,,,(S). If [dp,(w,), dp,(w2)) is not a positive basis, we say that p is orientationreversing at p. We now observe that both S and the unit sphere S2are embedded in R3. Thus, an orientation N on S induces an orientation N in S2.Let p E S be such that dN, is nonsingular. Since for a basis (w,, w2) in T,(S)
the Gauss map N will be orientationpreserving at p E S if K(p) > 0 and orientationreversing at p E S if K(p) < 0. Intuitively, this means the following (Fig. 321): An orientation of T,(S) induces an orientation of small closed curves in S around p ; the image by N of these curves will have the same or the opposite orientation to the initial one, depending on whether p is an elliptic or hyperbolic point, respectively. To take this fact into account we shall make the convention that the area of a region contained in a connected neighborhood V, where K # 0, and the area of its image by N have the same sign if K > 0 in V, and opposite signs if K < 0 in V (since V is connected, K does not change sign in V).
Figure 321. The Gauss map preserves orientation at an elliptic point and reverses it at a hyperbolic point.
The Gauss M a p in Local Coordinates
767
Now we can state the promised geometric interpretation of the Gaussian curvature K, for K # 0. PROPOSITION 2. Let p be a point of a surface S such that the Gaussian curvature K(p) # 0, and let V be a conlzectedneighb~rhoodof p where K does plot change sign. Then K(p) = lim A' Ah0 A where A is the area of a region B c V containing p, A' is the area of the image of B by the Gauss map N : S + S2, and the limit is taken through a sequence of regions B, that converges to p, in the sense that any sphere around p contains all B,, for n suficiently large. ProoJ The area A of B is given by (cf. Sec. 25)
!$.here x(u, v) is a parametrization in p, whose coordinate neighborhood contains V ( V can be assumed to be sufficientlysmall) and R is the region in the uv plane corresponding to B. The area A' of N(B) is
Using Eq. (I), the definition of K, and the above convention, we can write A'
=
JJ
R
K\ xu A x. 1 du dv.
Going to the limit and denoting also by R the area of the region R, we obtain
[J
lim (1 /R) KJxu x, J du d?, A'/R  R.,o A' = lim R lim A A R+O AIR lim(l/R)SJ ] x uA x.ldudv R+O
R
(notice that we have used the mean value theorem for double integrals), and Q.E.D. this proves the proposition. Remark. Comparing the proposition with the expression of the curvature
768
The Geometry o f the Gauss Map
of a plane curve C a t p (here s is the arc length of a small segment of C containing p, and a is the arc length of its image in the indicatrix of tangents; cf. Exercise 3 of Sec. 15), we see that the Gaussian curvature K is the analogue, for surfaces, of the curvature k of plane curves.
EXERCISES 1. Show that at the origin (0, 0,0)of the hyperboloid and H = 0.
z = axy
we have K
= a2
"2. Determine the asymptotic curves and the lines of curvature of the helicoid x = v cos U,y = v sin u, z = cu, and show that its mean curvature is zero.
"3. Determine the asymptotic curves of the catenoid
X(U,v)
= (cosh v
cos u, cosh v sin u, v).
4. Determine the asymptotic curves and the lines of curvature of z
= xy.
5. Consider the parametrized surface (Enneper's surface)
and show that
a. The coefficients of the first fundamenta1 form are
b. The coefficients of the second fundamentaI form are
c, The principal curvatures are
d. The lines of curvature are the coordinate curves. e. The asymptotic curves are u v = const., u  v
+
= const.
6 . ( A Surface with K = 1; the Pseudosphere.) *a. Determine an equation for the plane curve C, which is such that the segment of the tangent line between the point of tangency and some liner in the plane, which does not meet the curve, is constantly equal to 1 (this curve is called the tractrix; see Fig. 19). b. Rotate the tractrix C about the line r; determine if the "surface" of revolution thus obtained (the pseudosphere; see Fig. 322) is reguIar and find out a parametrization in a neighborhood of a regular point.
The Gauss Map in Local Coordinates
Rotation axis
Figure 322. The pseudosphere.
Figure 323
c. Show that the Gaussian curvature of any regular point of the pseudosphere is 7. (Surfaces of Revolution with Constant Curvature.) ( p ( v )cos u, p(v) sin u, ~ ( v ) ) is given as a surface of revolution with constant Gaussian curvature K. To determine the functions p and y, choose the parameter v in such a way that ( ~ 9 2 ( v / ' )= ~ 1 (geometrically, this means that v is the arc length of the generating curve (p(v),v/(v))).Show that
+
+
a. p satisfies q" K p = 0 and y is given by y = I d 1  ( P ' )dv; ~ thus, 0 < u < 271, and the domain of v is such that the last integral makes sense. b. All surfaces of revolution with constant curvature K = 1 which intersect perpendicularly the plane xOy are given by
where C is a constant (C = p(0)). Determine the domain of v and draw a rough sketch of the profiIe of the surface in the xz plane for the cases C = 1, C > 1, C < 1. Observe that C = 1 gives a sphere (Fig. 323). c. All surfaces of revohtion with constant curvature K =  1 may be given by one of the following types : 1. p(v)
= C cosh
y(v) =
1' 0
d l
v, 
C Zsinhz vdv.
The Geometry of the Gauss Map
2. ~ ( v = ) C sinh .z~, ~ ( v= ) J u 41  C2 cosh2 vdv. 0
Determine the domain of v and draw a rough sketch of the profile of the surface in the xz plane. d. The surface of type 3 in part c is the pseudosphere of Exercise 6.
e. The only surfaces of revolution with K the right circular cone, and the plane.
= 0 are the right circular cylinder,
8. (Contact of Order 2 2 of Surfaces.) Two surfaces S and f, with a common point p, have contact of order 2 2 at p if there exist parametrizations x(u, v) and %(u,71) in p of S and 3, respectively, such that
at p. Prove the following: *a. ~ e st and ?!, have contact of order 2 2 at p; x: U + S and 2 : u , be arbitrary parametrizations in p of S and 3, respectively; and f: V c R3 + R be a differentiable function in a neighborhood V o f p in R3. Then the partial derivatives of order < 2 off 0 ii: U 3 R are zero in 21(p) if and only if the partial derivatives of order < 2 off 0 x : U , R are zero in x'(p). *b. Let S and 3 have contact of order
> 2 at p. Let z = f (x,y), z =J ( X , ~be ) the
equations, in a neighborhood of p, of S and 3, respectively, where the xy plane is the common tangent plane at p = (0,O). Then the function f (x, y)  f (n,y) has all partial derivatives of order < 2, at (0, O), equal to zero. c. Let p be a point in a surface S c R3. Let Oxyz be a cartesian coordinate system for R3 such that 0 = p and the xy plane is the tangent plane of S at p. Show that the paraboloid
obtained by neglecting third and higherorder terms in the Taylor development around p = (0, O), has contact of order > 2 at p with S (the surface (*) is called the osculating paraboloid of S at p). *d. If a paraboloid (the degenerate cases of plane and parabolic cylinder are included) has contact of order 2 2 with a surface S at p, then it is the osculating paraboloid of S at p.
e. If two surfaces have contact of order 2 2 at p, then the osculating paraboloids of S and at p coincide. Conclude that the Gaussian and mean curvatures of S and at p are equal. f. The notion of contact of order > 2 is invariant by diffeomorphisms of R3;
The Gauss Map in Local Coordinates

171
that is, if S and S have contact of order 2 2 at p and q : R 3 R 3 is a diffeomorphism, then y(S) and p(3) have contact of order > 2 at q ( p ) . g. If S and f have contact of order 2 2 at p, then
where d is the length of the segment cut by the surfaces in a straight line normal to T,(S) = T,(S), which is at a distance r fromp. 9. (Contact of Curves.) Define contact of order 2 n (n integer 2 1) for regular curves in R3 with a common point p and prove that
a. The notion of contact of order 2 n is invariant by diffeomorphisms. b. Two curves have contact of order 2 1 at p if and only if they are tangent at p. 10. (Contact of Curves and Surfaces.) A curve C and a surface S, which have a common point p, have contact of order > n (n integer 2 1 ) at p if there exists a curve c S passing throughp such that C and have contact of order 2 n at p. Prove that a. If f ( x , y, z ) = 0 is a representation of a neighborhood of p in S and a ( t ) = (x(t),y(t), z(t))is a parametrization of C in p, with a(0) = p, then C and S have contact of order 2 n if and only if
where the derivatives are computed for t = 0 . b. If a plane has contact of order > 2 with a curve C a t p, then this is the osculating plane of C at p. c. If a sphere has contact of order 2 3 with a curve C at p, and a(s) is a parametrization by arc length of this curve, with a(0) = p, then the center of the sphere is given by
Such a sphere is caIled the osculating sphere of C at p. 11. Consider the monkey saddle S of Example 2. Construct the Dupin indicatrix at p = ( 0 , 0 , O ) using the definition of Sec. 32, and compare it with the curve obtained as the intersection of S with a plane parallel to T,(S) and close to p. Why are they no1 "approximately similar" (cf. Example 5 of Sec. 33)? G o through the argument of ~ x a m ~5l of e Sec. 33 and point out where it breaks down.
12. Consider the parametrized surface
sin u cos v , sin u sin v, cos u
u + log tan 2 + q(u)
The Geometry o f the Gauss Map
172
where p is a differentiable function. Prove that a. The curves v = const. are contained in planes which pass through the z axis and intersect the surface under a constant angle 8 given by cos B =
P'
41
+ (P')~
Conclude that the curves v = const. are lines of curvature of the surface. b. The length of the segment of a tangent line to a curve v = const., determined by its point of tangency and the z axis, is constantly equal to 1. Conclude that the curves v = const. are tractrices (cf. Exercise 6). 13. Let F: R3 ,R 3 be the map (a similarity) defined by F(p)
R3, c a positive constant. Let S c R3 be a regular surface and set F(S) = 3. Show that is a regular surface, and find formulas relating the Gaussian and mean curvatures, K and H, of S with the Gaussian and mean curvatures, K and fi, of 3. =
cp, p
E
14. Consider the surface obtained by rotating the curve y = x3,  1 < x < 1, about the line x = 1. Show that the points obtained by rotation of the origin ( 0 , O ) of the curve are planar points of the surface. *15. Give an example of a surface which has an isolated parabolic point p (that is, no other parabolic point is contained in some neighborhood of p). '16. Show that a surface which is compact (ie., it is bounded and closed in R3)has an elliptic point. 17. Define Gaussian curvature for a nonorientable surface. Can you define mean curvature for a nonorientable surface? 18. Show that the Mobius strip of Fig. 31 can be parametrized by
x(u, v)
=
((2

"). (
")
"2
v sin  sln u, 2  v sin  cos u, v cos 
2
2
and that its Gaussian curvature is K=

{ivz + (2

1 v sin ( ~ / 2 ) ) ~ 3 ~ '
*19. Obtain the asymptotic curves of the onesheeted hyperboloid x Z = 1.
+ y2  z2
*20. Determine the umbilical points of the elipsoid
*21. Let S be a surface with orientation N. Let V c S be an open set in S and let f : V c S + R be any nowherezero differentiable function in V. Let v, and vz be two differentiable (tangent) vector fields in V such that at each point of V, v, and v2 are orthonormal and vl A i?'z = N.
The Geuss Map in Locel Coordinetes
a. Prove that the Gaussian curvature K of V is given by
The virtue of this formula is that by a clever choice o f f we can often simplify the computation of K, as illustrated in part b. b. Apply the above resuIt to show that iff is the restriction of
to the ellipsoid
then the Gaussian curvature of the ellipsoid is
22. (The Hessian). Let h : S + R be a differentiable function on a surface S, and let p E S be a critical point of h (i.e., dh, 0). Let w E T,(S) and let

be a parametrized curve with a(0)
= p,
a'(0) = w. Set
a. Let x: U  S be a parametrization of S at p, and show that (the fact that p is a critical point of h is essential here)
Conclude that H,h: T,(S) 4 R is a welldefined (i.e., it does not depend on the choice of a ) quadratic form on Tp(S). H,h is called the Hessian of h at p. b. Let h: S + R be the height function of S relative to Tp(S); that is, h(q) = (q  p, N(p)), q G S. Verify that p is a critical point of h and thus that the Hessian Hph is well defined. Show that if w E Tp(S), I w [ = 1, then Hph(w) = normal curvature at p in the direction of w. Conclude that the Hessian at p of the height function relative to T,(S) is the secondfundamental form of S at p. 23. (Morse Functions on Surfaces.) A critical pointp E S of a differentiable function h: S + R is nondegenerate if the selfadjoint linear map Aph associated to the
174
The Geometry of the Gauss Map
quadratic form Hph (cf. the appendix to Chap. 3) is nonsingular (here Hph is the Hessian of h at p ; cf. Excercise 22). Otherwise, p is a degenerate critical point. A differentiable function on S is a Morse function if all its critical points are nondegenerate. Let h,: S c R3 + R be the distance function from S to r ; i,e.,
a. Show that p E S is a critical point of h, if and onIy if the straight line p r is normal to S at p. R. Let w t z Tp(S), ( w f = 1, and let b. Let p be a critical point of h,: S a :(€, f) + S be a curve parametrized by arc length with a(0) =p, ~ ' ( s = ) w. Prove that
where k, is the normal curvature at p along the directioa of w . Conclude that the orthonormal basis {el, ez>,where e and e2 are along the principal directions of Tp(S), diagonalizes the selfadjoint linear map Aph,. Conclude further thatp is a degenerate critical point of h, if and only if either h,(p) = l l k l or h,(p) = l/k2, where k l and kz are the principal curvatures at p. c. Show that the set
B = (r E R 3 ; h, is a Morse function] is an open and dense set in R3; here dense in R3 means that in each neighborhood of a given point of R3 there exists a point of B (this shows that on any regular surface there are "many" Morse firnctions).
24. (Local Convexity and Curvature). A surface S c R3 is locally convex at a point p 6 S if there exists a neighborhood V c S o f p such that V is contained in one of the closed halfspaces determined by Tp(S) in R 3 . If, in addition, V has only one common point with Tp(S), then S is called strictly locally convex at p. a. Prove that S is strictly locallyconvex at p if the principal curvatures of S at p are nonzero with the same sign (that is, the Gaussian curvature K(p) satisfies K(P) > 0). b. Prove that if S is locally convex at p, then the principal curvatures at padonot have different signs (thus, K(p) 2 0). c. To show that K 2 0 does not imply local convexity, consider the surface f(x, y) = x3(l y2), defined in the open set U = {(x, y) E R 2 ;y2 < $1. Show that the Gaussian curvature of this surface is nonnegative on U and yet the surface is not locally convex at (0,O) 6 U (a deep theorem, due to R. Sacksteder, implies that such an example cannot be extended to the entire RZ if we insist on keeping the curvature nonnegative; cf. Remark 3 of Sec. 56). *d. The example of part c is also very special in the following local sense. Let p be a point in a surface S, and assume that there exists a neighborhood V c S
+
175
Vector Fields
of p such that the principal curvatures on V do not have different signs (this does not happen in the example of part c). Prove that S is locally convex at p.
34. Vector Fields? In this section we shall use the fundamental theorems of ordinary differential equations (existence, uniqueness, and dependence on the initial conditions) to prove the existence of certain coordinate systems on surfaces. If the reader is willing to assume the results of Corollaries 2, 3, and 4 at the end of this section (which can be understood without reading the section), this material may be omitted on a first reading. We shall begin with a geometric presentation of the material on differentiaI equations that we intend to use. A vector Jield in an open set U c R2 is a map which assigns to each q E U a vector w(q) E R Z .The vector field w is said to be dzferentiable if writing q = (x, y) and w(q) = (a(x, y), b(x, y)), the functions a and b are differentiable functions in U. Geometrically, the definition corresponds to assigning to each point (x, y) E U a vector with coordinates a(x, y) and b(x, y) which vary differentiably with (x, y) (Fig. 324).
I
o
X
Figure 324
In what follows we shall consider only differentiable vector fields. In Fig. 325 some examples of vector fields are shown. Given a vector field w, it is natural to ask whether there exists a trajectory of this field, that is, whether there exists a differentiable parametrized curve a(t) = (x(t), y(t)), t E I, such that al(t) = w(a(t)). For instance, a trajectory, passing through the point (x,, yo), of the vector field w(x, y) = (x, y) is the straight line a(t) = (xoet,y,et), t E R, and a trajectory of w(x, y) = (y, x), passing through (x,, yo), is the circle yi. P(t) = ( r sin t, r cos t), t E R,r 2 = X;
+
?This section may be omitted on a first reading.
The Geometry o f the Gauss Map
Figure 325
In the language of ordinary differential equations, one says that the vector field w determines a system of differential equations,
and that a trajectory of w is a solution to Eq. ( I ) . The fundamental theorem of (local) existence and uniqueness of solutions of Eq. (1) is equivalent to the following statement on trajectories (in what follows, the letters I and J will denote open intervals of the line R, containing the origin 0 E R).
THEOREM 1. Let w be a vector field in an open set U c R2. Given p E U , there exists a trajectory a: I ,U of w (i.e., af(t) = w(a(t)), t E I) with a(0) = p. This trajectory is unique in the following sense: Any other trajectory /?: J U with P(O) = p agrees with a in I n J. t
An important complement to Theorem 1 is the fact that the trajectory passing through p "varies differentiably with p." This idea can be made precise as follows. THEOREM 2. Let w be a vector $eld in an open set U c R2. For each p E U there exist a neighborhood V c U of p, an interval I, and a mapping a : V x I + U such that 1. For afixed q E V , the curve a(q, t), t E I, is the trajectory of w passing through q; that is,
Vector Fields
2. a is diflerentiable. Geometrically Theorem 2 means that all trajectories which pass, for t = 0: in a certain neighborhood V of p may be "collected" into a single differentiable map. It is in this sense that we say that the trajectories depend differentiably on p (Fig. 326).
Figure 326
The map a is calIed the (local)Jow of w at p. Theorems 1 and 2 will be assumed in this book; for a proof, one can consult, for instance, W. Hurewicz, Lectures on Ordinary Diflerential Equations, M.I.T. Press, Cambridge, Mass., 1958, Chap. 2. For our purposes, we need the following consequence of these theorems. L E M M A . Let w be a vector field in an open set U c R2 and let p E U be such that w(p) # 0. Then there exist a neighborhood W c U of p and a diflerentiable function f : W R such that f is constant along each trajectory of w and df, # O.for all q E W . 3
Proof. Choose a cartesian coordinate system in R2 such that p = (0,O) and w(p) is in the direction of the x axis. Let a : V x I , U be the local flow at p, V c U, t E I, and let 8 be the restriction of a to the rectangle
(See Fig. 327.) By the definition of local flow, d&, maps the unit vector of the t axis into w and maps the unit vector of the y axis into itself. Therefore, d&, is nonsingular. It follows that there exists a neighborhood W c U of p, where &  I is defined and differentiable. The projection of &l(x,y) onto the y axis is a differentiable function 5 = f (x, y), which has the same
The Geometry of the Gauss Map
x
1
Figure 327
c).
value ( for all points of the trajectory passing through (0, Since dB, is nonsingular, W may be taken sufficiently small so that df, # 0 for all q E W. f is therefore the required function. Q.E.D. The function f of the above lemma is called a (local)$rst integral of w in a neighborhood of p. For instance, if w(x, y) = (y, x) is defined in R2, a first integral8 R2  ((0, 0)) ,R is f (x, y) = x2 y2.
+
Closely associated with the concept of vector field is the concept of field of directions. Afield of directions r in an open set U c R2 is a correspondence which assigns to each p E U a line r(p) in R2 passing through p. r is said to be diferentiable at p E U if there exists a nonzero differentiable vector field w, defined in a neighborhood V c U of p, such that for each q E V, w(q) # 0 is a basis of r(q); r is diferentiable in U if it is differentiable for every p E U. To each nonzero differentiable vector field w in U c R2,there corresponds a differentiable field of directions given by r(p) = line generated by w(p), p € U. By its very definition, each differentiable field of directions gives rise, locally, to a nonzero differentiable vector field. This, however, is not true globally, as is shown by the field of directions in R2  {(O, O)] given by the tangent lines to the curves of Fig. 328; any attempt to orient these curves in order to obtain a differentiable nonzero vector field leads to a contradiction. A regular connected curve C c U is an integral curre of a field of directions r defined in U c R2 if r(q) is the tangent line to C at q for all q E C. By what has been seen previously, it is clear that given a differentiable field of directions r in an open set U c R2, there passes, for each q E U, an integral curve C of r ; C agrees locally with the trace of a trajectory through q
Vector Fields
of the vector field determined in U by r. In what follows, we shall consider only differentiable fields of directions and shall omit, in general, the word differentiable. A natural way of describing a field of directions is as follows. We say that two nonzero vectors w , and w, at q E R2 are equivalent if w , = Aw, for some A E R, A + 0. Two such vectors represent the same straight line passing through g, and, conversely, if two nonzero vectors belong to the same straight line passing through q, they are equivalent. Thus, a field of directions r on an open set U c R2 can be given by assigning to each q E U a pair of real numbers (r,, r 2 ) (the coordinates of a nonzero vector belonging to r), where we consider the pairs (r,, r,) and (Ar,, Ar,), A + 0, as equivalent. In the language of differential equations, a field of directions r is usually given by
which simply means that at a point q = (x, y) we associate the line passing through q that contains the vector (b, a) or any of its nonzero multiples (Fig. 329). The trace of the trajectory of the vector field (by a) is an integral curve of r. Because the parametrization plays no role in the above considerations, it is often used, instead of Eq. (2), the expression
with the same meaning as before. The ideas introduced above belong to the domain of the local facts of R2, which depend only on the "differentiable structure" of R2. They can, therefore, be transported to a regular surface, without further difficulties, as follows.
DEFINITION 1. A vector field w in an open set U c S of a regular surface S is a correspondence which assigns to each p r U a vector w(p) E T,(S). The vector field w is differentiable at p E U if, for some parametrization
The Geometry of the Gauss Map
Figure 329. The differential equation adx
+ bdy = 0.
X(U,V) at p, the functions a(u, v) and b(u, v) given by
are differentiable functions at p; it is clear that this dejnition does not depend on the choice of x.
We can define, similarly, trajectories, field of directions, and integral curves. Theorems 1 and 2 and the lemma above extend easily to the present situation; up to a change of R2 by S, the statements are exactly the same. Example 1. A vector field in the usual torus T is obtained by parametrizing the meridians of T by arc length and defining w(p) as the velocity vector of the meridian through p (Fig. 330). Notice that I w(p) ( = 1 for all p E 7. It is left as an exercise (Exercise 2) to verify that w is differentiable. Example 2. A similar procedure, this time on the sphere S2and using the semimeridians of S2,yields a vector field w defined in the sphere minus the
181
Vector fields
two poles N and S. To obtain a vector field defined in the whole sphere, reparametrize all the semimeridians by the same parameter t ,  1 < t < I , and define v(p) = (1  t2)w(p) forp E S2 (N) U IS) and v(N) = v(S) = 0 (Fig. 331).
Figure 331
Example 3. Let S = ((x, y, z ) E R 3 ;z = x2  y 2 ] be the hyperbolic paraboloid. The intersection with S of the planes z = const. # 0 determines a family of curves {C,) such that through each point of S  ((0,0,O)) there passes one curve C,. The tangent lines to such curves give a differentiable field of directions r on S  ((0, 0, 0)). We want to find a field of directions r' on S  {(O, 0, O)] that is orthogonal to r at each point and to determine the integral curves of r'. r' is called the orthogonaljield to r, and its integral curves are called the orthogonalfamily of r (cf. Exercise 15, Sec. 25). We begin by parametrizing S by
The family (C,)is given by u2  v2 = const. # 0 (or rather by the image v'x, is a tangent vector of a regular parametrizaunder x of this set). If u'x, tion of some curve C,, we obtain, by differentiating u 2  v2 = const.,
+
Thus, (u', v') = (v, u). It follows that r is given, in the parametrization x, by the pair (v, u) or any of its nonzero multiples. Now, let (a(u,v), b(u, v)) be an expression for the orthogonal field r', in the parametrization x. Since
and r' is orthogonal to r at each point, we have
182
The Geometry o f the Gauss Map
Eav
or (1
+ F(bv + au) + Gbu = 0
+ 4u2)av  4uv(bv + au) + (1 $ 4v2)bu = 0.
I t follows that va
+ ub
= 0.
This determines the pair (a, 6) at each point, up to a nonzero multiple, and hence the field r'. To find the integral curves of r', let u'x, v'x, be a tangent vector of some regular parametrization of an integral curve of r'. Then (u', v') satisfies Eq. (3); that is, vu' tlv' = 0
+
+
It follows that the orthogonal family of (C,} is given by the intersections with S of the hyperbolic cylinders xy = const. # 0. The main result of this section is the following theorem. THEOREM. Let w, and wz be two vector Jields in an open set U c which are linearly independent at some point p E U . Then it is possible parametrize a neighborhood V c U of p in such a way that for each q E the coordinate lines of this parametrization passing through q are tangent the lines determined by w, (q) and w2(q).
S, to
V to
Proof. Let W be a neighborhood of p where the first integrals f l and f 2 of w, and w,, respectively, are defined. Define a map p: W 4R 2 by
Sincef,is constant on the trajectories of w , and (df,) + 0, we have at p
where a
= (df2),(wl)#
0, since w, and w2 are linearly independent. Similarly,
where 6 = (dfI),(w,) # 0. It follows that dp, is nonsingular, and hence that p is a local difleomorc R 2 of p(p) which is phism. There exists, therefore, a neighborhood
Vector Fields
183
mapped difTeomorphically by x = q  l onto a neighborhood V = x ( q of p ; that is, x is a parametrization of S at p, whose coordinate curves
are tangent at q to the lines determined by wl(q), w2(q),respectively. Q.E.D. It should be remarked that the theorem does not imply that the coordinate curves can be so parametrized that their velocity vectors are w,(q) and w2(q). The statement of the theorem applies to the coordinate curves as regular (point set) curves; more precisely, we have COROLLARY 1. Given two fields of directions r and r' in an open set U c S such that at p E U, r(p) rf(p), there exists a parametrization x in a neighborhood of p such that the coordinate curves of x are the integral curves of r and r'.
+
A first application of the above theorem is the proof of the existence of an orthogonal parametrization at any point of a regular surface. COROLLARY 2. For all p E S there exists a parametrization x(u, v) in a neighborhood V of p such that the coordinate curves u = const., v = const. intersect orthogonally for each q E V (such an x is called an orthogonal parametrization).

S at p, and define Proof: Consider an arbitrary parametrization 2 : 0 two vector fields w, = i y , w2 = (F/I?)ir, ii.in ir(fl), where E, F, G are the coefficients of the first fundamental form in 2. Since w,(q), w2(q) are orthogonal vectors, for each q E ji(u), an application of the theorem yields the required parametrization. Q.E.D.
+
A second application of the theorem (more precisely, of Corollary 1) is the existence of coordinates given by the asymptotic and principal directions. As we have seen in Sec. 33, the asymptotic curves are solutions of
In a neighborhood of a hyperbolic point p, we have eg  f < 0, and the lefthand side of the above equation can be decomposed into two distinct linear factors, yielding (Au'
+ Bvr)(Au' + Dv') = 0,
where the coefficients are determined by
(4)
The Geometry o f the Gauss M a p
184
The above system of equations has real solutions, since eg  f Eq. (4) gives rise to two equations:
< 0. Thus,
+ Bv' = 0, 0. Au' + Dv' Au'
=
Each of these equations determines a differentiable field of directions (for instance, Eq. (4a) determines the direction r which contains the nonzero vector (B,  A ) ) , and at each point of the neighborhood in question the directions given by Eqs. (4a) and (4b) are distinct. By applying Corollary 1, we see that it is possible to parametrize a neighborhood of p in such a way that the coordinate curves are the integral curves of Eqs. (4a) and (4b). In other words,
COROLLARY 3. Let p E S be a hyperbolic point of S. Then it is possible to parametrize a neighborhood of p in such a way that the coordinate curves of this parametrization are the asymptotic curves of S. Example 4. An almost trivial example, but one which illustrates the mechanism of the above method, is given by the hyperbolic paraboloid z = x 2  y2. As usual we parametrize the entire surface by
x(u, v)
= (u,
v, u 2  v2).
A simple computation shows that
Thus, the equation of the asymptotic curves can be written as
which can be factored into two linear equations and give the two fields of directions : rl: u1+v'=0, r2: u'  v'
= 0.
The integral curves of these fields of directions are given by the two families of curves : r , : u + v =const., r2: u

v
= const.
Vector Fields
785
+
Now, the functions f,(u, v ) = u v , fi(u, v ) = u  v are clearly first integrals of the vector fields associated t o r , and r,, respectively. Thus, by setting
we obtain a new parametrization for the entire surface z = x2  y 2 in which the coordinate curves are the asymptotic curves of the surface. I n this particular case, the change of parameters holds for the entire surface. I n general, it may fail to be globally onetoone, even if the whole surface consists only of hyperbolic points. Similarly, in a neighborhood of a nonumbilical point of S, it is possible to decompose the differential equation of the lines of curvature into distinct linear factors. By an analogous argument we obtain
COROLLARY 4. Let p E S be a nonumbiIica1 point of S. Then it is possi5le to parametrize a neighborhood of p in such a way that the coordinate curves of this parametrization are the lines of curvature of S .
EXERCISES 1. Prove that the differentiability of a vector field does not depend on the choice of
a coordinate system. 2. Prove that the vector field obtained on the torus by parametrizing all its merid
ians by arc length and taking their tangent vectors (Example 1) is differentiable. 3. Prove that a vector fieId w defined on a regular surface S c R3 is differentiable if and only if it is differentiable as a map w: S 4 R 3 . 4. Let S be a surface and x: U +S be a parametrization of S. Then
where a and b are differentiable functions, determines a field of directions r on x(U), namely, the correspondence which assigns to each x(u, v ) the straight line containing the vector b x ,  ax,. Show that a necessary and sufficient condition for the existence of an orthogonal field r' on x ( U ) (cf. Example 3 ) is that both functions Eb  Fa. Fb  Ga are nowhere simultaneously zero (here E, F, and G are the coefficients of the first fundamental form in x ) and that r' is then determined by (Eb  Fa)uf
+ (Fb

Ga).u'
= 0.
The Geometry o f the Gauss Map
186
5. Let S be a surface and x : U 4 S be a parametrization of S. If ac show that
+
a(u, v ) ( u ' ) ~ 2b(u, v)u'vl
+ c(u, v)(v'I2
=

b2 < 0 ,
0
can be factored into two distinct equations, each of which determines a field of directions on x ( U ) c S. Prove that these two fields of directions are orthogonal if and only if
6 . A straight line r meets the z axis and moves in such a way that it makes a constant angIe a # 0 with the z axis and each of its points describes a helix of pitch c # 0 about the z axis. The figure described by r is the trace of the parametrized surface (see Fig. 332) X(U,
v)
=
(v sin a cos u, v sin a sin u, v cos a
+ cu).
x is easily seen to be a regular parametrized surface (cf. Exercise 13, Sec, 25). Restrict the parameters (u, v ) to an open set U so that x ( U ) = S is a regular surface (cf. Prop. 2, Sec. 23). a. Find the orthogonal family (cf. Example 3) to the family of coordinate curves u = const. b. Use the curves u = const. and their orthogonal family to obtain an orthogonal parametrization for S. Show that in the new parameters (i,6)the coefficients of the first fundamental form are
L
I a
r
0
Figure 332
x
7. Define the derivative w(f) of a diferentiable function f: U c S vector field w in U by
where
a: I
+
+
R relative to a
S is a curve such that a(0) = q, ar(0)= w(q). Prove that
Vector Fields
187
a. w is differentiable in Uif and only if w(f ) is differentiable for all differentiable f i n U.
b. Let A and p be real numbers and g : U c S on U; then w(Af w(fg)
+ pf =
= Aw(f)
w(fk

R be a differentiable function
+ pw(f 1,
+ fw(g).
8. Show that if w is a differentiable vector fieId on a surface S and w(p) # 0 for some p E S , then it is possible to parametrize a neighborhood of p by x(u, v) in such a way that xu = w. 9. a. Let A : V 3 W be a nonsingular linear map of vector spaces V and W of dimension 2 and endowed with inner products ,) and ( ,), respectively. A is a similitude if there exists a real number 1 # 0 such that (Au,, Avz) = A(vl, u2) for all vectors v,, v2 E V. Assume that A is not a similitude and show that there exists a unique pair of orthonormal vectors el and e2 in V such that Ael, Ae2 are orthogonal in W.
= 0 for all t E I. We first want to find a parametrized curve p(t) such that (P'(t), wr(t)> = 0, t E I, and P(E)lies 011 the trace of x; that is,
for some realvalued function u curve p, one obtains
= u(t).
Assuming the existence of such a
pr = a r + urw + uwr; hence, since (w, wr)
It follows that u
= 0,
= u(t)
is given by u=
(a', w') (w',wr>
Thus, if we define p(t) by Eqs. (2) and (3), we obtain the required curve. We shall now show that the curve /3 does not depend on the choice of the
Ruled Surfaces and Minima/ Surfaces
19 1
directrix a for the ruled surface. P is then called the line of striction, and its points are called the central points of the ruled surface. To prove our claim, let CI be another directrix of the ruled surface; that is, let, for a11 ( t , v ) ,
for some function s
= s(t).
Then, from Eqs. (2) and (3) we obtain { a 1  a t ,w ' ) w ( w ' , w')
pfl=(a@)+
9
where B is the line of striction corresponding to d. On the other hand, Eq. (4) implies that a  a = (S  v)w(~). Thus, P  P = ( ( s  v) ((v  s)w ', w ' ) w = 0,
+
( W I ,
1
W I )
since ( w , w ' ) = 0. This proves our claim. We now take the line of striction as the directrix of the ruled surface and write it as follows: x(t, u)
= p(tj
+ uw(t).
(5)
With this choice, we have XI =
p'
+ uw',
X, =
p'
A w
and X,
/\
Since (w', w ) = 0 and (w', p') for some function IZ = IZ(t). Thus,
=
Xu = W
+ uw'
A w.
0, we conclude that P' A w

Aw'
It follows that the only singular points of the ruled surface (5) are along the line of striction u = 0, and they will occur if and only if A(t) = 0. Observe also that
where, as usual,
(P', w, w') is a short for (p'
A w, wf>.
The Geometry o f the Gauss Map
192
Let us compute the Gaussian curvature of the surface (5) at its regular points. Since
we have, for the coefficients of the second fundamental form,
hence (since g
= 0 we
do not need the value of e to compute K),
This shows that, at regular points, the Gaussian curvature K ofa ruled surface satisfies K < 0, and K is zero only along those rulings which meet the line of striction at a singular point. Equation (6) allows us to give a geometric interpretation of the (regular) central points of a ruled surface. Indeed, the points of a ruling, except perhaps the central point, are regular points of the surface. If A # 0, the function 1 K(u) I is a continuous function on the ruling and, by Eq. (6), the central point is characterized by the fact that I K(u) I has a maximum there. For another geometrical interpretation of the line of striction see Exercise 4. We also remark that the curvature K takes up the same values at points on a ruling that are symmetric relative to the central point (this justifies the name central). The function a(t) is called the distribution parameter of x. Since the line of striction is independent of the choice of the directrix, it follows that the same holds for a. If x is regular, we have the following interpretation of 1. The normal vector to the surface at (t, u) is
On the other hand
(A + O),
Therefore, if 8 is the angle formed by N(t, u) and N(t, 0), U tan 0 = .
A
Ruled Surfaces and Minima/ Surfaces
193
Thus, if 9 is the angle which the normal vector at a point of a ruling makes with the normal vector at the central point of this ruling, then tan 8 is proportional to the distance between these two points, and the coeficient of proportionality is the inverse of the distribution parameter. Example 3. Let S be the hypeibolic paraboloid z = k x y , k # 0. To show that S is a ruled surface, we observe that the lines y = z/tk, x = t, for each t # 0 belong to S. If we take the intersection of this family of lines with the plane z = 0, we obtain the curve x = t , y = 0, z = 0. Taking this curve as directrix and vectors w(t) parallel to the lines y = zltk, x = t, we obtain
This gives a ruled surface (Fig. 335)
the trace of which clearly agrees with S.
Figure 335. z
=
xy
as a ruled surface.
Since a'(t) = ( I , 0, 0), we obtain that the line of striction is a itself. The distribution parameter is
We also remark that the tangent of the angle 8 which w(t) makes with w(0) is tan 8 = tk.
The Geometry o f the Gauss Map
194
The last remark leads to an interesting general property of a ruled surface. If we consider the family of normal vectors along a ruling of a regular ruled surface, this family generates another ruled surface. By Eq. (7) and the last remark, the latter surface is exactly the hyperbolic paraboloid x = kxy, where llk is the value of the distribution parameter at the chosen ruling. Among the ruled surfaces, the developables play a distinguished role. Let us start again with an arbitrary ruled surface (not necessarily noncylindrical) ~ ( 2v) ,
= a(t)
generated by the family (a@),w(t)) with I w(t) I be developable if (w, w', a')

+ vw(i),
(8) 1. The surface (8) is said to
 0.
(9)
To find a geometric interpretation for condition (9), we shall compute the Gaussian curvature of a developable surface at a regular point. A computation entirely similar to the one made to obtain Eq. (6) gives
By condition (9), f

0; hence,
This implies that, at regular points, the Gaussian curvature of a developable surface is identically zero. For another geometric interpretation of a developable surface, see Exercise 6. We can now distinguish two nonexhaustive cases of developable surfaces: 1. w(t) A w'(t) = 0. This implies that wf(t) r 0. Thus, w(t) is constant and the ruled surface is a cylinder over a curve obtained intersecting the cylinder with a plane normal to w(t).
2. w(t) /\ wf(t) # 0 for all t E I. In this case wf(t) # 0 for all t E I. Thus, the surface is noncylindrical, and we can apply our previous work. Thus, we can determine the line of striction (2) and check that the distribution parameter
Ru/ed Surfaces and Minimal Surfaces
7 95

Therefore, the line of striction will be the locus of singular points of the developable surface. If Pf(t) # 0 for all t E I,it follows from Eq. (10) and the fact that (P', w') 0 that w is parallel to Dr.Thus, the ruled surface is the tangent surface of P. If P'(t) = 0 for all 2 E I, then the line of striction is a point, and the ruled surface is a cone with vertex at this point.
Of course, the above cases do not exhaust all possibilities. As usual, if there is a clustering of zeros of the functions involved, the analysis may become rather complicated. At any rate, away from these cluster points, a developable surface is a union of pieces of cylinders, cones, and tangent surfaces. As we have seen, at regular points, the Gaussian curvature of a developable surface is identically zero. In Sec. 58 we shall prove a sort of global converse to this which implies that a regular surface S c R3 which is closed as a subset of R3 and has zero Gaussian curvature is a cylinder. Example 4. (The Envelope of the Family of Tangent Planes Along a Curve of a Surface). Let S be a regular surface and a = a(s) a curve on S parametrized by arc length. Assume that a is nowhere tangent to an asymptotic direction. Consider the ruled surface
where by N(s) we denote the unit normal vector of S restricted to the curve X(S)(since a'($) is not an asymptotic direction, N'(s) # 0 for all s). We shall show that x is a developable surface which is regular in a neighborhood of v = 0 and is tangent to S along v = 0. Before that, however, let us give a geometric interpretation of the surface x. Consider the family (T,,,,(S)) of tangent planes to the surface S along the curve a(s). If As is small, the two planes T,,,,(S) and T,(,+A,,(S)of the family will intersect along a straight line parallel to the vector
If we let As go to zero, this straight line will approach a limiting position parallel to the vector lim N(s) A N(s f AS) = lim ~ ( A ~(N(s1 AS As0 = N(s) A N '(s).
As40
+ As)  N(s)) AS
The Geometry of the Gauss Map
196
This means intuitively that the rulings of x are the limiting positions of the intersection of neighboring planes of the family {T,(,,(S)).x is called the envelope of the family of tangent planes of S along a(s) (Fig. 336).
Figure 336
For instance, if a is a parametrization of a parallel of a sphere S2, then the envelope of tangent planes of S2 along a is either a cylinder, if the parallel is an equator, or a cone, if the parallel is not an equator (Fig. 337).
Figure 337. Envelopes of families of tangent planes along parallels of a sphere.
To show that x is a developable surface, we shall check that condition (9) holds for x. In fact, by a straightforward computation, we obtain
Ruled Surfaces and M i n / d Surfaces
1  ({N
I N' l2
A N', N")N, a ' )
=
This proves our claim. We shall now prove that x is regular in a neighborhood of v that it is tangent to S along a. In fact, at v = 0, we have
x , A x,
= a'
A ( N A N ' ) = ( N ' , a')m N
IN'I
= { N , a")
0. =0
and
N
IN I
where k, = k,(s) is the normal curvature of a . Since k,(s) is nowhere zero, this shows that x is regular in a neighborhood of v = 0 and that the unit normal vector of x at x(s, 0) agrees with N(s). Thus, x is tangent to S along v = 0, and this completes the proof of our assertions. We shall summarize our conclusions as follows. Let u(s) be a ciivvepayametrized by arc length on a surface S and assume that a is nowhere tangent to an asymptotic direction. Then the envelope (9) of the family of tangent planes to S along a is a developable surface, regular in a neighborhood of a(s) and tangent to S along a(s).
B, Minimal Surfaces A regular parametrized surface is caIled minimal if its mean curvature vanishes everywhere. A regular surface S c R3 is minimal if each of its parametrizations is minimal. To explain why we use the word minimal for such surfaces, we need to introduce the notion of a variation. Let x: U c R2 R3 be a regular parametrized surface. Choose a bounded domain D c U (cf. Sec. 25) and a differentiable function h : fi R, where fi is the union of the domain D with its boundary d D . The normal variation of x(d), determined by h, is the map (Fig. 338) given by,

+
For each fixed t
E
(6, e), the map x f : D
+
R3
The Geometry of the Gauss M a p
hN thN 0 thN
x +N
/
x(B)
(x  t h ~ ) ( B )
Figure 338. A normal variation of x(D).
is a parametrized surface with
Thus, if we denote by Et, F f , Gt the coefficients of the first fundamental form of x', we obtain
By using the fact that (xu, N u ) = e,
(xu, Nu?
+ (xu,N u )
=
2f,
(x,, N,)
=
g
and that the mean curvature H is (Sec. 33, Eq. (5))
we obtain EtGt  (Ft)Z= EG  F 2  2th(Eg = (EG 

FZ)(l  4thH)
2Ff
+ Ge) + R
+ R,
where lirn,,, (Rlt) = 0. It follows that if E is sufficiently small, xt is a regular parametrized surface. Furthermore, the area A(t) of xt(6) is
Ru/ed Surfaces and Minimal Surfaces
A(*) =
1
JEG'  (Ft)2dudv
D
where R = R/(EG  F2).It follows that if function and its derivative at t = 0 is
E
is small, A is a differentiable
We are now prepared to justify the use of the word minimal in connection with surfaces with vanishing mean curvature.
PROPOSITION 1. Let x: U + R3 be a regular parametrized surface and let D c U be a bounded domain in U . Then x is minimal if and only $ A'(0) = 0 for all such D and all normal variations of x(b). Proof. If x is minimal, H z 0 and the condition is clearly satisfied. Conversely, assume that the condition is satisfied and that H(q) # 0 for some q t D. Choose h: d R such that h(q) = H(q) and h is identically zero outside a small neighborhood of q. Then A'(0) < 0 for the variation determined by this h, and that is a contradiction. Q.E.D.

Thus, any bounded region x(6) of a minimal surface x is a critical point for the area function of any normal variation of ~ ( 6 ) It. should be noticed that this critical point may not be a minimum and that this makes the word minimal seem somewhat awkward. It is, however, a timehonored terminology which was introduced by Lagrange (who first defined a minimal surface) in 1760. Minimal surfaces are usually associated with soap films that can be obtained by dipping a wire frame into a soap solution and withdrawing it carefully. If the experiment is well performed, a soap film is obtained that has the same frame as a boundary. It can be shown by physical considerations that the film will assume a position where at its regular points the mean curvature is zero. In this way we can "manufacture" beautiful minimal surfaces, such as the one in Fig. 339. Remark I . It should be pointed out that not all soap films are minimal surfaces according to our definition. We have assumed minimal surfto be regular (we could have assumed some isolated singular points, but to go beyond that would make the treatment much less elementary). However, soap films can be formed, for instance, using a cube as a frame (Fig. 340), that have singularities along lines.
The Geometry of the Gauss Map
Figure 339
Figure 340
Remark 2. The connection'between minimal surfaces and soap films motivated the celebrated Plateau's problem (Plateau was a Belgian physicist who made careful experiments with soap films around 1850). The problem can be roughly described as follows: to prove that for each closed curve C c R3 there exists a surface S of minimum area with C as boundary. To make the problem precise (which curves and surfaces are allowed and what is meant by C being a boundary of S ) is itself a nontrivial part of the problem. A version of Plateau's problem was solved simultaneously by Douglas and Rad6 in 1930. Further versions (and generalizations of the problem for higher dimensions) have inspired the creation of mathematical entities which include at least as many things as soaplike films. We refer the interested reader to the Chap. 2 of Lawson [20] (references are at the end of the book) for further details and a recent bibliography of Plateau's problem.
It will be convenient to introduce, for an arbitrary parametrized regular
Ruled Surfaces and Minimal Surfaces
20 7
surface, the mean curvature vector defined by H = HN. The geometricaI meaning of the direction of H can be obtained from Eq. (12).Indeed, if u.e choose h = H , we have, for this particular variation,
This means that if we deform x(d) in the direction of the vector H , the area is initially decreasing. The mean curvature vector has another interpretation which we shall now pursue, since it has important implications for the theory of minimal surfaces. A regular parametrized surface x = x(u, u) is said to be isothermal if (xu,xu) = (x,, x,) and (xu,x,) = 0. PROPOSITION 2. Let x = x(u, v) be a regular parametrized surface and assume that x is isothermal. Then
where
A2 = (xu,xu) = (x,, x,).
Proof. Since x is isothermal, (xu,xu) = (xu,x,) and u = 0, Nu,  N,, = 0. By introducing the values of (I), we may write the above relations in the form
where Ai, B,, Ci, i
=
1,2, 3, are functions of E, F, G, e , f , g and of their
Intrinsic Geometry o f Surfaces
234
derivatives. Since the vectors xu,x,, N are linearly independent, (3a) impiies that there exist nine relations :
As an example, we shall determine the relations A , = 0, 3,= 0, C, = 0. By using the values of (I), the first of the relations (3) may be written
By using (I) again and equating the coefficients of xu, we obtain
Introducing the values of aij already computed (cf. Sec. 33) it follows that
  E eg  f 2 
EG F2
At this point it is convenient to interrupt out computations in order to draw attention to the fact that the above equation proves the following theorem, due to K. F. Gauss. THEOREMA EGREGIUM (Gauss). The Gaussian curvature K of a surface is invariant by local isometries.

In fact if x : U c R~ S is a parametrization at p S and if cp : V c S  S: where V c x(U) is a neighborhood of p, is a local isometry a t p, then y = xoq is a parametrization of S at cp(p). Since p is an isometry, the coefficients of the first fundamental form in the parametrizations x and y agree at corresponding points q and p(q), q E V; thus, the corresponding Christoffel symbols also agree. By Eq. (5), Kcan be computed at a point as a function of the Christoffel symbols in a given parametrization at the point. It follows that K(q) = K(cp(q)) for all q € v. The above expression, which yields the value of K in terms of the coefficients of the first fundamental form and its derivatives, is known as the Gauss formula. It was first proved by Gauss in a famous paper [I]. The Gauss theorem is considered, by the extension of its consequences,
The Gauss Theorem and the Equations o f Compatibility
235
one of the most important facts of differential geometry. For the moment we shall mention only the following corollary. As was proved in Sec. 42, a catenoid is locally isometric to a helicoid. It follows from the Gauss theorem that the Gaussian curvatures are equal at corresponding points, a fact which is geometrically nontrivial. Actually, it is a remarkable fact that a concept such as the Gaussian curvature, the definition of which made essential use of the position of a surface in the space, does not depend on this position but only on the metric structure (first fundamental form) of the surface, We shall see in the next section that many other concepts of differential geometry are in the same setting as the Gaussian curvature; that is, they depend only on the first fundamental form of the surface. It thus makes sense to talk about a geometry of the first fundamental form, which we call intrinsic geometry, since it may be developed without any reference to the space that contains the surface (once the first fundamental form is given). ?With an eye to a further geometrical result we come back to our computations. By equating the coefficients of xu in (4), we see that the relation A , = 0 may be written in the form
By equating also in Eq. (4) the coefficients of N, we obtain C, = 0 in the form
Observe that relation (5a) is (when F + 0) merely another form of the Gauss formula (5). By applying the same process to the second expression of (3), we obtain that both the equations A , = 0 and 3, = 0 give again the Gauss formula (5). Furthermore, C , = 0 is given by
Finally, the same process can be applied to the last expression of (3), yielding that C , = 0 is an identity and that A , = 0 and 3,= 0 are again Eqs. (6) and (6a). Equations (6) and (6a) are called MainardiCodazzi equations. The Gauss formuIa and the MainardiCodazzi equations are known under the name of compatibility equations of the theory of surfaces. ?The rest of this section will not be used until Chap. 5. If omitted, Exercises 7 and 8 should also be omitted.
236
Intrinsic Geometry of Surfaces
A natural question is whether there exist further relations of compatibility between the first and the second fundamentaI forms besides those already obtained. The theorem stated below shows that the answer is negative. In other words, by successive derivations or any other process we would obtain no further relations among the coefficients E, F, G , e, f, g and their derivatives. Actually, the theorem is more explicit and asserts that the knowledge of the first and second fundamental forms determines a surface locally. More precisely, THEOREM (Bonnet). Let E, F , G, e, f, g be dzerentiable functions, deJined in an open set V c R2, with E > 0 and G > 0. Assume that the given functions satisfy formally the Gauss and MainardiCodazzi equations and that E G  F2 > 0.Then, for every q E V there exists a neighborhood U c V of q and a diffeomorphism x : U , x(U) c R3 such that the regular surface x(U) c R3 has E, F, G and e, f, g as coeficieutts of the $first and secondfundamental forms, respectively. Furthermore, if U is connected and if
is another dzfleomorphism satisfying the same conditions, then there exist a translation T and a proper linear orthogonal transformation p in R 3 such that X = T 0 p o x .
A proof of this theorem may be found in the appendix to Chap. 4. For later use, it is convenient to observe how the MainardiCodazzi equations simplify when the coordinate neighborhood contains no umbilical points and the coordinate curves are lines of curvature ( F = 0 = f). Then, Eqs. (6) and (6a) may be written
By taking into consideration that F = 0 implies that
we conclude that the MainardiCodazzi equations take the following form:
The Gauss Theorem and the Equations of Compatibility
EXERCISES 1. Show that if x is an orthogonal parametrization, that is, F = 0, then
2. Show that if x is an isothermal parametrization, that is, E = G = l(u, u) and F = 0,then
+
where Ap denotes the Laplacian (dZpjdu2) (dZq1dv2) of the function p. Conclude that when E = G = (u2 v 2 e)2 and F = 0, then K = const. = 4c.
+ +
3. Verify that the surfaces
x(u, V) = (u cos v, u sin v, log u), 5(u, V) = (u cos v, u sin v, v), have equal Gaussian curvature at the points x(u, v) and t(u, v) but that the mapping t 0 x' is not an isometry. This shows that the "converse" of the Gauss theorem is not true. 4. Show that no neighborhood of a point in a sphere may be isometrically mapped into a plane. 5. If the coordinate curves form a Tchebyshef net (cf. Exercises 7 and 8, Sec. 25), then E = G = 1 and F = cos 0. Show that in this case
K = . 8 U V sin 0 6. Use Bonnet's theorem to show that there exists no surface x(u, V) such that E = G = = l , F  O a n d e = l , g = l,f=O.
7. Does there exist a surface x e =cos2u, f = O , g = I ?
= x(u, v)
with E
=
1, F
= 0,
G
8. Compute the Christoffel symbols for an open set of the plane
a. In cartesian coordinates. b. In polar coordinates. Use the Gauss formula to compute K in both cases. 9. Justify why the surfaces below are not painvise locally isometric:
a. Sphere. b. Cylinder. c. Saddle z = x2  yZ.
= cos2 u
and
Intrinsic Geometry o f Surfaces
238
44. Parallel Transport. Geodesics. We shall now proceed to a systematic exposition of the intrinsic geometry. To display the intuitive meaning of the concepts, we shall often give definitions and interpretations involving the space exterior to the surface. However, we shall prove in each case that the concepts to be introduced depend only on the first fundamental form. We shall start with the definition of covariant derivative of a vector field, which is the analogue for surfaces of the usual differentiation of vectors in the plane. We recall that a (tangent) vectorJield in an open set U c S of a regular surface S is a correspondence w that assigns to each p E U a vector w(p) r T,(S). The vector field w is diferentiable at p if, for some parametrization x(u, v) in p, the components a and b of w = ax, bx, in the basis {x,,%,I are differentiable functions at p. w is differentiable in U if it is differentiable for every p E U.
+
DEFINITION 1. Let w be a diferentiable vector field in an open set U c S and p E U . Let y E Tp(S). Consider a parametrized curve
with a(0) = p and al(0) = y, and let w(t), t E (  E , E ) , be the restriction of the vector field w to the curve a . The vector obtained by the normal projection of (dw/dt)(O) onto the plane T,(S) is called the covariant derivative at p of the vector Jield w relative to the vector y. This covariant derivative is genoted by (Dw/dt)(O) or (D,w)(p) (Fig.48).
Figure 48. The covariant derivative.
239
Para//el Transport. Geodesics
The above definition makes use of the normal vector of S and of a particular curve a , tangent to y at p. To show that covariant differentiation is a concept of the intrinsic geometry and that it does not depend on the choice of the curve a , we shall obtain its expression in terms of a parametrization x(u, v) of S in p. Let x(za(~),~ ( t ) = ) a(t) be the expression of the curve a and let
be the expression of w(t) in the parametrization x(u, v). Then
where prime denotes the derivative with respect to t . Since Dwldt is the component of dw/dt in the tangent plane, we use the expressions in (I) of Sec. 41 for xu,, xu,, and xu, and, by dropping the normal component, we obtain
Dwdt
(a'
+ T:,aul + T:,avl + r : 2 b ~ +' rizbv')x,
Expression (1) shows that Dwldt depends only on the vector (u', v') = y and not on the curve a. Furthermore, the surface makes its appearance in Eq. (I) through the Christoffel symbols, that is, through the first fundamental form. Our assertions are, therefore, proved. If, in particular, S is a plane, we know that it is possible to find a parametrization in such a way that E = G = 1 and F = 0. A quick inspection of the equations that give the Christoffel symbols shows that in this case the r$ become zero. In this case, it follows from Eq. (1) that the covariant derivative agrees with the usual derivative of vectors in the plane (this can also be seen geometrically from Def. 1). The covariant derivative is, therefore, a generalization of the usual derivative of vectors in the plane. Another consequence of Eq. (I) is that the definition of covariant derivative may be extended to a vector field which is defined only at the points of a parametrized curve. To make this point clear, we need some definitions.
DEFINITION 2. A parametrized curve a: [O,I] + S is the restriction to [0, 11 of a differentiable mapping of (0  6, 1 €1, 6 > 0, into S. Va(0) = p and a(l) = q, we say that a joins p to q. a i s regular $a'(t) + 0 for t E [O,I].
+
Intrinsic Geometry of Surfaces
240
In what follows it will be convenient to use the notation [O, I] er the specification of the end point 1 is not necessary.
=I
whenev

DEFINITION 3. Let a : I S be aparametrized curve in S . A vector field w along a is a correspondence that assigns to each t E I a vector
The vector Jield w is differentiable at to E I if for some parametrization X(U,V) in a(t,) the components a@),b(t) of w(t) = ax, bx, are diferentiable functions of t at to. w is differentiable in I if it is diflerentiable for every t E I.
+
An example of a (differentiable) vector field along a is given by the field a'(t) of the tangent vectors of a (Fig. 49).
Figure 49. The field of tangent vectors along a curve a.
DEFINITION 4. Let w be a diferentiable vector Jield along a : I + S. The expression (I) of (Dw/dt)(t), t E I, is well dejned and is called the covariant derivative of w at t. From a point of view external to the surface, in order to obtain the covariant derivative of a field w along a : 1 + S at t E I we take the usual derivative (dwldt)(t) of w in t and project this vector orthogonally onto the tangent plane T,,,,(S). It follows that when two surfaces are tangent along a parametrized curve a the covariant derivative of a field w along a is the same for both surfaces. If a(t) is a curve on S, we can think of it as the trajectory of a point which is moving on the surface. a f ( t )is then the speed and a"(t) the acceleration of a. The covariant derivative Da'ldt of the field a f ( t )is the tangential component of the acceleration aU(t).Intuitively Daf/dt is the acceleration of the point a(t) "as seen from the surface S."
24 1
Parallel Transport. Geodesics
DEFINITION 5. A vector Jield w along a parametrized curve a: I is said to be paralIel if Dw/dt = 0 far every t E I .
+
S
In the particular case of the plane, the notion of parallel field along a parametrized curve reduces to that of a constant field along the curve; that is, the length of the vector and its angle with a fixed dir ECt'ion are constant (Fig. 410). Those properties are partially reobtained on any surface as the following proposition shows.
Figure 410

PROPOSITION 1. Let w and v be parallel vector Jields along a : 1 S. Then (w(t), v(t)) is constant. In particular, I w(t) and I v(t) 1 are constant, and the angle between v(t) and w(t) is constant.
I
Proof. To say that the vector field w is parallel along a means that dwldt is normal to the plane which is tangent to the surface at a(t); that is,
On the other hand, v'(t) is also normal to the tangent plane at a(t). Thus,
that is, (v(t), w(t)> = constant.
Q.E.D.
Of course, on an arbitrary surface parallel fields may look strange to our R3intuition. For instance, the tangent vector field of a rneridian(parametrized by arc length) of a unit sphere S2is a parallel field on S2(Fig. 41 1). I n fact, since the meridian is a great circle on S2, the usual derivative of such a field is normal to S2.Thus, its covariant derivative is zero. The following proposition shows that there exist parallel vector fields along a parametrized curve a(t) and that they are completely determined by their values at a point to.
Intrinsic Geometry of Surfaces
1
I
Figure 411. ParalIel field on a sphere.
PROPOSITION 2. Let a: I + S be a parametrized curve in S and let wo E T,,,,,(S), to E I. Then there exists a unique parallel vector field w(t) along a(t), with w(to) = w,. An elementary proof of Prop. 2 will be given later in this section. Those who are familiar with the material of Sec. 36 will notice, however, that the proof is an immediate consequence of the theorem of existence and uniqueness of differential equations. Proposition 2 allows us to talk about parallel transport of a vector along a parametrized curve.

DEFINITION 6. Let a: I S be aparametrized curve and wo E T,,,,,(S), to E I. Let w be the parallel vector field along a , with w(to) = w,. The vector w(t,), t, E I, is called the parallel transport of wo along a at the point t , . It should be remarked that if a : I S, t E I , is regular, then the parallel transport does not depend on the parametrization of a(I). As a matter of fact, if P:J   t S, o E J is another regular parametrization for a(I), it follows from Eq. ( I ) that +
Dw
do
A
Dwdt d t do'
lEI,UEI.
Since dtldo f : 0, w(t) is parallel if and only if w(o) is parallel. Proposition 1 contains an interesting property of the parallel transport. Fix two points p, q E S and a parametrized curve a: I S with a(0) = p, a(1) = q. Denote by P,: T,(S) + T,(S) the map that assigns to each v E T,(S) its parallel transport along a at q. Proposition 1 says that this map is an isometry. +
243
Parallel Transport. Geodesics
Another interesting property of the paralIeI transport is that if two surfaces S and 3 are tangent along a parametrized curve a and w, is a vector of T,,,,,(S) = ~,,,,(3), then w(t) is the parallel transport of w, relative to the surface S if and only if w(t) is the parallel transport of w, relative to 3. Indeed, the covariant derivative Dwldt of w is the same for both surfaces. Since the parallel transport is unique, the assertion follows. The above property will allow us to give a simple and instructive example of parallel transport. Example 1. Let C be a parallel of colatitude p (see Fig. 412) of an oriented unit sphere and let w, be a unit vector, tangent to C at some point p of C. Let us determine the parallel transport of w, along C, parametrized by arc length s, with s = 0 at p.
3
Figure 412
orientation
Figure 413
Consider the cone which is tangent to the sphere along C. The angle ty at the vertex of this cone is given by y = (n/2)  p. By the above property, the problem reduces to the determination of the parallel transport of w,, along C, relative t o the tangent cone. The cone minus one generator is, however, isometric to an open set U c R2 (cf, Example 3, Sec. 42), given in polar coordinates by
Since in the plane the parallel transport coincides with the usual notion. we obtain, for a displacement s of p, corresponding to the central angle 8 (see Fig. 413) that the oriented angle formed by the tangent vector t(s) u'ith the parallel transport w(s) is given by 27z  8. I t is sometimes convenient to introduce the notion of a "broken which can be expressed as follows.
tune,
..
244

Intrinsic Gaornetry of Surfaces
DEFINITION 7. A map a : [0, I ] S is a parametrized piecewise regular curve if a is continuous and there exists a subdivision
of the interval [0, I ] in such a way that the restriction a I [ti, titl], i = 0, . . . ,k, is a parametrized regular curve. Each a I[ti, ti+,] is called a regular arc of a.
The notion of parallel transport can be easily extended t o parametrized piecewise regular curves. If, say, the initial value w, lies in the interval [ti, ti, 1], we perform the parallel transport in the regular arc a 1 [ti,ti+ as usuaI; if ti,, # I, we take w(ti,,) as the initial vaiue for the parallel transport in the next arc a ][ti,l, and so forth. Example 2.t The,previous example is a particular case of an interesting geometric construction of the parallel transport. Let C be a regular curve on a surface S and assume that Cis nowhere tangent to an asymptotic direction. Consider the envelope of the family of tangent planes of S along C (cf. Example 4, Sec. 35). In a neighborhood of C, this envelope is a regular surface C which is tangent to S along C. (In Example 1, C can be taken as a ribbon around C on the cone which is tangent to the sphere along C.) Thus, the parallel transport along C of any vector w E T'(S), p E S, is the same whether we consider it relative to S or to X. Furthermore, C is a developable surface; hence, its Gaussian curvature is identically zero. Now, we shall prove later in this book (Sec. 46, theorem of Minding) that a surface of zero Gaussian curvature is locally isometric to a plane. Thus, we can map a neighborhood V c C o f p into a plane P by an isometry p: V P. To obtain the parallel transport of w along V n C, we take the usual parallel transport in the plane of dy,(w) along p(C) and pull it back to C by d p (Fig. 414). This gives a geometric construction for the parallel transport along small arcs of C. We leave it as an exercise to show that this construction can be extended stepwise to a given arc of C. (Use the HeineBore1 theorem and proceed as in the case of broken curves.)

R2 of a plane along which the field of The parametrized curves y : I their tangent vectors y r ( t ) is parallel are precisely the straight lines of that plane. The parametrized curves that satisfy an analogous condition for a surface are called geodesics. +
?This example uses the materia1 on ruled surfaces of Sec. 35.
Parallel Transport. Geodesics
Figure 414. Parallel transport along C,
DEFINITION 8, A nonconstant, parametrized curve y : I 4 S is said to be geodesic at t E I if the field of its tangent vectors y'(t) is parallel along y at t ; that is,
\
y is a parametrized geodesic
if it is geodesic for
all t
E
I.
By Prop. 1, we obtain immediately that 1 y ' ( t ) 1 = const. = c # 0. Therefore, we may introduce the arc length s = ct as a parameter, and we concIude that the parameter t of a parametrized geodesic y is proportional to the arc length of y. Observe that a parametrized geodesic may admit selfintersections. (Example 6 will illustrate this; see Fig. 420.) However, its tangent vector is never zero, and thus the parametrization is regular.
246
Intrinsic Geometry of Surfaces
The notion of geodesic is clearIy local. The previous considerations allow us to extend the definition of geodesic to subsets of S that are regular curves.
DEFINITION 8a. A regular connected curve C in S is said to be a geodesic $ for every p E S, the parametrization a(s) of a coordinate neighborhood of p by the arc length s is a parametrized geodesic; that is, a'(s) is a parallel sector field along a(s) . Observe that every straight line contained in a surface satisfies Def. 8a. From a point of view exterior to the surface S, Def. 8a is equivalent to saying that aV(s) = k n is normal to the tangent plane, that is, parallel to the normal to the surface. In other words, a regular curve C c S (k + 0) is a geodesic if and only if its principa1 normal at each pointp E C is parallel to the normal to S at p. The above property can be used to identify some geodesics geometrically, as shown in the examples below.
Example 3. The great circles of a sphere S2 are geodesics. Indeed, the great circles C are obtained by intersecting the sphere with a plane that passes through the center 0 of the sphere. The principal normal at a point p E C lies in the direction of the line that connectsp to 0 because C i s a circle of center 0. Since S2is a sphere, the normal lies in the same direction, which verifies our assertion. Later in this section we shall prove the general fact that for each point p E S and each direction in Tp(S) there exists exactly one geodesic C c S passing through p and tangent to this direction. For the case of the sphere, through each point and tangent to each direction there passes exactly one great circle, which, as we proved before, is a geodesic. Therefore, by uniqueness, the great circles are the only geodesics of a sphere.
+
Example 4. For the right circular cylinder over the circle x2 y 2 = 1 , it is clear that the circles obtained by the intersection of the cylinder with planes that are normal to the axis of the cylinder are geodesics. That is so because the principal normal to any of its points is parallel t o the normal to the surface at this point. On the other hand, by the observation after Def. 8a the straight lines of the cylinder (generators) are also geodesics. To verify the existence of other geodesics on the cylinder C we shall consider a parametrization (cf. Example 2, Sec. 25) ~ ( uV) , = (cos u, sin u, v) of the cylinder in a point p
E
C, with x(0, 0)
= p.
In this parametrization,
Parallel Transport. Geodesics
247
a neighborhood of p in C is expressed by x(u(s), v(s)), where s is the arc length of C. As we saw previously (cf. Example 1, Sec. 42), x is a local isometry which maps a neighborhood U of (0, 0) of the uv plane into the cylinder. Since the condition of being a geodesic is local and invariant by isometries, the curve (u(s), v(s)) must be a geodesic in U passing through (0,O). But the geodesics of the pIane are the straight lines. Therefore, excluding the cases already obtained, ~ ( 3= ) as,
v(s)
=
bs,
a 2 + b2
=
1.
It follows that when a reguIar curve C (which is neither a circle or a line) is a geodesic of the cylinder it is locally of the form (Fig. 415) (COSas, sin as, bs), and thus it is a helix. In this way, all the geodesics of a right circular cylinder are determined.
Local isometry
Geodesic
Figure 415. Geodesics on a cylinder.
Observe that given two points on a cylinder which are not in a circle parallel t o the xy plane, it is possible to connect them through a n infinite number of helices. This fact means that two points of a cylinder may in general be connected through an infinite number of geodesics, in contrast to the situation in the plane. Observe that such a situation may occur only with geodesics that make a "complete turn," since the cylinder minus a generator is isometric to a plane (Fig. 416). Proceeding with the analogy with the plane, we observe that the lines, that is, the geodesics of a plane, are also characterized as regular curves of curvature zero. Now, the curvature of an oriented plane curve is given by the absolute value of the derivative of the unit vector field tangent t o the curve, associated to a sign which denotes the concavity of the curve in relation to the orientation of the plane (cf. Sec. 15, Remark I). T o take the sign into consideration, it is convenient t o introduce the following definition.
Intrinsic Geometry o f Surfaces
Figure 416. Two geodesics on a cylinder joining p and q.

DEFINITION 9. Let w be a diflerentiable Jield of unit vectors along a parametrized curve a : I S on an oriented surface S. Since w(t), t E I, is a unit vector field, (dw/dt)(t) is normal to w(t), and therefore
The real number 1= I(t), denoted by [Dwldt], is called the algebraic value of the covariant derivative of w at t. Observe that the sign of [Dwldt]depends on the orientation of S and that [Dwldt] = (dwldt, N A w). We should also make the general remark that, from now on, the orientation of S will play an essential role in the concepts to be introduced. The careful reader will have noticed that the definitions of parallel transport and geodesic do not depend on the orientation of S. In constrast, the geodesic curvature, to be defined below, changes its sign with a change of orientation of S. We shall now define, for a curve in a surface, a concept which is a n analogue of the curvature of plane curves.
DEFINITION 10. Let C be an oriented regular curve contained on an oriented surface S, and let a(s) be a parametrization of C , in a neighborhood of p E S, by the arc length s. The algebraic value of the covariant derivative [Dal(s)/ds] = kg of a1(s) at p is called the geodesic curvature of C at p. The geodesics which are regular curves are thus characterized as curves whose geodesic curvature is zero. From a point of view external to the surface, the absolute vaIue of the geodesic curvature kg of C at p is the absolute value of the tangential component of the vector al'(s) = kn, where k is the curvature of C at p and n is the normal vector of C at p. Recalling that the absolute value of the normal component of the vector kn is the absolute value of the normal curvature
Parallel Transpo f t . Geodesics
Figure 417
k, of C c S in p, we have immediately (Fig. 417)
For instance, the absolute value of the geodesic curvature k g of a parallel C of colatitude p in a unit sphere S2can be computed from the relation (see Fig. 418) sin2 p = k,Z k i = sin4 p k i ; that is, k t = sin2 p(1  sin2 p) = $ sin2 2p.
+
+
The sign of k g depends on the orientations of S2and C.
A further consequence of that external interpretation is that when two surfaces are tangent along a regular curve C, the absolute value of the geodesic curvature of C is the same relatively to any of the two surfaces.
Isin cos $1
Figure 418. Geodesic curvature of a parallel on a unit sphere.
250
Intrinsic Geometry o f Surfaces
Remark. The geodesic curvature of C c S changes sign when we change the orientation of either C or S. We shalI now obtain an expression for the algebraic value of the covariant derivative (Prop. 3 below). For that we need some preliminaries. Let v and w be two differentiable vector fields along the parametrized curve a: I  S, with I v(t) I = 1 w(t) ] = I, t E I. We want to define a differR in such a way that q(t), t E I, is a determination entiable function p : I of the angle from v(t) to w(t) in the orientation of S. For that, we consider the differentiable vector field G along a, defined by the condition that {v(t), G(t)) is an orthonormal positive basis for every t E I. Thus, w(t) may be expressed as

+
where a and b are differentiable functions in I and a 2 b2 = 1. Lemma 1 below shows that by fixing a determination p, of the angle from v(t,) to w(t,) it is possible to "extend it" differentiably in I, and this yields the desired function.
+
LEMMA 1. Let a and b be diferentiable functions in I with a2 b2 = 1 and p, be such that a(t,) = cos p,, b(t,) = sin p,. Then the diferentiable function
p
= p,
+ St (ab'  bat) dt to
is such that cos p(t)
= a(t),
sin p(t)
= b(t),
t E I, and p(to) = pa.
Proof. It suffices t o show that the function (a  cos p)2
+ (b

sin q ~ = ) ~2

2(a cos p
+ b sin p)
is identically zero, or that
+ bsinp
A
=
acosq
By using the fact that aa'
=
bbr and the definition of p, we easily obtain
A'
=
I.
+ b(cos p)p ' + a' cos p + b' sin p = bl(sin p)(a2 + b2) ~ ' ( C OqXa2 S + b2) 4 a'cos p + b'sin q = 0.
= a(sin 
p)p'

Therefore, A(t) = const., and since A(t,)
=
1, the lemma is proved.
Q.E.D.
We may now relate the covariant derivative of two unit vector fields along a curve to the variation of the angle that they form.
Parallel Transport. Geodesics
257
L E M M A 2. Let v and w be two diferentiable vector fields along the curve a : I + S , with (w(t)I =lv(t)l = 1, t E I. Then
where y, is one of the diflerentiable determinations of the angle from v to w, as given by Lemma I . ProoJ: We introduce the vectors @ w
= (cos q)v
=N
/\ v and iT
=N
r\ w. Then
+ (sin q ) ~ ,
(2,
+
A v (sin p)N /\ 6 = (cos q ) ~ (sin p)v.
fi = N A w = (cos q)N
(3)
By differentiating (2) with respect to t , we obtain
w ' = (sin q)q'v
+ (cos q)v' + (cos p)p% + (sin g>)v".
By taking the inner product of the last relation with G, using (3), and observing that ( v , 6) = 0, ( v , v') = 0, we conclude that (w', W)
+
= Q)'
+ (cos2 y,)(v', 6)
k (cos2 q)p'(sin2 q)(@',v ) (cos2 q)(v', 6)  (sin2 q)(@',v).
= (sin2 p ) p f
On the other hand, since (v, G )
= 0,
that is,
(v', 6) = {v, C'), we conclude that (w', i i )
= y'
+ (cos2 p + sin2 q)(vt, G) = q' + (v', G).
It follows that
[%I
= {w!,
I,+
=
r t + (v',
+ 2 + [217 =
since
which concludes the proof of the lemma.
Q.E.D.
An immediate consequence of the above lemma is the following observation. Let C be a regular oriented curve on S, a($) a parametrization by the
252
Intrinsic Geometry o f Surfaces
arc length s of C a t p E C, and v(s) a parallel field along a(s). Then, by taking w(s) = a'(s), we obtain D a '(s) kg($) = =3. ds In other words, the geodesic curvature is the rate of change of the angle that the tangent to the curve makes with a parallel direction along the curve. In the case of the plane, the parallel direction is fixed and the geodesic curvature reduces to the usual curvature. We are now able to obtain the promised expression for the algebraic value of the covariant derivative. Whenever we speak of a parametrization of an oriented surface, this parametrization is assumed t o be compatible with the given orientation.
PROPOSITION 3. Let x(u, v) be an orthogonal parametrization (that is, F = 0) of a neighborhood of an oriented surface S, and w(t) be a dzfferentiablefield of unit vectors along the curve x(u(t), v(t)). Then
where p(t) is the angle from xu to w(t) in the given orientation.
Proof. Let el = x,/,JE, e, = xu/,JC be the unit vectors tangent t o the coordinate curves. Observe that e , /\ e, = N, where N is the given orientation of S. By using Lemma 2, we may write
where e,(t) Now
= e,(u(t),
~ ( t ) )is the field e, restricted to the curve x(u(t), ~ ( t ) ) .
On the other hand, since F = 0, we have (xu,, xu> and therefore
Similarly,
=
&E,,
Para//e/ Transport. Geodesics
253
By introducing these relations in the expression of [Dwldtj, we finally obtain
which completes the proof.
Q.E.D.
As an application of Prop. 3, we shall prove the existence and uniqueness of the parallel transport (Prop. 2). Proof of Prop. 2. Let us assume initially that the parametrized curve a : I + S is contained in a coordinate neighborhood of an orthogonal parametrization x(u, v). Then, with the notations of Prop. 3, the condition of parallelism for the field w becomes
Denoting by p, a determination of the oriented angle from xu to w,, the field w is entirely determined by
which proves the existence and uniqueness of w in this case. If a(I) is not contained in a coordinate neighborhood, we shall use the compactness of I to divide a(I) into a finite number of parts, each contained in a coordinate neighborhood. By using the uniqueness of the first part of the proof in the noncmpty intersections of these pieces, it is easy to extend the Q.E.D. result to the present case.
A further application of Prop. 3 is the following expression for the geodesic curvature, known as Liouville's formula. PROPOSITION 4 (Liouville). Let a(s) be aparametrization by arc length of a neighborhood of apoint p E S of a regular oriented curve C on an oriented surface S. Let x(u, v) be an orthogonal parametrization of S in p and p(s) be the angle that X, makes with al(s) in the given orientation. Then
where (kg), and (kg), are the geodesic curvatures of the coordinate curves v = const. and u = const. respectively.
Intrinsic Geometry o f Surfaces
254
Proof. By setting w
= a'(s)
Along the coordinate curve v dulds = 1 / 1 / ; therefore,
in Prop. 3, we obtain
= const.
u = u(s), we have dv/ds = 0 and
Similarly,
By introducing these relations in the above formula for k,, we obtain
Since
JZ' ds
=
(a'(s),
3)
= cos p
and JB
2
=
sin q,
we finally arrive at
as we wished.
Q.E.D.
We shall now introduce the equations of a geodesic in a coordinate neighborhood. For that, let y: I + S be a parametrized curve of S and let x(u, v) be a parametrization of S in a neighborhood V of ~ ( t , ) ,t o E I. Let J c I be an open interval containing to such that y(J) c V. Let x(u(t), v(t)), t E J, be the expression of y: J + S in the parametrization x. Then, the tangent vector field yf(t), t E J, is given by
Therefore, the fact that w is parallel is equivalent to the system of differential equations
obtained from Eq. (1) by making a the coefficients of xu and x,.
= u'
and b
= v',
and equating to zero
255
Parallel Transport. Geodesics
In other words, y : I , S is a geodesic if and only if system (4) is satisfied for every interval J c I such that y(J) is contained in a coordinate neighborhood. The system (4) is known as the dtferential equations of the geodesics of S. An important consequence of the fact that the geodesics are characterized by the system (4) is the following proposition. PROPOSITION 5. Given a point p E S and a vector w E T,(S), w # 0, there exist an E. > 0 and a unique parametrizedgeodesic y : (  6 , ~ ) , S such that y(0) = p, y'(0) w.

In Sec. 45 we shall show how Prop. 5 may be derived from theorems on vector fields. Remark. The reason for taking w #.0 in Prop. 5 comes from the fact that we have excluded the constant curves in the definition of parametrized geodesics (cf. Def. 8). We shall use the rest of this section to give some geometrical applications of the differential equations (4). This material can be omitted if the reader wants to do so. In this case, Exercises 18,20, and 21 should also be omitted. Example 5. We shall use system (4) to study. locaIly the geodesics of a surface of revolution (cf. Example 4, Sec. 23) with the parametrization
x
=
f(v) cos u,
y
= f (v) sin
u,
z
= g(v).
By Example 1 of Sec. 41, the Christoffel symbols are given by
r:, = o,
rf1=  (f')' ff' t (go2'
r:2
ff' f
=7)
With the values above, system (4) becomes
f'g"(v')2 + (go2( u ' ) ~+ f(f' f')"Z ++ (g')2
ff ' ""

(f ')'
= 0.
We are going to obtain some conclusions from these equations. First, as expected, the meridians u = const. and v = v(s), parametrized by arc length s, are geodesics. Indeed, the first equation of (4a) is trivially satisfied by u = const. The second equation becomes
Intrinsic Geometry o f Surfaces
"
+
f ' f " 4 g'g" ( v ' ) ~= 0. (f 'I2 k (g')2
Since the first fundamental form along the meridian u yields ( ( f '1" (g'>"Xv')2 = we conclude that
=
+
= const.
v
= v(s)
1,
Therefore, by derivation,
or, since v' # 0, I 21'"

f'f"
+ g'g"(v')2;
(f'I2 + +g'I2
that is, along the meridian the second of the equations (4a) is also satisfied, which shows that in fact the meridians are geodesics. Now we are going to determine which parallels v = const. u = u(s), parametrized by arc length, are geodesics. The first of the equations (4a) gives u' = const. and the second becomes
In order that the parallel v = const., u = u(s) be a geodesic it is necessary that u' # 0. Since (f ')2 (g')2 + 0 and f # 0, we conclude from the above equation that f ' = 0. In other words, a necessary condition for a parallel of a surface of revolution to be a geodesic is that such a parallel be generated by the rotation of a point of the generating curve where the tangent is parallel to the axis of revolution (Fig. 419). This condition is clearly sufficient, since it implies that the normal line of the parallel agrees with the normal line to the surface (Fig. 419). We shall obtain for further use an interesting geometric consequence from the first of the equations (4a), known as Clairaut's relation. Observe that the first of the equations (4a) may be written as
+
(f '24')'
= f ""
+ 2ff
'u'v'
const.
= c.
hence, f2
~ = '
= 0;
Parallel Transport. Geodesics
Figure 419
On the other hand, the angle 0, 0 5 I9 5 7~12,of a geodesic with a parallel that intersects it is given by
where ( x u , xu) is the associated basis to the given parametrization. Since f = r is the radius of the parallel at the intersection point, we obtain Clairaut 's relation :
r cos I9
= const. =
I c 1.
In the next example we shall show how useful this relation is. See also Exercises 18, 20, and 21. Finally, we shall show that system (4a) m.ay be integrated by means of primitives. Let u = u(s), v = v(s) be a geodesic parametrized by arc length, which we shall assume not to be a meridian or a parallel of the surface. The ' const. = c + 0. first of the equations (4a) is then written as f 2 ~ = Observe initially that the first fundamental form along (u(s), ~ ( s ) ) ,
together with the first of the equations (4a), is equivalent to the second of the equations (4a). In fact, by substituting f 2u' =; c in Eq. (9,we obtain
Intrinsic Geometry o f Surfaces
hence, by differentiating with respect to s,
which is equivalent to the second equation of (4a), since (u(s), ~ ( s ) is ) not a parallel. (Of course the geodesic may be tangent to a parallel which is not a geodesic and then vf(s) = 0. However, Clairaut's relation shows that this happens only at isolated points.) On the other hand, since c + 0 (because the geodesic is not a meridian), we have ul(s) + 0. It follows that we may invert u = u(s), obtaining s = s(u), and therefore v = v(s(u)). By multiplying Eq. (5) by ( d s l d ~ ) ~we , obtain
or, by using the fact that ( d s l d ~ )=~ f4/c2,
that is,
hence, ucJ,J
1
(ff'12' +  c 2(g')' dv + const.
which is the equation of a segment of a geodesic of a surface of revolution which is neither a parallel nor a meridian. Example 6. We are going to show that any geodesic of a paraboloid of revolution z = x2 y 2 which is not a meridian intersects itself an infinite number of times. Let p, be a point of the paraboloid and let Po be the parallel of radius r , passing through p,. Let y be a parametrized geodesic passing through p, and making an angle 8, with Po. Since, by Clairaut's relation,
+
r cos 8
= const. = [
we conclude that 8 increases with r.
c 1,
n
0I 8 15.
Parallel Transport. Geodesics
259
Therefore, if we follow in the direction of the increasing parallels, 9 increases. It may happen that in some revolution surfaces y approaches asymptotically a meridian. We shall show in a while that such is not the case with a paraboloid of revolution. That is, the geodesic y intersects all the meridians, and therefore it makes an infinite number of turns around the paraboloid. On the other hand, if we follow the direction of decreasing parallels, the angle 8 decreases and approaches the value 0, which corresponds to a parallel of radius 1 c 1 (observe that if 8, + 0, I c ( < r). We shall prove later, in this book that no geodesic of a surface of revolution can be asymptotic to a parallel which is not itself a geodesic (Sec. 47). Since no parallel of the paraboloid is a geodesic, the geodesic y is actually tangent to the parallel of radius I c I at the point p,. Because 1 is a maximum for cos 8, the value of r will increase starting from p,. We are, therefore, in the same situation as before. The geodesic will go around the paraboloid an infinite number of turns, in the direction of the increasing r's, and it will clearly intersect the other branch infinitely often (Fig. 420),
v
Figure 420
Observe that if 8, = 0, the initial situation is that of the point p,. It remains to show that when r increases, the geodesic y meets all the meridians of the paraboloid. Observe initially that the geodesic cannot be tangent to a meridian. Otherwise, it would coincide with the meridian by the uniqueness part of Prop. 5. Since the angle 8 increases with r, if y did not cut all the meridians, it would approach asymptotically a meridian, say M. Let us assume that this is the case and let us choose a system of local coordinates for the paraboloid z = x2 y2, given by
+
x
=v
cos u,
y
sin u, z = v 2, O 0) of a hyperboloid of revolution x2 y 2  z2 = 1 and makes an angle 8 with the parallel passing through p in such a way that cos 8 = llr, where r is the distance from p to the z axis. Show that by following the geodesic in the direction of decreasing parallels, it approaches asymptotically the parallel x2 y2 = 1, z = 0 (Fig. 422).
+
+
*19. Show that when the differential equations (4) of the geodesics are referred to
the arc length then the second equation of (4) is, except for the coordinate curves, a consequence of the first equation of (4).
Parallel Transport. Geodesics
Figure 422
x
"20. Let T be a torus of revolution which we shall assume to be parametrized by (cf.
ExampIe 6, See. 22) X(U,
v)
= ((r cos LC
+ a) cos v, (r cos u + a) sin v, r sin u).
Prove that
a. If a geodesic is tangent to the paraIlel u = nj2, then it is entirely contained in the region of T given by
b. A geodesic that intersects the parallel u
also intersects the parallel u
=
cos
= 0 under
an angle 6 (0 < 6 < n/2)
n if
e < .aa + rr
21. Surfaces ofLiouville are those surfaces for which it is possible to obtain a system of local coordinates x(u, v ) such that the coefficients of the first fundamental form are written in the form
where U = U(u) is a function of u alone and V = V(v)is a function of v alone. Observe that the surfaces of Liouville generalize the surfaces of revolution and prove that (cf. Example 5) a. The geodesics of a surface of Liouville may be obtained by primitivation in the form
Intrinsic Geometry o f Surfeces
where c and c , are constants that depend on the initial conditions. b. If 6, 0 < 8 < n/2, is the angle which a geodesic makes with the curve v = const., then
(Notice that this is the analogue of Clairaut's relation for the surfaces of Liouville.) 22. ~ e St2 = {(x,y, Z ) E R 3 ;x2 y2 2 2 = 1) and let p E S2. For each piecewise regular parametrized curve a : [O, I ] + S2 with a(0) = a(1) = p, let P,: Tp(S2),T p ( S Zbe ) the map which assigns to each v E Tp(S2)its parallel transport along a back top. By Prop. 1, Pais an isometry. Prove that for every rotations R of Tp(S)there exists an a such that R = Pa. 23. Show that the isometries of the unit sphere
+ +
are the restrictions to S2 of the linear orthogonal transformations of R3.
45. The GaussBonnet Theorem and Its App/ications In this section, we shaIi present the GaussBonnet theorem and some of its consequences. The geometry involved in this theorem is fairly simple, and the difficulty of its proof lies in certain topological facts. These facts will be presented without proofs. The GaussBonnet theorem is probably the deepest theorem in the differential geometry of surfaces. A first version of this theorem was presented by Gauss i n a famous paper [I] and deaIs with geodesic triangles on surfaces (that is, triangles whose sides are arcs of geodesics). Roughly speaking, it asserts that the excess over n of the sum of the interior angles q l , q z ,q 3 of a geodesic triangle T is equal to the integral of the Gaussian curvature K over T ; that is (Fig. 423),

For instance, if K 0, we obtain that C q i = n, an extension of Thales' theorem of high school geometry to surfaces of zero curvature. Also, if K  1, we obtain that C qi  n = area (T) > 0. Thus, on a unit sphere, the sum of the interior angles of any geodesic triangle is greater than n, and the
The GaussBonnet Theorem,and Its Applications
Figure 423. A geodesic triangle.
excess over n is exactly the area of T.Similarly, on the pseudosphere (Exercise 6, Sec. 33) the sum of the interior angles of any geodesic triangle is smaller than n (Fig. 424). The extension of the theorem to a region bounded by a nongeodesic simple curve (see Eq. (1) below) is due to 0.Bonnet. To extend it even further, say, to compact surfaces, some topological considerations will come into play. Actually, one of the most important features of the GaussBonnet theorem is that it provides a remarkable relation between the topology of a compact surface and the integral of its curvature (see Corollary 2 below).
Figure 424
We shall now begin the details of a local version of the GaussBonnet theorem. We need a few definitions. S be a continuous map from the closed interval [0,1] Let a : [0, Z] into the regular surface S. We say that a is a simple, closed, piecewise regular, parametrized curve if
266
Intrinsic Geometry o f Surfaces
2. t, # t,, t,, t, E [0, 1), implies that a(t,) # a(t,). 3. There exists a subdivision
of [0, I] such that a is differentiable and regular in each [ti, ti+l], i = O , . . . , k. Intuitively, this means that a is a closed curve (condition I) without selfintersections (condition 2), which fails to have a welldefined tangent line only at a finite number of points (condition 3). The points a(t,), i 0, . . . , k, are called the vertices of a and the traces a([ti,ti,,]) are called the regular arcs of a. It is usual to call the trace a([O, I]) of a a closed piecewise regular curve. By the condition of regularity, for each vertex a(ti) there exist the limit from the left, i.e., for t < t i

and the limit from the right, i.e., for t
> ti,
lim aJ(t) = &'(ti tti
+ 0) # 0.
Assume now that S is oriented and let [Oi (,0 < 1 Of ( ( n, be the smallest determination of the angle from aJ(ti 0) to &'(ti 0). If / Qi I + n, we give 8, the sign of the determinant (aJ(ti 0), at(ti 0), N). This means that if the vertex a(ti) is not a "cusp" (Fig. 425), the sign of Oi is given by the orientation of S. The signed angle 8,, n < Oj < n, is called the external angle at the vertex a(ti). In the case that the vertex &(ti)is a cusp, that is, (0,( = n, we choose the sign of Oi as follows. By the condition of regularity, we can see that there exists a number E' > 0 such that the determinant (a'(ti  E ) , aJ(ti E), N)
+
+
+
f
Figure 425
The GaussBonnet Theorem and Its Applications
267
Figure 426. The sign of the external angle in the case of a cusp.
does not change sign for all 0 < E < E'. We give Oithe sign of this deterrninant (Fig. 426). S be a parametrization compatible with the orientation Let x: U c R2 of S. Assume further, that U is homeomorphic to an open disk in the plane. Let a: [O, 11 x(U) c S be a simple closed, piecewise regular, parai = 0, . . . , k. metrized curve, with vertices a(ti) and external angles @,, Let qi:[ti, ti+,] + R be differentiable functions which measure at each t E [ti,ti,,] the positive angle from xu to a'(t) (cf. Lemma 1, Sec. 44). The first topological fact that we shall present without proof is the following. t
THEOREM (of Turning Tangents). With the above notation
where the sign plus or minus depends on the orientation of a.
The theorem states that the total variation of the angle of the tangent vector to a with a given direction plus the "jumps" at the vertices is equal to 2n. An elegant proof of this theorem has been given by H. Hopf, Compositio Math. 2 (1935), 5062. For the case where a has no vertices, Hopf's proof can be found in Sec. 57 (Theorem 2) of this book. Before stating the local version of the GaussBonnet theorem we still need some terminology. Let S be an oriented surface. A region R c S (union of a connected open set with its boundary) is called a simple region if R is homeomorphic to a disk and the boundary dR of R is the trace of a simple, closed, piecewise regular, parametrized curve a: I S. We say then that a is positively oriented if

Intrinsic Geometry of Surfaces
268
for each a(t), belonging to a regular arc, the positive orthogonal basis {at(t),h(t)) satisfies the condition that h(t) "points toward" R; more precisely, for any curve p: 1 R with B(0) a(t) and Bt(0) + at(t), we have that {fi'(O),h(t)) > 0. Intuitively, this means that if one is walking on the curve a in the positive direction and with one's head pointing to N, then the region R remains to the left (Fig. 427). It can be shown that one of the two possible orientations of a makes it positively oriented.

+
Figure 427. A positively oriented boundary curve.
Now let x: U c R2 S be a parametrization of S compatible with its orientation and let R c x(U) be a bounded region of S. Iff is a differentiable function on S, then it is easily seen that the integral +
JJ
~(U,V),/EGF~~U~V
x' (R)
does not depend on the parametrization x, chosen in the class of orientation of x. (The proof is the same as in the definition of area; cf. Sec. 25.) This integral has, therefore, a geometrical meaning and is called the integral off over the region R. It is usual to denote it by
With these definitions, we now state the
GAUSSBONNET THEOREM (Local). Let x: U . S be an orthogonal parametrization (that is, F = 0), of an oriented surface S, where U c R 2 is homeomorphic to an open disk and x is compatible with the orientation of S.
The GaussBonnet Theorem and Its Appiications
269
Let R cz x(U) be a simple region of S and let a : T + S be such that dR = a(I). Assume that a is positively oriented, parametrized by arc length s, and let a(so), . . . , a(s,) and 00,. . . ,8, be, respectively, the vertices and the external angles of a. Then
where k,(s) is the geodesic curvature of the regular arcs of a and K is the Gaussian curvature of S. Remark. The restriction that the region R be contained in the image set of an orthogonal parametrization is needed only to simplify the proof. As we shall see later (Corollary 1 of the global GaussBonnet theorem) the above result still holds for any simple region of a regular surface. This is quite plausible, since Eq. (I) does not involve in any way a particular parametrizati0n.t ) the expression of a in the parametrizaProoJ: Let u = u(s), v = ~ ( s be tion x. By using Prop. 3 of Sec, 44, we have
where q i = pi($) is a differentiable function which measures the positive angle from xu to a'(s) in [s,, si+,]. By integrating the above expression in every interval [s,, si,,J and adding up the resuhs,
Now we use the GaussGreen theorem in the uv plane which states the following: u P ( u , v) and Q(u, v) are dzflerentiable functions in a simple region 4 c R2, the boundary of which is given by u = u(s), v = v(s), then
?If the truth of this assertion is assumed, applications 2 and 6 given below san be prexnted now.
Intrinsic Geometry o f Surfaces
It follows that
From the Gauss formula for F = 0 (cf. Exercise 1, Sec. 43), we know that
=
jj K do. R
On the other hand, by the theorem of turning tangents,
Since the curve a is positiveIy oriented, the sign should be plus, as can easily be seen in the particular case of the circle in a plane. By putting these facts together, we obtain
Q.E.D. Before going into a global version of the GaussBonnet theorem, we would like to show how the techniques used in the proof of this theorem may also be used to give an interpretation of the Gaussian curvature in terms of parallelism. To do that, let x : U ,S be an orthogonal parametrization at a point p E S, and let R c x(U) be a simple region without vertices, containing p in its interior. Let a: [0, I ] +x(U) be a curve parametrized by arc length s such that the trace of a is the boundary of R. Let wo be a unit vector tangent to S at a(0) and let w(s), s E [0, I ] , be the parallel transport of w , along a (Fig. 428). By using Prop. 3 of Sec. 44 and the GaussGreen theorem in the uu plane, we obtain
The GaussBonnet Theorem and Its Applications
Figure 428
where q = p(s) is a differentiable determination of the angle from xu to w(s). It follows that p(l)  q(0) = Aq is given by Ap
=
jj K d o . R
(2)
Now, Ap does not depend on the choice of w,, and it follows from the expression above that Aq does not depend on the choice of a(0) either. By taking the limit (in the sense of Prop. 2, Sec. 33)
& lim  K(P), R.P A(R) where A(R) denotes the area of the region R, we obtain the desired interpretation of K. To globalize the GaussBonnet theorem, we need further topological preliminaries. Let S be a regular surface. A region R c S is said to be regular if R is compact and its boundary dR is the finite union of (simple) closed piecewise regular curves which do not intersect (the region in Fig. 429(a) is regular, but that in Fig. 429(b) is not). For convenience, we shall consider a compact surface as a regular region, the boundary of which is empty. A simple region which has only three vertices with external angles a, # 0, i = 1,2, 3, is called a triangle. A triangulation of a regular region R c S is a finite family 3 of triangles Ti, i = I, . . . ,n, such that
1 . U:=,Ti = R. 2. If Tin T j # 6, then Tin Tjis either a common edge of Ti and T j or a common vertex of T, and TI. Given a triangulation 3 of a regular region R c S of a surface S, we shall denote by F the number of triangles (faces), by E the number of sides (edges),
Intrinsic Geometry of Surfaces
Figure 429
and by V the number of vertices of the triangulation. The number
is called the EulerPoincare' characteristic of the triangulation. The following propositions are presented without proofs. An exposition of these facts may be found, for instance, in L. Ahlfors and L. Sario, Riemann Surfaces, Princeton University Press, Princeton, N.J., 1960, Chap. 1.
PROPOSITION 1. Every regular region of a regular surface admits a triangulation. PROPOSITION 2. Let S be an oriented surface and {x,], a E A, a family of parametrizations compatible with the orientation of S. Let R c S be a regular region of S . Then there is a triangulation 3 of R such that every triangle T E 3 is contained in some coordinate neighborhood of the family {x,). Furthermore, if the boundary of every triangle of 3 is positively oriented, adjacent triangles determine opposite orientations in the common edge (Fig. 430).
v
Figure 430
PROPOSITION 3. If R c S is a regular region of a surface S, the EulerPoincari characteristic does not depend on the triangulation of R. It is convenient, therefore, to denote it by x(R).
The GaussBonnet Theorem end its Applicetions
273
The latter proposition shows that the EulerPoincarb characteristic is a topological invariant of the regular region R. For the sake of the applications of the GaussBonnet theorem, we shall mention the important fact that this invariant allows a topological classification of the compact surfaces in R 3 . It should be observed that a direct computation shows that the EulerPoincark characteristic of the sphere is 2, that of the torus (sphere with one "handle"; see Fig. 431) is zero, that of the double torus (sphere with two Sphere X = 2
/
Sphere with one handle X
=0
Sphere with two handles X = 2
......     ....... 2; ....:;
Torus
2Torus
Figure 431
handles) is 2, and, in generaI, that of the ntorus (sphere with n handles) is 2(n  1). The following proposition shows that this list exhausts all compact surfaces in R 3 , I
PROPOSITION 4. Let S c R3 be a compact connected surface; then one of the values 2,0, 2, . . . , 2n, . . . . is assumed by the EuZerPoincark characteristic x(S). Furthermore, if Sf c R3 is another compact surface and x(S) = x(S1), then S is homeomorphic to S'. In other words, every compact connected surface S c R3 is homeomorphic to a sphere with a certain number g of handles. The number
is called the genus of S. Finally, let R c S be a regular region of an oriented surface S and let 3 be a triangulation of R such that every triangle TjE 3, j = 1, ..., k, is contained in a coordinate neighborhood xj(Uj) of a family of parametrizations (x,], a E A, compatible with the orientation of S. Let f be a differentiable function on S. The following proposition shows that it makes sense to talk about the integral off over the region R.
274
intrinsic Geometry o f Surfaces
PROPOSITION 5, With the above notation, the sum
does not depend on the triangulation 3 or on the family (x,)of parametrizations of S. This sum has, therefore, a geometrical meaning and is called the integral off over the regular region R.It is usually denoted by
We are now in a position to state and prove the
GLOBAL GAUSSBONNET THEOREM. Let R c S be a regular region of an oriented surface and let C,, . . . , C, be the closed, simple, piecewise regular curves which form the boundary dR of R. Suppose that each Ci is positively oriented and let O,, . . . , 9 , be the set of all external angles of the curves C,, . . . , C,. Tlzen
where s denotes the arc length of Ci, and the integral over Ci means the sum of integrals in every regular arc of C,. Proof, Consider a triangulation 3 of the region R such that every triangle
TIis contained in a coordinate neighborhood of a family of orthogonal parametrizations compatible with the orientation of S. Such a triangulation exists by Prop. 2. Furthermore, if the boundary of every triangle of 3 is positively oriented, we obtain opposite orientations in the edges which are common to adjacent triangles (Fig. 432). By applying to every triangle the local GaussBonnet theorem and adding up the results we obtain, using Prop. 5 and the fact that each "interior" side is described twice in opposite orientations,
i\.here F denotes the number of triangles of 3, and O,,, O,,, Oj, are the external angles of the triangle T,. We shall now introduce the interior angles of the triangle T,, given by grjk = n  O,,. Thus,
The GaussBonnet Theorem and Its Applications
Figure 432
We shall use the following notation: Ee = number of external edges of 3, E, = number of internal edges of 3, V, = number of external vertices of 3, V, = number of internal vertices of 3.
Since the curves Ci are closed, Q induction that
3F
=
V,. Furthermore, it is easy to show by
= 2E,
+ E,
and therefore that
We observe now that the external vertices may be either vertices of some curve Ci or vertices introduced by the triangulation. We set Ve = V,, Vet, where V,, is the number of vertices of the curves Ci and Vetis the number of external vertices of the triangulation which are not vertices of some curve Ci. Since the sum of angIes around each internal vertex is 2n, we obtain
+
By adding nE, to and subtracting it from the expression above and taking into consideration that Ee = V,, we conclude that
276
intrinsic Geometry of Surfaces
By putting things together, we finally obtain
= 2nx(R).
Q.E.D.
Since the EulerPoincarC characteristic of a simple region is clearly 1, we obtain (cf. Remark 1) COROLLARY 1. I f R is a simple region of S, then
By taking into account the fact that a compact surface may be considered as a region with empty boundary, we obtain COROLLARY 2. Let S he an orientable compact surface; then
Corollary 2 is most striking. We have only to think af all possible shapes of a surface homeomorphic to a sphere to find it very surprising that in each case the curvature function distributes itself in such a way that the "total curvature," i.e., K d c , is the same for all cases. We shall present some applications of the GaussBonnet theorem below. For these applications (and for the exercises at the end of the section), it is convenient to assume a basic fact of the topology of the plane (the Jordan curve theorem) which we shall use in the following form: Every piecewise regular curve in the plane (thus without selfintersections) is the boundary of a simple region. I. A compact surface of positive curvature is homeomorphic to a sphere. The EulerPoincarC characteristic of such a surface is positive and the sphere is the only compact surface of R3 which satisfies this condition. 2. Let S be an orientable surface of negative or zero curvature. Then two geodesics y 1 and y, which start from a point p E S cannot meet again at a point q E S in such a way that the traces of y , and y, constitute the boundary of a simple region R of S. Assume that the contrary is true. By the GaussBonnet theorem (R is simple)
The GaussBonnet Theorem and Its Applications
277
where 8, and 8, are the external angles of the region R. Since the geodesics y , and 7, cannot be mutually tangent, we have 9,. < r , i = 1,2. On the other hand, K 5 0, whence the contradiction. When 8, = 9, = 0, the traces of the geodesics y , and y, constitute a simple closed geodesic of S (that is, a closed regular curve which is a geodesic), It follows that on a surface of zero or negative curvature, there exists no simple closed geodesic which is a boundary of a simple region of S. 3. Lek S be a surface homeomorphic to a cylinder with Gaussian curvature K < 0. Then S has at most one simple closed geodesic. Suppose that S contains one simple closed geodesic r. By application 2, and since there is a homeomorphism p of S with a plane P minus one point q E P,p ( r ) is the boundary of a simple region of P containing q. Assume now that S contains another simple closed geodesic We claim that does not intersect r. Otherwise, the arcs of q(T) and p(I") between two L ' ~ ~ n ~ e intersection ~ ~ t i ~ e "points, r , and r,, would be the boundary of a simple region, contradicting application 2 (see Fig. 433). By the above argu
r'.
r'
Figure 433
ment, q(T') is again the boundary of a simple region R of P containing q, the interior of which is homeomorphic to a cylinder. Thus, x(R) = 0. On the other hand, by the GaussBonnet theorem,
which is a contradiction, since K
< 0.
rl
4. I/there exist two simple closed geodesics and T, on a compact surface S ofpositive curvature, then r, and r, intersect. By application 1, S is homeomorphic t o a sphere. If TIand T2do not intersect, then the set formed by r, and r, is the boundary of a region R, the EulerPoincart characteristic of which is x(R) = 0.By the GaussBonnet theorem,
Intrinsic Geometry o f Surfaces
which is a contradiction, since K
> 0.
5. We shall prove the following result, due to Jacobi: Let a : I +R3 be a closed, regular, parametrized curve with nonzero curvature. Assume that the curve described by the normal vector n(s) in rhe unit sphere S2 (the normal indicatrix) is simple. Then n(I) divides S2 in two regions with equal areas. We may assume that a is parametrized by arc length. Let s' denote the arc length of the curve n = n(s) on S2.The geodesic curvature lgof n(s) is
where the dots denote differentiation with respect to i. Since
and
we obtain

 kz' ds k 2 $ r2 2F
z'k


d tanl(l)
ds
k
ds
s'
Thus, by applying the GaussBonnet theorem to one of the regions R bounded by n(1) and using the fact that K = I, we obtain
Since the area of S2is 471, the result follows.
6. Let T be a geodesic triangle (that is, the sides of T are geodesics) in an oriented surface S. Let 9,, 9,, 9, be the external angles of T and let p, = x  el, q2 = n  02, p3 = n  8, be its interior angles. By the GaussBonnet theorem,
The GaussBonnet Theorem and Its Applications
Thus,
It follows that the sum of the interior angles, C,LI yi, of a geodesic triangle is 1. Equal t o n if K = O . 2. Greater than n if K > 0. 3. Smaller than n if K < 0.
Furthermore, the difference cisely by
Ci3_,yi
 n (the excess of T ) is given pre
[ K d o , If K # 0 on T, this is the area of the image N ( T ) of T by T
the Gauss map M:S + S2(cf. Eq. (12), Sec. 33). This was the form in which Gauss himself stated his theorem: The excess of a geodesic triangle T is equal to the area of its spherical image N(T). The above fact is related to a historical controversy about the possibility of proving Euclid's fifth axiom (the axiom of the parallels), from which it follows that the sum of the interior angles of any triangle is equal to n. By considering the geodesics as straight lines, it is possible to show that the surfaces of constant negative curvature constitute a (local) model of a geometry where Euclid's axioms hold, except for the fifth and the axiom which guarantees the possibility of extending straight lines indefinitely. Actually, Hilbert showed that there does not exist in R3 a surface of constant negative curvature, the geodesics of which can be extended indefinitely (the pseudosphere of Exercise 6 , Sec. 33, has an edge of singular points). Therefore, the surfaces of R3 with constant negative Gaussian curvature do not yield a model to test the independence of the fifth axiom alone. However, by using the notion of abstract surface, it is possible to bypass this inconvenience and to buiId a model of geometry where all of Euclid's axioms but the fifth are valid. This axiom is, therefore, independent of the others. In Secs. 510 and 51 1, we shall prove the result of Hilbert just quoted and shall describe the abstract model of a noneuclidean geometry.
7. Vector Jields on surfaces."fet v be a differentiable vector field on an oriented surface S . We say that p E S is a singular point of v if c ( p ) = 0. The singular point p is isolated if there exists a neighborhood V of p in S such that v has no singular points in V other than p. To each isolated singular point p of a vector field v, we shall associate an ?This application requires the material of Sec. 34. If omitted, then Exercises 69 of this section should also be omitted.
280
Intrinsic Geometry o f Surfaces
integer, the index of v, defined as follows. Let x: U + S be an orthogonal parametrization at p = x(O, 0) compatible with the orientation of S, and let a : [0, 11 S be a simple, closed, piecewise regular parametrized curve such that a([O,I ] ) c x(U) is the boundary of a simple region R containing p as its only singular point. Let v = v(t), t E [O, I], be the restriction of v along a, and let = ~ ( t be ) some differentiable determination of the angle from xu to v(t), given by Lemma 1 of Sec. 44 (which can easily be extended to piecewise regular curves). Since a is closed, there is an integer I defined by

I is called the index of v at p. We must show that this definition is independent of the choices made, the first one being the parametrization x. Let w, E T,,,,(S)and let w(t) be the parallel transport of w, along a. Let ry(t) be a differentiable determination of the angle from xu to w(t). Then, as we have seen in the interpretation of K in terms of parallel transport (cf. Eq. (2)),
By subtracting the above relations, we obtain
Since ry  y does not depend on xu,the index I is independent of the parametrization x. The proof that the index does not depend on the choice of a is more technical (although rather intuitive) and we shall only sketch it. Let a, and a, be two curves as in the definition of index and Jet us show that the index of v is the same for both curves. We first suppose that the traces of a, and a, do not intersect. Then there is a homeomorphism of the regior bounded by the traces of a. and a, onto a region of the plane bounded by twc concentric circles C, and C, (a ring). Since we can obtain a family of concentric circles C, which depend continuously on t and deform C, into C , , we obtain a family of curves a,, which depend continuously on t and deform a, into a, (Fig. 434). Denote by It the index of v computed with the curve a,. Now, since the index is an integral, I, depends continuously on t , t E [0, 11. Being an integer, I, is constant under this deformation, and I, = I , , as we wished. If the traces of a, and a , intersect, we choose a curve sufficiently small so that its trace has no intersection with both a, and a, and then appl! the previous result.
The GaussBonnet Theorem and Its Applica tions
287
Figure 434
It should be noticed that the definition of index can stiIl be applied when p is not a singular point of v. It turns out, however, that the index is then zero. This follows from the fact that, since I does not depend on x,,, we can choose x, to be v itself; thus, q(t) = 0. In Fig. 435 we show some examples of indices of vector fields in the xy plane which have (0,O) as a singuIar point. The curves that appear in the drawings are the trajectories of the vector fields.
Figure 435
Now, let S c R3 be an oriented, compact surface and v a differentiable vector field with only isolated singular points. We remark that they are finite in number. Otherwise, by compactness (cf. Sec. 27, Property l), they have a limit point which is a nonisolated singular point. Let ( x u ) be a family of orthogonal parametrizations compatible with the orientation of S. Let 5 be a triangulation of S such that 1. Every triangle T E 3 is contained in some coordinate neighborhood of the family ( x u ] . 2. Every T E 3 contains at most one singular point. 3. The boundary of every T E 3 contains no singular points and is positively oriented. If we apply Eq. (1) to every triangle T E 3, sum up the results, and take into account that the edge of each T E 3 appears twice with opposite orientations, we obtain
Intrinsic Geometry of Surfaces
where Ii is the index of the singular point pi, i = 1, . . . , k. Joining this with the GaussBonnet theorem (cf. Corollary 2), we finally arrive at
Thus, we have proved the following:
POINCARE'S THEOREM. The sum of the indices of a diferentiable vector 5eld v with isolated singular points on a compact surface S is equal to the EulerPoincark characteristic of S. remarkable result. It implies that C 4 does not depend on v but only o n the topology of S. F o r instance, in any surface homeomorphic t o a sphere, all vector fields with isolated singularities must have the sum of their indices equal t o 2. In particular, n o such surface can have a differentiable vector field without singular points.
This is a
EXERCISES 1. Let S c R3 be a regular, compact, orientable surface which is not home
omorphic to a sphere. Prove that there are points on S where the Gaussian curvature is positive, negative, and zero. 2. Let T be a torus of revolution. Describe the image of the Gauss map of T and show, without using the GaussBonnet theorem, that
Compute the EulerPoincarC characteristic of T and check the above result with the GaussBonnet theorem.
r
3. Let S c R3 be a regular surface homeomorphic to a sphere. Let c S be a simple closed geodesic in S, and let A and B be the regions of S which have I as a common boundary. Let N: S + S 2 be the Gauss map of S. Prove that N ( A )
and N(B) have the same area.
4. Compute the EulerPoincark characteristic of a. An ellipsoid. *b. The surface S = {(x,y, z ) E R3; x 2 y4 z6
+ +
= 13.
5. Let C be a parallel of colatitude 9 on an oriented unit sphere S2, and let w o be a unit vector tangent to C at a point p E C (cf. Example 1, Sec. 44). Take the
283
The Exponential Map
parallel transport of w o along C and show that its position, after a compIete turn, makes an angle A9 = 2n(l  cos 9) with the initial position wo. Check that A9 = I lim Rp
A
= curvature
of S2,
where A is the area of the region R of S 2 bounded by C. 6. Show that (0, 0) is an isolated singular point and compute the index at (0, 0) of the following vector fields in the plane: *a. v = (x, y). b. v = (x, y). c. v = (x, y). "d. v = (x2  y2, 2xy). e. v = (x3  3xy2,y3  3x2~). 7. Can it happen that the index of a singular point is zero? If so, give an example. 8. Prove that an orientable compact surface S c R3 has a differentiable vector field without singular points if and only if S is homeomorphic to a torus. 9. Let C be a regular closed curve on a sphere S 2 . Let v be a differentiable vector field on S 2 such that the trajectories of v are never tangent to C. Prove that each of the two regions determined by C contains at least one singular point of v.
46. The Exponential Map. Geodesic Polar Coordinates In this section we shall introduce some special coordinate systems with an eye toward their geometric applications. The natural way of introducing such coordinates is by means of the exponential map, which we shall now describe. As we learned in Sec. 44, Prop. 5, given a point p of a regular surface S and a nonzero vector v E Tp(S) there exists a unique parametrized geodesic y : (6, E) + S, with y(0) = p and y'(0) = v. T o indicate the dependence of this geodesic on the vector v, it is convenient t o denote it by y(t, r ) = y.
L E M M A 1. If the geodesic y(t, v) is defied for t E (6, e), [hen the geodesic y(t, lv), 1 E R, Il f 0, is defined for t t (€/A, €/A), and ~ ( tav) , = ~ ( n t v). , Proof. Let u: (€/A, €/A) * S be a parametrized curve defined by u(t) = y(At). Then u(0) = y(O), ~'(0) = ;ly1(O), and, by the linearity of D (cf. Eq. (I), Sec. 4Z), Dar(rl~'(t) = 12Dr/(t)y'(t) = 0.
intrinsic Geometry o f Surfaces
284
If follows that a is a geodesic with initial conditions y(O), Ayl(0), and by uniqueness a(t) = y(t, Av)
= y(At,
Q.E.D.
v)
Intuitively, Lemma 1 means that since the speed of a geodesic is constant, we can go over its trace within a prescribed time by adjusting our speed appropriately. We shall now introduce the following notation. If v such that y(l v],v/]v I) = y(1, v) is defined, we set exp,(v)
= y(l,v)
= T,(S),
v
+ 0, is
and expp(0) =p.
Geometrically, the construction corresponds to laying off (if possible) a length equal to \ v ( along the geodesic that passes through p in the direction of v; the point of S thus obtained is denoted by expP(v) (Fig. 436).
Figure 437
Figure 436
For example, expp(v) is defined on the unit sphere S2for every v E Tp(S2). The points of the circles of radii n, 3n, . . . , (2n 1)n are mapped into the point q, the antipodal point of p. The points of the circles of radii 2n, 4n,. . . ,2nn are mapped back into p. On the other hand, on the regular surface C formed by the onesheeted cone minus the vertex, exp,(v) is not defined for a vector v E T,(C) in the direction of the meridian that connects p to the vertex, when [ vI 2 d and d is the distance from p to the vertex (Fig. 437). If, in the example of the sphere, we remove from S2the antipodal point ofp, then exp,(v) is defined only in the interior of a disk of T,(S2) of radius n and center in the origin. The important point is that exp, is always defined and differentiable in some neighborhood of p.
+
285
The Exponentid M a p
PROPOSITIONl. Given p E S there exists an 6 > 0 such that exp, is defined and diflerentiable in the interior B, of a disk of radius E of Tp(S), with center in the origin. ProoJ It is clear that for every direction of T,(S) it is possible, by Lemma 1, to take v sufficiently small so that the interval of definition of y(t, v) contains 1, and thus y(1, v) = exp,(v) is defined. To show that this reduction can be made uniformly in all directions, we need the theorem of the dependence of a geodesic on its initial conditions (see Sec. 47) in the following form : Given p E S there exist numbers e l > 0, c 2 > 0 and a diferentiable map
such that, for v E B,,, v # 0, t E (f2, f2), the curve y(t, v) is the geodesic of S with y(0, p) = p, yl(O, v) = v, and for v = 0, y(t, 0) = p. From this statement and Lemma 1, our assertion follows. In fact, since y(t, v) is defined for t 1 < E,, I v 1 < f , , we obtain, by setting A = 6,/2 in Lemma 1, that y(t, (e2/2)v) is defined for I t 1 < 2, 1 v 1 < 6,. Therefore, by taking a disk B, c T,(S), with center at the origin and radius 6 < 6,~,/2, we have that y(1, w) = exp, w, w E Be, is defined. The differentiability of Q.E.D. exp, in B, follows from the differentiability of y. An important complement to this result is the following:
PROPOSITION 2. exp,: B, c Tp(S) + S is a d~feomorphism in a neighborhood U c B, of the origin 0 of Tp(S). Proof. We shall show that the differential d(expp) is nonsingular at 0 E T,(S). To d o this, we identify the space of tangent vectors to T,(S) at 0 with T,(S) itself. Consider the curve a(t) = tv, v E T,(S). It is obvious that a(0) = 0 and a'O) = v. The curve (exp, 0 a)(t) = exp,(tv) has at t = 0 the tangent vector
It follows that (d expp)o(v)
= v,
which shows that d exp, is nonsingular at 0. By applying the inverse function theorem (cf. Prop. 3, Sec. 24), we complete the proof of the proposition. Q.E.D. It is convenient to call V c S a normal neighborhood of p E S if V is the image V = exp,(U) of a neighborhood U of the origin of T,(S) restricted to which exp, is a diffeomorphism.
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Intrinsic Geometry of Surfaces
Since the exponential map at p E S is a diffeomorphism on U, it may be used to introduce coordinates in V. Among the coordinate systems thus introduced, the most usual are 1. The normal coordinates which correspond to a system of rectangular coordinates in the tangent plane T,(S). 2. The geodesic polar coordinates which correspond to polar coordinates in the tangent plane T,(S) (Fig. 438).
Figure 438 Polar coordinates.
We shall first study the normal coordinates, which are obtained by choosing in the plane T,(S), p E S, two orthogonal unit vectors e, and e,. Since exp,: U +V c S is a diffeomorphism, it satisfies the conditions for a parametrization inp. If q E V, then q = exp,(w), where w = ue, ve, E U, and we say that q has coordinates (u, v). It is clear that the normal coordinates thus obtained depend on the choice of e,, e,. I n a system of normal coordinates centered in p, the geodesics that pass through p are the images by exp, of the lines u = at, v = bt which pass through the origin of T,(S). Observe also that at p the coefficients of the first fundamental form in such a system are given by E(p) = G(p) = 1, F(p) = 0.
+
Now we shall proceed to the geodesic polar coordinates. Choose in the plane T,(S), p E S, a system of polar coordinates (p, 8), where p is the polar
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287
radius and 8, 0 < 8 < 2n, is the polar angle, the pole of which is the origin 0 of T,(S). Observe that the polar coordinates in the plane are not defined in the closed halfline 1 which corresponds to 8 = 0. Set exp,(l) = L. Since exp,: U  1 V  L is still a diffeomorphism, we may parametrize the points of V  L by the coordinates (p, 8), which are called geodesic polar coordinates. We shall use the following terminology. The images by exp,: U V of circles in U centered in 0 will be called geodesic circles of V, and the images of exp, of the lines through 0 will be called radial geodesics of V. In V  L these are the curves p = const. and 8 = const., respectively. We shall now determine the coefficients of the first fundamental form in a system of geodesic polar coordinates.

PROPOSITION 3. Let x : U  I + V  L be a system of geodesic polar coordinates (p, 8). Then the coeficients E = E(p, 8), F = F(p, 8), and G = G(p, 8) of thefirst fundamental form satisfy the conditions E
=
1,
F
= 0,
lim G P0
= 0,
lim(JG), = 1. P'O
Proof By definition of the exponential map, p measures the arc length along the curve 8 = const. It follows immediately that E = 1. By introducing in the differential equation of a geodesic (Eq. (4), Sec. 44) the fact that 8 = const. is a geodesic, we conclude that ri, = 0. By using the first of the relations (2) of Sec. 43 that define the Christoffel symbols, we obtain 0 &E, = r i I E = rtl. 5
By introducing this relation in the second of the equations (2) of Sec. 43, we conclude that F, = 0, and, therefore, F(p, 8) does not depend on p. For each q E V, we shall denote by @(a)the geodesic circle that passes through q, where a E [0,2nj (if q = p , a(a) is the constant curve a(a) = p). We shall denote by y(s), where s is the arc Iength of y, the radial geodesic that passes through q. With this notation we may write
The coefficient F(p, 8) is not defined at p. However, if we fix the radial geodesic 8 = const., the second member of the above equation is defined for every point of this geodesic. Since at p, a(a) = p , that is, dalda = 0, we obtain lim F(p, 8) = lim P+O
Together with the fact that F does not depend on p, this implies that F = 0.
288
/ntrinsic Geometry of Surfaces
To prove the last assertion of the proposition, we choose a system of normal coordinates (ii, G) in p in such a way that the change of coordinates is given by ii=pcosO,
G=psinO,
p+O,
O 0, the general solution of Eq. (2) is given by
where A(8) and B(8) are functions of 8. That this expression is a solution of Eq. (2) is easily verified by differentiatiop. Since lim Jc = 0, we obtain A(8) = 0. Thus, 6'0
(JG),
and since lim ,)lf( 0
0

= B(8)JK
cos(J r p ) ,
1, we conclude that
Therefore, in this case,
3. Finally, if K < 0, the general solution of Eq. (2) is
fi= A(0)
cosh(Jp)
+ B(0)
sinh(mp).
By using the initial conditions, we verify that in this case
290
Intrinsic Geometry o f Surfaces
We are now prepared to prove Minding's theorem. Let V, and V, be normal neighborhoods of p , and p,, respectively. Let y be the linear isometry of Tp,(Sl)onto T,,(S,) given by p(e,) =f,, p(e,) =f,.Take a polar coordinate system (p, 0) in Tp,(S,) with axis I and set L, = expPl(l), L, = exp,, (~(1)).Let v : V , + V , be defined by
W \i7e claim that
= exp,,
0
q
0
expi,'.
w is the required isometry.
v
In fact, the restriction of y to V1  L, maps a poIar coordinate neighborhood with coordinates (p, 8 ) centered in p , into a polar coordinate neighborhood with coordinates (p, 8 ) centered in p,. By the above study of Eq. (2), the coefficients of the first fundamentaI forms at corresponding points are equal. By Prop. 1 of Sec. 42, y/ is an isometry. By continuity, y/ still preserves inner products at points of L , and thus is an isometry. It is immediate to check that dy(e,) =f,, dy(e2) =f2, and this concludes the proof. Q.E.D. Remark 2. In the case that K is not constant but maintains its sign, the has a nice intuitive meaning. Consider the expression f l =~ arc length L(p) of the curve p = const. between two close geodesics 8 = 8, and 8 = 8 , :
(+/c)),,
Assume that K
< 0. Since
lim (Jc)),
=
1 and
(
)=K
J G > 0,
P+O
the function L ( p ) behaves as in Fig. 439(a). This xeans that L(p) increases with p ; that is, as p increases, the geodesics 8 = 8, and 9 = 8 , get farther and farther apart (of course, we must remain in the coordinate neighborhood in question). On the other hand, if K > 0, L(p) behaves as in Fig. 439(b). The geodesics
(a) K
0
Figure 439. Spreading of close geodesics in a normal neighborhood.
29 f
The Exponentid/ M a p
8 = 8, and 8 = 8, may (case I) or may not (case 11) come closer together after a certain value of p, and this depends on the Gaussian curvature. For instance, in the case of a sphere two geodesics which leave from a pole start coming closer together after the equator (Fig. 440).
1
Figure 440
In Chap. 5 (Secs. 54 and 55) we shall come back to this subject and shall make this observation more precise. Another application of the geodesic polar coordinates consists of a geometrical interpretation of the Gaussian curvature K. To do this, we first observe that the expression of K in geodesic polar coordinates (p, 8), with center p E S, is given by
and therefore
Thus, recalling that lim JG P'O

0,
we obtain K(p) = lim

,+o
a3JG. dp3
On the other hand, by defining "/G and its successive derivatives with respect to p at p by its limit values (cf. Eq. (I)), we may write
292
Intrinsic Geometry o f Surfaces
where lim R(P, 6 ) = 0 , P0 P3 uniforn~lyin 8. By substituting in the above expression the values already known, we obtain
With this value for of radius p = r :
n,we compute the arc length L of a geodesic circle
27z6
L
= lirn 
0
o+c
, , / c ( r 9 8 ) d 8 = 2 n r   r723 K ( p ) + R l , 3
where
R = 0. lirn f ro
r
It follows that 3 2nr  L K(p) = lirn r...,o n r3 which gives an intrinsic interpretation of K ( p ) in terms of the radius r of a geodesic circle S,(p) around p and the arc lengths L and 2nr of S r ( P ) and exp; l(S,(p)), respectively. An interpretation of K ( p ) involving the area of the region bounded by Sr(p) is easily obtained by the above process (see Exercise 3). As a last application of the geodesic polar coordinates, we shall study some minimal properties of geodesics. A fundamental property of a geodesic is the fact that, locally, it minimizes arc length. More precisely, we have
PROPOSITION 4. Let p be a point on a surface S. Then, there exists a neighborhood W c S of p such that yy : I + W is aparametrized geodesic with y(0) = p, 7(tr) = 9, t, E I, and a $0, t,] S is a parametrized regular curve joining p to q, we have 4
zy < la, where 2, denotes the length of the curve a. Moreover, if I, of a coincides with the trace of a between p and q.
= Z,,
then the trace
293
The Exponential Map
Proof. Let V be a normal neighborhood of p, and let be the closed region bounded by a geodesic circle of radius r contained in V. Let (p, 8) be geodesic polar coordinates in w  L centered in p such that q E L. Suppose first that a((0, t,)) c fi  L, and set a(t) = (p(t), 8(t)). Observe initiaIly that
and equality holds if and only if 8' = 0; that is, 8 = const. Therefore, the length E,(c) of a between 6 and t ,  c satisfies
0 and equality holds if and only if 8 = const. and p' > 0. By making t I, and that equality holds if and in the expression above, we obtain that E, only if a is the radial geodesic 8 = const. with a parametrization p = p(t), where pr(t) > 0. It follows that if I, = I,, then the traces of a and y between p and q coincide. Suppose now that a((0, t,)) intersects L, and assume that this occurs for the first time at, say, a(t2). Then, by the previous argument, I, > I, between to and t,, and I, = Ey implies that the traces of a and y coincide. Since a([O, t , ] ) and L are compact, there exists a 6 2 t2 such that either a(?) is the Iast point where a((0, t,)) intersects L or a([< t , ] ) c L (Fig. 441). In any case, applying the above case, the conclusions of the proposition follow. +
>
Figure 441
Figure 442
Suppose finally that a([O, t , ] ) is not entirely contained in 6'.Let to E [0, t , ] be the first value for which a(to) = x belongs to the boundary of @. Let 7 be the radial geodesicpx and let d be the restriction of the curve a to the interval [0, to]. It is clear then that 2, 2 E, (see Fig. 442). By the previous argument, 1, 2 I,. Since q is a point in the interior of @, Q.E.D. I, > 1,. We conclude that 1, > I,, which ends the proof.
Intrinsic Geometry of Surfaces
294
Remark 3. For simplicity, we have proved the proposition for regular curves. However, it still holds for piecewise regular curves (cf. Def. 7, Sec. 44); the proof is entirely analogous and will be left as an exercise. &murk 4. The proof also shows that the converse of the last assertion of Prop. 4 holds true. However, this converse does not generalize to piecewise regular curves. The previous proposition is not true globaIly, as is shown by the example of the sphere. Two nonantipodal points of a sphere may be connected by two meridians of unequal lengths and only the smaller one satisfies the conclusions of the proposition. In other words, a geodesic, if sufficiently extended, may not be the shortest path between its end points. The following proposition shows, however, that when a regular curve is the shortest path between any two of its points, this curve is necessarily a geodesic,
PROPOSITION 5. Let a : I S be a regular parametrized curve with a parameter proportional to arc length. Suppose that the arc length of a between any two points t, z E I , is smaller than or equal to the arc length of any regular parametrized curve joining a(t) to a(z). Then a is a geodesic. 4
Proof. Let to E I be an arbitrary point of I and let W be the neighborhood of a(t,) = p given by Prop. 4. Let q = a ( t , ) E W. From the case of equality in Prop. 4, it follows that a is a geodesic in (to,t,). Otherwise a would have, between to and t , , a length greater than the radial geodesic joining a(?,) to a ( t , ) , a contradiction to the hypothesis. Since a is regular, we Q.E.D. have, by continuity, that a still is a geodesic in to.
1. Prove that on a surface of constant curvature the geodesic circles have constant
geodesic curvature. 2. Show that the equations of the geodesics in geodesic polar coordinates (E = 3, F = 0) are given by
3. If p is a point of a regular surface S, prove that
12nv2  A K ( p ) = lim rO
n
r4
9
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295
where K(p) is the Gaussian curvature of S at p, r is the radius of a geodesic circle S,(p) centered inp, and A is the area of the region bounded by S,(p).
4. Show that in a system of normal coordinates centered in p, all the Christoffel symbols are zero at p. 5, For which of the pair of surfaces given beIow does there exist a IocaI isometry ?
a. Torus of revolution and cone. b. Cone and sphere. c. Cone and cylinder. 6. Let S be a surface, let p be a point of S, and Iet Sl(p) be a geodesic circle around p, sufficiently small to be contained in a normal neighborhood. Let r. and s be two points of Sl(p), and C be an arc of S1(p)between r and s. Consider the curve expil(C) c T,(S). Prove that S1(p) can be chosen sufficiently small so that a. If K
> 0, then I(exp;I(C)) > l(C), where I(
) denotes the arc length of the
corresponding curve. b. If K < 0,then 1(expit(C)) < 1(C).
7. Let (p, 8) be a system of geodesic polar coordinates (E = 1, F = 0) on a surface, and let y(p(s), 8(s)) be a geodesic that makes an angle q ( s ) with the curves 6 = const. For definiteness, the curves 8 = const. are oriented in the sense of increasing p's and q is measured from 8 = const. to y in the orientation given by the parametrization (p, 8). Show that
"8. (Gauss Theorem on the Sum of the Internal Angles of a "Small" Geodesic Triangle.) Let b be a geodesic triangle (that is, its sides are segments of geodesics) on a surface S. Assume that A is sufficiently small to be contained in a normal neighborhood of some of its vertices. Prove directly (i.e., without using the GaussBonnet theorem) that
where K is the Gaussian curvature of S, and 0 < ai < n, i = 1, 2, 3, are the internal angles of the triangle A . 9. (A Local Isoperimetric Inequality for Geodesic Circles.) Let p E S and let S,(pl be a geodesic circle of center p and radius r. Let L be the arc length of S,(p) and A be the area of the region bounded by S,(p). Prove that
where K(p) is the Gaussian curvature of S at p and
R
lim  = 0. r+o r 4
296
/ntrinsic Geometry o f Surfaces
Thus, if K(p) > 0 (or < 0) and r is small, 4nA  L2 > 0 (or < 0). (Compare the isoperimetric inequality of Sec. 17.) S be two isometries of S. 10. Let S be a connected surface and let q, y:S Assume that there exists a point p cz S such that p(p) = ~ ( p and ) dpp(v) = dy,(v) for all v E T,(S). Prove that q(q) = y(q) for all q E S.
11. (Free Mobility of Small Geodesic Triangles.) Let S be a surface of constant Gaussian curvature. Choose points p,, p i E S and let V, V' be normal neighborhoods of p,, p:, respectively. Choose geodesic triangles p , , p2, p3 in V ,.n(geodesic means that the sides p,p2, ~ 2 ~p,pl 3 , are geodesic arcs) andp: ,p i , p: in V' in such a way that
(here I denotes the length of a geodesic arc). Show that there exists an isometry 8 : V +V' which maps the first triangle onto the second. (This is the local version, for surfaces of constant curvature, of the theorem of high school geometry that any two triangles in the plane with equal corresponding sides are congruen t.) 12. A diffeomorphism p: S14 S2 is said to be a geodesic mapping if for every geodesic C c S1of S1,the regular curve q(C) c S , is a geodesic of S2.If U is a neighborhood of p E S1, then q: U + S2is said to be a local geodesic mupping in p if there exists a neighborhood V of p(p) in S2such that q : U + V is a geodesic mapping.
a. Show that if q : S13 S, is both a geodesic and a conformal mapping, then q~ is a similarity; that is,
where 1 is constant.
+
b. Let S Z = {(x, y, Z) E R3; x2 f y2 z2 = 13 be the unit sphere, S= ((x, y, z) E S2;z < O) be its lower hemisphere, and P be the plane z = 1. Prove that the map (central projection) q : S 3 P which takes a point p E S to the intersection of P with the line that connects p to the center of S 2 is a geodesic mapping.
*c. Show that a surface of constant curvature admits a local geodesic mapping into the plane for every p E S.
13. (Beltrami's Theorem.) In Exercise 12, part c, it was shown that a surface S of constant curvature K admits a local geodesic mapping in the plane for every p E S. To prove the converse (Beltrami's theorem)If a regular connected surface S admits for every p E S a local geodesic mapping into the plane, then S has constant curvature, the following assertions should be proved :
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297
a. If v = u(u) is a geodesic, in a coordinate neighborhood of a surface parametrized by (u, v), which does not coincide with u = const., then

*b. If S admits a local geodesic mapping y : V R 2 of a neighborhood V of a point p E S into the plane R2, then it is possible to parametrize the neighborhood V by (u, u) in such a way that
*c. If there exists a geodesic mapping of a neighborhood V of p E S into a
piane, then the curvature K in V satisfies the relations
*d. If there exists a geodesic mapping of a neighborhood V of p plane, then the curvature K in V is constant.
E
S into a
e. Use the above, and a standard argument of connectedness, to prove Reltrami's theorem. 14. (The Holonomy Group.) Let S be a regular surface and p E S. For each piecewise regular parametrized curve a : [0, I] + S with a(0) = a(1) = p, let P,: Tp(S) TP(S)be the map which assigns to each v E TP(S)its parallel transport along a back top. By Prop. 1 of Sec. 44, P, is a linear isometry of Tp(S). If p :11, i] is another piecewise regular parametrized curve with B(Z) = B(i) = p, define the curve B a: [0, 1 f i] 4 S by running successively first a and then 8; that is, jl 0 a(s) = a(s) if s E [0, 11, and jl 0 a(s) = jl(s) if s E [I, i].

Q
a. Consider the set Hp(S) = {Pa: Tp(S) 3 T,(S); at1 a joining p top), ; where a is piecewise regular. Define in this set the operation Pfl0 P, = Pfl:, that is, Pfl Pa is the usual composition of performing first Pa and then Pfl. Prove that, with this operation, Hp(S) is a group (actually, a subgroup of the group of linear isometries of Tp(S)). Hp(S) is called the holonomy grorcp of S at p. b. Show that the holonomy group at any point of a surface with K = 0 reduces to the identity. 0
c. Prove that if S is connected, the holonomy groups Hp(S) and H,(S) at two
arbitrary points p, q E S are isomorphic. Thus, we can talk about the (abstract) holonomy group of a surface.
298
/ntrinsic Geometry of Surfaces
d. Prove that the holonomy group of a sphere is isomorphic to the group of
2 x 2 rotation matrices (cf. Exercise 22, Sec. 44).
47. Further Properties of G e o d e s i c s : Convex Neighborhoodst In this section we shall show how certain facts on geodesics (in particular, Prop. 5 of Sec. 44) follow from the general theorem of existence, uniqueness, and dependence on the initial condition of vector fields. The geodesics in a parametrization x(u, v) are given by the system
where the T& are functions of the local coordinates u and v. By setting u' = 5 and v' = q, we may write the above system in the general form
where F,(u, v, 5 , ~ = ) 5, F4@,v, 5, q) = q. It is convenient to use the following notation: (u, v, 5, q) will denote a point of R4 which will be thought of as the cartesian product R4 = R2 x R2; (u, v) will denote a point of the first factor and (5, q) a point of the second factor. The system (2) is equivalent to a vector field in an open set of R4 which is defined in a way entirely analogous to vector fields in R2(cf. Sec. 34). The theorem of existence and uniqueness of trajectories (Theorem 1, Sec. 34) still holds in this case (actually, the theorem hoIds for Rn;cf. S. Lang, Analyxis I, AddisonWesley, Reading, Mass., 1968, pp. 383386) and is stated as follows : Given the system (2) in an open set U c R4 and given apoint
there exists a unique trajectory a : (E, E) ,U of Eq. (2), with
?This section may be omitted on a first reading. Propositions 1 and 2 (the statements of which can be understood without reading the section) are, however, used in Chap. 5.
299
Further Properties of Geodesics
To apply this result to a regular surface S, we should observe that, given a parametrization x(u, v) in p E S, of coordinate neighborhood V, the set of pairs (q, v), q E V, E T,(S), may be identified to an open set V x R2 = U c R4. For that, we identify each Tq(S),q E V, with R2 by means of the basis {xu,x,). Whenever we speak about differentiability and continuity in the set of pairs (q, v) we mean the differentiability and continuity induced by this identification. Assuming the above theorem, the proof of Prop. 5 of Sec. 44 is trivial. Indeed, the equations of the geodesics in the parametrization x(u, v ) in p E S yield a system of the form (2) in U c R4. The fundamental theorem implies then that given a point q = (u,, v,) E V and a nonzero tangent vector V = (5,, q,) E Tq(S) there exists a unique parametrized geodesic

in V (where n(q, V) = q is the projection V x R2 V). The theorem of the dependence on the initial conditions for the vector field defined by Eq. (2) is also important. It is essentially the same as that for q,) E U , there exist a the vector fields of R2: Given a point p = (u,, v,, to, neighborhood V = V, x V2 of p (where V, is a neighborhood of (u,, v,) and V2 is a neighborhood of ( t o ,qo)), an open interval I, and a diflerentiable mapping a : I x V , x V2 ,U such that, fixed (u, v, (, q) = (q, v) E V, then a(t, q, v), t E I, is the trajectory of (2) passing through (q, v). To apply this statement to a regular surface S, we introduce a parametrization in p E S, with coordinate neighborhood V, and identify, as above, the set of pairs (q, v), q E V , v E Tq(S), with V x R2. Taking as the initial condition the pair (p, 0), we obtain an interval (e2, e2), a neighborhood V, c V of p in S, a neighborhood V2 of the origin in R2, and a differentiable map
such that if (q, v)
E
V, x V2, v # 0, the curve
is the geodesic of S satisfying y(O?q, v) = q, yl(O, q, v) = v, and if v = 0, this curve reduces to the point q. Here y = n o a , where n(q, v) = q is the projection U = V x R2 , V and m is the map given above. Back in the surface, the set V, x V2 is of the form
where V,(O) denotes a neighborhood of the origin in T,(S). Thus, if we restrict
300
intrinsic Geometry of Surfaces
to (€2, €2) x {p) x V2, we can choose { p ) x V2 = BE1c Tp(S), and obtain
THEOREM 1. Given p dzferentiable map
E
S there exist numbers t,
> 0, t, > 0
and a
+
such that for v E B,,, v 0, t E (E,, 6,) the curve t 4 y(t, v) is the geodesic of S with y(0, v) = p, yf(O, v) = v, and for v = 0, y(t, 0) = p.
This result was used in the proof of Prop. 1 of Sec. 46. The above theorem corresponds to the case wherep is fixed. To handle the general case, let us denote by B,(q) the domain bounded by a (small) geodesic circle of radius r and center q, and by &(q) the union of Br(q) with its boundary. Let E > 0 be such that B,(p) c V, . Let B,,,,(O) c Fq(0) be the largest open disk in the set Yq(0) formed by the union of Vq(0) with its limit points, and set t , = inf d(q), q E B,(p). Clearly, t , > 0. Thus, the set
is contained in V, x V2, and we obtain
THEOREM l a . Given p diferen t iable map
E
S, there exist positive numbers t , t , ,
E,
and a
where
such that y(t, q, 0)
= q,
and for v # 0 the curve
is the geodesic of S with y(0, q, v)
=
q, y'(0, q, v)
=
v.
Let us apply Theorem l a to obtain the following refinement of the existence of normal geodesics.
PROPOSITION 1. Given p E S there exist a neighborhood W of p in S and a number 6 > 0 such that for every q E W, exp, is a difeomorphism 011 B,(O) c T,(S) and exp, (B,(O)) 3 W ; that is, W is a normal neighborhood of all its points.
Further Properties of Geodesics
30 1
ProoJ Let V be a coordinate neighborhood of p. Let e, e,, e, and y: (e2, e,) X 7.l , V be as in Theorem la. By choosing e l < e,, we can make sure that, for (q, v) E 21, expq(v) = y(l v I, q, v) is well defined. Thus, we can define a differentiable map q : U , V x V by
We first show that dq is nonsinguIar at (p, 0). For that, we investigate how p transforms the curves in U given by
where w E T,(S) and a(t) is a curve in S with a(0) = p . Observe that the tangent vectors of these curves at t = 0 are (0, w) and (al(0), 01, respectively. Thus,
,
and dq , , takes linearly independent vectors into linearly independent vectors. Hence, d q ( , is nonsingular. It follows that we can apply the inverse function theorem, and conclude the existence of a neighborhood V of (p, 0) in U such that q maps V diffeomorphically onto a neighborhood of (p, p) in V x V. Let U c B,(p) and 6 > 0 be such that
,
Finally, let W c U be a neighborhood of p such that W x W c p('0). We claim that 6 and W thus obtained satisfy the statement of the theorem. In fact, since q is a diffeomorphism in '0, exp, is a diffeomorphism in B,(O), q E W. Furthermore, if q E W, then
and, by definition of q, exp,(B,(O))
3
W.
Q.E.D.
Remark 1. From the previous proposition, it follows that given two points q,, q, E W there exists a unique geodesic y of length less than 6 joining q, and q,. Furthermore, the proof also shows that y "depends differentiably" on q, and q, in the following sense: Given (q,, q,) E W x W, a unique v E T,,(S) is determined (precisely, the v given by p'(q,, q,) = (q,, v)) which depends differentiably on (q,, q,) and is such that yf(0) = v.
302
Intrinsic Geometry of Surfaces
One of the applications of the previous result consists of proving that a curve which locally minimizes arc length cannot be "broken." More precisely. we have PROPOSITION 2, Let a : I S be a parametrized, piecewise regular curve such that in each regular arc the parameter is proportional to the arc length. Suppose that the arc length between any two of its points is smaller than or equal to the arc length of any parametrized regular curve joining these points. Then a is a geodesic; in particular, a is regular everywhere. 3
ProoJ Let 0 = t o ( t , ( . . . ( t, (t,,, = 1 be a subdivision of [0, I ] = I in such a way that a 1 [ti,ti+1 ] , i = 0, . . . , k, is regular. By Prop. 5 of Sec. 46, a is geodesic at the points of (ti, ti+,). To prove that a is geodesic in ti, consider the neighborhood W, given by Prop. 1 , of a(?,). Let q1 = a(t,  f),qz = a(& E ) , e > 0, be two points of W , and let y be the radial geodesic of B,(q,) joining q , to q, (Fig. 443). By Prop. 4 of Sec. 46, extended to the piecewise regular curves, l(y) 5 l(a) between q , and q2. Together with the hypothesis of the proposition, this implies tha: l(y) = l(a). Thus, again by Prop. 4 of Sec. 46, the traces of y and a coincide. Therefore, a is geodesic in t , , which ends the proof. Q.E.D.
+
Figure 443
Figure 444
In Example 6 of Sec. 44 we have used the following fact: A geodesic y(t) of a surface of revolution cannot be asymptotic to apa&llel Po which is not itself a geodesic. As a further application of Prop. 1 , we shall sketch a proof of this fact (the details can be filled in as an exercise). Assume the contrary to the above statement, and let p be a point in the parallel Po. Let W and 6 be the neighborhood and the number given by Prop. 1, and let q E Po n W, q + p. Because y(t) is asymptotic to Po,the point p is a limit of points y(ti), where (ti]+ a,and the tangents of y at ti converge to the tangent of Po at p. By Remark 1 , the geodesic p(t) with length smaller than S joiningp to q must be tangent to Poat p. By Clairaut's relation
303
Further Properties of Geodesics
(cf. Example 5, Sec. 44), a small arc of p(t) around p will be in the region of W where y(t) lies. It follows that, sufficiently close to p, there is a pair of points in W joined by two geodesics of length smaller than 6 (see Fig. 444). This is a contradiction and proves our claim. One natural question about Prop. 1 is whether the geodesic of length less than 6 which joins two points q,, q, of W is contained in W. If this is the case for every pair of points in W, we say that W is convex. We say that a parametrized geodesic joining two points is minimal if its length is smaller than or equal to that of any other parametrized piecewise regular curve joining these two points. When W is convex, we have by Prop. 4 (see also Remark 3) of Sec. 46 that the geodesic y joining q, E W to 9, E W is minimal. Thus, in this case, we may say that any two points of Ware joined by a unique minimal geodesic in W. In general, however, W is not convex. We shall now prove that W can be so chosen that it becomes convex. The crucial point of the proof is the following proposition, which is interesting in its own right. As usual, we denote by B,(p) the interior of the region bounded by a geodesic circle S,(p) of radius r and center p.
PROPOSITION 3. For each point p E S there exists a positive number E with the following property: v a geodesic y(t) is tangent to the geodesic circle S,(p), r < f, at y(O), then, for t # 0 small, y(t) is outside B,(p) (Fig. 445).
Figure 445
Proof. Let W be the neighborhood of p given by Prop. 1. For each pair (q, v), q E W , v E T,(S), I v 1 = 1, consider the geodesic y(t, q, v) and se:. for a fixed pair (q, v) (Fig. 446),
Thus, for a fixed (q, v), F(t) is the square of the distance of the point
Intrinsic Geometry o f Swfeces
Figure 446
y(t, q, v) to p. Clearly, F(t, q, v) is differentiable. Observe that F(t, p, v)
=
I vt i2. Now denote by U1the set
and define a function Q: U1+R by
Since F is differentiable, Q is continuous. Furthermore, since
d2F  2(u(t), u "(t))
at2
+ 2(ut(t), ut(t)),
and at (p, v) u'(t)
= v,
u"(t)
= 0,
we obtain Q ( p , v)
= 2(v
l2
=2
>0
for all v E T'(S), ( v (
=
1.
It folIows, by continuity, that there exists a neighborhood V c W such t h a t Q ( q , v ) > O f o r a l I q E Vandv E T,(S)withIv[ = l . L e t ~ > O b e s u c h that B,(p) c V. We claim that this E satisfies the statement of the proposition. In fact, let r < E and let y(t, q, v) be a geodesic tangent to S,(p) at y(0) = q. By introducing geodesic polar coordinates around p, we see that (u(O), ~ ' ( 0 ) )= 0 (see Fig. 447). Thus, dF/dt(O) = 0. Since F(0, q, v) = r2, and (d2F/dt2)(0)> 0, we have that F(t) > r 2 f o r t # 0 small; hence, y(t) is outside B,(p). Q.E.D.
Further Properties o f Geodesics
Figure 447
We can now prove
PROPOSITION 4 (Existence of Convex Neighborhoods). For each point p E S there exists a number c > 0 such that B,(p) is convex; that is, any two points of Bc(p) can be joined by a unique n~inimalgeodesic in Bc(p). Proof. Let f be given as in Prop. 3. Choose 6 and Win Prop. 1 in such a way that S < €12. Choose c < 6 and such that B,(p) c W. We shall prove that B,(p) is convex. Let q,, q, E B,(p) and let y : I S be the geodesic with length less that 6 < €12 joining q , to q,. y(I) is clearly contained in B,(p), and we want to prove that y(I) is contained in B,(p). Assume the contrary. Then there is a point m E B,(p) where the maximum distance v of y(I) to p is attained (Fig. 448). In a neighborhood of m, the points of y(I) will be in B,(p). But this Q.E.D. contadicts Prop. 3.

EXERCISES "1. Let y and w be differentiablevector fields on an open set U c S. Let p E S and let a:1 + U be a curve such that a(0)= p , a'(0) = y. Denote by P,,,:
T,(,,(S) 4 T,(,,(S) the parallel transport along a from a(0)to a(?), t E I. Prove that
Intrinsic Geometry o f Surfaces
where the second member is the velocity vector of the curve P;,l,(w(&(t))) in T,(S) at t = 0. (Thus, the notion of covariant derivative can be derived from the notion of parallel transport.) 2. a. Show that the covariant derivative has the following properties. Let u, w, and y be differentiable vector fields in U c S, f: U + R be a differentiable function in S, y(f) be the derivative off in the direction of y (cf. Exercise 7, Sec. 34), and A, ,u be real numbers. Then
+
pD,(w). DY@V+ Pw) = AD,(u) 4 PD,(w); D,I,+,,(w) = AD,(w) 2. Dy(fv) = ~ ()v f fDy(v); DfY(v)= fDy(v). 3. Y((v, w)) = (Dyv, w) (v, Dyw). 4. DXp, = DxPxW, where x(u, v ) is a parametrization of S . *b. Show that property 3 is equivalent to the fact that the parallel transport along a given piecewise regular parametrized curve a: 1 S joining two points p, q E S is an isometry between T,(S) and T,(S). Show that property 4 is equivalent to the symmetry of the lower indices of the Christoffel symbols.
+
+

*c. Let U(U) be the space of (differentiable) vector fields in U
S and let D :U x U 3 CO (where we denote D(y,u) =Dy(v))be a map satisfying properties 14. Verify that Dy(v) coincides with the covariant derivative of the text. (In general, a D satisfying properties 1 and 2 is called a connection in U. The point of the exercise is to prove that on a surface with a given scalar product there exists a unique connection with the additional properties 3 and 4). c

*3. Let a: I = [O, 11 S be a simple, parametrized, reguIar curve. Consider a unit vector field v(t) along a, with (a'(t),v(t)) = 0 and a mapping x : R x I+S given
by X(S,t)
= eXpacr)(sv(t)),
s
E
R, t
E
I.
a. Show that x is differentiable in a neighborhood of I in R x I and that d x is nonsingular in (0, t), t
E
b. Show that there exists 6 Is1 < 6 .
I.
> 0 such that x is onetoone in the rectangle t
E
I,
(0, I), 1 s I < E , x is a parametrization of S, the coordinate neighborhood of which contains a((0, I)). The coordinates thus obtained are called geodesic coordinates (or Fermi's coordinates) of basis a. Show that in such a system F = 0, E = 1. Moreover, if a is a geodesic parametrized by the arc length, G(0, t) = 1 and G,(O, t ) = 0.
c. Show that in the open set t
E
d. Establish the following analogue of the Gauss lemma (Remark I after Prop. 3. Sec. 46). Let a: I + S be a regular parametrized curve and let y,(s), t E I, be a family of geodesics parametrized by arc length s and given by ; y,(O) = a(rr. {y:(O), a'(t)] is a positive orthogonal basis. Then, for a fixed i , sufficientl? small, the curve t + y,(i), t E I, intersects all y, orthogonally (such curve are called geodesic parallels).
Further Properties o f Geodesics
4. The energy E of a curve a : [a, b]

S is defined by
*a. Show that (I(a))z ( (b  a)E(a) and that equality holds if and only if t is proportional to the arc length. b. Conclude from part a that if y : [a, b] S is a minimal geodesic with y(a) = p, y(b) = q, then for any curve a : [a, b] S, joining p to q, we have E(y) I E(a) and equality holds if and only if a is a minimal geodesic.

5. Let y : [O,1] + S be a simple geodesic, parametrized by arc length, and denote by u and v the Fermi coordinates in a neighborhood of y([O, I]) which is given as u = 0 (cf. Exercise 3). Let u = y(v, t) be a family of curves depending on a parameter t ,  6 < t < F, such that y is differentiable and
Such a family is called a variation of y keeping the end points p and q fixed. Let E(t) be the energy of the curve y(v, t) (cf. Exercise 4); that is,
*a. Show that = 0,
E'(0)
,,
K = K(v) is the Gaussian curvature along y, and ' where q(v) = dyldt I, = denotes the derivative with respect to t (the above formulas are called rlze first and second variations, respectively, of the energy of 7; a more complete treatment of these formulas, including the case where y is not simple, will be given in Sec. 54). b. Conclude from part a that if K < 0, then any simple geodesic y : [O, I ] S is minimal relatively to the curves sufficiently close to and joining y(0) to y(1).

+
6. Let S be the cone z = kJx2 y2, k > 0, (x, y) # (0, 0), and let V c R2 be the open set of R2 given in polar coordinates by 0 < p < a,0 < 8 < 2nn sin P, where cotan p = k and n is the largest integer such that 2nn sin P < 2n (cf. Example 3, Sec. 42). Let q : V 4 S be the map
P(P, 8)
=
( p sin P cos
(),8
a. Prove that 9 is a local isometry.
p sin
P sin (Kb),8
p cos 8)
Intrinsic Geometry of Surfaces
308
"b. Let q E S. Assume that
p < n/6 and
let k be the largest integer such that 2nk sin p < n. Prove that there exist at least k geodesics that leaving from q return to q. Show that these geodesics are broken at q and that, therefore, none of them is a closed geodesic (Fig. 449). (P,E+ 2~ sin p ) = r ,
Figure 449
*c. Under the conditions of part b, prove that there are exactly k such geodesics. 7. Let a : I + R3 be a parametrized regular curve. For each t E I, let P(r) c R3 be a plane through a ( t ) which contains a'(t). When the unit normal vector N ( t ) of P(t) is a differentiable function of t and N'(t) # 0 , t E I, we say that the map t + (a(r),N ( t ) } is a diflerentiable family of tangent planes. Given such a family, we determine a parametrized surface (cf. Def. 2, Sec. 23) by
The parametrized surface x is called the envelope of the family { a ( t ) ,N ( t ) } (cf. Example 4, Sec. 35).
a. Let S be an oriented surface and let y : I + S be a geodesic parametrized by arc length with k(s) # 0 and z(s) st 0, s E I. Let N(s) be the unit normal vector of S along y. Prove that the enveIope of the family of tangent planes {~(s),N ( s ) )is regular in a neighborhood of y, has Gaussian curvature K 0, and is tangent to S along 7. (Thus, we have obtained a surface locally isometric to the plane which contains y as a geodesic.) b. Let a : I R3 be a curve parametrized by arc length with k(s) # 0 and z(s) r;t 0 , s E I, and let {a(s),n(s))be the family of its rectifying planes. Prove that the envelope of this family is regular in a neighborhood of a, has Gaussian curvature K 0, and contains a as a geodesic. (Thus, every curve is a geodesic in the envelope of its rectflying planes; since this envelope is locally isometric to the plane, this justifies the name rectifying plane.)


Appendix
Proofs of the
Fundamental Theorems of the Local Theory of Curves and Surfaces
In this appendix we shall show how the fundamental theorems of existence and uniqueness of curves and surfaces (Secs. 15 and 42) may be obtained from theorems on differential equations.
Proof of the Fundamental Theorem ofthe Local Theory of Curves (cf. statement in Sec. 15). The starting point is to observe that Frenet's equations dt ds
 = kn, kt dn 
ds

rb,
may be considered as a differential system in I x R9,
t, (549 5.5, 56) = n, (57, 5 8 7 59) = b, andJI.9 i = 1' . . . , 9, where (tl,t 2 Y 5 3 ) are linear functions (with coefficients that depend on s) of the coordinates ti.
intrinsic Geometry of Surfaces
370
In general, a differential system of type (la) cannot be associated to a "steady" vector field (as in Sec. 34). At any rate, a theorem of existence and uniqueness holds in the following form: Given initial conditions so E I, (5 . . . , ({g)o, there exist an open interval J z I containing so and a unique dzflerentiable mapping a : J .R9, with
where each fi, i = 1, . . . , 9, is calczdated in (s, a(s)) E J x R9. Furthermore, if the system is linear, J = I (cf. S. Lang, Analysis I, AddisonWesIey, Reading, Mass., 1968, pp. 383386). It follows that given an orthonormal, positively oriented trihedron (to,no, b,) in R3 and a value so E I, there exists a family of trihedrons {t(s), n(s), b(s)], s E J, with t(so) = to, n(so) = no, b(so) = b,. We shall first show that the family (t(s), n(s), b(s)) thus obtained remains orthonormal for every s E I. In fact, by using the system (1) to express the derivatives relative to s of the six quantities
( t , n),
(t, b),
(n, b),
(t, t),
(n, n),
(b, b)
as functions of these same quantities, we obtain the system of differential equations
d
(t, ds
b)
= k(n,
b)
+ z(t, n),
d  ( t , t) = 2k(t, n), ds d (n, n) = 2k(n, t)  2z(n, b), ds
It is easily checked that (t, n)
= 0,
(b, b) z 0,

(n, b) = 0, t2 1, n2 z 1, b2 = 1,
is a solution of the above system with initial conditions O,0,0, 1, 1, 1. B!
uniqueness, the family {t(s), n(s), b(s)] is orthonormal for every s claimed.
E
I, as we
Appendix :Fundamental Theorems of Curves and Surfaces
31 1
From the family (t(s), n(s), b(s)) it is possible to obtain a curve by setting a($) =
t(s) ds,
s
I,
E
where by the integral of a vector we understand the vector function obtained by integrating each component. It is clear that al(s) = t(s) and that a"(s) = kn. Therefore, k(s) is the curvature of a at s. Moreover, since a"'(s)
=
k'n
+ kn' = kln  k2t

kzb,
the torsion of a will be given by (cf. Exercise 3, Sec. 15)  (a1 A a", a"I)
k

 (t A kn, (k2t
k
+ k'n

kzb))

z;
a is, therefore, the required curve. We still have to show that a is unique up to translations and rotations of R3. Let a : I R3 be another curve with i ( s ) = k(s) and ?(s) = z(s), s E I , and let {i,, ri,, 6,) be the Frenet trihedron of a at so. It is clear that by a translation A and a rotation p it is possible to make the trihedron {i,,,A,,

coincide with the trihedron (to, no, b,) (both trihedrons are positive). By applying the uniqueness part of the above theorem on differential equations, Q.E.D. we obtain the desired result. Proofof the Fundamental Theorem ofthe Local Theory ofsurfaces (cf. statement in Sec. 43). The idea of the proof is the same as the one above; that is, we search for a family of trihedrons (xu,x,, N ) , depending on u and v, which satisfies the system xu, = T L x u xu, = r t xu, = r;2x, Nu Nu
a,1x, = a , 2% =
+ rTlx, + eN, + T:~X, tYN + + gN,
=
x,,,
E2x,
+ a2,x,, + a,2x,,
where the coefficients I?,: ajj, i, j = 1, 2, are obtained from E, F, G, e,f, g as if it were on a surface. The above equations define a system of partial differential equations in V x R9, (51), =f,(u, v, 51, . . 591,
Intrinsic Geometry of Surfaces
372
where 5 = (t1, 52, 53) = Xu, q = (547 5 5 7 t 6 > = Xv7 = (57, 58, 59) = N, andJ = 1, . . . , 15, are linear functions of the coordinates t,, j = 1, . . . , 9, with coefficients that depend on td and v. In contrast to what happens with ordinary differential equations, a system of type (2a) is not integrable, in general. For the case in question, the conditions which guarantee the existence and uniqueness of a local solution, for given initial conditions, are
A proof of this assertion is found in J. Stoker, Diflerential Geometry, WiIeyInterscience, New Y ork, 1969, Appendix B. As we have seen in Sec. 43, the conditions of integrability are equivalent to the equations of Gauss and MainardiCodazzi, which are, by hypothesis, satisfied. Therefore, the system (2a) is integrable. Let {t,q, be a solution of (2a) defined in a neighborhood of (u,, v,) E I/, with the initial conditions &u,, v,) = 5,, ~ ( u , v,) , = go, ((u,, vo) = Clearly, it is possible to choose the initial conditions in such a way that
c,.
5;
= E(lk0, vO),
II; = G(u0, v,),
ro) = F(u0, vo),
(to,
Pi = 1, (50,
c o > = (go,
(0)
=
0.
With the given solution we form a new system,
which is clearly integrable, since 5, = g,,. Let x :  R3 be a solution oi (4), defined in a neighborhood P of (u,, v,), with x(u0, vO)= po E R3. We shall show that by contracting and interchanging v and u, if necessar) x(v) is the required surface. We shall first show that the family (5, q , 51, which is a solution of (2a). has the following property. For every (u, v) where the solution is defined, R e have E,
0 ; that is, the Gaussian curvature in p is positive. 2. p is simuEtaneously apoint of local maximum for the function k , an a point of local minimum for the function k, ( k , > k,).
Then p is an umbilicalpoint of S. Proof. Let us assume that p is not an umbilical point and obtain a contradiction. If p is not an umbilical point of S, it is possible t o parametrize a neighborhood of p by coordinates (u, v) so that the coordinate lines are lines of curvature (Sec. 34). In this situation, F =f = 0, and the principal curvatures are given by e/E, g/G. Since the point p is not umbilical, we may assume, by interchanging u and v if necessary, that in a neighborhood of p
The Rigidity o f the Sphere
319
In the coordinate system thus obtained, the MainardiCodazzi equations are written as (Sec. 43, Eqs. (7) and (7a))
By differentiating the first equation of (1) with respect to v and using Eq. (2), we obtain E(k,), = %( k, 2
+ k,).
Similarly, by differentiating the second equation of (I) with respect to using Eq. (3),
11
and
On the other hand, when F = 0, the Gauss formula for K reduces to (Sec. 43, Exercise 1)
hence, 2KEG
= E,,
+ G,,
$
ME,
+ NG,,
(6)
where M = M(u, v) and N = N(u, v) are functions of (u, v), the expressions of which are immaterial for the proof. The same remark applies to @, fl, M ,and p,to be introduced below. From Eqs. (4) and (5) we obtain expressions for E, and G, which, after being differentiated and introduced in Eq. (6), yield
hence,
Since K > 0 and k , > k 2 at p, the first member of Eq. (7) is strictly nega
320
Global Differential Geometry
tive at p. Since k , reaches a local maximum at p and k , reaches a local minimum at p, we have
at p. However, this implies that the second member of Eq. (7) is positive or zero, which is a contradiction. This concludes the proof of Lemma 1. Q.E.D.
It should be observed that no contradiction arises in the proof if we assume that k , has a local minimum and k , has a local maximum a t p. Actually, such a situation may happen on a surface of positive curvature without p being an umbilical point, as shown in the following example.
Example 1. Let S be a surface of revolution given by (cf. Sec. 33, Example 4)
x = p(v) cos u,
y = p(v) sin u,
z = y(v),
0
< u < 2n,
where p(v)
=
C cos v,
y(v)
=
S
Jl

CZsin2 v dv,
y(0)
= 0.
We take I v 1 < sin'(l/C), so that y(v) is defined. By using expressions already 'known (Sec. 33, Example 4), we obtain E = C2 cos2 v, F = 0, G = 1, e = C cos v(J1
g= 

C 2 sin2 v),
C cos v J1  C2 sin2 v '
hence,
k
  e= 
IE
J1
C2 sin2 v C cos v 
k  g  
C

Ccosv J1  C 2 sin2v
Therefore, S has curvature K = k , k , = 1 > 0, positive and constant (cf. Exercise 7, Sec. 33). It is easily seen that k , > k, everywhere in S, since C > 1. Therefore, S has no umbilical points. Furthermore, since k , = (l/C) for v = 0, and
The Rigidity of the Sphere
k,
=
J1
 C2 sin2 v
C cos v
1 > C
for v # 0,
we conclude that k , reaches a minimum (and therefore k , reaches a maximum, since K = 1) at the points of the parallel v = 0. Incidentally, this example shows that the assumption of compactness in Theorem 1 is essential, since the surface S (see Fig. 52) has constant positive curvature but is not a sphere.
Figure 52
In the proof of Theorem I we shall use the following fact, which we establish as a lemma.
L E M M A 2. A regular compact surface S c R3 has at least one elliptic point.
Proof. Since S is compact, S is bounded. Therefore, there are spheres of R3, centered in a fixed point 0 E R3, such that $ is contained in the interior of the region bounded by any of them. Consider the set of all such spheres. Let r be the infimum of their radii and let C c R3 be a sphere of radius r centered in 0. It is clear that C and S have at least one common point, say p. The tangent plane to Z at p has only the common point p with S, in a neighborhood ofp. Therefore, C and S are tangent at p . By observing the normal sections at p, it is easy to conclude that any normal curvature of S at p is greater than or equal to the corresponding curvature of C at p. Therefore, K,(p) 2 K,(p) > 0, and p is an elliptic point, as we wished. Q.E.D. Proof of Theorem 1. Since S is compact, there is an elliptic point, by Lemma 2. Because K is constant, K > 0 in S. By compactness, the continuous function k , on S reaches a maximum at a point p E S (appendix to Chap. 5, Prop. 13). Since K = k,k2 is a positive
Global Differential G e o m efr y
322
constant, k, is a decreasing function of k,, and, therefore, it reaches a minimum at p. It folIows from Lemma 1 that p is an umbilical point; that is, ~ I ( P )= k,(p). Now let q be any given point of S. Since we assumed k,(q) > k,(q), we have that
Therefore, k,(q) = k,(q) for every q E S. It follows that a11 the points of S are umbilical points and, by Prop. 5 of Sec. 32, S is contained in a sphere or a plane. Since K > 0 , S is contained in a sphere C. By compactness, S is closed in C, and since S is a regular surface, S is open in C. Since C is connected and S is open and closed in C, S = C (appendix to Chap. 5, Prop. 5). Q.E.D. Therefore, the surface S is a sphere. Observe that in the proof of Theorem 1 the assumption that K = k,k, is constant is used only to guarantee that k, is a decreasing function of k,. The same conclusion follows if we assume that the mean curvature N = 4(k1 k,) is constant. This allows us to state
+
THEOREM la. Let S be a regular, compact, and connected surface with Gaussian curvature K > 0 and mean curvature H constant. Then S is a sphere.
The proof is entirely analogous to that of Theorem 1. Actually, the argument applies whenever k, = f(k,), where f is a decreasing function of k,. More precisely, we have THEOREM lb. Let S be a regular, compact, and connected surface of positive Gaussian curvature. If there exists a relation k , = f(k,) in S, where f is a decreasing function of k,, k, 2 k,, then S is a sphere. Remark 2. The compact, connected surfaces in R3 for which the Gaussian 0 are called ovaloids. Therefore Theorem l a may be stated as curvature K follows : An ovaloid of constant mean curvature is a sphere. On the other hand, it is a simple consequence of the GaussBonnet theorem that a n ovaloid is homeomorphic t o a sphere (cf. Sec. 45, application 1). H. Hopf proved that Theorem l a still holds with the following (stronger) statement: A regular surface of constant mean curvature that is homemorphic to a sphere is a sphere. A theorem due to A. AIexandroff extends this result further by replacing the condition of being homeomorphic to a sphere by compactness: A regular, compact, and connected surface of constant mean curvature is a sphere. A11 exposition of the abovementioned results can be found in Hopf [I I]. (References are listed at the end of the book.)
The Rigidity o f the Sphere
323
Remark 3. The rigidity of the sphere may be obtained as a consequence
of a general theorem of rigidity on ovaloids. This theorem, due to CohnVossen, states the following: Two isometric ovaloids difler by an orthogonal linear transformation of R3.A proof.of this result may be found in Chern [lo]. Theorem 1 is a typical result of global differential geometry, that is, information on local entities (in this case, the curvature) together with weak global hypotheses (in this case, compactness and connectedness) imply strong restrictions on the entire surface (in this case, being a sphere). Observe that the only effect of the connectedness is to prevent the occurrence of two or more spheres in the concIusion of Theorem 1. On the other hand, the hypothesis of compactness is essential in several ways, one of its functions being to ensure that we obtain an entire sphere and not a surface contained in a sphere.
EXERCISES 1. Let S c R3 be a compact regular surface and fix a point p, E R 3 ,po 6 S. Let d : S + R be the differentiable function defined by d(q) = I q  po 12,q E S. Since S is compact, there exists qo E S such that d(q,) 2 d ( q ) for all q E S. Prove that qo is an elliptic point of S (this gives another proof of Lemma I ) .
4
2. Let S c R3 be a regular surface with Gaussian curvature K > 0 and without umbilical points. Prove that there exists no point on S where H i s a maximum and K is a minimum. 3. (KazdanWarner's Remark.) Let S c
be an extended compact surface of revolution (cf. Remark 4, Sec. 23) obtained by rotating the curve R3
parametrized by arc length s E [O, I ] , about the z axis. Here q(O) = q(I) = 0 and q ( s ) > 0 for all s E [0, I ] . The regularity of S at the poles implies further that ~ ' ( 0=) 1, p'(I) =  1 (cf. Exercise 10, Sec. 23). We also know that the ) Example 4, Sec. 33). Gaussian curvature of S is given by K =  ~ " ( s ) / q ( s (cf. *a. Prove that
b. Conclude from part a that there exists no compact (extended)surface of revolution in R3 with monotonic increasing curvature.
The following exercise outlines a proof of Hopf's theorem: A regular surface with constant mean curvature which is homeomorphic to a sphere is a sphere (cf. Remark 2). Hopf's main idea has been used over and over again in recent work. The exercise requires some elementary facts on functions of complex variables.
324
Global Differential Geometry
4. Let U c R3 be an open connected subset of R2 and let x: U + S be an isothermal parametrization (i.e., E = G, F = 0; cf. Sec. 42) of a regular surface S. We identify R2 with the complex plane by setting u iv = c, (u, v) E R2, 5 E a=, is called the complex parameter corresponding to x. Let $: x(U) a= be the complexvalued function given by
a=
c
+
4
where e , f , g are the coefficients of the second fundamental form of S. a. Show that the MainardiCodazzi equations (cf. Sec. 43) can be written, in the isothermal parametrization x, as
and conclude that the mean curvature H of x(U) c S is constant if and only if $ is an analytic function of (i.e., = ($2)U, = ($2)U).
c
b. Define the "complex derivative"
and prove that $(c) = 2(xr7 Nc), where by 4, for instance, we mean the vector with complex coordinates
a=
+
a=
c. Let f: U c 4 V c be a onetoone complex function given by f(u iv) =x iy = q. Show that (x, y) are isothermal parameters on S (i.e., q is a complex parameter on S ) if and only iff is analytic and f '(c) # 0, E U. Let ) y = x f  l be the corresponding parametrization and define ~ ( q = 2(y,, N,). Show that on x(U) n y(V),
+
c
0
d. Let S2 be the unit sphere of R3. Use the stereographic projection (cf. Exercise 16, Sec. 22) from the poles N = (0,0, 1) and S = (0,0, 1) to cover S2by the coordinate neighborhoods of two (isothermal) complex parameters, 5 and q, with c(S) = 0 and q(N) = 0, in such a way that in the intersection Wof these coordinate neighborhoods (the sphere minus the two poles) q = 5l. Assume that there exists on each coordinate neighborhood anaIytic functions ~ ( c ) ~, ( q such ) that (*) holds in W. Use Liouville's theorem to prove that ~ ( 5 ) 0 (hence, ~ ( q = ) 0).

325
Complete Surfaces
e. Let S c R3 be a regular surface with constant mean curvature homeomorphic to a sphere. Assume that there exists a conformal diffeomorphism p: S + S 2 of S onto the unit sphere S 2 (this is a consequence of the uniformization theo
rem for Riemann surfaces and will be assumed here). Let 5" and if be the complex parameters corresponding under to the parameters 5 and q of S 2 given in part d. By part a, the function $([) = ((e  g)/2)  if is analytic. The simiIar function y(q) is also analytic, and by part c they are related by (*). Use part d to show that d(b = 0 (hence, y(if) = 0). Conclude that S is made up of umbilical points and hence is a sphere. This proves Hopf's theorem.
53. Complete Surfaces. Theorem of Hop fRinow All the surfaces to be considered from now on will be regular and connected, except when otherwise stated. The considerations at the end of Sec. 51 have shown that in order to obtain global theorems we require, besides the connectedness, some global hypothesis to ensure that the surface cannot be "extended" further as a regular surface. It is clear that the compactness serves this purpose. However, it would be useful to have a hypothesis weaker than compactness which could still have the same effect. That would allow us t o expect global theorems in a more general situation than that of compactness. A more precise formulation of the concept that a surface cannot be extended is given in the following definition.
DEFINITION 1. A regular (connected) surface S is said to be extendable if there exists a regular (connected) surface S such that S c as a proper subset. If there exists no such S, S said to be nonextendable. Unfortunately, the class of nonextendable surfaces is much too large t o allow interesting results. A more adequate hypothesis is given by
DEFINITION 2, A regular surface S is said to be complete when for every point p E S, any parametrized geodesic y : [0, 6) S of S, starting from p = y(O), may be extended into a parametrized geodesic 7: R + S, defined on the entire line R. In other words, S is complete when for every p E S the mapping exp,: T,(S) + S (Sec. 46) is de@ned for every v E T,(S). 4
We shall prove later (Prop. 1) that every complete surface is nonextendable and that there exist nonextendable surfaces which are not complete (Example 1). Therefore, the hypothesis of completeness is stronger than that
326
Global Differential Geometry
of nonextendability. Furthermore, we shall prove (Prop. 5) that every closed surface in R3 is complete; that is, the hypothesis of completeness is weaker than that of compactness. The object of this section is to prove that given two points p, q E S of a complete surface S there exists a geodesic joining p to q which is minimal (that is, its length is smaller than or equal to that of any other curve joining p to q). This fundamental result was first proved by Hopf and Rinow (H. Hopf, W. Rinow, " ~ b e den r Begriff der vollstandigen differentialgeometrischen Flachen," Comm. Math. Helv. 3 (1931), 209225). This theorem is the main reason the complete surfaces are more adequate for differential geometry than the nonextendable ones. Let us now look at some examples. The plane is clearly a complete surface. The cone minus the vertex is a noncomplete surface, since by extending a generator (which is a geodesic) sufficiently we reach the vertex, which does not belong to the surface. A sphere is a complete surface, since its parametrized geodesics (the traces of which are the great circles of the sphere) may be defined for every real value. The cylinder is also a complete surface since its geodesics are circles, lines, and helices, which are defined for all real values. On the other hand, a surface S  ( p } obtained by removing a point p from a complete surface S is not complete. In fact, a geodesic y of S should pass through p. By taking a point q, nearby p on y (Fig. 53), there exists a parametrized geodesic of S  ( p ) that starts from q and cannot be extended throughp (this argument will be given in detail in Prop. 1). Thus, a sphere minus a point and a cylinder minus a point are not complete surfaces.
PROPOSITION 1. A complete surface S is nonextendable. Proof. Let us assume that S is extendable and obtain a contradiction. To say that S is extendable means that there exists a regular (connected) surface 3 with S c 3. Since S is a regular surface, S is open in 5. The boundary (appendix to Chap. 5, Def. 4) Bd S of S in ,!? is nonempty; otherwise S = S u (S  S ) would be the union of two disjoint open sets S and ,!?  S, which contradicts the connectedness of (appendix to Chap. 5, Def. 10).
Complete Surfaces
327
+
Therefore, there exists a point p E Bd S, and since S is open in s , p S. Let P c S be a neighborhood of p in S such that every q t may be joined to p by a unique geodesic of S (Sec. 46, Prop. 2). Since p E Ed S, some q, E belongs to S. Let 7: [0, 11  be a geodesic of 3, with g(0) = p and p(1) = go. It is clear that a: [0, E ) S, given by a(t) = y(l  t), is a geodesic of S, with a(0) = q,, the extension of which to the line R would pass through p for t = 1 (Fig. 54). Since p S, this geodesic cannot be extended, which contradicts the hypothesis of completeness and concludes Q.E.D. the proof.

+
Figure 54
Figure 55
The converse of Prop. 1 is false, as shown in the following example.
Example 1, When we remove the vertex go from the onesheeted cone given by
we obtain a regular surface S. S is not complete since the generators cannot be extended for every value of the arc length without reaching the vertex. Let us show that ' S is nonextendable by assuming that S c 3, where f S is a regular surface, and by obtaining a contradiction. The argument consists of showing that the boundary of S in 3 reduces t o the vertex p, and  {pol c S. But that there exists a neighborhood p of p, in 3 such that this contradicts the fact that the cone (vertex po included) is not a regular surface in p, (Sec. 22, Example 5). First, we observe that the only geodesic of S, starting from a point p E S. that cannot be extended for every value of the parameter is the meridian (generator) that passes through p (see Fig. 55). This fact may easily be seen by using, for example, Clairaut's relation (Sec. 44, Example 5) and will be left as an exercise (Exercise 2).
328
Global Differential Geometry
Now let p E Bd S, where Bd S denotes the boundary of S in (as we have seen in Prop. 1, Bd S + 4). Since S is an open set in 3, p $ S. Let P be a neighborhood of p in 3 such that every point of may be joined t o p by Since p E Bd S, there exists q E V n S. Let a unique geodesic of S in 7 be a geodesic of joining p to q. Because S is an open set in 3, f agrees with a geodesic y of S in a neighborhood of q. Let p , be the first point of 7 that does not belong to S. By the initial observation, 7 is a meridian and p , is the vertex of S. Furthermore, p, = p ; otherwise there would exist a neighborhood of p that does not contain p,. By repeating the argument for that neighborhood, we obtain a vertex different fromp,, which is a contradiction. It follows that Bd S reduces to the vertex p,. Now let f i be a neighborhood of p , in 3 such that any two points of @ may be joined by a geodesic of S (Sec. 47, Prop. 1). We shall prove that  ( p , ) c S. In fact, the points of y belong to S. On the other hand, a point r E @which does not belong to y or to its extension may be joined to a point t of y, t # p,, t E @, by a geodesic a , different from y (see Fig. 541. By the initial observation, every point of a , in particular r, belongs to S. Finally, the points of the extension of y, except p,, also belong to S; otherwise, they would belong to the boundary of S which we have proved to be made up only of p , .
v.
Figure 56
In this way, our assertions are completely proved. Thus, S is nonextendable and the desired example is obtained. For what follows, it is convenient to introduce a notion of distance between two points of S which depends only on the intrinsic geometry of S and not on the way S is immersed in R 3 (cf. Remark 1, Sec. 42). Observe s that, since S c R3,it is possible to define a distance between two ~ o i n t of S as the distance between these two points in R3.However, this distance
Compfete Surfaces
329
depends on the second fundamental form, and, thus, it is not adequate for the purposes of this chapter. We need some preliminaries. A continuous mapping a : [a, b3 + S of a closed interval [a, b] c R of the line R onto the surface S is said to be a parametrized, piecewise dzferentiable curve joining a(a) to a(b) if there exists a partition of [a, b] by points a = to < t, < t, <   < t, < t,,, = b such that a is differentiable in [ti, ti+l],i = 0, . . . , k. The length /(a) of a is defined as
PROPOSITION 2. Given two points p, q E S of a regular (connected) surface S, there exists a parametried piecewise diferentiable curve joining p to q. Proof: Since S is connected, there exists a continuous curve a: [a, b] + S with a(a) = p, a(b) = q. Let t E [a, b] and let I, be an open interval in [a, b], containing t, such that a(I,) is contained in a coordinate neighborhood of a(t). The union u I,, t E [a, b], covers [a, b] and, by compactness, a finite number I,, . . . , I, still covers [a, b]. Therefore, it is possibIe to decompose I by points a = to < t, < . . < t, < t,,, = b in such a way that [ti, ti+1] is contained in some I j , j = 1, . . . , n. Thus, a(ti, t,,,) is contained in a coordinate neighborhood. Since p = a(to) and a(t,) lie in a same coordinate neighborhood x(U) c S, it is possible to join them by a differentiable curve, namely, the image by x of a differentiable curve in U c R2 joining xl(a(t,)) to x'(a(t,)). By this process, we join a(t,) to a(t,+,), i = 0, . . . , k, by a differentiable curve. This gives a piecewise differentiable curve, joining p = a(to) and q = a(t,,,), Q.E.D. and concludes the proof of the proposition. Now let p, q E S be two points of a regular surface S. We denote by ap,4 a parametrized, piecewise differentiable curve joining p to q and by l ( a , $ its length. ~ r o ~ o s i t i o2 nshows that the set of all such a , , is not empty. Thus, we can set the following:
DEFINITION 3. The (intrinsic) distance d(p, q) from the point p to the point q E S is the number
E
S
where the inf is taken over all piecewise diferentiable curves joining p to q .
PROPOSITION 3. The distance d dejned above has the following properties,
Global Differential Geometry
1 d(p, q)
= d(q,
+
p),
>
2 d(p, q) d(q, r) d(p, r), 3. d(p, q) 2 0, 4. d(p, q) = 0 if and only if p
= q,
where p, q, r are arbitrary points of S. Proof. Property 1 is immediate, since each parametrized curve a : [a, b] S, with a(a) = p, ajb) = q, gives rise to a parametrized curve&:[a, b] 4 S, defined by &(t) = a(a  t b). It is clear that &(a) = q, &(b) = p, and E(ap,q ) = l(Ep,q ) , Property 2 follows from the fact that when A and B are sets of real numbers and A E B then inf A )_ inf B. Property 3 follows from the fact that the infimum of positive numbers is positive or zero. Let us now prove property 4. Let p = q. Then, by taking the constant curve a : [a, b] S, given by a(t) = p, t E [a, b], we get l(a) = 0; hence, d(p, q) = 0. To prove that d(p, q) = 0 implies that p = q we proceed as follows. Let us assume that d(p, q) = inf l(a,,) = 0 and p # q. Let V be a neighborhood of p in S, with q y4 V, and such that every point of V may be joined to p by a unique geodesic in V. Let B,(p) c V be the region bounded by a geodesic circle of radius r, centered inp, and contained in V. By definition of infimum, given E > 0, 0 < E < r, there exists a parametrized, piecewise differentiable curve a : [a, b] + S joining p to q and with [(a) < E. Since a([a, b]) is connected and q $ B,, there exists a point to E [a, b] such that a(t,) belongs to the boundary of B,(p). It follows that E(a) > r > E, which is a contradiction. Therefore, p = q, and this concludes the proof of the proposition. Q.E.D. +
+
+
COROLLARY. I d(p, r)  d(r, q ) ] < d(p, q). It suffices to observe that d(p, r) d(r, 4) hence,
L d(p, q) 3 4 % r), L d(r, P) d(p, q);
+
d(p, 9) 2 d(p, r)  d(r, q) L d ( p , q).
PROPOSITION 4. If we let p, E S be a point of S, then the function f: S 4 R given by f(p) = d(p,, p), p E S, is continuous on S. Proof. We have to show that for each p E S, given 6 > 0, there exists 6 > 0 such that if q E Ba(p) n S, where B,(p) c R3 is an open ball of R3 centered at p and of radius 6, then I f (p)  f (q) ] = [ d(p,, p)  d(p,, q) I < E.
Complete Surfaces
33 1
Let E' < E be such that the exponential map exp, = T,(S) + S is a diffeomorphism in the disk B,/(O) c T,(S), where 0 is the origin of T,(S), and set exp(B,,(O)) = V. Clearly, V is an open set in S ; hence, there exists an open ball B,(p) in R3 such that B,(p) n S c V. Thus, if q E B,(p) n S,
Q.E.D.
which completes the proof.
Remark I. The readers with an elementary knowledge of topology will notice that Prop. 3 shows that the function d: S x S, R gives Sthe structure of a metric space. On the other hand, as a subset of a metric space, S c R3 has an induced metric 2 It is an important fact that these two metrics determine the same topology, that is, the same family of open sets in S. This follows from the fact that exp,: U c T,(S) 4 S is a local diffeomorphism, and its proof is analogous to that of Prop. 4. Having finished the preliminaries, we may now make the following observation.
PROPOSITION 5. A closed surface S c R3 is complete. S be a parametrized geodesic of S, y(0) = p E S, Proof. Let y : [0, E) which we may assume, without Ioss of generality, to be parametrized by arc S, length. We need to show that it is possible to extend y to a geodesic 7 : R defined on the entire line R. Observe first that when ?(so), so E R, is defined, then, by the theorem of existence and uniqueness of geodesics (Sec. 44, Prop. 5), it is possible to extend 7 to a neighborhood of soin R. Therefore, the set of all s E R where 7 is defined is open in R.If we can prove that this set is closed in R (which is connected), it will be possible to define 7 for all of R, and the proof will be completed. Let us assume that.? is defined for s < soand let us show that 7 is defined for s = so. Consider a sequence (s,) + so,with s, < so,n = 1, 2, . . . . We shall first prove that the sequence {T(s,)) converges in S. In fact, given E > 0, there exists no such that if n, m > no, then I s,  s, I < E. Denote by 2 the distance in R3, and observe that if p, q E S, then a(p, q) < d(p, q ) . Thus, 4
7
,7
s7
7
2ISn  Sm I
(O)be a disk in the tangent plane Ty(,,,(S), 0 of this tangent plane and contained in a neighborhood U',where exp,,,, is a diffeormophism. Let B,/(y(s0)) = exp,,,,,Bsl(0) and C' = Bd(&(y(s,)). If x' E E', the continuous function d(x', q) reaches a minimum at xb E C' (see Fig. 510). Then, as previously,
+
=
6'
+ d(xb, q).
Since Eq. (1) holds in so, we have that d(y(so),q)
Furthermore, since
we obtain from Eq. (2)
Figure 510
= r  so. Therefore,
Complete Surfaces
335
Observe now that the curve that goes from p to y(s,) through y and from y(so) to xi through a geodesic radius of B,,(y(s,)) has length exactly equal to so 6'. Since d(p, x',,)2 so 6', this curve, which joins p to xb, has minimal length. It follows (Sec. 46, Prop. 2) that it is a geodesic, and hence regular in all its points. Therefore, it should coincide with y; hence, xb = y(s 6'). Thus, Eq. (2) may be written as
+
+
+
+
which is Eq. (I) for s = s o 6'. This proves our assertion and concludes the proof.
Q.E.D.
COROLLARY 1. Let S be complete. Then for every point p E S the map S is onto S. exp,: T,(S) 3
This is true because if q E S and d(p, q) = r, then q = exp, rv, where v = yl(0) is the tangent vector of a minimal geodesic y parametrized by the arc length and joining p to q. COROLLARY 2. Let S be complete and bounded in the metric d (that is, there exists r > 0 such that d(p, q) < r for every pair p, q E S). Then S is compact. Proof. By fixing p E S, the fact that S is bounded implies the existence of a closed ball B c T,(S) of radius r, centered at the origin 0 of the tangent plane Tp(S), such that exp,(B) = exp,(T,(S)). By the fact that exp, is onto, we have S exp,(T,(S)) = expp(B). Since B is compact and exp, is conQ.E.D. tinuous, we conclude that S is compact.

From now on, the metric notions to be used will refer, except when otherwise stated, to the distance d in Def. 3. For instance, the diameter p(S) of a surface S is, by definition,
With this definition, the diameter of a unit sphere S2 is p(S2)
,=
n.
EXERCISES I. Let S c R3 be a complete surface and let F c S be a nonernpty, closed subset of S such that the complement S  F is connected. Show that S  F is a non
complete regular surface.
336
Global Differential Geometry
2. Let S be the onesheeted cone of Example 1. Show that, given p E S, the only geodesic of S that passes through p and cannot be extended for every value of the parameter is the meridian of S through p.
3. Let S be the onesheeted cone of Example 1. Use the isometry of Example 3 of Sec. 42 to show that any two points p, q E S (see Fig. 511) can be joined by a minimal geodesic on S.
Figure 511
4. We say that a sequence {p,) of points on a regular surface S c R3 converges to a point po E S in the (intrinsic) distance d if given E > 0 there exists an index no such that n 2 no implies that d(p,, po) < E . Prove that a sequence (p,) of points in S converges in d to po E S if and only if (p,] converges to po as a sequence of points in 3 3 (i.e., in the euclidean distance). "5. Let S c R3 be a regular surface. A sequence {p,) of points on S is a Cauchy sequence in rhe (intrinsic) distance d if given E > 0 there exists an index no such that when n, m 2 no then d(p,, p,) < E . Prove that S is complete if and only if every Cauchy sequence on S converges to a point in S.
"6. A geodesic 7 : [0, co) +S on a surface S is a ray issuing from ~ ( 0 if) it realizes the (intrinsic) distance between y(0) and y(s) for all s E [0, a). Let p be a point on a complete, noncompact surface S. Prove that S contains a ray issuing from
P*
7. A divergent curve on S is a differentiable map a : [O, a),S such that for every compact subset K c S there exists a to E (0, GO) with ~ ( t $4) K for t > t o (i.e,,
a "leaves"
every compact subset of S ) . The length of a divergenr curve is defined
as
Prove that S c R3 is complete i f and only if the length of every divergent curve is unbounded.
Complete Surfaces
337
*8. Let S and 3 be regular surfaces and let 9 : S + 3 be a diffeomorphism. Assume that ,!?is complete and that a constant c > 0 exists such that
for all p E S and all v E T,(S), where I and f denote the first fundamental forms of S and S, respectively. Prove that S is complete. "9. Let S1 c R 3 be a (connected) complete surface and S2 c R3 be a connected surface such that any two points of S2 can be joined by a unique geodesic. Let p: S1 + S2 be a local isometry. Prove that p is a global isometry.
"10. Let S c R3 be a complete surface. Fix a unit vector v E R, and let h: S + R be the height function h(p) = (p, v), p E S. We recall that the gradient of h is the (tangent) vector field grad h on S defined by
for all w
(grad h(p), w), = &,(w)
E
T,(S)
(cf. Exercise 14, Sec. 25). Let a(t) be a trajectory of grad h; i.e., a(t) is a curve on S such that a'(r) = grad h(a(t)). Prove that
a. I grad h(p) [ < 1 for all p E S. b. A trajectory a(t) of grad h is defined for all t
E
R.
The following exercise presumes the material of Sec. 35, part, B and an elementary knowledge of functions of complex variables.
{c
[c
11. (Osserman's Lemma.) Let D l = E C; I 5 1) be the unit disk in the complex plane C.As usual, we identify C = R2 by = u iv. Let x: D l 4 R3 be an isothermal parametrization of a minimal surface x(D,) c R3. This means (cf. Sec. 35, Part B) that
S > 0 is compact (Bonnet's theorem). The crucial point of the proof is to show that if K 2 6 > 0, a geodesic y joining two arbitrary points p, q E S and having length l(y) > n/JFis no longer minimal; that is, there exists a parametrized curve joining p and q, the length of which is smaller than l(y). Once this is proved, it follows that every minimal geodesic has length 1 nlrJ6; thus, S is bounded in the distance d Since S is complete, S is compact (Corollary 2, Sec. 53). We remark that, in addition, we obtain an estimate for the diameter of S, namely, p(S) n/J8. To prove the above point, we need to compare the arc length of a parametrized curve with the arc length of "neighboring curves." For this, we shalI introduce a number of ideas which are useful in other problems of differential geometry. Actually, these ideas are adaptations to the purposes of differential geometry of more general concepts found in calculus of variations. No knowledge of calculus of variations will be assumed. In this section, S will denote a regular (not necessarily complete) surface. We shall begin by making precise the idea of neighboring curves of a given curve.
0 such that if Ivl < 6, v E T,(,,(S), then exp,(,, v is well defined for all s E [0, El. In fact, for each p E a([O, 11) c S consider the neighborhood W , (a normal neighborhood of all of its points) and the number 6, > 0 given by Prop. 1 of Sec. 47. The union U , W , covers a([O, I]) and, by compactness, a finite number of them, say, W , , . . . , W,, stilI covers a([O, I]). Set 6 = min (a,, . . . ,6,), where 6, is the number corresponding to the neighborhood Wi,i = 1,. . . ,n. It is easily seen that 6 satisfies the above condition. Now let M = max V(s) E < 6/M, and define
,,,,,,, I
I,
h is clearly well defined. Furthermore, since
where y is the (differentiable) map of Theorem 1 of Sec. 47 (i.e., for t st 0, and V(s) # 0, y(1, a(s), t V(s)) is the geodesic y with initial conditions y(0) = a(s), y'(0) = V(s)), h is differentiable. It is immediately checked that h(s, 0) = a(s). Finally, the variational vector field of h is given by
and it is clear, by the definition of h, that if V(0)
=
V(1)
= 0, then
h is proper. Q.E.D.
We want to compare the arc length of a (= h,) with the arc length of h,. Thus, we define a function L: (6, E) + R by
The study of L in a neighborhood of t = 0 will inform us of the "arc length behavior" of curves neighboring a. We need some preliminary lemmas.
LEMMA 1. The function L deJined by Eq. ( I ) is diflerentiable in a neighborhood oft = 0; in such a neighborhood, the derivative of L may be obtained by dzferentiation under the integral sign.
Global Differential Geometry
342
Proof. Since a : [0, I]
3
S is parametrized by arc length,
It folloas, by compactness of [0, I ] , that there exists a 6 > 0, 6
E,
such that
Since the absolute value of a nonzero differentiable function is differentiable, the integrand of Eq. (I) is differentiable for I t [ < 6. By a classical theorem of calculus (see R. C. Buck, Advanced Calculus, 1965, p. 120), we conclude that L is differentiable for J t 1 < S and that L f ( t )=
l:dl@ I dt d s (
, ) ds.
Lemmas 2, 3, and 4 below have some independent interest.
LEMMA 2. Let w(t) be a diferentiable vectorfield along the parametrized curve a : [a, b] ,S and let f: [a, b] + R be a dzflerentiable function. Then
Proof. It suffices to use the fact that the covariant derivative is the tangential component of the usual derivative to conclude that (here ( ), denotes the tangential component of ( ))
Q.E.D.
L E M M A 3. Let vl(t) and w(t) be diferentiable vector jields along the parametrized curve a : [a, b] + S. Then
Proof. Using the remarks of the above proof, we obtain d & =
(h dt' w) + (v,
$)
=
((2); + W)
(up
(%)) Q.E.D.
Variations o f Arc Length
343
Before stating the next lemma it is convenient to introduce the following terminology. Let h: 10, 11 x (E, E ) + S be a differentiable map. A differentiable vector field along h is a differentiable map V: [0, 11 x (6, e) +S c R3 such that V(s, t) E T,,,,,,(S) for each (s, t) E [O, I] x (6, E). This generalizes the definition of a differentiable vector field along a parametrized curve (Sec. 44, Def. 2). For instance, the vector fields (dh/ds)(s, t) and, (dh/dt)(s, t), introduced above, are vector fields along h. If we restrict V(s, t) to the curves s = const., t = const., we obtain vector fields along curves. In this context, the notation (DV/dt)(s, t) means the covariant derivative, at the point (s, t), of the restriction of V(s, t) to the curve s = const. LEMMA 4. Let h: [0, I] x (c, Then
D dh
(s, ds dt
e) c R2 ,S be a dlflerentiable mapping. t)
D dh dt ds
=  (s,
t).
Proof. Let x: U + S be a parametrization of S at the point h(s, t), with parameters u, v, and let u
= h,(s,
t),
v
= hz(s,
t)
be the expression of h in this parametrization. Under these conditions, when (s, t) E h'(x(U)) = W, the curve h(s, to) may be expressed by
Since (dh/ds)(so, to) is tangent to the curve h(s, to) at s
= so, we
have that
By the arbitrariness of (so, to) G W, we conclude that  dh, dh ds xu as
ah2 + &, as
where we omit the indication of the point (s, t) for simplicity of notation. Similarly,  dh, dh Xu
dt
dt
+ Xu. dh2 dt
344
Global Differentiel Geometry
We shall now compute the covariant derivatives (D/ds)(dh/dt) and (D/dt)/(dh/ds) using the expression of the covariant derivative in terms of the Christoffel symbols IT; (Sec. 44, Eq. (I)) and obtain the asserted equality. For instance, the coefficient of xuin both derivatives is given by d2h,
+
dh dh E l $ &
dh, dh, dh, dh, dl22 + r bdh, dt ds tri2dt ds + ri2dt ds
The equality of the coefficientes of x, may be shown in the same way, thus concluding the proof. Q.E.D. We are now in a position to compute the first derivative of L at t and obtain
=0
PROPOSITION 2. Let h : [O,1] x E ) be a proper variation of the curve a : [0, I] 4 S and let V(s) = (dhldt)(s, 0), s E [O, 111, be the variational (PC,
vectorJieEd of h. Then
where A(s)
= (D/ds)(dh/ds)(s,
0).
Proof. If t belongs to the interval (6, 6) given by Lemma 1, then
By applying Lemmas 3 and 4, we obtain
Since J (dh/ds)(s, 0) I
=
1, we have that
where the integrand is calculated at (s, 0), which is omitted for simplicity of notation. According to Lemma 3,
Variations o f Arc Length
Therefore,
since (dh/dt)(O,0 ) = (dhldt) (1, 0 ) = 0 , due to the fact that the variation is proper. By recalling the definitions of A(s) and V(s), we may write the last expression in the form L'(0)
=
I
I
(A($), V ( s ) )ds.
Q.E.D.
0
Remark I. The vector A(s) is called the acceleration vector of the curve a, and its norm is nothing but the absolute value of the geodesic curvature of a. Observe that L'(0) depends only on the variational field V(s)and not on the variation h itself. Expression (2) is usually called the formula for the Jirst variation of the arc length of the curve a. Remark 2. The condition that h is proper was only used at the end of the proof in order to eliminate the terms
Therefore, if h is not proper, we obtain a formula which is similar to Eq. (2) and contains these additional boundary terms.
An interesting consequence of Prop. 2 is a characterization of the geodesics as solutions of a "variational problem." More precisely, PROPOSITION 3. A regular parametrized curve a : [0, 11 + S, where the parameter s c [0, I ] is the arc length of a, is a geodesic if and only if, for every S of a, L'(0) = 0. proper variation h : [0, 11 x ( E , E ) +
Proof. The necessity is trivial since the acceleration vector A(s) = (D/ds)(dalds) of a geodesic a is identically zero. Therefore, L'(0) = 0 for every proper variation. Suppose now that L'(0) = 0 for every proper variation of a and consider a vector field V(s) = f (s)A(s), where f : [0, I ] + R is a real differentiable function, with f (s) 0, f (0) = f (1) = 0, and A(s) is the acceleration vector of a. By constructing a variation corresponding to V(s),we have
>
Global Differential Geometry
Therefore, since f (s) 1 A($) l2
> 0, we obtain f (s) 1 A(s) l2
= 0.
We shall prove that the above relation implies that A(s)  0, s E [O, I]. In fact, if I A(so)I # 0, s o E (0, I), there exists an interval I = (so  E , so E ) such that ] A(s) 1 # 0 for s E I. By choosing f such that f (so) > 0, we contradict f (so)1 A(so)1 = 0. Therefore, I A(s) I = 0 when s E (0, I). By continuity, A(0) = A(1) = 0 as asserted. Since the acceleration vector of a is identically zero, a is geodesic. Q.E.D.
+
From now on, we shall only consider proper variations of geodesics y :[0,1] + S, parametrized by arc length; that is, we assume Lr(0) .= 0. To simplify the computations, we shall restrict ourselves to orthogonal variations; that is, we shall assume that the variational field V(s) satisfies the condition (V(s), yr(s)) = 0, s E LO, I ] . To study the behavior of the function L in a neighborhood of 0 we shall compute L"(0). For this computation, we need some lemmas that relate the Gaussian curvature to the covariant derivative. LEMMA 5. Let x: U 3 S be a parametrization at a point p E S of a regular surface S, with parameters u, v, and let K be the Gaussian curvature of S. Then
Proof. By observing that the covariant derivative is the component of the usual derivative in the tangent plane, we have that (Sec. 43)
By applying to the above expression the formula for the covariant derivative (Sec. 44, Eq. (I)), we obtain
We verify, by means of a similar computation, that
Variations of Arc Length
+ ~(r:,), + r:,r:, + r:2r:21~u. Therefore,
We now use the expressions of the curvature in terms of Christoffel symbols (Sec. 43, Eqs. ( 5 ) and (5a) and conclude that
= K{(xu, xu)xu 
= K(xu A
LEMMA 6. Let h: 10, I] x let V(s, t), (s, t) E 10, 11 x (  E , Then
(E,
e)
f),
be
(xu, xu)xu)
xu) A xu.
Q.E.D.
S be a dzflerentidble mapping and diferentiable vector Jield along h.
+
where K(s, t) is the curvature of S at the point h(s, t). Proof. Let x(u, v) be a system of coordinates of S around h(s, t) and let
be the expression of V(s, t) = V in this system of coordinates. By Lemma 2, we have D D  V = (axU bx,) ds ds
+
Therefore,
348
Global Differential Geometry
By a similar computation, we obtain a formula for (D/ds)(D/dt)V, which is given by interchanging s and t in the last expression. It follows that D D V  DV D= a dt ds
D D D Dx (dt ds " ds dt x")
ds dt
T o compute (D/dt)(D/ds)x,, we shall take the expression of h,
u = h , ( ~t), ,
v
= h,(s,
t),
in the parametrization x(u, v ) and write
Since the covariant derivative (D/ds)x, is the projection onto the tangent plane of the usual derivative (d/ds)xu,we have
 dh, D xu 
ds du
dh2 D
4 xu, ds dv
where T denotes the projection of a vector onto the tangent plane. With the same notation, we obtain
D xds D dt u={
dh2 D +( dt ds du ds &x"))
d dh, xu D
dh2 dhl D D
=
d2h, D
&% xu du
dh2 D D dv du xu).
+ Z ( Z & & X ~+dt
In a similar way, we obtain (D/ds)(D/dt)x,, which is given by interchanging s and t in the above expression. It follows that
Variations of Arc Length
where
A = ( d. h , dh, ds dt
dh, dh,) ds dt
By replacing xu for xu in the last expression, we obtain
By introducing the above expression in Eq. (3) and using Lemma 5, we conclude that
On the other hand, as we saw in the proof of Lemma 4, dh  xdh, ds Xu+&dh2 xo,
dh Xu dh, dt  dt
 
+ dh2 dt
Xu;
hence,
dh A dh = Ax, A ds
dt
x,.
Therefore,
Q.E.D. We are now in a position to compute L"(0).
PROPOSITION 4. Let h : [0, I ] x (  E , e ) + S be a proper orthogonal variation of a geodesic y : [0, I] + S parametrized by the arc length s E [O, I ] . Let V(s) = (dh/dt)(s,0) be the variational vectorfield of h. Then
where K(s)
= K(s,
0) is the Gaussian curvature of S at y(s)
Proof. As we saw in the proof of Prop. 2,
= h(s,
0).
350
Global Differential Geometry
for t belonging to the interval (6'6) the above expression, we obtain
given by Lemma I . By differentiating
Observe now that for t = 0, I (dh/ds)(s, 0) I
Since y is a geodesic, (D/ds)(dh/ds) orthogonal,
=0
=
for t
1. Furthermore,
= 0,
and since the variation is
It follows that L1I(O) =
[:dtd (D dhd t ' dhds} ds, ds

where the integrand is calculated at (s, 0). Let us now transform the integrand of Eq. (5) into a more convenient expression. Observe first that
D D dh dh) dt ds d t ' ds
On the other hand, for t

(D Ddh dh) ds dt dt ' ds
= 0,
since (D/ds)(dh/ds)(s, 0) = 0, owing to the fact that y is a geodesic. Moreover, by using Lemma 6 plus the fact that the variation is orthogonal, we
Variations of Arc
obtain (for t
L engfh
= 0)
D D ah d h ) dt ds dt ' ds
(D D dh d h ) ds dt dt' ds
By introducing the above values in Eq. (5), we have
Finally, since the variation is proper, (dh/dt)(O,t ) = (dh/dt)(l,t ) = 0, t E (8, 8). Thus, Q.E.D. Remark 3. Expression (4) is called the formula for the second variation of the arc length of y. Observe that it depends only on the variational field of h and not on the variation h itself. Sometimes it is convenient to indicate this dependence by writing LI(0). Remark 4. It is often convenient to have the formula (4) for the second variation written as follows:
Equation (4a) comes from Eq. (4), by noticing that V(0) = V ( l ) = 0 and that DV DV
D2V
ds
Thus,

f (Ev + KV, v ) d s ds o
=
J o
ds
+ KV, V ) ds.
Globs/ Differentia/ Geometry
352
The second variation L"(0) of the arc length is the tool that we need to prove the crucial step in Bonnet's theorem, which was mentioned in the beginning of this section. We may now prove
THEOREM ( B o ~ e t )Let . the Gaussian curvature K of a complete surface S satisfy the condition K>6>0. The~zS is compact and the diameter p of S satisfies the inequality
Proof. Since S is complete, given two points p, q E S, there exists, by the HopfRinow theorem, a minimal geodesic y of S joining p to q. We shall prove that the length 1 = d(p, q) of this geodesic satisfies the inequality
We shall assume that I > n/J6 and consider a variation of the geodesic y : [0, 11 + S, defined as follows. Let w, be a unit vector of T,,,,(S) such that (w,, yf(0)) = 0 and let w(s), s E LO, I ] , be the parallel transport of w, along y. I t is clear that I w(s)1 = 1 and that (w(s), y'(s)) = 0, s E [0, El. Consider the vector field V ( s )defined by
Since V(0) = V ( l ) = 0 and (V(s),y ' ( s ) ) = 0, the vector field V(s)determines a proper, orthogonal variation of y. By Prop. 4,
Since w(s) is a parallel vector field,
Thus, since I
> n/JT, so that
K
2 6 > n2/12,we obtain
Variations of Arc Length
=
$
I 1
cos 2n s ds 0 I
= 0.
Therefore, there exists a variation of y for which L"(0) < 0. However, since y is a minimal geodesic, its length is smaller than or equal to that of any curve joining p to q. Thus, for every variation of y we should have Lt(0) = 0 and L"(0) 2 0. We obtained therefore a contradiction, which shows that 1 = d(p, q) ( n/J& as we asserted. Since d(p, q) < n/JX for any two given points of S, we have that S is bounded and that its diameter p 5 7rlJT. Moreover, since S is complete and bounded, S is compact. Q.E.D. Remark 5. The choice of the variation V(s) = w(s) sin (n/l)s in the above proof may be better understood if we look at the second variation in the form (4a) of Remark 4. Since K > Z2/n2,we can write
Now it is easy to guess that the above V(s) makes the last integrand equal to zero; hence, LF(0) < 0. Remark 6. The hypothesis K 2 6 > 0 may not be weakened to K In fact, the paraboloid
> 0.
has Gaussian curvature K > 0, is compIete, and is not compact. Observe that the curvature of the paraboloid tends toward zero when the distance of the point (x, y) E R2 to the origin (0, 0) becomes arbitrarily large (cf. Remark 8 below). Rehark 7. The estimate of the diameter p _( n / J T given by Bonnet's theorem is the best possible, as shown by the example of the unit sphere: KG landp=n. Remark 8. The first proof of the above theorem was obtained by 0. Bonnet, "Sur quelquer proprittts des lignes gCodtsiques," C.R.Ac. Sc. Parts X L (1850), 1331, and "Note sur les lignes gtodCsiques,'? ibid. XLI (1851),
354
Global Differential Geometry
32. A formulation of the theorem in terms of complete surfaces is found in the article of HopfRinow quoted in the previous section. Actually, it is not necessary that K be bounded away from zero but only that it not approach zero too fast. See E. Calabi, "On Ricci Curvature and Geodesics," Duke Math. 3. 34 (1967), 667676; or R. Schneider, "Konvexe Flachen mit langSam abnehmender Kriimmung," Archiv der Math. 23 (1972), 650654 (cf. also Exercise 2 below).
1. Is the converse of Bonnet's theorem true; i.e., if S is compact and has diameter p i n/dS,is K > d ?
*2. (Kazdan Warner's Remark. cf. Exercise 10, Sec. 510.) Let S = (z = f(x, y ) ; (x, y) E R2] be a complete noncompact regular surface. Show that
lim ( inf K(x, y)) (0. I+
x2+y2>r
3. a. Derive a formula for the first variation of arc length without assuming that the variation is proper. b. Let S be a complete surface. Let y(s), s
R, be a geodesic on S and let d(s) be the distance d(y(s),p) from y(s) to a point p E S not in the trace of y. Show that there exists a point so E R such that d(so)(d(s) for all s E R and that the geodesic r joining p to y(so)is perpendicular to 7 (Fig. 514). E
Figure 514
c. Assume further that S is homeomorphic to a plane and has Gaussian curva
ture K _( 0. Prove that so (hence, r)is unique. 4. (Calculus of Variations.) Geodesics are particular cases of solutions to variational problems. In this exercise, we shall discuss some points of a simple, although
355
Variations of Arc Length
quite representative, variational problem. In the next exercise we shall make some applications of the ideas presented here. Let y = y(x), x E [xl, x2] be a differentiable curve in the xy plane and let a variation of y be given by a differentiable map y = y(x, t), t E (E, E). Here y(x, 0) = y(x) for all x E [xl, XZ],and y(xl, t) = y(xl), y(x2, t) = y(x2) for all t E (E, E) (i.e., the end points of the variation are fixed). Consider the integral
where F(x, y, y') is a differentiable function of three variables and y' = dylds. The problem of finding the critical points of I(?) is called a variationalproblem with integrand F.
a. Assume that the curve y
= y(x)
is a critical point of I(t) (i.e., dl/& t = 0). Use integration by parts to conclude that ( j = dI/dt)
=
0 for
Then, by using the boundary conditions, obtain
where q = (dy/dt)(x, 0). (The function q corresponds to the variational vector field of y(x, t).) b. Prove that if i(0) = 0 for all variations with fixed end points (i.e., for all 11 in (*) with tf(xl) = q(xz) = 0), then
Equation (* *) is called the EulerLagrange equation for the variational problem with integrand F. c. Show that if F does not involve explicitIy the variable x, i.e., F then, by differentiating yfFYf F, and using (a*) we obtain that
=
F(y, y'),
y'Fyt  F = const.
5. (Calculus of Variations; Applications.) a. (Surfaces of Revolution of Least Area.) Let S be a surface of revolution obtained by rotating the curve y = f(x), x E [x1, x2], about the x axis. Suppose
356
Global Differential Geometry
that S has least area among all surfaces of revolution generated by curves joining ( x , , f ( x l ) )to ( X Z , f(x2)). Thus, y = f ( x ) minimizes the integral (cf. Exercise 11 , Sec. 25)
for all variations y(x, t ) of y with fixed end points y(x,), y(x2). By part b of C y ' ) 2 satisfies the EulerLagrange equation (**). Exercise 4, F(y, y f ) = y d 1 Use part c of Exercise 4 to obtain that
+
yrFYf F = 
J l
1
  , 
Y
+(Y')~
c
= const.;
C
hence, y
1
=  cash(cx C
+ el),
cl
= const.
Conclude that ifthere exists a regular surface of revolution ofleast area connecting two given parallel circles, this surface is the catenoid which contains the two given circles as parallels. b. (Geodesics of Surfaces of Revolution.) Let
be a parametrization of a surface of revolution S. Let u = u(v) be the equation of a geodesic of S which is neither a parallel nor a meridian. Then u = u(v) is a critical point for the arc length integral ( F = 0)
+
Since E = f 2 , G = (f')2 (gf)2,we see that the EulerLagrange equation for this variational problem is
Notice that F does not depend on u. Thus, (d]dv)Fut= 0, and c
u'f2
= const. = F 
"'  J f
2(.f)2
+ ( f ' I 2 + ( g f ) 2
From this, obtain the following equation for the geodesic u Example 5, Sec. 44):
= u(v)
(cf.
Jacobi Fields and Conjugate Points
55. Jacobi Fields and Conjugate Points In this section we shall explore some details of the variational techniques which were used to prove Bonnet's theorem. We are interested in obtaining information on the behavior of geodesics neighboring a given geodesic y. The natural way to proceed is to consider variations of y which satisfy the further condition that the curves of the variation are themselves geodesics. The variational field of such a variation gives an idea of how densely the geodesics are distributed in a neighborhood of y. To simplify the exposition we shall assume that the surfaces are complete, although this assumption may be dropped with further work. The notation y: [0, 11 3 S will denote a geodesic parametrized by arc length on the complete surface S.
DEFINITION I. Let y : [O, I] S be a parametrized geodesic on S and S be a variation of y such that for every t E (6, E ) let h : [O, I] x (6, e ) the curve h,(s) = h(s, t), s E LO, I], is aparametrized geodesic (not necessarily parametrized by arc length). The variational field (dh/dt)(s, 0) = J(s) is called a Jacobi field along y. +
A trivial example of a Jacabi field is given by the field yl(s), s E [0, I], of tangent vectors to the geodesic y. In fact, by taking h(s, t) = y(s t), we have
+
We are particularly interested in studying the behavior of the geodesics neighboring y: [0, 1] , S, which start from y(O). Thus, we shall consider variations h : [0, I] x (E, E) S that satisfy the condition h(0, t ) = y(O), t E (E, 6 ) . Therefore, the corresponding Jacobi field satisfies the condition J(0) = O (see Fig. 515). Before presenting a nontrivial example of a Jacobi field, we shall prove that such a field may be characterized by an analytical condition. 4
PROPOSITION 1. Let J(s) be a JacobiJield along y : [O, 11 + S, s Then J satisjies the socalled Jacobi equation
where K(s) is the Gaussian curvature of S a t y(s).
E
LO, I ] .
Global Differential Geometry
Figure 515
Proof. By the definition of J(s), there exists a variation
h: [0, I ] x (6, E) + S of y such that (dh/dt)(s, 0) = J(s) and h,(s) is a geodesic, t follows that (D/ds)(dh/ds)(s, t) = 0. Therefore,
E
(E,
E).
It
On the other hand, by using Lemma 6 of Sec. 54 we have
Since (D/dt)(dh/ds)
= (D/ds)(dh/dt),
we have, for t
= 0,
D D ) 0. ds ds J(s) fK(s)(yf(s) /\ J(s)) A ~ ' ( 3 =
Q.E.D.
To draw some consequences from Prop. 1, it is convenient to put the Jacobi equation (1) in a more familiar form. For that, let e,(O) and e,(O) be unit orthogonal vectors in the tangent plane T,,,,,(S) and let e,(s) and e,(s) be the parallel transport of e,(O) and e,(O), respectively, along y(s). Assume that
for some functions a , = a,(s), a, = a,($). Then, by using Lemma 2 of the last section and omitting s for notational simplicity, we obtain
Jacobi Fields and Conjugate Points
On the other hand, if we write (7' A J) A y'
= Ate1 4
A2e2,
we have Itet
+ I2e2
= (7' =
A
ady'
+ a2e2)) A 7' A el) A Y' ady' A (ale,
Therefore, by setting ((y' A ei) A y', e j )
+
= a,,
i: j
=
e2) /\ y'.
1, 2, we obtain
It follows that Eq. (1) may be written

where all the elements are functions of s. Note that (la) is a system of linear, secondorder differential equations. The solutions (al(s), a2(s)) J(s) of such a system are defined for every s E [0, I ] and constitute a vector space. Moreover, a solution J(s) of (la) (or (I)) is completely determined by the initial conditions J(O), (DJ/ds)(O), and the space of the solutions has 2 x 2 = 4 dimensions. One can show that every vector field J(s) along a geodesic y: [0, El 4 S which satisfies Eq. (1) is, in fact, a Jacobi field. Since we are interested only in Jacobi fields J(s) which satisfy the condition J(0) = 0, we shall prove the proposition only for this particular case. We shall use the following notation. Let Tp(S),p E S, be the tangent plane to S at point p, and denote by (Tp(S)), the tangent space at v of Tp(S) conS, sidered as a surface in R3. Since exp,: T,(S)

We shall frequently make the following notational abuse: If v, w E Tp(S), then w denotes also the vector of (Tp(S)), obtained from w by a translation of vector v (see Fig. 516). This is equivalent to identifying the spaces Tp(S) and (Tp(S)), by the translation of vector v.
LEMMA I. Let p E S and choose v, w y :LO, I] S be the geodesic on S given by +
E
T,(S), with ] v l = 1. Let
Global Differential Geometry
Figure 516
Then, the vector field J(s) along y given by
is a Jacobifield. F'urthermore, J(0)
= 0, (DJ/ds)(O) = w.
ProoJ: Let t +v(t), t E (E, E ) , be a parametrized curve in T,(S) such that v(0) = v and (dv/dt)(O) = w. (Observe that we are making the notational abuse mentioned above.) Define (see Fig. 517)
The mapping h is obviously differentiable, and the curves s 4 h,(s) = h(s, t ) are the geodesics s + exp,(sv(t)). Therefore, the variational field of h is a Jacobi field along y. To compute the variationaI fieId (dh/dt)(s, 0), observe that the curve of Tp(S), s = so, t = 2, is given by t + s,v(t) and that the tangent vector to this curve at the point t = 0 is
It follows that
Jacobi Fields and Conjogate Points
Figure 517
The vector field J(s) = s(d exp,),,(w) is, therefore, a Jacobi field. It is immediate to check that J(0) = 0. To verify the last assertion of the lemma, we compute the covariant derivative of the above expression (cf. Lemma 2, Sec. 54), obtaining
Hence, at s
= 0,
Q.E.D.
PROPOSITION 2. If we let J(s) be a diflerentiable vector field along y: LO, I ] + S, s E [0, I], satisfying the Jacobi equation (I), with J(0) = 0, then J(s) is a Jacobi$eld along y. Proof. Let w = (DJ/ds)(O) and u = y'(0). By Lemma 1, there exists a Jacobi field s(d exp,),,(w) = j ( s ) , s E [O, I], satisfying
Then, J and are two vector fields satisfying the system (1) with the same initial conditions. By uniqueness, J(s) = j(s), s t [O, I]; hence, J is a Jacobi Q.E.D. field. We are now in a position to present a nontrivial example of a Jacobi field.
362
Global Differential Geometry
+ +
Example. Let S2 = ((x,y, Z) E R 3 ;xZ y2 z2 = 11 be the unit sphere and x(8, p) be a parametrization at p E S, by the colatitude 9 and thelongitude p (Sec. 22, Example 1). Consider on the parallel 9 = 4 2 the segment between p, = n/2 and p, = 3n/2. This segment is a geodesic y, which we assume to be parametrized by 47  y o = S. Let w(s) be the parallel transport along y of a vector ( 0 E T o ( ) , with / w(O)j = 1 and (w(O), y l ( 0 ) j = 0. We shall prove that the vector field (see Fig. 518)
is a_Jacobi field along y.
Figure 518. A Jacobi field on a sphere.
In fact, since J(0) = 0, it suffices to verify that J satisfies Eq. (1). By using the fact that K = 1 and w is a parallel field we obtain, sucessively,
+ K(y' A J ) A y' = (sin
D DJ ds ds
s)\u(s)
+ (sin s)w(s) = 0,
which shows that J is a Jacobi field. Observe that J(n)
= 0.
DEFINITION 2. Let y : LO, I] + S be a geodesic of S with y(0) = p. We say that the point q = y(s,), so E [O, I], is conjugate to p relative to the geodesic y if tlzere exists a Jacobi field J(s) which is not identically zero along y with J(0) = J(s,) = 0. As we saw in the previous example, given a point p E S2of a unit sphere S2,its antipodal point is conjugate to p along any geodesic that starts from p.
Jacobi Fields end Conjugete Points
363
However, the example of the sphere is not typical. In general, given a point p of a surface S, the "first" conjugate point q to p varies as we change the direction of the geodesic passing through p and describes a parametrized curve. The trace of such a curve is called the conjugate locus to p and is denoted by C(p). Figure 519 shows the situation for the ellipsoid, which is typical. The geodesics starting from a point p are tangent to the curve C(p) in such a way that when a geodesic 7 near y approaches y, then the intersection point of 7 and y approaches the conjugate point q of p relative to y. This situation was expressed in classical terminology by saying that the conjugate point is the point of intersection of two "infinitely close" geodesics.
Figure 519. The conjugate locus of an ellipsoid.
Remark I. The fact that, in the sphere S2, the conjugate locus of each point p E S2reduces ,to a single point (the antipodal point of p) is an exceptional situation. In fact, it can be proved that the sphere is the only such surface (cf. L. Green, "Aufwiedersehenflache," Ann. Math. 78 (1963), 289300). Remark 2. The conjugate locus of the general ellipsoid was determined by A. Braunmuhl, "Geodatische Linien auf dreiachsigen Flachen zweiten Grades," Math. Ann. 20 (1882), 557586. Compare also H. Mangoldt, "Geodatische Linien auf positiv gekrummten Flachen," Crelles Journ. 91 (188 I), 2352.
A useful property of Jacobi fields J along y: [0, I] when J(0)
= J(1) = 0,
then

S is the fact that
for every s E [0, 11. Actually, this is a consequence of the following properties of Jacobi fields.
PROPOSITION 3. Let J,(s) and J,(s) be Jacobifields along y : [0, I] s E [0, Iji. Then

S,
Globel Differential Geometry
Proof. It suffices to differentiate the expression of the statement and apply Prop. 1 (s is omitted for notational convenience):
" "",  K{(~' 
D DJ, J ~) (J1,) ds ds J,) A y', J,)

DJ + (+, s
DJ ?) s

((y' A J2) A Y',J1))
(% ,ds ds
052)
= O
Q.E.D.
PROPOSITION 4. Let J(s) be a Jacobi field along y :[O, 11 + S, with
Then (J(s),y'(s))=O,
SE[O,Z].
Proof. We set J,(s) = J(s) and J,(s) the previous proposition and obtain
DJ (z, y'(s))
= yl(s)
(which is a Jacobi field) in
= const. = A.
Therefore,
hence, (J(s), yl(s))
= As
I B,
where B is a constant. Since the linear expression As s,, s2 E [0, I], s1 # s,, it is identically zero.
+ B is zero for Q.E.D.
COROLLARY. Let J(s) be a Jacobi field along y: [0, I ] , S, with J(0) = ~ ( 1= ) 0. Then (J(s), yr(s)) = 0, s E [0, 11. We shall now show that the conjugate points may be characterized by the behavior of the exponential map. Recall that when q : S, 3 S, is a differentiable mapping of the regular surface S, into the regular surface s,, a point p E S, is said to be a critical point of p if the linear map
Jacobi Fields and Conjugate Points
is singular, that is, if there exists v E Tp(S,), v # 0, with dp,(v)
= 0.
PROPOSITION 5. Let p, q E S be two points of S and let y: [0, 11 S be a geodesic joining p = y(0) to q = exp,(lyf(0)). Then q is conjugate to p relative to y if and only zfv = ly'(0) is a critical point of exp,: T,(S) + S. +
Proof. As we saw in Lemma 1, for every w E T,(S) (which we identify with (T,(S)),) there exists a Jacobi field J(s) along y with
and
J(0
= [((d exp,),(w)].
If v E T,(S) is a critical point of exp,, there exists w E Tp(S)),, w # 0, with (d expp),(w) = 0. This implies that the above vector field J(s) is not identically zero and that J(0) = J(1) = 0; that is, y(1) is conjugate to y(0) relative t o y. Conversely, if q = y(1) is conjugate to p = y(0) relative to y, there exists a Jacobi field j(s), not identically zero, with j(0)  j(l) = 0. Let (Dj,ds)(0) = w # 0. By constructing a Jacobi field J(s) as above, we obtain, by uniqueness, j ( s ) = J(s). Since
we conclude that (d expp),(w) of exp,.
= 0,
with w # 0. Therefore v is a critical point Q.E.D.
The fact that Eq. (1) of Jacobi fields involves the Gaussian curvature K of S is an indication that the "spreading out" of the geodesics which start from a point p E S is closely related to the distribution of the curvature in S (cf. Remark 2, Sec. 46). It is an elementary fact that two neighboring geodesics starting from a point p E S initially pull apart. In the case of a sphere or an ellipsoid (K > 6 > 0) they reapproach each other and become tangent to the conjugate locus C(p).In the case of a plane they never get closer again. The following theorem shows that an "infinitesimal version" of the situation for the plane occurs in surfaces of negative or zero curvature. (See Remark 3 after the proof of the theorem.)
THEOREM. Assume that the Gaussian curvature K of a surface S satisfies the condition K < 0. Then, for every p E S, the conjugate locus of p is empty. In short, a surface of curvature K (0 does not have conjugate points.
366
Global Differential Geometry
Proof. Let p E S and let 7: [0, I] ,S be a geodesic of S with y(0) = p. Assume that there exists a nonvanishing Jacobi field J(s), with J(0) = J(1) = 0. We shall prove that this gives a contradiction. In fact, since J(s) is a Jacobi field and J(0) = J(1) = 0, we have, by the corollary of Prop. 4, that (J(s), y'(s)> = 0, s E [O, I ] . Therefore,
since K (0. It follows that
Therefore, the function (DJ/ds, J) does not decreaseain the interval [O, I ] . Since this function is zero for s = 0 and s = I, we conclude that
(g,
~ ( s ) )= 0,
s E [O, I].
Finally, by observing that
we have I J l2 = const. Since J(0) = 0, we conclude that 1 J(s) I that is, J is identically zero in [0, I]. This is a contradiction.
= 0, s E
[O, 11; Q.E.D.
Remark 3. The theorem does not assert that two geodesics starting from a given point will never meet again. Actually, this is false, as shown by the closed geodesics of a cylinder, the curvature of which is zero. The assertion is not true even if we consider geodesics that start from a given point with "nearby directions." It suffices to consider a meridian of the cylinder and to observe that the helices that follow directions nearby that of the meridian meet this meridian. What the proposition asserts is that the intersection point of two "neighboring" geodesics goes to "infinity" as these geodesics approach each other (this is precisely what occurs in the cylinder). In a classical terminology we can say that two "infinitely close" geodesics never meet. In this sense, the theorem is an infinitesimal version of the situation for the plane.
An immediate consequence of Prop. 5, the above theorem, and the inverse function theorem is the following corollary.
Jacobi Fields and Conjugate Points
367
COROLLARY. Assume the Gaussian curvature K of S to be negative or zero. Then for every p E S, the mapping exp, : T,(S) +S
is a local difleomorphisrn. We shall use later the following lemma, which generalizes the fact that, in a normal neighborhood of p, the geodesic circles are orthogonal to the radial geodesics (Sec. 46, Prop. 3 and Remark 1). LEMMA 2 (Gauss). Let p E S be a point of a (complete) surface S and let u E Tp(S) and w t (T,(S)),. Then
where the identification T,(S) Proof. Let 1 = 1 u 1, v given by
= (Tp(S)), is being used.
= u/(u (
and let y : [0, 11
y(s)=exp,(sv),

S be a geodesic of S
seP,~l.
Then yf(0) = v. Furthermore, if we consider the curve s +s v in T,(S) which passes through u for s = 1 with tangent vector v (see Fig. 520), we obtain
Figure 520
Consider now a Jacobi field J along y, given by J(0) (cf. Lemma I). Then, since y(s) is a geodesic,
= 0,
(DJ/ds)(O) = w
Global Differential Geometry
and since J is a Jacobi field, d b(Y'(S),
DJ =)
= (? '(s),
= 0. E) ds
It follows that
hence (since J(0) = 0) ( ~ ' ( 4 J(s)) ,
=
cs
T o compute the constant C, set s equal to I in Eq. (3). By Lemma 1, J(0 = l(d exp,)u(wI Therefore,
From Eq. (2) we conclude that
(Y '(I), %(I))
=
c = (Y'(o),
g(0))(v, =
w).
By using the value of C, we obtain from the above expression
(u,W)
= ((d
~ x P ~ ) ~ ((du e) ,x ~ ~ ) ~ ( w ) ) .
Q.E.D.
EXERCISES 1. a. Let y : [O, I ] +S be a geodesic parametrized by arc length on a surface S and let J(s) be a Jacobi field along y with J(0) = 0, (J'(O), ~'(0))= 0. Prove that (J(s), y'(s)) = 0 for all s E [0, I]. b. Assume further that IJr(0)I = 1 . Take the parallel transport of e,(O) = yf(0) and of e,(O) = J'(0) along y and obtain orthonormal bases {el(s), e2(s)]for all T,(,,(S), s G [O, I]. By part a, J(s) = u(s)e2(s) for some function u = u(s). Show that the Jacobi equation for Jcan be written as
with initial conditions u(0)
=
0, u'(0)
=
1.
2. Show that the point p = (0, 0, 0) of the paraboloid z gate point relative to a geodesic y(s) with y(0) = p.
= x2 f y2
has no conju
Jacobi Fie/ds and Conjugate Points
369

3. (The Comparison Theorems.) Let S and 5 be complete surfaces. Let p E S, j7 E S" and choose a linear isometry i: T,(S) T@). Let y: [0, co) + S be a geodesic on S with y(0) = p, [ y'(0) [ = 1, and let J(s) be a Jacobi field along y
with J(0) = 0, (J1(0), y'(0)) = 0,l J1(0)I = 1. By using the linear isometry i, construct a geodesic J : [O, m) + with J(0) = 6, Jf(0) = i(yl(0)), and a Jacobi field Talong J with f(0) = 0, f'(0) = i(J'(0)) (Fig. 521). Below we shall describe two theorems (which are essentially geometric interpretations of the classical Sturm comparison theorems) that allow us to compare the Jacobi fields J and J f r o m a "comparison hypothesis" on the curvatures of S and 3.
s
Figure 521
a. Use Exercise I to show that J(s) = v(s)e2(s),f(s) = u(s)c2(s),where u = u(s), v = v(s) are differentiabIe functions, and e,(s) (respectively, Z2(s)) is the parallel transport along y (respectively, 7) of J1(0) (respectively, T(0)). Conclude that the Jacobi equations for J a n d yare
respectively, where K and I? denote the Gaussian curvatures of S and 3 *b. Assume that K(s) < I?(s), s E [0, a]. Show that
0=
1"
(~(v"f Kv)  v(u"
0
=[uu~vu~I;
+
+ I?u)~))ds (*I
(K@M~s.
Conclude that if a is the first zero of u in (0, m ) (ie., u(a)
= 0 and u(s)
> 0 in
Global Differential Geometry
370
>
(0, a)) and b is the first zero of v in (0, co), then b a. Thus, ifK(s) 5 k(s) for all s, the first conjugate point of p relative to y does not occur before the first conjugafepoint of f5 relative to 7. This is called the first comparison theorem. *c. Assume that K(s)
< k(s),
s E [0, a). Use (*) and the fact that L( and vare positive in (0, a) to obtain that [uv'  vuf]; > 0. Use this inequality to show that v(s) 2 U(S)for all s E (0, a). Thus, $K(s) ( R(s) for all s before thefirst conjugatepoint of 7, then / J(s) I > ~?(s)1 for all such s. This is called the second comparison theorem (of course, this includes the first one as a particular case; we have separated the first case because it is easier and because it is the one that we use more often).
d. Prove that in part c the equality v(s) if K(s) = k(s), s E [0, a).
= u(s)
holds for all s
E
[0, a) if and only
4. Let S be a complete surface with Gaussian curvature K < KO,where KO is a positive constant. Compare S with a sphere S2(Ko)with curvature KO(that is, set, in Exercise 3, 3 = S2(Ko) and use the first comparison theorem, Exercise 3, part b) to conclude that any geodesic y : [0, co) S on S has no point conjugate to y(0) in the interval (0, n/,q'K).


5. Let S be a complete surface with K 2 K1 > 0, where K is the Gaussian curvature S has a point of S and K , is a constant. Prove that every geodesic y : 10, co) conjugate to ~ ( 0 in ) the interval (0, n/,q'K]. *6. (Sturm's Oscillation Theorem.) The following slight generaIization of the first comparison theorem (Exercise 3, part b) is often useful. Let S be a complete surface and y : [0, co) + S be a geodesic in S. Let J(s) be a Jacobi field along y with J(0) = J(so) = 0, so E (0, a ) and J(s) # 0, for s E (0, so).Thus, J(s) is a normal field (corollary of Prop. 4). It follows that J(s) = v(s)e2(s), where v(s) is a solution of v"(s)
+ K(s)v(s)
=
0,
s
[O, a ) ,
E
and e2(s) is the parallel transport of a unit vector at T,(,,(S) normal to ~'(0). Assume that the Gaussian curvature K(s) of S satisfies K(s) L(s), where L is a differentiable function on [0, co). Prove that any solution of
0" by
PROPOSITION 7. Let S be a complete surface with Gaussian curvature K ( 0. Then the map exp,: T,(S) +S, p E S, is a covering map.
386
G/oba/ Differentia/ Geometry
Proof. Since we know that exp, is a local diffeomorphism, it suffices (by Prop. 6) to show that exp, has the property of lifting arcs. Let a : [O,l] 3 S be an arc in S and also let v E T,(S) be such that exp, v = a(0). Such a v exists since S is complete. Because exp, is a local diffeomorphism, there exists a neighborhood U of v in T,(S) such that exp, restricted to U is a diffeomorphisrn. By using exp; in exp,(U), it is possible to define & in a neighborhood of 0. Now let A be the set of t E [O, I ] such that & is defined in [0, t]. A is nonempty, and if &(to)is defined, then & is defined in a neighborhood of t o ; that is, A is open in [O, I ] . Once we prove that A is closed in [0, 11, we have, by connectedness of [0, I], that A = [O, I] and a may be entirely lifted. The crucial point of the proof consists, therefore, in showing that A is closed in [0, 11. For this, let to E [O, I] be an accumulation point of A and Itn] be a sequence with {t,) .to, t, E A, n = 1,2, . . .. We shall first prove that a"(t,) has an accumulation point. Assume that &(t,) has no accumulation point in T,(S). Then, given a closed disk D of T,(S), with center &(O),there is an no such that &(tnll)$ D. It follows that the distance, in T,(S), from &(O) to &(t,) becomes arbitrarily large. Since, by Lemma 1, exp,: T,(S) + S increases lengths of the vectors, it is clear that the intrinsic distance in S from a(0) to a(t,) becomes arbitrarily large. But that contradicts the fact that the intrinsic distance from a(0) to &(to)= lim,n,,, a@,) is finite, which proves our assertion. We shall denote by q an accumulation point of a"(t,). Now let V be a neighborhood of q in T,(S) such that the restriction of exp, to V is a diffeomorphism. Since q is an accumulation point of (&(tn)), there exists an n, such that &(t,,) E V. Moreover, since a is continuous, theie exists an open interval I c [0, I], to E I, such that a(I) c exp,(V) = U. By using the restriction of exp; ' in U it is possible to define a lifting of a in I , with origin in &(t,,). Since exp, is a local diffeomorphism, this lifting coincides with & in [0, to) n I a n d is therefore an extension of a" to an interval containing to. Thus, the set A is closed, and this ends the proof of Prop. 7.
Q.E.D.

Remark I. It should be noticed that the curvature condition K < 0 was S is a lengthincreasing local difused only to guarantee that exp,: T,(S) S, is a feomorphism. Therefore, we have actually proved that if p : S, local difleornorphism of a complete surface S, onto a surface S,, which is lengthincreasing, then p is a covering map.

The following proposition, known as the Hadamard theorem, describes the topological structure of a complete surface with curvature K _( 0.
Covering Spaces; The Theorems o f Hadamard
387
THEOREM 1 (Hadamard). Let S be a simply connected, complete surface, with Gaussian curvature K (0. Then exp,: T,(S) + S, p E S, is a dzfeomorphism; that is, S is difeornorphic to a plane. ProoJ: By Prop. 7 , exp,: T,(S) + S is a covering map. By the corollary of Prop. 5, exp, is a homeomorphism. Since exp, is a local diffeomorphism, Q.E.D. its inverse map is differentiable, and exp, is a diffeomorphism.
We shall now present another geometric application of the covering spaces, also known as the Hadamard theorem. Recall that a connected, compact, regular surface, with Gaussian curvature K > 0, is called an ovaloid (cf. Remark 1, Sec. 52).

THEOREM 2 (Hadamard). Let ,S be an ovaloid. Then the Gauss map N:S S2 is a difleomorphism. In particular, S is dzfeomorphic to a sphere. Proof. Since for every p E S the Gaussian curvature of S, K = det(dN,), is positive, N is a local diffeomorphism. By Prop. 1, N is a covering map. Since the sphere S2 is simply connected, we conclude from the corollary of Prop. 5 that N: S +S2is a homeomorphism of S onto the unit sphere S 2 . Since N is a local diffeomorphism, its inverse map is differentiable. Therefore, N is a diffeomorphism. Q.E.D. Remark 2. Actually, we have proved somewhat more. Since the Gauss map N is a diffeomorphism, each unit vector v of R3 appears exactly once as a unit normal vector to S. Taking a plane normal to v, away from the surface, and displacing it parallel to itself until it meets the surface, we conclude that S lies on one side of each of its tangent planes. This is expressed by saying that an ovaloid S is locally convex. It can be proved from this that S i s actually the boundary of a convex set (that is, a set K c R3 such that the line segment joining any two points p, q E K belongs entirely to K). Remark 3. The fact that compact surfaces with K > 0 are homeomorphic to spheres was extended to compact surfaces with K )0 by S. S. Chern and R. K. Lashof ("On the Total Curvature of Immersed Manifolds," Michigan Math. J. 5 (1958), 512). A generalization for complete surfaces was first obtained by J. J. Stoker ("Uber die Gestalt der positiv gekriimnten offenen Flache," Compositio Math. 3 (1936), 5889), who proved, among other things, the following: A complete surface with K > 0 is homeomorphic to a sphere or aplane. This result still holds for K 0 if one assumes that at some point K > 0 (for a proof and a survey of this problem, see M. do Carmo and E. Lima, "lsometric immersions with Nonnegative Sectional Curvatures," Boletim da Soc. Bras. Mat. 2 (1971), 922)
>
Global Differential Geometry
1. Show that the map n : R  S1= ((x, y) E R 2 ;x2 by n(t)= (cos t, sin t), t E R, is a covering map.
+ y2
=
1) that is given
2. Show that the map n : R 2  {0, 0) t R2  (0,O) given by
is a twosheeted covering map. 3. Let S be the helicoid generated by the normals to the helix (cos t, sin t, bt). Denote by L the z axis and let n : S  L +R2  (0,O) be the projection n(x, y, z) = (x, y). Show that n is a covering map.
4. Those who are familiar with functions of a complex variable will have noticed that the map n in Exercise 2 is nothing but the map n(z) = z2 from C  {O} onto  {O];here is the complex plane and z E C. Generalize that by proving that the map n :  (01  (0) given by n(z) = zn is an nsheeted covering map.
a=
a= c

5. Let B c R3 be an arcwise connected set. Show that the following two properties are equivalent (cf. Def. 3): 1. For any pair of points p, q E B and any pair of arcs a,: 10, I]&+ B, a l : [0, I] + B, there exists a homotopy in B joining a, to a,. 2. For any p E B and any arc a: [0, I] + B with a(0) = a(l) = p (that is, a is a closed arc with initial and end point p) there exists a homotopy joining a to the constant arc a(s) = p, s E [O, I].

6. Fix a point p, E R2 and define a family of maps y,: R2 R2, t E [0, 11, by p r ( ~= ) tpo (1  f)p,p E R2. Notice that po(p) = p , yI(p) = P O . Thus, y, is a continuous family of maps which starts with the identity map and ends with the constant mappo. Apply these considerations to prove that R2 is simply connected.
+
7. a. Use stereographic projection and Exercise 6 to show that any closed arc on a sphere S 2 which omits at least one point of S2is homotopic to a constant arc. b. Show that any closed arc on S 2 is homotopic to a closed arc in S2 which omits at least one point. c. Conclude from parts a and b that S 2 is simply connected. Why is part b necessary ? 8. (Klingenberg's Lemma.) Let S R3 be a complete surface with Gaussian curvature K KO,where KOis a nonnegative constant. Let p, q E S and let yo and y l be two distinct geodesics joiningp to q, with [(yo) I .I(y,); here I( ) denotes the length of the corresponding curve. Assume that yo is homotopic to y l ;i.e., there exists a continuous family of curves a,, t E [O, 11, joining p to q with
0 there exists a t ( ~ such and El(,, contains points at a distance ( E from the boundary of B. Thus, w
d. Choose in part c a sequence of E's,{E,) 3 0, and consider a converging sub. the existence of a curve a,,, bo E [O, 11, such sequence of ( t ( ~ , ) )Conclude that
Global Differential Geometry
9. a. Use Klingenberg's lemma to prove that if S is a complete, simply connected surface with K (0, then exp,: T,(S) + S is onetoone. b. Use part a to give a simple proof of Hadamard's theorem (Theorem 1).
*lo. (Synge's Lemma.) We recall that a differentiable closed curve on a surface S is a differentiable map a: 10, I] 3 S such that a and all its derivatives agree at 0 and 1. Two differentiable closed curves ao, a l : [O, I] .+ S are freely homotopic if there exists a continuous map H : [0, I] x [0, 11 t S such that H(s, 0) = aa(s), H(s, 1) = a,(s), s E [O, I ] . The map H i s called afree homofopy (the end points are not fixed) between a. and a,. Assume that S is orientable and has positive Gaussian curvature. Prove that any simple closed geodesic on S is freely homotopic to a closed curve of smaller length. 11. Let S be a complete surface. A point p E S is called a pole if every geodesic y : [O, CO) 4S with ~ ( 0 = ) p contains no point conjugate t o p relative to y . Use the techniques of Klingenberg's lemma (Exercise 8) to prove that if S is simply connected and has a pole p, then exp, : T,(S) S is a diffeomorphism. t
57 G/oba/ Theorems for Curves; The FaryMilnor Theorem In this section, some global theorems for closed curves will be presented. The main tool used here is the degree theory for continuous maps of the circle. To introduce the notion of degree, we shall use some properties of covering maps developed in Sec. 56. Let St = {(x, y ) E R 2 ;x2 y2 = I] and let n : R S1be the covering of S1by the real line R given by
+
Z(X) =.(COSx, sin x),
3
x
E
R.
Let p : S' + S1 be a continuous map. The degree of y, is defined as follows. We can think of the first S1 in the map y, : S1 t S1 as a closed interval [O, I ] with its end points 0 and E identified. Thus, y, can be thought of as a continuous map y,: 10, 11 + S1, with p(0) = p(I) = p E S1.Thus, y, is a closed arc at p in S1 which, by Prop. 2 of Sec. 56, can be lifted into a unique arc @: [O, 11 3 R, starting at a point x E R with n(x) = p. Since x(@(O))= n(@(l)), the difference @ ( E )  @(O)is an integral multiple of 2n. The integer deg p given by @ ( I )  @(0)= (deg y,)2n is called the degree of p. Intuitively, deg y, is the number of times that p : 10, 21 .+ S1 "wraps" 10, I ] around Sf (Fig. 530). Notice that the function @: [O, I ] + R is a con
39 1
G/oba/ Theorems for Curves
Figure 530
tinuous determination of the positive angle that the fixed vector p(0)  0 makes with q(t)  0, t E [0, I], 0 = (0,O)e. g., the map n : S1+ S1 described in Example 4 of Sec. 56, Part A, has degree k. We must show that the definition of degree is independent of the choices of p and x. First, deg p is independent of the choice of x. In fact, let x , > x be a point in R such that n(x,) = p, and let @,(t)= @(t)$ (x,  x), t E [0, I]. Since x1  x is an integral multiple of 2n, @, is a lifting of p starting at x,. By the uniqueness part of Prop. 2 of Sec. 56, @, is the lifting of starting at x,.Since
the degree of is the same whether computed with x or with x,. Second, deg p is independent of the choice of p E S1. In fact, each point p, E S1, except the antipodal point of p, beIongs to a distinguished neighborhood U, ofp. Choose x,, in the connected component of 7r1(Ul)containing x, such that n(x,) = p , , and let @, be the lifting of
starting at x,. Clearly, 1 @,(O) @(0)I < 27r. It follows from the stepwise process through which liftings are constructed (cf. the proof of Prop. 2, Sec. (I)  @(I)I < 2n. Since both differences @(I) @(0),@ , ( I )  @,(0) 56) that I @, must be integral multiples of 2n;, their values are actually equal. By continuity, the conclusion also holds for the antipodal point of p, and this proves our claim. The most important property of degree is its invariance under homotopy. More precisely, let p l , p2 : S1' S1 be continuous maps. Fix a point p E S1, thus obtaining two closed arcs at p, p l , p2: [0, E] . S1, ~ ~ ( = 0 ~) ~ ( =0 p.) If tpl and p2 are homotopic, then deg p1 = deg 01,. This follows immediately
392
Global Differential Geometry
from the fact that (Prop. 4, Sec. 56) the Iiftings of p1 and y,, starting from a fixed point x E R are homotopic, and hence have the same end points. It should be remarked that if y,: [O, 11 3 S1is differentiable, it determines differentiable functions u = a(t), b = b(t), given by y,(t) = (a(t), b(t)), which satisfy the condition a 2 I b2 = 1. In this case, the lifting p, starting at @, = x, is precisely the differentiable function (cf. Lemma 1, Sec. 44) @(O= do
+ Sto (abf

bar) dt.
This follows from the uniqueness of the lifting and the fact that cos @ ( I ) =a(t), sin @(t)= b(t), @(O) = @,. Thus, in the differentiable case, the degree of p can be expressed by an integral,
In the latter form, the notion of degree has appeared repeatedly in this book. For instance, when v: U c R2 3 R2, U 3 S1, is a vector field, and (0, 0) is its only singularity, the index of v at (0, 0) (cf. Sec. 45, Application 5) may be interpreted as the degree of the map y,: S1+ S1 that is given by $ 4 ~=)v(p)lt v(p)I, P S1. Before going into further examples, let us recall that a closed (differentiR3 (or R2, if it is a plane curve) able) curve is a differentiable map a : [0, I] such that the components of a , together with all its derivatives, agree at 0 and 1. The curve a is regular if a'(t) # 0 for all t E [O, I], and a is simple if whenever t1 t,, t,, t2 E [0, I), then a@,) a(t2). Sometimes it is convenient to assume that a is merely continuous; in this case, we shall say explicitly that a is a continuous closed curve. 4
+
+

Example 1 (The Winding Number of a Curve). Let a : [0, I] R2 be a plane, continuous closed curve. Choose a point p, E R2, p0 6 a([O, I]), and let y,: [0, I] + S1be given by
Clearly ~ ( 0 = ) y,(E), and y, may be thought of as a map of S1into S 1 ; it is called the position map of a relative to p,. The degree of y, is called the winding number (or the index) of the curve a relative t o p , (Fig. 531). Notice that by moving p, along an arc P which does not meet a([O, I]) the winding number remains unchanged. Indeed, the position maps of a relative to any two points of p can clearly be joined by a homotopy. It follows that the winding number of a relative to q is constant when q runs in a connected component of R2  a([O, I]).
Global Theorems for Curves
cu(t)  t'o
S'
Figure 531
Example 2 (The Rotation Index of a Curve). Let a : [0, I] +R2 be a regular plane closed curve, and let y, : [0, I] .S1 be given by
Clearly y, is differentiable and p(0) = ~ ( 1 ) y, . is called the tangent map of a , and the degree of y, is called the rotation index of a. Intuitively, the rotation index of a closed curve is the number of complete turns given by the tangent vector field along the curve (Fig. 127, Sec. 17). It is possible to extend the notion of rotation index to piecewise regular curves by using the angles at the vertices (see Sec. 45) and to prove that the rotation index of a simple, closed, piecewise regular curve is &1 (the theorem of turning tangents). This fact is used in the proof of the GaussBonnet theorem. Later in this section we shall prove a differentiable version of the theorem of turning tangents. Our first global theorem will be a differentiable version of the socalled Jordan curve theorem. For the proof we shall presume some familiarity with the material of Sec. 27. THEOREM 1 (Differentiable Jordan Curve Theorem). Let a : [0, I] + R2 be a plane, regular, closed, simple curve. Then R2  al([O, 11) has exactly two connected components, and a([O, I]) is their common boundary.
Proox Let N,(a) be a tubular neighborhood of a([O, El). This is constructed in the same way as that used for the tubular neighborhood of a compact surface (cf. Sec. 27). We recall that N,(a) is the union of open normal segments I,(t), with length 26 and center in a(t). Clearly, N,(a)  a([O, 11) has two connected components T I and T,. Denote by w(p) the winding number of a relative to p E R2  a([O, 11). The crucial point of the proof is to show that if both p, and p2 belong to distinct connected components of Nc(a)  a([O, I]) and to the same I,(to), to E [0, I ] , then w(p,)  w(p,) = rt 1, the sign depending on the orientation of a.
394
Global Differential Geometry
Choose points A = a(t,), D = a@,), t, < to < t,, so close to to that the arc A D of a can be deformed homotopically onto the polygon ABCD of Fig. 532. Here BC is a segment of the tangent line at a(t), and BA and C D are parallel to the normal line at &(to). Tubular
,neighborhood
3
Orientation of the plane

Figure 532
Let us denote by 8: [0, I] R2 the curve obtained from a by replacing the ) A and arc A D by the polygon ABCD, and let us assume that P(0) = ~ ( i = that P(t,) = D. Clearly, w(p,) and w(p,) remain unchanged. S1be the position maps of p relative to p , , p,, respecLet p,, p,: [0, i] R be their liftings from a fixed tively (cf. Example I), and let Q,, @,: [0, i] point, say O E R. For convenience, let us assume the orientation of p to be given as in Fig. 532. We first remark that if t t [t,, i],the distances from a(t) to bothp, andp, remain bounded below by a number independent of t, namely, the smallest of the two numbers dist(p,, Bd N,(a))and dist(p,, Bd N,(a)). It follows that the angle of a(t)  p, with a(r)  p, tends uniformly to zero in [I,, i] as p, approaches p, . Now, it is clearly possible to choose p, and p, so close to each other that


395
Global Theorems for Curves
pl(t3)  gl(0) = R  c I , and P2(t3) @,(0) = (R smaller than n/3. Furthermore,
+ c,),
with c , and c,
By the above remark, the first term can be made arbitrarily small, say equal to < 7~13,if pl is sufficiently close to p,. Thus,
fS
where 6 < 7~ ifpl is sufficiently close top,. It follows that w(p,)  w(p2) = I, as we had claimed. It is now easy to complete the proof. Since w(p) is constant in each connected component of R2  a([O, El) = W, it follows from the above that there are at least two connected components in W. We shall show that there are exactly two such components. In fact, let C be a connected component of W. Clearly Bd C # q5 and Bd C c a([O, I]). On the other hand, if p E a([O, El), there is a neighborhood of p that contains only points of a([O, I]), points of TI, and points of T, (TI and T, are the connected components of N,(a)  a([O, I])). Thus, either TI or T, intersect C. Since C is a connected component, C 3 TI or C 3 T,. Therefore, there are at most two (hence, exactly two) connected components of W. Denote them by C , and C,. The argument also shows that Q.E.D. Bd C1 = &([O,El) = Bd C2. The two connected components given by Theorem 1 can easily be distinguished. One starts from the observation that if p, is outside a closed disk D containing a([O, El) (since [0, I] is compact, such a disk exists), then the winding number of a relative t o p, is zero. This comes from the fact that the Iines joining p, to a(t), t E [O, El, are all within a region containing D and bounded by the two tangents from p, to the circle Bd D.Thus, the connected component with winding number zero is unbounded and contains all points outside a certain disk. Clearly the remaining connected component has winding number A 1 and is bounded. It is usual to call them the exterior and the interior of a , respectively. Remark I . A useful complement to the above theorem, which was used in the applications of the GaussBonnet theorem (Sec. 49, is the fact that the interior of a is homeomorphic to an open disk. A proof of that can be found in J. J. Stoker, Diflerential Geometry, Wiieyinter science, New York, 1969, pp. 4345.
396
Global Differential Geometry
We shall now prove a differentiable version of the theorem of turning tangents.
THEOREM 2. Let P : [0, 11 ,R 2 be a plane, regular, simple, closed curve. Then the rotation index of is Ilfi 1 (depending on the orientation of P). Proof. Consider a line that does not meet the curve and displace it parallel to itself until it is tangent to the curve. Denote by E this position of the line and b y p a point of tangency of the curve with l. Clearly the curve is entirely R2 on one side of I (Fig. 533). Choose a new parametrization a: [0, I ] for the curve so that a(0) = p. Now let +
be a triangle, and define a "secant map" y: T +S1 by a(t2)  a(t1) v(t17t 2 ) = I a(tp)  a(tl) 1
for t1
+ t2, (t,, t2) E T  [(o, z)}
Figure 533
Since a is regular, ry is easily seen to be continuous. Let A = (0, O), B = (0, I), C = (I, 1) be the vertices of the triangle T. Notice that ry restricted to the side AC is the tangent map of a, the degree of which is the rotation number of a. Clearly (Fig. 533), the tangent map is homotopic to the restriction of ry to the remaining sides AB and BC. Thus, we are reduced to show that the degree of the latter map is 41 1.
397
G/oba/ Theorems for Curves
Assume that the orientations of the plane and the curve are such that the oriented angle from a'(0) to a'(O) is n. Then the restriction of y to AB covers half of S1 in the positive direction, and the restriction of v to BC covers the remaining half also in the positive direction (Fig. 533). Thus, the degree of ty restricted to AB and BC is 1. Reversing the orientation, we shall obtain  1 for this degree, and this completes the proof. Q.E.D.
+
The theorem of turning tangents can be used to give a characterization of an important class of curves, namely the convex curves. A plane, regular, closed curve a: [0, 11 R2 is convex if, for each t E [O, I ] , the curve lies in one of the closed halfplanes determined by the tangent line at t (Fig. 534; cf. also Sec. 17). If a is simple, convexity can be expressed in terms of curvature. We recall that for plane curves, curvature always means the signed curvature (Sec. 15, Remark 1). 4
Convex curve
Nonconvex curves
Figure 534
PROPOSITION 1. A plane, regular, closed curve is convex if and only if it is simple and its curvature k does not change sign. Proof. Let q : [0, 11 ,S1 be the tangent map of a and @: [O, I] ,R be the lifting of p starting at 0 E R. We first remark that the condition that k does not change sign is equivalent to the condition that (5 is monotonic (nondecreasing if k 2 0, or nonincreasing if k 0). Now, suppose that a is simple and that k does not change sign. We can orient the plane of the curve so that k 2 0. Assume that a is not convex. Then there exists to E [O,l] such that points of a([O, I]) can be found on both sides of the tangent line T a t a(to). Let n = n(t,) be the normal vector at to,and set
Since [O, l] is compact and both sides of T contain points of the curve, the "height function" h, has a maximum at t, + to and a minimum at t , # t o .
Global Differential G e o m e t r y
398
The tangent vectors at the points to, t,, t, are all parallel, so two of them, say af(t0),af(t1),have the same direction. It follows that p(to) = p(tl) and, by Theorem 2 (a is simple), @(to)= @(t,). Let us assume that t, > to. By the above remark, @ is monotonic nondecreasing, and hence constant in [to,t,]. This means that a([t,, t,]) c T. But this contradicts the choice of T and shows that a is convex. Conversely, assume that a is convex. We shalI leave it as an exercise to show that if a is not simple, at a selfintersection point (Fig. 535(a)), or nearby it (Fig. 535(b)), the convexity condition is violated. Thus, a is simpIe.
Figure 535
We now assume that a is convex and that k changes sign in [0, I]. Then there are points t,, t , E [0, 11, t, < t,, with @(t,)= @(t,) and @ not constant in [I,, t2I. We shall show that this leads to a contradiction, thereby concluding the proof. By Theorem 2, there exists t3 E [O, I ) with p(t3) = p(t,). By convexity, two of the three parallel tangent lines at a@,),a(t,), a(t,) must coincide. Assume this to be the case for a(t,) = p, a(t,) = q, t3 > t l . We claim that the arc of a between p and q is the line segment pq. In fact, assume that r # q is the last point for which this arc is a Iine segment (r may agree with p). Since the curve lies in the same side of the line pq, it is easily seen that some tangent T near p will cross the segment in an interior point (Fig. 536). Thenp and q lie on distinct sides of T. That is a contradiction and proves our claim. It follows that the coincident tangent,lines have the same directions; that
Figure 536
399
Global Theorems for Curves
is, they are actually the tangent lines at a(t,) and a@,).Thus, @ is constant in [t,, t,], and this contradiction proves that k does not change sign in LO, 11. Q.E.D. Remark 2. The condition that a is simple is essential to the proposition, as shown by the example of the curve in Fig. 534(c). Remark 3. The proposition should be compared with Remarks 2 and 3 of Sec. 56; there it is stated that a similar situation hoIds for surfaces. It is to be noticed that, in the case of surfaces, the nonexistence of selfintersections is not an assumption but a consequence. Remark 4. It can be proved that a plane, regular, closed curve is convex if and only if its interior is a convex set K c RZ (cf. Exercise 4). We shall now turn our attention to space curves. In what follows the word curve will mean a parametrized regular curve a : LO, I] R3 with arc length s as parameter. If a is a plane curve, the curvature k(s) is the signed curvature of a (cf. Sec. 15); otherwise, k(s) is assumed to be positive for all s E [0, 21. It is convenient to call t
the total curvature of a. Probably the bestknown global theorem on space curves is the socalled Fenchel's, theorem.
THEOREM 3 (Fenchel's Theorem). The total curvature of a simple closed curve is 2 2n, and equality holds $and only if the curve is a plane convex curve. Before going into the proof, we shall introduce an auxiliary surface which is also useful for the proof of Theorem 4. The tube of radius r around the curve a is the parametrized surface
where n = n(s) and b = b(s) are the normal and the binormal vector of or: respectively. It is easily checked that
We assume that r is so small that rk, < 1, where k, < max I k(s) 1, s Then x is regular, and a straightforward computation gives
E
[O, I].
G/oba/ Differential Geometry
+
(n cos v b sin v), X, A xu = r(1  rk cos v)N, N, A N , = k cos v(~2cos v b sin v) = k N cos v k cos v xu A x,. r(1  rk cos V) N
=
+
It follows that the Gaussian curvature K K(s,
V) =

= K(s,
v) of the tube is given by
k cos v r(l  rk cos v)'
Notice that the trace T of x may have selfintersections. However, if a is simple, it is possible to choose r so small that this does not occur; we use the compactness of [0, El and proceed as in the case of a tubular neighborhood constructed in Sec, 27. If, in addition, a is closed, T is a regular surface homeomorphic to a torus, also called a tube around a. In what follows, we assume this to be the case. Proof of Theorem 3. Let T be a tube around a , and let R c T be the region of T'where the Gaussian curvature of T i s nonnegative. On the one hand,
On the other hand, each halfline L through the origin of R3 appears at least once as a normal direction of R. For if we take a plane P perpendicular to L such that P n T = 4 and move P parallel to itself toward T (Fig. 537), it will meet T for the first time at a point where K 2 0. It follows that the Gauss map N of R covers the entire unit sphere S2at least once; hence,
ISRK d a 2 4n. Therefore, the total curvature of a is 2 2n,
and we have proved the first part of Theorem 3. Observe that the image of the Gauss map N restricted to each circle s = const. is onetoone and that its image is a great circle r, c S2. We shall denote by I',*c r, the closed halfcircle corresponding to points where K > 0. Assume that a is a plane convex curve. Then all T,* have the same end points p, q, and, by convexity, r,,n I?,, = (p} U {q} for s, # s2,s,, s2 E [O, I ) . By the first part of the theorem, it follows that
K da
= 47z;
hence,
S S R
the total curvature of a is equal to 271. Assume now that the total curvature of a is equal to 271. By the first part
Global Theorems for Curves
Figure 537
of the theorem,
15 K d o
= 4n.
We claim that all T: have the same end
R
points p and q. Otherwise, there are two distinct great circles T,,, T,,, s, arbitrarily close to s, that intersect in two antipodal points which are not in N(R n Q), where Q is the set of points in T with nonpositive curvature. It follows that there are two points of positive curvature which are mapped by N into a single point of S 2 . Since N is a local diffeomorphism at such points and each point of S2 is the image of at least one point of R, we conclude that
ISRKO > 471, a contradiction. 
By observing that the points of zero Gaussian curvature in T a r e the intersections of the binormal of a with T, we see that the binormal vector of a is parallel to the line pq. Thus, a is contained in a plane normal to this line. We finally prove that a is convex. We may assume that a is so oriented that its rotation number is positive. Since the total curvature of a is 271, n.e have
Global Differential Geometry I
2n=J
o
I
l k l d s 2 J o kds.
On the other hand,
where J = {s E [0, El; k(s) 2 0). This holds for any plane closed curve and follows from an argument entirely similar to the one used for R c T i n the beginning of this proof. Thus, I
kds 0
Therefore, k
=
J l k l d s = 2n. 0
2 0, and a is a plane convex curve.
Q.E.D.
Remark 5. It is not hard to see that the proof goes through even if a is not simple. The tube wilI then have selfintersections, but this is irrelevant to the argument. In the last step of the proof (the convexity of a), one has to observe that we have actually shown that a is nonnegatively curved and that its rotation index is equal to 1. Looking back at the first part of the of Prop. 1, one easily sees that this implies that a is convex. We want to use the above method of proving Fenchel's theorem to obtain a sharpening of this theorem which states that if a space curve is knotted (a concept to be defined presently), then the total curvature is actually greater than or equal to 471. A simple closed continuous curve C c R3 is unknotted if there exists a homotopy H: S1 x I+ R3, I = [O, 11, such that
and
H(S1 x (0))
= S1
H(S1 x {I))
=
C;
H(S1 x {t)) = C, c R3
is homeomorphic to S1for all t E [O, I]. Intuitively, this means that C can be deformed continuously onto the circle S1 so that all intermediate positions are homeomorphic to S1.Such a homotopy is called an isotopy; an unknotted curve is then a curve isotopic to S'. When this is not the case, C i s said to be knotted (Fig. 538). THEOREM 4 (FaryMilnor). The total curvature of a knotted simple closed curve is greater than or equal to 4 ~ .
Proof. Let C = a([O, l ] ) ,let T be a tube around a, and let R c T be the region of T where K 2 0. Let b = b(s) be the binormal vector of a, and let
G/oba/ Theorems for Curves
Unknotted
Knotted
Figure 538
v E R3 be a unit vector, v # b(s), for all s E [0, I]. Let h,: [0, 11 +R be the height function of a in the direction of v; that is, h,(s) = (a(s)  O,v), s E [0, I]. Clearly, s is a critical point of /I, if and only if v is perpendicular to the tangent line at a(s). Furthermore, at a critical point,
since v # b(s) for all s and k > 0. Thus, the critical points of h, are either maxima or minima. Now, assume the total curvature of a to be smaller than 4s. This means that
We claim that, for some v, 6 b([O, I]), h,, has exactly two critical points (since 10, I] is compact, such points correspond to the maximum and minimum of h,,). Assume that the contrary is true. Then, for every v $ b([O, El), h, has at least three critical points. We shall assume that two of them are points of minima, s , and s, the case of maxima being treated similarly. Consider a plane P perpendicular to v such that P n T = 4, and move it paralIeI to itself toward T. Either h,(s,) = h,(s,) or, say, h,(s,) < h,(s,). In the first case, P meets T at points q, # q,, and since v $ b([O, I]), K(q,) and K(q,) are positive. In the second case, before meeting a@,),P will meet T a t a point q, with K(q,) > 0. Consider a second plane F, parallel to and at a distance r above P (r is the radius of the tube T). Move 13 further up until it reaches a(s2); then P will meet T a t a point q, # q, (Fig. 539). Since s, is a point of minimum and v $ b([0,I]), K(q,) > 0. In any case, there are two distinct points in T with K > 0 that are mapped by N into a single point of
S2.This contradicts the fact that J J K do < 8s, and proves our claim. R Let s , and s, be the critical points of h,,, and let P, and P2 be planes perpendicular to v, and passing through a(s,) and a(s,), respectively. Each plane parallel to v, and between P, and P2will meet C in exactly two points. Joining these pairs of points by line segments, we generate a surface bounded by C
Global Differential Geometry
which is easily seen to be homeomorphic to a disk. Thus, C is unknotted, and Q.E.D. this contradiction completes the proof.
EXERCISES 1. Determine the rotation indices of curves (a), (b), (c),and (d) in Fig. 540.
Figure 540
Global Theorems for Curves
405

2. Let a(t) (x(t), y(t)), t E [0, I ] , be a differentiable plane closed cun,e. Let PO= (xO,yo) E RZ, (x0,yo) $ a([O, I]), and define the functions
a. Use Lemma 1 of Sec. 44 to show that the differentiable function
is a determination of the angle that the x axis makes with the position vector ~> Po I. (a([>  P O ) I I ~ (b. Use part a to show that when a is a differentiable closed plane curve, the winding number of a relative to po is given by the integral
w
1
I
2R
0
(ob'  bd) dr.
=
3. Let a : [0, I] + R 2 and p : [O, I ] + R2 be two differentiable plane closed curves, and let po E R2 be a point such that po $ a([O, I]) and p, $ P([O, 11). Assume that, for each t E [0, I], the points a(t) and P(t) are closer than the points a(?) and p, ; i.e.,
1 a(t)  P(t) I < I a(t)  Po I. Use Exercise 2 to prove that the winding number of a relative to po is equal to the winding number of P relative to g o . 4. a. Let C be a regular plane dosed convex curve. Since C is simple, it determines, by the Jordan curve theorem, an interior region K c R2. Prove that K is a convex set (i.e., given p, q E K, the segment of straight linegq is contained in K; cf. Exercise 9, Sec. 17). b. Conversely, let C be a regular plane curve (not necessarily closed), and assume that C is the boundary of a convex region. Prove that C is convex. 5. Let C be a regular plane, closed, convex curve. By Exercise 4, the interior of C is a convex set K. Letpo E K, po 4 C. a. Show that the line which joins po to an arbitrary point q E C is not tangent to C at q. b. Conclude from part a that the rotation index of Cis equal to the winding number of C relative to po. c. Obtain from part b a simple proof for the fact that the rotation index of a closed convex curve is f1.
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Global Differential Geometry

6. Let a : [0, I] R 3 be a regular closed curve parametrized by arc length. Assume 1 for all s E [0, I]. Prove that 1 2 2 and that 1 = 2n if and that 0 # ] k(s) 1 I only if a is a plane convex curve.
7. (Schur9sTheorem for Plane Curves.) Let a: [0, I] + R 2 and 8 : [O, I] + R Zbe two plane convex curves parametrized by arc length, both with the same length I. Denote by k and the curvatures of a and E , respectively, and by d and d the lengths of the chords of a and &, respectively; i.e,,
Assume that k(s) 2 k(s), s E [0, I ] . We want to prove that d(s) ( d(s), s E 10, I] (i.e., if we stretch a curve, its chords become longer) and that equality holds for s E [O, I] if and only if the two curves differ by a rigid motion. We remark that the theorem can be extended to the case where E is a space curve and has a number of applications. Compare S. S. Chern [lo]. The following outline may be helpful.
a. Fix a point s = s l . Put both curves a(s) = (x(s), y(s)), &(s) = (X(s), Y(s)) in , and &(s,),lie on the x the lower halfplane y 5 0 so that a(O), ~ ( s , ) &(0), axis and x(sl) > x(O), X(sl) > X(0) (see Fig. 541). Let so E [O, s,] be such that a'(s,) is parallel to the x axis. Choose the function O(s) which gives a differentiable determination of the angle that the x axis makes with ar(s) in such a way that &so) = 0. Show that, by convexity, n < 6) < n.
Figure 541
b. Let 8(s), 8(so) = 0, be a differentiable determination of the angle that the x axis makes with a'(s). (Notice that &'(so) may no longer be parallel to the x axis.) Prove that Q(s) 1 6 0 ) and use part a to conclude that
For the equality case, just trace back your steps and apply the uniqueness theorem for plane curves.
a : R t R 2 be a regular plane curve parametrized by arc length. Assume that a satisfies the following conditions:
8. (Stoker's Theorem for Plane Curves.) Let
407
Global Theorems for Curves
1 . The curvature of
a is strictly positive.
2. lim I a(s) I
that is, the curve extends to infinity in both directions.
=
m;
s++m
3. a has no selfintersections. The goal of the exercise is to prove that the total curvature of a is (n. The following indications may be helpful. Assume that the total curvature is > n and that a has no selfintersections. To obtain a contradiction, proceed as follows :
a. Prove that there exist points, say, p = a(O), q = a(sl), s l > 0, such that the tangent lines Tp, T, at the points p and q, respectively, are parallel and there exists no tangent line parallel to Tp in the arc a([O, sJ). b. Show that as s increases, a(s) meets T, at a point, say, r (Fig. 542). c. The arc a((my 0)) must meet T, at a point t between p and r. d. Complete the arc tpqr of a with an arc without selfintersections joining r to t , thus obtaining a closed curve C. Show that the rotation index of C is >_ 2. Show that this implies that a has selfintersections, a contradiction.
P
Figure 542
'9. Let a : [O,1] 3 S be a regular closed curve oh a sphere S = {(x, y, Z ) E R 3; x 2 y 2 Iz2 = 13. Assume that a is parametrized by arc length and that the curvature k(s) is nowhere zero. Prove that
+
(The above integral is actualIy a sufficient condition for a nonplanar curve to lie on the surface of a sphere. For this and related results, see H. Geppert, "Sopra
408
Global Differential Geometry
una caractterizzazione della sfera," Ann. di Mat. Pura ed App. XX (1941), 5966; and B. Segre, "Una nuova caracterizazzione della sfera," Atti Accad. Naz. dei Lincei 3 (1947), 420422.)
58. Surfaces of Zero Gaussian Curvature We have already seen (Sec. 46) that the regular surfaces with identically zero Gaussian curvature are locally isometric to the plane. In this section, we shall look upon such surfaces from the point of view of their position in R3 and prove the following global theorem.
THEOREM. Let S c R3 be a complete surface with zero Gaussian curvature. Then S is a cylinder or a plane. By definition, a cylinder is a regular surface S such that through each point p E S there passes a unique line R(p) c S (the generator through p) which satisfies the condition that if q + p, then the lines R(p) and .R(q) are parallel or equal. It is a strange fact in the history of differential geometry that such a theorem was proved only somewhat late in its development. The first proof came as a corollary of a theorem of P. Hartman and' L. Nirenberg ("On Spherical Images Whose Jacobians Do Not Change Signs," Amer. J. Math. 81 (1959), 901920) dealing with a situation much more general than ours. Later, W. S. Massey ('"urfaces of Gaussian Curvature Zero in Euclidean Space," Tohoku Math. J. 14 (I962), 7379) and J. J. Stoker ("Developable Surfaces in the Large," Cornm. Pure and Appl. Math. 14 (1961), 627635) obtained elementary and direct proofs of the theorem. The proof we present here is a modification of Massey's proof. It should be remarked that Stoker's paper contains a slightly more general theorem. We shall start with the study of some local properties of a surface of zero curvature. Let S c R3 be a regular surface with Gaussian curvature K =. 0. Since K = klk2, where k, and k , are the principal curvatures, the points of S are either parabolic or planar points. We denote by P the set of planar points and by U = S  P the set of parabolic points of S. P is dosed in S. In fact, the points of P satisfy the condition that the mean curvature H = +(k, k,) is zero. A point of accumulation of P has, by continuity of H, zero mean curvature; hence, it belongs to P. It follows that U = S  P is open in S. An instructive example of the relations between the sets P and U is given in the following example.
+
Surfaces o f Zero
Gaussian Curvature
409
Example I. Consider the open triangIe ABC and add to each side a cylindrical surface, with generators parallel to the given side (see Fig. 543). It is possible to make this construction in such a way that the resulting surface is a regular surface. For instance, to ensure regularity along the open segment BC,it suffices that the section FG of the cylindrical band BCDE by a plane normal to BC is a curve of the form
exp
(

.
2>:
Observe that the vertices A, B,C of the triangle and the edges BE,CD,etc., of the cylindrical bands do not belong to S.

Figure 543
The surface S so constructed has curvature K 0. The set P is formed by the closed triangle ABC minus the vertices. Observe that P is closed i n S but not in R3. The set U is formed by the points which are interior to the cylindrical bands. Through each point of U there passes a unique line which will never meet P. The boundary of P is formed by the open segments AB, BC,and CA. In the following, we shall prove that the relevant properties of this example appear in the general case. First, let p E U. Since p is a parabolic point, one of the principal directions at p is an asymptotic direction, and there is no other asymptotic direction at p. We shall prove that the unique asymptotic curve that passes throush p is a segment of a line.
PROPOSITION 1. The unique asymptotic line that passes through a parabolic point p E U c S of a surface S of curvature K = 0 is an (open) segment of a (straight) line in S.
410
Global Differential Geometry
Proof. Since p is not umbilical, it is possible to parametrize a neighborhood V c U of p by x(u, v) = x in such a way that the coordinate curves are lines of curvature. Suppose that v = const. is an asymptotic curve; that is, it has zero normal curvature. Then, by the theorem of Olinde Rodrigues (Sec. 32, Prop. 4), Nu = 0 along v .= const. Since through each point of the neighborhood V there passes a curve v = const., the relation N , = 0 holds for every point of V. It follows that in V
(x, N),
=
(xu, N ) 4 {x, Nu)
= 0.
Therefore, (x, N)
= yl(v),
(1)
where p(v) is a differentiable function of v alone. By differentiating Eq. (1) with respect to v, we obtain
On the other hand, N , is normal to N and different from zero, since the points of V are parabolic. Therefore, N and Nu are linearly independent. Furthermore, Nu, = Nu, = 0 in V. We now observe that along the curve v = const. = v, the vector N(u) = No and Nu(u) = (Nu), = const. Thus, Eq. (I) implies that the curve x(t/, v,) belongs to a plane normal to the constant vector No, and Eq. (2) implies that this curve belongs to a plane normal to the constant vector (Nu),. Therefore, the curve is contained in the intersection of two planes (the intersection exists since No and (Nu), are linearly independent); hence, it is a segment of a line. Q.E.D.
Remark. It is essential that K = 0 in the above proposition. For instance, the upper parallel of a torus of revolution is an asymptotic curve formed by parabolic points and it is not a segment of a line. We are now going to see what happens when we extend this segment of line. The following proposition shows that (cf. Example 1) the extended line never meets the set P ;either it "ends" at a boundary point of S or stays indefinitely in U. It is convenient to use the following terminology. An asymptotic curve passing through a point p t S is said to be maximal if it is not a proper subset of some asymptotic curve passing through p.
PROPOSITION 2 (Massey, loc. cit.). Let r be a maximal asymptotic line passing through a parabolic point p E U cr S of a surface S of curvature K = 0 and let P c S be the set of planar points of S. Then r n P = 4.
411
Surfaces of Zero Gaussian Curvature
The proof of Prop. 2 depends on the following local lemma, for which we use the MainardiCodazzi equations (cf. Sec. 43).
L E M M A 1, Let s be the arc length of the asymptotic curve passing through a parabolic point p of a surface S of zero curvature and let H = H(s) be the mean curvature of S along this curve. Then, in U ,
Proof of Lemma 1. We introduce in a neighborhood V c U of p a system of coordinates (u, V) such that the coordinate curves are lines of curvature and the curves v = const. are the asymptotic curves of V. Let e , f , and g be the coefficients of the second fundamental form in this parametrization. Since f = 0 and the curve v = const., u = u(s) must satisfy the differential equation of the asymptotic curves
we conclude that e given by
= 0.
Under these conditions, the mean curvature H is
By introducing the values F =f = e = 0 in the MainardiCodazzi equations (Sec. 41, Eq. (7) and (7a)), we obtain
From the first equation of (4) it follows that E, = 0. Thus, E = E(u) is a function of u alone. Therefore, it is possible to make a change of parameters: v'
= v,
1.7
=
,/E(u)
du.
We shaII still denote the new parameters by u and v. u now measures the arc length along v = const., and thus E = 1. In the new parametrization ( F = 0, E = 1) the expression for the Gaussian curvature is
412
Global Differential Geometry
Therefore,
JC
= cI(v)u
+ cz(~),
where c,(v) and c,(v) are functions of v alone. On the other hand, the second equation of (4) may be written (g # 0)
hence, g 'c , ( v ) J F , where c,(v) is a function of v. By introducing Eqs. (5) and (6) into Eq. (3) we obtain
Finally, by recalling that u = s and differentiating the above expression with respect to s, we conclude that
Q.E.D. Proof of Prop. 2. Assume that the maximal asymptotic line r passing through p and parametrized by arc length s contains a point q E P. Since r is connected and U is open, there exists a point po of r, corresponding to so, such that p, E P and the points of r with s < so belong to U. On the other hand, from Lemma 1, we conclude that along r and for S
< So,
where a and b are constants. Since the points of P have zero mean curvature, we obtain 1
H(p,) = 0 = lim H(s) a+So
= lim s+s,
1
as
which is a contradiction and concludes the proof.
+ b' Q.E.D.
Let now Bd(U) be the boundary of U in S ; that is, Bd(U) is the set of points p E S such that every neighborhood of p in S contains points of U and points of S  U = P. Since U is open in S, it follows that Bd(U) c P. Furthermore, since the definition of a boundary point is symmetric in U
Surfaces of Zero Gaussian Curvature
and P, we have that
The following proposition shows that (just as in Example 1) the set Bd(U) = BdfP) is formed by segments of straight lines. PROPOSITION 3 (Massey). Let p E Bd(U) c S be a point of the boundary of the set U of parabolic points of a surface S of curvature K =. 0. Then through p there passes a unique open segment of line C(p) c S. Furthermore, C(p) c Bd(U); that is, the boundary of U is formed by segments of lines.

Proof. Let p E Bd(U). Since p is a limit point of U, it is possible to choose a sequence {p,], pn E U, with lim,,, p, p. For every p,, let C(p,) be the unique maximal asymptotic curve (open segment of a line) that passes through pn (cf. Prop. 1). We shall prove that, as n + 00, the directions of C(p,) converge to a certain direction that does not depend on the choice of the sequence {p,]. In fact, let C c R3 be a sufficiently small sphere aroundp. Since the sphere C is compact, the points {q,) of intersection of C(pn)with C have at least one point of accumulation q E C, which occurs simultaneously with its antipodal point. If there were another point of accumulation r besides q and its antipodal point, then through arbitrarily near points p, and p, there should pass asymptotic lines C(pn) and C(p,) making an angle greater than
thus contradicting the continuity of asymptotic lines. It follows that the lines C(p,) have a limiting direction. An analogous argument shows that this limiting direction does not depend on the chosen sequence (p,] with limn,, p, = p, as previously asserted. Since the directions of C(p,) converge and p, 4 p, the open segments of lines C(p,) converge to a segment C(p) c S that passes through p. The segment C(p) does not reduce itself to the point p. Otherwise, since C(p,) is maximal, p E S would be a point of accumulation of the extremities of C(pn), which do not belong to S (cf. Prop. 2). By the same reasoning, the segment C(p) does not contain its extreme points. Finally, we shall prove that C(p) c Bd(U). In fact, if q E C(p), there exists a sequence
Then q E U u Bd(U). Assume that q $ Bd(U). Then q E U, and, by the continuity of the asymptotic directions, C(p) is the unique asymptotic line
414
Global Differential Geometry
that passes through q. This implies, by Prop. 2, that p E U, which is a contradiction. Therefore, q E Bd(U), that is, C(p) c Bd(U), and this concludes the proof. Q.E.D. We are now in a position to prove the global result stated in the beginning of this section. Proof of the Theorem. Assume that S is not a plane. Then (Sec. 32, Prop. 5) S contains parabolic points. Let U be the (open) set of parabolic points of S a n d P be the (closed) set of planar points of S. We shall denote by int P, the interior of P, the set of points which have a neighborhood entirely contained in P. int P is an open set in S which contains only planar points. Therefore, each connected component of int P is contained in a plane (Sec. 32, Prop. 5). We shall first prove that if q E S and q $ int P, then through q there passes a unique line R(q) c S, and two such lines are either equal or do not intersect. In fact, when q E U,then there exists a unique maximal asym'ptotic line r passing through q. r is a segment of line (thus, a geodesic) and r n P = 4 (cf. Props. 1 and 2). By parametrizing r by arc length we see that r is not a finite segment. Otherwise, there exists a geodesic which cannot be extended to all values of the parameter, which contradicts the completeness of S. Therefore, r is an entire line .R(q), and since r n P = 4, we conclude that R(q) c U. It follows that when p is another point of U, p $ R(q), then R(p) n R(q) = 4. Otherwise, through the intersection point there should pass two asymptotic lines, which contradicts the asserted uniqueness. On the other hand, if q E Bd(U) = Bd(P), then (cf. Prop. 3) through q there passes a unique open segment of line which is contained in Bd(U). By the previous argument, this segment may be extended into an entire line R(q) c Bd(U), and if p E Bd(U), p $ R(q), then R(p) n Rfq) = 4. Clearly, since U is open, if q E U and p E Bd(U), then R(p) n R(q) = 4. In this way, through each point of S  int P = U u Bd(U) there passes a unique line contained in 'S  int P, and two such lines are either equal or do not intersect, as we claimed. If we prove that these lines are parallel, we shall concIude that Bd(U) (= Bd(P)) is formed by parallel lines and that each connected component of int P is a n open set of a plane, bounded by two parallel lines. Thus, through each point t c int P there passes a unique line R(t) c int P parallel to the common direction. It follows that through each point of Sthere passes a unique generator and that the generators are parallel, that is, S is a cylinder, as we wish. To prove that the lines passing through the points of U u Bd(U) are parallel, we shall proceed in the following way. Let q E U u Bd(U) and p E U. Since S is connected, there exists an arc a : [0, I ] + S, with a(0) = p,
Jacobi's Theorems
475

a(1) = q. The map exp, : T,(S) S is a covering map (Prop. 7, Sec. 56) and T,(S) be a local isometry (corollary of Lemma 2, Sec. 56). Let &: [O,1] the lifting of a , with origin at the origin 0 E T,(S). For each &(t), with exp, 8(t) = a(t) E U U Bd(U), let r, be the lifting of R(a(1)) with origin at iZ(t). Since exp, is a local isometry, r, is a line in T,(S). We shall prove that when a(tl) + a(tz), t , , tz E [0, I], the lines r,, and r,, are parallel. In fact, if u E r,, n r,,, then +
which is a contradiction. So far we have not defined R(a(t)) when a(t) E int P. This will now be done. When E(t) is such that exp, &(t) = a(t) E int P, we draw through &(t) a line r in T,(S) parallel to the common direction we have just obtained. It is clear that exp,(r) c int P, and since exp,(r) is a geodesic, exp,(r) is an entire line contained in S. In this way, the line R(a(t)) is defined for every t E [0, 13. We shall now prove that the lines R(a(t)), t F [O, I], are parallel lines. In fact, by the usual compactness argument, it is possible to cover the interval [O, 11 with a finite number of open intervals I,, . . . , I, such that E(Ii) is contained in a neighborhood Vi of a([,), ti E Ii, where the restriction of exp, is an isometry in Vi. Observe now that when t,, t, E Ii and a(t,) + a(t2) then R(a(t,)) is parallel to R(a(t,)). In fact, since rll is parallel to rt, and exp, is an isometry in Vi, the open segment exp,(rtl n Vi)is parallel to exp,(rt, n V,); this means that the lines exp, r,, = R(a(t,)) and exp, r,, = R(a(tz)) have paralIe1 open segments and are therefore parallel. By then using the decomposition qf 10, I ] by I , , . . . ,I,, we shall prove, step by step, that the lines R(a(t)) are parallel. In particular, the line R(q) is parallel to R(p). If s is another point in U u Bd(U), then, by the same argument, R(s) is parallel to R(p) and hence parallel to R(q). In this way, it is proved that all the lines that pass through U u Bd(U) are parallel, and this concludes the proof of the theorem. Q.E.D.
59. Jacobi's Theorems It is a fundamental property of a geodesic y (Sec. 46, Prop. 4) that when two points p and q of y are sufficiently close, then y minimizes the arc length between p and q. This means that the arc length of y betweenp and q is smaller than or equal to the arc length of any curve joining p to q. Suppose now that we follow a geodesic y starting from a point p. It is then natural to ask how far the geodesic y minimizes arc length. In the case of a sphere, for instance, a geodesic y (a meridian) starting from a point p minimizes arc length up to
476
Global Differential Geometry
the first conjugate point o f p relative to y (that is, up to the antipodal point of p). Past the antipodal point of p, the geodesic stops being minimal, as we may intuitively see by the following considerations. A geodesic joining two points p and q of a sphere may be thought of as a thread stretched over the sphere and joining the two given points. When the n arc pq is smaller than a semimeridian and the points p and q are kept fixed, it is not possible to move the thread without increasing its length. On the n other hand, when the arcpq is greater than a semimeridian, a small displacement of the thread (with p and q fixed) "loosens" the thread (see Fig. 544).
Figure 544
In other words, when q is farther away than the antipodal point of p, it is n possible to obtain curves joining p to q that are close to the geodesic arc pq and are shorter than this arc. Clearly, this is far from being a mathematical argument. In this section we shall begin the study of this question and prove a result, due to Jacobi, which may be roughly described as follows. A geodesic y starting from a point p minimizes arc length, relative to "neighboring" curves of y, only up to the "first" conjugate point of p relative to y (more precise statements will be given later; see Theorems 1 and 2). For simplicity, the surfaces in this section are assumed to be complete and the geodesics are parametrized by arc length.

We need some preliminary results. S of a The foIlowing lemma shows that the image by exp,: T,(S) segment of line of T,(S) with origin at 0 E T,(S) (geodesic starting from p) is minimal relative to the images by exp, curves of T,(S) which join the extremities of this segment. More precisely, let
Jacobi's Theorems
and let
7: [O, 11
3
T p ( S )be the line o f T p ( S )given by

T p ( S )be a differentiable parametrized curve o f T p ( S ) ,with Let &: [0, I ] &(0) 0, &(I) = u, and &(s)# 0 i f s # 0. Furthermore, let (Fig. 545)
a($>= ~ X P E(s) , and y(s) = exp, F(s).
Figure 545
L E M M A 1. With the above notation, we have 1. l(a) 2 l ( y ) , where I ( ) denotes the arc length of the corresponding curve. In addition, ifa'(s) is not a criticalpoint of exp,, s of a and y are distinct, then
E
10, E ) ] , and ifthe traces
Proof. Let &(s)/l&(s) 1 = r, and let n be a unit vector o f T p ( S ) ,with (r, n) = 0. In the basis {r, n) of T,('S) we can write (Fig. 545) &'(s)= ar
+ bn,
where
a b
r), (E1(s),n).
= (@'(s),

4 18
Global Differential Geometry
By definition
Therefore, by using the Gauss lemma (cf. Sec. 55, Lemma 2) we obtain (a'@), al(s))
= a2
+ c2,
where
c2 = b2 I (d expp)acs,(n)1.' It follows that a2.
On the other hand,
Therefore,
and this proves part 1. T o prove part 2, let us assume that E(a)
= l(y).
I
J a (al(s), at(s))
lj2
ds
=
J
Then
I
a ds, 0
and since (al(s), a ' ( ~ ) ) ' / 2 ~ a, the equality must hold in the Iast expression for every s
E
[O, I]. Therefore,
Since @(s)in not a critical point of exp,, we conclude that b = 0. It follows that the tangent lines to the curve 6i all pass through the origin O of Tp(s). Thus, & is a line of T,(S) which passes through 0. Since &(I) = 7(1), the lines & and 7 coincide, thus contradicting the assumption that the traces of a and y are distinct. From this contradiction it foIlows that /(a) > E(y), which proves Q.E.D. part 2 and ends the proof of the lemma.
419
Jacobi's Theorems
We are now in a position to prove that if a geodesic arc contains no conjugate points, it yields a local minimum for the arc length. More preciselj., we have

THEOREM 1 (Jacobi.) Let y : [O, 1 1 3 S, y(0) = p, be a geodesic withoztt conjugate points; that is, exp,: Tp(S) S is regular at the points of the iii~e F(s) = syl(0) of Tp(S), s E [0, I]. Let h : [0, I] x (6, 6) + S be a proper variation of y. Then
1. There exists a 8 > 0, 6 ( E, such that
if t
E (6,
6),

where L(t) is the length of the curve h,: [0, I] S that is given bj* h,(s) = h(s, t). 2. If, in addition, the trace of h, is distinct from the trace of y , L(t) > L(0). Proof ,The proof consists essentially of showing that it is possible, for every t t (6, 6), to lift the curve h, into a curve & of T,(S) such that &*(o)= 0, ht,(1) = 7(1) and then to apply Lemma 1. Since exp, is regular at the points of the line 7 of T,(S), for each s E [O, I] there exists a neighborhood Us of p(s) such that exp, restricted to Us is a diffeomorphism. The family (Us], s E [O, I], covers T([O, I]), and, by compactness, it is possible to obtain a finite subfamily, say, U , , . . . , U, which still covers y([O, I]). It follows that we may divide the interval [0, I] by points
c Ui, i = 1, . . . , n. Since h is continuous and in such a way that y([si, [s,, si+,] is compact, there exists 6, > 0 such that

Let 8 = rnin(a1, . . . , 8,). For t E (8, a), the curve h,: [O,1] 3 5' may be lifted into a curve 5,: [0, I] T,(S), with origin &(0) = 0, in the follo~i,in_e way. Let s E [s,, s2]. Then

where exp;' is the inverse map of exp, : U , V,. By applying the same technique we used for covering spaces (cf. Prop. 2, Sec. 56), we can extend %, for all s E [0, I] and obtain &(I) = r)(l). In this way, we conclude that y(s) = exp, ?(s) and that h,(s) = exp, /i:(s),
420
Global Differential Geometry
t E (6, a), with h(0) = 0, &,(l) = y(1). We then apply Lemma 1 to this situation and obtain the desired conclusions. Q.E.D.
Remark I . A geodesic y containing no conjugate points may well not be minimal relative to the curves which are not in a neighborhood of y. Such a situation occurs, for instance, in the cylinder (which has no conjugate points), as the reader will easily verify by observing a closed geodesic of the cylinder. This situation is related to the fact that conjugate points inform us only about the differential of the exponential map, that is, about the rate of "spreading out" of the geodesics nieghboring a given geodesic. On the other hand, the global behavior of the geodesics is controlled by the exponential map itself, which may not be globally onetoone even when its differential is nonsingular everywhere. Another example (this time simply connected) where the same fact occurs is in the ellipsoid, as the reader may verify by observing the figure of the ellipsoid in Sec. 55 (Fig. 519). The study of the locus of the points for which the geodesics starting from p stop globally minimizing the arc length (called the cut locus of p) is of fundamental importance for certain global theorems of differential geometry, but it will not be considered in this book.
We shall proceed now to prove that a geodesic y containing conjugate points is not a local minimum for the arc length; that is, "arbitrarily near" to y there exists a curve, joining its extreme points, the length of which is smaller than tbat of y. We shall need some preliminaries, the first of which is an extension of the definition of variation of a geodesic to the case where piecewise differentiable functions are admitted. DEFINITION I. Let y : [0, I ] 
S be a geodesic of S and let
be a continuous map with
h is said to be a broken variation of y $ there exists a partition
of [0, 11 such that
h:[si, si+,] x (E, 6)
S, i
= 0,
I,.
.. ,n
 1,
Jacobi's Theorems
42 1
is dzferentiable. The broken variation is said to be proper if h(0, t) h(7, t) = y(I) for every t E (c, 6).
= y(O),
The curves h,(s), s E [0, I], of the variation are now piecewise differentiable curves. The variational vector field V(s) = (dh/dt)(s, 0) is a picccwise R3 is a continuous map, differentiable vector field along y; that is, V: [0, I] differentiable in each [ti, ti+,I. The broken variation h is said to be orthogonal if (V(s), yl(s)) = 0, s E [0, I ] . In a way entirely analogous to that of Prop. I of Sec. 54, it is possible to prove that a piecewise differentiable vector field V along y gives rise to a broken variation of y, the variational field of which is V. Furthermore, if


the variation can be chosen to be proper. Similarly, the function L: (  6 , E) R (the arc length of a curve of the variation) is defined by
By Lemma 1 of Sec. 54, each summand of this sum is differentiable in a neighborhood of 0. Therefore, L is differentiable in (6, 6) if 6 is sufficiently small. The expression of the second variation of the arc length (Lf'(0)), for proper and orthogonal broken variations, is exactly the same as that obtained in Prop. 4 of Section 54, as may easily be verified. Thus, if V is a piecewise differentiable vector field along a geodesic y: [0, I ] S such that

0, define a vector field Y, along y by

4
y,
= J(s)
+ qZ(s),
= rlZ(s)
s s
E E
[0, so], [so, 11.
Y,, is a piecewise differentiable vector field orthogonal to y . Since Y,(O) = Y,(l) = 0, it gives rise to a proper, orthogonal, broken variation of y . We shall compute L"(0) = I(Y,, Y,). For the segment of geodesic between 0 and so,we shall use the bilinearity of I and Lemma 2 to obtain
where I,, indicates that the coresponding integral is taken between 0 and so. By using I t o denote the integral between 0 and I and noticing that the integral is additive, we have
Observe now that if q = q0 is sufficiently small, the above expression is negative. Therefore, by taking Y,,, we shall obtain a proper broken variation, with L"(0) < 0. Since L'(0) = 0, this means that 0 is a point of local maximum for L; that is, there exists 6 > 0 such that if t E (6, 6), t # 0, then Q.E.D. L(t) < L(0).
424
Global Differential Geometry
Remark 3. Jacobi's theorem is a particular case of the Morse index theorem, quoted i n Remark 2. Actually, the crucial point of the proof of the index theorem is essentially an extension of the ideas presented i n the proof of Theorem 2.
EXERCISES 1. (Bonnet's Theorem.) Let S be a complete surface with Gaussian curvature K 2 6 > 0. By Exercise 5 of Sec. 55, every geodesic y : [0, oo) ,S has a point conjugate to y(0) in the interval (0, nl481. Use Jacobi's theorems to show that this implies that S is compact and that the diameter p ( S ) (n / 4 6 (this gives a new proof of Bonnet's theorem of Sec. 54).
2. (Lines on Complete Surfaces.) A geodesic y : [   a , m) 4 S is called a line if its length realizes the (intrinsic) distance between any two of its points. a. Show that through each point of the complete cylinder x 2 y 2 = 1 there passes a Iine. b. Assume that S is a complete surface with Gaussian curvature K > 0. Let y :( oo, m) 4 S be a geodesic on S and let J(s) be a Jacobi field along Y given by 0, that it has a differentiable inverse. Thus, $ is a diffeomorphism, and we can induce an inner product in R: by setting
To compute this inner product, we observe that
d  d dx
dv'
d

d
p
6
du'
hence,
R: with this inner product is isometric to H, and it is sometimes called the Poincare' halfplane. To determine the geodesics of H, we work with the PoincarC halfplane and make two further coordinate changes. First, fix a point (x,, 0) and set (Fig. 550)
x 0

x,
=p
cos 0,
y
=p
sin 0,
< 0 < n, 0 < p < +m. This is a diffeomorphism of R: into itself, and
G/obal Differential Geometry Parallels to
Y through P
"0
Figure 550
Next, consider the diffeomorphism of R: given by (we want to change 8 into a parameter that measures the arc length along p = const.)
which yields

By looking again at the differential equations for the geodesics (F = 0, G I), we see that p , = p = const. are geodesics. (Another way of finding the geodesics of R: is given in Exercise 8.) Collecting our observations, we conclude that the lines and the halfcircles which are perpendicular to the axis y > 0 are geodesics of the Poincari halfplane R t . These are all the geodesics of R:, since through each point q E R: and each direction issuing from q there passes either a circle tangent to that line and normal to the axis y = 0 or a vertical line (when the direction is vertical). The geometric surface R: is complete; that is, geodesics can be defined for all values of the parameter. The proof of this fact will be left as an exercise (Exercise 7; cf. also Exercise 6). It is now easy to see, if we define a straight line of R: to be a geodesic, that all the axioms of Euclid but the axiom of paraIlels hold true in this geometry. The axiom of parallels in the Euclidean plane P asserts that from a point not in a straight line r c P one can draw a unique straight line
433
Abstract Surfacas
r' c P that does not meet r. Actually, in R;L, from a point not in a geodesic y we can draw an infinite number of geodesics which do not meet y.
The question then arises whether such a surface can be found as a regular surface in R 3 . The natural context for this question is the following definition. DEFINITION 6. A dlerentiable map p : S + R3 of an abstract surface S into R 3 is an immersion $the dlerential dpp: T,(S) 4 Tp(R3)is injective. If, in addition, S has a metric ( , ) and
p is said to be atz isometric immersion.
Notice that the first inner product in the above relation is the usual inner product of R 3 , whereas the second one is the given Riemannian metric on S. This means that in an isometric immersion, the metric "induced" by R 3 on S agrees with the given metric on S. Hilbert's theorem, to be proved in Sec. 51 1, states that there is no isometric immersion into R3 of the complete hyperbolic plane. I n particular, one cannot find a model of the geometry of Lobachewski as a regular surface in R3. Actually, there is no need to restrict ourselves to R 3 . The above definition of isometric immersion makes perfect sense when we replace R3 by R4 or, for that matter, by an arbitrary Rn.Thus, we can broaden our initial question, and asli: For what values of n is there an isometric immersion of the complete hyperbolicplane into R n ?Hilbert's theorem say that n 2 4. As far as we know, the case n = 4 is still unsettled. Thus, the introduction of abstract surfaces brings in new objects and illuminates our view of important questions. In the rest of this section, we shall explore in more detail some of the ideas just introduced and shall show how they Iead naturaIly to further important generalizations. This part will not be needed for the understanding of the next section. Let us look into further examples. Example 4. Let R2 be a plane with coordinates (x, y ) and T,, ,: R2 +R2 be the map (translation) T,n,,(x, y) = (x f m, y n), where m and are integers. Define an equivalence relation in R2 by (x, y ) (xl,y l ) if there exist integers m , n such that T,,.(x, y) = (x,, y ,). Let T be the quotient space of R2 by this equivalence relation, and let n : R2 ,T be the natural projection map n(x, y ) = {T,,.(x, y ) ; all integers m , n ] . Thus, in each open unit square whose vertices have integer coordinates, there is only one representative of
+
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434
T, and T may be thought of as a closed square with opposite sides identified. (See Fig. 551. Notice that all points of R2 denoted by x represent the same point p in T.) R 2 be a family of parametrizations of R2, where i, Let i,: U, c R2 is the identity map, such that U, n T,,,(U,) = $ for all m, n. Since T,,, is a diffeomorphism, it is easily checked that the family (U*,noi,) is a differentiable structure for 7: T is called a (differentiable) torus. From the very definiT is a differentiable map tion of the differentiable structure on T,n : R2 and a local diffeomorphism (the construction made in Fig. 551 indicates that T is diffeomorphic to the standard torus in R3). Now notice that T,,, is an isometry of R2 and introduce a geometric (Riemannian) structure on T as follows. Let p E T and u E T,(T). Let q l , q2 E R2 and w,, w2 E R2 be such that n(ql) = n(qz) = p and dn,,(w,) = dn,,(w2) = U. Then ql 9, ;hence, there exists T,, ,such that T,, ,(q,) = q,, d(T,,,),,(w,) = w,.Since T,,, is an isometry, 1 w , I = w2 1. Now, define the length of u in T,(T) by I u ]= I dn,(w,) I = I w, 1. By what we have seen, this is well defined. Clearly this gives rise to an inner product ( , ), on T,(T) for each p E T. Since this is essentjally the inner product of R 2 and n is a local diffeomorphism, ( , ), varies differentiably with p. +
t

Figure 551 The torus.
435
Abstract Surfaces
Observe that the coefficients of the first fundamental form of T, in any of the parametrizations of the family (U,, 72 0 i,), are E = G = 1, F = 0, Thus: this torus behaves localIy like a Euclidean space. For instance, its Gaussian curvature is identically zero (cf. Exercise 1, Sec. 43). This accounts for the name flat torus, which is usually given to T with the inner product just described. Clearly the flat torus cannot be isometrically immersed in R3, since, by compactness, it would have a point of positive curvature (cf. Exercise 16, Sec. 33, or Lemma I, Sec. 52). However, it can be isometrically immersed in R4. In fact, let F : A2 + R 4 be given by 1 F(x, y) = (cos 272x, sin 272x, cos 271~1,sin 27zy). 272
+
+
R4 Since F(x m, y n) = F(x, y) for all m, n, we can define a map p : T by p(p) = F(q), where q E n'(p). Clearly, p 0 n = F, and since 72: R2 ,T is a local diffeomorphism, p is differentiable. Furthermore, the rank of dp is equal to the rank of dF, which is easily computed to be 2. Thus, p is an immersion. To see that the immersion is isometric, we first observe that if e, = (1, O), e, = (0, 1) are the vectors of the canonical basis in R2, the vectors dn,(e,) =f,, dn,(e,) =f,, q E R2, form a basis for T,(,,(T). By definition of the inner product on T, (h,fj) = (e,, e,), i, j = 1,2. Next, we compute +
aF = dF(e,) = (sin dx d F dF(e,)
au
=
2nx, cos 272x, 0, O),
(0, 0, sin 272y, cos 272y),
and obtain that (dF(ei), dF(e
,>>
(pi,
ej) = (A.7 fj).
Thus, (ddJ;J, d d f j ) )
=
(ddd72(ei)), d ~ ( d d e j ) >= > (f,,fj).
It follows that p is an isometric immersion, as we had asserted. Rn It should be remarked that the image p(S) of an immersion p: S may have selfintersections. In the previous example, p : T + R4 is onetoone, and furthermore p is a homeomorphism onto its image. It is convenient to use the following terminology. +

DEFINITION 7. Let S be an abstract surface. A dzflerentiable map p: S Rn is an embedding if p is an immersion and a homeomorphism onto its image.
436
Global Differential Geometry
For instance, a regular surface in R3 can be characterized as the image of an abstract surface S by an embedding yl: S R3. This means that only those abstract surfaces which can be embedded in R3 could have been detected in our previous study of regular surfaces in R3.That this is a serious restriction can be seen by the example below. +
Example 5, We first remark that the definition of orientability (cf. Sec. 26, Def. 1) can be extended, without changing a single word, to abstract surfaces. Now consider the real projective plane P2 of Example 1. We claim that P 2 is nonorientable. To prove this, we first make the following general observation. Whenever an abstract surface S contains an open set M diffeomorphic to a Mobius strip (Sec. 26, Example 3), it is nonorientable. Otherwise, there exists a family of parametrizations covering S with the property that all coordinate changes have positive Jacobian; the restriction of such a family to M will induce an orientation on M which is a contradiction. Now, P 2 is obtained from the sphere S 2 by identifying antipodal points. Consider on S 2 a thin strip B made up of open segments of meridians whose centers lay on half an equator (Fig. 552). Under identification of antipodal points, B clearly becomes an open Mobius strip in P 2 .Thus, P 2 is nonorientable.
Figure 552. The projective plane contains a Mobius strip.
By a similar argument, it can be shown that the Klein bottle K of Example 2 is also nonorientable. In general, whenever a regular surface S c R3 is symmetric relative to the origin of R3,identification of symmetric points gives rise to a. nonorientable abstract surface. It can be proved that a compact regular surface in R3 is orientable (cf. Remark 2, Sec. 27). Thus, P 2 and K cannot be embedded in R3, and the same happens to the compact orientable surfaces generated as above. Thus, we miss quite a number of surfaces in R3. P 2 and K can, however, be embedded in R4. For the Klein bottle K, consider the map G: R2  R4 @en by G(U,v) = ((r cos ~r t a) cos U , (r cos v U r sin v cos , r sin v sin 2
+ a) sin
UY
437
Abstract Surfaces
+
Notice that G(u, v) = G(u 2mn, 2nn  v), where m and n are integers. Thus, G induces a map y of the space obtained from the square
by first reflecting one of its sides in the center of this side and then identifying opposite sides (see Fig. 553). That this is the Klein bottle, as defined in Example 2, can be seen by throwing away an open half of the torus in which antipodal points are being identified and observing that both processes lead to the same surface (Fig. 553).
Klein bottle immersed in R3 with selfintersections
Figure 553
Thus,
I,U
is a map of K into R4. Observe further that
It follows that G = 0 n,0 n, where n : R2 + T is essentially the natural projection on the torus T (cf. Example 4) and n,:T + K corresponds to identifying "antipodal" points in T. By the definition of the differentiable structures on T and K, n and n, are local diffeomorphisms. Thus, y : K R4 is differentiable, and the rank of ~ I , Uis the same as the rank of dG. The latter is easily computed to be 2 ; hence, tp is an immersion. Since Kis compact and y is onetoone, I,u~is easily seen to be continuous in yl(K). Thus, y is an embedding, as we wished. For the projective plane P 2 , consider the map F: R3 R 4 given by

+
Let S 2 c R3 be the unit sphere with center in the origin of R3. It is clear that the restriction yl = F/S2 is such that ~ ( p = ) y(I)). Thus, y induces a map
Global Differential Geometry
438
To see that 01 (hence, (5) is an immersion, consider the parametrization x of S2 given by x(x, y) = (x, y, dl  x 2  y2), where x 2 y 2 1. Then
+
+
9 0 x(x, y)
=
(x2  y2, xy, xD, yD),
D
=
dl  x 2  yZ.
It is easily checked that the matrix of d(9ox) has rank 2. Thus, (5 is an immersion. To see that (3 is onetoone, set
+
It suffices to show that, under the condition x2 + y 2 z2 = 1, the above equations have onIy two solutions which are of the form (x, y, z) and (x, y, 2). In fact, we can write xZd= bc,
y2c = bd,
z2b = cd,
x2  y Z = a,
x2 + y Z
+ z2 = 1
where the first three equations come from the last three equations of (2). Now, if one of the numbers b, c, d is nonzero, the equations in (3) will give x2,y2, and z2, and the equations in (2) will determine the sign of two coordinates, once given the sign of the remaining one. If b = c = d = 0, the equations in (2) and the last equation of (3) show that exactly two coordinates will be zero, the remaining one being & I . In any case, the solutions have the required form, and (5 is onetoone. By compactness, q~ is an embedding, and that concIudes the example. If we look back to the definition of abstract surface, we see that the number 2 has played no essential role. Thus, we can extend that definition to an arbitrary n and, as we shall see presently, this may be useful.
DEFINITION la. A differentiable manifold of dimension n is a set M together with a family of onetoone maps x,: U, + M of open sets U, c Rn into M such that
2. For each pair a, p with x,(U,) n x,(U,) = W .f 4, we have that x i l(W), xp '(W) are open sets in Rnand that x i 'ox,, x i ' ox, are dzferentiable maps.
439
Absrtact Surfaces
3. The family (U,, x,) is maximal relative to conditions 1 and 2.
A family (U,, x,) satisfying conditions 1 and 2 is called a drfferentiable structure on M. Given a differentiable structure on M we can easily complete it into a maximal one by adding to it all possible parametrjzations that, together with some parametrization of the family {U,, x,), satisfy condition 2. Thus, with some abuse of language, we may say that a differentiable manifold is a set together with a differentiable structure. Remark. A family of open sets can be defined in M by the following requirement: V c M is an open set if for every a, x i l ( V n x,(U,)) is an open set in Rn.The readers with some knowledge of point set topology wiIl notice that such a family defines a natural topology on M. In this topology, the maps xu are continuous and the sets x,(U,) are open in M. In some deeper theorems on manifolds, it is necessary to impose some conditions on the natural topoIogy of M. The definitions of differentiable maps and tangent vector carry over, word by word, to differentiable manifolds. Of course, the tangent space is now an ndimensional vector space. The definitions of differential and orientability also extend straightforwardly to the present situation. In the following example we shall show how questions on twodimensional manifolds lead naturally into the consideration of higherdimensional manifolds.
Example 6 (The Tangent Bundle). Let S be an abstract surface and let T(S) = ((p, w), p G S, w E T,(S)). We shall show that the set T(S) can be given a differentiable structure (of dimension 4) to be called the tangeut bund[e of S, Let (U,, xu) be a differentiable structure for S. We shall denote by (u,, v,) the coordinates of U,, and by (dldu,, dldv,} the associated bases in the T(S) by tangent planes of x,(U,). For each a, define a map y,: U, x R2 +
Y,(~Wu,, x , y )
=
(
d du,
x,(u,, v,), x
+ ydv,
>
,
(x,y)
E
R2.
Geometrically, this means that we shall take as coordinates of a point (p, w) E T(S) the coordinates u,, v, o f p plus the coordinates of w in the basis (dldua, a/av,1We shalI show that ( U , x R2, y,) is a differentiable structure for T ( S ) . Since U, x,(U,) = S and (dx,),(R2) = T (S), q E U,, we have that
,,
Global Differential Geometry
440
and that verifies condition 1 of Def. la. Now let
(P,w)
E yU(Uux R2) n y,(U,
x R".
Then
where g, E Uu, 9, E U,, w,, w, E R2. Thus,
Since x i 1 0 xu is differentiable, so is d ( x i l 0 x,). It follows that y j 0 y, is differentiable, and that verifies condition 2 of Def. la. The tangent bundle of S is the natural space to work with when one is dealing with secondorder differential equations on S. For instance, the equations of a geodesic on a geometric surface S can be written, in a coordinate neighborhood, as (cf. Sec. 47)
The classical "trick" of introducing new variables x the above to the firstorder system
= u',
y
= v'
to reduce
may beinterpreted as bringing into consideration the tangent bundle T ( S ) , with coordinates (u, v, x,y) and as looking upon the geodesics as trajectories of a vector field given locally in T(S) by (4.). It can be shown that such a vector field is well defined in the entire T(S); that is, in the intersection of two coordinate neighborhoods, the vector fieIds given by (4) agree. This field (or rather its trajectories) is called thegeodesicflow on T ( S ) .It is a very natural object to work with when studying global properties of the geodesics on S . By looking back to Sec. 47, it will be noticed that we have used, in a disguised form, the manifold T ( S ) . Since we were interested only in local properties, we could get along with a coordinate neighborhood (which is essentially an open set of R4). However, even this local work becomes neater when the notion of tangent bundle is brought into consideration. Of course, we can also define the tangent bundle of an arbitrary ndimen
Abstract Surfaces
447
sional manifold. Except for notation, the details are the same and will be left as an exercise. We can also extend the definition of a geometric surface to an arbitrary dimension.
DEFINITION 5a. A Riemannian manifold is an ndimensional diferentiable man$old M together with a choice, for each p E M , of an inner product ( , ),in T,(M) that variesd~ferentiablywitlzp inthefollowiizgsense. For some (hence, all) parametrization x u : U, ,M with p E x,(U,), the functions
are dzferentiable at x i (p); here (u,, . . . , u,) are the coordinates of U, c Rn. The differentiable family (( ), p E M j is called a Riemannian structure (or Riemannian metric) for M. Notice that in the case of surfaces we have used the traditional notation gll = E,9 1 2 = g 2 1 = F, g 2 2 = G' The extension of the notions of the intrinsic geometry to Riemannian manifolds is not so straightforward as in the case of differentiable manifolds. First, we must define a notion of covariant derivative for Riemannian manifolds. For this, let x : U + M be a parametrization with coordinates ( u l , . . . , u,) and set xi = d/dui. Thus, gij = ( x i , x j ) . We want to define the covariant derivative Dwv of a vector field v relative to a vector field w. We would like Dwv to have the properties we are used to and that have shown themselves to be effective in the past. First, it should have the distributive properties of the old covariant derivative. Thus, if u, v, w are vector fields on M and f,g are differentiable functions on M, we want
where dfldu, for instance, is a function whose value at p E M is the derivative (f 0 a)'(O) of the restriction of f to a curve a: (6, E) ,M, a(0) = p , ar(0) = u. Equations ( 5 ) and (6) show that the covariant derivative D is entirely determined once we know its values on the basis vectors
where the coefficients
are functions yet to be determined.
Global Differential Geometry
442
Second, we want the
r&to be symmetric in i and j(r&= rFi);that is, DXrxj= DXjxi
for all i, j
(?
Third, we want the law of products to hold; that is,
From Eqs. (7) and (8), it follows that
or, equivalently,
+
Since det(g,) 0, we can solve the last system, and obtain the r,kias functions of the Riemannian metric g,, and its derivatives (the reader should compare the system above with system (2) of Sec. 43). If we think of gij as a matrix and write its inverse as gij, the solution of the above system is
Thus, given a Riemannian structure for M , there exists a unique covariant derivative on M (also called the LeviCivita connection of the given Riemannian structure) satisfying Eqs. (5)(8). Starting from the covariant derivative, we can define parallel transport, geodesics, geodesic curvature, the exponential map, completeness, etc. The definitions are exactly the same as those we have given previously. The notion of curvature, however, requires more elaboration. The following concept, due to Riemann, is probably the best analogue in Riemannian geometry of the Gaussian curvature. Let p E M and let a c Tp(M) be a twodimensional subspace of the tangent space Tp(M). Consider all those geodesics of M that start from p and are tangent to a. From the fact that the exponential map is a local diffeomorphism at the origin of Tp(M), it can be shown that small segments of such geodesics make up an abstract surface S containing p . S has a natural geometric structure induced by the Riemannian structure of M. The Gaussian curvature of S at p is called the sectional curvatrrre K(p, a ) of M at p along a. It is possible to formalize the sectional curvature in terms of the LeviCivita connection but that is too technical to be described here. We shall only
443
Abstract Surfaces
mention that most of the theorems in this chapter can be posed as natural questions in Riemannian geometry. Some of them are true with littIe or no modificatio~lof the given proofs. (The HopfRinow theorem, the Bonnet theorem, the first Hadamard theorem, and the Jacobi theorems are all in this class.) Some others, however, require further assumptions to hold true (the second Hadamard theorem, for instance) and were seeds for further developments. A full development of the above ideas would lead us into the realm of Riemannian geometry. We must stop here and refer the reader to the bibliography at the end of the book.
EXERCISES 1. Introduce a metric on the projective plane P 2 (cf. ExampIe 1 ) so that the natural projection n : S2+P 2 is a local isometry. What is the (Gaussian) curvature of
such a metric? 2. (The Infinite Mobius Sir@.) Let
be a cylinder and A : C + C be the map (the antipodal map) A(x, y, z) = (x,  y , z). Let M be the quotient of C by the equivalence relation p  A(p), and let n : C + M be the map n(p) = {p, A(p)), p E C. a. Show that M can be given a differentiable structure so that n is a local diffeomorphism (M is then called the infinite Mobius strip). b. Prove that M is nonorientable. c. Introduce on M a Riemannian metric so that n is a local isometry. What is the curvature of such a metric? 3. a. Show that the projection n: S2,PZfrom the sphere onto the projective plane has the following properties: (1) n is continuous and n(S2) = P2; (2) each point p E P 2 has a neighborhood U such that nl(U) = Vl u V2, where Vl and V2 are disjoint open subsets of S2, and the restriction of n to each Vi,i = 1, 2, is a homeomorphism onto U. Thus, n satisfies formally the conditions for a covering map (see Sec. 56, Def. 1) with two sheets. Because of this, we say that S2 is an orientable double covering of P2. b. Show that, in this sense, the torus T is an orientable double covering of the KIein bottle K (cf. Example 2) and that the cylinder is an orientable double covering of the infinite Mobius strip (cf. Exercise 2).
4. (The Orientable Double Covering). This exercise gives a general construction for the orientable double covering of a nonorientable surface. Let S be an abstract, connected, nonorientable surface. For each p E S, consider the set B of all bases of T,(S) and call two bases equivalent if they are related by a matrix with
Global Differential Geometry
positive determinant. This is clearly an equivalence relation and divides B into two disjoint sets (cf. Sec. 14). Let 0, be the quotient space of B by this equivalence relation. 0, has two eIements, and each element 0, E 0, is an orientation of T,(S) (cf. Sec. 14). Let iT: be the set
To give 5 a differentiable structure, let {U,, x,) be the maximal differentiable structure of S and define fi,: Ua > 5 by
where (u,, 21,) E U, and [dldu,, dldv,] denotes the element of 0, determined by the basis {dldu,, dldv,). Show that a. {U,, 9,) is a differentiable structure on and that 5 with such a differentiable structure is an orientable surface. b. The map x : 5 + S given by x(p, 0,) = p is a differentiable surjective map. Furthermore, each point p E S has a neighborhood U such that r l ( U ) = VI U V2, where Vl and V2 are disjoint open subsets of and x restricted to each 6,i = 1, 2, is a diffeomorphism onto U. Because of this, S is called an orientable doubie covering of's.
5. Extend the GaussBonnet theorem (see Sec. 45) to orientable geometric surfaces and apply it to prove the following facts : a. There is no Riemannian metric on an abstract surface T diffeomorphic to a torus such that its curvature is positive (or negative) at all points of T. b. Let T and S 2 be abstract surfaces diffeomorphic to the torus and the sphere, respectively, and let p: T + S2be a differentiable map. Then p has at least one critical point, i.e., a point p E Tsuch that dy, = 0. 6. Consider the upper halfplane R t (cf. Example 3) with the metric
Show that the lengths of vectors become arbitrarily large as we approach the boundary of R$ and yet the length of the vertical segment
approaches 2 as 6 + 0. ConcIude that such a metric is not complete.
"7. Prove that the Poincark halfplane (cf. Example 3) is a complete geometric surface. Conclude that the hyperbolic plane is complete. 8. Another way of finding the geodesics of the Poincark halfplane (cf. Example 3) is to use the EulerLagrange equation for the corresponding variational problem (cf. Exercise 4, Sec. 54). Since we know that the vertical lines are geodesics, we
445
Abstract Surfaces
can restrict ourselves to geodesics of the form y the critical points of the integral (F = 0)
= y(x).
Thus, we must look for
since E = G = l/y2. Use Exercise 4, Sec. 54, to show that the solution to this variationa1 problem is a family of circles of the form
9. Let 3 and S be connected geometric surfaces and let x : ? !, 4 S be a surjective differentiable map with the following property: For each p E S, there exists a where the V,'s are open disneighborhood U of p such that xdl(U) = (Java, joint subsets of and x restricted to each V, is an isometry onto U (thus, 7t is essentially a covering map and a local isometry). a. Prove that S is complete if and only if 3 is complete. b. Is the metric on the infinite Mobius strip, introduced in Exercise 2, part c, a
compIete metric? 10. (Kazdan Warner's Results.)
a. Let a metric on R 2 be given by
Show that the curvature of this metric is given by
b. Conversely, given a function K ( x , y) on R2, regard y as a parameter and let JZ be the solution of (*) with the initial conditions
Prove that G is positive jn a neighborhood of (x,, y ) and thus defines a metric in this neighborhood. This shows that every diferentiable function is locally the curvature of some (abstract) metric. *c. Assume that K ( x , y) (0 for all (x,y ) E R2.Show that the solution of part b satisfies
J G (x, y) 2 4 
=
1
for all x.
Thus, G(x, y) defines a metric on all of R2. Prove also that this metric is complete. This shows that any nonpositive diferentiable function on R 2 is the curvature of some complete metric on R2. If we do not insist on the metric being complete, the result is true for any differentiable function K on R2. Compare
Global Differential G e o m e t r y
J. Kazdan and F. Warner, "Curvature Functions for Open 2Manifolds," Ann. of Math. 99 (1974), 203219, where it is also proved that the condition
on K given in Exercise 2 of Sec. 54 is necessary and suficiept for the metric to be complete.
51 1. Hilbert's Theorem Hilbert's theorem can be stated as follows.
THEOREM. A complete geometric surface S with constant negative curvature cannot be isometrically immersed in R3. Remark 1. Hilbert's theorem was first treated in D. Hilbert, " ~ b e r Flachen von konstanter Gausscher Kriimung," Trarzs. Amer. Math. Soc. 2 (1901), 8799. A different proof was given shortly after by E. Holmgren, "Sur les surfaces h courbure constante negative," C. R. Acad. Sci. Paris 134 (1902), 740743. The proof we shall present here follows Hilbert's original ideas. The local part is essentially the same as in Hilbert's paper; the global part, however, is substantially different. We want to thank J. A. Scheinkman for helping us to work out this proof and M. Spivak for suggesting Lemma 7 below. We shall start with some observations. By multiplying the inner product by a constant factor, we may assume that the curvature K =  1. Moreover, since exp,: T,(S) + S is a local diffeomorphism (corollary of the theorem of Sec. 55), it induces an inner product in T,(S). Denote by S' the geometric surface T,(S) with this inner product. If y : S 3 R3 is an isometric immersion, the same,holds for q = y.0 exp,: S' R3. Thus, we are reduced to proving that there exists no isometric immersion q : S' R3 of a plane S' with an inner product such that K =  1.

L E M M A 1. The area of St is infinite. Proof. We shall prove that Sf is (globally) isometric to the hyperbolic plane H. Since the area of the latter is (cf. Example 3, Sec. 510)
this will prove the lemma. Let p E H, p' E S', and choose a linear isometry y: T,(H) + Tp,(Sf)between their tangent spaces. Define a map q : H + S' by q = exp, o y 0 exp; '.
Hilbert ' s Theorem
447
Since each point of H is joined to p by a unique minimal geodesic, o,. is we11 defined. We now use poIar coordinates (p, 8) and (p', 8') around p and p', respectively, requiring that O, maps the axis 8 = 0 into the axis 8' = 0. By the results of Sec. 46, O, preserves the first fundamental form; hence, it is Iocally an isometry. By using the remark made after Hadamard's theorem, we conclude that p is a covering map. Since S' is simpIy connected, p is a homeomorphism, and hence a (global) isometry. Q.E.D. For the rest of this section we shall assume that there exists an isometric immersion o,:Sr+ R3, where S' is a geometric surface homeomorphic to a plane and with K =  1. To avoid the difficulties associated with possible selfintersections of q(Sr), we shall work with S' and use the immersion O, to induce on S' the local extrinsic geometry of q(S') c R3. More precisely, since O, is an immersion, for each p E S' there exists a neighborhood V' c S' of p such that the restriction O, [V' = @ is a diffeomorphism. At each @(q)E yl(Vf), there exist, for instance, two asymptotic directions. Through yl, these directions induce two directions at q E S', which will be called the asymptotic directions on S' at q. In this way, it makes sense to talk about asymptotic curves on S', and the same procedure can be applied to any other locd entity ofp(S'). We now recall that the coordinate curves of a parametrization constitute a Tchebyshef net if the opposite sides of any quadrilateral formed by them have equal length (cf. Exercise 7, Sec. 25). If this is the case, it is possible to reparametrize the coordinate neighborhood in such a way that E = 1, F = cos 8, G = 1, where 8 is the angle formed by the coordinate curves, (Sec. 25, Exercise 8). Furthermore, in this situation, K = (8,,/sin 8) (Sec. 43, Exercise 5).
L E M M A 2. For each p E S' there is a parametrization x: U c R2 +Sr, p E x (U), such that the coordinate curves of x are the asymptotic curves of x(U) = V' and form a Tchebyshef net (we shall express this by saying that the asymptotic curves of V' form a Tchebyshef net). Proof. Since K < 0, a neighborhood V' c S' of p can be parametrized by x(zr, v) in such a way that the coordinate curves of x are the asymptotic curves of V'. Thus, if e , f , and g are the coefficients of the second fundamental form of S' in this parametrization, we have e = g = 0. Notice that we are using the above convention of referring to the second fundamental form of S' rather than the second fundamental form of p(S) c R3. Now in p(V') c R3, we have
G/oba/ Differentid Geometry
Nu A Nv hence, setting D
= JEG
( N A N,),
=
K(xu A xu);
 F2,
 (N
N u ) = 2KDN.
A Nu), = 2(N,
Furthermore,
N A Nu
aI {(xuA x.1
=
 (1 fx,
1
A Nu1
{(xu,N J x u  (xu, Nu)x.J
 ex,),
D
and, similarly,
1 N A Nu = ji(gx. fx,). Since K
= 1 =
= g = 0,
( f =ID2)and e
N/\N,=&x,,
we obtain
NANv==A~v;
hence,
2KDN
=
2DN
=
Ax,, I Z x,,
=
&2x,,.

It follows that xu, is parallel to N ; hence, E, = 2{x,,, xu) = 0 and G, 2(xu,, xu) = 0. But E, = G, = 0 implies (Sec. 25, Exercise 7) that the coordinate curves form a Tchebyshef net. Q.E.D. LEMMA 3. Let V' c Sf be a coordinate neighborhood of S' such that the coordinate curves are the asymptotic curves in V'. Then the area A of any quadrilateral formed by the coordinate curves is smaller than 2n.
Proof. Let (G,6)be the coordinates of V'. By the argument of Lemma I, the coordinate curves form a Tchebyshef net. Thus, it is possible to reparametrize V' by, say, (u, v ) so that E = G = 1 and F = cos 0. Let R be a quadrilateral that is formed by the coordinate curves with vertices (uI, v,), (u,, v l ) , (u2,v2),( u l ,v 2 )and interior angles a,, a,, a 3 ,a,, respectively (Fig. 554). Since E = G = I, F = cos 8, and B,, = sin 8, we obtain
A
=
J
d~ = R
J
R
sin e du dv
= 0(u,, v l )  O(uz,v,) = a,
since ai < 7t.
+ a,

=
J R o,, du dv
+ 8(u2,v z )  O(ul,
( n  a,)  (I( a,)
=
21,)
C a,  2n < 2n,
i= 1
Q.E.D.
Hilb eft's Theorem
Figure 554

So far the considerations have been local. We shall now define a map S' and show that x is a parametrization for the entire S', x: R2 The map x is defined as foIlows (Fig. 555). Fix a point 0 E S' and choose orientations on the asymptotic curves passing through 0. Make a definite choice of one of these asymptotic curves, to be called a,, and denote the other one by a,. For each (s, t) E R2, lay off on a , a length equal to s starting from 0. Let p' be the point thus obtained. Through p' there pass two asymptotic curves, one of which is a,. Choose the other one and give it the orientation obtained by the continuous extension, along a,, of the orientation of a2. Over this oriented asymptotic curve lay off a length equal to t starting from p'. The point so obtained is x(s, t).
Figure 555
x(s, t ) is well defined for all (s, t ) E R2. In fact, if x(s, 0) is not defined, there exists s, such that a,(s) is defined for all s < s, but not for s = s,. Let q = lim,,,, a,(s). By completeness, q E S t . By using Lemma 2, we see that a,(s,) is defined, which is a contradiction and shows that x(s, 0) is defined for all s E R. With the same argument we show that x(s, t ) is defined for all t E R.
450
Global Differential Geometry
Now we must show that x is a parametrization of S'. This will be done through a series of lemmas. LEMMA 4. For aJixed t , the curve x(s, t), tic curve with s as arc length.
 oo < s < 00,is an asympto
Proof. For each point x(s', t') E S f , there exists by Lemma 2 a "rectangular" neighborhood (that is, of the form to < t < t,, s, < s < s,) such that the asymptotic curves of this neighborhood form a Tchebyshef net. We first remark that if for some to,t, < to < t,, the curve x(s, to),s, < s < s,, is an asymptotic curve, then we know the same holds for every curve x(s, f), to < i < t,. In fact, the point x(s, i) is obtained by laying off a segment of length i from x(s, 0) which is equivalent to laying off a segment of length f  to from x(s, to).Since the asymptotic curves form a Tchebyshef net in this neighborhood, the assertion follows. x ( ~ , 1t i f
Figure 556
Now, let x(s,, t , ) E S' be an arbitrary point. By compactness of the segment x(s,, t), 0 t ( t,, it is possible to cover it by a finite number of rectangular neighborhoods such that the asymptotic curves of each of them form a Tchebyshef net (Fig. 556). Since x(s, 0) is an asymptotic curve, we iterate the previous remark and show that x(s, t , ) is an asymptotic curve in a neighborhood of s,. Since (s,, t , ) was arbitrary, the assertion of the lemma Q.E.D. follows.
0 there exists to E [O, 11 such that if t < to, then Iv,(a,(l))  v,(ao(l))l < E . Thus, if to is small enough, we have u,(or,(l)) = v, (a,(l)) for t < to. Since [O, 11 is compact, we can extend this argument stepwise to all t E [0, 11. Hence, v,(cc,(l)) = v,(ao(l)) .

452
Global Differential Geometry
This proves our claim and concludes the proof of the lemma.
Q.E.D.
LEMMA 8. x is injective.

Proqf We want to show that x(so, to) = x(s,, t,) implies that (so, to) ($1, t l ) . We first assume that x(s,, to) = x ( s ~ to), , s l > so, and show that this leads to a contradiction. By Lemma 7, an asymptotic curve cannot intersect itself unless the tangent lines agree at the intersection point. Since x is a local diffeomorphism, there exists an E > 0 such that
By the same reason, the points of the curve x(so, t) for which
form an open and closed set of this curve; hence, x(so, t ) = x(s,, t) for all t. Moreover, by the construction of the map x, x(so a, t,) = x(s, a, to), 0 ( a < s,  s o ; hence, x(so $ a, t) x(s, a, t) for all t. Thus, either

+
+
+
1. ~ ( $ 0to) , # x(so, t) for all t > to, or 2. There exists t = t, > to such that x(so, to) = x(so, t,); by a similar argument, we shall prove that x(s, to b) = x(s, t , b) for all s, 0 jb 5 t,  to.
+
+
In case I, x maps each strip of R2 between two vertical lines at a distance s,  soonto S' and identifies points in these lines with the same t. This implies that S' is homeomorphic to a cylinder, and this is a contradiction (Fig. 558). In case 2, x maps each square formed by two horizontal lines at a distance s,  s o and two vertical lines at a distance t,  to onto S and identifies '
Figure 558
Hiibert ls Theorem
453
corresponding points on opposite sides of the boundary. This implies that S' is homeomorphic to a torus, and this is a contradiction (Fig. 559). By a similar argument, we show that x(s,, to) x(s,, t,), t, > to, leads to the same contradiction.

Figure 559
We now consider the case x(s,, 2,) = x(s,, t,), s, > so, t, > t,. By using the fact that x is a local diffeornorphism and the connectedness of S', we see that x maps onto Sfa strip of RZbetween two lines perpendicular to the vector (s,  so,t,  to) E RZ and at a distance J(s,  so)' (t,  tJ2 apart. We can now consider cases 1 and 2 as in the previous argument and showthat Sfis then homeomorphic either to a cylinder or to a torus. In any case, this Q.E.D. is a contradiction.
+
The proof of Hilbert7s theorem now follows easily.

Proof of the Theorem. Assume the existence of an isometric immersion y : S R3, where S is a complete surface with K r 1. Let p E S and denote by S' the tangent plane T,(S) endowed with the metric induced by exp,: T,(S) , S. Then p = yoexp,: S' , R3 is a n isometric immersion and Lemmas 5, 6, and 8 show the existence of a parametrization x: R 2 + S' of the entire S' such that the coordinate curves of x are the asymptotic curves of S' (Lemma 4). Thus, we can cover S' by a union of "coordinate quadrilaterals" Q,, with Q, c Q,,,. By Lemma 3, the area of each Q, is smaller than 27c. On the other hand, by Lemma 1, the area of S' is unbounded. This is a contradiction and concludes the proof. Q.E.D. Remark 2. Hilbert's theorem was generalized by N. V. Efimov, "Appearance of Singularities on Surfaces of Negative Curvature," Math. Sb. 106 (1954). A.M.S. Translations. Series 2, Vol. 66, 1968, 154190, who proved the following conjecture of CohnVossen: Let S be a complete surface with curvature K satisfying K j S < 0. Thevl there exists no isometric immersiovl of S into R3.Efimov7sproof is very long, and a shorter proof would be desirable.
Global Diffarentiel Geometry
454
A n excellent exposition of Efimov's proof can be found i n a paper by T. Klotz Milnor, "Efirnov's Theorem About Complete Immersed Surfaces of Negative Curvature," Adva~cesiu: Mathematics 8 (1972), 474543. This paper also contains another proof of Hilbert's theorem which holds for surfaces of cIass C2. F o r further details o n immersion of the hyperbolic plane see M. L. Gromov and V. A. Rokhlin, "Embeddings and Immersions in Riemannian Geometry," Russia~Math. Surveys (1.970), 1.57, especially p. 15.
EXERCISES 1. (Stoker's Remark.) Let S be a complete geometric surface. Assume that the
Gaussian curvature K satisfies K (6 < 0. Show that there is no isometric immersion q : S + R3 such that the absolute value of the mean curvature H is bounded. This proves Efimov's theorem quoted in Remark 2 with the additional condition on the mean curvature. The following outline may be useful: a. Assume such a q exists and consider the Gauss map N: y(S) c R3 + S2, where S2is the unit sphere. Since K # 0 everywhere, N induces a new metric ( , ) on S by requiring that N 0 p : S +S2 be a local isometry. Choose coordinates on S so that the images by p of the coordinate curves are lines of curvature of p(S). Show that the coefficients of the new metric in this coordinate system are
where E, F (= O), and G are the coefficients of the initial metric in the same system. b. Show that there exists a constant M > 0 such that k: I Mykg (M. Use the fact that the initial metric is complete to conclude that the new metric is also complete. c. Use part b to show that S is compact; hence, it has points with positive curvature, a contradiction. 2. The goal of this exercise is to prove that there is no regular complete surface of revolution S in R3 with K _( 6 < 0 (this proves Efimov's theorem for surfaces of revolution). Assume the existence of such an S c R3. a. Prove that the only possible forms for the generating curve of S are those in Fig. 560(a) and (b), where the meridian curve goes to infinity in both directions. Notice that in Fig. 560(b) the lower part of the meridian is asymptotic to the z axis. b. Parametrize the generating curve ( ~ ( s )y/(s)) , by arc length s E R so that y/(O) = 0. Use the relations p" f Ky = 0 (cf. Example 4, Sec. 33, Eq. (9)) and K I 6 < 0 to conclude that there exists a point so r LO, oo) such that =
=.
+
Hilberf's Theorem
455
c. Show that each of the three possibilities to continue the meridian (q(s), ly(s)) of S past the point po = (q(s,,), I,Y(s~) (described in Fig. 560(c) as I, 11, and 111) leads to a contradiction. Thus, S is not complete.
Figure 560

3. (T. K. Milnor's Proof of Hilbert 's Theorem.) Let S be a pIane with a complete 1. Assume that there exists an isometric metric gl such that its curvature K
immersion q : S 3R3. TO obtain a contradiction, proceed as follows: a, Consider the Gauss map N: q(S) c R3 + S2and let gz be the metric on S obtained by requiring that N 0 q : S + S2be a local isometry. Choose local coordinates on S so that the images by q of the coordinate curves are the asymptotic curves of q(S). Show that, in such a coordinate system, gl can be written as du2
+ 2 cos 8 du dv + dv2
and that g, can be written as du2
+

2 cos 8 du dv
+ dv2.
b. Prove that g3 = +(gl g2) is a metric on S with vanishing curvature. Use the fact that g, is a complete metric and 3g3 2 g l to conclude that the metric g3 is complete. c. Prove that the plane with the metric g3 is globally isometric to the standard (Euclidean) plane R2. Thus, there is an isometry q : S , R2. Prove further that q maps the asymptotic curves of S, parametrized by arc length, into a rectangular system of straight lines in RZ,parametrized by arc length. d; Use the global coordinate system on S given by part c, and obtain a contradiction as in the proof of Hilbert's theorem in the text.
A ppen dix
pointset
T O ~ O I O ~  ,
of Euclidean Spaces
In Chap. 5 we have used more freely some elementary topological properties of Rn. The usual properties of compact and connected subsets of Rn, as they appear in courses of advanced calculus, are essentially all that is needed. For completeness, we shall make a brief presentation of this material here, with proofs. We shall assume the material of the appendix to Chap. 2, Part A, and the basic properties of real numbers.
A. Preliminaries Here we shalI complete in some points the material of the appendix to Chap. 2, Part A. In what follows U c Rnwill denote an open set in Rn. The index i varies in the range 1,2, . . . , m, . . . , and if p = ( x , , . . . , x,), q = ( y , , . . . , y,), I p  q I will denote the distance from p to q ; that is,
DEFINITION 1. A sequence p,, . . . , pi, . . . E Rnconverges to p, E Rn ifgiven E > 0, there exists an index i, of the sequence such that pi E B,(po) for all i > i,. In this situation, p, is the limit of the sequence {pi}, and this is denoted by {pi] 3 p,. Convergence is related to continuity by the following proposition.
Appendix :PointSet Topology of Euclidean Spaces
457
PROPOSITION 1. A map F: U c Rn Rm is covltivluous a t p, E U if avld ovlly iffor each convergivlg sequevlce {pi) 4 p, ivl U, the sequence {F(pi)] covlverges to F(p,).
ProoJ Assume F to be continuous at pa and let E. > 0 be given. By continuity, there exists 6 > 0 such that F(B,(p,)) c B,(F(p,)). Let (pi] be a sequence in U, with {pi) po E U. Then there exists in correspondence with 6 an index i, such that pi F: R,(p,) for i > i,. Thus, for i > i,,

which implies that { F ( p i ) ) .F(p,). Suppose now that F is not continuous at p,. Then there.exists a number E > 0 such that for every 6 > 0 we can find a point p E B,(p,), with F(p) $ B,(F(p,)). Fix this E, and set 6 = 1, 1/2, . . . , lli, . . . ,thus obtaining a sequence {pi] which converges to p,. However, since F(pi) $ B,(F(p,)), Q.E.D. the sequence {F(pi)) does not converge to F(p,). DEFINITION 2. A poivlt p E Rn is a limit point of a set A c Rnif every vleighborhood of p irz Rncovltains ovle poivlt of A distivlct from p.
To avoid some confusion with the notion of limit of a sequence, a limit point is sometimes called a cluster pmht or an accumulatiovl poivlt. Definition 2 is equivalent ta saying that every neighborhood V of p contains infinitely many points of A. In fact, let q , # p be the point of A given by the definition, and consider a ball. B,(p) c V so that q, $ R,(p). Then there is a point q, # p, q, E A n B,(p). By repeating this process, we obtain a sequence {qi}in V, where the q, E A are all distinct. Since {q,) +p, the argument also shows that p is a limit point of A if and only if p is the limit of some sequence of distinct points in A. Example 1. The sequence 1, 112, 113, . . . , l/i, . . . converges to 0. The sequence 312, 413, . . . , i lli, . . . converges to 1. The "intertwined" sequence 1, 312, 112,413, 113, . . . , 1 (lli), lli, . . . does not converge and has two limit points, namely 0 and 1 (Fig. A51).
+
+
Figure AS1
It should be observed that the limit p, of a converging sequence has the property that any neighborhood of p , contains all but a finite number of
458
G/oba/ Differential Geometry
points of the sequence, whereas a limit point p of a set has the weaker property that any neighborhood o f p contains infinitely many points of the set. Thus, a sequence which contains no constant subsequence is convergent if and only if, as a set, it contains only one limit point. An interesting example is given by the rational numbers Q . It can be proved that Q is countable; that is, it can be made into a sequence. Since arbitrarily near any real number there are rational numbers, the set of limit points of the sequence Q is the real line R.
DEFINITION 3. A set F c R n is dosed if every limit poivlt of F belovlgs to F. The closure of A c Rn devloted by A, is the uvlion of A with its limit poivlts. Intuitively, F is closed if it contains the limit of all its convergent sequences, or, in other words, it is invariant under the operation of passing to the limit. It is obvious that the closure of a set is a closed set. It is convenient to make the convention that the empty set $ is both open and closed. There is a very simple relation between open and closed sets.
PROPOSITXON 2. F Rn  F of F is open.
c
Rn is closed if avld ovlly if the complemevlt
ProoJ Assume F to be closed and let p E Rn  F. Since p is not a limit point of F, there exists a ball B,(p) which contains no points of F. Thus, B, c Rn F ; hence Rn F is open. Conversely, suppose that Rn  F is open and that p is a limit point of F. We want to prove that p E F. Assume the contrary. Then there is a ball B,(p) c Rn F. This implies that B,(p) contains no point of F and contraQ.E.D. dicts the fact that p is a limit point of F. Continuity can also be expressed in terms of closed sets, which is a consequence of the following fact.
PROPOSITION 3. A map F: U c Rn Rm is contivluous if and ovlly i f f o r each opevl set V c Rm, F S 1 ( V )is an open set. +
ProoJ: Assume F to be continuous and Iet V c Rm be an open set in Rm. If F S 1 ( V )= 6, there is nothing to prove, since we have set the convention that the empty set is open. If F  ' ( V ) # $, let p E F  ' ( y ) . Then F ( p ) E V, and since V is open, there exists a ball B,(F(p)) c V. By continuity of F, there exists a ball B,(p) such that
Appendix :PointSet Topology o f Euclidean Spaces
Thus, B,(p) c F P 1 ( V ) hence, ; F  l ( V ) is open. Assume now that F P 1 ( V )is open for every open set V c Rm. Let p E U and E > 0 be given. Then A = FW1(B,(F(p)))is open. Thus, there exists S > 0 such that Ba(p) c A. Therefore,
hence, F is continuous in p.
Q.E.D.
COROLLARY. F: U c Rn 3 Rm is covltivluous if and only if for ever}' closed set A c Rm, F  l ( A ) is a closed set. Example 2, Proposition 3 and its corollary give what is probably the best way of describing open and closed subsets of Rn. For instance, let f: R 2 R be given by f (x,y ) = ( x 2 / a 2 ) ( y2/b2) 1. Observe that f is continuous, 0 E R is a closed set in R, and (0, +m) is an open set in R. Thus, the set +
is closed in R 2 , and the sets
are open in R2. On the other hand, the set A
=
+
((x,y ) E R 2 , x2 y2 < I} U I(X, y ) E R 2 ; x 2 y 2 = 1 , x > 0, y
+
> OI)
is neither open nor closed (Fig. A52). The last example suggests the following definition.
DEFINITION 4. Let A c Ru. The boundary Bd A of A is the set of poivlts p ivl R n such that every vleighborhood of p contaivls poivlts ivl A and poivlts ivl Rn  A.
+
Thus, if A is the set of Example 2, Bd A is the circle x2 y 2 = 1. Clearly, A c Rn is open if and only if no point of Bd A belongs to A, and B c R" is closed if and only if all points of Bd B belong to B. A final remark on these preliminary notions: Here, as in the appendix to Chap. 2, definitions were given under the assumption that Rn was the "am
Global Differential Geometry
Figure A52
bient" space. It is often convenient, as already remarked in the appendix to Chap. 2, to extend such definitions to subsets of an arbitrary set A c Rn. To do that, we adopt the following definition.
DEFINITION 5. Let A c Rn.We say that V c A is ail open set in A if there exists an open set U in Rn such that V = U n A. A neighborhood of p E A in A is an open set in A containing p. With this notion of "proximity" in A, it is a simple matter to extend the previous definitions to subsets of A and to check that the propositions already proved still hold with the new definitions. Now we want to recall a basic property of the reaI numbers. We need some definitions.
DEFINITION 6. A subset A c R of the real line R is bounded above i f there exists M E R such that M > a for all a E A. The number M is called an upper bound for A. When A is bounded above, a supremum or a least upper bound of A, sup A (or 1.u.b. A) is an upper bound M which satisfies the following condition: Given E > 0, there exists a E A such that M  E < a. By changing he sign of the above inequalities, we define similarly a lower bound for A and an infimum (or a greatest lower bound) of A, inf A (or g.1.b. A). AXIOM OF COMPLETENESS OF REAL NUMBERS. Let A c R be nonernpty and bounded above (below). Then there exists sup A (inf A). There are several equivalent ways of expressing the basic property of compIeteness of the realnumber system. We have chosen the above, which, although not the most intuitive, is probably the most effective one. It is convenient to set the following convention. If A c R is not bounded
Appendix :PointSet Topology of Euclidean Spaces
467
above (below), we say that sup A = +m (inf A = a) With . this convention the above axiom can be stated as follows: Every vlovlempty set of real numbers has a sup and an inf.
Example 3. The sup of the set (0, 1) is 1, which does not belong to the set. The sup of the set
is 2. The point 2 is an isolated point of B ; that is, it belongs to B but is not a limit point of B. Observe that the greatest limit point of B is 1, which is not sup B. However, if a bounded set has no isolated points, its sup is certainly a limit point of the set. One important consequence of the completeness of the real numbers is the following "intrinsic" characterization of convergence, which is actually equivaIent to completeness (however, we shall not prove that). L E M M A 1. Call a sequence {xi) of real numbers a Cauchy sequence f givevl E < 0, there exists i, such that [ xi  xj I < E for all i, j > i,. A sequence is convergevlt if and only if it is a Cauchy sequence. ProoJ: Let {xi}+ x0. Then, if E > 0 is given, there exists i, such that I xi  X, I < ~ / for 2 i > io. Thus, for i,.j > i,, we have
hence, {xi) is a Cauchy sequence. Conversely, let (x,) be a Cauchy sequence. The set { x i ) is clearly a bounded set. Let a, = inf{xi), b, = sup{xi). Either, one of these points is a limit point of {xi) and then (xi] converges to this point, or both are isolated points of the set Ex,). In the latter case, consider the set of points in the open interval (a,, b,), and let a, and b, be its inf and sup, respectively. Proceeding in this way, we obtain that either (xi) converges or there are two bounded sequences a , < a, < and b, > b, >  .  .Let a = sup{a,) and b =inf(b,). Since {xi}is a Cauchy sequence, a = b, and this common value x, is the unique Q.E.D. limit point of { x i } .Thus, {xi) 4 x,.  . =
This form of completness extends naturally to Euclidean spaces.
DEFINITION 7. A seqzrevlce {pi), pi E Rn,fs a Cauchy sequence if given E > 0, there exists avl index i, such that the distavlce I pi  pj I < E for all i, j > i,.
462
Global Differential Geometry
PROPOSITION 4. A sequence (pi], pi E R", converges if and only fi it is a Cauchy sequence. ProoJ: A convergent sequence is clearly a Cauchy sequence (see the argument in Lemma 1). Conversely, let { p i ) be a Cauchy sequence, and consider its projection on the j axis of R",j .= 1, . . . , n. This gives a sequence of real numbers (x,,} which, since the projection decreases distances, is again a Cauchy sequence. By Lemma 1, {xi,),xi,. It follows that {pi] + p0 = Q.E.D. { X I 0 , X20, . . . xno). 9
6. Connected Sets DEFINITION 8. A covltinuoz~scurve oc: [a, b] arc in A joining a(a) to a(b).
+
A c Rn is called an
DEFINITION 9. A c Rn is arcwise connected if, given two points p, q E A, there exists m arc in A joining p to q.
Earlier in the book we have used the word connected to mean arcwise connected (Sec. 22). Since we were considering only regular surfaces, this can be justified, as will be done presently. For a general subset of Rn,however, the notion of arcwise connectedness is much too restrictive, and it is more convenient to use the following definition. DEFINITION 10. A c Rn is connected whevl if is not possible to write A = U , u U,, where U, and U, are nonempty open sets ivl A and U, n U2 =
6. Intuitively, this means that it is impossible to decompose A into disjoint pieces. For instance, the sets U , and F, in Example 2 are not connected. By taking the complements of U , and U,, we see that we can replace the word "open" by "closed" in Def. 10. PROPOSITION 5. Let A c R n be covlnected and let B c A be si'multaneously open and closed in A. Then either B = $ or B  A.

Proof. Suppose that B # 4 and B # A and write A B U ( A  B). Since B is closed in A, A  B is open in A, Thus, A is a union of disjoint, nonvoid, open sets, namely B and A  B. This contradicts the connectedness Q.E.D. of A .
The next proposition shows that the continuous image of a connected set is connected.
Appendix : PointSet Topology of Euclidean Spaces
PROPOSITION 6. Let F: A c Rn necdtetl. Then F(A) i.s chonnected.
+
463
R" be continuous and A be con
Proof: Assume that F(A) is not connected. Then F(A) = U ,u U,, where U , and U, are disjoint, nonvoid, open sets in F(A). Since F is continuous, FI(U,), F'(U2) are also disjoint, nonvoid, open sets in A . Since A = F (U,) u F'(U,), this contradicts the connectedness of A. Q.E.D. For the purposes o f t his section, it is convenient to extend the definition of interval as follows :
DEFINITION 11. An interval of the real line R is any of the sets a < x < b a s x s b , a < x < b, a < x < b , x r R . The cases a = b, a =  x b = +co are not excluded, so that an interval may be a point, a halfZine, 01 R itself. PROPOSITION 7. A c R is connectrd if and only
if A
is an intervnl.
Proof. Let A c R be an interval and assume that A is not connected. We shall arrive at a contradiction. Since A is not connected, A = U, U U,, where U, and U, are nonvoid. disjoint, and open in A. Let a , E U,, b, E U, and assume that a , < b,. By b,)/2, we dividing the closed interval [a,, b,] = I, by the midpoint (a, obtain two intervals, one of which, to be denoted by I,, has one of its end points in U , and the other end point in U,. Considering the midpoint of I2 and proceeding as before, we obtain a n interval I, c I, c I,. Thus, we obtain a family of closed intervals I, 3 I, =, . . . 3 In3 . .  whose lengths approach zero. Let us rewrite Ii = [ci, d,]. Then c l ( c, (  . ( c, ( . . . , dn . . . Let c = sup{c,) and d = infld,). Since and dl d2 2 . di  ci is arbitrarily small, c = d. Furthermore, any neighborhood of c contains some Iifor i sufficiently large. Thus, c is a limit point of both U, and U,. Since U, and U2 are closed, c E U, n U,, and that contradicts the disjointness of U, and U,. Conversely, assumc that A is connected. If A has a single element, A is trivially an interval. Suppose that A has at least two elements, and let n = inf A, b = sup A, a + b. Clearly, A c [a, b]. We shall show that (a, b) c A, and that implies that A is an interval. Assume the contrary; that is, there exists t, a < t < b, such that t $ A. The sets A n (00, t ) = V1, A n ( t , $03) = V, are open in A = V , u V,. Since A is connected, one of this implies both that b $ A these sets, say, V2, is empty. Since b E (t, +a), and b is not a limit point of A. This contradicts the fact that b = sup A. In the same way, if V, = 4, we obtain a contradiction with the fact that Q.E.D. a = inf A.
+
>
> >
464
Global Differential Geometry
PROPOSITION 8. Let f: A c Rn+ R be continuous and A be connected. Assume that f ( q ) # 0 for all q E A. Then f does not change sign in A. Pro05 By Prop. 5, f ( A ) c R is connected. By Prop. 7, f ( A ) is an interval. By hypothesis, f ( A ) does not contain zero. Thus, the points in f ( A ) all have Q.E.D. the same sign.
PROPOSITION 9. Let A c Rn be arcwise connected. Then A is connected. Proof. Assume that A is not connected. Then A = U, u U2, where U,, U2 are nonvoid, disjoint, open sets in A. Let p E U1, q E U2. Since A is arcwise connected, there is an arc a: [a, bJ A joining p to q. Since a is continuous, B = a([a,b ] ) c A is connected. Set V, =  B n U,, V, = B n U2. Then B = V, u V,, where V, and V2 are nonvoid, disjoint, open sets in B, and that is a contradiction. Q.E.D. 4
The converse is, in general, not true. However, there is an important special case where the converse holds.
DEFINITION 12. A set A c Rn is locally arcwise connected if for each p E A and each neighborhood V of p in A there exists an arcwise connected neighborhood U c V of p in A. Intuitively, this means that each point of A has arbitrarily small arcwise connected neighborhoods. A simple example of a locally arcwise connected set in R3 is a regular surface. In fact, for each p E S and each neighborhood W o f p in R3, there exists a neighborhood V c W of p in R3 such that V n S is homeomorphic to an open disk in R2; since open disks are arcwise connected, each neighborhood W n S of p E S contains an arcwise connected neighborhood. The next proposition shows that our usage of the word connected for arcwise connected surfaces was entirely justified.
PROPOSITlON 10. Let A c Rnbe a locally arcwise connected set. Then A is connected i f and only if it is arcwise connected. Proof. Half of the statement has already been proved in Prop. 9. Now assume that A is connected. Let p E A and let A , be the set of points in A that can be joined to p by some arc in A . We claim that A , is open in A . In fact, let q E A , and let a : [a, b] + A be the arc joining p to q. Since A is locally arcwise connected, there is a neighborhood V of q in A such that q can be joined to any point r E V by an arc 8: [b, c] + V (Fig. A53).
Figure A53
It follows that the arc in A,
joins q to r, and this proves our claim. By a similar argument, we prove that the complement of A , is also open in A. Thus, A , is both open and closed in A . Since A is locally arcwise connected, A , is not empty. Since A is connected, A , = A. Q.E.D. ExampIe 4. A set may be arcwise connected and yet fail to be locally arcwise connected. For instance, let A c R 2 be the set made up of vertical Iines passing through (lln, 0), n = I , . . . , plus the x and y axis. A is clearly arcwise connected, but a small neighborhood of (0, y), y # 0, is not arcwise connected. This comes from the fact that although there is a "long" arcjoining any two points p, q E A, there may be no short arc joining these points (Fig. A54).
Figure A54
Global Differential Geometry
C. Compact Sets DEFINITION 13. A set A c Rnis bounded f i t is contained in some ball of R". A set K c Rnis compact if it is closed and bounded.
We have already met compact sets in Sec. 27. For completness, we shall prove here properties 1 and 2 of compact sets, which were assumed in Sec. DEFINITION 14. An open cover of a set A c Rnis a family of open sets {U,], a E a such that U, U, = A. When there are onlyjinitely many U, in the family, we say that the cover is finite. If the subfamily (Up], P E 63 c a, still covers A, that is, U,, Up = A, we say that CUB}is a subcover of (U,). PROPOSITION 11. For a set K c R n the following assertions are equivalent: I . K is compact. 2. (HeineBorel). Every open cover of K has afinite subcover. 3. (BolzanoWeierstrass). Every injinite subset of K has a limit point in K . Proof. We shaIl prove 1 =, 2 =, 3 =, 1. 2: Let (U,), a E a, be an open cover of the compact K, and assume 1 that (U,] has no finite subcover. We shall show that this leads to a contradiction. Since K is compact, it is contained in a closed rectangular region
B
=
{(xI, . . . , x,)
E
Rn;a j (x j (b j ,
j
=
1, . . . , n].
+
Let us divide B by the hyperplanes x j = ( a j bj)/2(for instance, if K c R2, B is a rectangle, and we are dividing B into 22 = 4 rectangles). We thus obtain 2" smaller closed rectangular regions. By hypothesis, at least one of these regions, to be denoted by B,, is such that B , n K is not covered by a finite number of open sets of {U,]. We now divide B , in a similar way, and, by repeating the process, we obtain a sequence of closed rectangular regions (Fig. A55)
which is such that no Bi n K is covered by a finite number of open sets of {U,) and the length of the largest side of Bi converges to zero. We claim that there existsp E n Bi.In fact, by projecting each Bi on the j axis of R",j = 1, . . . , n, we obtain a sequence of closed intervals
Appendix: PointSet Topology of Euclidean Spaces
Figure AS5
Since (bji  a,,) is arbitrarily small, we see that
a j = sup{aji] = inflb,,]
= b,;
hence,
ni
B,, as we claimed. Thus, p = (al, . . . , a,) E Now, any neighborhood of p contains some Bi for i sufficiently large; hence, it contains infinitely many points of K. Thus, p is a limit point of K, and since K i s closed, p E K. Let U , be an element of the family (U,) which contains p. Since U , is open, there exists a ball B,(p) c U,. On the other hand, for i sufficiently large, Bic B,(p) c U,. This contradicts the fact that no B, n K can be covered by a finite number of U,'s and proves that 12. 2 =+ 3. Assume that A c K is an infinite subset of K and that no point of .K is a limit point of A. Then it is possible, for each p E K, p 4 A, to choose a neighborhood Vp o f p such that V, n A = I$ and for each q E A to choose a neighborhood W, of q such that W , n A = q. Thus, the family (V,, W,), p E K  A, q E A, is an open cover of K. Since A is infinite and the omission of any W, of the family leaves the point q uncovered, the family (V,, W,] has no finite subcover. This contradicts assertion 2. 3  1 : We have to show that K is closed and bounded. K is closed, because if p is a limit point of K, by considering concentric. balls Bl,i(p) = B,, we obtain a sequence p , E B1  B2, p 2 E B2  B3, . . . , p i E Bi  Bif . . . which has p as a limit point. By assertion 3, p E K. iY is bounded. Otherwise, by considering concentric balls Bi(p), of radius 1, 2, . . . , i, . . . , we will obtain a sequence p , E B,, p 2 E B2  B1, . . . , p, E Bi  Bi,, . . . with no limit point. This proves that 3  1. Q.E.D.
468
Global Differential Geometry
The next proposition shows that a continbous image of a compact set is compact.
PROPOSITION 12. Let F: K c Rn, Rm be continuous and let K be compact. Then F(K) is compact. Proof. If F(K) is finite, it is trivially compact. Assume that F(K) is not finite and consider aminfinite subset (F(p,)] c F(K), p, E K. Clearly the set (p,} c K is infinite and has, by compactness, a limit point q E K. Thus, there exists a sequence p,, . . . ,pi, . . . , q, pi E {p,}. By the continuity of F, the sequence F(pi) + F(q) E F(K) (Prop. I). Thus, (F(p,)] has a limit point F(q) E F(K); hence, F(K) is compact. Q.E.D.

The following is probably the most important property of compact sets.
PROPOSITION 13. Let f: K c R" R be a continuousfunction defined on a compact set K. Tlzen tlzere exists p,, p, E K such that +
that is, f reaclzes a maximum at p, and a minimum at p,. Proof. We shall prove the existence of p , ; the case of minimum can be treated similarly. By Prop. 12, f(K) is compact, and hence closed and bounded. Thus, there exists sup f (K) = x,. Since f (K) is closed, x, E f (K). It follows that there exists p , E K with x I = f(pl). Clearly, f(p) 5 f ( p l ) = x , for all Q.E.D. p E K. Although we shall make no use of it, the notion of uniform continuity fits so naturally in the present context that we should say a few words about it. A map F : A c Rn+ Rm is uniformly continuous in A if given 6 > 0, there exists 6 > 0 such that F(B,(p)) c B,(F(p)) for all p E A . Formally, the difference between this definition and that of (simple) continuity is the fact that here, given 6, the number 6 is the same for all p E B, whereas in simple continuity, given 6, the number 6 may vary withp. Thus, uniform continuity is a global, rather than a local, notion. It is an important fact that on compact sets the two notions agree. More precisely, let F: K c Rn,Rm be continuous and K be compact. Then F is uniformly continuous in K. The proof of this fact is simple if we recall the notion of the Lebesgue number of a n open cover, introduced in Sec. 27. In fact, given 6 > 0, there exists for each p c K a number 6!p) > 0 such that F(B,,,,(p)) c B,;,(F(p)).
Appendix: PointSef Topology o f Euclidean Spaces
469
The family (B,,,,(p), p E K ] is an open cover of K. Let 6 > 0 be the Lebesgue number of this family (Sec. 27, property 3). If q E B,(p), p E K, then q andp belong to some element of the open cover. Thus, I F(p)  F(q) I < 6. Since q is arbitrary, F(B,(p)) c B,(F(p)). This shows that 6 satisfies the definition of uniform continuity, as we wished.
DmConnected Components When a set is not connected, it may be split into its connected components. To make this idea precise, we shall first prove the following proposition.
PROPOSITION 14. Let C, c Rnbe a family of connected sets such that
Then
U C,
=C
is a connected set.
a
Proof. Assume that C = U, u U,, where U, and U2 are nonvoid, disjoint, open sets in C , and that some point q E n,C, belongs to U,. Let q E U,. Since C = U,C, a n d p E n,C,, there exists some C, such that p, q E C,. Then C, n U1 and C, n U, are nonvoid, disjoint, open sets in C,. This Q.E.D. contradicts the connectedness of C, and shows that Cis connected.
DEFINITION 15. Let A c Rn and p E A. The union of all connected subsets of A which contain p is called the connected component of A containing P.
By Prop. 14, a connected component is a connected set. Intuitively the connected component of A containing p E A is the largest connected subset of A (that is, it is contained in no other connected subset of A that containsp). A connected component of a set A is always closed in A. This is a consequence of the following proposition. PROPOSITION 15. Let C c A c R" be a connected set. Then the closure of C in A is connected. Proof. Let us suppose that C = U1 U U2, where U1, U2 are nonvoid, Since C 3 C, the sets C n U, = V,, C h U, = V2 disjoint, open sets in are open in C, disjoint, and V , u V, = C. We shalI show that V, and V 2 are nonvoid, thus reaching a contradiction with the connectedness of C. Let p E U 1 .Since U , is open in C, there exists a neighborhood W of p in A such that W n c U,. Since p is a limit of C, there exists q E W n C c W n c U,. Thus, q E C n u, = V , , and V , is not empty. In a similar way, it can be shown that V , is not empty. Q.E.D.
c.
470
Global Differential Geometry
COROLLARY. A connected component C c A c Rn of a set A is closed in A.
c
c,
In fact, if # C, there exists a connected subset of A, namely which contains C properly. This contradicts the maximality of the connected component C. In some special cases, a connected component of set A is also an open set in A .
PROPOSITION 16. Let C c A c Rn be a connected component of a locally arcwise connected set A. Then C is open in A. Proof. Let p E C c A. Since A is locally arcwise connected, there exists an arcwise connected neighborhood V of p in A. By Prop. 9, V is connected. Since C is maximal, C 3 V ;hence, Cis open in A. Q.E.D.
Bibliograpby an d Comments
The basic work of differential geometry of surfaces is Gauss' paper "Disquisitiones generales circa superficies curvas," Cornm. Soc. Goftingen Bd 6, 18231827. There are translations into several languages, for instance, 1. Gauss, K. F., General Investigations of Curved Surfaces, Raven Press, New York, 1965. We believe that the reader of this book is now in a position to try to understand that paper. Patience and openmindedness will be required, but the experience is most rewarding. The classical source of differential geometry of surfaces is the fourvolume treatise of Darboux : 2. Darboux, G., Thkorie des Surfaces, GauthierViIlars, Paris, 1887, 1889, 1894, 1896. There exists a reprint published by Chelsea Publishing Co., Inc., New York. ,
This is a hard reading for beginners. However, beyond the wealth of information, there are still many unexplored ideas in this book that make it worthwhile to come to it from time to time. The most influential classical text in the EngIish language was probably 3. Eisenhart, L. P.? A Treatise on the Dlflerential Geometry of Curves and Surfaces, Ginn and Company, Boston, 1909, reprinted by Dover, New York, 1960.
An excellent presentation of some intuitive ideas of classical differential geometry can be found in Chap. 4 of
472
Bibliography and Comments
4. Hilbert, H., and S. CohnVossen, Geometry and Imagination, Chelsea Publishing Company, Inc., New York, 1962 (translation of a book in German, first published in 1932).
Below we shall present, in chronological order, a few other textbooks. They are more or less pitched at about the level of the present book. A more complete list can be found in [9], which, in addition, contains quite a number of global theorems.
5. Struik, D. J., Lectures on Classical DifSerential Geometry, AddisonWesley, Reading, Mass., 1950. 6. Pogorelov, A. V., Dzfferential Geometry, Noordhoff, Groningen, Netherlands, 1958.
7. Willmore, T. J., An Introduction to Dzfferential Geometry, Oxford University Press, Inc., London 1959. 8. O'Neill, B., Elementary Dzfferential Geometry, Academic Press, New York, 1966.
9. Stoker, J. J., Dzflerential Geometry, WileyInterscience, New York, 1969.
A clear and elementary exposition of the method of moving frames, not treated in the present book, can be found in [8]. Also, more details on the theory of curves, treated briefly here, can be found in [5], [6], and [9]. Although not textbooks, the following references should be included. Reference [lo] is a beautiful presentation of'some global theorems on curves and surfaces, and [ l l ] is a set of notes which became a classic on the subject. 10. Chern, S. S., Curves and Surfaces in Euclidean Spaces, Studies in Global Geometry and Analysis, MAA Studies in Mathematics, The Mathematical Association of America, 1967.
11. Hopf, H., Lectures on Diflerential Geometry in the Large, notes published by Stanford University, 1955. For more advanced reading, one should probably start by learning something of differentiable manifolds and Lie groups. For instance,
12. Spivak, M., A Comprehensive Introduction to Differential Geometry, Vol. I, Brandeis University, 1970. 13. Warner, F., Foundations of Differentiable Manifolds and Lie Groups, Scott, Foresman, Glenview, Ill., 1971. Reference [12] is a delightfulreading. Chapters 14 of [13] provide a short and efficient account of the basics of the subject. After that, there is a wide choice of reading material, depending on the reader's tastes and interests. Below we include a possible choice, by no means unique. In El61 and [17] one can find extensive lists of books and papers.
Bibliography and Comments
473
14. Berger, M., P. Gauduchon, and E. hlazet, Le Spectre d'une Vurie'te' Riemannienne, Lecture Notes 194, Springer, Berlin, 1971. 15. Bishop, R. L., and R. J. Crittenden, Geometry of Manifolds, Academic Press, New York, 1964. 16. Cheeger, J., and D. Ebin, Comparison Theorems in Riemannian Geometry, NorthHolland, Amsterdam, 1974. 17. Helgason, S., Diferential Geometry and Symmetric Spaces, Academic Press, New York, 1963. 18. Kobayashi, S., and K. Nomizu, Foundations of Diferential Geometry, Vols. I and 11, WileyInterscience, New York, 1963 and 1969. 19. Klingenberg, W., D. Gromoll, and W. Meyer, Riemannsche Geometrie im Grossen, Lecture Notes 55, SpringerVerlag, Berlin, 1968. 20. Lawson, B., Lectures on Minimal Submanifolds, Monografias de Matemitica, IMPA, Rio de Janeiro, 1973. 21. Milnor, J., Morse Theory, Princeton University Press, Princeton, N. J., 1963. 22. Spivak, M., A Comprehensive Introduction to D~flerentialGeometry, Vol. 11, Brandeis University, 1970. The theory of minimal submanifolds, 1201 and the references therein; the problems associated with the spectrum, 1141; and the topological behavior of positively curved manifolds, [161 and [19], are only three of the many interesting topics of presentday differential geometry.
Hints and Answers
SECTION 73 2. a. a(t) = ( t  sin t, 1  cos t ) ; see Fig. 17. Singular points: t is any integer.
=
2nn, where n
7. b. AppIy the mean value theorem to each of the functions x, y, z to prove that the ) h, k vector (a(t h)  a(t k))j(h  k ) converges to the vector ~ ' ( tas +0. Since a f ( t )f 0 , the line determined by a(t h), a(t k ) converges to the line determined by a'(t).
+
+
8. By the definition of integral, given
+
+
6
> 0, there exists a 6' > 0 such that if
IP]< S', then
On the other hand, since a' is uniformly continuous in [a,b], given E > 0, there then exists 6" > 0 such that if t, s E [a,b] with It  s ]
I
C (ti 1

ti) I af(ti)11
Si
where ti, _( si _( ti. Together with the above, this gives the required inequality.
Hints and Answers
SECTION 74 2. Let the points po = (x,, yo, zo) and p = (x, y, z) belong to the plane P. Then axo byo $ czo d = 0 = ax by cz d. Thus, a(x  x0) b(y  yo) c(z  z,) = 0. Since the vector (x  x,,y  yo, z  zo) is parallel to P, the vector (a, b, c) is normal to P. Given a point p = (x, y, z) E P, the distance p from the plane P to the origin 0 is given by p = [ pI cos 0 = (p v)/l v 1, where 0 is the angle of Op with the normal vector v. Sincep . v = d,
+
+
+ + +
+
+

3. This is the angle of their n o r W vectors.
4. Two planes are parallel if and only if their normal vectors are parallel. 6. v l and v z are both perpendicular to the line of intersection. Thus, v l A v2 is parallel to this line. 7. A plane and a line are parallel when a normal vector to the plane is perpendicular to the direction of the line. 8. The direction of the common perpendicular to the given lines is the direction of u A v. The distance between these lines is obtained by projecting the vector r = (x,,  x l , yo  yl, z0  zl) onto the common perpendicular. Such a projection is clearly the inner product of r with the unit vector (u A ,v)/l u A v I.
SECTION 75 kn, a"' = kn' $ k'n 4. Differentiate a(s) + I(s)n(s) = const., obtaining 2. Use the fact that
a'
=
t, a"
=
=
k2t $ krn  kzb.
It follows that z = 0 (the curve is contained in a plane) and that 1= const. = Ilk. 7. a. Parametrize u by arc length. b. Parametrize a by arc length s. The normal lines at sl and s2 are
respectively. Their point of intersection will be given by values of t and z such that
Use Taylor's formula n(s2) = n(s,) 4 (s2  sl)n'(sl) $ R, and let s2 + sl to concIude that a'(sl) = fn'(sl), where t is the limiting common value of t and z as s2 + sl.It follows that i = ilk.
477
Hints and Answers
13. To prove that the condition is necessary, differentiate three times 1 a(s)12 = const., obtaining a(s) = Rn 4 R'Tb. For the sufficiency, differentiate P(s) = a(s) Rn  R'Tb, obtaining
+
Pf(s)= t
+ R(kt
 zb) 4 R'n
 (TRf)'b
On the other hand, by differentiating R2 0
=
2RR'
R'n
=
+ (TR')')b.
(Rz
+ (TR')2= const., one obtains
+ 2(TR')(TR')' = T2R'( R z + (TR')'),
since k' # 0 and z # 0. Hence, P(s) is a constant p,, and
15. Since b' = zn is known, ] z 1 = 16').Then, up to a sign, n is determined. Since t = i z A b and the curvature is positive and given by t' = kn, the curvature can also be determined. 16. First show that
I
Thus, a(s) ds = arc tan (klr);hence, k/z can be determined; since k is positive, this also gives the sign of z. Furthermore, 1 n' l2 = I kt  zb l2 = k 2 z 2 is also known. Together with k/z, this suffices to determine k 2 and T ~ .
+
17. a. Let a be the unit vector of the fixed direction and let 0 be the constant angle. Then t a = cos 8 = const., which differentiated gives n a = 0. Thus, a = t cos 8 b sin 8, which differentiated gives k cos 0 z sin 8 = 0, or k/z = tan 8 = const. Conversely, if k / z = const. = tan 0 = (sin 0lcos a), we can retrace our steps, obtaining that t cos 6' b sin 0 is a constant vector a. Thus, t a = cos 0 = const. b. From the argument of part a, it follows immediately that t a = const. implies that n a = 0 ; the last condition means that n is parallel to a plane normal to a. Conversely, if n a = 0,then (dtlds) . a = 0 ; hence, t a = const. c. From the argument of part a, it follows that t a = const. implies that b . a = const. Conversely, if b a = const., by differentiation we find that n.a=O.

+
+
+


18. a. Parametrize
a by arc length s and differentiate a
=
a 4 rn with respect to s,
obtaining
da  (1  rk)t ds
+ r'n  rzb.
Since dalds is tangent to k , (da/ds).n = 0 ; hence, r'
= 0.
478
Hints and Answers
b. Parametrize a by arc lengths, and denote by S and i the arc length and the unit tangent vector of a. Since di/ds = (dfldF)(di/ds), we obtain that
hence, t
f
= const. = cos
0. Thus, by using that a = a
+ rn, we have
From these two relations, it follows that 1  rk 
rz
const.
=
B. r
Thus, setting r = A, we finally obtain that Ak + Bz = 1. Conversely, let this last relation hold, set A = r, and define Then, by again using the reIation, we obtain
dB ds
 = ( 1  rk)t

rzb
= z(Bt
a
=
u
+ rn.
 rb).
Thus, a unit vector f of a is (Bt  r b ) / J m = i. It follows that dilds = ((Bk  rz)/d=z)n. Therefore, E(s) = hn(s), and the normal lines of a and a at s agree. Thus, a is a Bertrand curve. c. Assume the existence of two distinct Bertrand mates B = a Fn, & = a ?n. By part b there exist constants c l and cz so that 1  Fk = c,(Fz), 1  fk = cz(fz). Clearly, c, # c2. Differentiating these expressions, we obtain k' = z'cl, k' = z'c,, respectively. This implies that k' = z' = 0. Using the uniqueness part of the fundamental theorem of the local theory of curves, it is easy to see that the circular helix is the only such curve.
+
+
SECTION 16 1. Assume that s = 0, and consider the canonical form around s = 0. By condition 1, P must be of the form z = cy, or y = 0. The pIane y = 0 is the rectifying plane, which does not satisfy condition 2. Observe now that if 1 s 1 is sufficiently small, y(s) > 0, and z(s) has the same sign as s. By condition 2, c = z/y is simultaneously positive and negative. Thus, P is the plane z = 0. 2. a. Consider the canonical form of a(s) = (x(s), y(s), z(s)) in a neighborhood of cz = 0 be the plane that passes through a(O), a(0 + h,), s = 0. Let ax by a(0 h,). Define a function F(s) = ax(s) by(s) cz(s) and notice that F(0) = F ( h l ) F(hz) = 0. Use the canonical form to show that F'(0) = a ,
+
+ +

+
+
Hints and Answers
479
F"(0) = bk. Use the mean value theorem to show that as h,, h2 + 0, then a  4 0 and b + 0 . Thus, as hl, h2 +0 the plane ax by = cz = 0 approaches the pIane z = 0, that is, the osculating plane.
+
1. No. Use the isoperimetric inequality.
P
2. Let S be a circle such that A Bis a chord of S and one of the two arcs a and determined by A and B on S 1 ,say dl, has length I. Consider the piecewise C1 closed curve (see Remark 2 after Theorem 1) formed by P and C. Let be fixed and C vary in the family of all curves joining A to B with length 1. By the isoperimetric inequality for piecewise C curves, the curve of the family that bounds the largest area is S1.Since is fixed, the arc of circle a is the solution to our problem.
P
P
4. Choose coordinates such that the center 0 is at p and the x and y axes are directed along the tangent and normal vectors at p, respectively. Parametrize C by arc length, a(s) = (x(s), y(s)), and assume that a(0) = p. Consider the (finite) Taylor's expansion
where lim,,o R/s2 = 0. Let k be the curvature of a at s
where R
= (R,,
= 0,
and obtain
R,) and the sign depends on the orientation of a. Thus,
5. Let 0 be the center of the disk D.Shrink the boundary of D through a family of concentric circles until it meets the curve C at a point p. Use Exercise 4 to show that the curvature k of C at p satisfies I k 1 2 llr. 8. Since a is simple, we have, by the theorem of turning tangents,
Since k(s) _( c, we obtain
9. By the Jordan curve theorem, a simple closed curve C bounds a set K. If K is not convex, there are points p, q E K such that the segment p< contains points
480
Hints a n d Answers
that do not belong to K, and% meets C a t a point r, r # p, q. Use the argument given in the middle of the proof of the fourvertex theorem to show that the line L determined byp and q is tangent to C at the pointsp, q, r and that the segment E i s contained in C c K. This is a contradiction.
11. Observe that the area bounded by H i s greater than or equal to the area bounded by C and that the length of H is smaller than or equal to the length of C. Expand Hthrough a family of curves parallel to H (Exercise 6) until its length reaches the' length of C. Since the area either remains the same or has been further increased in this process, we obtain a convex curve H' with the same length as C but bounding an area greater than or equal to the area of C.
(See Fig. 140.)
5. No. x is not onetoone. 11. b. To see that x is onetoone, observe that from z one obtains i u . Since cosh v > 0, the sign of u is the same as the sign of x. Thus, sinh v (and hence v) is determined.
13. x(u, v) = (sinh u cos v, sinh u sin v, cosh v). 15. Eliminate t in the equations x/a p(t) (0, 0, t ) to q(t) = (a, t, 0).

= ylt = (z

t)/t of the line joining
17. c. Extend Prop. 3 for plane curves and apply the argument of Example 5. 18. For the first part, use the inverse function theorem. To determine F, set u = p2, v =/+an q, w = tan2 6. Write x = f (p, 8) cos p, y = f (p, 8) sin p, where f is t6 be determined. Then / '
It follows that f
=p
cos 8, z = p sin 8. Therefore,
19. No. For C, observe that no neighborhood in R2 of a point in the vertical arc can be written as the graph of a differentiable function. The same argument applies to S.
Hints and Answers
SECTION 23 1. Since A 2 = identity, A
= A1.
5. d is the restriction to S of a function d: R3 d(x, Y, z)
=

+
C(x  ~ 0 ) O,~
R:
+ (2  zO)2)1'2, , zo). (x, Y, 2) f ( ~ 0Yo,
8. If p = (x, y, z), F(p) lies in the intersection with H of the line t + (tx, ty, z), t > 0.Thus,
Let U be R3 minus the z axis. Then F : U c: R3 ,R 3 as defined above is differentiable.
13. Iff is such a restriction, f is differentiable (Example 1). To prove the converse, let x: U+= R3 be a parametrization of S in p. As in Prop. 1, extend x to F: U x R + R3. Let W be a neighborhood of p in R 3 on which F1 is a diffeomorphism. Define g: W + R by g(q) = f 0 x 0 n 0 Fdl(q), q E W, where n : U x R 3 U is the natural projection. Then g is differentiable, and the restriction g J W C? S = f. 16. F is differentiable in S2  { N ] as a composition of differentiable maps. To prove that F i s differentiable at N, consider the stcreographic projection ns from the south pole S = (0, 0, 1) and set Q = ns 0 F 0 nil : , (of course, we are identifying the plane z = 1 with c ) . Show that n, n i l :  (0) + is given by nN 0 7ci1(c)= 4/[. Conclude that
c
0
hence, Q is differentiable at l, = 0.Thus, F = n i l N.
0
Q
0
a: c
c
n, is differentiable at
SECTION 24 1. Let U(t) = (x(t), y(t), z(t)) be a curve on the surface passing through p o = (XO,YO,ZO)for t = 0. Thus, f (x(f), ~ ( t )z(t)) , = 0; hence, f,x'(O) fly'(0) f,zf(0) = 0, where all derivatives are computed at go.This means that all tangent vectors at po are perpendicular to the vector (f,, f,,f,), and hence the desired equation.
+
+
4. Denote by f ' the derivative of f(y/x) with respect to t = y/x. Then z, = f  b/x) f', z, =f '. Thus, the equation of the tangent plane at (x,, yo) is z = xof $ (f  (yo/xo)f')(x  xo) f '(y  yo), where the functions are computed at (x,, yo). It follows that if x = 0, y = 0, then z = 0.
+
Hints and Answers
482
12. For the orthogonality, consider, for instance, the first two surfaces. Their normals are parallel to the vectors (2x  a, 2y, 2z), (2x, 2y  by22). In the intersection of these surfaces, ax = by; introduce this relation in the inner product of the above vectors to show that this inner product is zero. 13. a. Let a(t) be a curve on S with a(0) dfp(w)
a'
= z(,
PO/
It follows that p is a critical point o f f if and only if (w, p w E Tp(S).
 po>=
0 for all
14. a. f (t) is continuous in the interval (  a ,c), and lim,, f (t) = 0, lim ,, f(t) = +a. Thus, for some t t (coy c), f (t ,) = 1. By similar arguments, we find real roots t 2 E (c, b), t, t (b, a). b. The condition for the surfaces f (t ,) = 1,f (t,) = 1 to be orthogonal is
This reduces to
which follows from the fact that t # t2 and f(t,)
 f(t2) =
0.
17. Since every surface is locally the graph of a differentiable function, S1is given by f(x, y, z) = 0 and S2 by g(x, y, z) = 0 in a neighborhood of p ; here 0 is a regular value of the differentiable functions f and g. In this neighborhood of p, S1 n S 2 is given as the inverse image of (0,O) of the map F: R3 4 R 2 : F ( q ) = (f(q), g(q)). Since S, and S2 intersect transversally, the normal vectors (f,, f,,f,) and (g,, g,, g,) are linearly independent. Thus, (0,O) is a regular value of F and S1 fl S2is a regular curve (cf. Exercise 17, Sec. 22). 20. The equation of the tangent plane at (xo,yo, z,) is
zzo xxo YYo z +  +  7  1 

The line through 0 and perpendicular to the tangent plane is given by
From the last expression, we obtain
+
x2a2 b2y2 + c2z2    a2x2  y2b2    z2c2 xxo YYo zzo xxo k YYo I zzo 
From the same expression, and taking into account the equation of the ellips
Hints and Answers
oid, we obtain
Again from the same expression and using the equation of the tangent plane, we obtain
The righthand sides of the three last equations are therefore equal, and hence the asserted equation. 21. Imitate the proof of Prop. 9 of the appendix to Chap. 2. 22. Let r be the fixed line which is met by the normals of S and letp E S. The plane PI,which containsp and r, contains all the normals to S a t the points of PI n S. Consider a plane P2passing throughp and perpendicular to r. Since the normal through p meets r, P2 is transversal to T,(S); hence, P2 n S is a regular plane curve C in a neighborhood of p (cf. Exercise 17, Sec. 24). FurthermoreP1 n P2 is perpendicular to T J S ) n P 2 ; hence, PI n P2is normal to C. It follows that the normals of C all pass through a fixed point q = r n P2;hence, C is contained in a circle (cf. Exercise 4, Sec. 15). Thus, everyp E S has a neighborhood contained in some surface of revolution with axis r. By connectedness, S is contained in a fixed one of these surfaces.
8. Since d E / d v = 0,E = E(u) is a function of u alone. Set ii = J J B du. Similarlj, G = G(v) is a function of v alone, and we can set 2i = Jc dv. Thus, ii and 6 measure arc lengths aIong the coordinate curves, whence I? = = 1, = cos 0.
9. Parametrize the generating curve by arc length.
13. Since the osculating plane is normal to N, N' = zn and, therefore, .t2 = 1 N'12 = k4 cos2 8 kg sin2 6, where 8 is the angle of el with the tangent to the curve. Since the direction is asymptotic, we obtain cos2 8 and sin28 as functions of k l and k2, which substituted in the expression above yields z2 = klk2.
+
14. By setting 1 1 = A 1
~
and 2 L2 = A2N1we have that
11,  n21 = ~ I ( ~ , N ~ )N + ,ZN ~ ) N I I =A ,' ? + A;  2A1A2cos 8.
Hints end Answers
On the other hand,
16. Intersect the torus by a plane containing its axis and use Exercise 15. 18. Use the fact that if
0
= 27z/m, then
which may be proved by observing that
and that the expression under the summation sign is the sum of a geometric progression, which yields
sin(2mO  8) = 1. sin 8
19. a. Express t and h in the basis {el, e 2 ) given by the principal directions, and compute (dN(t),h). b. Differentiate cos 0 = (N, n), use that dN(t) = k,t z,h, and observe that ( N , b) = (h, N ) = sin 0, where b is the binormal vector.
+
20. Let S , , S2, and S3 be the surfaces that pass throughp. Show that the geodesic torsions of C1 = S2 n S3 relative to S2 and S3 are equal; it will be denoted by z l . ' ~ i r n i l a r ~zy2, denotes the geodesic torsion of C2 = S1 n S3and T 3 that of Sl n S 2 . Use the definition of z , to show that, since C , , Cz, C3 are painvise orthogonal, zl z 2 = 0, z 2 z 3 = O1 z 3 z 1 = 0. It follows that z , = z2 = 7 3 = 0.
+
+
+
SECTION 33 2, Asymptotic curves: u
= const., v = const.
Iog (v
3. u
+ v = const.
+ d v 2 + c2)
Lines of curvature:
u
= const.
u  v = const.
6. a. By taking the line r as the z axis and a normal to r as the x axis, we have that
Hints and Answers
By setting x
= sin
8, we obtain
8 d o = log tan 2
If z(nj2) = 0, then C
+ cos 6 + C.
= 0.
8. a. The assertion is clearly true if x = x l and 2 = are parametrizations that satisfy the definition of contact. If x and iare arbitrary, observe that x = X I 0 h, where h is the change of coordinates. It follows that the partial derivatives off 0 x = f 0 xl 0 h are linear combinations of the partial derivatives off 0 X I . Therefore, they become zero with the latter ones. b. Introduce parametrizations x(x, y) = (x, y, f (x, y)) and g(x, y ) = (x, y,f(x, y)), and define a function h(x, y, z) = f (x,y)  z. Observe that It follows from part a, applied the function h, h 0 x = 0 and h 0 2 = f that f  f has partial derivatives of order (I2 equal to zero at (0,O). d. Since contact of order 2 2 implies contact of order 2 1, the paraboloid passes through p and is tangent to the surface at p. By taking the plane T,(S) as the xy plane, the equation of the paraboloid becomes fa
f (x, y) = ax2
+ 2bxy + cy2 + dx + ey.
Let z =f (x, y) be the representation of the surface in the pIane Tp(S).By using part b, we obtain that d = c = 0, a = f,,, b = f,,, c = kf,.
15. If there exists such an example, it may locally be written in the form z =f ( x , y), with f (0,O) = 0, f,(O, 0) = f,(O, 0) = 0. The given conditions require that f f ;y 0 at (0,O) and that f,,&  f 0 if and only if (x, y) = (0,O). By setting, tentatively, f (x, y) = a(x) fi(y) xy, where a(x) is a function of x alone and P(y) is a function of y alone, we verify that a,, = cos x, By, = cos y satisfy the conditions above. It follows that
:,+ +

:, +
+
is such an example.
16. Take a sphere containing the surface and decrease its radius continuously. Study the normal sections at the point (or points) where the sphere meets the surface for the first time. 19. Show that the hyperboloid contains two oneparameter families of lines which are necessarily the asymptotic lines. To find such families of lines, write the equation of the hyperboloid as
and show that, for each k # 0, the line x (l/k)(l  y) belongs to the surface.
+ z = k(1 + y),
x z =
486
Hints and Answers
20. Observe that (xIa2, ylb2, z/c2) = f N for some function f and that an umbilical point satisfies the equation
for every curve ~ ( t = ) (x(t), y(t), z(t)) on the surface. Assume that z # 0, multiply this equation by zlc2, and eliminate z and dz/dt (observe that the equation holds for every tangent vector on the surface). Four umbilical points are found, namely,
The hypothesis z 21. a. Let dN(vl)
=
b. Show that f N
=0
av,
=
d(fN)(vi) i
=
does not yield any further umbilical points.
+ bv,, dN(v2)
=
cvl
+ dv2. A direct computation yields
(x/a2,y/b2, z/c2) = W, and observe that =
(3.
3)
62' c2
1,2. By choosing vi so that vl A v2
where X
=
where vi =
(x, y, z), and therefore ( W, X)
=
(ai, Pi, yi),
N, concIude that
=
1.
24. d. Choose a coordinate system in R3 SO that the origin 0 is at p E S, the xy plane agrees with T,(S), and the positive direction of the z axis agrees with the orientation of S at p. Furthermore, choose the x and y axes in T,(S) along the principal directions at p. If V is sufficiently small, it can then be represented as the graph of a differentiable function
where D is an open disk in R2 and
We can assume, without loss of generality, that k l 2 0 and k2 2 0 on D, and we want to prove that f(x, y) 2 0 on D. Assume that, for some (3,y) E D, f(3,y) < 0. Consider the function tI 1. Since hb(0) = 0, there exists a t , 0 I t < 1, h,(t) = f (t2, ty), 0 I such that h:(t ,) < 0, Let p1 = (t ,2,t,?, f (t12, fly)) E S, and consider the height function hl of Vrelative to the tangent plane T,,(S) a t p l . Restricted to
,
,
Hints and Answers
487
the curve a(t)=(tZ, t.7,f (tx, t?)), this height function is h,(t)=(a(t)  p l y N , ) , where N, is the unit normal vector atp,. Thus, h:(t) = (~"(t), N,), and, at t = tl,
But h:(tl) = (~"(t,), N,) is, up to a positive factor, the normal curvature at p in the direction of a'(t ,). This is a contradiction.
SECTION 34 10. c. Reduce the problem to the fact that if A is an irrational number and m and n run through the integers, the set {Am n} is dense in the real line. To prove the last assertion, it suffices to show that the set {Am n] has arbitrarily small positive elements. Assume the contrary, show that the greatest lower bound of the positive elements of {am n] still belongs to that set, and obtain a contradiction.
+
+
+
11, Consider the set {ai: Ii 3 I ] of trajectories of w,'with ai(0) = p , and set I = (,Ii Ii.By uniqueness, the maximal trajectory a : I U may be defined by setting a(t) = ~ , ( t )where , t E Ii. 12. For every q E S, there exist a neighborhood U of q and an interval (  6 , E ) , 6 > 0, such that the trajectory u(t), with a(0) = q, is defined in (  6 , E). By compactness, it is possible to cover S with a finite number of such neighborhoods. Let 6, = minimum of the corresponding f's. If ~ ( t is) defined for t < to and is not defined for to, take t l E (0, to), with I to  tl I < e0/2. Consider the trajectory P(t) of w, with P(t,) = ~ ( t , )and , obtain a contradiction.
SECTION 42

3. The "only if" part is immediate. To prove the "if" part, Ietp E S and v E T,(S), u # 0. Consider a curve a: (6, E ) S, with ~ ' ( 0 )= u. We claim that I dy,(af(0)) j = I ~ ' ( o ) ]Otherwise, . say, 1 dy,(~'(O)) I > 1af(O)1, and in a neighborhood J of 0 in (€, €), we have I dy,(al(t)) 1 > I U'(t) I. This implies that the length of a ( J ) is greater than the length of y 0 u(J), a contradiction.
6. Parametrize a by arc length s in a neighborhood of t o . Construct in the plane a curve with curvature k = k(s) and apply Exercise 5. 8. Set 0 = (0,0, O), G(0) = p,, and G(p)  po = F(p). Then F: R3 + R3 is a map such that F(0) = 0 and 1 F(p) I = I G(p)  G(0) 1 = Ip 1. This implies that F preserves the inner product of R3. Thus, it maps the basis
onto an orthonormal basis, and i f p = C aif;,, i = 1,2,3, then F(p) Therefore, F is linear.
= CaiF(f;).
488
Hints and Answers
11. a. Since F is distancepreserving and the arc length of a differentiable curve is the limit of the lengths of inscribed polygons, the restriction F I S preserves the arc length of a curve in S. c. Consider the isometry of an open strip of the plane onto a cylinder minus a generator. 12. The restriction of F ( x , y, z) = (x, y,  2 ) to Cis an isometry of C (cf. Exercise 1I), the fixed points of which are (1, 0, 0) and ( 1,0,0).
17. The loxodromes make a constant angle with the meridians of the sphere. Under Mercator's projection (see Exercise 16) the meridians go into parallel straight lines in the plane. Since Mercator's projection is conformal, the loxodromes also go into straight lines. Thus, the sum of the interior angles of the triangle in the sphere is the same as the sum of the interior angles of a rectilinear plane triangle.
SECTION 44 6. Use the fact that the absolute value of the geodesic curvature is the absolute value of the projection onto the tangent plane of the usual curvature.
8. Use Exercise 1, part b, and Prop. 5 of Sec. 32. 9. Use the fact that the meridians are geodesics and that the parallel transport preserves angles.
+
10. Apply the relation k i k,2 = k2 and the Meusnier theorem to the projecting cylinder. 12. Parametrize a neighborhood of p E S in such a way that the two families of geodesics are coordinate curves (Corollary 1, Sec. 34). Show that this implies that  F = 0, E, = 0 = G,. Make a change of parameters to obtain that F = 0, E=G=l. A
13. Fix two orthogonal unit vectors v(p) and w(p) in T,(S) and parallel transport them to each point of V. Two differentiable, orthogonal, unit vector fields are thus obtained. Parametrize V in such a way that the directions of these vectors are tangent to the coordinate curves, which are then geodesics. Apply Exercise 12. 16. Parametrize a neighborhood of p E S in such a way that the lines of curvature are the coordinate curves and that v = const. aye the asymptotic curves. It follows that e, = 0, and from the MainardiCodazzi equations, we conclude that E, = 0. This implies that the geodesic curvature of v = const. is zero. For the example, look at the upper parallel of the torus. 18. Use Clairaut's relation (cf. Example 5 ) . 19. Substitute in Eq. (4) the Christoffel symboIs by their values as functions of E, F, and G and differentiate the expression of the first fundamental form:
20. Use Clairaut's relation.
Hints and Answers
SECTION 45 4. b. Observe that the map x = 2, y = (T),, z = ( Z ) 3 gives a homeomorphism of the sphere x 2 y 2 z2 = 1 onto the surface (2)2 ( J ) 4 + ( Z ) 6 = 1.
+ +
+
6. a. Restrict v to the curve a(t) = (cos t, sin t), t E [0, 2x1. The angle that v(t) forms with the x axis is t. Thus, 2x1 = 27~;hence, I = 1. d. By restricting v to the curve a(t) = (cos t, sin t), t E [O, 2x1, we obtain v(t) = (cos2 t  sin2 t, 2 cos t sin t ) = (cos 2t, s i n 2t). Thus, I = 2.
SECTION 46 8, Let (p, 8) be a system of geodesic polar coordinates such that its pole is one of the vertices of A and one of the sides of A corresponds to 8 = 0. Let the two other sides be given by 6 = and p = h(8). Since the vertex that corresponds to the pole does not belong to the coordinate neighborhood, take a small circle of radius 6 around the pole. Then
Observing that
K ~ =G ( d z ) , ,
and that lim ( d c ) ,
limit enclosed in parentheses is given by
€+a
= I,
we have that the
By using Exercise 7, we obtain

12. c. For K 0, the problem is trivial. For K > 0, use part b. For K < 0, consider a coordinate neighborhood V of the pseudosphere (cf. Exercise 6, part b, Sec. 33), parametrized by polar coordinates (p, 0); that is, E = I, F = 0, G = sinh2 p. Compute the geodesics of V; it is convenient to use the change of coordinates tanh p = llw, p # 0, 8 = 8, so that
and the other Christoffel symbols are zero. It follows that the nonradial w = 0, where w = w(8). Thus, geodesics satisfy the equation (d2wld92)
+
Hints and Answers
490
w
=
~ c o s +6 B s i n O ; t h a t i s A tanh p cos 6 f B tanh p sin 8
=
1.
Therefore, the map of V into R 2 given by
5 = tanh p cos 6, (5, q)
E
q
=
tanh p sin 6,
R2, is a geodesic mapping.
13. b. Define x = q1: q(U) c R2 + S. Let v = v(u) be a geodesic in U. Since q is a geodesic mapping and the geodesics of R2 are lines, then dZvldu2 = 0. By bringing this condition into part a, the required result is obtained. c. Equation (a) is obtained from Eq. (5) of Sec. 43 using part b. From Eq. (5a) of Sec. 43 together with part b we have
By interchanging u and v in the expression above and subtracting the results, we obtain (r!,), = (rf,),, whence Eq. (b). Finally, Eqs. (c) and (d) are obtained from Eqs. (a) and (b), respectiveIy, by interchanging u and v. d. By differentiating Eq. (a) with respect to v, Eq. (b) with respect to u, and subtracting the results, we obtain
the expression above yields
By taking into account the values of EKv

FKu
=
K(Ev

F,)
t K(Ev

F,)
Similarly, from Eqs. (c) and (d) we obtain FK, K, = K, = 0.
=

0.
GK,
= 0,
whence
SECTION 47 1. Consider an orthonormal basis {e e23at T, ,,,(S) and take the parallel transport of el and e2 along a , obtaining an orthonormal basis (el(t), e2(t)}at each T,,,](S). Set w(a(t)) = wl(t)el(t) w2(t)e2(t). Then D,w = w',(0)el w',(0)e2 and the second member is the velocity of the curve wl(t)el w2(t)e2in T,(S) at t = 0.
+
+
+
2. b. Show that if (t,, t2) c I is small and does not contain "break points of a," then the tangent vector field of a((tl, t2)) can be extended to a vector field y in a neighborhood of a(@,, t,)). Thus, by restricting v and w to a, property 3 becomes
which implies that paralIeI transport in a ] (t,, t2)is an isometry. By compactness,
49 1
Hints and Answers
this can be extended to the entire I. Conversely, assume that parallel transport is an isometry. Let a be the trajectory of y through a point p E S. Restrict v and w to a . Choose orthonormal basis { e l ( t ) ,e z ( t ) )as in the solution of Exercise 1 , and set u(t) = u , e l uze2, w(t) = w l e l wze2.Then property 3 becomes the "product rule" :
+
+
c. Let D be given and choose an orthogonal parametrization x(u, u). Let y = ylxu yzx,, w = wlxu w2x,. From properties 1 , 2, and 3, it follows that D,w is determined by the knowledge of DXUxu,Dx,%, Dx,%. Set DXUx,= Ailxu A f l x , , DXUx, = A ! z x , At2x,, DXux,= Ai2xu A , ~ , x , . From property 3 it folIows that the A: satisfy the same equations as the (cf. Eq. (21, Sec. 43). Thus, A: = T,kj,which proves that D,v agrees with the operation "Take the usuaI derivative and project it onto the tangent plane."
+
+
+
+
+
r$
3. a. Observe that
b. Use the fact that x is a local diffeomorphism to cover the compact set I with a family of open intervals in which x is onetoone. Use the HeineBore1 theorem and the Lebesgue number of the covering (cf. Sec. 27) to globalize the result. c. To show that F = 0 , we compute (cf. property 4 of Exercise 2)
because the vector field dxlds is parallel along t
= const.
Since
F does not depend on s. Since F(0, t ) = 0, we have F = 0. d. This is a consequence of the fact that F = 0.
4. a. Use Schwarz's inequality,
with f
I
1 and g
=
I da/dt j.
5. a. By noticing that E(r) =
I 0
{(duldv)' f G(y(v,r), v ) ] dv, we obtain (we write
y(v, t ) = ~ ( vt ), ,for convenience)
Hints and Answers
Since, for t = 0, duldv Furthermore,
Hence, by using G,,
=
=0
and dG/du
= 0,
we have proved the first part.
 2 ~ d cand noting that
E"(0)
=2
S'{in,o);
~
+/G ~
=
d:. 2
1 for t
= 0,
we obtain
)
0
6. b. Choose 6 > 0 and coordinates in R3 3 S SO that ~ ( p6,) = q. Consider the 6 2nk sin B) = #rk. points (p, 6 ) = r , (p, 6 2n sin p ) = r l , . . . , (p,Taking E sufficientIy small, we see that the line segments r o r l , . . . , rorkbelong to Vif 2zk sin p < z (Fig. 449). Since q is a local isometry, the images of these segments will be geodesics joining q to q, which are clearly broken at q (Fig. 449). c. It must be proved that each geodesic y: [O,1] ) q is the + S with y(0) = ~ ( 1 = image by q of one of the line segments r o r l , . . . ,r z referred to in part b. For some neighborhood U c V of ro, the restriction 1 U = Q is an isometry. Thus, @1 0 y is a segment of a halfline L starting at ro. Since p(L) is a geodesic which agrees with y([O, I]) in an open interval, it agrees with y where y is defined. Since y(1) = q, L passes through one of the points ri: i = 1, . . . , k, say rj, and so y is the image of G.
+
+
SECTION 52
+
3. a. Use the relation q'' = Kq to obtain (qf2 Kq2)' = K'q2. Integrate both sides of the last relation and use the boundary conditions of the statement.
SECTION 53 5. Assume that every Cauchy sequence in d converges and let y(s) be a geodesic parametrized by arc length. Suppose, by contradiction, that y(s) is defined for s < so but not for s = so. Choose a sequence Isn) + so. Thus, given > 0, there exists no such that if n, m > no, Is,  s, I < 6. Therefore,
and {y(s,)] is a Cauchy sequence in d. Let {y(s,)) + p, E S and let W be a neighborhood of p , as given by Prop. 1 of Sec. 47. If m, n are sufficientIylarge,
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Hints and Answers
the small geodesic joining y(s,) to y(s,) clearIy agrees with 7. Thus, y can be extended through po, a contradiction, Conversely, assume that S is complete and let {p,] be a Cauchy sequence in d of points on S. Since d is greater than or equal to the Euclidean distance d, {an]is a Cauchy sequence in d. Thus, {p,] converges t o po E R3. Assume, by contradiction, that po $ S. Since a Cauchy sequence is bounded, given E > 0 there exists an index no such that, for all n > no, the distance d(pn,,p,) < E. By the HopfRinow theorem, there is a minimal geodesic yn joining p,, top, with length < E. As n 3 a,yn tends to a minimal geodesic y with length I E. Parametrize y by arc Iength s. Then, since po $ S, y is not defined for s = E. This contradicts the completeness of S.
6. Let {p,) be a sequence of points on S such that d(p, p,)
+ co. Since
S is com
plete, there is a minimal geodesic y,(s) (parametrized by arc 1ength)joiningp to pn with ~ ~ ( = 0 p. ) The unit vectors y:(O) have a limit point v on the (compact)
unit sphere of T,(S). Let y(s) = exp, sv, s 2 0. Then y(s) is a ray issuing fromp. T o see this, notice that, for a fixed so and n sufficiently large, limn,, y,(so) = y(so). This follows from the continuous dependence of geodesics from the initiaI conditions. Furthermore, since d is continuous,
But if n is large enough, d(p, yn(so))= so.Thus, d(p, ?(so)) = so, and y is a ray.
8. First show that if d and d denote the intrinsic distances of S and 3, respectively, then d(p, q) 2 cd(p(p), q(q)) for all p, q E S. Now let {p,] be a Cauchy sequence in d of points on S. By the initial remark, {q(pn)3is a Cauchy sequence in d. Since is complete, {~(p,)]t p(po). Since q is continuous, {pJ +p o Thus, every Cauchy sequence in d converges; hence S is complete (cf. Exercise 5). 9.
QI is onetoone: Assume, by contradiction, that p1 # pz E S1 are such that q(pl) = q(pz) = q. Since S1is complete, there is a minimal geodesic y joining p, top2. Since q is a Iocal isometry, q 0 y is a geodesic jo\iningq to itself with the same length as y. Any point distinct from q on q 0 y can be joined to q by two geodesics, a contradiction. q is onto: Since q is a local diffeomorphism, q(S1) c S2 is an open set in S2. We shall prove that q(Sl) is also closed in S2;since S2is connected, this will imply that q(S,) = Sz.If q(Sl) is not closed in S2, there exists a sequence ) . {q(pn)] is a nOnconfp(pn)], Pn Sl?such that (~(pn)]+PO 6 ~ ( ~ 1Thus, verging Cauchy sequence in q(S1). Since q is a onetoone local isometry, {p,) is a nonconverging Cauchy sequence in S1,a contradiction to the completeness of SI. 10. a. Since d d ;ii(hOdQ) = = ; i i < ~ ( tv)) , = 0.
r+m x 2 + y 2 2 r 4
Then there exists R
> 0 such that if (x, y) $ D, where D = {(x, y) E R 2 ; x2 + y 2 < R2],
then K ( x , y) 2 c. Thus, by taking points outside the disk D, we can obtain arbitrarily large disks where K(x, y) >_ c > 0. This is easily seen to contradict Bonnet's theorem.
SECTION 55 3. b. Assume that a > b and set s = b in relation (*). Use the initial conditions and the facts v7(b)< 0,u(b) > 0, uv 2 0 in [O, b] to obtain a contradiction, C. From [uvf  vuf]; 0 , one obtains v'/v 2 u'/u; that is, (log v)' 2 (log u)'. Now, let 0 4 so s I a, and integrate the last inequality between so and s to obtain
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