Two-Dimensional Geometries: A Problem-Solving Approach 1470447606, 9781470447601

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Sally

The

SERIES

Pure and Applied UNDERGRADUATE TEXTS

Two-Dimensional Geometries A Problem-Solving Approach

C. Herbert Clemens

34

Two-Dimensional Geometries A Problem-Solving Approach

Sally

The

Pure and Applied UNDERGRADUATE TEXTS • 34

SERIES

Two-Dimensional Geometries A Problem-Solving Approach

C. Herbert Clemens

EDITORIAL COMMITTEE Gerald B. Folland (Chair) Jamie Pommersheim

Steven J. Miller Serge Tabachnikov

2010 Mathematics Subject Classification. Primary 51F10, 51F20, 51F25, 51-01, 51K10.

For additional information and updates on this book, visit www.ams.org/bookpages/amstext-34

Library of Congress Cataloging-in-Publication Data Names: Clemens, C. Herbert (Charles Herbert), 1939- author. Title: Two dimensional geometries : a problem-solving approach / C. Herbert Clemens. Other titles: 2 dimensional geometries Description: Providence, Rhode Island : American Mathematical Society, [2019] | Series: Pure and applied undergraduate texts ; volume 34 | Includes bibliographical references and index. Identifiers: LCCN 2018045838 | ISBN 9781470447601 (alk. paper) Subjects: LCSH: Geometry, Descriptive. | AMS: Geometry – Metric geometry – Absolute spaces. msc | Geometry – Metric geometry – Congruence and orthogonality. msc | Geometry – Metric geometry – Orthogonal and unitary groups. msc | Geometry – Instructional exposition (textbooks, tutorial papers, etc.). msc | Geometry – Distance geometry – Synthetic differential geometry. msc Classification: LCC QA501 .C6575 2019 | DDC 516/.154–dc23 LC record available at https://lccn.loc.gov/2018045838

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established to ensure permanence and durability. Visit the AMS home page at https://www.ams.org/ 10 9 8 7 6 5 4 3 2 1

24 23 22 21 20 19

Contents

Chapter 1. Introduction

1

1.1. Design of this book

1

1.2. Parts V–VII: How many two-dimensional geometries are there?

3

1.3. Parts IV–VII: Some needed multivariable calculus and linear algebra facts

4

1.4. References and notation

5

Part I. Neutral geometry Chapter 2. Euclid’s postulates for plane geometry 2.1. Neutral geometry

9 9

2.2. Sum of angles in a triangle in NG

12

2.3. Are there rectangles in NG?

14

Part II. Euclidean (plane) geometry Chapter 3. Rectangles and cartesian coordinates

19

3.1. Euclid’s Fifth Postulate, the Parallel Postulate

19

3.2. The distance formula in EG

21

3.3. Law of Sines and Law of Cosines

22

3.4. Dilations in EG

24

3.5. Similarity in EG

26

Chapter 4. Concurrence and circles in Euclidean geometry

29

4.1. Concurrence theorems in EG, Ceva’s theorem

29

4.2. Properties of circles in EG

32

4.3. Circles and sines and cosines

34 v

vi

Contents

4.4. Cross-ratio of points on a circle 4.5. Ptolemy’s theorem

35 37

Part III. Spherical geometry Chapter 5. Surface area and volume of the R-sphere in Euclidean three-space 41 5.1. Volumes of pyramids 41 5.2. Magnification principle 42 5.3. Relation between volume and surface area of a sphere 43 5.4. Surface area 43 5.5. Areas on spheres in Euclidean three-space 46 Part IV. Usual dot-product for three-dimensional Euclidean space Chapter 6. Euclidean three-space as a metric space 6.1. Points and vectors in Euclidean three-space 6.2. Curves in Euclidean three-space and vectors tangent to them 6.3. Surfaces in Euclidean three-space and vectors tangent to them

51 51 56 59

Chapter 7. Transformations 7.1. Rigid motions of Euclidean three-space 7.2. Orthogonal matrices 7.3. Linear fractional transformations

61 61 63 64

Part V. K-geometry Chapter 8. Changing coordinates 8.1. Bringing the North Pole of the R-sphere to (0, 0, 1) 8.2. K-geometry: Euclidean lengths and angles in (x, y, z)-coordinates 8.3. Congruences, that is, rigid motions

71 71 73 76

Chapter 9. Uniform coordinates for the two-dimensional geometries 9.1. The two-dimensional geometries in (x, y, z)-coordinates 9.2. Central projection 9.3. Stereographic projection 9.4. Relationship between central and stereographic projection coordinates

79 79 83 89 94

Part VI. Return to spherical geometry Chapter 10.1. 10.2. 10.3. 10.4.

10. Spherical geometry from an advanced viewpoint Rigid motions in spherical geometry Spherical geometry is homogeneous Lines in spherical geometry Central projection in SG

99 99 101 104 107

Contents

10.5. Stereographic projection in SG

vii

108

Part VII. Hyperbolic geometry Chapter 11. The curvature K becomes negative 11.1. The world sheet and the light cone 11.2. Hyperbolic geometry is homogeneous 11.3. Lines in hyperbolic geometry 11.4. Central projection in HG 11.5. Stereographic projection in HG

113 113 115 119 122 128

Definitions

133

Bibliography

139

Index

141

Chapter 1

Introduction

1.1. Design of this book It has often been said that “mathematics is not a spectator sport”. This truism is very much evident in the writing of this book. It is written so as to guide you through the entire story, yet permit you whenever appropriate to construct the mathematical story for yourself. The idea is to do some mathematics yourself rather than just observe it done by others. This “doing mathematics oneself” takes the form of exercises with enough help (hints) provided so that the “doing” is not so onerous as to get in the way of the story itself. Strong evidence has been provided by students of mathematics over many centuries that such guided “doing” is indispensable for full understanding and retention. When the book is used as a course text, the instructor is encouraged to make available the exercise statements, one to a page, so that students can write out solutions (in correctable form), correct any errors, and insert the corrected exercise solution at the end of the appropriate section in the book. Parts I–III of the book presuppose a working knowledge of high school geometry and basic trigonometry. They contain the mathematical material appropriate for a one-quarter geometry course for preservice or inservice middle school mathematics teachers. Parts I–III followed by Part IV contain appropriate mathematical material for a one-semester geometry course for preservice or inservice high school mathematics teachers. For more advanced students, a rapid review of the material in Parts I–III followed by Parts IV–VII can comprise a one-semester upper division Advanced Geometry course. Parts V–VII tell the unified story of all the two-dimensional geometries, one for each real number K. It is in this part of the story that the notion of the group of rigid motions enters and manifests some of its power.

1.1.1. One and only one two-dimensional geometry for each real number. In the end, a geometry will be seen to be completely characterized by its definition 1

2

1. Introduction

of distance, called its “K-dot-product”, and the group of rigid motions that preserve that K-dot-product. As we have already mentioned, there will turn out to be one and only one geometry for each real number K. If K > 0, the geometry will be that of the sphere of radius K −1/2 . If K = 0, the geometry will be the Euclidean geometry that you were introduced to in high school. If K < 0, the geometry will be one of the so-called hyperbolic geometries, the one whose distance and area formulas have scaling factors of |K|−1/2 and |K|−1 , respectively. Besides telling their own beautiful story, Parts IV–VII can also serve as a bridge to Curves and Surfaces in Euclidean Three-space, the standard first course in elementary Riemannian geometry. Parts IV–VII treat the special case of twodimensional geometries that are homogeneous, that is, that look the same at all their points and in all directions. To treat these geometries efficiently, we introduce the notion of changing coordinates for the geometry without changing the geometry itself. The idea is to use a particular coordinate system, for example, spherical coordinates, to calculate a particular quantity or property, for example, distance. A different coordinate system for the same geometry, for example, stereographic projection coordinates, may be more useful for calculating, for example, measure of angles. That same notion of “different coordinates to treat different properties” also allowed geometers to treat surfaces and higher-dimensional smooth spaces that look different at different points, ones that often cannot be treated at all their points using a single set of coordinates. For our purposes in this book, though, a geometry will be a two-dimensional set or space that “looks the same” at all its points. That is, the set admits a distancepreserving transformation that takes any given point to any other given point and takes any given direction at the first point to any given direction at the other point. It is my hope and intention in writing this little book that you engage with and enjoy this uniform way of understanding all two-dimensional geometries as much as I did! 1.1.2. Special message to current or future teachers of high school geometry. As mentioned above, Parts I–III of this book are especially relevant to your teaching of the subject. Look especially closely at the treatment of congruence (rigid motion), similarity (dilation), circles, expressing geometric properties with equations, and geometric measurement and dimension, and compare them with the high school geometry sections of the Common Core State Standards in Mathematics. The latter can be found at http://www.corestandards.org/Math/Content/HSG/introduction. A useful companion course to one based on this book, one that might be called Geometry for Teaching, would explicitly make the connections between the material covered as in this book and what you do (or will do) in your high school geometry classroom. The idea is not that the material we will cover in Parts I–III will tell you how to teach that material, but rather that the treatment given here will give you the depth and breadth of geometric understanding that will allow you to design what you teach and bring it into your classroom in ways that those who lack that understanding cannot.

1.2. Parts V–VII: How many two-dimensional geometries are there?

3

1.1.3. Acknowledgments. It is my pleasure to thank Dr. Bart Snapp of the Mathematics Department of Ohio State University for his contributions to this book, especially to the exercises, where his many suggestions have greatly improved their precision and richness. Finally I wish to thank the Ohio State Mathematics Department itself for the opportunity to regularly teach the advanced geometry course from which this book evolved over a 15-year period.

1.2. Parts V–VII: How many two-dimensional geometries are there? As a young mathematician I was introduced to the classic Leçons sur la Géométrie des Espaces de Riemann, written by the great French geometer Élie Cartan. Early in his treatise on geometries in all dimensions, Cartan presents the case of twodimensional geometries, in particular, those two-dimensional geometries that look the same at all points and in all directions. (For example, a cylinder looks the same at each of its points but not in all directions emanating from any one of its points, whereas a sphere looks the same at all points and in all directions.) As we mentioned above, it turns out that there is one and only one such geometry for each real number K, called the curvature of the geometry. The case K = 0 is the (flat) Euclidean geometry that you learned in high school. These 25 pages of Cartan’s book (Chapter VI, §i–v) so captivated me that I have returned to them regularly throughout my career and have adapted and taught them many times at the advanced undergraduate level. They form the basis for Parts V–VII of this book. To me they tell one of the most beautiful and satisfying stories in all of geometry, one which exemplifies a fundamental principle of all great mathematics, namely that, using the tools at hand but in a slightly novel way, the clouds part and one sees that objects and relationships that seemed so different are in fact parts of a single elegant story! When K > 0 it turns out that the K-geometry is the geometry of the sphere of radius R = 1/K 1/2 that we can “see” as a subset of Euclidean three-space R3 . But the geometries with K < 0 are not so easy to visualize. They are the so-called “hyperbolic” geometries. In fact it took mathematicians a couple thousand years to realize that they existed at all! It turns out that the secret to understanding all the two-dimensional geometries, including the ones with K < 0, in a unified way is to simply rescale the third coordinate in R3 and, using these “unusual” coordinates (x, y, z), to look at each two-dimensional geometry as the solution set to the equation   K x2 + y 2 + z 2 = 1. However, the idea of changing coordinates without changing the underlying geometry described by those coordinates is a challenging one that did not come into mathematics until a couple of centuries ago. In Chapter 9 we introduce the main coordinate change that we are going to use, namely the above-mentioned rescaling of the third coordinate in Euclidean three-space. This change is central to the second half of the book. However, before we begin our unified study of two-dimensional geometries in the second half of this book, we devote Chapter 8 to a review of the concepts from several variable calculus and linear algebra that we will need. We briefly indicate these next.

4

1. Introduction

1.3. Parts IV–VII: Some needed multivariable calculus and linear algebra facts Parts IV–VII of this book will suppose familiarity with several variable calculus and the linear algebra of matrices and how they are used to define linear transformations. Also it will often be useful to consider a vector, for example, V = (a, b, c), as a 1 × 3 matrix, writing   (V ) = a b c , or as a 3 × 1 matrix, writing

⎛

⎞ a t (V ) = ⎝ b ⎠ . c

This will allow us, for example, to write the scalar product of two vectors V • V  = (a, b, c) • (a , b , c ) = aa + bb + cc as a product of matrices (V ) · (V  ) = t



a b

c



⎞ a · ⎝ b ⎠ . c ⎛

Also you will need basic facts about 2 × 2 and 3 × 3 matrices, such as how to add and multiply them and basic properties of their determinants. From several variable calculus you will in particular need to remember and apply two rules, the Chain Rule (Theorem 1) for differentiable functions of several variables, written in matrix notation, and the Substitution Rule (Theorem 2). Theorem 1 (Chain Rule). Given differentiable mappings (y1 (x1 , . . . , xm ) , . . . , yn (x1 , . . . , xm )) and (z1 (y1 , . . . , yn ) , . . . , zp (y1 , . . . , yn )) , partial derivatives satisfy the matrix equation (∂zk /∂xi ) = (∂zk /∂yj ) · (∂yj /∂xi ) . Proof. The rule comes from the fact that if I pick one of the functions zk (y1 , . . . , yn ) and pick one of the variables xi and vary it (and leave all the other xi constant), I am left with finding the formula for dz (y1 (x) , . . . , yn (x)) dx that turns out to be n  ∂z dyj (y1 (x) , . . . , yn (x)) · (x) . ∂yj dx j=1

1.4. References and notation

5

This last expression can be written in matrix form as

∂z ∂y1

(y1 (x) , . . . , yn (x)) . . .

∂z ∂yn

(y1 (x) , . . . , yn (x))

⎛

⎞ (x) · ⎝ ... ⎠. dyn dx (x) dy1 dx

Setting z = zk and x = xi , we have exactly the matrix multiplication formula for ∂zk /∂xi in the theorem.  Exercise 1. a) Write out Theorem 1 in the case in which n = m = p = 1 to be sure that you recognize how it is just a several variable formulation of the most important theorems from your introductory course in calculus. b) Write the full 2 × 2 matrix form of the Chain Rule when n = m = p = 2. Theorem 2 (Substitution Rule). Given a mapping (y1 (x1 , . . . , xn ) , . . . , yn (x1 , . . . , xn )) between corresponding regions Ry in (y1 , . . . , yn )-space and Rx in (x1 , . . . , xn )space, and given a function f (y1 , . . . , yn ) on Ry ,  f (y1 , . . . , yn ) · dy1 . . . dyn Ry  = Rx f (y1 (x1 , . . . , xn ) , . . . , yn (x1 , . . . , xn )) · Dyx · dx1 . . . dxn , where Dyx = det (∂yj /∂xi ) . Exercise 2. a) Write out Theorem 2 in the case in which n = 1 to be sure that you recognize how it is just a several variable formulation of the most important theorem from your introductory course in integral calculus.   b) Integrate sin y12 + y22 over the region in the (y1 , y2 )-plane bounded by the unit circle. Hint: Use the Substitution Rule to change (y1 , y2 ) to polar coordinates.

1.4. References and notation As help, at some points in the text and in some of the exercises, a more complete treatment of a particular topic can be found in [MJG] or [DS]. The corresponding topics in these texts are referenced. For example, [MJG, 311] refers to page 311 in the Greenberg book and [DS, 59ff] refers to page 59 and those pages just following page 59 in the Davis–Snider book. Some final remarks about notation in this book: The letters EG will always mean Euclidean (usually plane but occasionally three-dimensional) geometry, the letters SG will always mean spherical geometry, and the letters HG will always mean hyperbolic geometry. One further kind of geometry, which we call neutral geometry, will be explained in the book and denoted by NG.

Part I

Neutral geometry

Chapter 2

Euclid’s postulates for plane geometry

2.1. Neutral geometry We now turn our attention to plane (or “flat”) two-dimensional geometry. In Western civilization, the primary source of our understanding of this geometry comes from Euclid’s Elements. The treatise is of transcendent importance well beyond geometry itself, because it is among the first examples, and perhaps the most influential single example, of formal deductive logical reasoning. Certain fundamentals called axioms are postulated or “given”, providing the platform on which a “geometry” is built, that is, a mathematical entity modeling a physical “reality”. Its properties are arrived at by applying the laws of logic to the given fundamentals. As a starting point, we will take as given the notions of point, line, and plane and of containment and intersection, as in, for example, the statement that two lines in a plane may or may not intersect. Euclid gives five axioms for plane geometry, the first four of which seem to be “obvious” reflections of physical reality. In paraphrased form, they are: Axiom (E1). Through any point P and any other point Q in the plane, there lies a unique line. Axiom (E2). Given any two segments AB and CD, there is a segment AE such that B lies on AE and |CD| = |BE|. (We might define a segment as the set of points between two distinct points on a line, but that still leaves “between” undefined. This axiom and definition stuff is not as easy as it may seem!) (NB: In this book we use the notation |AB| and d (A, B) interchangeably to denote the distance between two points A and B. Axiom (E3). Given any point P and any positive real number r, there exists a (unique) circle of radius r and center P . (Said another way, along any line through P , you will encounter exactly two points at distance r from P .) 9

10

2. Euclid’s postulates for plane geometry

Axiom (E4). All right angles are congruent. (A right angle is defined as follows. Let C be the midpoint on the segment AB. Let E be any point not equal to C. The angle ∠ACE is called a right angle if ∠ACE is congruent to ∠BCE [MJG, 17– 18].) Definition 1. If we are only given E1–E4, we will call our geometry neutral geometry (NG). Next we have one of the most challenging definitions. Definition 2. In NG, two distinct lines are called parallel if and only if they don’t intersect. We will call the set of points on a line that lie on one side of a given point a ray. We call the given point the origin of the ray. So giving a ray is the same thing as giving a point in the plane (the origin of the ray) and a vector or direction emanating from that point (the direction along the ray). We call two rays in the plane parallel if they lie on parallel lines and if they both lie on the same side of the transversal line passing through their origins. Strictly speaking, an angle in the plane is the union of two ordered rays with a common origin and the choice of one of the two connected regions into which the union of the rays divides the plane.

We often denote angles by ∠BAC, where A is the common origin, B is a point along one of the rays, called the initial ray, and C is a point along the other ray, called the final ray. The choice of the region is either clear from the context or explicitly given. Again strictly speaking, we should be careful about the distinction between an angle and its measure. The angle is the union of two rays, and its measure is a number (in this book that number is always the size of the angle in degrees or radians). Roughly speaking, two angles are congruent if and only if they have the same measure. In some situations, however, the measure has a sign: positive if the chosen region is on the counterclockwise side of the initial ray, or negative if the chosen region is on the clockwise side of the initial ray. In these situations congruence may require that the signed measures be the same. That said, it is often the case (even in this book) that it is asserted that two angles are equal when what is precisely meant is that they are congruent, that is, of equal measure. Exercise 3. Given an angle ∠BAC, show by drawings the two regions into which it divides the plane. Show how the (signed) measure of the angle depends on which

2.1. Neutral geometry

11

region you pick and on which is the initial ray and which is the final ray of the angle. One implicit assumption of two-dimensional neutral (and Euclidean) geometry is the existence of (a group of) rigid motions or congruences. That is, it is assumed ˆ and that given any point Aˆ and any vector (i.e., direction) Vˆ emanating from A, ˆ ˆ ˆ given any second point B and any vector W emanating from B, there is a “rigid ˆ such that motion” M ˆ ˆ 1) M takes Aˆ to B, ˆ takes the direction Vˆ to the direction W ˆ. 2) M ˆ is a rigid motion means that Saying that M ˆ leaves the distance between them un1) for any two points Aˆ and Aˆ , M changed, that is, ˆ ˆ ˆ ˆ ˆ ˆ M A M A = A A ,

ˆ the angle between M ˆ Vˆ  2) for any two rays Vˆ  and Vˆ  emanating from A,

ˆ Vˆ  is the same as the angle between Vˆ  and Vˆ  . and M Later in the book we will use Euclidean coordinates and matrices to make the notion of congruences and rigid motions more explicit and mathematically rigorous. Exercise 4. Using a sketch on grid paper or an algebraic formulation in the Euclidean plane, a) give a concrete example of a rigid motion that takes (1, 2) to (3, 5), b) modify your answer in a) so that the tangent vector (1, 0) emanating from (1, 2) goes to the tangent vector (0, 1) emanating from (3, 5). Exercise 5 (NG). Think back to high school days and write the triangle congruence rules SSS, SAS, and ASA, in other words “side-side-side”, “side-angle-side”, and “angle-side-angle”. Be careful with your wording for each rule so that two triangles can be moved onto each other by a rigid motion if and only if they satisfy that rule. Exercise 6. Give a counterexample to show that there is no universal SSA law. Although it is a bit tedious, you can derive the usual rules for congruent triangles (SSS, SAS, ASA) using only E1–E4. Thus these laws hold in any neutral geometry, that is, in any geometry satisfying E1–E4. Exercise 7. a) Given an angle ∠BAC and a ray R, use E1–E4 to produce an angle congruent to ∠BAC but with R as one of its sides. b) Use a) and E1–E4 to prove the SAS rule for congruent triangles. Exercise 8 (NG). Suppose in the following diagram that |BD| = |CD| and |AD| = |ED|.

12

2. Euclid’s postulates for plane geometry

a) Show that ∠ADB and ∠CDE have the same measure and are therefore congruent. (A pair of opposite angles where two lines cross is called a vertical angle pair. Your argument will show that the two opposite angles in any vertical angle pair are congruent.) b) Show that triangle BDA and triangle CDE are congruent [MJG, 138]. Exercise 9. In ABC below, an exterior angle at vertex C is formed by a side of the triangle that contains C and the outward extension of the other side of the triangle containing C. Show in neutral geometry that the measure of either exterior angle of the triangle at C is greater than the measure of either remote interior angle [MJG, 199].

Hint: Compare this figure with the figure in Exercise 8. Exercise 10. Use Exercise 9 to show that the sum of any two angles of a triangle is less than 180°. Hint: Begin with the fact that the sum of the measures of the interior angle at C and either exterior angle at C is 180°. Exercise 11 (NG). Show in NG that if two lines cut by a transversal line have a pair of congruent alternate interior angles, then they are parallel [MJG, 117]. Hint: Suppose the assertion is false for some pair of lines. Find a triangle that violates the conclusion of Exercise 9.

2.2. Sum of angles in a triangle in NG We will now reason to one of the fundamental results about neutral geometry, one that puzzled mathematicians for many centuries, in fact until the discovery of hyperbolic geometry about two centuries ago. That discovery showed that there was more than one geometry that satisfied the axioms of neutral geometry, and that

2.2. Sum of angles in a triangle in NG

13

attempts to show that Euclid’s Fifth Postulate was a consequence of the four axioms of neutral geometry were futile. There was another geometry, namely hyperbolic geometry, that satisfied E1–E4. The thing that separates hyperbolic geometry from Euclidean (plane) geometry is the sum of the angles in a triangle. (If you had a good geometry course in high school, you may remember that you had to use Euclid’s Fifth Postulate in order to show that the sum of the angles in a triangle was 180°. But more on that later.) There is one important fact about the sum of the angles in a triangle that you can prove in NG, that is, without invoking Euclid’s Fifth Postulate. We will in fact accomplish that in this section. Exercise 12 (NG). For the diagram in Exercise 8, show that the sum of the angles in ACE is the same as the sum of the angles in ACB. Exercise 13 (NG). Suppose that in NG the sum of the angles in a triangle ABC is (180 + x) ° with x > 0. For the ABC in Exercise 8, show that one of the angles of ACE is no more than half the size of ∠BAC, yet by Exercise 12 the sum of the angles in a triangle ABC is still (180 + x) °. Hint: In the diagram in Exercise 8, this new “smaller” angle may or may not have vertex A [MJG, 125–127]. Exercise 14 (NG). Suppose that in NG the sum of the angles in a triangle ABC is (180 + x) ° with x > 0. Let α denote the measure of ∠BAC. a) Rename the vertices of ACE in Exercise 13 to get the same triangle denoted as A2 B2 C2 but with A2 denoting the vertex (either A or E) that has the smaller angle. Repeat the above constructions to construct a triangle ABC in which the sum of the angles is (180 + x) °, but one of its angles is less than or equal to 1 α. 4 b) Repeat the construction in Exercise 13 over and over again n times to construct a triangle with the sum of its angles still equal to (180 + x) ° but such that one of its angles has size less than 1 α. 2n Exercise 15 (NG). Suppose that in NG the sum of the angles in a triangle ABC is (180 + x) ° with x > 0. Show that there is a positive integer n0 so that if you repeat the construction in Exercise 13 n0 times, the result will be a triangle An0 Bn0 Cn0 with the sum of its angles still equal to (180 + x) ° but with one of its angles having measure less than x [MJG, 125–127]. Hint: Remember that x > 0 is fixed once and for all at the beginning of the argument in Exercise 13. Then use that limn→∞ 21n α = 0. Exercise 16 (NG). a) For An0 Bn0 Cn0 in Exercise 15, explain why the sum of two of its interior angles would have to be greater than 180°. b) Use Exercise 10 to conclude that you cannot have a triangle with two angles summing to more than 180° [MJG, 124].

14

2. Euclid’s postulates for plane geometry

So we have a contradiction to the hypothesis that the sum of the angles in some triangle ABC is (180 + x) ° with x > 0. We review the argument. Theorem 3. In NG, the sum of the interior angles in any triangle is no greater than 180°. Proof. We argue by contradiction. Start with ABC as in Exercise 13 in which the sum of the angles is (180 + x) ° with x > 0. Suppose the measure of the angle at A is denoted by α. By Exercise 13 there exists a triangle A(1) B (1) C (1) such that the sum of its angles is (180 + x) ° and the measure of the angle at the vertex A(1) is less than or equal to α2 . By Exercise 14 there is a triangle A(n) B (n) C (n) such that the sum of its angles is (180 + x) ° and the measure αn of the angle at the vertex A(n) is less than or equal to 2αn . If n is sufficiently big, α < x. 2n So, for that value of n, if βn is the measure of the angle of A(n) B (n) C (n) at the vertex B (n) and γn is the measure of the angle of A(n) B (n) C (n) at the vertex C (n) , then we have the two relations α αn < n < x, 2 αn + βn + γn = 180 + x. So βn + γn = 180 + (x − αn ) > 180. But this is impossible by Exercise 9.



Exercise 17 (NG). Show the following: a) The sum of the interior angles in any quadrilateral is no greater than 360°. b) The sum of the interior angles of an n-gon is no greater than (n − 2) · 180°. Hint: Divide the n-gon into an r-gon and an s-gon with r + s = n + 2 and use complete induction.

2.3. Are there rectangles in NG? Suppose we have the following figure in NG where |AB| = |CD| and ∠ABC and ∠BCD are both right angles.

If we add the line segment AD to the figure, do we obtain a rectangle, that is, a quadrilateral in which all four interior angles are right angles? The next two exercises will lead us to the answer. Remember that the SSS, SAS, and ASA rules are true in neutral geometry, but we only know that the sum of interior angles in a triangle is less than or equal to 180° in neutral geometry.

2.3. Are there rectangles in NG?

15

Exercise 18 (NG). a) If we add the line segment AD to the figure above, show that ∠BAD is congruent to ∠ADC. b) Use Exercise 17 to show that the two congruent angles in a) must each have measure less than or equal to 90°. Now suppose that in the above figure we extend the side AB to A B so that |A B| > |CD|, and then construct the line segment A D: 

Exercise 19 (NG). Show that the measure of ∠BA D is less than the measure of ∠A DC. Hint: Use Exercises 18 and 9. The last two exercises let us conclude the following. Let ABCD be a quadrilateral with right angles ∠ABC and ∠BCD. (We denote polygons by naming their vertices in counterclockwise order [MJG, 164– 165].) Then |AB| = |CD| implies that ∠BAD = ∠ADC, |AB| > |CD| implies that ∠BAD < ∠ADC, |AB| < |CD| implies that ∠BAD > ∠ADC. Exercise 20. Use the three implications just above and pure logic to show that ∠BAD < ∠ADC implies that |AB| > |CD|, ∠BAD = ∠ADC implies that |AB| = |CD|, ∠BAD > ∠ADC implies that |AB| < |CD|. Hint: If the first implication is false, then ∠BAD < ∠ADC and either |AB| < |CD| or |AB| = |CD|. If |AB| < |CD|, we have already shown that ∠BAD > ∠ADC, so we have a contradiction!

Part II

Euclidean (plane) geometry

Chapter 3

Rectangles and cartesian coordinates

3.1. Euclid’s Fifth Postulate, the Parallel Postulate We are finally ready to introduce Euclid’s fifth and final axiom, the so-called Parallel Postulate. Axiom (E5). Through a point not on a line there passes one and only one parallel line. NG together with E5 is called Euclidean geometry (EG). As mentioned above, we will see later that there is another geometry HG that satisfies all the postulates of NG but not E5. In it, an infinite number of parallel lines will pass through a given point not on a line. Furthermore the sum of the interior angles of a triangle will always be less than 180° [MJG, 134]! Now suppose we have two parallel lines, L and M , in the Euclidean plane (EG). Suppose we have a third line that is transversal to L and M , that is, a line that meets L at a point A and meets M a point B. Using Exercise 11, construct a line M  through B that is parallel to L. Exercise 21 (EG). a) Show that if two parallel lines L and M are cut by a transversal line, opposite interior angles are equal. Hint: Concentrate on the words “only one” in E5. b) Show that if two parallel lines L and M are cut by a transversal line, corresponding angles are equal. Hint: Use a) and the fact that vertical angles are equal. Exercise 22. a) Show that if a line T crosses one of two parallel lines L and M , it must cross the other. b) Show that if two distinct lines L and M are both parallel to another line, then they are parallel to each other. 19

20

3. Rectangles and cartesian coordinates

Exercise 23. a) Show that angles ∠BAC and ∠B  A C  in the Euclidean plane are equal (i.e., have the same measure) if their corresponding rays are parallel. Hint: Construct a transversal through A and A . b) Show that ∠BAC and ∠B  A C  are equal (i.e., have the same measure) if    ∠B A C can be rotated around A to obtain an angle ∠B  A C  with corresponding rays parallel to those of ∠BAC. Exercise 24 (EG). Use the “only one” assertion in E5 together with what we have established about neutral geometry to show that in EG the sum of the interior angles of any triangle is 180°. Exercise 25 (EG). Show that in EG the sum of the interior angles of a quadrilateral is 360°. Exercise 26 (EG). a) Show that in EG, given any positive real numbers a and b, there exists a rectangle with adjacent sides of lengths a and b. Hint: Construct a third side of length a perpendicular to the side of length b; that is, the two sides meet in a right angle. Then use Exercises 18a) and 25. b) If we define the area inside the rectangle with adjacent sides of lengths a and b as a · b, show that the area of any right triangle is 12 (base) · (height). c) Show that the area of (i.e., inside) any triangle is 12 (base) · (height). Hint: Find two right triangles such that the area of the given rectangle is the sum or difference of the areas of the right triangles. Exercise 27 (EG). Show that there is a cartesian coordinate system on EG; that is, the set of points of EG are in one-to-one correspondence with the set of pairs of real numbers. Write your solution as if you are explaining this to an eighth grader. Important distinction: We can use cartesian coordinates to define the position of a point in the plane. Thus (1, 3) can indicate the point one unit to the right and three units up from the origin in the cartesian plane. It is when we use (1, 3) in this sense that we will call (1, 3) a point. But we can also use cartesian coordinates to indicate a displacement of points in the cartesian plane. So (2, −3) can be used to indicate a displacement of points by moving them two units to the right and three units down in the plane. It is when we use (2, −3) in this latter sense that we call (2, −3) a vector. We will also use an initial point (a, b) and a final point (a , b ) to define the vector (a − a, b − b) that represents the displacement of any point (x, y) to the point (x + a − a, y + b − b) and, in particular, the displacement of the point (a, b) to the point (a , b ). Definition 3 (EG). The slope of a line in the cartesian (x, y)-plane is the (signed) tangent of the angle that the line makes with the x-axis, unless the line is parallel to the x-axis, in which case its slope is zero, or perpendicular with the x-axis, in which case we say that its slope is ±∞. (Notice that the notion of slope is always defined with respect to a given cartesian coordinate system on the Euclidean plane.)

3.2. The distance formula in EG

21

Exercise 28 (EG). Show that the slope of a line in a cartesian plane is obtained by taking any two points (x1 , y1 ) and (x2 , y2 ) and computing y2 − y1 x2 − x1 for any two distinct points (x1 , y1 ) and (x2 , y2 ) on the line. Exercise 29 (EG). a) Show that two lines in a cartesian plane are parallel if and only if they have the same slopes. Hint: Use Exercise 21. b) Show that two lines L and M intersecting at a point P are perpendicular if and only if the product of their slopes is −1 (or one line has slope zero and the other has slope ±∞). Hint: Suppose that the slopes are not zero or ±∞. Draw the line N parallel to the x-axis through P . Show that the following two statements are equivalent: i) the right triangle with one side along N and its hypotenuse along L and of length one is congruent to the right triangle with a side on N and a length one hypotenuse along M , ii) L and M are perpendicular.

3.2. The distance formula in EG It is the existence of a cartesian coordinate system in EG that allows us to define the distance between points as

(3.2.1) d ((a1 , b1 ) , (a2 , b2 )) = (a2 − a1 )2 + (b2 − b1 )2 , giving rigorous mathematical meaning to a concept that the ancient Greeks were never able to describe precisely, namely the similarity of figures in EG. For that we will require the notion of a dilation or magnification in EG. And we need a cartesian coordinate system to describe dilation precisely, a reality backed up by the fact that similarities do not exist in HG or SG. (Try drawing two triangles that are similar but not congruent on a perfectly spherical balloon!) There is an issue that arises at this point that we will “sweep under the rug” for a while but return to later in this book. Namely, there is a different cartesian coordinate system for each ray in the plane, namely the cartesian coordinate system that assigns (0, 0) to the origin of the ray and makes the ray itself the positive xaxis. Implicit in (3.2.1) is the assumption (to be used now but justified later) that the distance between two points in the plane does not depend on the cartesian coordinate system that we impose on the plane as long as the unit distance along the respective x-axes is the same. Exercise 30 (EG). State the Pythagorean theorem in EG and use Exercise 26 to prove it. Hint: For the given right triangle, draw a cartesian coordinate system with (0, 0) at the vertex of the right triangle and with the x-axis containing one side of the right angle and the y-axis containing the other side. In that cartesian plane, draw a square with vertices (0, 0) , (a + b, 0) , (0, a + b) , and (a + b, a + b). Inside that

22

3. Rectangles and cartesian coordinates

square, draw the square with vertices (a, 0) , (a + b, a) , (b, a + b) , and (0, b). Show by rearranging the triangles in your drawing that the area of the big square is the area of the little square plus the area of four right triangles, each of area ab 2 . (Such an argument is one that depends least on the particular choice of cartesian coordinate system.) Exercise 31. Use the Pythagorean theorem to justify the definition (3.2.1). Exercise 32 (Triangle inequality). Show that for any three points (a1 , b1 ), (a2 , b2 ), and (a3 , b3 ) in the cartesian plane, d ((a1 , b1 ) , (a3 , b3 )) ≤ d ((a1 , b1 ) , (a2 , b2 )) + d ((a2 , b2 ) , (a3 , b3 )) . Hint: Write what you need to show in terms of (x1 , x2 ) := (a2 − a1 , b2 − b1 ) , (y1 , y2 ) := (a3 − a2 , b3 − b2 ) , (x1 + y1 , x2 + y2 ) = (a3 − a1 , b3 − b1 ) . Then use that 2

(x1 y2 − x2 y1 ) ≥ 0.

3.3. Law of Sines and Law of Cosines We next review some basic facts about triangles that are established by dividing the given triangle OP1 P2 (as pictured below) into two right triangles, OP P1 and P1 P P2 , and applying the Pythagorean theorem. That is, in the picture below ∠OP P1 has an angle of 90°.

Theorem 4 (Law of Sines). In the triangle above sin (∠P1 OP2 ) sin (∠P1 P2 O) = . |P1 P2 | |P1 O| Proof. Use sin (∠P1 OP2 ) = sin (∠P1 P2 O) =

|P1 P | |P1 O| , |P1 P | |P1 P2 | ,



and algebra. Notice that the above theorem implies that for any ABC, sin (∠BAC) sin (∠ABC) sin (∠ACB) = = . |BC| |AC| |AB|

3.3. Law of Sines and Law of Cosines

23

Exercise 33. The next theorem is called the Law of Cosines. As you work through the proof, construct a diagram or picture for each step. Theorem 5 (Law of Cosines). The angle ϑ = ∠P1 OP2 in the triangle above satisfies the relation 2

2

2

|P1 P2 | = |OP1 | + |OP2 | − 2 |OP1 | · |OP2 | · cos ϑ. Proof. By the Pythagorean theorem |P1 P2 |2 − |P2 P |2 = |P P1 |2 2

2

= |OP1 | − |OP |

|P1 P2 |2 = |OP1 |2 + |P2 P |2 − |OP |2 = |OP1 |2 + (|P2 P | + |OP |) (|P2 P | − |OP |) = |OP1 |2 + |OP2 | (|P2 P | − |OP |) 2

= |OP1 | + |OP2 | (|OP2 | − 2 |OP |) , but |OP | = |OP1 | · cos ϑ.  Exercise 34. Give a quick proof of the triangle inequality using the Law of Cosines. Notice that the above theorem implies that for any ABC with interior angle ϑ at vertex C, |AB|2 = |AC|2 + |BC|2 − 2 |AC| · |BC| · cos ϑ. −−→ −−→ Corollary 1. The angle ϑ = ∠P1 OP2 between two vectors OP1 and OP2 is given by the formula

⎞ ⎛ |P1 P2 |2 − |OP1 |2 + |OP2 |2 ⎠. ϑ = arccos ⎝ −2 |OP1 | · |OP2 |

The expression |P1 P2 |2 − |OP1 |2 + |OP2 |2 in the equation just above is a bit complicated. If we write it out in cartesian coordinates, we will see that we can simplify it quite a bit: −−→ OP1 = (a, b) −−→ OP2 = (c, d) −−−→ P1 P2 = (c − a, d − b) 2 2 2 |P1 P2 | = (c − a) + (d − b)

  2 2 2 2 2 |P1 P2 | − |OP1 | + |OP2 | = (c − a) + (d − b) − c2 + a2 + d2 + b2

2 2 2 |P1 P2 | − |OP1 | + |OP2 | = −2 (ac + bd) . This leads us to the following definition.

24

3. Rectangles and cartesian coordinates

Definition 4. If

−−→ OP1 = (a, b) , −−→ OP2 = (c, d) , −−→ −−→ then we define the dot-product of the vectors OP1 and OP2 emanating from the same point O as −−→ −−→ OP1 • OP2 = (a, b) • (c, d) = ac + bd.

As a consequence we have the more elegant formula  −−→ −−→  OP1 • OP2 ϑ = arccos |OP1 | · |OP2 | −−→ −−→ for the angle ϑ formed between the two vectors OP1 and OP2 . The Law of Cosines also lets us derive a formula for the area inside a triangle given the lengths a, b, and c of its three sides. Namely, if ϑ is the interior angle between the sides corresponding to a and b, then   −2ab · cos ϑ = c2 − a2 + b2 2   4a2 b2 · cos2 ϑ = c2 − a2 + b2 2     4a2 b2 · 1 − sin2 ϑ = c2 − a2 + b2    2 4a2 b2 − c2 − a2 + b2 = 16 · Area2 . Exercise 35. Do algebra on the left-hand side of the last equation above to derive Heron’s formula  Area = s (s − a) (s − b) (s − c), . where s = a+b+c 2

3.4. Dilations in EG We next consider transformations F of the Euclidean plane that are not rigid motions but still retain strong metric properties. These transformations will be oneto-one (also called injective); that is, if the point A does not equal the point B, then the point F (A) does not equal the point F (B). Also they will be onto (also called surjective); that is, for every point B in the plane, there is a point A in the plane such that F (A) = B. (Transformations that are both injective and surjective are called one-to-one correspondences.) Definition 5 (EG). A dilation is a one-to-one correspondence of the cartesian plane to itself that a) fixes one point called the center of the dilation, b) takes each line through the fixpoint to itself, c) multiplies all distances by a fixed positive real number called the magnification factor of the dilation. Definition 6. Given a point (x0 , y0 ) in the (cartesian) plane and a positive real number r, we define a mapping D with center (x0 , y0 ) and magnification factor r by the formula (3.4.1)

D (x, y) = (x0 , y0 ) + r (x − x0 , y − y0 ) .

3.4. Dilations in EG

25

  We will also denote the output D (x, y) of the dilation as x, y and write   (3.4.2) x, y = (x0 , y0 ) + r (x − x0 , y − y0 ) . What follows is a sequence of exercises that can be done just using the definition of dilation and the properties of Euclidean geometry that we have already derived. Alternatively they can be done by noting that   D (x, y) = T(x0 ,y0 ) ◦ D0 ◦ T(−x0 ,−y0 ) (x, y) , where D0 is a dilation with (x0 , y0 ) = (0, 0) and T(a,b) (x, y) = (x + a, y + b), and proving the requisite properties for D0 and T(a,b) . The advantage of this last approach is that the formulas D0 (x, y) = (rx, ry) and T(a,b) (x, y) = (x + a, y + b) are easy to work with. Exercise 36 (EG). Using cartesian coordinates for the plane, show that the mapping D defined in (3.4.1) satisfies all three conditions in the definition of a dilation with magnification factor r and center (x0 , y0 ). Hint: For the second condition, use a parametric representation for a line through (x0 , y0 ). Exercise 37 (EG). Show (using the Substitution Rule, Theorem 2 if you wish) that a dilation with magnification factor r multiplies all areas by a factor of r 2 . Exercise 38 (EG). a) Show that the inverse mapping of a dilation is again a dilation with the same center but with magnification factor r −1 .   Hint: Use (3.4.2) and solve for (x, y) in terms of x, y . b) Show that a dilation D takes lines to lines. Hint: Start with a line ax + by = c   in the x, y -plane. Show that it comes from a line in the (x, y)-plane by substitution using (3.4.2). Or start with a line ax + by = c in the (x, y)-plane and use the formula for D−1 you derived in a). Exercise 39 (EG). Show that a dilation takes any line to a line parallel (or equal) to itself. Hint: Compute slopes. Exercise 40 (EG). Show that a dilation by a factor of r takes any vector to r times itself. Hint: Realize the vector as the difference of two points. Exercise 41 (EG). Show that a dilation of the plane preserves angles. Hint: Use Exercises 23a) and 39.

26

3. Rectangles and cartesian coordinates

3.5. Similarity in EG Definition 7 (EG). Two plane figures are congruent if there is a rigid motion of the plane that takes one to a figure that coincides with the other. Two plane figures are similar if there is a dilation of the plane that takes one to a figure that is congruent to the other. We write ABC ∼ A B  C  to denote that these two triangles are similar (where the order of the vertices tells us which vertices correspond). Theorem 6 (EG). a) If two triangles are similar, then corresponding sides are proportional. b) If corresponding sides of two triangles are equal, then the two triangles are similar. Proof. a) Given that ABC ∼ A B  C  , there is a dilation D with magnification factor r that takes ABC to a triangle A B  C  that is congruent to A B  C  . By the definition of dilation |A B  | = r · |AB| , |B  C  | = r · |BC| , |A C  | = r · |AC| . Since A B  C  is congruent to A B  C  , we have by SSS that |A B  | = |A B  | , |B  C  | = |B  C  | , |A C  | = |A C  | . So by algebra

| A B  |

= r,

|B  C  |

= r,

|A  C  |

= r.

|AB|

|BC| |AC|

So the lengths of corresponding sides are proportional; that is, the ratios of corresponding sides are all equal to the same constant r. b) Suppose that corresponding sides of the triangles ABC and A B  C  are proportional; that is, | A B  | |AB| = r, |B  C  | |BC| = r, A | C| |AC| = r. Let D be a dilation with center A and magnification factor r. So D takes ABC to a triangle A B  C  such that |A B  | = r · |AB| = |A B  | , |B  C  | = r · |BC| = |B  C  | , |A C  | = r · |AC| = |A C  | .

3.5. Similarity in EG

27

So, by SSS, A B  C  is congruent to A B  C  . Therefore ABC and A B  C  satisfy the definition of similarity since the dilation D takes ABC to a triangle  that is congruent to A B  C  . Exercise 42 (EG). a) Show that if two triangles are similar, then corresponding angles are equal. Hint: You have to start from the supposition that the two triangles satisfy the definition of similar triangles. b) Show that if corresponding angles of ABC and A B  C  are equal, then ABC and A B  C  are similar. Hint: You have to start from the supposition that corresponding angles of the two triangles have equal measure, then dilate ABC with a dilation for which r = |A B  |/|AB|. Finally use ASA to show that the dilation of ABC is congruent to A B  C  . Exercise 43 (EG). Show that two triangles are similar if corresponding sides are parallel. Hint: Use Exercise 23a). Exercise 44. Show that two triangles are similar if corresponding sides are perpendicular. Hint: Extend one of the rays of the first triangle until it crosses the corresponding ray of the second triangle.

Chapter 4

Concurrence and circles in Euclidean geometry

4.1. Concurrence theorems in EG, Ceva’s theorem Before leaving (plane) Euclidean geometry, we will visit two more of its many sets of memorable properties, ones that you may or may not have seen in high school. The first collection of these are called concurrence theorems—these theorems relate the measures of the three sides (or angles) of a triangle to the fact that three lines constructed using those measures are concurrent; that is, the three lines pass through a common point. Exercise 45 (EG). Denote the measure or area of a triangle ABC as |ABC|. Show that, in the diagram below, |AF C| |AF X| |AF | = = . |F B| |CF B| |XF B|

Exercise 46 (EG). Use Exercise 45 to show by pure algebra that (4.1.1)

|AXC| |AF | = . |F B| |CXB|

Hint: |AXC| = |AF C| − |AF X|, etc. 29

30

4. Concurrence and circles in Euclidean geometry

Exercise 47 (EG). For three concurrent segments AD, BE, and CF as given in Exercise 45, use Exercise 46 to show that |AF | |BD| |CE| · · = 1. |F B| |DC| |EA| Hint: Use (4.1.1) with side BC replacing AB and (4.1.1) with side CA replacing AB. This last result together with its converse, which we will show next, is called Ceva’s theorem [MJG, 287–288]. Theorem 7 (Ceva’s theorem, EG). Line segments AD, BE, and CF from vertices of a triangle to points on their opposite sides pass through a common point if and only if |AF | |BD| |CE| · · = 1. |F B| |DC| |EA| Proof. By Exercise 47 we only have to show the converse direction. Namely, suppose |AF | |BD| |CE| · · =1 |F B| |DC| |EA| but that the three segments do not pass through a common point.

We argue by contradiction. Let F  be the point segment AB so that segment AF  passes through the intersection point of segments AD and BE. Then |AF  | |BD| |CE| · · = 1, |F  B| |DC| |EA| again by Exercise 47. Then by algebra |AF  | |AF | (4.1.2) = ·  |F B| |F B| But if, for example, F  is to the left of F along the segment AB, then |AF  | is strictly smaller than |AF | and |F  B| is strictly larger than |F B|. So |AF  | |AF | < ·  |F B| |F B| By the same reasoning, if F  is to the right of F along the segment AB, then |AF  | is strictly larger than |AF | and |F  B| is strictly smaller than |F B|. So |AF | |AF  | > · |F  B| |F B|

4.1. Concurrence theorems in EG, Ceva’s theorem

31

Either possibility contradicts (4.1.2). So the point F  must be the same as the point F.  Exercise 48 (EG). A median of a triangle is a line segment from a vertex to the midpoint of the opposite side. Show that the medians of any triangle meet in a common point. Hint: Use Ceva’s theorem. For our final concurrence theorem, we will use two of the facts about similar triangles that we proved previously, namely that if corresponding angles are equal, then triangles are similar, and if corresponding triangles are similar, then corresponding sides are proportional. Furthermore, we will start from pairs of triangles for which we only know that two of their corresponding angles are equal. But that will be enough in EG, because the third angle of each triangle must then have measure 180° minus the sum of the measures of the other two, and so the third pair of corresponding angles must also be equal. Theorem 8 (EG). The three altitudes of a triangle are concurrent.

Proof. The altitude to a vertex of a triangle is perpendicular to the side opposite that vertex. Therefore by the property that two pairs of corresponding angles are equal, we have the following similarities: CEB ∼ CDA, AF C ∼ AEB, BDA ∼ BF C. From these three similarities, we get three proportionalities of sides: |CE| |CB| |AF | |AC| |BD| |BA|

= = =

|CD| |CA| , |AE| |AB| , |BF | |BC| .

Using these equalities we have by algebra

=

|CE| |CA| |AF | |AB| |BD| |BC| |CB| · |CD| · |AC| · |AE| · |BA| · |BF | |CE| |AF | |BD| 1 1 1 1 · |CD| · 1 · |AE| · 1 · |BF | = 1

Now apply (the “converse direction” of) Ceva’s theorem.

· 1 · 1. 

32

4. Concurrence and circles in Euclidean geometry

4.2. Properties of circles in EG Our final topic before leaving Euclidean geometry is circles. We include this partly for its own interest and partly because the properties we visit here will be useful later on. Again we explore the topic through a sequence of exercises (with hints to their solutions included to ease the way). We begin with perhaps the most basic fact of all about circles in EG. Exercise 49 (EG). The circle of radius one has (interior) area that we denote as π. Use this to reason to the fact that the circle of radius one has circumference 2π.

Hint: Slice up the circle into slices of equal size as illustrated in the above picture. Approximate a rectangle by rearranging those slices. Notice that the greater the number of slices, the better the approximation. Use the fact that the area covered by the slices is independent of the number of slices to conclude that it must be the same as the area of the limiting rectangle. Next we turn to some facts about chords in circles and angles inscribed in circles. Exercise 50 (EG). On the circle with center O below,

show that ∠AXB = (1/2)(∠AOB). Hint: OAX is isosceles.

4.2. Properties of circles in EG

33

Exercise 51 (EG). On the circle with center O below,

show that ∠AXB = (1/2)(∠AOB). Hint: Draw the diameter through O and X, use Exercise 50, and add. Exercise 52 (EG). On the circle with center O below,

show that ∠AXB = (1/2)(∠AOB). Hint: Draw the diameter through O and X, use Exercise 50, and subtract. We can summarize the results of the last three exercises into the following theorem. Theorem 9. The measure of any angle inscribed in a circle is one-half of the measure of the corresponding central angle. Exercise 53 (EG). Use similar triangles and the previous exercises to show that |AX| · |XB| = |A X| · |XB  | in the figure below.

Hint: Draw AB  and A B.

34

4. Concurrence and circles in Euclidean geometry

Exercise 54 (EG). Use similar triangles and the previous exercises to show that |AX| · |XB| = |A X| · |XB  | in the figure below.

Hint: Draw AB  and A B.

4.3. Circles and sines and cosines Lemma 1. i) The set of points in the cartesian plane equidistant from two given points A = (a1 , a2 ) and B = (b1 , b2 ) forms a straight line. ii) That straight line is perpendicular to the line through A and B. Proof. i) The equation expressing the equality of the two distances is

(x − a1 )2 + (y − a2 )2 = (x − b1 )2 + (y − b2 )2 . We square both sides, (x − a1 )2 + (y − a2 )2 = (x − b1 )2 + (y − b2 )2 , and simplify: −2a1 x + a21 − 2a2 y + a22 = −2b1 x + b21 − 2b2 y + b22 (a2 +a2 )−(b2 +b2 ) (a1 − b1 ) x + (a2 − b2 ) y = 1 2 2 1 2 . ii) Pick any point C on the above line except the point D lying on the line through A and B. By SSS ADC is congruent to the triangle BDC.  The line given by Lemma 1 is called the perpendicular bisector of the line segment AB. Exercise 55. Show that given any three noncollinear points in the cartesian plane, there is a unique circle passing through the three points. Hint: The three noncollinear points are the three vertices of a unique triangle. Pick any two sides of that triangle and form their perpendicular bisectors. Show that the center of the circle must be the point at which those perpendicular bisectors intersect.

4.4. Cross-ratio of points on a circle

35

Exercise 56 (Extended Law of Sines). In the diagram below, O is the center of the circle and d is its diameter.

Show that sin α 1 = . |AB| d Hint: Bisect AB at a point C. Compare the inscribed angle to the central angle to show that |AC| |BC| = sin α = . |OA| |OB|

4.4. Cross-ratio of points on a circle We have just seen that given three noncollinear points in the plane, there is a unique circle passing through the three points. But how about four points in the plane, no three of which are collinear? The issue we will explore in this section is the question of finding a numerical condition determined by the four points that tells us exactly when they all lie on a single circle. The relevant mathematical quantity determined by four points lying on a circle (or by four points lying on a line) is called the cross-ratio of the four points. Exercise 57 (EG). In the diagram below, show that |AB| sin α sin (∠AOB) = = . |CB| sin β sin (∠COB)

Hint: Use the Extended Law of Sines. Alternatively notice that by Theorem 9 m (∠BAO) + m (OCB) = 180°

36

4. Concurrence and circles in Euclidean geometry

so that sin (∠BAO) = sin (∠OCB) . Now use the Law of Sines. Exercise 58 (EG). Show that if, in the above figure, B moves along the circle to the other side of C, it is still true that |AB| sin (∠AOB) = . |CB| sin (∠COB) Exercise 59 (EG). In the diagram

show that

|A B  | sin (∠A OB  ) sin (∠B  A O) = ÷ |C  B  | sin (∠C  OB  ) sin (∠B  C  O)

[MJG, 266–267]. Exercise 60 (EG). Show that if, in the above figure, B  moves along the line to the other side of C  , it is still true that |A B  | sin (∠A OB  ) sin (∠B  A O) = ÷ . |C  B  | sin (∠C  OB  ) sin (∠B  C  O) These last two exercises allow us to define the cross-ratio of four points on a circle. Definition 8 (EG). For a sequence of four (ordered) points A, B, C, and D on a circle, we define |AB| |AD| ÷ (A : B : C : D) = |CB| |CD| which we call the cross-ratio of the ordered sequence of the four points. Similarly for a sequence of four (ordered) points A , B  , C  , and D on a line, we define (A : B  : C  : D ) =

|A B  | |A D | ÷ |C  B  | |C  D |

which we call the cross-ratio of the ordered sequence of the four points. Note that Definition 17 is just a refinement of the definition of (A : B  : C  : D ) just above. In Definition 17 we are keeping track of the signs of the terms in the quotients whereas (A : B  : C  : D ) is always non-negative.

4.5. Ptolemy’s theorem

37

Exercise 61. a) Show that in the figure

we have the equality (A : B : C : D) = (A : B  : C  : D ) . Hint: Use Exercises 57–60. b) What happens in a) if we move B to the other side of C? We say that the cross-ratio is invariant under stereographic projection.

4.5. Ptolemy’s theorem Given any three noncollinear points in the Euclidean plane, there is one and only one circle that passes through the three points. (How do you construct it?) We have already seen with a few examples that given four noncollinear points A, B, C, and D in the plane, it is not always true that there is a circle that passes through all four. A famous theorem of classical Euclidean geometry gives the exact condition that there is a circle that passes through all four. Theorem 10 (Ptolemy). If the ordered sequence of points A, B, C, and D all lie on a circle, then |AC| · |BD| = |AD| · |BC| + |AB| · |CD| .

That is, if an ordered sequence of points A, B, C, and D all lie on a circle, then the product of the diagonals of the quadrilateral ABCD is the sum of the products of pairs of opposite sides. Proof. We need to check that |AC| · |BD| = |AD| · |BC| + |AB| · |CD|

38

4. Concurrence and circles in Euclidean geometry

or, similarly, we need to check that |AB| · |CD| |AC| · |BD| =1+ . |AD| · |BC| |AD| · |BC| That is, we need to check that (A : C : B : D) = 1 + (A : B : C : D) . But by Exercise 61 this is the same as checking that (A : C  : B  : D ) = 1 + (A : B  : C  : D ) for the projection of the four points onto a line from a point O on the circle. But that is the same thing as showing that |A B  | · |C  D | |A C  | · |B  D | =1+   ,     |A D | · |B C | |A D | · |B  C  | which is the same thing as showing that |A C  | · |B  D | = |A D | · |B  C  | + |A B  | · |C  D | . Now check this last equality by high school algebra.



Exercise 62. a) Draw four points in the Euclidean plane, no three of which are collinear, that cannot lie on a single circle. With the help of a ruler, show that the distances between pairs of the four points violate the equality in Ptolemy’s theorem. b) Draw four points in the Euclidean plane that do lie on a single circle. With the help of a ruler, show that the distances between pairs of the four points satisfy the equality in Ptolemy’s theorem. Exercise 63 (Addition Law for Sines). The diameter of the circle below is one. Use Ptolemy’s theorem and the Extended Law of Sines to show the Addition Law for Sines, sin (α + β) = sin α · cos β + cos α · sin β.

Hint: Examine the central angles associated to the chords |AD| and |BC|.

Part III

Spherical geometry

Chapter 5

Surface area and volume of the R-sphere in Euclidean three-space

5.1. Volumes of pyramids We are now going to study the geometry of the R-sphere in Euclidean three-space. We will first study this geometry in the usual way, namely using Euclidean (ˆ x, yˆ, zˆ)coordinates. (You may ask why there are “hats” over the coordinates (x, y, z). That is because we want to reserve the notation (x, y, z) for another “non-Euclidean” set of coordinates on three-dimensional space that we will study in later chapters.) We start with a relationship that shows why there is a factor of 1/3 in many formulas for volumes in three-dimensional Euclidean geometry, just like there is a factor of 1/2 in many formulas for areas in two-dimensional Euclidean geometry. Exercise 64 (EG). Show that an r × r × r cube can be constructed from three equal pyramids with r × r square bases. Conclude that the volume of each pyramid is 1/3 the volume of the cube, namely 1 3 r . 3 Hint: Suppose the cube had a hollow interior and infinitely thin faces. Put your (infinitely tiny) eye at one vertex of the cube and look inside. How many faces of the cube can you see? We next want to show why any pyramid with an r × r square base and a vertical altitude r has the same volume. That is, if we put the vertex of the pyramid anywhere in a plane parallel to the base and at distance r, the volume is unchanged. 41

42

5. Surface area and volume of the R-sphere in Euclidean three-space

This fact is an example of Cavalieri’s principle: Shearing a figure parallel to a fixed direction does not change the n-dimensional measure of an object in Euclidean n-space. (Think of a stack of very thin books.) Exercise 65. Show that Cavalieri’s principle is true for the pyramid using several variable calculus. Hint: Put the base of the pyramid P so that its vertices are (0, 0), (r, 0), (0, r), and (r, r) in three-dimensional Euclidean space. Consider the transformation ⎛ ⎞ 1 0 0   x, yˆ, zˆ) ⎝ 0 1 0 ⎠ , x ˆ, yˆ, zˆ = (ˆ a b 1 and notice that

 P

⎛  1 ⎝ 0 dˆ xdˆ y dˆ z= P a

⎞ 0 0 1 0 ⎠ dˆ xdˆ y dˆ z. b 1

5.2. Magnification principle Magnification principle: If an object in Euclidean n-space is magnified by factors of r1 , . . . , rn , its n-dimensional measure is multiplied by r1 , . . . , rn . Exercise 66. Prove the magnification principle using several variable calculus. Hint: Use the transformation ⎞ ⎛ r1 . . . 0   x ˆ1 , . . . , x x1 , . . . , x ˆn = (ˆ ˆn ) ⎝ . . . . . . . . . ⎠ 0 . . . rn   and the Substitution Rule (Theorem 2) to compare dˆ x1 . . . dˆ xn with dˆ x1 . . . dˆ xn . P

P

Exercise 67 (EG). Use this magnification principle to justify the volume formula (1/3)B · h for any pyramid with rectangular base of area B and vertical altitude h. Now suppose we have any pyramid with any shaped base of area B and any vertical altitude h.

Approximate the base as close as you want (i.e., ε-close) by tiling its interior with rectangles. Let t denote the sum of the areas of these rectangles. From the other side approximate the base as close as you want (i.e., ε-close) by covering it entirely with rectangles. Let T denote the sum of the areas of these rectangles.

5.4. Surface area

43

Exercise 68 (EG). a) Why is the area B of the base of the pyramid caught between B − ε and B + ε? b) Show that the volume V of the pyramid is caught between (1/3) · t · h and (1/3) · T · h. Exercise 69 (EG). a) Argue that given any positive real number ε, however small, the volume V of the pyramid is caught between (1/3)·(B − ε)·h and (1/3)·(B + ε)·h. b) Show the formula V = (1/3) · B · h for the volume of a pyramid, where B is the area of the base and h is the vertical height.

5.3. Relation between volume and surface area of a sphere Think of a disco-ball. Its surface is approximately a sphere, but that surface is made up of tiny flat mirrors. Exercise 70 (SG). a) Why can you think of the disco-ball as being made up of pyramids, with each pyramid’s base as one of the tiny mirrors and vertex at the interior point O at the center of the disco-ball? b) Argue that the volume of the disco-ball is (1/3) times the distance h from a mirror to O times the sum of the areas of all the mirrors. Exercise 71 (SG). a) Argue that as the mirrors are made to be smaller and smaller, 1) the sum of the areas of the mirrors approaches the surface area of a sphere, 2) the distance h approaches the radius R of that sphere, 3) the volume of the disco-ball approaches the volume of the region whose boundary is the sphere. b) Use limits to conclude that for a sphere of radius R in Euclidean three-space, the relation between the volume V of the region bounded by the sphere and the surface area S of the sphere is given by the formula R·S . V = 3 Our goal in the next section is to compute the surface area of the sphere of radius R in Euclidean three-space. The formula just above will then let us compute the volume of the region whose boundary is the same sphere.

5.4. Surface area To compute the surface area of the sphere of radius R in three-dimensional Euclidean space, we will show that its surface area is equal to the surface area of something we can lay out flat. This argument goes way back to the great physicist and mathematician, Archimedes of Alexandria, in the second century B.C. To follow his argument, we have to begin by computing the area of a “lamp shade” or “collar”. We think of a circular collar in the following figure as approximated by an arrangement of trapezoids.

44

5. Surface area and volume of the R-sphere in Euclidean three-space

To achieve this, we approximate the bottom circle of the collar by an inscribed regular n-gon whose vertices are the points of intersection with the slant lines in the figure. Similarly approximate the top circle by an inscribed regular n-gon positioned directly above the bottom one, again with vertices given by the points of intersection with the slant lines. Complete a side of the bottom n-gon and the side of the top n-gon directly above it to a trapezoid by adjoining the two slant lines in the figure that connect endpoints. Let bn denote the length of a side of the bottom regular n-gon, and let tn denote the length of a side of the top n-gon. Then the trapezoid has area   bn + tn · hn , 2 where hn is the vertical height of the trapezoid. The collar is approximated by the union of these n trapezoids, so the area of the collar is approximated by the sum of the areas of the n congruent trapezoids, namely  n·

bn + tn 2



 · hn =

n · bn + n · tn 2

 · hn .

As n goes to infinity, the area of the approximation approaches the area of the collar. But lim n · bn = cb ,

n→∞

lim n · tn = ct ,

n→∞

lim hn = s,

n→∞

where cb is the circumference of the bottom circle, ct is the circumference of the top circle, and s is the slant height of the collar as shown in the above figure. We conclude that the area of the collar is (5.4.1)

cb + ct · s = π · (rb + rt ) · s 2 = 2π · ra · s,

where rb and rt are the radii of the respective circles and ra is the average of the two. Theorem 11 (SG). The surface area of the sphere of radius R is the same as the surface area of the label of the smallest can into which the sphere will fit.

5.4. Surface area

45

Namely, the surface area of the sphere of radius R is 2πR · 2R = 4πR2 . Proof (SG). Slice the picture above into n horizontal slices. Approximate the piece of the surface of the sphere between the ith pair of successive slices by a collar Ci . Let a (Ci ) denote the area of Ci , let ri denote its average radius, and let si denote its slant height. Then the surface area of the sphere is approximately n  2π · ri · si , (5.4.2) i=1

n at least if the slices are pretty thin. Also the approximate area i=1 (Ci ) approaches the exact surface area of the sphere as the slices get thinner and thinner. Next let hi denote the vertical height of the label on the can between the ith pair of successive slices. The area of the label is exactly n  2π · hi . i=1

The relationship between each ri , si , and hi is given by the picture below.

By Exercise 44 the two triangles in the above diagram are similar, so corresponding sides are proportional. That is, hi si

= rRi ri · si = hi · R. So by (5.4.2) Approx. area of sphere = 2π ·

n  i=1

ri · si = 2π ·

n  i=1

hi · R = Area of label.

n As the horizontal n slices are made thinner and thinner, 2π· i=1 hi ·R doesn’t change, whereas 2π · i=1 ri · si converges to the exact surface area of the sphere. 

46

5. Surface area and volume of the R-sphere in Euclidean three-space

5.5. Areas on spheres in Euclidean three-space 5.5.1. Lunes. In the picture we have shaded in an “α-lune” on the R-sphere in Euclidean three-space.

The lune has two vertices. They are at opposite (antipodal) points on the R-sphere; that is, the line in Euclidean three-space that joins the two vertices runs through the center of the sphere. The angle at a vertex of the lune is α radians. Exercise 72 (SG). Explain why the area of the α-lune is 2α · R2 . 5.5.2. Spherical triangles and n-gons. If a triangle on the sphere of radius R has interior angles with radian measures α, β, and γ, it can be covered three times by lunes as shown in the figure below.

Notice that one lune has vertex at the vertex A of the spherical triangle ABC and angle equal to the interior angle α of the triangle at that vertex. The second lune has vertex at the vertex B of the spherical triangle ABC and angle equal to the interior angle β of the triangle at that vertex. The third lune has vertex at the vertex C of the spherical triangle ABC and angle equal to the interior angle γ of the triangle at that vertex. Each lune has an “opposite” lune of the same angle and therefore of the same area. In more formal language, we put the center of the sphere at the origin in Euclidean three-dimensional space with Euclidean coordinates (ˆ x, yˆ, zˆ). Then the opposite lune is just the image of the given one under the reflection ⎛ ⎞ −1 0 0   x, yˆ, zˆ) · ⎝ 0 −1 0 ⎠ x ˆ, yˆ, zˆ = (ˆ 0 0 −1

5.5. Areas on spheres in Euclidean three-space

47

through the center of the sphere. The other vertex of each lune is a vertex of the “opposite” spherical triangle A B  C  . The opposite spherical triangle has the same area as the given one, since the three lunes that cover it have the same areas as the corresponding lunes covering the original spherical triangle. Theorem 12 (SG). If the sphere shown above has radius R, the area of the spherical triangle is given by the formula R2 ((α + β + γ) − π) , that is, |K|

−1

((α + β + γ) − π) .

Proof. The three lunes cover the triangle three times. The three opposite lunes cover the opposite triangle three times. Each lune has the same area as its opposite lune, and spherical triangle ABC has the same area as its opposite spherical triangle A B  C  . If you take all six lunes together, they cover each of the two triangles three times and everything else exactly once. So   2· 2πR2 α+2πR2 β +2πR2 γ = 4πR2 +2 · (Area (ABC))+2 · (Area (A B  C  )) (Area(ABC))+(Area(A B  C  )) R2 (πα + πβ + πγ) = πR2 + 2 (πα + πβ + πγ) = πR2 + (Area (ABC)) Area (ABC) = R2 ((α + β + γ) − π) .  Exercise 73. Give a formula for the area of any spherical n-gon. Hint: Divide the spherical n-gon into spherical triangles.

Part IV

Usual dot-product for three-dimensional Euclidean space

Chapter 6

Euclidean three-space as a metric space

6.1. Points and vectors in Euclidean three-space NB: We have arrived at the point in this book at which we will pick up the pace a bit. We will take our story to a more abstract level and will consequently employ more of the mathematics you have learned in previous courses. The pay-off will be quite substantial. As we move higher up the hill, the view will become more and more spectacular, and the view from the top of the hill will be cohesive and complete. Namely, from that viewpoint we will see that all two-dimensional geometries “play by the same rules” and fit into a single coherent whole, with each geometry assigned its place by a single real number called its curvature. Every two-dimensional geometry has its “curvature number”, and every real number has its unique two-dimensional geometry. So from this point on, we embark on a coherent study of all two-dimensional geometries (spheres, the plane, and hyperbolic spaces). Each one of these geometries looks the same at each of its points and also looks the same in every direction emanating from any of its points. But to study them all at the same time and in a uniform way we will need to visualize them all as different surfaces lying in some common three-dimensional space. We begin with ordinary three-dimensional Euclidean space x, yˆ, zˆ) : x ˆ, yˆ, zˆ ∈ R} , R3 = {(ˆ where there is a standard way to measure distance between two points ˆ 1 = (ˆ X x1 , yˆ1 , zˆ1 ) , ˆ X2 = (ˆ x2 , yˆ2 , zˆ2 ) , 51

52

namely (6.1.1)

6. Euclidean three-space as a metric space



2 2 2 ˆ1, X ˆ 2 = (ˆ d X x2 − x ˆ1 ) + (ˆ y2 − yˆ1 ) + (ˆ z2 − zˆ1 ) .

As we will see, formula (6.1.1) is compatible with “infinitesimal distances” on such objects as spheres   (ˆ x, yˆ, zˆ) ∈ R3 : x ˆ2 + yˆ2 + zˆ2 = R2 of a fixed radius R, since spheres can be faithfully represented as smooth surfaces in ordinary Euclidean three-space, and, from the perspective of an extremely tiny observer on any smooth surface, the surface “looks flat”. Said otherwise, as the size of the observer contracts to a point, the surface looks more and more like a plane to that observer. However, there is one disconcerting fact about studying the geometry of spheres in this way. Namely, as the radius R of the sphere approaches infinity, the geometry of the R-sphere at any point does look more and more like plane geometry, but on the other hand, that “limit” of plane geometry is “out at infinity”. So in order to study all the two-dimensional geometries, including plane geometry and the hyperbolic geometries, in a uniform way we will have to change the coordinate system we use, or, what will turn out to be the same thing, we will have to change the distance formula slightly for each geometry. We will do that in later sections, but we first want to review some of the basic properties of the ordinary Euclidean three-space that you learned about in calculus. We write (ˆ x, yˆ, zˆ) for our ordinary Euclidean coordinates. We will want to reserve the notation (x, y, z) for some new coordinates that we will eventually put on the “same” objects. These new coordinates will be chosen to keep the north and south poles from going to infinity as the radius R of a sphere increases without bound. This change of viewpoint will eventually let us go non-Euclidean or, in the language of Buzz Lightyear, let us go “to infinity and beyond”. The idea will be similar to the change from rectangular to polar coordinates for the plane that you encountered in calculus, only easier. When you see (ˆ x, yˆ, zˆ) in what follows, that means that distance between points is measured by formula (6.1.1). One more thing—in Euclidean three-space it will again be important thoughout to make the distinction between points and vectors. Although each will be represented by a triple of real numbers we will use ˆ = (ˆ X x, yˆ, zˆ) to denote points, that is, position in Euclidean three-space, and

Vˆ = a ˆ, ˆb, cˆ to denote vectors, that is, displacement, by which we mean the amount and direction by which a given point is being moved. So vectors always indicate motion from an explicit (or implicitly understood) point of reference. In matrix notation we can think of ˆ 1 = (ˆ ˆ2 − X x2 − x ˆ1 , yˆ2 − yˆ1 , zˆ2 − zˆ1 ) V =X

6.1. Points and vectors in Euclidean three-space

53



ˆ2 − X ˆ 1 . Then we can write the formula for the distance X ˆ 2 in Euclidean three-space in terms of the matrix ˆ 1 and X between two points X product





t ˆ2 − X ˆ2 − X ˆ ˆ ˆ1 · X ˆ1 X d X1 , X2 = .

as a 1 × 3 matrix

So we can define the length of any vector as just the Euclidean distance between the vector’s initial point and its endpoint. There are various operations we can perform on one or more vectors when we think of them as emanating from the same point in Euclidean three-space. The first is the dot-product of two vectors. Definition 9. The dot-product of two vectors

ˆ1 , ˆb1 , cˆ1 , Vˆ1 = a

Vˆ2 = a ˆ2 , ˆb2 , cˆ2 emanating from the same point in three-dimensional Euclidean space is defined as the real number given by the formula a ˆ1 a ˆ2 + ˆb1ˆb2 + cˆ1 cˆ2 or in matrix notation as 

a ˆ1

ˆb1

cˆ1



⎛

⎞ a ˆ2 · ⎝ ˆb2 ⎠ . cˆ2

The dot-product is denoted as Vˆ1 • Vˆ2 or in matrix notation as

t Vˆ1 · Vˆ2 .

So we can define distance in terms of the dot-product as





ˆ1, X ˆ2 − X ˆ2 − X ˆ2 = ˆ1 • X ˆ1 . X (6.1.2) d X

Exercise 74. Give the formula for the length Vˆ of a vector Vˆ = a ˆ, ˆb, cˆ in three-dimensional Euclidean space in terms of the dot-product.



ˆ1 , ˆb1 , cˆ1 and Vˆ2 = a ˆ2 , ˆb2 , cˆ2 Lemma 2. The angle ϑ between two vectors Vˆ1 = a emanating from the same point in Euclidean three-space satisfies the relation (6.1.3) Vˆ1 • Vˆ2 = Vˆ1 · Vˆ2 · cos ϑ [DS, 30ff].

54

6. Euclidean three-space as a metric space

Proof. Multiplying out using the definition and algebraic properties of the dotproduct, 2



ˆ V2 − Vˆ1 = Vˆ2 − Vˆ1 • Vˆ2 − Vˆ1 2 2

= Vˆ1 + Vˆ2 − 2 Vˆ1 • Vˆ2 .



Now consider the triangle with vertices (0, 0, 0), a ˆ1 , ˆb1 , cˆ1 , and a ˆ2 , ˆb2 , cˆ2 . Then Theorem 5 implies that 2



ˆ V2 − Vˆ1 = Vˆ2 − Vˆ1 • Vˆ2 − Vˆ1 2 2 = Vˆ1 + Vˆ2 − 2 Vˆ1 · Vˆ2 · cos ϑ. Comparing this equation with the previous one we obtain that Vˆ1 • Vˆ2 = Vˆ1 · Vˆ2 · cos ϑ.  The significance of Lemma 2 is that the measure of angles between vectors depends only on the definition of the dot-product. Exercise 75. Lemma 2 does not distinguish between the two angles that the vectors determine in the plane that they span. What can you say about the cosine of the two angles formed by two vectors Vˆ1 and Vˆ2 ? That is, what is the relation between the formula using ϑ and the formula using (360° − ϑ)?

ˆ1 , ˆb1 , cˆ1 Corollary 2. The formula for the angle ϑ between two vectors Vˆ1 = a

and Vˆ2 = a ˆ2 , ˆb2 , cˆ2 in three-dimensional Euclidean space depends only on the dot-products of the two vectors with themselves and with each other. Namely, ⎛ ⎞ ˆ1 • Vˆ2 V ϑ = arccos ⎝ ⎠ . Vˆ1 · Vˆ2 In fact it is also true that the formula for the area of the parallelogram determined by two vectors Vˆ1 and Vˆ2 depends only on the dot-products of the two vectors with themselves and with each other. You will see this by answering the following exercises. Exercise 76. Show that the area of the parallelogram determined by Vˆ1 and Vˆ2 emanating from the same point in Euclidean three-space is given by ˆ ˆ (6.1.4) V1 · V2 · sin ϑ. Lemma 3. The area of the parallelogram determined by Vˆ1 and Vˆ2 emanating from the same point in Euclidean three-space is also given by  Vˆ1 • Vˆ1 Vˆ2 • Vˆ1 Vˆ1 • Vˆ2 Vˆ2 • Vˆ2 .

6.1. Points and vectors in Euclidean three-space

55

Proof. From Exercise 76 the square of the area of the parallelogram is

Vˆ1 • Vˆ1 · Vˆ2 • Vˆ2 · sin2 ϑ.   Substitute 1 − cos2 ϑ for sin2 ϑ to obtain







 Vˆ1 • Vˆ1 Vˆ2 • Vˆ2 − cos2 ϑ Vˆ1 • Vˆ1 Vˆ2 • Vˆ2 .

2 Now use (6.1.3) to rewrite the second term as Vˆ1 • Vˆ2 .



Exercise 77. Show that we have the following equality of matrices:  ⎛ ⎞  t t

Vˆ1 Vˆ1 • Vˆ1 Vˆ2 • Vˆ1 . =⎝ ⎠· Vˆ2 Vˆ1 Vˆ1 • Vˆ2 Vˆ2 • Vˆ2 Vˆ2 Again, the significance of Lemma 3 is that to compute areas, we only need to know how to compute dot-products—the definition of the dot-product of the vectors completely determines the calculation of the area of the parallelogram they generate. We finish this section with one other related fact.



Lemma 4. The area of the parallelogram determined by two vectors Vˆ1 = a ˆ1 , ˆb1 , cˆ1

and Vˆ2 = a ˆ2 , ˆb2 , cˆ2 emanating from the same point in Euclidean three-space is given by the length of the cross-product   ˆb cˆ cˆ a ˆ1 ˆb1 ˆ1 a , Vˆ1 × Vˆ2 = ˆ1 1 , 1 . cˆ2 a ˆ2 a ˆ2 ˆb2 b2 cˆ2 Proof. We use some facts from linear algebra. First of all, develop the determinant

Vˆ1

Vˆ2

ˆ V1 × Vˆ2



of the 3 × 3 matrix whose rows are Vˆ1 , Vˆ2 , and Vˆ1 × Vˆ2 along the third row. Then writing out both sides of the equation,

Vˆ1 2

ˆ ˆ V2 (6.1.5) = V1 × Vˆ2 ,

ˆ V1 × Vˆ2 we conclude that they are equal. On the other hand, Vˆ 1 ˆ V2 = 0, ˆ V1

56

6. Euclidean three-space as a metric space

and developing the left-hand determinant along the third row, we conclude that Vˆ1 is perpendicular to Vˆ1 × Vˆ2 . Similarly Vˆ2 is perpendicular to Vˆ1 × Vˆ2 . Finally, the absolute value of the determinant of a 3 × 3 matrix

Vˆ1

Vˆ2

ˆ V1 × Vˆ2 is the volume of the parallelepiped determined by the row vectors of the matrix. But that volume is

ˆ ˆ V1 · V2 · sin ϑ · Vˆ1 × Vˆ2 ,

since Vˆ1 · Vˆ2 · sin ϑ is the area of the base of the parallelepiped and Vˆ1 × Vˆ2 is perpendicular to both V1 and V2 . So using (6.1.5)

Vˆ1 2



ˆ ˆ ˆ ˆ Vˆ2 = V1 × Vˆ2 . V1 · V2 · sin ϑ · V1 × Vˆ2 =

ˆ V1 × Vˆ2 So



ˆ ˆ V1 · V2 · sin ϑ = Vˆ1 × Vˆ2 

[DS, 42–47].

6.2. Curves in Euclidean three-space and vectors tangent to them Definition 10. A smooth curve in three-dimensional Euclidean space is given by a differentiable mapping ˆ : [b, e] → R3 , X t → (ˆ x (t) , yˆ (t) , zˆ (t)) from an interval [b, e] on the real line. We shall sometimes use the notation ˆ (t) . (ˆ x (t) , yˆ (t) , zˆ (t)) = X ˆ (t) must have the additional property that the tangent vector The mapping X 

 dˆ ˆ x dˆ y dˆ z dX a ˆ (t) , ˆb (t) , cˆ (t) = , , = dt dt dt dt is not the zero vector for any t in [b, e]. Just as displacement is sometimes measured by showing how a given point is displaced, as in ˆ2 − X ˆ 1 = (ˆ Vˆ = X x2 − x ˆ1 , yˆ2 − yˆ1 , zˆ2 − zˆ1 ) , rate of displacement can be expressed as the instantaneous velocity of a point moving along a curve, as in  ˆ (t)  dˆ x (t) dˆ y (t) dˆ z (t) dX = , , Vˆ = dt dt dt dt [DS, 30ff].

6.2. Curves in Euclidean three-space and vectors tangent to them

57

Exercise 78. a) Give two examples of smooth curves, ˆ 1 (s) = (ˆ X x1 (s) , yˆ1 (s) , zˆ1 (s)) , ˆ 2 (t) = (ˆ X x2 (t) , yˆ2 (t) , zˆ2 (t)) , neither of which is a straight line, in three-dimensional Euclidean space. Do this so that the two curves pass through a common point and go in distinct tangent directions at that point. Please choose curves so that none of the coordinate functions of s or t are constant functions [DS, 71ff]. b) Compute the tangent vectors of each of the two curves at each of their points. c) For the two curves you defined in a), what are the coordinates of the point in Euclidean three-space at which the two curves intersect? d) Use the dot-product formula to compute the angle ϑ between (the tangent vectors to) your two example curves in a) at the point at which the curves intersect [DS, 20–21]. 6.2.1. Length of a smooth curve in Euclidean three-space. Exercise 79. Compute the length of the tangent vector  ⎛     ˆ ˆ  dˆx dˆy dˆz  ⎜ dX dX • = l(t) =  dt dt dt · ⎝ dt dt

dˆ x dt dˆ y dt dˆ z dt

⎞ ⎟ ⎠

to each of your two example curves in Exercise 78 at each of their points. ˆ (t), t ∈ [b, e], in Euclidean three-space Definition 11. The length L of a curve X is obtained by integrating the length of the tangent vector to the curve, that is,  e l (t) dt. L= b

Said otherwise, for any curve ˆ (t) = (ˆ X x (t) , yˆ (t) , zˆ (t)) , b ≤ t ≤ e, its length is

       t  e  e  ˆ ˆ d X d X   · dt =  dt dt b b

 where

ˆ dX dt

⎛ dˆ x dt

dˆ y dt

dˆ z dt

 ⎜ ·⎝

dˆ x dt dˆ y dt dˆ z dt

⎞ ⎟ ⎠dt,

t

ˆ ˆ X X · ddt is the length of the tangent vector ddt (t) to the curve at

ˆ (t) [DS, 82]. the point X Notice that the length of any curve only depends on the definition of the dotproduct. That is, if we know the formula for the dot-product of the tangent vectors with themselves, we know (the formula for) the length of any curve.

58

6. Euclidean three-space as a metric space

Our first example is the path (6.2.1)

(ˆ x (t) , yˆ (t) , zˆ (t)) = (R · sin t, 0, R · cos t) , 0 ≤ t ≤ π.

Notice that this path lies on the sphere of radius R. Exercise 80. Write the formula for the tangent vector to the path (6.2.1) at each point using (ˆ x (t) , yˆ (t) , zˆ (t))-coordinates. Show that the length of this path is Rπ. Exercise 81. Compute the length of each of your two example curves in Exercise 78. Warning: In this last exercise, you may be confronted with an integral that ˆ 1 (t) happens to describe an you cannot compute. For example, if your curve X ellipse that is not circular, it was proved in the nineteenth century that no formula involving only the standard functions from calculus will give you the length of your path from a fixed beginning point to a variable ending point on the ellipse. If that kind of thing occurs, go back and change the definitions of your curves in Exercise 78 until you get two curves for which you can exactly compute the length of your path from a fixed beginning point to a fixed ending point [DS, 81–82]. Suppose now that the curve is moved by a linear transformation   ˆ x, yˆ, zˆ) · M x ˆ, yˆ, zˆ = (ˆ ˆ such that given by a matrix M ˆ ·M ˆ t = I, M where I denotes the identity matrix. ˆ (t) in Euclidean coordinates is Recalling that the length of a curve X  b

e

   t   dX ˆ ˆ dX  · dt, dt dt

and that the transformation takes the tangent vector the length of the moved curve is

ˆ dX dt

to



⎞t

⎞ ⎛ ⎛ ˆ ˆ ˆ ˆ e d X · M d X · M  ⎠·⎝ ⎠ dt. ⎝ dt dt b



ˆ dX dt

ˆ , we see that ·M

6.3. Surfaces in Euclidean three-space and vectors tangent to them

59

But

 

⎞t

⎞ ⎛ ⎛   t  e d X  e ˆ ·M ˆ ˆ ·M ˆ d X  dX ˆ ˆ  d X  ˆ · ˆ dt ⎝ ⎠·⎝ ⎠ dt = ·M ·M dt dt dt dt b b      e  dX ˆt ˆ dX  t ˆ ˆ ·M · M · = dt dt dt b    t  e  dX ˆ ˆ dX  = dt. · dt dt b

ˆ The same argument shows that for any two vectors Vˆ1 and Vˆ2 and any matrix M such that ˆ ·M ˆ t = I, M



ˆ • Vˆ1 · M ˆ . Vˆ1 • Vˆ2 = Vˆ1 · M Corollary 3. If a curve ˆ (t) = (ˆ X x (t) , yˆ (t) , zˆ (t)) , b ≤ t ≤ e, ˆ , its length is moved by a (linear) transformation given by an orthogonal matrix M is unchanged.

6.3. Surfaces in Euclidean three-space and vectors tangent to them Definition 12. A smooth surface in three-dimensional Euclidean space is given by a differentiable mapping ˆ : U → R3 , X t → (ˆ x (s, t) , yˆ (s, t) , zˆ (s, t)) ˆ (s, t) must have the additional property from a region U ⊆ R2 . The mapping X that the tangent vectors   ˆ dˆ x dˆ y dˆ z dX , , = ds ds ds ds and   ˆ dˆ x dˆ y dˆ z dX , , = dt dt dt dt are linearly independent at each point; that is, the area a (s, t) of the parallelogram that they span is never zero. 6.3.1. Area of a smooth surface in Euclidean three-space. ˆ (s, t), (s, t) ∈ U ⊆ R2 , in Euclidean Definition 13. The area A of the surface X three-space is obtained by integrating the area of the parallelogram spanned by the two tangent vectors to the surface at each of its points; that is,  a (s, t) dsdt A= U

60

6. Euclidean three-space as a metric space

[DS, 82]. Notice that by Exercise 3 the area of any surface only depends on the definition of the dot-product. That is, if we know the formula for the dot-product, we know (the formula for) the area of any surface. Furthermore, if the surface is moved by an orthogonal transformation, since dot-products are unchanged, we have that areas a (s, t) of the parallelogram spanned by the two tangent vectors to the surface at each of its points are unchanged. So the area of the surface is unchanged. Exercise 82. Use spherical coordinates x ˆ (s, t) = R · sin s · cos t, yˆ (s, t) = R · sin s · sin t, zˆ (s, t) = R · cos s and the definition just above to compute the area of an α-lune on the sphere of radius R.

Chapter 7

Transformations

7.1. Rigid motions of Euclidean three-space Consider the following mapping of Euclidean three-space to itself:     ˆ, x ˆ yˆ zˆ = x ˆ yˆ zˆ · M (7.1.1) ˆ is an invertible 3 × 3 matrix. Then by matrix multiplication where M     ˆ −1 ˆ yˆ zˆ · M x ˆ yˆ zˆ = x so that this mapping is one-to-one and onto. Definition 14. The mapping of Euclidean three-space to itself given by the rule (7.1.1) is called a rigid motion if the distance between any two points in Euclidean ˆ 1 and three-space is left unchanged by the mapping; that is, for any two points X ˆ X2 in Euclidean three-space,





ˆ1 · M ˆ, X ˆ2 · M ˆ =d X ˆ1, X ˆ2 . d X ˆ 1 and X ˆ2 We saw in formula (6.1.2) that the square of the distance between X is just the dot-product of the vector ˆ2 − X ˆ1 Vˆ = X ˆ will leave distances with itself. So the transformation given by the matrix M unchanged if and only if, for all vectors Vˆ ,



ˆ • Vˆ · M ˆ = Vˆ • Vˆ , Vˆ · M that is,





t t ˆ · Vˆ · M ˆ = Vˆ · Vˆ . Vˆ · M

We can rewrite this requirement as

t t ˆ ·M ˆ t · Vˆ = Vˆ · Vˆ Vˆ · M (7.1.2) 61

62

7. Transformations

for all vectors Vˆ . So condition (7.1.2) is certainly ⎛ 1 0 ˆ ·M ˆt = I = ⎝ 0 1 (7.1.3) M 0 0

satisfied for all vectors Vˆ if ⎞ 0 0 ⎠. 1

Lemma 5. A linear transformation   ˆ x ˆ, yˆ, zˆ = (ˆ x, yˆ, zˆ) ·M preserves the lengths of all vectors if and only if it also preserves the dot-product between any two vectors. Proof. Since the length of a vector is given by its dot-product with itself, any transformation that preserves the dot-product preserves the lengths of all vectors. In the other direction, let Vˆ1 and Vˆ2 be any two vectors, and let ˆ = Vˆ2 − Vˆ1 . W Then by hypothesis



ˆ ·M ˆ ˆ ·M ˆ • W ˆ •W ˆ = W W







ˆ • Vˆ2 − Vˆ1 · M ˆ Vˆ2 − Vˆ1 • Vˆ2 − Vˆ1 = Vˆ2 − Vˆ1 · M

ˆ ˆ ˆ ˆ ˆ ˆ



V2 • V 2 − 2V2 • V1 + V1 • V1



ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ + Vˆ1 · M ˆ • Vˆ1 · M ˆ = V2 · M • V2 · M − 2 V2 · M • V1 · M



ˆ • Vˆ1 · M ˆ . −2Vˆ2 • Vˆ1 = −2 Vˆ2 · M Therefore





ˆ • Vˆ1 · M ˆ . Vˆ2 • Vˆ1 = Vˆ2 · M 

Corollary 4. i) If a linear transformation   ˆ x ˆ, yˆ, zˆ = (ˆ x, yˆ, zˆ) · M preserves dot-products, then it preserves angles between vectors. ii) Since every linear transformation that preserves lengths of vectors preserves dot-products, every linear transformation that preserves lengths preserves areas of parallelograms. Proof. i) This follows immediately from Lemma 2. ii) This follows immediately from Lemma 3. ˆ is such that Exercise 83. Suppose M ⎛

⎞ 1 2 0 ˆ ·M ˆt = ⎝ 2 1 0 ⎠. M 0 0 1



Find a vector Vˆ = a ˆ, ˆb, cˆ such that

ˆ ·M ˆ t · Vˆ t = Vˆ · Vˆ t . Vˆ · M



7.2. Orthogonal matrices

63

7.2. Orthogonal matrices Lemma 6. A linear transformation ˆ Vˆ → Vˆ · M preserves the lengths of all vectors if and only if ˆ ·M ˆ t = I. M Proof. We have seen in Lemma 5 that if the linear transformation ˆ Vˆ → Vˆ · M ˆ preserves dot-products; that is, for all preserves the lengths of all vectors, then M vectors Vˆ1 and Vˆ2 ,



ˆ • Vˆ2 · M ˆ . Vˆ1 • Vˆ2 = Vˆ1 · M ˆ preserves dot-products, then all In the other direction, we will show that if M t ˆ ˆ the off-diagonal entries in M · M equal zero and all the diagonal entries equal one. Let   ˆ = (mij ) , M ˆ t = mtij = mji M for 1 ≤ i, j ≤ 3. Let Vˆ1 be the vector whose entries are all zero except for the ith entry which is one, and let Vˆ2 be the vector whose entries are all zero except for the jth entry which is one. Then Vˆ1 • Vˆ2 is one if i = j and zero otherwise. But



ˆ • Vˆ2 · M ˆ Vˆ1 • Vˆ2 = Vˆ1 · M and





 ˆ • Vˆ2 · M ˆ = mi1 Vˆ1 · M =

=

 

mi1

mi1

mi2

mi2

mi2 mi3

mi3

  mi3 • ⎛ mj1  · ⎝ mj2 ⎛ mtj3 m1j  · ⎝ mt2j mt3j

mj1 ⎞

mj2

mj3



⎠ ⎞ ⎠,

ˆ ·M ˆ t . So and this last expression is the (i, j)th entry in the matrix product M t ˆ ·M ˆ = I. M  ˆ satisfying Definition 15. A matrix M ˆ ·M ˆt = I M is called an orthogonal matrix [DS, 316–321]. Exercise 84. Show that the matrix ⎛ cos ϑ sin ϑ ˆ = ⎝ − sin ϑ cos ϑ M 0 0

⎞ 0 0 ⎠ 1

is orthogonal. Can you describe geometrically what this rigid motion is doing to the points in Euclidean three-space?

64

Exercise 85. Show that the matrix ⎛ cos ψ ˆ =⎝ 0 M − sin ψ

7. Transformations

⎞ 0 sin ψ 1 0 ⎠ 0 cos ψ

is orthogonal. Can you describe geometrically what this rigid motion is doing to the points in Euclidean three-space? ˆ forms a group. Exercise 86 (SG). Show that the set of orthogonal matrices M That is, show that a) the product of two orthogonal matrices is orthogonal, b) matrix multiplication (when defined) is associative (recall from linear algebra), c) the identity matrix is orthogonal, ˆ −1 of an orthogonal matrix M ˆ is orthogonal. d) the inverse matrix M Hint: Write

ˆ ·M ˆ −1 = I = M ˆ ·M ˆt M and use matrix multiplication to reduce to showing that the transpose of an orthogonal matrix is orthogonal [MJG, 311]. To sum up, the major theme of Part IV is that just by knowing the dot-product of vectors, we can compute all the major geometric quantities: length, area, volume, and angle. In other words, knowing the formula for the dot-product completely determines the properties of the geometry. Or said another way, we can stretch or bend or dent an object in space and thereby change its geometric properties, or we can accomplish the same thing by leaving the object fixed and changing the definition of the dot-product of two vectors emanating from the various points on the object.

7.3. Linear fractional transformations We end this chapter with the study of transformations of the (extended) real line whose graphs are not straight lines but only “fractionally linear”. Not surprisingly they will be called linear fractional transformations. They will help us answer the question of when four points on one line in the plane can be the shadows of four points on another line in the plane if we place the sun at a point on the plane that lies on neither of the two lines (or we place the sun at infinity). Perhaps not surprisingly this notion is very closely related to the notion of perspective in painting—that is, how do you faithfully render depth on a flat canvas? It will turn out that the first answer to the above question is that one ordered set of four collinear points can be the shadow of another ordered set of four collinear points if and only if the two sets have the same signed cross-ratio! Earlier, when we defined the cross-ratio of four distinct points lying on a circle or on a line, we defined it in terms of the distances between pairs of the four points, and therefore it was always a positive real number. But to make the answer to the above shadow question precisely correct, we will have to refine our definition of the cross-ratio of an ordered set of four points on a line so that the cross-ratio carries

7.3. Linear fractional transformations

65

a plus or minus sign—the old definition will just be the absolute value of the new one. It will also turn out that the second answer to the above question is that one ordered set of four collinear points can be the shadow of another ordered set of four collinear points if and only there is a linear fractional transformation that takes the first set of four points in order to the second set of four points. But first we need to understand what a linear fractional transformation is. Start with the set of points on the real number line with coordinate t and add one extra point called t = ∞. Call the resulting set R. You can think of the set R in the following way. Identify t with the unique line through the origin in the (x, y)-plane having slope t. Then the y-axis is identified with t = ∞. Thus the set R is in one-to-one correspondence with the set of all lines through the origin in the (x, y)-plane. Exercise 87. a) Show that the transformation    d (x, y) → x, y = (x, y) · c

b a



is a one-to-one, onto (linear) transformation of R2 as long as d b (7.3.1) c a = 0. b) For the transformation in a), show that every line through the origin in  (x, y)-space is sent to a line through the origin in x, y -space. The slope t of the line through (0, 0) and (x, y) is of course t = xy . What is the slope t of the line through (0, 0) and x, y ? Show that (7.3.2)

t=

at + b . ct + d

Definition 16. Functions (7.3.2) for which the condition (7.3.1) holds are called linear fractional transformations. Exercise 88. Show that a linear fractional transformation R → R, at + b ct + d is one-to-one and onto. What is its inverse function? (Your answer should show that the inverse function is also a linear fractional transformation.) Hint: Algebraically solve for t in terms of t. Then graph at + b t= ct + d in the (t, t)-plane. If c = 0, show that the graph is a straight line with nonzero slope and ∞ → ∞. If c = 0, show that the graph has exactly one horizontal asymptote where t → ∞ and one vertical asymptote where t → ∞. t → t =

66

7. Transformations

Exercise 89. Show that the set of linear fractional transformations forms a group under the operation of composition of functions: a) Show that the composition of two linear fractional transformations is again a linear fractional transformation. b) Since compositions of functions are always associative, compositions of linear fractional transformations are automatically associative. c) Show that the identity map from R to R is a linear fractional transformation. d) Show that the inverse function of a linear fractional transformation is a linear fractional transformation. That is, show that for any linear fractional transformation f , there is an (inverse) linear fractional transformation g such that f ◦ g = identity transformation, g ◦ f = identity transformation. Exercise 90. a) Show that for any three distinct fixed points t2 , t3 , and t4 , the function of t given by the formula t3 − t4 t − t2 t − t2 t − t4 t= = ÷ t3 − t2 t − t4 t3 − t2 t3 − t4 is a linear fractional transformation. That is, show that it is a function of the form (7.3.2) for which the condition (7.3.1) holds. b) Show that the linear fractional transformation in a) takes t2 to zero, takes t3 to one, and takes t4 to ∞. Exercise 91. Show that any linear fractional transformation (7.3.2) that leaves zero, one, and ∞ fixed is the identity map. A corollary of these last two exercises follows. Corollary 5. If f (t) and g (t) are linear fractional transformations and f (ti ) = g (ti ) ¯ then for all ∈ R, ¯ for three distinct points t2 , t3 , t4 ∈ R, f (t) = g (t) . Proof. The hypothesis implies that g −1 ◦ f leaves t2 , t3 , and t4 fixed. Let h (t) be the linear fractional transformation defined in Exercise 90a). Then   h ◦ g −1 ◦ f ◦ h−1 leaves zero, one, and ∞ fixed. So by Exercise 91, for all t we have    −1  −1 (t) = t  h ◦−1g ◦ f −1◦h h ◦ g  ◦ f ◦ h (h (t)) = h (t) −1   h−1g (f (t))  = h (t) −1 h h g (f (t)) = h−1 (h (t)) = t g −1  (f (t)) =t f (t) = g g −1 (f (t)) = g (t) . 

7.3. Linear fractional transformations

67

Theorem 13. Suppose that you are given a sequence of four distinct points t1 , t2 , t3 , ¯ and another sequence of four distinct points s1 , s2 , s3 , and s4 in R. ¯ Then and t4 in R there is a linear fractional transformation (7.3.3)

f (t) =

at + b ct + d

such that f (ti ) = si for i = 1, 2, 3, 4 if and only if s1 − s2 s1 − s4 t1 − t2 t1 − t4 ÷ = ÷ s3 − s2 s3 − s4 t3 − t2 t3 − t4 [MJG, 288]. Proof. By Exercise 90 the linear fractional transformation t − t4 t − t2 ÷ f (t) = t3 − t2 t3 − t4 takes t2 to zero, takes t3 to one, and takes t4 to ∞. Similarly the inverse of the linear fractional transformation t − s2 t − s4 g (t) = ÷ s3 − s2 s3 − s4 takes zero to s2 , takes one to s3 , and takes ∞ to s4 . So the linear fractional transformation g −1 ◦ f has the property that  −1  g ◦ f (ti ) = si for i = 2, 3, 4. So by the definitions of f and g  −1  g ◦ f (t1 ) = s1 if and only if

s1 − s2 s1 − s4 t1 − t2 t1 − t4 ÷ = ÷ . s3 − s2 s3 − s4 t3 − t2 t3 − t4 On the other hand, if h is any linear fractional transformation that has the property that h (ti ) = si

for i = 2, 3, 4, then g ◦ h ◦ f −1 leaves zero, one, and ∞ fixed and therefore is the identity map by Exercise 91. So f =g◦h h = g −1 ◦ f.  Definition 17. The cross-ratio (t1 : t2 : t3 : t4 ) of four (ordered) points t1 , t2 , t3 , and t4 is defined by t1 − t2 t1 − t4 ÷ . (t1 : t2 : t3 : t4 ) = t3 − t2 t3 − t4

68

7. Transformations

Exercise 92. If for two ordered sets of points {sk } and {tk }, (s1 : s2 : s3 : s4 ) = (t1 : t2 : t3 : t4 ) , explicitly define the linear fractional transformation f such that for k = 1, . . . , 4, f (sk ) = tk . Hint: Begin with

and solve for s.

s − s2 s − s4 t − t2 t − t4 ÷ = ÷ s3 − s2 s3 − s4 t3 − t2 t3 − t4

Part V

K-geometry

Chapter 8

Changing coordinates

The famous nineteenth-century German mathematician Bernhard Riemann first exploited and codified the idea that all geometry depends only on the definition of the dot-product into the full mathematical theory that we today suitably call Riemannian geometry. In this part of the book, we will take our first small step into Riemannian geometry. Namely, given any fixed real number K, we will introduce a K-dot-product on three-dimensional space and study that K-dot-product on vectors tangent to the solution set of the equation   K x2 + y 2 + z 2 = 1. When K > 0, we will see that we are studying spherical geometry for the Euclidean sphere of radius K −1/2 ; when K = 0, we are just studying Euclidean plane geometry; and when K < 0, we are studying the hyperbolic geometries. The advantage of proceeding in this way is that we can derive many of the needed formulas by doing a single computation that works for all K!

8.1. Bringing the North Pole of the R-sphere to (0, 0, 1) We are now ready to introduce a slightly different set of coordinates for R3 , threedimensional Euclidean space. To see why we do this, suppose we are standing at the North Pole N = (0, 0, R) of the sphere (8.1.1)

x ˆ2 + yˆ2 + zˆ2 = R2

of radius R. As R increases (but we stay our same size), the sphere around us becomes more and more like a flat plane surface. However, it can never get completely flat because we are zooming out the positive zˆ-axis, and we would have to be “at infinity” for our surface to become exactly flat. We remedy that unfortunate situation by considering another copy of R3 whose coordinates we denote as (x, y, z), 71

72

8. Changing coordinates

and we make the following rule in order to pass between the two R3 s: (8.1.2)

x ˆ = x, yˆ = y, zˆ = Rz.

We think of the (x, y, z)-coordinates as simply being a different set of addresses for the points in Euclidean three-space. For example, (x, y, z) = (0, 0, 1) tells me that the point in Euclidean three-space that I’m referring to is (ˆ x, yˆ, zˆ) = (0, 0, R) = N, and the sphere of radius R in Euclidean three-space is given by R2 = x ˆ2 + yˆ2 + zˆ2 = x2 + y 2 + R 2 z 2 , that is, by the equation (8.1.3)

1=

 1  2 x + y2 + z2 . R2

The quantity 1 R2 is called the curvature of the R-sphere. So in (x, y, z)-coordinates, as R goes to infinity, K goes to zero. Therefore formula (8.1.3) is rewritten as   (8.1.4) 1 = K x2 + y 2 + z 2 K=

and therefore goes to 1 = z2 as R goes to infinity. So, in (x, y, z)-coordinates, our “R-geometry” does indeed go to something finite and flat as R goes to infinity, namely the set given by the formula z = ±1 which is in fact (two copies of) a plane! Exercise 93. a) Sketch the solution set in (x, y, z)-coordinates representing the sphere ˆ2 + yˆ2 + zˆ2 = 22 R2 = x of radius two in Euclidean three-space. b) Sketch the solution set in (x, y, z)-coordinates representing the sphere R2 = x ˆ2 + yˆ2 + zˆ2 = 102 of radius ten in Euclidean three-space. c) Sketch the solution set in (x, y, z)-coordinates representing the sphere R2 = x ˆ2 + yˆ2 + zˆ2 = 10−2 of radius 10−1 in Euclidean three-space.

8.2. K-geometry: Euclidean lengths and angles in (x, y, z)-coordinates

73

8.2. K-geometry: Euclidean lengths and angles in (x, y, z)-coordinates To prepare ourselves to do hyperbolic geometry, which has no satisfactory model in Euclidean three-space, we will “practice” by doing spherical geometry (which does have a completely satisfactory model in Euclidean three-space) using these “slightly strange” (x, y, z)-coordinates. Throughout the remainder of this course we will gradually discover that the same rules that govern spherical geometry, expressed in (x, y, z)-coordinates, also govern flat and hyperbolic geometry! In all three cases, the space in (x, y, z)-coordinates that we will study is   (8.2.1) 1 = K x2 + y 2 + z 2 . If K > 0, the geometry we will be studying is the geometry of the Euclidean sphere of radius 1 R= √ . K If K = 0, we will be studying flat (plane) geometry considered in R3 as the plane given by z = 1. If K < 0, we will be studying hyperbolic geometry. In all cases, the number K is called the curvature of the geometry. In short, for all K, we want to compute lengths, shortest paths, angles, areas, and volumes in (x, y, z)-coordinates, but, at least for non-negative K, we want the answers to be the same as they were before when we computed them in the usual Euclidean coordinates. That is, we want the usual answers for the sphere of radius x, yˆ, zˆ)-coordinates when K > 0, and the usual plane geometry answers R = √1K in (ˆ when K = 0. Exercise 94. a) Suppose we have functions (ˆ x (x, y, z) , yˆ (x, y, z) , zˆ (x, y, z)) , where x = f (t) , y = g (t) , z = h (t) . State the Chain Rule (Theorem 1) for dˆ x =, dt dˆ y =, dt dˆ z =. dt b) Rewrite the Chain Rule in matrix notation: 

dˆ x dt

dˆ y dt

dˆ z dt



=



dx dt

dy dt

dz dt



  ·

.

74

8. Changing coordinates

Exercise 95. Recalling that R is a positive constant, use (8.1.2) and the Chain ˆ (t) = (ˆ Rule to show that for any path X x (t) , yˆ (t) , zˆ (t)) in Euclidean three-space, dˆ x dx = , dt dt dy dˆ y = , dt dt dz dˆ z =R . dt dt Exercise 96. Use matrix multiplication [DS, 307] and Exercise 95 to show that ⎛ ⎞ 1 0 0 ˆ (t)  dX (t)  dX ⎝ 0 1 0 ⎠, = dt dt 0 0 R ⎞  ⎛ 1 0 0 ˆ (t) dX dX (t) ⎝ 0 1 0 ⎠. = dt dt 0 0 R−1 NB: This last computation shows that if

Vˆ1 = a ˆ1 , ˆb1 , cˆ1 ,

Vˆ2 = a ˆ2 , ˆb2 , cˆ2 are tangent vectors in (ˆ x, yˆ, zˆ)-coordinates and V1 = (a1 , b1 , c1 ) , V2 = (a2 , b2 , c2 ) are their transformations into (x, y, z)-coordinates, then ⎛ ⎞ 1 0 0 Vˆ1 = (V1 ) ⎝ 0 1 0 ⎠ , 0 0 R ⎛ ⎞ 1 0 0 Vˆ2 = (V2 ) ⎝ 0 1 0 ⎠ , 0 0 R and

t Vˆ1 • Vˆ2 = Vˆ1 · Vˆ2 ⎛ ⎛ ⎞⎛ 1 0 0 = (V1 ) ⎝ 0 1 0 ⎠ ⎝(V2 ) ⎝ 0 0 R ⎛ ⎞⎛ 1 0 0 1 0 = (V1 ) ⎝ 0 1 0 ⎠ ⎝ 0 1 0 0 R 0 0 ⎞ ⎛ 1 0 0 0 ⎠ (V2 )t . = (V1 ) ⎝ 0 1 0 0 K −1

⎞⎞t 1 0 0 0 1 0 ⎠⎠ 0 0 R ⎞ 0 0 ⎠ (V2 )t R

8.2. K-geometry: Euclidean lengths and angles in (x, y, z)-coordinates

75

This last computation says that we can compute the Euclidean dot Vˆ1 • Vˆ2 without ever referring to Euclidean coordinates. We incorporate that fact into the following definition. Definition 18. We define the “K-dot-product” of vectors in (x, y, z)-coordinates as follows: ⎞ ⎛ 1 0 0 0 ⎠ (V2 )t V1 •K V2 = (V1 ) ⎝ 0 1 (8.2.2) 0 0 K −1 ⎛ ⎞⎛ ⎞ 1 0 0 a2   0 ⎠ ⎝ b2 ⎠ . = a1 b1 c1 ⎝ 0 1 c2 0 0 K −1 So, suppose we have a curve on the R-sphere in Euclidean three-space, but it is given to us in X (t) = (x (t) , y (t) , z (t))-coordinates. Then we can calculate the length of that curve in Euclidean three-space by the formula  e dX dX •K dt. dt dt b Exercise 97. Use Lemma 3 to show that if we have any two vectors in Euclidean three-space that are tangent to the R-sphere at some point on it, but the two vectors are given to us in (x, y, z)-coordinates as V1 = (a1 , b1 , c1 ) , V2 = (a2 , b2 , c2 ) , then the area of the parallelogram spanned by those two vectors in Euclidean threespace is  ⎞ ⎛     1 0 0 V1 •K V1 V2 •K V1  (V1 )    t t =  ⎝ 0 1 0 ⎠ · (V1 ) (V2 ) . V1 •K V2 V2 •K V2 (V2 ) · 0 0 K −1 Moral of the story: The dot-product rules! That is, if you know the dotproduct, you know everything there is to know about a geometry: lengths, areas, angles, everything. The set (8.2.1) continues to make sense even when K is negative. And the definition of the K-dot-product also makes sense for tangent vectors to that set when K is negative. (We will need to check later on that when K is negative, the K-dot-product of a nonzero tangent vector to the set (8.2.1) with itself is still positive. This is no longer obvious.) The geometry we get when the constant K is negative is called a hyperbolic geometry. If K = 0, the geometry is called a non-Euclidean geometry. So all the non-Euclidean two-dimensional geometries are either spherical or hyperbolic. Coming attractions: A big idea is that in hyperbolic geometry K −1 in (8.2.2) becomes negative, so that the third coordinate of the velocity vector (a, b, c), that is, the c-direction, actually makes the velocity smaller than it is for the vector (a, b, 0). That is, if the c-direction is the time direction and the a-direction and b-direction are space directions, then the faster something is going with relation to the fixed observer, the smaller it appears. This idea is central to Einstein’s Theory of Special Relativity.

76

8. Changing coordinates

8.3. Congruences, that is, rigid motions 8.3.1. Formula in (x, y, z)-coordinates for rigid motions of Euclidean threespace. We now wish to figure out how to write the transformation (7.1.1) in (x, y, z)-coordinates. This is a simple substitution problem: 

(8.3.1)

x ˆ



yˆ zˆ

=



x ˆ

yˆ zˆ



ˆ ·M

⎞ 1 0 0 x ˆ yˆ zˆ = x y Rz = x y z · ⎝ 0 1 0 ⎠ 0 0 R ⎛ ⎞ 1 0 0       x ˆ yˆ zˆ = x y Rz = x y z · ⎝ 0 1 0 ⎠ . 0 0 R 











⎛

So we have the diagram ⎛



x ˆ



⎜ ·⎜ ⎝

 yˆ zˆ ∈ R3 ˆ ↓ ·M

x ˆ

yˆ zˆ

⎛ ⎜ ·⎜ ⎝



⎞

1 0 0 0 1 0 0 0 R ←−

⎟ ⎟ ⎠

1 0 0 0 1 0 0 0 R−1 −→

Exercise 98. Starting from the equality Euclidean coordinates, explain why ⎛ 1     x y z = x y z ·⎝ 0 0



x

⎞ ⎟ ⎟ ⎠



 y z ∈ R3 ↓ ·M =?

x

y

z



.

(8.3.1) describing the transformation in ⎞ ⎛ ⎞ 0 0 1 0 0 ˆ ·⎝ 0 1 1 0 ⎠·M 0 ⎠. 0 R 0 0 R−1

So, if we let ⎞ 0 0 1 0 ⎠, 0 R−1

⎛

⎞ ⎛ 1 0 0 1 ˆ ·⎝ 0 M =⎝ 0 1 0 ⎠·M 0 0 R 0 then (8.3.2)



x

y

z



=



x

y

z



· M.

That is, M is the matrix that gives the transformation (7.1.1) in (x, y, z)-coordinates. Now how would we check whether a transformation given in (x, y, z)-coordinates by a matrix M preserves distances in Euclidean three-space? Again, starting from

8.3. Congruences, that is, rigid motions

(7.1.3) this is just a substitution problem: ˆ ·M ˆt = I M ⎛ ⎞ ⎛ 1 0 0 ˆ ·⎝ M =⎝ 0 1 0 ⎠·M 0 0 R ⎛ ⎞ ⎛ 1 0 0 1 ⎝ 0 1 0 ⎠·M ·⎝ 0 0 0 R−1 0

77

⎞ 1 0 0 0 1 0 ⎠ 0 0 R−1 ⎞ 0 0 ˆ. 1 0 ⎠=M 0 R

Exercise 99. Finish the matrix algebra computations just above to show that the condition that a transformation M in (x, y, z)-coordinates preserves distances in Euclidean three-space is the condition that ⎛ ⎞ ⎞ ⎛ 1 0 0 1 0 0 0 ⎠ · Mt = ⎝ 0 1 0 ⎠. (8.3.3) M ·⎝ 0 1 −1 0 0 K 0 0 K −1 This is the condition (in (x, y, z)-coordinates) that implies that the transformation taking the path (x(t), y(t), z(t)) to the path (x(t), y(t), z(t)) · M preserves lengths of tangent vectors at corresponding points. Therefore, by integrating, the (total) length of the curve {(x(t), y(t), z(t)) · M : b ≤ t ≤ e} is the same as the total length of the curve {(x(t), y(t), z(t)) : b ≤ t ≤ e}. Exercise 100. Check that (8.3.3) is the correct condition by showing that any 3 × 3 matrix M that satisfies (8.3.3) also satisfies ((V ) · M ) •K ((V ) · M ) = V •K V, where V = X2 − X1 . That is, the transformation given in (x, y, z)-coordinates by a matrix M that satisfies (8.3.3) preserves the K-dot-product.

Chapter 9

Uniform coordinates for the two-dimensional geometries

9.1. The two-dimensional geometries in (x, y, z)-coordinates In Chapter 8 we changed the coordinates on Euclidean three-space so that the equations for the sphere of radius R became   (9.1.1) 1 = K x2 + y 2 + z 2 , where K=

1 . R2

In these new (x, y, z)-coordinates, which we will call K -coordinates, the set of points (x, y, z) satisfying (9.1.1) when K > 0 matched up in one-to-one fashion with the R-sphere   (ˆ x, yˆ, zˆ) ∈ R3 : x ˆ2 + yˆ2 + zˆ2 = R2 in the usual coordinates (ˆ x, yˆ, zˆ) of three-dimensional Euclidean space. We made this change because working in (x, y, z)-coordinates will let us see what happens to the geometry when K becomes zero or even a negative number. Recall that when K becomes zero, the radius R of the sphere becomes infinite, so we have lost our ability to compute anything. But, as we can see from equation (9.1.1), the set of points defined by that equation still makes sense when K becomes zero or even a negative number. In this chapter we will do K-geometry for any K so that any of the results we obtain will apply no matter whether K is positive or negative. We will apply these results for K positive in Chapter 10 and for K negative in Chapter 11, the final chapter of this book. If we have a curve (ˆ x (t) , yˆ (t) , zˆ (t)) lying in the R-sphere in Euclidean space, then for all t ∈ [b, e], 2 2 2 x ˆ (t) + yˆ (t) + zˆ (t) = R2 .

79

80

9. Uniform coordinates for the two-dimensional geometries

Differentiating both sides with respect to t we obtain 2ˆ x (t)

dˆ y dˆ z dˆ x + 2ˆ y (t) + 2ˆ z (t) = 0, dt dt dt

which we can rewrite as (ˆ x (t) , yˆ (t) , zˆ (t)) •

ˆ (t) dX =0 dt

ˆ (t) if [DS, 105ff]. Said another way, vectors Vˆ are tangent to the R-sphere at X and only if ˆ (t) • Vˆ = 0 X [DS, 106, 109]. Repeating the same calculation in (x, y, z)-coordinates, the corresponding curve (x (t) , y (t) , z (t)) lies in the set (9.1.1) so that

2 2 2 1 = K x (t) + y (t) + z (t)   dy dx dz + 2y (t) 0 = K 2x (t) + 2z (t) . dt dt dt That is, a vector V = (a, b, c) is tangent to the set ⎛ 2K 0 (9.1.2) (x (t) , y (t) , z (t)) · ⎝ 0 2K 0 0

(9.1.1) if and only if ⎞ 0 0 ⎠ · V t = 0. 2

Exercise 101. For K = 0, show that the condition (9.1.2) on V is exactly the same condition as (x (t) , y (t) , z (t)) •K V = 0. We will call the set of (x, y, z) satisfying (9.1.1) K-geometry. Its tangent vectors at a point (x, y, z) in the set are the vectors V = (a, b, c) such that (x, y, z) •K V = 0. If you get nervous using these weird coordinates to compute things that are clearer in (ˆ x, yˆ, zˆ)-coordinates, just go through each construction in Part V in the special case K = 1 first. In that special case (x, y, z) = (ˆ x, yˆ, zˆ) , and your calculations reduce to the usual ones on the unit sphere in ordinary Euclidean three-space. 9.1.1. Computing rigid motions in (x, y, z)-coordinates. We are now going to study K-geometry using only (x, y, z)-coordinates. If we have a curve X (t) = (x (t) , y (t) , z (t)) on the surface given in K-coordinates as   (9.1.3) 1 = K x2 + y 2 + z 2 , we have seen that we should measure its length L by the formula  e l (t) dt, (9.1.4) L= b

9.1. The two-dimensional geometries in (x, y, z)-coordinates

81

where dX dX •K , dt dt and that we should measure angles ϑ between tangent vectors V1 and V2 at a point on the surface by the formula   V1 •K V2 ϑ = arccos , |V1 |K · |V2 |K

(9.1.5)

2

l (t) =

where |V |2K = V •K V. We now want to explore the condition that a transformation   x, y, z = (x, y, z) · M takes the surface (9.1.3) to itself and preserves the length of any curve (x (t) , y (t) , z (t)) lying on the surface. Rewriting the transformation as (X) = (X) · M, formulas (9.1.4) and (9.1.5) show that all we have to worry about is that dX dX dX dX •K = •K dt dt dt dt for all values t of the parameter of the curve. But     dX dX = ·M dt dt by the product rule since M is a constant matrix. So the transformation given by ˆ will preserve the length of any path and will preserve the measure of the matrix M any angle if (9.1.6) ⎞ ⎞ ⎛ ⎛     t   t  1 0 0 1 0 0 dX dX dX dX ⎠ ⎠ ⎝ ⎝ 0 0 · · . ·M · 0 1 ·M = · 0 1 dt dt dt dt 0 0 K −1 0 0 K −1 Exercise 102. a) Show that this last equality is ⎞ ⎛ ⎛ 1 0 0 0 ⎠ · Mt = ⎝ (9.1.7) M ·⎝ 0 1 0 0 K −1

always true if ⎞ 1 0 0 0 1 0 ⎠. 0 0 K −1

  b) Show that if M satisfies the identity (9.1.7), then the transformation x, y, z = (x, y, z) · M takes the set of points (x, y, z), such that   1 = K x2 + y 2 + z 2 ,   to the set of points x, y, z , such that   1 = K x2 + y 2 + z 2 . That is, M gives a one-to-one, onto mapping of K-geometry to itself.

82

9. Uniform coordinates for the two-dimensional geometries

Hint: For K = 0, write the equation ⎛ 1   x y z ·⎝ 0 0

  1 = K x2 + y 2 + z 2 in matrix notation as ⎞ ⎛ ⎞ x 0 0 1 1 0 ⎠·⎝ y ⎠= . K 0 K −1 z

Definition 19. A 3 × 3 matrix M is called K-orthogonal if ⎞ ⎞ ⎛ ⎛ 1 0 0 1 0 0 0 ⎠. 0 ⎠ · Mt = ⎝ 0 1 M ·⎝ 0 1 0 0 K −1 0 0 K −1 Definition 20. A K-distance-preserving transformation of K-geometry is called a K-rigid motion or a K-congruence. So K-orthogonal matrices give K-rigid motions. Exercise 103. For K = 0, show that the set of K-orthogonal matrices M forms a group. That is, show that a) the product of two K-orthogonal matrices is K-orthogonal, b) matrix multiplication is associative, c) the identity matrix is K-orthogonal, d) the inverse matrix M −1 of a K-orthogonal matrix M is K-orthogonal. Hint: Write ⎞ ⎛ ⎛ ⎞ 1 0 0 1 0 0 0 ⎠ · Mt · ⎝ 0 1 0 ⎠ , M · M −1 = I = M · ⎝ 0 1 0 0 K −1 0 0 K and use matrix multiplication to reduce to showing ⎞ ⎛ ⎛ 1 0 0 1 ⎝ 0 1 0 ⎠ · Mt · ⎝ 0 0 0 K −1 0

that

⎞ 0 0 1 0 ⎠ 0 K

is K-orthogonal. 9.1.2. Why use K-coordinates? We saw that we could measure the usual ˆ (t) on the usual Euclidean R-sphere just in terms of Euclidean lengths of curves X the formulas X (t) for their paths in (x, y, z)-coordinates using the K-dot-product, since lengths depended only on lengths of tangent vectors and ˆ (t) ˆ (t) dX dX (t) dX dX (t) • = •K . dt dt dt dt where ⎞ ⎛    t 1 0 0 dX (t) dX (t) dX (t) dX (t) ⎠ ⎝ 0 1 0 •K = · . · dt dt dt dt 0 0 K −1 In other words, the usual geometry of the sphere of radius R is simply the geometry of the set (9.1.1) with K = 1/R2 and with lengths (and areas) given by the Kdot-product. Said another way, we can do all of spherical geometry in (x, y, z)coordinates. All we need is the set (9.1.1) and the K-dot-product. But the set

9.2. Central projection

83

(9.1.1) continues to exist even if K = 0 or K < 0, and the K-dot-product formula continues to make sense even if K < 0. In short we have the following table:

(9.1.8)

Spherical (K > 0) x ˆ2 + yˆ2 + zˆ2 = R2 ˆ ˆ V 2• V 2  1 = K x + y + z2 V1 •K V2

Euclidean (K = 0)

Hyperbolic (K < 0)

  1 = K x2 + y 2 + z 2

  1 = K x2 + y 2 + z 2 V1 •K V2

This table tells us that “there is something else out there”, that is, some other type of two-dimensional geometry beyond plane geometry and spherical geometry. But the gap in the bottom row of the table is a bit disturbing. If we can’t express the usual dot-product in plane geometry as the K-dot-product for K = 0, we can’t pass smoothly from spherical geometry through plane geometry to hyperbolic geometry using (x, y, z)-coordinates. We now examine two ways to produce coordinates uniformly for spherical, plane, and hyperbolic geometry that overcome this difficulty.

9.2. Central projection 9.2.1. Central projection coordinates. Let’s project K-geometry, that is, the set (9.2.1)

  1 = K x2 + y 2 + z 2 ,

onto the set z=1 using the origin O = (0, 0, 0) as the center of projection. If K > 0, the picture is

84

9. Uniform coordinates for the two-dimensional geometries

If K < 0, the picture is

That is, r · (xc , yc , 1) = (x, y, z) .

(9.2.2) So

r=z and, from equation (9.2.1),



2 2 K (rxc ) + (ryc ) + r 2 = 1 r2 =

K

(x2c

1 . + yc2 ) + 1

Notice that when K < 0, this last formula only makes sense when   K x2c + yc2 > −1 (9.2.3) −1 x2c + yc2 < . K Exercise 104. a) For the projection of the set (9.2.1) onto the z = 1 plane with center of projection O, write (xc , yc ) as a function of (x, y, z). b) For the projection of the set (9.2.1) onto the z = 1 plane with center of projection O, write (x, y, z) as a function of (xc , yc ). Exercise 104b) gives us a parametrization of the K-geometry   K x2 + y 2 + z 2 = 1 by a region in the (xc , yc )-plane. Exercise 105. Show that a) when K > 0, the entire (xc , yc )-plane parametrizes the upper hemisphere, b) when K = 0, the (xc , yc )-plane coincides with the plane z = 1, c) when K < 0, the interior of a circle of radius √1 in the (xc , yc )-plane |K|

parametrizes the entire K-geometry.

9.2. Central projection

85

9.2.2. Rigid motion in central projection coordinates. Suppose now we have a K-rigid motion   x, y, z = (x, y, z) · M of K-geometry, given by a K-orthogonal ⎛ m11 M = ⎝ m21 m31

matrix

⎞ m13 m23 ⎠ . m33

m12 m22 m32

To see what this K-rigid motion looks like in simply do the matrix multiplication: ⎛ m11   x, y, z = (x, y, z) · ⎝ m21 m31

central projection coordinates we ⎞ m13 m23 ⎠ m33

m12 m22 m32

= ((m11 x + m21 y + m31 z) , (m12 x + m22 y + m32 z) , (m13 x + m23 y + m33 z)) . Then we substitute into the formulas for central projection coordinates: x (9.2.4) xc = z m11 x + m21 y + m31 z = m13 x + m23 y + m33 z m11 (x/z) + m21 (y/z) + m31 = m13 (x/z) + m23 (y/z) + m33 m11 xc + m21 yc + m31 = m13 xc + m23 yc + m33 and similarly yc = Finally we write

 

=

m12 xc + m22 yc + m32 . m13 xc + m23 yc + m33  xc , yc = Mc (xc , yc )

m11 xc + m21 yc + m31 m12 xc + m22 yc + m32 , m13 xc + m23 yc + m33 m13 xc + m23 yc + m33

 .

9.2.3. Length and angle in central projection coordinates. We next explore lengths and angles in K-geometry in terms of the coordinates (xc , yc ). Exercise 106. For the K-geometry coordinates X (xc , yc ) = (x (xc , yc ) , y (xc , yc ) , z (xc , yc )) , use the formulas you derived in Exercise 104b) to calculate     ∂X ∂X dX = dxc + dyc . ∂xc ∂yc That is, calculate the 2 × 3 matrix  ∂x ∂y Dc =

∂xc ∂x ∂yc

∂xc ∂y ∂yc

∂z ∂xc ∂z ∂yc



⎛ =⎝

∂X ∂xc ∂X ∂yc

⎞

⎠.

86

9. Uniform coordinates for the two-dimensional geometries

Hint: Use logarithmic differentiation: dx = d (rxc ) = xc dr + rdxc r −1 dx = xc dln (r) + dxc ,



dy dz than , , and similarly for y and z since it is easier to compute r −1 dx dt dt dt

dx dy dz dt , dt , dt . Next use that       2dln (r) = dln r 2 = −dln K x2c + yc2 + 1     1 d K x2c + yc2 + 1 =− 2 2 K (xc + yc ) + 1 = −r 2 K (2xc dxc + 2yc dyc ) . Exercise 107. Now suppose we have a path (xc (t) , yc (t)) , a ≤ t ≤ b in the (xc , yc )-plane, that is, in the central projection plane (xc , yc , 1) . Use the formula you derived in Exercise 104b) to write the corresponding path x (xc (t) , yc (t)) , y (xc (t) , yc (t)) , z (xc (t) , yc (t))   in the K-geometry space of (x, y, z) such that K x2 + y 2 + z 2 = 1. Exercise 108. For the path (x (t) , y (t) , z (t)) in Exercise 107 lying on the set (9.2.1), use the Chain Rule (Theorem 1) from calculus of several variables to compute     dxc (t) dyc (t) dx dy dz , , , = · Dc . dt dt dt dt dt This last exercise allows us to do something very nice. Namely, not only can we now use the coordinates (xc , yc ) for our geometry but we can also compute the K-dot-product in terms of these coordinates. By the several variable Chain Rule     dx dy dz dxc dyc , , , = · Dc . dt dt dt dt dt So 

dx dy dz , , dt dt dt



 •K

dx dy dz , , dt dt dt

=



dxc dt



dyc dt

⎞ ⎛ dx ⎞ dt 1 0 0 dy dy ⎟ dz ⎝ 0 1 ⎠⎜ 0 = dx ⎝ dt ⎠ dt dt dt dz 0 0 K −1 dt ⎞ ⎛   1 0 0 dxc  dt 0 ⎠ · Dct · · Dc · ⎝ 0 1 . dyc −1 dt 0 0 K 



⎛

Exercise 109. Compute the 2 × 2 matrix ⎞ ⎛ 1 0 0 0 ⎠ · Dct Pc = Dc · ⎝ 0 1 0 0 K −1

9.2. Central projection

87

that gives the K-dot-product in (xc , yc )-coordinates. That is, use matrix multiplication to show that   2  4 −r r 1 − r 2 Kx2c Kxc yc   Pc = . −r 4 Kxc yc r 2 1 − r 2 Kyc2 Hint: For example

so that



  ∂ln (r) ∂x = r xc + 1 = −r 3 Kx2c + r, ∂xc ∂xc   ∂y ∂ln (r) = r yc = −r 3 Kxc yc , ∂xc ∂xc   ∂ln (r) ∂z =r = −r 3 Kxc ∂xc ∂xc ∂x ∂y ∂z , , ∂xc ∂xc ∂xc



 •K

∂x ∂y ∂z , , ∂xc ∂xc ∂xc



= r 6 K 2 x4c − 2r 4 Kx2c + r 2 + r 6 K 2 x2c yc2 + r 6 Kx2c   = r 6 K 2 x4c + r 6 K 2 x2c yc2 + r 6 Kx2c − 2r 4 Kx2c + r 2   = r 4 Kx2c − 2r 4 Kx2c + r 2 = r 2 1 − r 2 Kx2c . So, if K > 0 and you have a path on the sphere of radius R = K −1/2 in Euclidean three-space given in (xc , yc )-coordinates as (xc (t) , yc (t)) for t ∈ [b, e], you can trace back everything we have done with coordinate changes to see that the length of the path on the sphere of radius R = K −1/2 in Euclidean three-space is given by  e

l (t) dt, b

where

 2

l (t) = =



dxc dyc , dt dt dxc dt





dyc dt

•c 

 dxc dyc , dt dt  

· Pc ·

dxc dt dyc dt

.

Of course if K > 0, we again have Euclidean angles ϑ between vectors Vˆ1 and Vˆ2 tangent to the R-sphere at some point computed by Vˆ1 • Vˆ2 = Vˆ1 · Vˆ2 · cos ϑ = V1c •c V2c . Exercise 110. Show that when K = 0, the formulas for the dot product •c are exactly the ordinary formulas for (flat) plane geometry. This is a big (mathematical) deal! For the first time we have a formula that works for all real numbers K. That is, we have a single formula for lengths and a single formula for angles that hold for every two-dimensional geometry, no matter

88

9. Uniform coordinates for the two-dimensional geometries

its curvature, since the matrix Pc still makes sense when K = 0 and when K becomes negative. So now we have Spherical (K > 0) x ˆ2 + yˆ2 + zˆ2 = R2 ˆ ˆ V • V  1 = K x2 + y 2 + z 2 V1 •K V2 V1c •c V2c

Euclidean (K = 0)

Hyperbolic (K < 0)

  1 = K x2 + y 2 + z 2

  1 = K x2 + y 2 + z 2 V1 •K V2 V1c •c V2c

V1c •c V2c

where t

V1c •c V2c = (V1c ) · Pc · (V2c ) . 9.2.4. Plane sections in central projection coordinates. It will turn out that the shortest paths between two points in K-geometry will be cut out by the plane containing the two points and the origin (x, y, z) = (0, 0, 0). Exercise 111. Explain why a curve in K-geometry cut out by a plane through (0, 0, 0) in (x, y, z)-coordinates corresponds to a line in the (xc , yc )-coordinate plane. Hint: See (9.2.2) relating (x, y, z)-coordinates with (xc , yc )-coordinates. 9.2.5. Area in central projection coordinates. Suppose you are given a region Gc in the (xc , yc )-coordinate plane. Also suppose that K > 0. If you trace back everything we have done with coordinate changes, you can see how Gc gives you ˆ on the sphere of radius R = K −1/2 in Euclidean three-space via the a region G formulas (ˆ x, yˆ, zˆ) = (x, y, Rz) = r · (xc , yc , R)   yc R xc , , . =  K (x2c + yc2 ) + 1 K (x2c + yc2 ) + 1 K (x2c + yc2 ) + 1 Now in section 6.3 we have seen the formula in several variable calculus for comˆ on the sphere of radius R in Euclidean three-space puting the area of the region G in terms of the parameters (xc , yc ) [DS, 49, 231]. It is    ˆ dX ˆ dX (9.2.5) a ˆ , dxc dyc dxc dyc Gc

ˆ dX ˆ dX where a ˆ dx , is the (Euclidean) area of the parallelogram spanned by the two c dyc vectors

ˆ dX dxc

and

ˆ dX dyc

in Euclidean three-space. Thus   ˆ dX ˆ ˆ dX dX ˆ dX , a ˆ = · · sin (ϑ) , dxc dyc dxc dyc

where ϑ is the angle between the two vectors.

9.3. Stereographic projection

89

Exercise 112. Using Lemma 3 and Exercise 97 show that 2 dXˆ dXˆ dXˆ dXˆ  ˆ ˆ dX • • dX , = dxˆc dxˆc dyˆc dxˆc a ˆ dX • dX dX • dX dxc dyc dxc dyc dyc dyc dX •K dX dX •K dX dxc dxc dyc dxc = dX dX dX dX dx •K dy dyc •K dyc c c ⎛ ⎞ ⎛ ⎞ dX 1 0 0 t dxc dX 0 ⎠ = ⎝ ⎠ ⎝ 0 1 dxc dX −1 0 0 K dyc





t dX dyc

= |Pc | . Exercise 113. Use Exercise 109 to show that  2 ˆ dX ˆ dX 1 a ˆ , = r6 = 2 dxc dyc (K (xc + yc2 ) + 1)3 as a function of (xc , yc ). Hint: Notice that the matrix Dc in Exercise 106 is simply the 2 × 3 matrix whose dX dX and dy . So referring to Exercise 109, we know that rows are the vectors dx c c  dXˆ dXˆ dXˆ dXˆ      4 −r r 2 1 − r 2 Kx2c Kxc yc  dxc • dxc dyc • dxc  = . ˆ ˆ ˆ ˆ dX dX dX dX −r 4 Kxc yc r 2 1 − r 2 Kyc2 dxc • dyc dyc • dyc Since all these computations can be extended to K-geometry for all K, we define the K-area of a region Gc in the (xc , yc )-coordinate plane by first computing dX dX and dy at each point of Gc as the K-area of the parallelogram spanned by dx c c   dX dX dX dY · · sin (ϑK ) , aK = dxc dyc dxc K dyc K   dX dX dX dX  dx •K dx • K dyc dxc c =  dXc dX dX dX dx •K dy dy •K dy c

c

and then integrating this area over Gc to get    dX dY aK , dxc dyc AK (Gc ) = dxc dyc Gc    dX dX dX  dx  c •K dxc dyc •K = dX dX dX dx •K dy Gc dy •K c

c

c

c

dX dxc dX dyc

c

dxc dyc .

9.3. Stereographic projection 9.3.1. Stereographic projection coordinates. On the other hand we can project the set   1 = K x2 + y 2 + z 2 onto the set z=1

90

9. Uniform coordinates for the two-dimensional geometries

using the “South Pole” S = (0, 0, −1) as the center of projection. If K > 0, the picture is

If K < 0, the picture is

In both cases the equation relating the point (x, y, z) on K-geometry and the projected point in the plane z = 1 is (9.3.1)

ρ · (xs , ys , (1 − (−1))) = (x, y, (z − (−1))) .

So substituting for (x, y, z) in the equation   1 = K x2 + y 2 + z 2 we obtain z+1 , 2 z = 2ρ − 1,

ρ=

and, from equation (9.2.1),

K (ρxs )2 + (ρys )2 + (2ρ − 1)2 = 1

2 2 K (ρxs ) + (ρys ) + 4ρ2 − 4ρ = 0   ρK x2s + ys2 + 4ρ = 4 1 ρ= K 2 . 2 4 (xs + ys ) + 1

9.3. Stereographic projection

91

Notice that when K < 0, this last formula only makes sense when  K 2 xs + ys2 > −1 4 4 −4 = . x2s + ys2 < K |K| Exercise 114. a) For the projection of the set (9.2.1) onto the z = 1 plane with center of projection S, write (xs , ys ) as a function of (x, y, z). b) For the projection of the set (9.2.1) onto the z = 1 plane with center of projection S, write (x, y, z) as a function of (xs , ys ). 9.3.2. Length and angle in stereographic projection coordinates. Exercise 115. Suppose we have a path X (xs (t) , ys (t)) = (x (xs (t) , ys (t)) , y (xs (t) , ys (t)) , z (xs (t) , ys (t))) lying on the set (9.2.1) given in terms of its projection (xs (t) , ys (t)) in the plane z = 1. Use the formula you derived in Exercise 114b) and the Chain Rule (Theorem 1) to find the 2 × 3 matrix

⎞ ⎛ ∂X

∂xs Ds = ⎝ ⎠ ∂X ∂ys

such that



dx dy dz , , dt dt dt



 =

dxs (t) dys (t) , dt dt

 · Ds .

Hint: Use logarithmic differentiation: dx = d (ρxs ) = xs dρ + ρdxs ρ−1 dx = xs dln (ρ) + dxs , and similarly for y. Also 

  K 2 2 dln (ρ) = −dln xs + ys + 1 4    K 2 1 2 =−K 2 d xs + ys + 1 2 4 4 (xs + ys ) + 1 K = −ρ (2xs dxs + 2ys dys ) . 4 This last exercise allows us to do something very nice. Namely, not only can we now use the coordinates (xs , ys ) for our geometry but we can also compute the K-dot-product in terms of these coordinates:     dx dy dz dxs dys , , , = · Ds dt dt dt dt dt

92

9. Uniform coordinates for the two-dimensional geometries

so that 

dx dy dz , , dt dt dt



 •K

dx dy dz , , dt dt dt

=





dys dt

dxs dt

⎞ ⎛ dx ⎞ dt 1 0 0   dy dy ⎟ dz ⎝ 0 1 ⎠⎜ 0 = dx ⎝ dt dt dt dt ⎠ dz 0 0 K −1 dt ⎞ ⎛   1 0 0 dxs  dt 0 ⎠ · Dst · · Ds · ⎝ 0 1 . dys dt 0 0 K −1 ⎛

Exercise 116. Use matrix multiplication to compute the 2 × 2 matrix ⎞ ⎛ 1 0 0 0 ⎠ · Dst , Ps = Ds · ⎝ 0 1 0 0 K −1 that is, to compute the K-dot-product in (xs , ys )-coordinates. (You may be surprised at the answer! It is quite simple and only involves the quantity ρ.) So, if K > 0 and you have a path on the sphere of radius R = K −1/2 in Euclidean three-space given in (xs , ys )-coordinates as (xs (t) , ys (t)) for t ∈ [b, e], you can trace back everything we have done with coordinate changes to see that the length of the path on the sphere of radius R = K −1/2 in Euclidean three-space is given by  e

l (t) dt, b

where

 2

l (t) = =



dxs dys , dt dt dxs dt





dys dt

•s 

 dxs dys , dt dt  

· Ps ·

dxs dt dys dt

,

and that the measure ϑ of an angle between vectors Vˆ1 and Vˆ2 on the R-sphere is computed by   t V1s · Ps · (V2s ) arccos . |V1s |s |V2s |s Notice that the matrix Ps still makes sense when K = 0 and when K becomes negative. Exercise 117. Write the formula for the K-dot-product in (xs , ys )-coordinates when K = 0. Does it look familiar? So we do have Spherical (K > 0) x ˆ2 + yˆ2 + zˆ2 = R2 ˆ ˆ V 2• V 2  1 = K x + y + z2 V1 •K V2 V1c •c V2c V1s •s V2s

Euclidean (K = 0)

Hyperbolic (K < 0)

  1 = K x2 + y 2 + z 2

  1 = K x2 + y 2 + z 2 V1 •K V2 V1c •c V2c V1s •s V2s

V1c •c V2c V1s •s V2s

9.3. Stereographic projection

93

where V1s •s V2s = (V1s ) · Ps · (V2s )t . Of course if K > 0, we again have Euclidean angles ϑ between vectors Vˆ1 and Vˆ2 tangent to the R-sphere at some point computed by Vˆ1 • Vˆ2 = Vˆ1 · Vˆ2 · cos ϑ = V1s •s V2s . 9.3.3. Stereographic projection is conformal. One of the most striking uses of stereographic projection coordinates for K-geometry is the fact that stereographic projection is conformal; that is, angles between intersecting paths in K-geometry are exactly the same as the ordinary plane geometry angles between the corresponding paths in the Euclidean (xs , ys )-plane. To see this, from Exercise 116 we know that for tangent vectors V1 and V2 emanating from the same point on the K-geometry,  2  ρ 0 t s s s V1 •K V2 = V1 •s V2 = (V1 ) · · (V2s ) . 0 ρ2 So since we define angles ϑ in K-geometry by the formula |V1 |K · |V2 |K · cos ϑ = V1 •K V2 = V1s •s V2s , then

V2 ϑ = arccos |V1V|1 •K ·|V2 | K

K

V1s •s V2s |V1s |s ·|V2s |s ρ2 (V1s )·(V2s )t  = arccos  s 2 ρ (V1 )·(V1s )t · (V2s )·(V2s )t s (V1 )·(V2s )t  = arccos  s , (V1 )·(V1s )t · (V2s )·(V2s )t

= arccos

that is, just the usual Euclidean angle between the vectors V1s and V2s in the (xs , ys )plane. 9.3.4. Plane sections of K-geometry. We return to the fact that it will turn out that the shortest paths between two points in K-geometry will be cut out by the plane containing the two points and the origin (x, y, z) = (0, 0, 0). What will these shortest paths look like in (xs , ys )-coordinates? Exercise 118. Suppose we intersect K-geometry (9.4.1) with a plane ax + by + z = 0. a) Show that the equation for the resulting path in stereographic projection coordinates is   2  2  4 K + a2 + b2 2a 2b + ys − = . xs − K K K2 Hint: Use formula (9.3.1) to relate (x, y, z)-coordinates with (xs , ys )-coordinates. b) What is the equation for the resulting path in stereographic projection coordinates if we intersect the K-geometry with a plane given by ax + by = 0,

94

9. Uniform coordinates for the two-dimensional geometries

that is, a plane containing the z-axis? 9.3.5. Area in stereographic projection coordinates. Suppose you are given a region Gs in the (xs , ys )-coordinate plane. Also suppose that K > 0. If you trace back everything we have done with coordinate changes, you can see how Gs gives ˆ on the sphere of radius R = K −1/2 in Euclidean three-space via the you a region G formulas (ˆ x, yˆ, zˆ) = (x, y, Rz) = ρ · (xs , ys , R (2ρ − 1))   2  2 1− K x + y ys xs s s 4 , K 2 ,R · = K 2 . 2 2 2 2 1+ K 4 (xs + ys ) + 1 4 (xs + ys ) + 1 4 (xs + ys ) Now we saw in section 6.3 that there is a formula in several variable calculus for ˆ on the sphere of radius R in Euclidean threecomputing the area of the region G space in terms of the parameters (xs , ys ) [DS, 49, 231]. It is    ˆ dX ˆ dX dxs dys , a ˆ , dxs dys Gc

ˆ dX ˆ dX is the (Euclidean) area of the parallelogram spanned by the two where a ˆ dx , s dys vectors

ˆ dX dxs

and

ˆ dX dys

in Euclidean three-space. That is,   ˆ dX ˆ ˆ dX dX ˆ dX , = a ˆ · · sin ϑ, dxs dys dxs dys

where ϑ is the angle between the two vectors

ˆ dX dxs

Exercise 119. As in Lemma 3 show that  2 dXˆ dXˆ ˆ dX ˆ dX s • dxs a ˆ , = dx dXˆ • dXˆ dxs dys dxs dys dX •K dX s dxs = dx dX dX dx • K dys s

ˆ dX dys .

and ˆ dX dys ˆ dX dys

ˆ dX dxs ˆ dX dys

• •

dX dys dX dys

•K



dX dxs dX •K dy s

.

Now notice the matrix Ds in Exercise 115 is simply the 2 × 3 matrix whose dX dX and dy . rows are the vectors dx s s Exercise 120. Use Exercise 116 to show that  2 ˆ dX ˆ dX 1 a ˆ , = ρ4 =  4 . K dxs dys (x2 + y 2 ) + 1 4

s

s

9.4. Relationship between central and stereographic projection coordinates 9.4.1. Two addresses for the same point in K-geometry. Finally we should compare the relationship between the two kinds of coordinates for the set     (9.4.1) (x, y, z) ∈ R3 : 1 = K x2 + y 2 + z 2

9.4. Relationship between central and stereographic projection coordinates

95

that we have been exploring, namely central projection coordinates x xc = , (9.4.2) z y yc = z and stereographic projection coordinates 2x xs = (9.4.3) , z+1 2y ys = . z+1 To do this we use the exercises in which we wrote (x, y, z) in the set (9.4.1) as functions of (xc , yc ) and (xs , ys ), respectively. Namely, xc x=  , K (x2c + yc2 ) + 1 yc y= , 2 K (xc + yc2 ) + 1 1 , z= 2 K (xc + yc2 ) + 1 and xs , 1 + (x2s + ys2 ) ys , y= 2 + y2 ) 1+ K (x s s 4   2 2 1− K 4 xs + ys z= . K 1 + 4 (x2s + ys2 )

x=

K 4

The rest is simple algebra [MJG, 258]. Exercise 121. Do the algebra to write the explicit formulas for (xs (xc , yc ) , ys (xc , yc )) and (xc (xs , ys ) , yc (xs , ys )) . 9.4.2. When K is negative, there are asymptotic cones. Notice that if K < 0, the equation of K-geometry becomes   z 2 − |K| x2 + y 2 = 1. Thus K-geometry forms a two-sheeted hyperboloid with the z-axis as major axis. The hyperboloid is obtained by rotating the hyperbola (9.4.4)

z 2 − |K| x2 = 1

in the (x, z)-plane around the z-axis. We consider only the sheet on which z is positive as forming the K-geometry. Now the asymptotes of this last hyperbola are the pair of lines



1/2 1/2 z + |K| x z + |K| x = z 2 − |K| x2 = 0.

96

9. Uniform coordinates for the two-dimensional geometries

Rotating the asymptotes around the z-axis we obtain the asymptotic cone   z 2 − |K| x2 + y 2 = 0 for K-geometry. Since the entire K-geometry lies inside the asymptotic cone, the central projection coordinates (xc , yc ) only correspond to points in K-geometry when (xc , yc , 1) lies inside the asymptotic cone, that is, when   1 − |K| x2c + yc2 > 0. Said otherwise, the circle of radius |K|−1/2 around (0, 0) in the (xc , yc )-plane captures inside it all the points (x, y, z) of K-geometry under central projection. We call the circle of radius |K|−1/2 around (0, 0) the circle at infinity. The points on the circle at infinity are not points of the geometry. In fact they lie at infinite distance from the point (0, 0), as we will see. Exercise 122. a) Show that the slopes of the asymptotes to the hyperbola (9.4.4) are the limits as x goes to ±∞ of the slopes of the lines through (0, 0) and the point (x, z) on the hyperbola. b) Show that the slopes of the asymptotes to the hyperbola (9.4.4) are the limits as x goes to ±∞ of the slopes of the lines through (0, −1) and the point (x, z) on the hyperbola. c) Use b) to compute the radius of the disk around (0, 0) in the (xs , ys )-plane that captures all the points (x, y, z) of K-geometry under stereographic projection.

Part VI

Return to spherical geometry

Chapter 10

Spherical geometry from an advanced viewpoint

10.1. Rigid motions in spherical geometry ˆ denote an 10.1.1. Computing rigid motions in Euclidean coordinates. Let M invertible 3 × 3 matrix. We begin by returning to the condition that the (linear) transformation   ˆ x ˆ, yˆ, zˆ = (ˆ x, yˆ, zˆ) · M takes the R-sphere to itself. Exercise 123 (SG). Show that the transformation (10.1.3) takes the R-sphere to itself if ˆ ·M ˆ t = I, M ˆ satisfying this where I is the 3 × 3 identity matrix. (Recall that a matrix M condition is called an orthogonal matrix.) ˆ (t) = (ˆ If we have a curve X x (t) , yˆ (t) , zˆ (t)) on the R-sphere, we have seen that we measure its length L by the formula  e (10.1.1) L= l (t) dt, b

where ˆ ˆ dX dX dX dX • = •K , dt dt dt dt and that we measure angles ϑ between tangent vectors Vˆ1 and Vˆ2 at a point on the R-sphere by the formula ⎞ ⎛   ˆ ˆ V1 •K V2 V1 • V2 , ϑ = arccos ⎝ ⎠ = arccos |V1 |K · |V2 |K Vˆ1 · Vˆ2

(10.1.2)

2

l (t) =

99

100

10. Spherical geometry from an advanced viewpoint

where 2

|V |K = V •K V. Rewriting the transformation as (10.1.3)

ˆ ·M ˆ, ˆ =X X

we next wish to check that if ˆ ·M ˆ t = I, M then not only does the transformation take the R-sphere to itself but it in fact preserves the length of any curve lying on the R-sphere. Formulas (10.1.1) and (10.1.2) show that all we have to worry about is that ˆ dX ˆ ˆ dX ˆ dX dX • = • dt dt dt dt for all values t of the parameter of the curve. But (10.1.4)

ˆ ˆ dX dX ˆ = ·M dt dt

ˆ is a constant matrix. So the transformation given by by the product rule since M ˆ the matrix M will preserve the length of any path since  t  t ˆ ˆ ˆ ˆ ˆ ˆ dX dX d X d X d X dX ˆ · ˆ • = ·M ·M · = . (10.1.5) dt dt dt dt dt dt The transformation will also preserve the measure of any angle since for any two paths X (t) and Y (t ) crossing at a point,   t t ˆ ˆ ˆ ˆ dX d Y d Y d X ˆ · ˆ ·M · ·M = dt dt dt dt at that point. Definition 21. A transformation of any space or geometry that preserves distances and angles is called a rigid motion, a congruence, or an isometry. ˆ corresponds to a rigid motion So, by Exercise 123, every orthogonal matrix M of the R-sphere. (It can be shown that every rigid motion of the R-sphere is given by an orthogonal matrix, but we will not treat that subtlety in this book.) 10.1.2. Orthogonal and K-orthogonal matrices. Recalling again the fact that the Euclidean R-sphere is a K-geometry with K = 1/R2 , we should compare the K-orthogonal transformations M in Definition 19 with the orthogonal ones just above. Exercise 124 (SG). Referring to Definition 19, show that the transformations   ˆ x, yˆ, zˆ) · M x ˆ, yˆ, zˆ = (ˆ and

  x, y, z = (x, y, z) · M

10.2. Spherical geometry is homogeneous

give the same rigid motion of ⎛ 1 M =⎝ 0 0

101

the R-sphere if ⎞ ⎛ 0 0 1 ˆ ·⎝ 0 1 0 ⎠·M 0 R 0

⎞ 0 0 1 0 ⎠. 0 R−1

Exercise 125 (SG). Show that the matrix ⎛ ⎞ cos ϑ sin ϑ 0 ⎝ − sin ϑ cos ϑ 0 ⎠ 0 0 1 is both orthogonal and K-orthogonal and gives the same transformation of the Euclidean R-sphere. Describe geometrically what this transformation is doing to the R-sphere. Exercise 126 (SG). Show that the matrix ⎛ ⎞ cos ϕ 0 sin ϕ ˆ =⎝ 0 1 0 ⎠ M − sin ϕ 0 cos ϕ is orthogonal and that the matrix ⎛ cos ϕ 0 M =⎝ −R · sin ϕ

⎞ 0 R−1 · sin ϕ ⎠ 1 0 0 cos ϕ

is the K-orthogonal matrix describing the same transformation of the R-sphere. Describe geometrically what this transformation is doing to the R-sphere.

10.2. Spherical geometry is homogeneous 10.2.1. Moving a point to the North Pole by a rigid motion. As the heading suggests, we are next going to move any point on the R-sphere to the North Pole by a rigid motion. However, we are going to describe the entire process in (x, y, z)coordinates, that is, in K-geometry. This will allow us to use all the computations we did in Part V since SG is a K-geometry in the sense of Part V. Recall that in (x, y, z)-coordinates, the equation for the R-sphere becomes   (10.2.1) K x2 + y 2 + z 2 = 1 with

1 , R2 and the Euclidean dot-product is given by the K-dot-product. Again, if you get nervous using these weird coordinates to compute things that are clearer in (ˆ x, yˆ, zˆ)coordinates, just go through the constructions in the case K = 1 first. In that special case (x, y, z) = (ˆ x, yˆ, zˆ) , and your calculations above (as well as all those in Part V) reduce to the usual ones on the unit sphere in ordinary Euclidean three-space. K=

So, first of all, in (x, y, z)-coordinates the North Pole is the point N = (0, 0, 1) .

102

10. Spherical geometry from an advanced viewpoint

Suppose we start with a point X0 = (x0 , y0 , z0 ) in the geometry, that is, satisfying equation (10.2.1). Exercise 127 (SG). Write an explicit K-rigid motion of the type ⎛ ⎞ cos ϑ sin ϑ 0 M1 = ⎝ − sin ϑ cos ϑ 0 ⎠ 0 0 1 that takes the point X0 to a point X1 = (x1 , 0, z0 ). Hint: Start from the identity −y0 x0  · x0 +  2 · y0 = 0 x20 + y02 x0 + y02 and the desired identity sin ϑ · x0 + cos ϑ · y0 = 0, and conclude that there is a ϑ so that x0 , cos ϑ =  2 x0 + y02 −y0 sin ϑ =  2 . x0 + y02 Exercise 128 (SG). Write an explicit K-rigid motion of the type ⎛ ⎞ cos ϕ 0 R−1 · sin ϕ ⎠ 0 1 0 M2 = ⎝ −R · sin ϕ 0 cos ϕ that takes the point X1 = (x1 , 0, z0 ) to N = (0, 0, 1). Using these last two exercises we conclude that the transformation   x, y, z = (x, y, z) · (M1 · M2 ) is a K-rigid motion (why?) and that N = (x0 , y0 , z0 ) · (M1 · M2 ) (why?). 10.2.2. Moving a (point, direction) to any other (point, direction) by a rigid motion. Let V2 = (a2 , b2 , 0) be a tangent vector to K-geometry at the North Pole N . Exercise 129 (SG). Write an explicit K-rigid motion of the type ⎛ ⎞ sin ϑ 0 cos ϑ M3 = ⎝ − sin ϑ cos ϑ 0 ⎠ 0 0 1

10.2. Spherical geometry is homogeneous

103

that takes V2 to the vector   

2 2 a2 + b2 , 0, 0 = V2 •K V2 , 0, 0 . Why does the transformation given by M3 leave the North Pole N fixed? Now suppose we have any point X0 = (x0 , y0 , z0 ) in K-geometry and any K-tangent vector V0 = (a0 , b0 , c0 ) at that point. Exercise 130 (SG). Explain in words why the K-rigid motion   x, y, z = (x, y, z) · (M1 · M2 · M3 ) constructed over √ the last couple  of sections takes the point X0 to N and the tangent V0 •K V0 , 0, 0 . vector V0 to Now suppose that (X0 , V0 ) gives a point X0 in K-geometry and a tangent direction V0 to K-geometry at X0 . Suppose that (X0 , V0 ) gives another point in K-geometry and a tangent direction to K-geometry at X0 . Finally suppose that V0 and V0 have the same length, that is, V0 •K V0 = V0 •K V0 . As above, find a K-rigid motion given by M = (M1 · M2 · M3 ) √  taking X0 to the North Pole and V0 to V0 •K V0 , 0, 0 . Similarly find a K-rigid motion given by M  = (M1 · M2 · M3 ) 

taking X0 to the North Pole and V0 to V0 •K V0 , 0, 0 . Exercise 131 (SG). Explain in words why the K-rigid motion given by −1

M · (M  ) takes (X0 , V0 ) to (X0 , V0 ).

By completing this exercise we have shown that K-geometry looks the same at each point and in each direction at that point. That is, we have shown that each K-geometry is homogeneous.

104

10. Spherical geometry from an advanced viewpoint

10.3. Lines in spherical geometry 10.3.1. Spherical coordinates, a shortest path from the North Pole. Next we will figure out the shortest path you can take between two points on the Euclidean R-sphere, that is, what paths serve as “lines” on the R-sphere. Again we will do our calculation using only (x, y, z)-coordinates (since, as we have seen in (9.1.8), we won’t have (ˆ x, yˆ, zˆ)-coordinates when we get to hyperbolic geometry). For our purposes, it will be convenient to use yet another set of coordinates for K-geometry for K > 0, namely what are commonly known as spherical coordinates: x (σ, τ ) = R · sin σ · cos τ, (10.3.1)

y (σ, τ ) = R · sin σ · sin τ, z (σ, τ ) = cos σ.

Exercise 132 (SG). Show that these spherical coordinates do actually parametrize the R-sphere, that is, that

K x (σ, τ )2 + y (σ, τ )2 + z (σ, τ )2 ≡ 1 for all (σ, τ ). Notice that you can write a path on the R-sphere by giving a path (σ (t) , τ (t)) in the (σ, τ )-plane. In fact, you can use σ as the parameter t and just write (10.3.2)

(σ, τ (σ)) ,

where τ is a function of σ. To write a path that starts at the North Pole, just write (σ, τ (σ)) , 0 ≤ σ ≤ ε and demand that τ (0) = 0. If you want the path to end on the plane y = yˆ = 0, demand additionally that τ (ε) = 0. But if we are going to describe paths on the R-sphere by paths in the (σ, τ )-plane, we are going to need to figure out the K-dot-product in (σ, τ )-coordinates so that we can compute the lengths of paths in these coordinates. Exercise 133 (SG). a) Referring to (10.3.1), compute the 2 × 3 matrix   Dsph =

dx dσ dx dτ

dy dσ dy dτ

dz dσ dz dτ

.

b) Show that if a path in K-geometry is given by a path (σ (t) , τ (t)) in the (σ, τ )-plane,     dσ dτ dx dy dz , , , = · Dsph . dt dt dt dt dt

10.3. Lines in spherical geometry

105

c) For two paths in K-geometry given by paths (σ1 (t) , τ1 (t)) and (σ2 (t) , τ2 (t)) in the (σ, τ )-plane, use a) and b) to show that    t dˆ x1 dˆ y1 dˆ z1 dˆ x2 dˆ y2 dˆ z2 , , , , · dt dt dt dt dt dt ⎞ ⎛    t 1 0 0 dx2 dy2 dz2 dx1 dy1 dz1 ⎠ ⎝ 0 1 0 , , , , · = · dt dt dt dt dt dt 0 0 K −1    −1 t   dσ1 dτ1 dσ2 dτ2 0 K , , = . · · 0 K −1 sin2 σ dt dt dt dt 

d) Explain why the formula       −1 dσ2 dτ2 dσ1 dτ1 dσ1 dτ1 K , , , •sph = · 0 dt dt dt dt dt dt

0 K −1 sin2 σ

  t dσ2 dτ2 , · dt dt

allows us to compute the ordinary Euclidean dot-product of two tangent vectors to the R-sphere in Euclidean space if we just know the values of the vectors in the (σ, τ )-plane that correspond to the two tangent vectors. Exercise 134 (SG). Show that the length L of any path on the R-sphere given by (σ, τ (σ)) , 0 ≤ σ ≤ ε with τ (0) = 0 and τ (ε) = 0 is given by the formula   t   ε  dτ dτ 1 0 L=R· dσ. 1, 1, 0 sin2 σ dσ dσ 0 Hint: Notice that the exercise makes the first coordinate σ the parameter for the curve, that is, σ (t) = t, dσ dt ≡ 1. This last formula for L lets us figure out the shortest path from N = (R · sin 0 · cos 0, R · sin 0 · sin 0, cos 0) to (R · sin ε, 0, cos ε) = (R · sin ε · cos 0, R · sin 0 · sin 0, cos ε) . Since

 L=R·

ε

 1 + sin2 σ ·

0



dτ dσ

2 dσ

dτ is identiand sin2 σ is positive for almost all σ ∈ [0, ε], L is minimal only when dσ cally zero. But this means that τ (σ) is a constant function. Since τ (0) = 0, this means that τ (σ) is identically zero. So we have shown the following result.

Theorem 14 (SG). The shortest path on the R-sphere from the North Pole to a point (x, y, z) = (R · sin ε, 0, cos ε) is the path lying in the plane y = 0.

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10. Spherical geometry from an advanced viewpoint

10.3.2. Shortest path between any two points. We next prove the theorem that shows that the shortest path on the surface of the earth from Rio de Janeiro to Los Angeles is the one cut on the surface of the earth by the plane that passes through the center of the earth, through Rio, and through Los Angeles. That is usually the route an airplane would take when flying between the two cities. Theorem 15 (SG). Given any two points X1 = (x1 , y1 , z1 ) and X2 = (x2 , y2 , z2 ) in K-geometry, the shortest path between the two points is the path cut out by the two equations   K x2 + y 2 + z 2 = 1, ⎛ ⎞ x y z ⎝ ⎠ (10.3.3) x1 y1 z1 = 0, x2 y2 z2 that is, the plane containing (0, 0, 0), X1 , and X2 . Proof. By Exercise 131 there is a K-rigid motion X =X ·M

  that takes X1 to the North Pole N and X2 to K −1/2 sin ε, 0, cos ε for some ε. That is, X2 · M = (R · sin ε, 0, cos ε) for some ε since all points in K-geometry with y = 0 can be written as (R · sin ε, 0, cos ε) for some ε. Furthermore, the transformation is linear so it takes the plane through X1 , X2 , and (0, 0, 0) to the plane through X1 , X2 , and (0, 0, 0), that is, to the plane given by y = 0. Said otherwise, ⎛ ⎞ x y z ⎝ ⎠ = 0 0 0 1 R · sin ε 0 cos ε if and only if y = 0. Since M is a K-rigid motion it must take the shortest path from X1 to X2 to the shortest path from X1 · M = N to X2 · M = (R · sin ε, 0, cos ε). But we already know that the shortest path from X1 · M to X2 · M is the one cut out by the plane y = 0. On the other hand, it is the path given by equation (10.3.3) that passes through X1 to X2 and goes to the path given by y = 0 under the rigid motion. So  it must be the shortest path between X1 and X2 . The path given by equation (10.3.3) is called the great circular arc between X1 and X2 . Definition 22. A line in SG will be a curve that extends infinitely in each direction and has the property that given any two points X1 and X2 on the path, the shortest path between X1 and X2 lies along that curve. Lines in SG are usually called great circles on the R-sphere. They are the intersections of the R-sphere with planes through (0, 0, 0).

10.4. Central projection in SG

107

Letting X2 approach X1 along the great circular arc joining X1 and X2 , we see that the solution set to equation (10.3.3) does not change. Taking a limit as X2 approaches X1 , this set can also be expressed as the solution set of the equation ⎛ ⎞ x y z ⎝ x1 y1 z1 ⎠ = 0, a1 b1 c1 where (a1 , b1 , c1 ) is a tangent vector at the point X1 pointing in the direction of X2 .

10.4. Central projection in SG 10.4.1. Central projection preserves lines. We all probably realize that you can’t make a perfect map of the world; that is, you can’t make a map so that angles on the map are equal to the corresponding angles on the sphere and straight lines on the map correspond to great circular arcs on the sphere. We do the next best thing—we make two maps of the sphere, one that has the property that angles are faithfully represented and the other for which straight lines on the map correspond to the shortest paths on the sphere. We start with a simple way to make a map for which straight lines on the map correspond to the shortest paths on the sphere. The map coordinates we use to do this are the central projection coordinates we learned about in Part V. Now SG is a K-geometry in the sense of Part V since, in (x, y, z)-coordinates, the equation for the R-sphere becomes   K x2 + y 2 + z 2 = 1 with

1 , R2 and the Euclidean dot-product on the R-sphere is given in (x, y, z)-coordinates by the K-dot-product. So all the computations of Part V hold for spherical geometry as long as we understand that we are computing it in (x, y, z)-coordinates. K=

Exercise 135. Show that central projection of a point on the R-sphere in (ˆ x, yˆ, zˆ)space to the plane zˆ = R is the same as central projection of the corresponding point in (x, y, z)-coordinates to the plane z = 1. Hint: Recall (9.2.2) and write the corresponding relation rˆ (ˆ xc , yˆc , R) = (ˆ x, yˆ, zˆ) in (ˆ x, yˆ, zˆ)-coordinates. Conclude that rˆ = r. (Why?) Exercise 136 (SG). Show that lines (i.e., shortest paths in SG) correspond under central projection to straight lines in (xc , yc )-coordinates. Hint: See Exercise 111 and Theorem 15. Or just write the equation for a line in (xc , yc )-coordinates and substitute (9.4.2). Then reverse the process. 10.4.2. Spherical area computed in central projection coordinates. Recall that in Part V we learned how to compute the K-area in K-geometry of a region G given by a region Gc in the (xc , yc )-plane. That is, we learned how to compute ˆ on the R-sphere given by a region Gc in the (xc , yc )-plane. the area of a region G

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10. Spherical geometry from an advanced viewpoint

Exercise 137 (SG). Explain why we know from Exercise 113 in Part V that if ˆ on the R-sphere is parametrized by a region Gc in (xc , yc )-coordinates, a region G ˆ is given by the formula then the area Aˆ of G    2  −3/2 ˆ K xc + yc2 + 1 A= dxc dyc . Gc

10.5. Stereographic projection in SG 10.5.1. Stereographic projection preserves angles. We now turn to a simple way to make a map of the R-sphere in such a way that the measure of any angle on the map is exactly the same as the measure of the corresponding angle on the R-sphere. The map coordinates that do the job are the stereographic projection coordinates that we learned about in Part V. Exercise 138 (SG). a) Compute the stereographic projection of a point on the R-sphere in (ˆ x, yˆ, zˆ)-space to the plane zˆ = R. b) Show that the coordinates (ˆ xs , yˆs ) of the stereographic projection of a point on the R-sphere in (ˆ x, yˆ, zˆ)-space to the plane zˆ = R are the same as the coordinates (xs , ys ) of the stereographic projection of the corresponding point in (x, y, z)coordinates to the plane z = 1. Hint: Reduce to showing that     2x 2ˆ x 2ˆ y 2y , = , . R zˆ + R zˆ + R z+1 z+1 Exercise 139 (SG). a) Show that stereographic projection is conformal, that is, that the angle between two paths through a point on the R-sphere in (ˆ x, yˆ, zˆ)-space is the same as the usual (Euclidean) angle between the corresponding two paths through the corresponding point in the (xs , ys )-plane. Hint: See subsection 9.3.3. b) Draw a picture of an angle between two paths through a point on the Euclidean R-sphere and the stereographic projection of that angle onto the plane zˆ = R. Try to give an intuitive geometric explanation for why it should have the same measure as the original angle. Hint: Let L denote the line through (ˆ x, yˆ, zˆ) and (xs , ys , R). Compare the angle between L and the tangent plane to the R-sphere at (ˆ x, yˆ, zˆ) to the angle between L and the plane zˆ = R. (In fact, by rotational symmetry, you can assume that yˆ = ys = 0.) 10.5.2. Areas of spherical triangles in stereographic projection coordinates. Use Exercise 118b) to show that lines in SG become circles under stereographic projection unless the line in SG passes through the North Pole (in which case it corresponds to a line through (xs , ys ) = (0, 0) in the (xs , ys )-plane). Suppose a spherical triangle T corresponds to a region Ts in (xs , ys )-coordinates and the vertices of T correspond to (xs , ys ) = (−2, 0), (xs , ys ) = (2, 0), and (xs , ys ) = (0, 2). So one side of Ts lies on the line ys = 0. Exercise 140. a) Use Exercise 118b) to compute the equations for the other two sides of Ts .

10.5. Stereographic projection in SG

109

b) In the (xs , ys )-plane, draw Ts as accurately as you can when K = 4, then when K = 14 , then finally when K = 0, that is, for the sphere of infinite radius. c) Use subsection 5.5.2 and the fact that stereographic projection is conformal to compute the area of T in all cases in b). Hint: Your job will be easier if you notice that the ys -axis divides Ts into two congruent isosceles triangles. Then the only calculation you will need to do is to calculate the radian measure of the angle at the vertex (xs , ys ) = (2, 0) of Ts . dys by implicit differentiation of (Why?) To calculate the interior angle, calculate dx s

dys radians. the first quadrant equation in a), then take π − arctan dx s Exercise 141. Explain why we know from Exercise 120 in Part V that in all of the cases in Exercise 140 the area of the spherical triangle T is also given by the formula  1   dxs dys . K 2 2 2 Ts 1 + 4 (xs + ys )

Part VII

Hyperbolic geometry

Chapter 11

The curvature K becomes negative

11.1. The world sheet and the light cone We now turn to the case in which the radius R of the Euclidean R-sphere goes to infinity and beyond! Of course that doesn’t make any sense in (ˆ x, yˆ, zˆ)-space but if we look at the R-sphere in (x, y, z)-coordinates, it makes perfect sense because there the equation of the R-sphere is (11.1.1)

  K x2 + y 2 + z 2 = 1

for K = R12 , so that R going to infinity means that K goes to zero, and “beyond” simply means that K becomes negative. We have seen that all we need to have a geometry with lengths, angles, areas, and congruences is to have a smooth set and a dot-product between vectors tangent to that set. Now if K becomes negative, our geometry becomes a hyperboloid of two sheets (obtained by rotating a hyperbola in the (x, z)-plane with major axis the z-axis around that axis). We will still have a “K-geometry” so all the calculations of Part V are still valid. This means that we have already done much of the work for hyperbolic geometry. So this final chapter will be shorter, and the work that we do will so closely parallel that for spherical geometry that it will be presented as a sequence of exercises, where the hints are the analogies with what we have already done in the spherical geometry case. So that we have a connected universe, we only consider the “top” sheet (where z > 0) as our K-geometry. (In special relativity, this sheet might be called something like the “world sheet”.) If, instead of rotating a hyperbola around the z-axis which gives our K-geometry, we rotate the asymptotes of the hyperbola around the z-axis, we obtain a cone given by the equation   K x2 + y 2 + z 2 = 0. 113

114

11. The curvature K becomes negative

(Again this might be called something like the “light cone”.) This cone will play a major role in what follows. Indeed it is the role of this cone that is the source of the major differences arising in the case of hyperbolic geometry. There is one potential problem we need to worry about when K < 0, and it concerns the lengths of tangent vectors. Namely, our formulas for lengths involve taking the square root of the dot-product of a tangent vector with itself, so that dot-product had better be non-negative (or zero only if the tangent vector itself is the zero-vector). Our K-dot-product is given by the formula ⎞ ⎛ ⎛ ⎞ a 1 0 0   a b c ·⎝ 0 1 0 ⎠ · ⎝ b ⎠. c 0 0 K −1 So, when K < 0, it seems entirely possible that some tangent vector V has the property that V •K V < 0. (Indeed that will always happen if c is sufficiently big and x and y are sufficiently small [MJG, 241–242].) 11.1.1. Nonzero tangent vectors in HG have positive lengths. Exercise 142. Suppose that a vector V emanates from (0, 0, 0) in (x, y, z)-space. a) Show that V •K V = 0 if and only if V points in the direction of the light cone. b) Show that V •K V < 0 if and only if V points in a direction inside the light cone. c) Show that V •K V > 0 if and only if V points in a direction outside the light cone. Hint: Use the fact that the (Euclidean) angle ϑ that the light cone makes with the plane z = 0 is given by taking any point (x, y, z) on the light cone with z > 0 and computing z 1/2 = |K| . tan ϑ =  x2 + y 2 Now our world sheet lies inside the light cone but tangent vectors to it point outside the light cone. That is what saves our K-dot-product, as we see in the next lemma. Lemma 7 (HG). Let V = (a, b, c) denote a vector that is tangent to our Kgeometry, that is, to the set (11.1.1). Then V •K V ≥ 0 and V •K V = 0 if and only if V = 0. Proof. If c = 0, then the assertion of the lemma is obviously true. So we can assume c = 0. Notice that since V is assumed to be a tangent vector at (x, y, z), this means that (x, y, z) is not the North Pole so that x2 + y 2 > 0. Next, replacing V with the case in which

1 c

(V ) just multiplies V •K V by V = (a, b, 1) ,

1 c2 ,

so it suffices to consider

11.2. Hyperbolic geometry is homogeneous

and we must show that

115

 a2 + b2 + K −1 > 0. Since V is tangent to our K-geometry at some point (x, y, z), we know by Exercise 101 that (x, y, z) •K V = 0, that is, z = 0. ax + by + K On the other hand   K x2 + y 2 + z 2 = 1. Substituting the expression for z given by the former equation into the latter gives   2 K x2 + y 2 + K 2 (ax + by) = 1. 

On the other hand (ay − bx)2 ≥ 0 2

2

(ay) + (bx) ≥ 2abxy so that



  K x2 + y 2 + K 2 (ax)2 + (by)2 + (ay)2 + (bx)2 ≥ 1      K x2 + y 2 + K 2 a2 + b2 x2 + y 2 ≥ 1   1 . K −1 + a2 + b2 ≥ 2 2 K (x + y 2 ) 

11.2. Hyperbolic geometry is homogeneous 11.2.1. Rigid motions in (x, y, z)-coordinates. Now HG is a K-geometry in the sense of Part V since, in (x, y, z)-coordinates, the equation for the K-geometry becomes   (11.2.1) K x2 + y 2 + z 2 = 1 with K < 0, and the K-dot-product is a valid dot-product. If we have a curve X (t) = (x (t) , y (t) , z (t)) on the K-geometry given (in K-coordinates) as   1 = K x2 + y 2 + z 2 , we have seen that we measure its length L by the formula  e l (t) dt, (11.2.2) L= b

where dX dX •K , dt dt and that we measure angles ϑ between tangent vectors Vˆ1 and Vˆ2 at a point by the formula   V1 •K V2 ϑ = arccos , |V1 |K · |V2 |K

(11.2.3)

2

l (t) =

116

11. The curvature K becomes negative

where 2

|V |K = V •K V. We again want to explore the condition that a transformation   x, y, z = (x, y, z) · M preserves the length of any curve (x (t) , y (t) , z (t)) lying on the K-geometry. Rewriting the transformation as X = X · M, all we have to worry about is that dX dX dX dX •K = •K . dt dt dt dt So, referring to Definition 19, a transformation given by a matrix M will preserve the length of any path and will preserve the measure of any angle if M is K-orthogonal. Exercise 143 (HG). Show that the matrix ⎛ ⎞ cos ϑ sin ϑ 0 ⎝ − sin ϑ cos ϑ 0 ⎠ 0 0 1 is K-orthogonal. Describe geometrically what this transformation is doing to the K-geometry. For our second K-rigid motion in HG we will need a pair of functions   eϕ − e−ϕ eϕ + e−ϕ , sinh σ = cosh ϕ = 2 2 that parametrize the unit hyperbola z 2 − x2 = 1 in the same way that (cos ϕ, sin ϕ) parametrize the unit circle. That is, cosh2 ϕ − sinh2 ϕ ≡ 1, and if z02 − x20 = 1, then there is a ϕ such that z0 = cosh ϕ and x0 = sinh ϕ. Exercise 144 (HG). Show that the matrix ⎛ ⎞ 1/2 cosh ϕ 0 |K| · sinh ϕ ⎝ ⎠ 0 1 0 −1/2 · sinh ϕ 0 cosh ϕ |K| is K-orthogonal. Describe geometrically what this transformation is doing to the the K-geometry.

11.2. Hyperbolic geometry is homogeneous

117

11.2.2. Moving a point and vector to the North Pole by a rigid motion. So, first of all, the North Pole is the point N = (0, 0, 1) . Suppose we start with a point X0 = (x0 , y0 , z0 ) in the geometry, that is, satisfying equation (11.2.1). Exercise 145 (HG). Write an explicit K-rigid motion ⎛ ⎞ cos ϑ sin ϑ 0 M1 = ⎝ − sin ϑ cos ϑ 0 ⎠ 0 0 1 that takes the point X0 to a point X1 = (x1 , 0, z0 ). Exercise 146 (HG). Write an explicit K-rigid motion ⎞ ⎛ cosh ϕ 0 |K|1/2 · sinh ϕ ⎠ 0 1 0 M2 = ⎝ −1/2 · sinh ϕ 0 cosh ϕ |K| that takes the point X1 = (x1 , 0, z0 ) to N = (0, 0, 1). Hint: Notice that

2 Kx21 + z02 = 1 = − − |K|1/2 · x1 + z02 . So there is a ϕ with cosh ϕ = z0 and sinh ϕ = − |K|

1/2

· x1 .

Try that ϕ in M2 . Using these last two exercises, we conclude that the transformation   x, y, z = (x, y, z) · (M1 · M2 ) is a K-rigid motion (why?) and that N = (x0 , y0 , z0 ) · (M1 · M2 ) (why?). Let V2 = (a2 , b2 , 0) be a tangent vector to K-geometry at the North Pole N . Exercise 147 (HG). Write an explicit K-rigid motion ⎛ ⎞ sin ϑ 0 cos ϑ M3 = ⎝ − sin ϑ cos ϑ 0 ⎠ 0 0 1

118

11. The curvature K becomes negative

that takes V2 to the vector   

2 2 a2 + b2 , 0, 0 = V2 •K V2 , 0, 0 . Why does the transformation given by M3 leave the North Pole N fixed? 11.2.3. Moving a (point, direction) to any other (point, direction) by a rigid motion. Now suppose we have any point X0 = (x0 , y0 , z0 ) in K-geometry and any K-tangent vector V0 = (a0 , b0 , c0 ) at that point. Exercise 148 (HG). Explain why the K-rigid motion   x, y, z = (x, y, z) · (M1 · M2 · M3 ) constructed over √ the last couple  of sections takes the point X0 to N and the tangent vector V0 to V0 •K V0 , 0, 0 . Now suppose that (X0 , V0 ) gives a point X0 in K-geometry and a tangent direction V0 to K-geometry at X0 . Suppose that (X0 , V0 ) gives another point in K-geometry and a tangent direction to K-geometry at X0 . Finally suppose that V0 •K V0 = V0 •K V0 . As above, find a K-rigid motion given by M = (M1 · M2 · M3 ) √  taking X0 to the North Pole and V0 to V0 •K V0 , 0, 0 . Similarly find a K-rigid motion given by M  = (M1 · M2 · M3 ) 

taking X0 to the North Pole and V0 to V0 •K V0 , 0, 0 . Exercise 149 (HG). Explain why the K-rigid motion given by −1

M · (M  ) takes (X0 , V0 ) to (X0 , V0 ).

By completing this exercise we have shown that HG looks the same at each point and in each direction at that point. That is, we have shown that HG is homogeneous.

11.3. Lines in hyperbolic geometry

119

11.3. Lines in hyperbolic geometry 11.3.1. Hyperbolic coordinates, the shortest path from the North Pole. Next, we will figure out the shortest path you can take between two points in HG. Again we will do our calculation using only (x, y, z)-coordinates (since, as we have seen in (9.1.8), we don’t have (ˆ x, yˆ, zˆ)-coordinates). The (x, y, z)-coordinates for SG, namely x (σ, τ ) = R · sin σ · cos τ, y (σ, τ ) = R · sin σ · sin τ, z (σ, τ ) = cos σ, won’t work this time because they involve R which has gone off to infinity. Fortunately there are hyperbolic coordinates   eσ − e−σ eσ + e−σ , sinh σ = cosh σ = 2 2 that parametrize the “unit” hyperbola just like (cos σ, sin σ) parametrize the unit circle. So we define x (σ, τ ) = |K|−1/2 · sinh σ · cos τ, y (σ, τ ) = |K|−1/2 · sinh σ · sin τ, z (σ, τ ) = cosh σ. Exercise 150 (HG). Show that these hyperbolic coordinates do actually parametrize the K-geometry, that is, that

K x (σ, τ )2 + y (σ, τ )2 + z (σ, τ )2 ≡ 1 for all (σ, τ ). Again notice that you can write a path on the R-sphere by giving a path (σ (t) , τ (t)) in the (σ, τ )-plane. In fact, you can use σ as the parameter t and just write (σ, τ (σ)) , where τ is a function of σ. To write a path that starts at the North Pole, just write (σ, τ (σ)) , 0 ≤ σ ≤ ε and demand that τ (0) = 0. If you want the path to end on the plane y = 0, demand additionally that τ (ε) = 0. But if we are going to describe paths on HG by paths in the (σ, τ )-plane, we are going to need to figure out the K-dot-product in (σ, τ )-coordinates so that we can compute the lengths of paths in these coordinates.

120

11. The curvature K becomes negative

Exercise 151 (HG). a) Compute the 2 × 3 matrix Dhyp such that     dσ dτ dx dy dz , , , = · Dhyp dt dt dt dt dt when a path in K-geometry is given by a path in the (σ, τ )-plane. Hint: By the Chain Rule from several variable calculus   dx dσ dx dτ

Dhyp =

dy dσ dy dτ

dz dσ dz dτ

.

b) Use a) to compute the K-dot-product in (σ, τ )-coordinates, namely     dσ1 dτ1 dσ2 dτ2 , , •hyp dt dt dt dt     dx1 dy1 dz1 dx2 dy2 dz2 , , , , = •K dt dt dt dt dt dt ⎞ ⎛    t 1 0 0 dx2 dy2 dz2 dx1 dy1 dz1 ⎠ ⎝ 0 1 0 , , , , · = · dt dt dt dt dt dt 0 0 K −1 ⎞ ⎛    t 1 0 0 dσ1 dτ1 dσ2 dτ2 t ⎠ ⎝ 0 1 0 , , · Dhyp · = . · Dhyp · dt dt dt dt 0 0 K −1 Exercise 152 (HG). Show that the length L of any path in our K-geometry given by (σ, τ (σ)) , 0 ≤ σ ≤ ε with τ (0) = 0 and τ (ε) = 0 is given by the formula      t  ε  dτ dτ 1 0 −1/2 dσ. 1, · 1, · L = |K| 0 sinh2 σ dσ dσ 0 This last formula for L lets us figure out the shortest path from N = (sinh 0 · cos 0, R · sinh 0 · sin 0, cosh 0) to



−1/2 −1/2 −1/2 · sinh ε, 0, cosh ε = |K| · sin ε · cos 0, |K| · sin 0 · sin 0, cosh ε . |K| Since −1/2

L = |K|

 ε  dτ 2 · 1 + sinh2 σ · dσ dσ 0

dτ is and sinh σ is positive for almost all σ ∈ [0, ε], L is minimal only when dσ identically zero. But this means that τ (σ) is a constant function. Since τ (0) = 0, this means that τ (σ) is identically zero. So we have shown the following result. 2

11.3. Lines in hyperbolic geometry

121

Theorem 16 (HG). The shortest path in K-geometry from the North Pole to a −1/2 point (x, y, z) = |K| · sinh ε, 0, cosh ε is the path lying in the plane y = 0. The K-length of that shortest path is |K|−1/2 · ε. 11.3.2. Shortest path between any two points. Theorem 17 (HG). Given any two points X1 = (x1 , y1 , z1 ) and X2 = (x2 , y2 , z2 ) in K-geometry, the shortest path between the two points is the path cut out by   K x2 + y 2 + z 2 = 1 and the plane (11.3.1)

⎛ x ⎝ x1 x2

y y1 y2

⎞ z ⎠ z1 = 0, z2

that is, the plane containing (0, 0, 0), X1 , and X2 . Proof. Let V1 = (a1 , b1 , c1 ) be the unit tangent vector at X1 to the curve cut out by the plane (11.3.1) of K-length one. Then (x, y, z) = (a1 , b1 , c1 ) also satisfies equation (11.3.1), and so the equation for that plane can also be written ⎛ ⎞ x y z ⎝ x1 y1 z1 ⎠ = 0. (11.3.2) a1 b1 c1 By Exercise 131 there is a K-rigid motion M that takes X1 to the North Pole N and V1 to (1, 0, 0). So M takes the plane (11.3.2) to the plane given by the equation ⎛ ⎞ x y z ⎝ 0 0 1 ⎠ = 0, 1 0 0 namely the plane y = 0. So X2 · M must also lie in the plane y = 0 since X2 lies in the plane (11.3.2). So

X2 · M = |K|−1/2 · sinh ε, 0, cosh ε for some ε since all points in K-geometry with y = 0 can be written as

−1/2 · sinh ε, 0, cosh ε for some ε. Since M is a K-rigid motion, it must |K| take the shortest path from X1 to X

2 to the shortest path from X1 · M = N to X2 · M = |K|−1/2 · sinh ε, 0, cosh ε . And we already know that the shortest path from X1 · M to X2 · M is the one that lies in the plane y = 0. But that path comes from the path cut out by the plane given by equation (11.3.2), that is, the plane given by equation (11.3.1).  The path given by equation (11.3.1) is called the great hyperbolic arc between X1 and X2 .

122

11. The curvature K becomes negative

Exercise 153 (HG). Explain why lines in HG extend infinitely in each direction. Hint: There is a K-rigid

motion that takes any two points to (0, 0, 1) and −1/2 |K| · sinh ε, 0, cosh ε for some ε > 0. Why does that mean that the Kdistance between the two points is equal to |K|−1/2 · ε? Definition 23. A line in HG will be a curve that extends infinitely in each direction and has the property that given any two points X1 and X2 on the path, the shortest path between X1 and X2 lies along that curve. Lines in HG are the intersections of the K-geometry with planes through (0, 0, 0). The length of the shortest path between two points in K-geometry will be called the K-distance.

11.4. Central projection in HG 11.4.1. The edge of the universe. Again HG is a K-geometry in the sense of Part V since, in (x, y, z)-coordinates, the equation for the K-geometry is   (11.4.1) K x2 + y 2 + z 2 = 1 with K < 0, and the K-dot-product is a valid dot-product. So all the calculations in Part V hold, in particular (9.2.3). So the (xc , yc )-coordinates that parametrize the entire K-geometry are such that 1 . x2c + yc2 < |K| We call the circle 1 x2c + yc2 = |K| the edge of the universe. (The (xc , yc )-coordinates are called Klein coordinates and the disk of radius |K|−1/2 is called the Klein model for HG in honor of the famous German geometer, Felix Klein.) Exercise 154. a) Use formula (9.2.4) for K-rigid motions to check that for K-rigid motions, ⎛ ⎞⎛ ⎞ x 1   ⎠⎝ y ⎠ = 0 x y z ⎝ 1 −1 K z whenever ⎛ ⎞⎛ ⎞ 1 x   ⎠⎝ y ⎠ = 0 x y z ⎝ 1 K −1 z and vice versa. b) Recall that m13 x + m23 y + m33 z = z for the K-rigid motion given by the matrix M in (x, y, z)-coordinates. If K < 0, use a) to explain why (x, y, z) lies on  the light cone if and only if x, y, z lies on the light cone. c) Use z = m13 x + m23 y + m33 z

11.4. Central projection in HG

123

to conclude that the plane given by m13 x + m23 y + m33 z = 0 intersects the light cone only at (0, 0, 0). d) Show that the line given by m13 xc + m23 yc + m33 = 0 never intersects the edge of the universe in central projection coordinates (nor in stereographic projection coordinates), no matter the (negative) value of K. We will use this last exercise in what follows since it implies that a K-rigid motion in (xc , yc )-coordinates takes any tangent line to the edge of the universe circle to another tangent line to the edge of the universe circle.

11.4.2. Lines go to chords. Again all the calculations in Part V hold, in particular Exercise 118a). We conclude that lines in HG correspond to chords on the Klein (xc , yc )-disk that connect two points on the edge of the universe. Exercise 155 (HG). a) Explain why the K-line y = 0 is given by the xc -axis and the North Pole N is given by (xc , yc ) = (0, 0).

−1/2 b) Explain why the point |K| · sinh ε, 0, cosh ε in the K-geometry is given by the point  (xc , yc ) =

|K|−1/2 ·

 eε − e−ε , 0 . eε + e−ε

why the K-distance between (xc , yc ) = (0, 0) and (xc , yc ) = c) Explain −1/2 eε −e−ε −1/2 · eε +e−ε , 0 is |K| · ε. |K| Hint: Convert the statement to a statement about the distance between two points in (x, y, z)-coordinates, a distance that we have computed previously. Exercise 156 (HG). Use (xc , yc )-coordinates to show that HG satisfies the four Euclidean postulates E1, E2, E3, and E4. Thus hyperbolic geometry is a neutral geometry (NG). 11.4.3. K-perpendicularity in the Klein model for HG. Suppose we are given any three distinct points P  , R , and Q on the edge of the universe of the Klein K-disk. We construct the line L through P  and Q and mark a point A on it as shown in the following figure.

124

11. The curvature K becomes negative

We know that there is a K-rigid motion Mc that takes A to (0, 0) and L to the xc -axis. (Why?) Viewed as a transformation of the entire (xc , yc )-plane, this transformation takes the tangent line of the universe at P  to the tangent to the edge

−1/2 line to the edge of the universe at − |K| , 0 and the tangent line to the edge

of the universe at Q to the tangent line to the edge of the universe at |K|−1/2 , 0 .



Since the tangent lines at − |K|−1/2 , 0 and |K|−1/2 , 0 are vertical, the point S  must have gone to infinity under the K-rigid motion, and so the line through A and R must go to the yc -axis. Exercise 157. Explain why there is a K-rigid motion Mc that takes any three points P  , R , and Q in order along the edge of the universe to any other three points P  , R , and Q in order along the edge of the universe. Hint: Use the fact that the set of K-rigid motions forms a group under the composition operation. Exercise 158 (HG). Explain why the above discussion implies that the angles ∠P  A R and ∠Q A R must both be K-right angles; that is, their K-measures must each be 90°. So the line segments P  Q and A R are K-perpendicular [MJG, 238– 239]. Hint: You may there is a K-rigid motion that

the fact that since

need to use −1/2 −1/2 , 0 and |K| , 0 and leaves (0, 0) fixed, the xc -axis interchanges − |K| and the yc -axis are K-perpendicular. Exercise 159 (HG). Use the previous exercise and the fact that A can be any point along the chord P  Q in the figure above to explain why the Klein model is not conformal; that is, it does not faithfully represent the measure of angles in HG. Exercise 160. Use (xc , yc )-coordinates to show that HG does not satisfy Euclid’s postulate E5. That is, through a point not on a line, it is not true that there passes a unique parallel (i.e., nonintersecting) line. 11.4.4. Computing K-distances using cross-ratio in Klein coordinates. In fact, the tool that will let us compute all K-distances in (xc , yc )-coordinates is the

11.4. Central projection in HG

125

cross-ratio from Definition 17. Let dK (Ac , Bc ) denote the K-distance between two points Ac and Bc in the Klein K-disk. Now we know that    eε − e−ε dK (0, 0) , |K|−1/2 · ε , 0 = |K|−1/2 · ε. e + e−ε To see what this has to do with cross-ratio, we begin by computing the cross-ratio   ε −ε −1/2 −1/2 e − e −1/2 0 : − |K| : |K| · ε : |K| e + e−ε



−1/2 eε −e−ε −1/2 given by the two points (0, 0), |K| · eε +e−ε , 0 and the two points −|K| ,0

and |K|−1/2 , 0 where the xc -axis intersects the edge of the universe. Exercise 161 (HG). a) Draw a picture of the Klein K-disk, the edge of the universe, and the four points on the xc -axis. b) Show that     ε −ε eε − e−ε −1/2 −1/2 e − e −1/2 0 : − |K| : |K| · ε : |K| : 1 . = 0 : −1 : e + e−ε eε + e−ε In particular, notice that the computation doesn’t depend on K. c) Show that   eε − e−ε : 1 = e−2ε . 0 : −1 : ε e + e−ε From this exercise we conclude that    ε −ε −1/2 e − e · ε ,0 (11.4.2) dK (0, 0) , |K| e + e−ε   ε −ε |K|−1/2 −1/2 −1/2 e − e −1/2 = : |K| · ε : |K| · ln 0 : − |K| . 2 e + e−ε Now suppose we are given any two points Ac and Bc in the Klein K-disk. They determine a line (11.4.3)

αxc + βyc + γ = 0

and points Pc and Qc where that line meets the edge of the universe as shown in the figure below.

We are now ready to prove the following theorem.

126

11. The curvature K becomes negative

Theorem 18 (HG). For any two points Ac and Bc on the Klein K-disk, the K-distance dK (Ac , Bc ) between them is given by the formula |K|−1/2 · |ln (xc (Ac ) : xc (Pc ) : xc (Bc ) : xc (Qc ))| , 2 where Pc and Qc are the endpoints of the chord through Ac and Bc . (Compare with [MJG, 268].)   Proof. We know that there is a K-rigid motion xc , yc = Mc (xc , yc ) of the Klein

−1/2 eε −e−ε disk that takes Ac to (0, 0) and Bc to some point |K| · eε +e−ε , 0 on the positive xc -axis. From (9.2.4) we know that m11 xc + m21 yc + m31 . xc = m13 xc + m23 yc + m33 dK (Ac , Bc ) =

But from (11.4.3) we know that for our four points Ac , Bc , Pc , and Qc , αxc + βyc + γ = 0 −αxc − γ yc = . β So if we calculate Mc only for these four points, we have

m11 xc + m21 −αxβc −γ + m31

xc = m13 xc + m23 −αxβc −γ + m33



m21 γ α + m − x m11 − m21 c 31 β β

. = m23 γ m23 α m13 − β xc + m33 − β That is, the function xc → xc is a linear fractional transformation! So by Theorem 13,   (xc (Ac ) : xc (Pc ) : xc (Bc ) : xc (Qc )) = xc (Ac ) : xc (Pc ) : xc (Bc ) : xc (Qc )   ε −ε −1/2 −1/2 e −e −1/2 : |K| · ε −ε : |K| = 0 : −|K| e +e = e−2ε . Therefore

   ε −ε −1/2 e − e dK (Ac , Bc ) = dK (0, 0) , |K| · ε ,0 e + e−ε  −1/2  |K| eε − e−ε −1/2 · ln 0 : − |K|−1/2 : |K|−1/2 · ε = : |K| 2 e + e−ε =

|K|−1/2 · |ln (xc (Ac ) : xc (Pc ) : xc (Bc ) : xc (Qc ))| . 2 

Exercise 162. For K = −1, calculate the K-distance between the two points given in (xc , yc )-coordinates by (0, 0) and (1/2, 0).

11.4. Central projection in HG

127

11.4.5. Areas of hyperbolic lunes. Finally, there is one K-area computation that is convenient to do in Klein coordinates. Namely, suppose that we take vertices (0, 0), the ordinary triangle T c in the (xc , yc )-plane with

−1/2 −1/2 −1/2 −1/2 |K| cos β, |K| sin β , and |K| cos β, − |K| sin β . Notice that two of the three vertices lie on the edge of the universe of the Klein K-disk and that the K-angle at the third vertex is α = 2β. We will call the interior of this triangle, or any K-rigid motion of it, an α-lune. So we wish to compute the K-area of an α-lune. Since HG is a K-geometry, we know from Exercise 113 that this area AK (α) is given by the formula  1 dxc dyc . AK (α) = 2 3/2 Tc (1 − |K| (x2 c + yc )) Exercise 163. Show that AK (α) = |K|

−1

(π − α) .

Hint: Use the substitution xc = |K|

1/2

xc ,

yc = |K|

1/2

yc

to reduce the computation to the computation in the case that |K| = 1. Then use polar coordinates to get  β  ϑ=β  r= cos cos ϑ 1 r · dr dϑ. A−1 (α) = r=0 ϑ=−β (1 − r 2 )3/2 Then do the substitution u = 1 − r2 , du = −2rdr to compute

cos β  r= cos ϑ

r=0

1 r (1−r 2 )3/2

· dr. In the final step use the substitution t = sin ϑ

to reduce to an integral of the form 

a−1

 2 dt. 1 − at

Definition 24. For K < 0, let TK denote the region in K-geometry given by the triangle Tc whose area is calculated in Exercise 163. An α-lune in K-geometry is any region in K-geometry that is obtained from the region TK by a K-rigid motion. So the K-area of any α-lune is |K|−1 (π − α).

128

11. The curvature K becomes negative

11.5. Stereographic projection in HG 11.5.1. The Poincaré K-disk and the edge of the universe. Under stereographic projection, the center of projection is the South Pole (0, 0, −1). So if K < 0 and a point (x, 0, z) on the K-geometry goes out the hyperboloid to infinity while continuing to lie on the (x, z)-plane, the line joining (0, 0, −1) to that point becomes parallel to an asymptote of the hyperbola Kx2 + z 2 = 1. 1/2

in the (x, z)-plane, and therefore that So the line approaches a line of slope ± |K| 1/2 limit line is the line z = ± |K| x − 1. Therefore the intersection of that line with ±2 the line z = 1 in the (x, z)-plane approaches the point with x = |K| 1/2 . Therefore under stereographic projection, the edge of the universe is given by the circle x2s + ys2 =

4 . |K|

The interior of this circle, that is, the image of K-geometry under stereographic projection, is called the Poincaré model of hyperbolic geometry, of course after the famous geometer, Henrí Poincaré. Again, since HG is a K-geometry, all the rules of Part V apply. So by Exercise 118b), lines in the K-geometry are given by circles of the form    2  2 4 K + a2 + b2 2a 2b + ys − = (11.5.1) xs − K K K2 in the Poincaré K-disk. The darker arc below

represents the K-line ax + by + 1 = 0 in the Poincaré K-disk. 11.5.2. Stereographic projection preserves angles. Exercise 164 (HG). a) Show that stereographic projection is conformal, that is, that the measure of K-angles between K-lines on K-geometry is just the ordinary Euclidean measure of angles formed by their (usually circular) stereographic projections. Hint: See subsection 9.3.3.

11.5. Stereographic projection in HG

129

b) For K = −1, construct the K-line in (xs , ys )-coordinates that meets the K-line (xs − 2)2 + (ys − 2)2 = 4 √  √  perpendicularly at the point 2 − 2, 2 − 2 . Exercise 165. Show that in K-geometry for any K, the angle between two tangent vectors at the North Pole is the same as the ordinary Euclidean angle between the two corresponding tangent vectors in the (xc , yc )-plane, and that angle is also the same as the ordinary Euclidean angle between the two corresponding tangent vectors in the (xs , ys )-plane. To get a more precise idea of what K-lines look like under stereographic projection, again consider equation (11.5.1) and the picture below it. The equations of the circles in the picture are (11.5.2) and (11.5.3)

x2s + ys2 =

4 |K|

   2  2 4 K + a2 + b2 2a 2b xs − + ys − = . K K K2

  2b Construct a third circle whose diameter is the line segment from (0, 0) to 2a K, K ,   2  a 2  b 2  a 2 namely the circle xs − K + ys − Kb = K + K which can be rewritten (11.5.4)

x2s + ys2 −

2a 2b xs − ys = 0. K K

Lemma 8. The circles (11.5.2), (11.5.3), and (11.5.4) all pass through two common points. Proof. From (11.5.2) and (11.5.3) we get by addition that    2  2 4 K + a2 + b2 2a 2b 4 2 2 + xs + ys + xs − + ys − = . K K |K| K2 Simplifying this last equation and dividing both sides by two, we obtain equation (11.5.4). So the two points P  and Q in picture (11.5.1) that satisfy both equations (11.5.2) and (11.5.3) also satisfy equation (11.5.4).  The lemma tells us that the angle formed by the segments (0, 0) P  and  2a 2b  P  K , K is a right angle since it is an inscribed angle in the circle (11.5.4) whose associated central angle is a diameter of that circle. But (0, 0) P  is a radius of circle     2b  2a , 2b is a (11.5.2) and so P  2a K , K is tangent to circle (11.5.2). Similarly P K K radius of circle (11.5.3) and so (0, 0) P  is tangent to circle (11.5.3). So we conclude the following theorem. Theorem 19 (HG). In the Poincaré model for K-geometry, the K-lines are represesented by circular arcs that meet the edge of the universe perpendicularly.

130

11. The curvature K becomes negative

11.5.3. Infinite triangles in the Poincaré K-disk. By Exercise 118a) and 118b), lines in HG become circles under stereographic projection unless the line in HG passes through the North Pole (in which case it corresponds to a line through (xs , ys ) = (0, 0) in the (xs , ys )-plane). Suppose a hyperbolic triangle T corresponds to a region Ts in (xs , ys )-coordinates and the vertices of T correspond to (xs , ys ) = (−2, 0), (xs , ys ) = (2, 0), and (xs , ys ) = (0, 2). So one side of Ts lies on the line ys = 0. Exercise 166 (HG). a) Use Exercise 118b) to compute the equations for the other two sides of Ts . b) In the (xs , ys )-plane, draw Ts as accurately as you can when K = − 14 , then when K = −1. The area of a hyperbolic triangle T is given by the formula  1   dxs dys . K 2 2 2 Ts 1 + 4 (xs + ys ) However, we do not yet have a way to calculate the area numerically for any given triangle T . The last topic in this book will remedy that situation. Analogous to the case of spherical triangles, we start from the fact that we do know the area of α-lunes. From Exercise 163 the K-area of an α-lune with one vertex at (0, 0) in (xc , yc )-coordinates is |K|

−1

(π − α) .

Exercise 167. Draw the α-lune in Exercise 163 in (xs , ys )-coordinates. Since rigid motions preserve K-lengths and angles, and since stereographic projection is conformal, the ordinary Euclidean measure of the angle of the figure in Exercise 167 is α. Since rotation of the (xc , yc )-plane around (0, 0) is a K-rigid motion, this formula holds for any K-lune with vertex at (0, 0). Now by Exercise 165 we can represent the same lune using the same angle in the (xs , ys )-plane. Below is a picture in the (xs , ys )-plane of some of these K-lunes.

11.5. Stereographic projection in HG

131

Exercise 168 (HG). Use Exercise 163 to show that in the above picture the Karea that lies in the union of the α-lune and the β-lune but does not lie in the (α + β)-lune is |K|−1 π. Exercise 169. Move the vertex of the α-lune in Exercise 167 to any other point of K-geometry by a rigid motion in (xs , ys )-coordinates. Draw the resulting figure (that we will continue to call an α-lune). Definition 25 (HG). An infinite K-triangle is the figure given in stereographic projection coordinates by the stereographic projection of three K-lines such that any two meet the edge of the universe at a common point. Exercise 170. a) (HG) Use Exercise 157 to show that the area of (the interior of) any infinite triangle has K-area −1

|K|

· π.

For example, if K = −1, we have

b) Use a) to give a formula for the K-area of any infinite n-gon in HG, that is, a figure described by a set of n disjoint K-lines that is the limit of a family of finite n-gons, all of whose vertices have gone to infinity. In particular, what is the area of any infinite hexagon? Hint: Divide the infinite n-gon into infinite triangles. 11.5.4. Areas of polygons in HG. Consider the picture below in the Poincaré model for HG. Locate the infinite hexagon whose vertices are the points at infinity on the lines expending the sides of the hyperbolic triangle with interior angles α, β, and γ.

132

11. The curvature K becomes negative

Exercise 171 (HG). a) Use the picture

and remarks just above to explain why the K-area of the hyperbolic triangle is |K|−1 · (π − (α + β + γ)) . Hint: Locate α-lunes, two β-lunes, and two γ-lunes in the picture and notice that they cover the hyperbolic triangle three times. b) Use a) to give a formula for the K-area of a hyperbolic n-gon.

Definitions

Angle: A figure formed by two rays, called the sides of the angle, sharing a common endpoint, called the vertex of the angle. “Angle.” Wikipedia: The Free Encyclopedia. Wikimedia Foundation, Inc. 13 July 2018. Web. 30 July 2018. Area: The quantity that expresses the extent of a two-dimensional figure or shape. “Area.” Wikipedia: The Free Encyclopedia. Wikimedia Foundation, Inc. 25 July 2018. Web. 30 July 2018. Axiom: An axiom or postulate is a statement that is taken to be true, to serve as a premise or starting point for further reasoning and arguments. “Axiom.” Wikipedia: The Free Encyclopedia. Wikimedia Foundation, Inc. 22 July 2018. Web. 30 July 2018. Bijective mapping: A mapping that is both one-to-one and onto. Circle: The set of all points on a two-dimensional surface that are at a given distance from a given point called the center. Congruence: The relationship between two figures that have the same size and shape. Coordinate system: A system which uses one or more numbers, or coordinates, to uniquely determine the position of the points or other geometric elements on a manifold such as Euclidean space. “Coordinate system.” Wikipedia: The Free Encyclopedia. Wikimedia Foundation, Inc. 1 July 2018. Web. 30 July 2018. Counterexample: A counterexample is an exception to a proposed general rule or law showing that it does not apply in certain cases. 133

134

Definitions

“Counterexample.” Wikipedia: The Free Encyclopedia. Wikimedia Foundation, Inc. 1 July 2018. Web. 30 July 2018. Curvature: The amount by which a geometric object such as a surface deviates from being a flat plane, or a curve from being straight as in the case of a line, but this is defined in different ways depending on the context. “Curvature.” Wikipedia: The Free Encyclopedia. Wikimedia Foundation, Inc. 26 May 2018. Web. 30 July 2018. Differentiable: The property of a function of one or several variables that says that its graph is smooth at each point at which it is defined. Dilation: A self-transformation of the plane that fixes one point called the center of the dilation and moves every other point without changing its direction from the center but changing its distance from the center by a fixed positive ratio called the magnification factor of the dilation. Dimension: The dimension of a mathematical space (or object) is informally defined as the minimum number of coordinates needed to specify any point within it. “Dimension.” Wikipedia: The Free Encyclopedia. Wikimedia Foundation, Inc. 30 June 2018. Web. 30 July 2018. Distance: Distance is a numerical measurement of how far apart objects are. “Distance.” Wikipedia: The Free Encyclopedia. Wikimedia Foundation, Inc. 10 June 2018. Web. 30 July 2018. Geodesic: The shortest path between two points on a flat or curved space. More precisely, it is a path with the property that if two of its points are sufficiently close, the shortest path between them in the space is the segment they define on the given path. Geometry: A branch of mathematics concerned with questions of shape, size, relative position of figures, and the properties of space. “Geometry.” Wikipedia: The Free Encyclopedia. Wikimedia Foundation, Inc. 10 June 2018. Web. 30 July 2018. (If a particular space or set of points and rules for distance and angle are specified, this particular collection of objects and rules can be called a “geometry”.) Homogeneous: The property that defines a geometry that is the same at all points and in all directions. Injective or One-to-one: The property of a function or mapping f that f (x) = f (y) implies that x = y. Isometry: A distance-preserving transformation between...spaces, usually assumed to be bijective. “Isometry.” Wikipedia: The Free Encyclopedia. Wikimedia Foundation, Inc. 3 May 2018. Web. 31 July 2018.

Definitions

135

Light cone: The path that a flash of light, emanating from a single event (localized to a single point in space and a single moment in time) and traveling in all directions, would take through spacetime. “Light cone.” Wikipedia: The Free Encyclopedia. Wikimedia Foundation, Inc. 12 May 2018. Web. 31 July 2018. Limit: The value that a function (or sequence) “approaches” as the input (or index) “approaches” some value. “Limit.” Wikipedia: The Free Encyclopedia. Wikimedia Foundation, Inc. 13 June 2018. Web. 31 July 2018. Line or Straight line: A concept introduced by ancient mathematicians to represent straight objects (i.e., having no curvature) with negligible width and depth. Lines are an idealization of such objects. Until the 17th century, lines were defined in this manner. In a Cartesian plane...lines can be described algebraically by linear equations. “Line (geometry).” Wikipedia: The Free Encyclopedia. Wikimedia Foundation, Inc. 21 May 2018. Web. 31 July 2018. Linear transformation: Let V and W be vector spaces over the field of real numbers. A function f :V →W is said to be a linear transformation if for any two vectors v ∈ V and w ∈ V , f (v + w) = f (v) + f (w), and for any vector v ∈ V and real number c, f (c · v) = c · f (v). “Linear map.” Wikipedia: The Free Encyclopedia. Wikimedia Foundation, Inc. 25 March 2018. Web. 30 July 2018. Line segment: A part of a line that is bounded by two distinct end points, and contains every point on the line between its endpoints. “Line segment.” Wikipedia: The Free Encyclopedia. Wikimedia Foundation, Inc. 24 April 2018. Web. 31 July 2018. Orthogonal: Two Euclidean vectors are orthogonal if they are perpendicular, i.e., they form a right angle. “Orthogonality.” Wikipedia: The Free Encyclopedia. Wikimedia Foundation, Inc. 9 July 2018. Web. 30 July 2018. Parallel: In geometry, parallel lines are lines in a plane which do not meet; that is, two lines in a plane that do not intersect or touch each other at any point are said to be parallel. By extension, a line and a plane, or two planes, in three-dimensional Euclidean space that do not share a point are said to be parallel. However, two lines in three-dimensional space which do not meet must be in a common plane to be considered parallel; otherwise they are called skew lines. Parallel planes are planes in the same three-dimensional space that never meet.

136

Definitions

“Parallel (geometry).” Wikipedia: The Free Encyclopedia. Wikimedia Foundation, Inc. 26 April 2018. Web. 30 July 2018. Perpendicular: The property of being perpendicular (perpendicularity) is the relationship between two lines which meet at a right angle (90 degrees). “Perpendicular.” Wikipedia: The Free Encyclopedia. Wikimedia Foundation, Inc. 23 July 2018. Web. 30 July 2018. Plane: A flat, two-dimensional surface that extends infinitely far in every direction. Projection: A mapping from a space onto a subspace of one less dimension with the property that the set of points mapping to a given point is a line through the given point in such a way that all the lines are parallel or all the lines pass through a common point in the larger space. Ray: A subset of a line consisting of all points on the line on one side of a fixed point called the vertex of the ray. Rigid motion: A bijective isometry. Similarity: Two geometrical objects are called similar if they both have the same shape, or one has the same shape as the mirror image of the other. More precisely, one can be obtained from the other by uniformly scaling (enlarging or reducing). “Similarity (geometry).” Wikipedia: The Free Encyclopedia. Wikimedia Foundation, Inc. 22 July 2018. Web. 31 July 2018. Sphere: The set of points that are all at the same distance r from a given point...in a three-dimensional space. “Sphere.” Wikipedia: The Free Encyclopedia. Wikimedia Foundation, Inc. 12 July 2018. Web. 31 July 2018. Surjective or Onto: The property of a function or mapping f : X → Y that for all y ∈ Y , there is an x ∈ X such that f (x) = y. Symmetry: A geometric object has symmetry if there is an “operation” or “transformation” (...an isometry...) that maps the figure/object onto itself; i.e., it is said that the object has an invariance under the transform. “Symmetry (geometry).” Wikipedia: The Free Encyclopedia. Wikimedia Foundation, Inc. 27 July 2018. Web. 31 July 2018. Transformation: A function f that maps a set X to itself, i.e., f : X → X. “Transformation (function).” Wikipedia: The Free Encyclopedia. Wikimedia Foundation, Inc. 21 July 2018. Web. 31 July 2018. Vector: [A mathematical object that] specifies the change in position of a point relative to a previous position. “Vector (mathematics and physics).” Wikipedia: The Free Encyclopedia. Wikimedia Foundation, Inc. 18 April 2018. Web. 31 July 2018. Vertex: A vertex (plural: vertices or vertexes) is a point where two or more curves, lines, or edges meet.

Definitions

137

“Vertex.” Wikipedia: The Free Encyclopedia. Wikimedia Foundation, Inc. 21 May 2018. Web. 30 July 2018. Volume: The quantity that expresses the extent of a three-dimensional figure or shape. World sheet: A two-dimensional manifold [in string theory] which describes the embedding of a string in spacetime. “Worldsheet.” Wikipedia: The Free Encyclopedia. Wikimedia Foundation, Inc. 5 January 2018. Web. 30 July 2018.

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139

Index

Addition Law for Sines, 38 addresses for the same point in K-geometry, 94 angle, 10, 54, 81, 99, 115 area, 54, 127 area in central projection coordinates, 88 area in stereographic projection coordinates, 94 area of a smooth surface, 59 axiom, 9 cartesian coordinate system, 20 Cavalieri’s principle, 42 central projection, 83 central projection coordinates, 83 Ceva’s theorem, 30 Chain Rule, 4, 73 chord, 123 Common Core State Standards in Mathematics, 2 concurrence theorems, 29 conformal, 93, 128 congruence, 11, 100 cross-ratio, 124 cross-ratio of points on a circle, 35 cube, 41 curvature, 3, 72, 73 determinant, 55 dilation, 24 disco-ball, 43 displacement, 52 distance, 21, 51

dot-product, 53 edge of the universe, 122, 124 EG, 5 Euclid’s Elements, 9 Euclidean coordinates, 11 Euclidean plane, 20 Euclidean three-space, 51 exterior angle, 12 group, 64, 66, 82 HG, 5 homogeneous, 2, 101, 103, 115 hyperbolic geometry, 2 hyperbolic triangle, 132 hypotenuse, 21 infinite hexagon, 131 infinite K-triangle, 131 instantaneous velocity, 56 K-congruence, 82 K-coordinates, 79 K-distances, 124 K-dot-product, 75 K-geometry, 3, 73, 83 K-lines, 129 K-orthogonal, 82 K-orthogonal matrices, 100 K-orthogonal transformations, 100 K-perpendicularity, 123 K-right angles, 124 K-rigid motion, 82, 116 Klein model, 123

141

142

Law of Cosines, 23 Law of Sines, 22 length, 75, 105 length and angle in central projection coordinates, 85 length and angle in stereographic projection coordinates, 91 length of a smooth curve, 57 light cone, 113 line, 9, 104, 106, 122 linear algebra, 4 linear fractional transformation, 65, 126 linear transformation, 62 lines in hyperbolic geometry, 119 lines in spherical geometry, 104 lune, 46, 127, 130 magnification factor, 24 magnification principle, 42 matrix notation, 53 measure, 10 n-gon, 14, 47, 132 neutral geometry, 10 NG, 5 North Pole, 71 orthogonal matrix, 63 parallel, 10 Parallel Postulate, 19 parallelogram, 54, 75 perpendicular, 20 plane geometry, 9 plane section, 88, 93 Poincaré K-disk, 128 Poincaré model, 128, 131 point, 52 position, 52 Ptolemy’s theorem, 37 pyramid, 41 Pythagorean theorem, 21 quadrilateral, 14 R-sphere, 72 ray, 10 rectangle, 14, 20 regular n-gon, 44 Riemannian geometry, 2, 71 right angle, 10 rigid motion, 11, 61, 80, 99, 100

Index

rigid motion in central projection coordinates, 85 segment, 9 SG, 5 shortest path, 104, 106, 121 similar, 26 slope, 20 smooth curve, 56 spherical area, 107 spherical coordinates, 60 spherical geometry, 99 spherical triangles, 46, 108 SSS, SAS, and ASA, 11 stereographic projection, 89, 108 stereographic projection coordinates, 89 Substitution Rule, 5 surface area, 43 surface area of the sphere, 44 tangent vector, 56 transformations, 61 transversal line, 12, 19 triangle, 12 triangle inequality, 22 vector, 52 vertical angle, 12 volume of a pyramid, 43 world sheet, 113

Published Titles in This Series 34 33 32 31

C. Herbert Clemens, Two-Dimensional Geometries: A Problem-Solving Approach, 2019 Brad G. Osgood, Lectures on the Fourier Transform and Its Applications, 2019 John M. Erdman, A Problems Based Course in Advanced Calculus, 2018 Benjamin Hutz, An Experimental Introduction to Number Theory, 2018

30 Steven J. Miller, Mathematics of Optimization: How to do Things Faster, 2017 29 Tom L. Lindstrøm, Spaces, 2017 28 Randall Pruim, Foundations and Applications of Statistics: An Introduction Using R, Second Edition, 2018 27 Shahriar Shahriari, Algebra in Action, 2017 26 25 24 23

Tamara J. Lakins, The Tools of Mathematical Reasoning, 2016 Hossein Hosseini Giv, Mathematical Analysis and Its Inherent Nature, 2016 Helene Shapiro, Linear Algebra and Matrices, 2015 Sergei Ovchinnikov, Number Systems, 2015

22 21 20 19

Hugh L. Montgomery, Early Fourier Analysis, 2014 John M. Lee, Axiomatic Geometry, 2013 Paul J. Sally, Jr., Fundamentals of Mathematical Analysis, 2013 R. Clark Robinson, An Introduction to Dynamical Systems: Continuous and Discrete, Second Edition, 2012 18 Joseph L. Taylor, Foundations of Analysis, 2012 17 Peter Duren, Invitation to Classical Analysis, 2012 16 Joseph L. Taylor, Complex Variables, 2011

15 Mark A. Pinsky, Partial Differential Equations and Boundary-Value Problems with Applications, Third Edition, 1998 14 Michael E. Taylor, Introduction to Differential Equations, 2011 13 Randall Pruim, Foundations and Applications of Statistics, 2011 12 John P. D’Angelo, An Introduction to Complex Analysis and Geometry, 2010 11 10 9 8 7 6 5 4

Mark R. Sepanski, Algebra, 2010 Sue E. Goodman, Beginning Topology, 2005 Ronald Solomon, Abstract Algebra, 2003 I. Martin Isaacs, Geometry for College Students, 2001 Victor Goodman and Joseph Stampfli, The Mathematics of Finance, 2001 Michael A. Bean, Probability: The Science of Uncertainty, 2001 Patrick M. Fitzpatrick, Advanced Calculus, Second Edition, 2006 Gerald B. Folland, Fourier Analysis and Its Applications, 1992

3 Bettina Richmond and Thomas Richmond, A Discrete Transition to Advanced Mathematics, 2004 2 David Kincaid and Ward Cheney, Numerical Analysis: Mathematics of Scientific Computing, Third Edition, 2002 1 Edward D. Gaughan, Introduction to Analysis, Fifth Edition, 1998

The first half of the text covers plane geometry without and with Euclid’s Fifth Postulate, followed by a brief synthetic treatment of spherical geometry through the excess angle formula. This part only requires a background in high school geometry and basic trigonometry and is suitable for a quarter course for future high school geometry teachers. A brief foray into the second half could complete a semester course. The second half of the text gives a uniform treatment of all the complete, simply connected, two-dimensional geometries of constant curvature, one geometry for each real number (its curvature), including their groups of isometries, geodesics, measures of lengths and areas, as well as formulas for areas of regions bounded by polygons in terms of the curvature of the geometry and the sum of the interior angles of the polygon. A basic knowledge of real linear algebra and calculus of several (real) variables is useful background for this portion of the text.

For additional information and updates on this book, visit www.ams.org/bookpages/amstext-34

AMSTEXT/34

Sally

The

SERIES

This series was founded by the highly respected mathematician and educator, Paul J. Sally, Jr.

Photo Credit: John Schoger

This book on two-dimensional geometry uses a problemsolving approach to actively engage students in the learning process. The aim is to guide readers through the story of the subject, while giving them room to discover and partially construct the story themselves. The book bridges the study of plane geometry and the study of curves and surfaces of non-constant curvature in three-dimensional Euclidean space. One useful feature is that the book can be adapted to suit different audiences.