46243922 Geometry From a Differentiable Viewpoint

John McCleary

Geometry from a Differentiable Viewpoint

Geometry from a Differentiable Viewpoint JOHN McCLEARY Vassar College

CAMBRIDGE UNIVERSITY PRESS

PUBLISHED BY THE PRESS SYNDICATE OF THE UNIVERSITY OF CAMBRIDGE

The Pitt Budding, Trumpington Street, Cambridge CB2 IRP, United Kingdom CAMBRIDGE UNIVERSITY PRESS

The Edinburgh Building, Cambridge CB2 2RU, United Kingdom 40 West 20th Street, New York, NY 10011-4211, USA 10 Stamford Road, Oakleigh, Melbourne 3166, Australia ® Cambridge University Press 1994

This book is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 1994 Reprinted 1996, 1997

Printed in the United States of America Typeset in Garamond

A catalogue record for tlus book is available from the British Library Library of Congress Cataloguing-in-Publication Data is available ISBN 0-521-41430-X hardback ISBN 0-521-42480-1 paperback

To my sisters Marv Ann, Denise, and Rose

Contents

page ix

Introduction PART A

I. 2.

3.

4.

5.

Prelude and themes: Synthetic methods and results Spherical geometry Euclid Euclid's theory of parallels Appendix. The Elements: Book 1

3

10 16

21

The theory of parallels Uniqueness of parallels Equidistance and boundedness of parallels On the angle sum of a triangle Similarity of triangles Non-Euclidean geometry 1 The work of Saccheri The work of Gauss, Bolyai, and Lobachevskii Non-Euclidean geometry 11 The circumference of a circle

24 24 26 28 31

34 34 39

45 56

PART B

Development: Differential geometry 6.

Curves

Early work on plane curves (Huygens, Leibniz. Newton, Euler) The tractrix Directed curvature Digression: Involutes and evolutes

7.

Curves in space Appendix: On Euclidean rigid motions

8.

Surfaces The tangent plane The first fundamental form Area

8°i.

Map projections Stereographic projection Vii

viii

Contents

Central projection Mercator projection Lambert's cylindrical projection Azimuthal projection Sample map projections 9.

126 126 127 131

131

134

Metric equivalence of surfaces Special coordinates

11.

Geodesics

13.

124

Curvature for surfaces Euler's work on surfaces The Gauss map

10.

12.

123

145 151

157

Euclid revisited I: The Hopf-Rinow theorem The Gauss-Bonnet theorem Euclid revisited I1: Uniqueness of lines Compact surfaces A digression on curves Constant-curvature surfaces Euclid revisited III: Congruences The work of Minding

165 171

175

176

180 186 191

192

PART C

Recapitulation and coda 14.

Abstract surfaces

201

Hilbert's theorem

203

Abstract surfaces

206 217 220 224 227

15.

Modeling the non-Euclidean plane The Beltrami disk The Poincar6 disk The Poincar6 half-plane

16.

Epilog: Where from here? Manifolds (differential topology) Vector and tensor fields Metrical relations (Riemannian manifolds) Curvature Covariant differentiation

242 243 247 249 252 261

Riemann's Habilitationsvortrag: On the hypotheses which lie at the foundations of geometry Appendix: Notes on selected exercises

269 279

Bibliography

297

Symbol index Name index

303 304

Subject index

305

Introduction ArEaMETPHTOE MHAEIE E!ElTn. Over the entrance to Plato's Academy

One of the many roles of history is to tell a story. The history of the Parallel Postulate is a great story - it spans more than two millennia, stars an impressive cast of characters, and contains some of the most beautiful results in all of mathematics. My immodest goal for this book is to tell this story. Another role of history is to focus our attention and so to provide a thread of unity through a parade of events, people, and ideas. My goal grows small and quite modest before all of Geometry, especially its recent history. A more modest goal then is to provide a focus in which to view the standard tools of differential geometry, and in so doing offer an exposition, motivated by the history, that prepares the reader for the modern, global foundations of the subject. In recent years, it has become a luxury to offer a course in differential geometry in an undergraduate curriculum. When such a course exists, its students often arrive with a modern introduction to analysis, but without having seen geometry since high school. In the United States geometry taught in high schools is generally elementary Euclidean geometry based on Hilbert's axiom scheme. The beautiful world of non-Euclidean geometry is relegated to a footnote, enrichment material, or a "cultural" essay. This is also the case in most current introductions to differential geometry. The modern subject turns on problems that have emerged from the new foundations that are far removed from the ancient roots of geometry. When we teach the new and cut off the past, students are left to find their own way to a meaning of geometry in differential geometry, or to identify their activity as something different, unconnected.

This book is an attempt to carry the reader from the familiar Euclid to the state of development of differential geometry at the beginning of the twentieth century. One narrow

thread that runs through this vast historical period is the search for a proof of Euclid's Postulate V. the Parallel Postulate, and the eventual emergence of a new and non-Euclidean geometry. In the course of spinning out this tale, another theme enters - the importance of properties of a surface that are intrinsic, that is, independent of the manner in which the surface is embedded in space. This idea, emphasized by Gauss, provides an analytic key concerning the properties that are really geometric, and it introduces new realms to explore. The book is written in sonata-allegro form. Pan A opens with a prelude - a small dose of spherical geometry, in which some of the important ideas of non-Euclidean geometry are touched on. One of the main themes of the sonata is given in Chapters 2 and 3, which ix

X

Introduction

treat Book I of Euclid's Elements, one of the most important works of Western Culture, and the criticism of Euclid's theory of parallels. The rest of the main themes are found in Chapters 4 and 5, which are an exposition of synthetic non-Euclidean geometry as introduced by Saccheri, Gauss, Bolyai, and Lobachevskii. I have tried to follow the historical path in Part A which is based on the accounts in Gray (1979) and Rosenfeld (1988).

A word is in order here about the level of rigor employed in the text. If there were a real-valued measure of rigor, then it would be seen to be monotonically increasing through Part A. In Chapter I. I take a deliberately informal tone to invite the reader to glimpse a new world without the burden of exacting detail. What fails to be completely proved here will be established in Part B with the methods of differential geometry. I have taken the same tone as Euclid, his critics, and the founders of non-Euclidean geometry in the rest of Part A. with a little anachronistic help from Hilbert for those seeking rock-solid foundations. The intuitive power of the geometers of the ancient to premodern era is remarkable from the hindsight of more formal studies. The excitement of their achievements goes further without the burden of complete rigor. What remains unresolved at the end of Part A is the existence of a concrete representation

of non-Euclidean geometry, that is, a rigorous model. The development section of this sonata is dedicated to making this possible. One of the key features leading to the founding

of non-Euclidean geometry is the introduction of analysis into the development of the foundations. Trigonometric formulas such as the Lobachevskii-Bolyai theorem (5.8) and their proofs reveal the basic role that analytic notions play in geometry. Trust in analysis is what convinced Lobachevskii and Bolyai of the correctness of their ideas, and it is to analysis that we turn in order to obtain the means to build the non-Euclidean plane. Part B begins with curves, a success story through the introduction of appropriate coordinates and measures such as curvature and torsion. There is a brief interlude in Chapter 6 where the story of involutes, evolutes. and Hugyens's clock is told. Though it is out of the flow of the main story, Hugyens's work is paradigmatic fordifferential geometry -questions of an applied nature press the geometer to find new ways to think about basic notions. R3. Another interlude follows on map Chapter 8 presents the basic theory of surfaces in projections, a particular application of the definitions and apparatus associated to a surface, in this case, the sphere. Chapters 9 and 10 develop the analog of curvature for curves on a surface. This curvature, the Gaussian curvature, is shown to be independent of the manner

in which the surface lies in R3, that is, it is an intrinsic feature of the surface. A new theme emerges from these considerations - in order to know what is truly geometric about a surface, look to what is intrinsic. Guided by this idea, in Chapter I1 we introduce geodesics, that is, "lines" on a surface. Integrals of the Gaussian curvature are computed in Chapter 12 leading to the Gauss-Bonnet theorem and its global consequences. In Chapter 13 we finally

obtain an analytic recipe for our desired non-Euclidean plane - it is a complete, simply connected surface of constant negative Gaussian curvature. The final part of the sonata is a recapitulation of themes from Pan A. The development spills over to this section in the discussion of Hilbert's theorem - there are no complete, simply connected surfaces of constant negative Gaussian curvature in R3. This leads us to consider something more general. abstract surfaces, an idea that is possible if the intrinsic properties are our guide. Chapter 15 is the climax of the sonata where the models

Introduction

xi

of non-Euclidean geometry are constructed and their properties discussed. Chapter 16 is a coda on the theme of the intrinsic; Riemann's visionary lecture of 1854 is discussed along with the developments it motivated in differential geometry up to the turn of the century. This includes the modem idea of an n-dimensional manifold, Riemannian and Lorentz metrics, vector fields and tensor fields, the Riemann curvature tensor, covariant differentiation, and Levi-Civita parallelism. Chapter 16 is followed by a translation of Riemann's Habilitationsvortrag, "On the Hypotheses which Lie at the Foundations of Geometry." My translation is based on Michael Spivak's found in Spivak (1970, vol. 2). My thanks go to him for permission to use it. I have cleaned up some of the language and restored Riemann's rhetorical structure. Exercises follow each chapter. Those marked with an asterisk have a solution in the final appendix prepared by Jason Cantarella. The ideal of presenting a strict chronology of ideas throughout the book would have limited the choice of topics, and so I have chosen anachronism in the pursuit of clarity. I have also chosen to restrict my attention to functions that are smooth, that is, differentiable to all orders, though this restriction is not required to prove most of the theorems. The interested reader should try to find the degree of differentiability needed for each construction or theorem to obtain the most general results. This concession is to uniformity and simplicity in the hopes that only the most geometric details remain.

How to use this book This book started as a semester-long course at Vassar College, first taught this way in 1982 (my thanks to Becky Austen, Mike Homer, and Abhay Puri for making it a good experience). Since then I have added sections, details, and digressions that make it impossible to do all of the book in a semester. In order to use this book in one term, I recommend the following choices: Chapters 1, 4, and 5 (Chapters 2 and 3 as reading assignments). Chapter 6 through the fundamental theorem. Chapter 7 (Appendix as a reading assignment). Chapters 8, 9, and 10. Chapter I 1 up to the statement of the Hopf-Rinow theorem, skip the proof of Theorem 11.6. Chapter 12, stop before Jacobi's theorem (12.10).

Chapter 13. Chapter 14, skip the proof of Hilbert's theorem.

Chapter 15 with the discussion of stereographic projection

from Chapter 8'. At a pace averaging a chapter a week, these choices fill a 13-week semester. One could also make all of Chapters 1 through 5 a reading assignment, and concentrate on differential geometry. This serves the student who is learning the subject in order to study general relativity. The short course requires the reader to be acquainted with multivariable calculus, with elementary linear algebra including determinants and inner products, and with advanced

xii

Introduction

calculus or real analysis through compactness. A nodding acquaintance with differential equations would be nice but is not required. A student with strong courses in multivariable calculus and in linear algebra may take on faith the bits from classical analysis, such as convergence criteria and the extreme-value principle, and comfortably read the text. To read further into all of the nooks and crannies of the book some acquaintance with point set topology is also recommended.

Acknowledgments Many folks have offered their encouragement, time, and advice during the preparation of this book. Most thanks are due to my grammatical and mathematical conscience, Jason Cantarella, who combed the text for errors of expression and exposition. A small part of his effort was supported by a grant from the Ford Foundation and by research positions at Vassar College. Thanks to MaryJo Santagate for taking my handwritten class notes and providing me text files to rewrite. Diane Winkler helped in the preparation of the index. David Ellis had the courage to try the unedited text when he taught differential geometry his advice was very helpful, as was the advice of Bill Massey, Larry Smith, Jeremy Shor, and my anonymous editors. My thanks to Harvey Flad, Ruth Gornet, Erwin Kreyszig, and John Stillwell for hard-to-get materials that filled in gaps in my understanding. Thanks to John Jones for introducing me to the publisher, to Lauren Cowles of Cambridge University Press for her patience and understanding, and to Greg Schreiber for extending the work of a copy editor far beyond this author's expectations. Special thanks to Jeremy Gray whose books, articles, and conversations have provided much inspiration in this project. Finally, thanks to my wife, Carlie Graves, for her love and for tolerating another book project.

PART A

Prelude and themes Synthetic methods and results

1 11

Spherical geometry Geometry is the art of good reasoning from poorly drawn figures. Anonymous

The word geometry is of Greek origin, yewuEtpia, to "measure the earth" In antiquity geometric techniques were used by the Egyptians to assess taxes fairly according to area. The geometry chronicled and developed by Euclid (ca. 300 B.C.) in his great work The Elements

begins with the geometry of the plane - the abstract field of a farmer. Before discussing Euclid's work and its later generalizations, let us make a short detour into spherical geometry

to "measure the Earth;' idealized as a sphere. The earliest science of astronomy and the need to measure time accurately by the sun led to the development of this geometry. Basic plane geometry is concerned with points and lines, with their incidence relations, and congruences. To study such basic notions on the sphere, we need to know what a "line" or a "line segment" means.

Definition 1.1. A great circle on a sphere is the intersection of that sphere with a plane passing through the center of the sphere (for example, the equator and the lines of constant longitude). Early geometers understood that great circles share many formal properties with lines in the plane making them a natural choice for lines. For example, given any pair of nonantipodal points on the sphere, there is a unique great circle joining that pair of points; to construct it. take the intersection of the sphere and the plane containing the pair of points and the center of the sphere. Following Archimedes (287-212 B.c.) we could define a line to be a curve that is the shortest path joining any two points that lie on it. In fact, great circles satisfy this criterion (as we prove in Chapter I I ). To get an intuitive sense of why this is true pin down a piece of a rubber band on an orange and snap it tight. The rubber band will seek the curve of least energy, which, in this case, is the curve of shortest length. On a spherical orange, the resulting curve will be part of a great circle.

A spherical triangle. 3

4

Prelude and themes: Synthetic methods and results

Interpreting great circles as lines, we can measure the angle between two intersecting great circles as the angle formed by the intersection of the defining planes of the great circles with the plane tangent to the sphere at the point of intersection. With this definition of angle we can form triangles on the sphere whose interior angles sum to greater than n. In fact, every triangle of great-circle segments has interior angle sum greater than jr. To see this we study the area of a triangle on a sphere. Area is a subtle concept that is best treated via integration (see Chapter 8). The properties we need to define area for polygonal regions on the sphere are few, however, and we take them as given:

(I) The sphere of radius R has area 47r R2. (2) The area of a union of nonoverlapping regions is the sum of their areas. (3) The areas of congruent regions are equal. (4) The ratio of the area enclosed by two great circles to the area of the whole sphere is the same as the ratio of the angle between them to 27r.

Proposition 1.2. On a sphere of radius R, a triangle AABC with interior angles of, fi. y has area given by: area(AABC) = R2(a + fi +y - ir). PROOF (due to Euler, 1781). Let the term lune refer to one of the regions between two great circles and the antipodal points where they cross. Antipodal points or figures are those that correspond to opposite ends of a diameter of the sphere. A lune is given by an angle B at the center of the sphere between the two defining planes, which is also the angle between the great circles. Assumption (4) implies

area of the lune =

B

2n

4n R2 = 20 R2.

Three lunes.

We assume our triangle lies in one hemisphere - if not, subdivide it into smaller triangles and argue on each piece. A triangle is the intersection of three lunes. Extending the three lunes giving AABC to the rest of the sphere gives rise to antipodal limes and an antipodal copy of the triangle, AA'B'C'. Taking AA BC along with the three lunes that determine it, we cover a hemisphere counting the area of A A BC three times. This gives the equation

2trR2 = 2aR2 + 20R2 + 2y R2 - 3 area(AABC) + area(AABC),

/. Spherical geometry

5

and so area(AABC) = R2(a + fi + y - n). We call the value a + fi + y - n the angle excess of the spherical triangle. Since every triangle on the sphere has area, every triangle on the sphere has interior angle sum greater than n. If the radius of the sphere is very large and the triangle very small in area - as a triangle of human dimensions on this planet would be - there is negligible angle excess.

Astronomers track the positions of heavenly objects and timekeepers calculate the positions of the sun and moon. For most methods of calculation an analytic expression for points

on the sphere is desirable. Denote the sphere of radius R, centered at the origin in R3. by ER. In terms of rectangular coordinates, ER is the set of points (x, y, z) satisfying x2 + y2 + z2 = R2. The familiar spherical coordinates for Spherical coordinates.

ER give the representation

ER = ((R cos >Jr sin B, R sin* sin 9, R cos O) 10 < * < 27r.0 < 0 < 7r).

From this representation we can develop spherical trigonometry as a tool. Trigonometry is concerned with the relations between lengths of sides and angles of triangles. The length of a path between two points along a great circle is easy to define; it is the measure in radians of the angle made by the radii at each point multiplied by the radius of the sphere. Notice that this length is unchanged when we rotate the sphere around some axis or reflect the sphere across a great circle. We now prove an important theorem in spherical trigonometry.

Theorem 1.3 (the spherical Pythagorean theorem). Fora right triangle DA BC on a sphere of radius R with right angle at vertex C and sides of length a, b, and c (as shown in the diagram) c a b cos R = cos R cos R

PROOF. By rotating the sphere we can arrange the point A to have coordinates (R, 0, 0) and the point C to lie in the xyplane. The point B then has spherical coordinates (fl, n/2 -a) where a and fi are the central angles determined by the sides BC and AC. respectively. With these choices we have A right triangle.

A = (R, 0, 0),

B= (Rcosflcosa, Rsinficosa, Rsina). C = (R cos fl, R sin fl; 0).

Let y denote the central angle subtending A B. From the elementary properties of the dot

6


product on 1R3 we compute the cosine of the angle between vectors A and B:

AB

cosy =

IIAII

=

R2cosacos$

= cos a cos /i.

R2

IIBII

If we express angles a, Q, and y in radians, we get a = a/ R, i = b/ R. and y = c/ R. and so the theorem is proved. We call this the Pythagorean theorem because it relates the hypotenuse of a right spherical triangle to its sides. To see a connection with the classical Pythagorean theorem (see Chapter 2) recall the Taylor series for the cosine: x2 cosx=l- 2 +--x4

x6

-6!+....

The spherical Pythagorean theorem gives the equation

2R2+..._rl_22+...1(I_ R2+...1 a'-b'` a2 b`=I_2R2-z22+4R4 +....

Subtract I and multiply by -2R2 to obtain 2

c+

stuff RZ

,

=a +b `+ 2

other stuff R2

Let R go to infinity and we deduce the classical Pythagorean theorem. Since the Earth is a sphere of such immense radius compared to everyday phenomena, ordinary triangles would seem to obey the classical Pythagorean theorem. In the previous proof we used the dot product on R3 and the embedding of the sphere in IR3. We go further with this idea and prove another of the classical formulas of spherical trigonometry which relates the sides and angles of a spherical triangle.

Theorem 1.4 (the spherical sine theorem). Let A A BC be a spherical triangle on a sphere of radius R. Let a, b, and c denote the lengths of the sides in radians, and let L A, L B. and LC denote the interior angles at each vertex. Then sin(a/R) _ sin(b/R) _ sin(c/R)

sin/A

sin LB

sin LC

PROOF. We first treat the case of a right triangle. We restrict our attention to triangles lying entirely within a quarter of a hemisphere. Suitable modifications of the proof can be made for larger triangles.

1. Spherical geometry

7

Let DA BC be a right triangle with right angle at C. Extend the radius OA to OA' where BA' is the tangent to the great circle segment BA. Similarly extend OC to OC' where BC' is tangent to the great circle segment BC. It follows immediately that AOBC' and AO BA' are right triangles and L A' BC' is L A BC = L B. Notice that the plane A'BC' is tangent to the sphere at B and so the planes A'BC' and OBC are perpendicular, that is, lines in each plane that intersect on the line of intersection of the planes and are perpendicularto that line are perpendicular to each other. Since L ACB is a right angle, planes OBC and OAC are also perpendicular. We leave it to the reader to show that planes A'BC' and OAC meet in a line A'C' perpendicularto OC'. Thus AOC'A' is a right triangle with right angle at C'. The classical Pythagorean theorem now gives

(OA')2 = (BA')2 + R2. (OC')2 = (BC')2 + R2, and (OA')2 = (A'C')2 + (OC')2. These equations imply that

(BA')2 = (A'C')2 + (BC')2, and so AA' BC' is a right triangle with right angle at C'. Let a, fi, and y denote the angles at the center of the sphere determined by L BOC, L AOC, and LAO B, respectively. By the definition of the trigonometric functions we find sin # =

and so

sin(b/R)

sin LB sin(a/R) sin LA

A'C' BA' A'C' = = sin L B sin OA' BA' OA' y

= sin y. By reversing the roles of A and B we obtain in a similar fashion

= sin y, and so

sin(a/R) sin LA

sin(b/R)

= sin LB


8

For an arbitrary triangle we can construct the spherical analog of an altitude to reduce the relation for two sides to the case of a right triangle. For example, in the adjoining figure we insert the altitudes from A and B. By the right angle case we see Altitudes

sin(a/R) sin LC = sin(p/R) = sin(c/R) sin LA. From the other altitude we find

sin(b/R) sin LC = sin(q/R) = sin(c/R) sin(rr - L B) = sin(c/R) sin L'8. The theorem follows.

From formulas such as the last two theorems a key result in spherical trigonometry can be proved: There are six related pieces of data determining a spherical triangle, namely, the three sides and three angles. Given any three pieces of these data, one can determine the other three. It follows that similar triangles on a sphere are congruent. The solution of this problem was important to early astronomers (see Rosenfeld (1988) for more details). The lessons to be learned from this short visit to a geometry different from the geometry

of our school days set the stage for the rest of the book. The sphere has some striking geometric properties that differ significantly from those enjoyed in the plane. For example. angle sum and area are intimately related on the sphere. The Pythagorean theorem is shared by the sphere and the plane in an analytic fashion by viewing the plane as a sphere of infinite radius. Other analytic tools, such as trigonometry, coordinates, and the embedding of the sphere in R3, open up the geometry of this surface for us. All of these ideas will return to guide us in later chapters.

Exercises 1.1

Prove that two great circles bisect one another.

1.2'

The sphere of radius one can be coordinatized as the set of points (x, y, :) in 1R3 satisfying

x2 + rz + z2 = 1, or as the set of points (1, >y, 0) in spherical coordinates, with 0 < >' < 2n, and 0 < 0 < r. In these two coordinate systems, determine the distance along great circles between two arbitrary points on the sphere as a function of the coordinates.

1.3'

If three planes n i , f12, and fl3 meet at a point P and fl; is perpendicular to both fl 1 and f13. show that R 1 meets 112 and n3 in a pair of perpendicular lines.

1.4

Show that the points equidistant on one side of a fixed great circle on a sphere do not necessarily determine a great circle. Give necessary and sufficient conditions to obtain a

great circle. 1.5

A pole of a great circle is one of the endpoints of a diameter of the sphere perpendicular to the plane of the great circle, for example, the North and South Poles arc poles to the

1. Spherical geometry

9

equator. Prove that through a given point, not on a given great circle and not the pole of the great circle, there is a unique great circle through the given point perpendicular to the given great circle. 1.6

Show that an isosceles triangle on the sphere has base angles congruent.

1.7'

Show that the circumference of a circle on the sphere of radius R of radius p is given by L = 27rR sin(p/R). What happens when the radius of the sphere goes to infinity?

1.8

Suppose AA BC is a right triangle on the sphere of radius R with right angle at vertex C.

Prove the formula cos A = sin B cos(a/R). 1.9

Show that there are no similar triangles on the sphere that are not congruent. This is the congruence criterion for spherical triangles, Angle-Angle-Angle.

2 Euclid There has never been, and till we see it we never shall believe that there can be, a system of geometry worthy of the name which has any material departures ... from the plan laid down by Euclid.

A. DeMorgan (October 1848)

The prehistory of geometry was a practical matter. It consisted of facts and rules that could be applied to determine the positions of the sun and stars or to measure land areas. Merchants brought these ideas from Egypt and Babylonia to ancient Greece where the prevailing ideas of systematic thought changed geometry from a tool to a deductive discipline of the mind. Centuries of contemplation and careful reorganization culminated in Euclid's Elements (Euclid 1956), whose thirteen surviving books summarized the mathematics of his day and influenced all subsequent generations. The Elements proceeds by the axiomatic method - definitions and axioms are presented first, then propositions are shown to follow from these assumptions and from each other through logical deduction. When later mathematicians sought a model for the rigorous development of mathematical ideas, they turned to Euclid. Let us begin with some definitions from the Elements:

(I) A point is that which has no part. (2) A line is a breadthless length. (3) The extremities of a line are points.

(4) A straight line is a line which lies evenly with the points on itself. (5) A surface is that which has length and breadth only. (6) The extremities of a surface are lines. (7) A plane surface is a surface which lies evenly with straight lines on itself.

Euclid does little more here than to recall the familiar. For example, as stated here, the definition of point holds little meaning - Heath, the editor of Euclid (1956). explains it as a device to avoid circularity. The modern view is to take such terms as point, line, and plane as undefined and give axioms to determine the relations among them. One can begin with a set S (space) and two classes of subsets, C, the lines, and P. the planes. A point P. that is, an element of S. lies on a line ! in .C if P E 1. and a line ! lies in a plane n in P if ! c Il. After some further definitions, Euclid sets out his assumptions about these objects in the five postulates and five common notions. We discuss these next. A list of the propositions of Book 1 (1.1-1.48) follows in the appendix to this chapter. 10

2. Euclid

Postulate I.

11

To draw a straight line from any point to any point.

The modem concept that makes this postulate precise is given by the following structure:

Definition 2.1. A set S together with collections of subsets C, of lines, and P. of planes. is an Incidence geometry if the following axioms of incidence hold:

(1) For any P, Q in S, there is an I in C with P E I and Q E 1. If P Q. I is unique. We denote the line determined by P and Q by ". (2) If I is in C, then there are points P, Q in S, P E 1, Q E I and P 0 Q. (3) For any plane n E P there are points P. Q. R in n so that if P E 1. Q E 1, then R¢ 1. These axioms are due to David Hilbert (1862-1943) (see Hilbert 1899). They are strong enough to capture the notion of incidence for points and lines in Book I of the Elements. There are further axioms for points, lines, and planes in space. Notice the uniqueness clause in axiom (1); Euclid tacitly assumes this, and he applies uniqueness in the proofs of certain propositions (for example, Proposition 1.4). The axiomatic approach proceeds without interpretation of the basic terms. This allows for different realizations of the axioms. For example, certain finite sets with subsets and relations are incidence geometries and their properties are used to study problems in combinatorics and algebra. In Postulate I Euclid described a tool for drawing, the ideal straight edge. In Postulate II Euclid tells us more about a straight edge.

Postulate U. To produce a finite straight line continuously in a straight line. In order to give the modern version of Postulate II we need to introduce two further primitive notions: Given three distinct points P. Q, R lying on a line 1. we want to say when Q lies between P and R. This relation is denoted B(P, Q, R) and can be defined set-theoretically as a ternary relation, B C S x S x S, satisfying certain axioms (to be given later) which make abstract the notion of betweenness. A line segment PQ C I is defined as the subset

PQ ={R Eli R = PorR=QorB(P,R.Q)). The other primitive notion is that of congruence of line segments, denoted PQ = A B. The axioms for betweenness imply that the endpoints determine a line segment as a subset of a line, so we can think of congruence of line segments as a four-place relation E(P, Q; A, B) satisfying certain axioms. We write PQ - A B whenever the relation E(P. Q; A, B) holds.

If we take these primitive notions as given, then the modern version of Postulate II becomes

Given two line segments PQ and AB there is a point R on PQ so that QR - AB and Q is between P and R.

12

Prelude and themes:

Synthetic methods and results

The notions of betweenness and congruence for Postulate 11 are realized if the set S is equipped with a distance function, d: S x S -+ R. making S a metric space. Recall that a distance function determines a metric space if the following conditions are satisfied:

(1) For all P and Q in S. d(P, Q) = d(Q, P). (2) For all P and Q in S. d(P, Q) 0 and d(P, Q) = 0 if and only if P = Q. (3) (The triangle inequality) For all P. Q, and R in S, d(P, Q) +d(Q, R) > d(P, R). In the presence of a metric space structure on S betweenness can be defined as the ternary relation

B(P, Q. R) if and only if d(P, Q) +d(Q, R) = d(P, R).

Congruence of line segments can be defined by PQ - AB if and only if d(P, Q) = d (A, B). An incidence geometry along with a distance function is called a metric geometry (see Millman and Parker (1977) or Mac Lane (1959) for an alternate foundation of geometry based on metric postulates).

Alternatively, if we have a notion of betweenness and congruence for an incidence geometry (S, C, P), then we can ask for a measure of length. This is a real-valued function 0 defined on the set of line segments in S satisfying the properties:

(1) For all line segments PQ, O(PQ) > 0. (2) If PQ = P'Q', then O(PQ) = O(P'Q'). (3) If B(P, Q, R). then O(PQ) + O(QR) = O(PR). This notion is equivalent to a metric-spacc structure. One can prove that any two measures

of length are proportional and, given a particular line segment AB, there is a measure of length 00 with O0(AB) = I. The chosen segment AB is the unit of the measure 00. Restricting 00 to a line I E G, together with a pair of points 0, 0' E 1. gives a function Dot, : I -' R that determines a ruler or coordinate system on 1. The point 0 is the origin of the coordinate system and the ray 00' determines the positive part of the line. In later chapters we will assume the existence of rulers on lines in order to introduce functions defined on lengths, that is, on congruence classes of line segments. For a proof of the existence of measures of length, the interested reader may consult Borsuk and Szmielew (1960) (especially Chapter III, §§9-10). Postulates I and II describe the ideal straight edge (not a calibrated ruler). The Greeks allowed another ideal tool for geometric constructions.

Postulate III. To describe a circle with any center and distance. In the modern parlance Postulate III reads

Given a point 0 and another point A there is a circle with center 0 and radius OA, determined by the set of points Q such that OQ = OA. This postulate allows construction with a compass that is idealized to have any finite radius. Without further description we must take Euclid's compass to be collapsible, that is, one cannot simply move a radius from place to place. This hindrance is overcome by Proposition

2. Euclid

13

1.2 of the Elements in which a clever construction permits one to move a compass without collapsing the arms. The next postulate deals with angles. We first recall a few more of Euclid's definitions:

(8) A plane angle is the inclination to one another of two lines in the plane that meet one another and do not lie in a straight line. (9) And when the lines containing the angle are straight, the angle is called rectilineal. (10) When a straight line set up on a straight line makes adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands. (11) An obtuse angle is an angle greater than a right angle. (12) An acute angle is an angle less than a right angle.

Postulate IV. All right angles are congruent. In order to make the definition of right angle and Postulate IV precise, we need a notion of congruence of angles. Euclid imagined congruence of figures to be a motion in space of one figure that superimposes it on another figure such that corresponding points and line segments coincide. Bertrand Russell (1872-1970), in his 1902 article for the Encyclopedia Britannica Supplement, points out that "(a)ctual superposition ... is not required; all that

is required is the transference of our attention from one figure to another." We take a congruence, a "transference of attention;' to mean a mapping 0: S - S where (1) S is the underlying set of an incidence or metric geometry. (2) 0 is bijective. (3) 0 takes lines to lines and planes to planes. And if (S, G, P. d) is a metric geometry,

then d(O(P), 0(Q)) = d(P. Q). Hilbert takes congruence of angles to be a primitive notion satisfying certain axioms and deduces Postulate IV as a theorem. From the modem viewpoint Euclid's Postulate IV is a statement about the existence of enough congruences to compare every pair of right angles. For example, to superimpose adjacent right angles, we can reflect one right angle through space onto its neighbor. Euclid assumes the existence of a congruence that we call reflection across a line in his definition of right angle. This implies the "homogeneity of space." The motion of a figure determined by two line segments at right angles onto another that preserves angles and distances implies that close to the vertices of the angles the planes containing the right angles look the same. On a more general surface it may not be the case that right angles are congruent in the sense of superposition. Consider the case of a torus (an inner tube): The meridians (vertical slices) and the innermost and outermost horizontal circles on the torus may be taken for lines. There is no "motion" of the torus to itself, however, such that the right angle at the point X of intersection of a given meridian with the innermost circle and the one at the point Y of intersection of the given meridian with the outermost circle are superimposed, and so they are not congruent. Homogeneity of a general surface will be made precise in Chapter 10.


14

Euclid worked deductively from his definitions, common notions, and Postulates I through IV to derive the first 28 propositions of Book I. Let us consider a few of these propositions and discuss his methods and assumptions further.

Proposition I.I. Given a finite straight line one can construct an equilateral triangle with the given line segment as base. PROOF.

Let AB be the given line segment. By Postulate III

there is a circle with center A and radius A B as well as a circle

with center B and radius BA. Since AB is contained in both circles they meet at a point C. By Postulate I we can form the line segments AC and BC and so a triangle. Now BC is a radius and so it is congruent to BA = AB: AC is a radius and so it is congruent to AB. Thus an equilateral triangle is constructed. REMARKS.

(I) Notice how the axiomatic method is applied here. only postulates and

definitions are used. (2) Euclid assumes that the constructed circles must meet at the point C. Criticism of this assumption goes back before the fifth century A.D. (see Proclus (1970)). The problem is overcome in Hilbert's scheme (Hilbert 1970) by requiring the axiom of continuity. This axiom and a related statement, Pasch's axiom, guarantee the existence of points of intersections for many figures in Euclid.

Axiom of continuity (Archimedes). Given line segments AB and CD there are points = An_i An and B lies AI A2 = A1, A2, ... , An lying on Ah such that CD = AAt between A and An.

Pasch's axiom. If a line in a plane enters a triangle in that plane through at most one vertex of the triangle, then the line exits the triangle at a point on one of the sides not already cut by the given line.

To illustrate why these axioms are necessary, suppose the plane is simply the set of ordered pairs Q x Q. If we take A = (0,0) and B = (1.0) in Proposition 1.1, then the

point C = (i,

) in the proof is missing. The axiom of continuity provides the basis for the completeness of the real number line, the property of R that all Cauchy sequences of real numbers converge to a real number. With this property the planes of Euclid are like R2 and completeness implies the existence of the desired points.

2. Euclid

Is

REMARK. The existence of rulers on lines requires the axiom of continuity. By using the fact that we can bisect a line segment (Proposition 1.10) and Postulate 11, we can apply the axiom of continuity to coordinatize a line by making points correspond to dyadic expansions of real numbers. The notion of Dedekind cuts may also be used to coordinatize lines (see Borsuk and Szmielew (1960)). It is also possible to introduce a measure of angles, that is, protractors. Such a measure satisfies properties analogous to those satisfied by a measure of lengths. Once again the axiom of continuity plays a role in the construction of such measures. One of the highlights of Book I is the following useful theorem.

Proposition 1.16 (the exterior angles theorem). In any triangle, if one of the sides is extended, the exterior angle formed is greater than either of the opposite interior angles.

Consider the extraordinary figure. Given AABC, extend BC along to D. We PROOF.

want to show LBAC is less than LACD. Let M be the midpoint of AC (which can be constructed as in 1.10) and join B to M. Extend

BM to a point E so that BM = ME (Pos-

tulate II). Now CM = MA, BM - ME, and LAMB = LCME (these are vertical angles and hence congruent by 1.15). By the congruence criterion Side-Angle-Side (1.4). LAMB - ACME, and in particular, LBAM = LECM. Since E is on the ray BM and D is on the ray LECM is inside LDCM and so LBAM is less than LMCD. Thus LBAC is less than LACD. To argue that LACD exceeds LA BC, construct the analogous diagram extending AC. Though we are stepping out of Euclid for a moment, we next derive an important theorem for the theory of parallels. The assumptions are in place for the proof of this result, and the method is analogous to the proof of Proposition 1.16.

Theorem 2.2 (the Saccheri-Legendre theorem). The sum of the angles interior to a triangle is less than or equal to two right angles.

PROOF. Suppose we have a triangle AA BC and suppose the sum of the interior angles is two right angles plus an angle congruent to LBAZ. Suppose LBAC is the least of the three interior angles. As in the proof of 1.16, let M be the midpoint of BC and construct the


16

-4

----4

ray AM. Mark DI on AM so that AM

MD1, and so ACMA - ABMD1. Notice that

L B A M + LCAM = L B AC so one of these angles is less than or equal to i L BAC, where the meaning of i L BAC is clear. Thus either L BADI or L B Di A is less than 1 L BAC. Now observe

LBAC+ LABC+ LBCA = LBADI + LBD1A+ LABC + LMBDi = LBADI + LBDi A+ LABD,. Thus AABDi has the same angle sum as AABC but has one angle less than half the smallest angle of AABC. If we iterate this procedure n times, we obtain a triangle whose smallest angle is less than (112")Z BAC. The axiom of continuity maybe applied to angles, and so repeated halving produces an angle less than L BAZ. The triangle constructed at this stage still has angle sum two right angles plus LBAZ. Thus the two remaining angles in the triangle must sum to more than two right angles which contradicts 1.17 - the sum of any two interior angles in a triangle is less than two right angles.

From this result it is clear that the first four postulates of Euclid cannot be satisfied by spherical geometry, since triangles on the sphere have angle sum greater than two right angles. One of the points of departure for the sphere is Postulate 11; it is not possible to define a natural notion of "between" on a great circle and so the modern version of the postulate cannot be realized. The uniqueness assertion of Postulate I also fails on the sphere; when two antipodal points are chosen there are infinitely many great circles joining such points.

Euclid's theory of parallels Book I of the Elements breaks naturally into two parts at Proposition 1.27. It is here that Euclid begins to construct a theory of parallels and area in order to prove the Pythagorean theorem (1.47). Euclid defines parallel lines as follows:

(23) Parallel straight lines are straight lines which, being in the same plane and being produced indefinitely in both directions, do not meet one another in either direction. The first important result about parallels in the Elements is the following proposition.

Proposition 1.27 (the alternate interior angles theorem). if a straight line falling on two straight lines makes the alternate interior angles congruent to one another, the straight lines will be parallel to one another. E

A

PROOF. Let the line' cross lines Ah and t

B

so that L BEF is congruent to LCFE. Suppose

Y

X

C

F

D

AmeetsonthesideofBandDatX.Consider A EFX. Now LCFE is an exterior angle and hence it is greater than LFXE and LXEF

(1.16). But LXEF = LBEF = LCFE. This is a contradiction. We can argue similarly that A and t and C.

do not meet on the side of A

2. Euclid

17

The converse of Proposition 1.27 is Proposition 1.29 and it is the first case in the Elements that Euclid calls upon the last of his assumptions.

Postulate V. That, if a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than two right angles. It is immediate that this Postulate is different in nature than the other four. It is simply more difficult to state than the others and it uses loose words such as "indefinitely." Postulate V became the focus of criticism of later generations of mathematicians who accused Euclid

of introducing an "unproved theorem" to simplify proofs. This criticism is the source of centuries of interesting geometry and it is the focus of much of the rest of the book. Euclid develops the theory of parallels in order to construct figures whose areas can be compared. He introduces a different use of the word "equals" (in 1.35) to mean "of equal area." Along the way he shows that the interior angle sum of a triangle is two right angles, that corresponding angles and segments of a parallelogram are congruent, and that parallelograms may be constructed with area that of a given triangle. After showing that squares may be constructed on a given base, he reaches the climax of Book I:

Proposition 1.47 (the Pythagorean theorem). In a right-angled triangle the square on the side subtending the right angle is equal in area to the squares on the sides containing the right angle.

Refer to the figure for notation. Given the right triangle A ABC, construct the squares on each of the sides (1.46). Using the fact that the sum of the angles in a square is four right angles and that they are all congruent, we see that rA is the line AZ and PROOF.


1g

Ah is the line . Now LDCB + LBCA = LKCA + LBCA and sides DC = CB and AC = CK. By Side-Angle-Side (1.4) we have ACKB = OCDA. If AM is an altitude from A (that is, AM is perpendicular to BC) consider the rectangles CDNM and 0CKHA. By 1.41, CKHA is twice ACKB in area and CDNM is twice,6CDA. Since ECKB - t CDA, oCKHA and CDNM have the same area. Similarly oABFG has the same area as 0MNEB and the result follows by putting CDNM and 0MNEB together.

Reading the propositions of Book I that follow 1.27, one is struck by the role that parallels

play in the determination of relations among the areas of elementary figures. This use of areas is in part due to the lack of symbolic algebra in Greek mathematics. The square of a number was interpreted as the area of a square with side the given number. There are many other proofs of the Pythagorean theorem, some of which are based on simple algebraic formulas. Before leaving Euclid let us record the rest of Hilbert's system for describing planar geometry. This provides the modern means to make Euclid's work precise as well as an example of contrasting axiomatic systems. Fix a set S with classes of subsets G of lines and P of planes. Assume that (S, L;, P)

is an incidence geometry. Let us assume that we have relations B C S x S x S and E C S x S x S x S whose properties are intended to encode our ideas of betweenness and congruence. If a triple of points (P, Q, R) lies in B. we say that B(P. Q, R) holds, or that Q lies between P and R. If a four-tuple of points lies in E, we say that E(P, Q; A, B) holds, and we denote this by PQ = AB.

Axioms of order. (1) If P. Q. R lie on a line l and B(P, Q, R) holds, then B(R, Q. P) holds. (2) If P and R lie on a line 1, then there are points Q and S on I such that B(P. Q, R) and B(P, R. S) hold. (3) If P, Q, R lie on a line 1, then either B(P, Q, R) or B(Q. R, P) or B(R, P. Q) holds.

(4) Any jour points P. Q. R, S on I can always be so arranged that B(P. Q. R) and B(P, Q, S) holds, as well as B(P. R, S) and B(Q, R, S). (5) Pasch's axiom.

Axioms of congruence.

(1) P Q

P Q always holds. If P. Q lie on a line I and P' lies on I', then there is a point Q' on 1' on a given side of P' such that P Q = P'Q'.

(2) IfPQ=AB and PQ=CD,then AB=CD. (3) If P, Q, R lie on a line !, P', Q', R' lie on a line I'. B(P, Q, R) and B(P'. Q', R')

hold, and PQ = P'Q' and QR = Q'R', then PR - P'R'. (4) LPOQ - LPOQ and LPOQ = LQOP always hold. In a plane we have an angle L P 0 Q and in another plane a ray O'P' along with a choice of half-plane

2. Euclid

H

19

-)

determined by the line O'P'. Then there is a unique ray 0'Q' in the chosen halfplane such that L PO Q = L P'O' Q'.

(5) if L POQ - L P'O'Q' and L POQ - L P"O"Q", then L P'O'Q' - L P-0- Q-. (6) Given two triangles DABC and AA'B'C', suppose AB = A'B'. AC = A'C', and LBAC = LB'A'C', then LABC - LA'B'C'and LACB = LA'C'B'(Side-AngleSide).

The first few exercises treat the relationship between these axioms and Euclid's work. Hilbert gave five groups of axioms - incidence, order, continuity, congruence, and parallels - in his scheme to put geometry on firm foundations. He proved that all of the axioms are independent and so all are required. The axiom for parallel lines will be discussed in the next chapter.

Exercises 2.1

From the axioms for an incidence geometry show that two lines I and I', 1 # I'. either share a point in common or no points in common.

2.2

Using only the notions of incidence and order how do you define a ray or half-line, Ok C I? Prove that a line lying in a plane divides the plane into two distinct regions called half-planes.

2.3

From the congruence axioms prove 1.4, that is, if given AA BC and tA'B'C' such that

AB = A'B', AC = A'C', and LBAC = LB'A'C'. then BC = B'C'. Prove the other congruence criteria - Angle-Angle-Side. Angle-Side-Angle (1.26). and Side-Side-Side (1.8) - from the axioms. 2.4

0'

OQ = LP'O'Q' and a ray lying within LPOQ, prove that there is a ray O'X' lying in L P'O'Q' such that L POX L P'O'X' and (a) Given congruent angles

LXOQ = LX'O'Q'. (b) Given L POQ - L P'O'Q' and L QO R = L Q'O'R', then, if Q lies in L POR and Q' lies in L P'O' R', prove that L PO R = L P' O' R'. (c) From (a) and (b) prove Euclid's Postulate IV. 2.5

Consider the sets P = (A, B. C, D. E, F. G) and G = (ABC. CDE. EFA, AGD. BGE, CGF, BDF) with the obvious relation E. Show that this forms an incidence geometry. (Try to draw a picture of this finite geometry - it is known as the Fano plane. It encodes the multiplicative relations among the unit Cayley numbers that are not equal to 1.) Prove or disprove: In an incidence geometry, each line has at least three points on it.

2.6

Let S denote the set of all lines through the origin in R3. Taking a plane through the origin as a set of lines. we can define G as the set of planes through the origin in 1R3 with

E meaning I C P. Show that this set S, the set of lines L. and E so interpreted give an incidence geometry (with one plane, all of 1R3). 2.7

From Postulates I through IV prove Propositions 1.9,1.10, 1.1 1, and 1.18 in the appendix to the chapter. You may assume any propositions previous to the one you are proving.

2.8

Assuming Postulates I through IV and Proposition 1.29 prove Postulate V.

20


2.9'

Prove that the segment joining the midpoints of two sides of a triangle has length less than or equal to half the third side. For the proof did you need Postulate V?

2.10

Suppose that two right triangles have congruent hypotenuses and one corresponding side congruent. Show that the triangles are congruent. Does this hold for general triangles?

2.11

Prove that the internal bisectors of the angles of a triangle meet in a point. For the proof did you need Postulate V? (Hint: Take two bisectors that meet in a point. Join the third vertex to this point and consider the segments perpendicular to the sides through this point. Now apply the preceding exercise.)

2.12

In a circle form a triangle with one side a diameter and the third vertex on the circle. Show that this is a right triangle. Does your proof require Postulate V?

2.13' Assume the existence of a unit length and, using straight edge and compass, give constructions for the sum and product of the lengths of two line segments, the reciprocal of the length of a line segment, and the square root of the length of a line segment. (Hint: Assume the relations between sides of similar triangles.)

Appendix The Elements: Book I

Common Notions 1. Things which are equal to the same thing are also equal to each other. 2. If equals are added to equals, then the wholes are equal. 3. If equals are subtracted from equals, then the remainders are equal. 4. Things which coincide with one another are equal to one another. 5. The whole is greater than the parts.

Propositions Proposition 1. 1. Given a line segment one can construct an equilateral triangle with the given line segment as base. Proposition 1.2. Given a point and a line segment, one can construct a line segment with the given point as endpoint and congruent to the given line segment. Proposition 1.3. Given two unequal line segments, one can construct on the greater a line segment congruent to the lesser.

Proposition 1.4 (Side-Angle-Side). If two triangles have the two sides congruent to two sides respectively, and the angles enclosed by the two congruent sides congruent, then the triangles are congruent.

Proposition 1.5 (pons asinorum). In an isosceles triangle the angles at the base are congruent. If one of the congruent sides is extended to form a new triangle and the other side is extended by the same length, then the angles under the base are congruent. Proposition 1.6. If in a triangle two angles are congruent, then the sides opposite the congruent angles are congruent.

Proposition 1.7. Given a triangle, there cannot be another triangle constructed on the same base to a different point which is congruent to the first triangle with the sides in the same order.

Proposition 1.8 (Side-Side-Side). If two triangles have corresponding sides congruent, then they also have angles enclosed by corresponding sides congruent. Proposition 1.9. (With straight edge and compass) one can bisect a given angle. Proposition 1.10. One can bisect a given line segment.

Proposition 1.1 1. One can construct a line perpendicular to a given line at a given point on the line. Proposition 1.12. To a given line and a point not on the line one can construct a line through the point perpendicular to the given line.

Proposition 1.13. If a line meets another line, on one side they form angles which sum to two right angles. 21

22


Proposition 1.14. If two lines through a point on a given line form adjacent angles with that line which sum to two right angles, then the two lines are the same line. Proposition 1.15. If two lines meet, they make congruent vertical angles.

Proposition 1.16 (the exterior angles theorem). In any triangle, if one of the sides is extended, the exterior angle formed is greater than either of the opposite interior angles. Proposition 1.17. In any triangle two angles taken together are less than two right angles.

Proposition I.18. In any triangle the greater side is opposite the greater angle. Proposition 1.19. In any triangle the greater angle is opposite the greater side.

Proposition 1.20 (the triangle inequality). In any triangle two sides taken together are greater than the remaining one. Proposition 1.21. If a triangle is constructed inside a given triangle on its base, then the two sides of the new triangle are less than the corresponding sides of the given triangle and they form a greater angle.

Proposition 1.22. Given three line segments so that any two taken together are greater than the third, there is a triangle with sides congruent to the given line segments.

Proposition 1.23. At a point on a given line one can construct an angle congruent to a given angle. Proposition 1.24. If two triangles have two sides congruent, then the greater included angle is opposite the greater base. Proposition 1.25. If two triangles have two sides congruent, then the greater base is opposite the greater included angle.

Proposition 1.26 (Angle-Side-Angle, Angle-Angle-Side). If two triangles have two angles and one side congruent, then the triangles are congruent. Proposition 1.27 (the alternate interior angles theorem). If a line falling on two straight lines makes the alternate interior angles congruent to one another, the straight lines are parallel to one another.

Proposition 1.28. If a line falling on two straight lines makes the exterior angle congruent to the interior and opposite angle on the same side, the straight lines are parallel to one another. Proposition 1.29. A line cutting two parallel lines makes alternate interior angles congruent, the exterior angle congruent to the interior and opposite angle, and the interior angles on one side congruent to two right angles. Proposition 1.30. Lines parallel to a given line are parallel to each other. Proposition 1.31. Through a given point not on a given line one can construct a line parallel to the given line.

Proposition 1.32. In any triangle, if one of the sides is extended, the exterior angle is congruent to the two opposite interior angles taken together, and the three interior angles sum to two right angles.

Appendix. The Elements: Book/ Proposition 1.33. The line segments joining corresponding endpoints of congruent and parallel line segments are themselves congruent and parallel. Proposition 1.34. In a parallelogram the opposite sides and angles are congruent and the diameter bisects the area. Proposition 1.35. Parallelograms on the same base and in the same parallels have equal areas. Proposition 1.36. Parallelograms on congruent bases and in the same parallels have equal areas.

Proposition 1.37. Triangles which are on the same base and in the same parallels have equal areas. Proposition 1.38. Triangles which are on congruent bases and in the same parallels have equal areas. Proposition 1.39. Triangles on the same base and of equal areas are on the same parallels.

Proposition 1.40. Triangles on congruent bases on the same side of a given line and of equal areas are also in the same parallels. Proposition 1.41. If a parallelogram has the same base with a triangle and they are in the same parallels, the parallelogram has twice the area of the triangle. Proposition 1.42. In a given angle, one can construct a parallelogram of area equal to the area of a given triangle.

Z7

Proposition 1.43. In any parallelogram the complements of the parallelogram about the diameter are equal in area.

Proposition 1.44. On a given line segment in a given angle one can construct a parallelogram of area equal to a given triangle.

Proposition 1.45. One can construct in a given angle a parallelogram of area equal to a given quadrilateral. Proposition 1.46. One can construct a square on a given line segment.

Proposition 1.47 (Pythagorean theorem). In a right-angled triangle the square on the hypotenuse is equal in area to the sum of the squares on the sides. Proposition 1.48. If in a triangle the square on one of the sides is equal in area to the sum of the squares on the remaining two sides of the triangle, then the angle contained by the two sides is a right angle.

23

3 The theory of parallels This ought even to be struck out of the Postulates altogether; for it is a theorem ... the converse of it is actually proved by Euclid himself as a theorem .... It is clear then from this that we should seek a proof of the present theorem, and that it is alien to the special character of postulates.

Proclus (410-85)

The most reliable information about Euclid and early Greek geometry is based on the commentaries of Proclus whose objections to Postulate V are stated in the quote. To its author and early readers the Elements provided an idealized description of physical space. From this viewpoint it is natural to understand the objections to Postulate V. The phrase "if produced indefinitely" strains the intuition based on constructions with compass and straight edge. Furthermore, Euclid studiously avoided using Postulate V for the first 28 propositions of Book 1. The first application was to prove 1.29, the converse of 1.27 and 1.28. Several of the previous propositions are related to their neighbors as converses with proofs that simply observe the contradiction to the earlier statement were the converse false. Proposition 1.29 does not yield to this logic. Why introduce such an unnatural statement to prove it? To eliminate this "blemish" on Euclid's great work, subsequent generations heeded Proclus's call and either sought a proof of Postulate V from the other assumptions, or they tried to replace it with a more self-evident assumption. Some of the titles of their efforts indicate their intentions, for example, Treatise that heals doubts raised by parallel lines (Nasir al-Din al-TOs 1201-74) and Euclid restored (Giovanni Alfonso Borelli 1608-79). In most of these efforts an assumption was made that was equivalent to Postulate V. later generations, seeking to avoid this fate, corrected the work of their predecessors only to commit petitio principii (begging the question, that is, assuming what you want to prove) themselves. Their efforts fall into certain categories which reveal the properties of parallels. In later chapters where we consider negations of Postulate V. we will derive properties of the associated geometry from the equivalent statements. We now turn to some of these ideas.

Uniqueness of parallels In a book on parallels that is said to have been written by Claudius Ptolemy (100-78), Proposition 1.29 is proved without the benefit of Postulate V. The proof was recorded by Proclus (1970). 24

3. The theory of parallels

25

Proposition 1.29. A line cutting two parallel lines makes alternate angles congruent, the exterior angle congruent to the interior and opposite angle, and the interior angles on the same side congruent to two right angles. PROOF (Ptolemy). A straight line which cuts parallels must make the sum of the interior angles on the same side equal to, greater than, or less than two B A F/ right angles. "Greater than" is not allowed because, following the figure, if L BFG and L DG F sum to more than two right angles, D so also must LA FG and LCG F because A and CMG are no more parallel than FB and G. "Less than" is argued similarly. Proclus identified the source of Ptolemy's petitio principii which has since become known as Mayfair's axiom (from a 1795 English edition of Euclid by John Playfair).

Playfair's axiom. Through a point not on a line there is one and only one line through the point parallel to the given line.

FI

A

C

IG

D

In the figure we see how two parallels through the point F. if they existed, would ruin Ptolemy's argument. In fact, Playfair's axiom is equivalent to Postulate V. and so Ptolemy had actually assumed Postulate V in his "proof."

Theorem 3.1. Playfair's axiom is equivalent to Postulate V. PROOF. In one direction, suppose a pair of lines, A-b and t. is cut by another line so that the interior angles on one side sum to less than two right angles. Construct line ti through F so that LG FM is congruent to L FE B, which is possible by 1.23. By the alternate interior

angles theorem (1.27), ti is parallel to A1. By Playfair's axiom, n is not parallel and furthermore, D is inside L EFM. Therefore tb meets A on the side of B and M. C

F

M

L n

A

E

B

In the other direction, if we are given a point F not on a line A construct (1.12) the line

rMso that EF is perpendicular to rM and EF is perpendicular to Ah. By Proposition 1.27, rM is parallel to A4-h. Suppose tb is any line through F. If t 54 t, makes an angle on one side less than a right angle. By Postulate V. C meets Ah. Hence rM is the unique parallel.

26


Equidistance and boundedness of parallels Proclus himself gave a "proof" of Postulate V, in which he asserts: If from one point two straight lines forming an angle be produced indefinitely, the distance between the said lines produced indefinitely will exceed any finite magnitude. From this he observes that if any straight line cuts one of two parallels, it will also cut the other. The argument is as follows: Let Ah be parallel to andC, cut A1 at F. If F and Fh are produced indefinitely, the distance between them will exceed any magnitude, including the interval between the parallels. Therefore, when they are at a distance greater

than the gap between Ah and n, T will cut t .

Postulate V can now be proved by arguing as follows: If LGFP and L DPF sum to less than two right angles, construct the line n such that L BFP and L DPF sum to two right angles. Proposition 1.28 implies that Ah is parallel to and so T-6 cuts a parallel at the point F. Thus FFG meets n on the side of the lesser angles. The unspoken assumption made by Proclus here is the following statement:

If two parallels that have a common perpendicular are produced indefinitely then the perpendicular from a point of one upon the other remains finite (in fact, bounded).

In the sixth century the Byzantine commentator Simplicius revealed a proof of the fifth postulate due to Aghanis who asserts that equidistant straight lines exist. This assumption is stronger than Proclus's and can be stated The set of points equidistant from a given straight line is also a straight line.

Proclus's initial assumption was known as the "Philosopher's Principle" to medieval Islamic geometers. The philosopher is Aristotle (384-322 n.c.) who wrote on parallels as a source of logical errors. The Jesuit priest Christopher Clavius (1537-1612) asserted that it needed proof and he offered that it is false of other species of curves (such as a hyperbola and a line parallel but close to its asymptote). We record a proof of Proclus's assertion here. Proposition 3.2. If A A C M is a right triangle with right angle at C and B on AM is the midpoint of that segment, then if BD is the perpendicular to AC meeting at D. B D is not greater than half of CM.

PROOF (Saccheri 1733. Proposition XX). Extend DB to DH with BH = BD. If BD is greater than half of CM, then DH is greater than CM. Extend CM to CK = DH.

27


By the congruence criterion Side-Angle-Side, ADBAiscongruenttotHBMandsoDA = HMand LB H M is a right angle. Hence LDHK is greater than a right angle. The reader can now prove (or look in Chap-

ter4) that LCKH is congruent to LDHK. This gives us a quadrilateral with angle sum greater than four right angles, which is not possible by the Saccheri-Legendre

theorem (Theorem 2.2). Hence twice BD is less than or equal to CM. Corollary 3.3. Given an angle L MAC, if AM and A C are extended indefinitely, then the distance between them will exceed any given finite length.

PROOF. Double AM to A Q and construct the perpendic-

ular to At from Q to P. Then PQ > 2C M. By iterated doubling and the axiom of continuity, we can construct a perpendicular from M to AC that exceeds any given length. The assumptions that parallel lines were a fixed distance

C

P

apart or that they were no more than a bounded distance

apart find their way into many of the attempts to prove Postulate V. Equidistance may have figured in Archimedes's definition of parallels (Rosenfeld 1988, p. 41) and it was the basis for proofs of Postulate V by Posidonius (ca. 135-50 n.c.) and by Ibn Sind (980-1037). Boundedness of parallels figures in Proclus's attempt as well as a more subtle proof by ' Umar Khayyam (1027-1123, of Ruba' iyat fame).

Theorem 3.4. That the set of points equidistant from a given line forms a line is equivalent to Postulate V.

PROOF. From Postulates I through V. one can prove Euclid's proposition 1.34, which asserts that, in a parallelogram, the opposite sides and angles are congruent to each other. Construct

P perpendicular ton and ) perpendicular to k. Also construct (1.12) V through S perpendicular to )8 (the unique parallel). This meets " at Q to form a parallelogram. Thus PQ = RS (that is, construct any pair of perpendiculars from V to n and they are congruent by 1.34). S

P

in

n

P

R

I

In the other direction, it suffices to prove Playfair's axiom from our assumption. Suppose m is the line of points equidistant from I through P. and n is any other line through P. Since

28


m and n form an angle at P we can now argue, as Proclus, that the lines m and n diverge. Thus n meets I and n is not parallel to 1, that is, m is the unique parallel to I through P. Notice that the assertion, that the locus of points equidistant from a given line forms a line, fails on a sphere where great circles are lines.

On the angle sum of a triangle Much of the literature from the golden age of Greece reached us through the translation and preservation of texts by the people of medieval Islam. The distinguished geometers of this culture also tried to improve Euclid by proving Postulate V or replacing it with a more reasonable assumption. The key to many of their attempts is the assumption:

C

H

n

D

if Ah and n are two straight lines such that successive perpendiculars from A-h to t always make with Ah unequal angles always acute on the side toward A and obtuse on the side toward B, then the lines approach continually nearer in the direction of the acute angles and diverge in the direction of the obtuse angles, and the perpendiculars will diminish and increase in the respective directions. Furthermore, the converse holds.

This assumption is found in one form or another in the work of Thabit ibn Qurra (83690 1), 'Umar Khayyiim, and Nasir al-Din al-Tusi. The most important consequence of this assumption is the following:

Lemma 3.5. Under the previous assumption, if AC and B D are at right angles to A B and AC = BD and CD is joined, then LACD and LBDC are right angles. C.

i

D

PROOF. Suppose not. Then LACD is either acute or ob-

tuse. In either case, by the assumption, AC could not be congruent to BD because the sides AC and BD are diminishing or increasing . n n A B Notice that this can be applied to show that there are triangles for which the three interior angles sum to two right angles. Take the diagonal of the quadrilateral discussed in the lemma and apply the Saccheri-Legendre theorem. We obtain two triangles each with angle sum two right angles. This fact is the key to the derivation of Postulate V in many of the Islamic efforts.

Theorem 3.6. That the sum of the angles interior to a triangle is two right angles is equivalent to Postulate V.

29


PROOF. We take Postulate V in the form of Playfair's axiom. Let

be the unique parallel

through C to Ah. By uniqueness and 1.29, LACE = LCAB and LBCD = LCBA. Thus LCAB + LCBA + LACB = LECA + LACB + LBCD = two right angles. cut by a line ACS with A perpendicular In the other direction, we are given lines 1, to n and LCAB acute. Let GI be on AB and construct GI H, perpendicular to AC' at Ht. Now Hi falls on the side of V toward C. If AH, is congruent to or exceeds AC, then we are done by Pasch's axiom.

n

Suppose H, falls between A and C. Construct points H2, H3, .... H on A with H and AH greater than AC. This is possible by the axiom AHi = HI H2 = = GIG2 of continuity. Construct points G2, G3, ... , G along A with AGI Let G2 K be the perpendicular to A_ from GZ at K. We show that K = H2. Let AL = G, H, be perpendicular to AC and join LG,. Since the angle sum in a triangle is two /G, HI A. right angles we see that L ALG, = L H, G, L = a right angle, and AAG, L

Thus AH, = LG,. Let M on G2K be such that KM = G, H,. By the same argument L K MG I = L HI G I L = a right angle and G I Al = H, K. Since L, G, . and M are now seen to be collinear, L LG, A = LG2G, M and so by Angle-Angle-Side, AG I MG2 - LAG, L A

andH,K - GIM - LGI - AH, =H,H2.Therefore K=H2. to

By iterating the argument we prove that Gk HA. and in particular. G H,,, is perpendicular is inside DAG N and n is parallel tot . Pasch's axiom implies Thus t

n leaves LAGnH through side AG,,, that is, t

meets A1

This proof is credited by Wallis and Sacchcri to Nasir al-Din al-Tusi. but Rosenfeld (1988) argues that it was due to a later geometer, perhaps al-Tusi's son. Lemma 3.5 allows us to construct certain triangles with angle sum equal to two right angles. To establish the full strength of Theorem 3.6 we prove the following theorem.

Theorem 3.7 (Three Musketeers theorem). If there exists one triangle with interior angle .sum equal to two right angles, then every triangle has angle sum two right angles.

If AABC has angle sum two right angles and from B we construct an interior altitude BT. then we can consider ABTC and ABTA. Both are right triangles and PROOF.

30


LBTC + LTCB + LCBT + LBTA + LTAB + LABT = four right angles. By the Saccheri-Legendre theorem this can only be so if

LBTC + LTCB + LCBT = LBTA + LTAB + LABT = 2 right angles. Therefore, if one triangle has interior angle sum equal to two right angles, so does a right triangle. We now show that every right triangle has angle sum two right angles.

M

S

B

A

T

N

Construct S, as in the figure shown, so that S is not on the same side of the line A as T

and AS - BT, BS - AT. Then AA BT - 6BAS and AT BS is a rectangle. If AA MN is obtained by marking off n copies of AT along A and n copies of AB along by filling in with rectangles we get that AAMN has angle sum two right angles. Now let A PQR be any right triangle and construct AA'QT' T' congruent to A A N M so that A' Q > P Q and T' Q > R Q. The triangle AA'QT' has angle sum two right angles and so

angle sum AA'QT' = angle sum AA'PT'

+ angle sum APT'Q - two right angles (from LA'PT + LT'PQ). The Saccheri-Legendre theorem forces the equation

angle sum AA'PT' = angle sum APT'Q = two right angles. Similarly the angle sum of A PT'Q equals the angle sum of A PQR plus the angle sum of APR T' minus two right angles and, as before, the angle sum of A PQR is two right angles. Thus every right triangle has interior angle sum two right angles. Since an arbitrary triangle can be divided into two right triangles by an interior altitude, the theorem is proved.

This theorem gives a simple criterion for the failure of Postulate V: If any triangle has angle sum strictly less than two right angles, then all do. Also, if one wants to prove Postulate V from Postulates I through IV, then it suffices to produce one triangle with angle sum two right angles. Saccheri tried in vain to prove this (Saccheri 1733). His efforts will be discussed in the next chapter.


31

Similarity of triangles We next consider one more noteworthy attempt to prove the fifth postulate due to John Wallis

(1676-1703). His approach to the blemish on Euclid's work was to render Postulate V a theorem from Postulates I through IV plus another assumption that was more self-evident. He proposed the following.

Wallis's postulate. To every triangle, there exists a similar triangle of arbitrary magnitude. That is, given a triangle & ABC and any line segment DE there is a point F so that AABC is similar to ADEF, that is, LABC = LDEF, LBCA = LEFD, and LCAB = LFDE. Wallis knew that he had assumed Postulate V in his postulate, which he considered more natural. Euclid's Postulate III provides similar circles of arbitrary radius. Wallis's postulate is the analogous statement for triangles, and hence for polygons.

Theorem 3.8. Wallis's postulate is equivalent to Postulate V.

PROOF. In one direction, suppose we have lines A and cut by ACS so that L BAC and L DCA sum to less than two right angles. We "slide" Ah along A6 (parallel transport?) to

the point C as follows - choose points Bt on A and Ai on A as in the figure. To any point P on AC construct the triangle &A PQ similar to AA iABI by Wallis's postulate. The line PZ is parallel to A4 by the alternate interior angles theorem. At C. the line of so constructed forms supplementary angles LACH and LCAB. Thus the line n enters the angle LACH. By continuity we know that somewhere in the slide, the image of A must cross t , say as Cl E, where E is on5Now t. construct the triangle with base AC similar to AC1 CE. This new triangle has sides that are segments along A and t and so these lines meet.

32


In the other direction we present a proof due to Saccheri (1733, Scholion III). Suppose we are given L1 A BC and segment DE. By Euclid's 1.23 construct

Dk andESsothat LEDR-LBACandLDESLABC. By I.17 these base angles sum to less than

two right angles and so by Postulate V, 5 meets ES at say F. It suffices to show L DFE - LACB. Let N be on DF so that NF - AC. By Playfair's BD E axiom there is only one line through N parallel to DE. By Postulate V it meets EF at a point, say M. Since is parallel to and Postulate V holds, LNMF = LDEF. By Angle-Angle-Side ANMF = ABC and so

LACB - LNFM LDFE and AABC is similarto ADEF. This theorem is startling in its consequences. Suppose Postulate V were false. Then no similar triangles could exist and Angle-Angle-Angle is a congruence criterion. As J. H. Lambert (1728-77) observed, AAA allows one to have an absolute unit of length the need for a standard meter or foot is eliminated by solving the problem:

Construct an equilateral triangle with a given interior angle.

If we agree on 7r/4 as the given angle, then the side of an equilateral triangle with such an angle, which could be constructed anywhere, could be a standard of length, and geometric construction would replace the need for a Bureau of Standards. To contrast with more familiar notions, we use an absolute unit, "1 /360 of a circle;' to measure angles (the degree). Similarly, given a line segment, we could construct an equilateral triangle on it by Proposition 1.1. The amount that this differed from n in angle sum could be fed into a function which gave out the length of a side, since the angle defect of an equilateral triangle completely characterizes it by AAA. The complete lack of similarity has a major drawback - no one could carry a reliable street map for it would have to be the size of the city. Architectural plans would need to be the same size as the building! Many other equivalents to Postulate V can be given. In the following theorem we mention some and summarize our previous efforts: Theorem 3.9. Postulate V is equivalent to each of the following:

(I) Playfair's axiom. (2) The set of points equidistant fmm a given line forms a line. (3) The interior angles of a triangle sum to two right angles. (4) Wallis's postulate. (5) The converse of the alternate interior angles theorem. (6) The perpendicular bisectors of the sides of a triangle are concurrent. (7) There exists a point equidistant fmm any three noncollinear points

(8) If C is on a circle with diameter AB and C does not lie on AB, then LACB is a right angle. (9) There exists a rectangle.


33

(10) There exists an acute angle such that every point in the interior of the angle is on a line intersecting both rays of the angle away from the vertex. (11) Any pair of parallel lines has a common perpendicular. (12) The Pythagorean theorem.

Exercises 3.1

Prove that the sum of two angles interior to a triangle is less than or equal to the remote exterior angle.

3.2' Prove the assertions of the last theorem that Postulate V is equivalent to (a) the existence of a rectangle (a quadrilateral with four interior right angles), (b) if C is on a circle with diameter A B and C does not lie on AB, then L ACB is a right angle, and (c) any pair of parallel lines has a common perpendicular. 3.3

Suppose all triangles have an angle defect, that is, rr- the interior angle sum of any triangle is a positive number. Let S(AABC) = a - LABC - LCAB - LBCA denote the angle defect. Suppose that a triangle is subdivided into two nonoverlapping triangles (for example, by an interior altitude); then show that the angle defect of the large triangle is the sum of the angle defects of the constituent parts.

3.4'

Criticize the following "proof" of Postulate V (due to Legendre, 1833) and determine which of the equivalents of Postulate V he assumed: "Given DABC, construct on BC the triangle LBCD F

congruent to AABC with LDBC = LBCA, LDCB LCBA. Then draw through D any line which cuts A and Ah. If the angle sum of DABC is jr - S. then the angle sum of AAEF is less than or equal to rr - 28. Repeating this process n times gives a triangle with angle sum less than or equal to n - 2"S. If S > 0, then n can be chosen large enough that 2"6 > n; this leads to a contradiction. Thus the angle sum of the triangle must be Yr." 3.5

On the sphere with great circles as lines we have seen several results that show that Postulates I through V cannot hold. Show how the various equivalents of Postulate V fail on the sphere and interpret their failure as a fact about the geometry of the sphere.

4 Non-Euclidean geometry I The non-Euclidean geometry throughout holds nothing contradictory. C. F Gauss (12 July. 1831)

Between Euclid's time and 1829. the year Lobachevskii's On the Principles of Geometry appeared, most of the critics of the Elements were concerned with the "purification" of Euclid's work from its perceived imperfections. So strong was the conviction that Postulate V depended on Postulates I through IV that some of them did not see in their work the basis for a new geometry. We begin this chapter with one such critic, Girolamo Saccheri (1667-1733). Further in the chapter the viewpoint changes when we consider the work of Gauss, Bolyai, and Lobachevskii, the founders of a "new geometry." What distinguishes this chapter from the next is the role of analysis in the discussion, as well as the role of space (three-dimensional). In this chapter we restrict ourselves, for the most part, to synthetic (not analytic) arguments and arguments in the plane.

The work of Saccheri Girolamo Saccheri was a Jesuit priest and professor at the University of Pavia. His Euclides

ab omni naevo vindicates ("Euclid vindicated of every flaw," Saccheri (1733)) marks a triumph of logic in the pursuit of a proof of Postulate V. It also contains the beginning of the study of non-Euclidean geometry, disguised by a flaw in Saccheri's work. The goal is to prove that Postulates I, 11, ill, and IV imply V. This can be accomplished through a reductio ad absurdum argument; assume Postulates I, II, Ill, and IV and the negation of Postulate V and then reason to a contradiction. Saccheri's great insight is his idea of a general theory of parallels with precise versions of the negation of Postulate V. The device that allows this generality is the Saccheri quadrilateral. In fact, this figure had been introduced much earlier in the work of 'Umar Khayyam. It also figured in a work by Giordano Vitale (1633-1711) who gave another petitio principii proof of Postulate V in his 1686 book Euclides restituto. Let A B be a line segment and suppose AD and BC are line segments with AD = BC, AD perpendicular to AB and BC perpendicular to AB. Join the points C and D to form the Saccheri quadrilateral ABCD. We first consider LADC and L BCD. The references to proposition numbers, scholia, and such in this section are to the original statements in Saccheri (1733).

34

4. Non-Euclidean geometry !

35

Proposition 4.1 (Proposition I). In the Saccheri quadrilateral A BC Don the base A B.

LADC - LBCD. C

PROOF. Consider the line segments AC and B D. Since A D- BC and AB - BA, by the congruence criterion Side-Angle-Side, A A C B is congruent to ABDA, and so AC - BD. Now by Side -Side-Side, 6BCD = AA DC and the proposition follows.

B

A

We need the next observation in a later result.

Proposition 4.2 (Proposition II). Let A BC D be a Saccheri quadrilateral on the base A B and let M and N be the midpoints of A B and CD, respectively. Then MN is perpendicular to A B and MN is perpendicular to CD. D

N

A

M

C

I/N

PROOF.

Join AN and BN. By Side-Angle-Side,

LADN -

ABCN. By Side-Side-Side we find AAMN - ABMN and so 1A MN = LBMN and MN is perpendicular to AB. By joining B

D to M and C to M the same argument applies to show that MN is perpendicular to CD.

Saccheri considers all possible cases of such a quadrilateral. They are LADC is a right angle: LADC is an obtuse angle: LADC is an acute angle:

HRA, the hypothesis of the right angle. HOA, the hypothesis of the obtuse angle. HAA, the hypothesis of the acute angle.

In the previous chapter we proved that HRA is equivalent to Postulate V so we may take HOA or HAA as negations of Postulate V. The three musketeers theorem (Theorem 3.7) implies that if one of HRA, HOA, or HAA holds for one quadrilateral, it holds for all.

Theorem 4.3 (Proposition III). Let ABCD be a Saccheri quadrilateral on the base AB. Under the assumption HRA. HOA, or HAA we have AB = CD, AB > CD, or A B < CD, respectively, and the angle sum of a triangle is =, >, or < two right angles, respectively.

N

U

A

1nM

C T

PROOF.

Let M and N denote the midpoints of AB and CD, re-

spectively. We consider the assumption HAA since HRA is known

and the argument for HOA is similar. Suppose CD < AB, then

CN < BM and so extend CD to T with NT - MB. Then

L MBT >a right angle. But, in the Saccheri quadrilateral NMBT, L BT N - L T BM by Proposition 4.1. Now L BT N is opposite the exterior angle L BCN. By HAA, L BCN is acute. Since L BT N = L T BM > a right angle we get a contradiction to the exterior angles theorem (Euclid 1.16). Thus AB < CD.


36

Now suppose we have AABC. a right triangle with right an-

D

C

A

B

gle at B. Construct AD perpendicular to AB with AD = BC. Assuming HAA, CD > AB. Thus LDAC > LACB since the greater side subtends the greater angle (Euclid 1.25). But LCA B + LDAC =a right angle. Therefore, the angle sum of AA BC is less than two right angles.

Saccheri, to prove Postulate V from the other four postulates, attempts to prove that HOA and HAA lead to contradictions. As we have already seen, HOA contradicts the SaccheriLegendre theorem (Theorem 2.2) and so HOA is disallowed. Saccheri then began "a lengthy battle against HAA which alone opposes the truth of the axiom" (Saccheri 1733, p. 13). He proved some remarkable theorems along the way. C

X

Consider two parallel lines A and 5, and construct lines AU and YV perpendicular to --h from X. Y on

D

Y

Consider L U XY and L V Y X. If both angles are acute, then,

H

IV

by continuity. A and

kD

b e tween the

segments

have a common perpendicular

UX an d VY If one o f /" VV or

V I

LV YX is right and the other is acute (by HAA), then CD and A already share a common perpendicular. Suppose LUXY is acute and LV YX is obtuse. If we move V away from U along A then LV YX may change to a right angle (giving a common perpendicular) or remain obtuse.

Theorem 4.4 (Proposition XXIII). In the preceding diagram, and assuming HAA, if as V moves away from U along Ah, L V Y X remains obtuse and A is parallel to

then n is asymptotic to Al. PROOF.

W

V

V'

If. for some V'Y' perpendicular to Al to the right of

V, we have V'Y' > V Y, then let IV be on V'Y' with V'W 25 V Y. By Proposition 4.1 L V YW = LV'W Y and by assumption

LV YW is acute. However, LV'Y'Y is obtuse and the exterior angles theorem implies LV'WY > LV'Y'Y which is absurd. Since V'Y' = V Y implies that there is a common perpendicular

to A and t contrary to our assumption, then we see that V Y > V'Y'.

It remains to show that V'Y' can be made smaller than any fixed line segment. For this we use the fact that, under HAA, all triangles, and hence all quadrilaterals, have an angle defect. We also assume that we can measures angles. By Theorem 4.3 the sum of the interior angles in a triangle is less than two right angles and so the sum of the interior angles of a quadrilateral is less than four right angles. Denote the angle defect of a quadrilateral

KLMN by

d(KLMN) = 27r - (L KLM + LLMN + LMNK + LNKL).

4. Non-Euclidean geometry l

37

Suppose there is a length, say of a segment ST, such that V Y > ST for every choice of V to the right of U. Along the ray A mark the points V1. V2, ... , V,, with UVI = V1 V2 = Vn-I Vn - ST. Construct the line segments V1 Y1, V2 Y2, ... , Vn Yn perpendicular to Al with Y1, Y2, ... , Yn on t h. Consider the angle defect of the quadrilateral UVVY,X. The basic property of angle defect is that it sums like area (Exercise 3.3) and so

S(UVnYnX)=S(UV1Y1X)+ +S(Vn_1VnYnYn-1) Since the vertical segments are perpendicular to A we see that

0 < S(UVYYX) < tr. Since each of the constituent quadrilaterals has angle defect greater than 0, there must be some Vk Vk+1 Yk+l Yk with S(Vk Vk+1 Yk+l Yk) < 7r/n. Since VkYk > ST we can always

construct a quadrilateral VkS'T'Vk+l with VkS' = T'Vk+l = VkVk+I = ST andeverysuch quadrilateral is congruent to any other. Let S (Vk S' T' Vk+l) = p > 0. We now choose n large enough so that tr/n < p. Since S(Vk Vk+l Yk+l Yk) = S (Vk S' T' Vk+l) + S (S' Yk Yk+I T'), we get S(S'Yk Yk+I T') < 0, a contradiction. This completes the proof.

In the figure notice that LVnYnC < LVn-1 Y,,-IC. It cannot be equal since this implies HRA and it cannot be greater because LVnYnC + LVn_IYn_1D would be greater than two right angles, which would violate HAA. Saccheri shows that asymptotic lines as in Theorem 4.4 exist under HAA.

Theorem 4.5 (Proposition XXXII). Given a point P not on a line A1 there are three classes of lines through P:

(1) lines meeting Ah, (2) lines with a common perpendicular to A

and

(3) lines without a common perpendicular to Al and hence asymptotic to A , PROOF. We can construct a line n with common perpendicular A P to Al by Euclid I.11. Suppose PD' is a line through P with common perpendicular B'D' to Al, and suppose

PD" is a line so that

LAPD' < LAPD" < a right angle.


38

By the exterior angles theorem L PD" D' is acute

and LA PD" is acute; by continuity of the change of angle, PD" must have a common perpendicular with A1 somewhere between A and Y. Suppose P1 is a line with P' meeting Ah at Y.

If P His another line with 0 < LAPX' < LAPX, then PX' is trapped inside A A P Y and, by Pasch's axiom, exits the triangle. Clearly it must exit through B

AY.

Let PAL be a line through P so that if P{' is another line and L A PW is less than L A PZ,

meets A1, that is. PAL is the upper limit of lines through P meeting A41. Let PZ' be the line so that if Pte' is any line with L A PV acute and L A PV > L A PZ', then PV has a common perpendicular with A First notice PZ' has no common perpendicular, for if it did one could find PV with LAPV < LAPZ' and satisfying the criteria for PZ'. PP'{

Claim: Pd = PZ'. Suppose PL 0 PZ' Since LZPZ' is acute, the distance

between Pl and PZ' can be made as large as desired (Corollary 3.3). So choose S' on PZ and Son Pal so that at SS' > AP. Construct the perpendicular from S' to T. Since PZ' is asymptotic to AP, AP > S'T. Now the distance between a line and a point is given by the perpendicular and since S' is on the opposite side of Pal from T.

A

S'T > SS'. Therefore, A P > S'T > AP, a contradiction. Thus PL = PZ'. Saccheri now concludes "the hypothesis of the acute angle is absolutely false; because it is repugnant to the nature of straight lines" (Proposition XXXIII). He reasoned as follows: The lines PAL and Ab are both "perpendicular" to the "line at infinity" at the same point (being asymptotic). Two lines cannot be perpendicular from the same point to a given line without being the same line and so the line PL cannot exist. After a tour de force of flawless logic, he attributes properties at infinity that are only true in finite ranges. The flaw in his work was observed by J. H. Lambert (1786). From Saccheri's work we have a wealth of theorems true for the system of geometry determined by the first four postulates with HAA; these include:

(I) Angle sums of triangles are less than two right angles. (2) There exist parallel lines without a common perpendicular which are therefore asymptotic. (3) Lines through a point not on a line fall into three classes (as in Theorem 4.5). His conviction that Euclidean geometry was the only true geometry clouded his vision to a new world he had effectively explored. This new world waited for almost another 100 years for giants to lay claim to it.

4. Non-Euclidean geometry/

39

The work of Gauss, Bolyai, and Lobachevskii We skip over the very interesting work of the next hundred years, slighting Legendre, Lambent, Taurinus, Schweikart, F. Bolyai, and others (see Bonola (1955) or Gray (1979)) to get to Carl-Friedrich Gauss (1777-1855). His contributions are found in two brief memoranda (Gauss, 1870, VIII. pp. 202-8), letters, and unpublished notes. The following is an excerpt from a letter that expresses his feelings on the investigations: The assumption that the sum of three angles is less than 180° leads to a curious geometry, quite different from ours, but thoroughly consistent ... I to Taurinus, 8 November /8241.

This sentiment distinguishes his work from his predecessors. It denies the absolute nature of Euclid's geometry. Gauss's unwillingness to publish his work left the public introduction of this "curious geometry" to Janos Bolyai (1802-60) and Nicolai 1. Lobachevskii (17921850) whose most striking contributions we will consider in the next chapter. We begin with a different definition of parallel:

Definition 4.6. Given a line A and a point P not on A1, let A P be the perpendicular

from P to Al A line "is parallel to A1 through P if for any line Iq with S lying in BA PQ and 0 < LAPS < LA PQ, it follows that 11 meets Al. If we consider the set of lines through the point P, the parallel is the "first" line in the set of lines through P, sweeping up from not meeting A4-h. Notice that without Postulate V we do not know if this first line is unique (Playfair's axiom), and so it depends on which direction we sweep the lines. To include this distinction we speak of parallel lines in a direction (here A h) according to this definition. Those lines through P that do not intersect A will be called nonlntersectingiines to distinguish them from these parallels.

Theorem 4.7. In a given direction, "being parallel" is an equivalence relation.

(a) We first show parallel to then, for any point P between P and Q. P'Q is parallel to Al, that is, parallelism is well defined PROOF (Gauss, 1870, VIII, pp. 202-5

for lines in a direction. Consider any line through

H

P', P'C, such that C is inside BA P'Q.

40


Join P to C and by assumption Pt meets A *-h. But now P'C has entered a triangle and, by Pasch's axiom, must exit through

(b) Next we show that if " is parallel to A1 then Ah is parallel to Let R be on P . Let AI be any line through A with S lying in QPAB. Choose ABC so that L RAC - i LSAB; then either A meets " or it doesn't. Suppose it does at point D. Let E be such that R E - RD, E # D. Then LEAD = L SA B. Choose D k with L A D F = L A ED. The line F is parallel to Ah and so DDD' meets A-h at say G. Let H be on " so that EH - DG; then by Side-Angle-Side, AA EH is congruent to AA DG. so that AR is perpendicular to

Thus LEAH =LDAG. Now LEAN = LEAD+LDAH = LSAB+LDAH = LDAG which implies that LSAB = LHAB and A coincides with AS. We leave it to the reader to prove the case when A does not meet P

.

(c) Finally we show that if Ah is parallel to " and n is parallel to UV, then A-h is parallel to

Case 1. "is between Ah and UV . Suppose AS is a line toward Band S inside BAUV. Since Al is parallel to, AS meets " at a point, say M. Let T be on A past M.

Since "is parallel to Ti,, 17 meets UP and so AS meets UV. B A

P

M

Q

Q

r

B

T

S

A

U

Case 2. Ah is between " and UV. If Al is not parallel to UV, there is a line U with Ah parallel to U{'. Now W is not inside UV BA since "is parallel to UV, which implies that UI' meets ", a contradiction. Also W is not outside UV BA since U--& parallel to A and A parallel to implies by Case I that ('ii is parallel to ", and so UV meets P, a contradiction. From this notion of parallel we define a function of a perpendicular line segment to a given line:

Definition 4.8. Theangle of parallelism fl (A P) is the angle between a parallel and a perpendicular through P to Al at A.

p R(AP)

Under the assumption HRA, I1(AP) equals a right angle for any AP; HAA implies that fl (AP) is strictly less than a right angle. We record some elementary properties of P(AP) here. In the next chapter we will determine an analytic expression involving

A

fl(AP).

(I) fl(AP) depends only on the length of AP. That is. if AP - CR, then fl(AP) _ ll(CR). To see this, suppose that n(AP) < I1(CR).

4. Non-Euclidean geometry 1

41

Then there is a line R1 so that LC R T- L A PQ and a point, say U. Now let E be on A with A E = CU and consider the triangle AA PE. Since

AP = CR and LPAE = LRCU, by Side-AngleSide, RAPE = ACRU. Then LAPE = LCRU, which equals the angle of parallelism, and the lines and P coincide, which is impossible. The case for fl(AP) > fl(CR) is symmetric.

Suppose we have chosen a unit length and there is a measure of length on line segmen

This property of the angle of parallelism allows us to define a function n :

l0, oo) -

1O,s/21 given by fl (x) = fl(AP), where AP is any line segment of length .r in the measure. (2) If HAA holds and A P > A R. then Il (A P) < fl(AR), that is, the angle of parallelism

is a monotone decreasing function of length. To see this, suppose that fl (AP) = fl(AR).

Let M denote the midpoint of PR. Construct MQ' perpendicular to " and extend it to MS'. Since Il(AP) = fl(AR), LMPQ' = LMRS'. We also have AMR - MP and L PMQ' = L RMS', so APQ'M = L RS'M. It follows that MS' is perpendicular to . is parallel to. share a common perpendicular. Since This implies that n and under HAA this cannot happen.

Suppose I1(AP) > fl(AR), then there is a line through P. say P' with LRPT LARS. Then LAPT < LAPQ and P meets Ah, and hence )q. By the alternate interior angles theorem (1.27), however, n does not meet

.

P Q'

M

Q

S.

S

A

(3) In fact, any acute angle is the angle of parallelism for some length. To see this, fix the

angle L BA X. Suppose R, S, is perpendicular to A at R, and St lies on. The triangle AA R, SI has some angle defect, say S > 0. Mark R2 along Ah with R, R2 = A Ri. If the at S2, then R2 S2 > R, S1 and we can construct the perpendicular to Ah from R2 meets point S, along R2S2 such that R2S, = R1 S. It follows that AAR2S2 contains two copies of LIAR, S, and so the angle defect of AAR2S2 is at least 28. If every perpendicular to A1 within L BAX meets Al, then we can iterate this procedure to construct a triangle with angle defect greater than any nS and so greater than two right angles. which is impossible. Hence L BA X is the angle of parallelism for some length.


42

We now introduce an object which has the property that it is absolute, that is, it does not depend on HAA or HRA. With it we can organize many properties of parallels. We begin with a pencil of lines, which is a set of lines satisfying one of two properties - the pencil of all lines through a given point and the pencil of all lines parallel to a given line. We could consider the pencil of parallel lines as the set of lines passing through the same point "at infinity" to gain an analogy to the pencil of lines through a point. Notice that every point in the plane lies on a unique line in a pencil of parallels, and every point other than the given point P lies on a unique line in the pencil of

lines through P.

Definition 4.9. Given a pencil of lines, two points X and Y are said to correspond with respect to the pencil, if the segment XY makes congruent angles with the unique lines in the pencil through the points. If one has a pencil of parallel lines, the set of all points corresponding to a given point is called a horocycle. If one has a pencil of lines through a given point, then it is a simple consequence of the converse of pons asinorum (Euclid 1.7) that the set of points corresponding to a given point forms a circle. A horocycle may be thought of as a "circle with center at infinity." We leave the proof of the following proposition to the reader.

Proposition 4.10. A horocyele is a line if and only if HRA holds. One way of characterizing the points on a horocycle is through the angle of parallelism. If A and B are on the horocycle, the perpendicular bisector of the segment A B lies in the pencil of parallel lines that determine the horocycle. The corresponding angles at A and B have measure fl (AM) = fl (BM), where M is the midpoint of AB. This will be useful in Chapter 5. We say that two horocycles are concentric if they are determined by the same pencil of parallel lines. The statements in the following lemma are elementary to prove and are left to the reader. Henceforth we will assume that we have chosen an interval as unit length and that all segments have a length. Curves have lengths that are limits of polygonal approximations.

Lemma 4.11. (1) To equal chords there correspond equal arcs of the horocycle. To the longer there corresponds a longer arc. (2) Segments of lines in the pencil intercepted between two concentric horocycles are congruent. (3) The arcs of concentric horocycles intercepted between two lines in the pencil decrease in the direction of parallelism. With these ideas we arrive at the following result.

4. Non-Euclidean geometry/

43

Theorem 4.12. The ratio of the lengths of concentric arcs of two homcvcles intercepted between two lines in the pencil is expressible in terms of an exponential function of the distance between these arcs.

PROOF. We first show that the ratio depends only on the distance. Denote the length of an

arc along the horocycle between points A, B by AB. First suppose BC =

m

AB. Then the n congruent segments along the parallels AA', BB', and CC' induce equal subdivisions of

A'B' and B'C'. We then have B'C' =

m n

A'B', and so

AB

n BC

A'B'

m B'C'

BC

=

B'C'

'

n

therefore the ratio AB/A'B' depends only on A A'. If A B and BC are related by an irrational number, a limiting process obtains the desired ratio. B"

B'

sl

s3

.s2

A'

Y

A..

Now suppose sl = AB, S2 = A'B', and S3 = A"B". We can write s,/s2 = f(x) and s2/s3 = f (y), where x and y denote the lengths AA' and A'A", respectively. It follows that .s, /s3 = f (x + y) and we obtain the equation

f (x)

f(Ax) - I lim f(x + Ax) - f(x) -= f(x) ex-O f'(x) = ex-O Ax Ax


44

By the definition of f (x), as Ax goes to 0, f (Ax) goes to 1, and we write, for some value of k, lim

ax-.0

f (AX) - 1 Ax

=

I

k

Then f(x) = f(x) and so f(x) = e /' or AB =

A'B'eAA'l

k The value I/k is chosen with malice of forethought, we will see in Chapter 13 how it figures in other situations. Before introducing trigonometry and the further properties of the angle of parallelism, we move on to some fundamental constructions due to Lobachevskii and Bolyai. Their original insight was to work in space rather than simply in a fixed plane.

Exercises 4.1

A Lambert quadrilateral is one in which three of the interior angles arc right angles. Show that the fourth angle is a right angle if and only if HRA holds. If HAA holds, show that the sides adjacent to the acute angle are greater than their respective opposite sides.

4.2'

Prove that if 1 and m are nonintersecting lines with a common perpendicular, then, under HAA, I and m diverge.

4.3

Complete the proof of part (b) in Theorem 4.7 where you consider a ray A that does

not meet the line ". 4.4

Assuming HAA show that for every acute angle there is a line parallel to both rays forming the angle.

4.5'

Given a pencil of lines parallel to a given line, if P. Q, and R are points, and P and Q correspond and Q and R correspond with respect to this pencil, then show that P and R correspond as well.

4.6'

Prove Proposition 4.10.

4.7

Prove the three statements of Lemma 4.11 on properties of horocycles.

4.8

There is yet another notion of a pencil of lines - the pencil of lines perpendicular to a given line. Show that the set of points that correspond with respect to this pencil is the curve of points equidistant to the given line. Observe that HRA holds if and only if this curve is a line.

4.9

Fix a line Al and a perpendicular segment OA. Suppose I is a line through O. and M is a point on I in the direction of X. Suppose that the perpendicular to A' through M

meets A ' at P. Show that the length of PM is a continuous function of the length of the segment OM. (Hint: Use Saccheri quadrilaterals.)

5 11

Non-Euclidean geometry II I have created a new universe from nothing. All that I have hitherto sent you compares to this only as a house of cards to a castle.

J. Bolyai (3 November, 1823)

The work of Gauss, J. Bolyai, and Lobachevskii differed from the work of their predecessors in that they were forging a new geometry, not repairing an old one. They also developed

methods in a three-dimensional space where HAA holds in each plane leading them to formulas in hyperbolic trigonometry. Let us assume that we are in a space. like the familiar three-dimensional space with points, lines, and planes. We assume that the properties of an incidence geometry (Definition 2.1), Hilbert's axioms of order and congruence, and Postulates I-IV hold on any plane. However, we are uncertain of the status of Postulate V. In order to develop the geometry of space we need to explore certain relations between pairs of lines, lines and planes, and pairs of planes. To begin, two lines that do not meet in such a space may be skew or they may be coplanar, that is, they lie in the same plane. If the latter occurs, then we can speak of the lines being nonintersecting or parallel if they satisfy the criterion of Definition 4.6.

Lemma 5.1. Suppose 11 and 12 are two lines parallel in space and T1 and T2 are two planes with It on T1, 12 on T2, and Ti 96 T2. If m is the intersection of Ti and T2, m # 11, and m 96 12, then m is parallel to 11 and 12Q

M

Notice that we have not assumed HAA or HRA. We first prove that m does not intersect It or PROOF.

Q S

12

It

S

B A

A'

12. Let To denote the plane on which 11 and 12 lie. If m meets 11 at a point P, then 12 and P lie on both the planes To and T2, which contradicts the assumption that T1 96 T2. Similarly m does not meet 12. Now let Q and Q' lie on m, let A lie on 11 = AA',

H

and let B lie on 12 = BB', so that AQ is perpendicular to 11, and QB is perpendicular to 12. Suppose QS is a line through Q on T1 so that L AQS < LA QQ' and QS lies in A'AQQ'. Consider the plane determined by B, S, and Q which meets To in a line BY in the direction of parallelism of l1 and 12. Thus the ray BY, which is a sort of projection of QS onto To, meets It, and it meets 11 where QS meets 11, that is, in the point representing the intersection of the planes To, T1, and QBS. Thus m is parallel to 11. Similarly m is parallel to 12. 45

46


Corollary 5.2. If in space 11 is parallel to 12 and 12 is parallel to 13, then !i is parallel to 13.

The proof is left as an exercise to the reader. We next develop certain relations between lines and planes in space.

DeSnitlon 5.3. A line ! Is perpendicular to a plane T if ! meets Tat a point P and every line on T passing through P is perpendicular to 1. If a line ! is perpendicular to a plane T, then we also say that the plane T is perpendicular to the line 1.

Lemma 5.4. If a line 1 meets a plane T at a point P, then I is perpendicular to T if and only if there are two lines m i and m2 lying on T, passing through P, and perpendicular to 1.

PROOF. One direction is obvious. Suppose n is a line lying on T passing through P. Choose points RI on m r and R2 on m2 so that n enters the triangle ARi PR2 at P and exits through a point Son R1 R2. Choose points Qi and Q2 on ! with P between Q1 and Q2 and PQr = Q2 P. Because m r and m2 are both perpendicular to 1. the right triangles formed satisfy

OQ1 PRi = AQ2PRi, and OQ1 PR2 - LQ2PR2. This implies RI Q1 - RI Q2 and R2Qi = R2Q2. By the congruence criterion Side-SideSide, ARI Q1 R2 = ARI Q2R2. This implies that LQr Ri R2 = LQ2 Ri R2 and LQr R2Rr

LQ2R2Ri. By Side-Angle-Side ARuSQu = AR1SQ2 and so SQi = SQ2. Thus A Q1 SQ2 is an isosceles triangle with the segment PS a bisector of the base Q i Q2. Hence PS is perpendicular to !.

The lemma allows us to construct planes perpendicular to a given line and lines perpendicular to a given plane.

Corollary 5.5. Given a line I and a point P, then there is a unique plane T containing P and perpendicular to 1. PROOF. We prove the case when the point P lies on the line l. The case P ' I is similar and is left to the reader. Because we have an incidence geometry there is some point Qj not on


47

the given line !. Let Tt be the plane generated by the line ! and the point Qt. On Ti there is a line 1i perpendicular to ! through P. Now the assumption of an incidence geometry further implies the existence of another point Q2 not lying in the plane T1. Let T2 be the plane generated by / and Q2. On T2 let 12 be the line perpendicular to ! through P. The plane T is the plane generated by !i and 12. If T' is another plane containing P and perpendicular to

!. then choose a point Q in T' that is not in T. Then the plane T" generated by I and Q meets the plane T in a line n and the plane T' in a line n'. However, because Q lies on n', n 54 n'. This gives two lines on the plane T" passing through P and both perpendicular to 1, which is impossible. Thus T is the unique plane containing P perpendicular to !. The dual problem of constructing a line perpendicular to a given plane passing through a given point is solved with the help of the following result.

Lemma 5.6. Suppose I is a line perpendicular to a plane T and P = I fl T. If m is any line on T and rA the line on T perpendicular to m with A on m, then if B # P is any other point on 1, then the line segment A B is perpendicular to m.

Let Q 54 A be any other point on m and Q' be on m with A between Q and Q' and A Q = AQ'. By Side-Angle-Side triangles O PAQ and APAQ' are congruent and so PQ = PQ'. Again by Side-Angle-Side the triangles APBQ and APBQ' are congruent from which it follows that B Q = B Q'. Finally triangles ABA Q and ABAQ' are congruent by Side-Side-Side and so LBAQ = LBAQ', that is, AB meets m at right angles. PROOF.

Corollary 5.7. Given a plane T and a point P there is a unique line I perpendicular to T and passing through P. If P E T, then let !t be any line on T passing through P. By Corollary 5.5 there is a plane Ti containing P and perpendicular to !i. On Ti let 1 be the line through P

PROOF.

that is perpendicular to the line m given by the intersection of T and T1. This line is also perpendicular to !i and so I is perpendicular to T. If !' is another line perpendicular to T passing through P, then let T" denote the plane determined by ! and 1'. This plane meets T in a line m and m has the property that 1 and


48

1' are both perpendicular to m at the point P in the plane T". Since this is impossible. I is unique.

If P ¢ T, then suppose m is any line on T. On the plane determined by P and m form the line segment PQ with Q on m and PQ perpendicular to m. On T construct the line n perpendicular to m through the point Q. In the plane determined by n and P let A P be the line segment perpendicular to n with A on n. By construction the plane of P, A. and Q is perpendicular to the line m. Now apply Lemma 5.6 with P and Q reversed to see that the line I = AP is perpendicular to the plane T. The uniqueness of I follows from the uniqueness of the line perpendicular to T through A E T. From the relation of perpendicularity of a line and a plane we develop the idea of planes being perpendicular. Definition 5.8. A plane T is perpendicular to a plane T' if there is a line I on T' that is perpendicular to T.

Lemma 5.9. A plane T is perpendicular to a plane T' if and only if T' is perpendicular to T.

PROOF. Suppose ! is a line lying on T' with ! perpendicular to T and ! meets T at a point P. Let m = T n T' denote the line that is the intersection of T and T' and let n be the line on T passing through P perpendicular tom. Since n is a line through P it is perpendicular to !. Since the plane T' is determined by the lines ! and m, n is perpendicular to T' and so T' is perpendicular to T.

Some further properties of perpendicular planes and lines in space are stated in the following proposition. The proofs are left to the reader.

Proposition 5.10. (1) If T and T' are perpendicular planes and ! is a line on T, then 1 is perpendicular to T' if and only if ! is perpendicular to the line m = T n T'. (2) If 1 i and 12 are lines both perpendicular to a plane T. then 11 and 12 are coplanar and nonintersecting.

49


Proposition 5.11. Given a line ! and a plane T with I not lying on T there is a unique plane T' containing I and perpendicular to T. Given two distinct planes T1 and T2, there is a plane T" perpendicular to both T1 and T2. Let A be any point on I that does not lie in T. By Corollary 5.7 there is a line m passing through A and perpendicular to T. The plane T' determined by ! and m is perpendicular to T. By the uniqueness of the line m and Proposition 5.10 the plane T' is PROOF.

the only plane perpendicular to T and containing !. From a point A in T1 that is not in T2 construct the line 1 through A perpendicular to T2. If I n T2 = B, then construct the line m through B perpendicular to T1. If ! = m, then any plane containing I is perpendicular to both T1 and T2. If 154 m, then the plane determined by 1 and m is perpendicular to Ti and T2.

Definition 5.12. If T' is the unique plane containing a line ! and perpendicular to a given plane T, then the line m = T f1 T' is called the perpendicular projection of l onto T. A line ! is parallel to a plane T if ! is parallel to its perpendicular projection onto T. Two planes are parallel if the intersections with some common perpendicular plane are parallel lines.

Notice that two planes that intersect cannot be parallel; a common perpendicular is perpendicular to the line of intersection. Thus parallel planes do not intersect. We next prove the analog of Playfair's axiom with lines as points and planes for lines.

rr

--------

AI

Y

------------------

I

------------------

nt

n Theorem 5.13. Through a line I parallel to a plane T there is only one plane that contains I and is parallel to T. PROOF. Let fl be the unique plane containing ! and perpendicular to T. Let in = n n T so 1 is parallel tom. Let T' be the plane containing ! perpendicular to fl. By definition T' is parallel to T. Suppose A is a point on I and A B the line segment on Il perpendicular tom at B. Suppose AA' and BB' determine the direction of parallelism of! and m. To preserve the

50


direction of parallelism consider the plane perpendicular to fl containing A B. This plane separates space into two half-spaces. The half-space containing A' and B' is the direction where intersections will occur between planes. Suppose U is another plane containing I. Let Z denote the point on U with BZ perpendicular to U at Z. We first observe that under HAA Z lies in the half-space of A' and B' between T and T'. Consider the plane determined by Z and m. The intersection of this plane with U is a line that is parallel to ! and m by Lemma 5.1. Furthermore the angle LB'BZ = fl(BZ) is acute which puts Z in the half-space of A' and B'.

Let ZY denote the line segment on U with Y on 1 and ZY perpendicular to !. To show that U meets T we consider the plane Il' determined by A B and Z; fl' is perpendicular to

be the intersection of fl' and T; also Al = Il' fl U. We show that L BAZ < LBAY = fl(AB). It then follows that AAlr meets -81 from the definition of the angle of parallelism. To compare the angles join B to Y and consider the triangles ABAY and ABAZ. Since BZ is perpendicular to U, ABZY is a right triangle and so, by Euclid's U. Let BB

Proposition 1.18, BZ < BY. By Lemma 5.6 LBYA is a right angle. In the plane of ABYA

construct a triangle ABZ'A = ABZA. Since BZ < BY, LAYB = LAZ'B, and both triangles are on the base AB, a judicious use of Euclid's 1.19 implies that L BAZ < L BAY and the theorem is proved.

Observe that Theorem 5.13 is a fact in Absolute Geometry, that is, it can be proved under the assumption of HAA or HRA. We use this fact in conjunction with the spatial analog of the horocycle. Consider the pencil of lines in space parallel to a given line in a given direction. Two points in space are said to correspond if they correspond in the plane determined by the two parallels through them.

F A

B

rn

5. Non-Euclidean geometry !!

51

Definition 5.14. The set of points corresponding to a given point in space with respect to a given pencil of parallel lines is called a horosphere. A reasonable property to be expected of the horosphere is that it be "homogeneous" in the sense that it can be defined by any of its points. This is made precise by the following result. Proposition 5.15. Two points in space corresponding to a third correspond to each other.

In Lobachevskii's Theory of Parallels this is Proposition 34 and Proposition 10 in Bolyai's The Science Absolute of Space (see Bonola (1955)). Suppose A corresponds to B PROOF.

H

HHHH

and A corresponds to C. Let D be the midpoint of A B and E the midpoint of AC. Let AA',

BB', CC', DD', and EE' be lines through these points in the pencil of parallels. By the definition of correspondence we have that

LD'DA=LD'DB=LE'EA=LE'EC=aright angle. Consider the plane containing ABC and the plane containing the lines AA' and B'B; the angle between them is fl (a) for some length a. Let FD be of length a, perpendicular to

H

H

A B. and lying in the plane of A BC. The line FF' perpendicular to the plane of A BC is part of the pencil of lines since it is parallel to DD' by the choice of the length of FD. It follows

4-

H

4-

that FF' is parallel to EE' and so they are coplanar. The line AC is perpendicular to EE' which lies in a plane perpendicular to the plane of ABC containing AC. It follows that AC is

perpendicular to any line through E in the plane of EE'F'F, including EF. Construct AF, BF, and CF. By repeated application of Side-Angle-Side we find that A F = BF - CF. Construct the perpendicular bisector from F to BC (ABFC is isosceles) to the point G and consider the intersection of the plane BB'C'C with the plane FF'G, which we denote by GG'. Since FFFF" is parallel to BY and CC', we get that GG is parallel to BY and hence to CC'. But LG'GB = LG'GC = a right angle, so B and C correspond. From Proposition 5.15 and the properties of parallelism we can take any line in the pencil of parallels as the axis of the horosphere and any point as the initial point to define it (all other points on it correspond to the initial point).

52


If we intersect a plane containing one of the parallels in the defining pencil with the horosphere, the intersection is a horocycle. Furthermore, any two distinct points on the horosphere determine a unique horocycle. If we interpret "line" as horocycle on the "plane;' that is, the horosphere, Theorem 5.13 is Playfair's axiom and so the geometry of horocycles and points on the horosphere is Euclidean. This result was established before the works cited earlier. It is due to F. L. Wachter (1792-1817), a student of Gauss. For the next result choose some line segment for a unit length and so determine a measure of length on line segments.

Theorem 5.16 (the Lobachevskii-Bolyal theorem). There is a constant k for which the following equation holds when the length of the line segment A P is given by the real number x:

tan(n(x)/2) =

-xlk.

(.

Except for the determination of the constant k, this formula gives an analytic expression to compute n (x). Notice that a nice addition formula also follows immediately from it: tan

(n (x2 y)1

tan

(n2w) tan

Also HRA corresponds to the case k ll= oo.

Before we begin a proof of Theorem 5.16 we record some elementary facts about the hyperbolic trigonometric functions sinhx, cosh x, and tanhx. These functions arose as analogs of the ordinary trigonometric functions with which one can express the area under a semicircle as an integral f a - x 2 dx. Area under a hyperbola is given by the integral f x - a dx whose solution shares many properties with ordinary trigonometric functions. In 1768, J. H. Lambert produced a comprehensive study of the hyperbolic

trigonometric functions containing the work of Euler and the Bemoullis. Formally introduce a complex variable into the Maclaurin series for ex to get

ei) =_ l +iy -

/

- iY3 3!

Y2 2!

v4

4!

a

2

J

5

\

=c\\osy+isiny. It follows from these formulas that et"+e-;,,

cosy=

2

:sing

2

By analogy with these formulas, we introduce the hyperbolic trigonometric functions ex -x ex _ e-x cosh x = , , and tanh x = sinh x e 2 cosh x It is interesting to note that 2 cos(i y) = cosh y, and sin(iy) = i sinh(y).

The analysis that supports our formal manipulations is classical and can be found in any good complex analysis book (for example, Ahlfors (1966)).

5. Non-Euclidean geometry Il

53

Lemma 5.17. Let 01 and OO' be perpendicular lines and OB the horocycle segment at right angles through 0 with axis . lfu = OA. v = AB. and s = OB denote the lengths of each segment, then evl k = cosh(u/k). PROOF. Consider the diagram. Here TT is parallel to both VX and -0 which are taken to be perpendicular. Choose some length u > 0. Along 01 let A denote the point with OA of length u. Let AA be the line parallel to O' through A. Let NN be perpendicular to AA and parallel to O0 Since LOAA' = 11(OA) =fl(u) and LNAY = LOAA'

we have AN = OA. Similarly let U be on Y) with OU = OA, U # A, and UU' the parallel to 01 through U in the direction of X. Let UM have length u and construct MM

perpendicular to UU and parallel to'. This determines concentric horocycles N D6 C, and ME. Construct the perpendiculars to T T' from N. O, and M, to obtain points D', C', and E', respectively. By the construction ND' = OC' = ME', all of length x for which

17(x) = r/4. It follows from Lemma 4.11 that ND = OC = ME. We denote their common length by a.

Now apply the formula in Theorem 4.12. That is, the ratio of the lengths of concentric arcs

of two horocycles is given by AB = A'B'e-`l t. where t is the distance between the arcs.

H

H

Let B denote the point along AA' where OC crosses. Let V denote the point corresponding

to B along UU'. Write BN = u + v. and lets denote the length of OB; we then have or - s = e_(u+v)/k. a = (a - s)et°+Olk or a Since JIM = u - v and we have the concentric arcs V C = a + s and ME = a, then we also have a + s = ae(Q-v)l k, or (a + s)/a = e(u-v)lk Thus 2 = a + s + or - s = e_(u+v)lk a or Multiplying through by evl k gives

+ e"''''.

eulk

evl k =

This proves the lemma.

2e-ul k

= cosh(u/k).


54

Corollary 5.18. s = a tanh(u/k). PROOF. By the foregoing discussion we see that

2 s = a+ S- a- s = e(u-v)/k - e-(u+v)/k a

or

_

or

(eu/k

-

Therefore,

S eu/k

a

= S cosh(u/k) =

a

e-u/k)e-v/k.

/k - -u/k = sinh(u/k)

-e"

2

and so s = or tanh(u/k).

We now complete the proof of Theorem 5.16. Let DABC be a right triangle in the (hyperbolic or Euclidean) plane with the right angle LACB. Let AA' be the line in space per-

H

pendicular to the plane of DABC. Lines CC' BHB' are the lines through C and B in the and

H

pencil of lines in space parallel to AA' in the di-

rection AA'. These lines meet the horosphere through A associated to this pencil at points Cl and B1, giving a triangle DABC ( on the horosphere. The geometry of the horosphere is Euclidean and so ordinary trigonometry applies to AAB(C(. First observe that LAC( B( is a right angle since the planes ABC, A A'C'C, and CC'B'B

are all at right angles. We next record the relations between s(= B)Ci, S2 = ACi, and

s3=AB( in terms ofa=BC,b=AC,andc=AB. From our previous lemmas we have s2 = a tanh(b/k) and s; = a tanh(c/k), where or is the length defined in the proof of Lemma 5.17. To express si consider the piece of the concentric horocycle CB2:

Si = CB2e-CC,/k = (a

tanh(a/k))e-CC,/k.

Now CC( corresponds to v in Lemma 5.17 for the figure ACC( and so

eCCi/k = cosh(b/k),

and

tanh(a/k)

s( =a cosh(h/k)

The Pythagorean theorem on the horosphere yields s3 = si + sz. This gives us

(tanh(c/k))2 =

(tanh(a/k))2

+ (tanh(b/k))2.

The fundamental relation for the hyperbolic trigonometric functions is given by

cosh2 x - sinh2 x = I.


Dividing by cosh2 x and rearranging we obtain tanh2 x = I find Z

tank (c/k) =

55

- coshZ x 1

Substituting we

coshZ(a/k) - I cosh2(b/k) - 1 cosh2(a/k)cosh2(b/k) + cosh2(b/k)

=I-

1

cosh2(a/k) cosh2(b/k)

We also know that tanh2(c/k) = I -

and so we have proved cosh clk)

Theorem 5.19 (the hyperbolic Pythagorean theorem). cosh(c/k) = cosh(a/k)cosh(b/k).

Now let us consider LCAB = LCIABI. Since s2 = s3cos(LCAB), we have a tanh(b/k) = a tanh(c/k) cos(LCAB) or

cos(LCAB) =

tanh(b/k) tanh(c/k)

Similarly, had we begun with BY perpendicular to the plane of AA BC, we would have gotten

cos(LABC) =

tanh(a/k) tanh(c/k)

Consider the ordinary trigonometric relation si = s3 sin(LCAB). It follows that

a tanh(a/k) cosh(b/k)

= a tanh(c/k) sin(LCAB)

or

sin(LCAB) =

tanh(a/k) cosh(b/k) tanh(c/k)

Look at our right triangle on the plane as in the adjacent figure. As a goes to infinity, so also does c and LCAB tends to 11(b). Since b xlimo

tanhx = lym

-co ex +e-

x=

71

C

a

B

we see then

cos 11(b) = lim cos(LCAB) = lim

aoo

and

sin 11(b) = lim

tanh(b/k)

a-.oo tanh(c/k)

= tanh(b/k),

tanh(a/k) cosh(b/k) tanh(c/k) = cosh(b/k) 1

1,

56


Finally, to prove the Lobachevskii-Bolyai theorem, we apply the half-angle formula

tan(0/2) =

I - cos9 sin o

tan(11(b)/2) =

I - cos Il(b) sin fl(b)

-

I - tanh(b/k) = cosh(b/k) - sinh(b/k) (1/cosh(b/k))

- eb/k + e-b/k _ eb/k - e-b/k = e_blk 2

2

The circumference of a circle As an application of the trigonometry associated to the geometry of Lobachevskii and Bolyai, we compute the circumference of a circle of radius R. The resulting formula can be compared to a formula in Chapter 13 where the mysterious constant k will be determined. We begin with the non-Euclidean analog of the spherical sine theorem (Theorem 1.4).

This result appears in the 1825 work of F. A. Taurinus (1794-1874), in the work of Lobachevskii, and in a particularly interesting form as Proposition 25 of J. Bolyai's paper (in Bonola (1955)).

Theorem 5.20. In a triangle DABC with sides of length c for AB. b for AC, and a for BC we have

sinh(a/k) _ sinh(b/k) _ sinh(c/k) sin LA

sin LB

sin LC

PROOF. Like the proof of the spherical sine theorem, we prove the case of a right triangle and leave the general case to the reader. Suppose that the right angle is at C. In the discussion after the proof of the hyperbolic Pythagorean theorem we proved that

sin LA =

_

tanh(a/k) cosh (b/k) tanh(c/k)

sinh(a/k) cosh(a/k) cosh(b/k) tanh(c/k)

_ sinh(a/k) sinh(c/k) '

Rearranging the terms gives one of the equations. Because there is nothing special about choosing the vertex to be A. the other equation holds by relabeling.

The relations just proved are just what we need to compute the perimeter of a regular n-gon inscribed in a circle of radius R. Let S denote half of the length of a side of the regular inscribed n-gon. A right triangle is formed by the center of the circle, the midpoint of a side, and the adjacent vertex. The interior angle at the center is n/n for this triangle, and Theorem 5.20 implies

sin(tr/n) It follows that

sinb(R/k) sin(tr/2)

sin(tr/n) sinh(R/k).

5. Non-Euclidean geometry II

57

The strategy to get the circumference is due to Archimedes - we compute the limit, limn..,. 2nSn, of the perimeter of the regular inscribed n-gon. Recall that lim x-.0

sinh x X

= I.

This follows easily from the Maclaurin series sinh x = x + 7T + exists, then it follows from the elementary theory of limits that

+

If limn 2nSn

lim 2nSn =n-.00 lim 2nk(Sn/k) = n-.o0 lim 2nksinh(Sn/k).

N-00 We compute

lim 2nSn = lim 2nksinh(Sn/k)

R-00

n-00

= lim 2nk sin(n/n) sinh(R/k) n-+00

= 27rksinh(R/k) lim

sin(n/n)

n-- (n/n)

= 2rrksinh(R/k). Theorem 5.21 (Gauss 1831). The circumference of a circle of radius R is given by

2nk sinh(R/k). If we denote the circumference of a circle of radius R by circum(R), then notice that the formula leads to a series expression:

circum(R) = 2nksinh(R/k)

=2nk

/

3

R5

k+6k3+120k5+

=2nR+

it R3 3k2

+

When k goes to infinity, we obtain the familiar formula for the circumference of a circle. The term 7rR3/3k2 will be identified in Chapter 13. We pause here to reflect on our achievements. From a workable definition of parallel lines in space we have taken some elementary notions and derived formulas in a manner consistent with analysis. These formulas contain an unknown, k, and they coincide with Euclidean formulas when k goes to oo. If k # oo, we have developed a non-Euclidean geometry and, because our formulas do not lead to contradictions, we are led to abandon the idea that Postulate V follows from Postulates I through IV. Thus, through the power of the axiomatic method allied with analysis, we have reached a turning point in the 2000year-long history of Postulate V. What remains to finish the story is a concrete realization of non-Euclidean geometry, a mathematical object consisting of "points:' "lines:'and "planes" which are related in the way developed by Gauss, Lobachevskii, and Bolyai. For this goal we turn for support and foundations to deeper analysis and its relation to geometry.


58

Exercises 5.1

The generalized Pythagorean theorem takes the form cosh(c/k) = cosh(a/k)cosh(b/k).

Show that ask -+ oo this gives the Pythagorean theorem. This implies that if a, b, c are small in comparison to k, then the Pythagorean theorem is very closely approximated. 5.2

Under HRA show that any convex quadrilateral tiles the plane, that is, by repeating the figure the plane can be covered without overlapping. Under HAA show that any convex polygon with an even number of sides with angle sum equal to a submultiple of 27r tiles the plane. What happens when there is an odd number of sides?

5.3

In the astronomical diagram shown let p denote the parallax of Sirius, which is I ", 24. Sirius

From the Lobachevskil-Bolyai theorem calculate an upper bound for a/k.

Sun

5.4'

a

Earth

Prove Lobachcvskil's formula for a triangle AABC with the length of side denoted by the opposite vertex:

cos(LBAC)cosfl(b)cosfl(c)+

sin Il (b) sin f1(c) sin fl (a)

= I.

5.5

Prove Theorem 5.20 for a general triangle, AABC.

5.6'

From Theorem 5.20 ( previous exercise) deduce the following construction for f1(x ): Let AE and BD be perpendicular to A B. Let A E have length x. Let DE be perpendicular to AE. If DE meets AE at E, then let AO = DE with 0 on BD. Then LOAE = 1(x).

5.7

Prove Corollary 5.2 and Proposition 5.10.

5.8

If a is a side and y the interior angle of a regular n-gon prove that

z cos

5.9

y

a

sin 2 cosh -. 2

If AA BC is given, show that the perpendicular bisectors of three sides belong to a pencil of lines (that is, either concurrent, parallel, or perpendicular to another line).

5.10 With HAA, say that a figure made up of three mutually parallel lines encloses a finite area, say A. We prove a theorem of Gauss in this exercise that there is a constant A such that the area of an arbitrary triangle is ,l()r - a.- fi - y), where a, fi. y are the measures of the interior angles of the triangle. Following Gauss in a letter to F. Bolyai (1832), introduce the function f(m), which is the area subtended by the supplement to angle 0


59

and the parallel to both rays (figure A). Then f (¢) + f (x - 0) = A as in figure B. Figure

C shows how f(4,) + f(>G) + f(tr - 0 - i) = A. Now deduce that f(4)) + f(*) = f(m + ,'), which can only be satisfied by a linear function f (O) = X¢ + c. Since f(O) = 0, c = 0. Figure D shows how to introduce the triangle.

5.11

Another approach to Gauss's theorem of the previous exercise is to consider angle defect. Show that angle defect is proportional to area for triangles by showing that angle defect is additive, that is, if a triangle is decomposed into subtriangles, the angle defect of the large triangle is the sum of the angle defects of the parts. Show that this implies that there is an upper bound to the area of any triangle. In particular, a figure made up of three mutually parallel lines encloses a finite area.

PART B

Development Differential geometry

6 Curves I have been occupied with a new discovery ... in order to make my clock even more exact .... What, however, I never had expected I would discover, I have now hit upon, the undoubtedly true shape of curves ... I determined it by geometric reasoning. C. Huygens (6 December 1659)

The history of differential geometry begins with the subject of curves. Notions such as tangents to curves can be found in Euclid, Archimedes, and Appolonius. The calculus is founded on geometric ideas and so it is natural to find investigations of curves among the topics treated by the pioneers of analysis, Newton, Leibniz, the Bernoullis, and Euler. We begin our consideration of differential geometry with the basic definitions and prop-

erties of curves in R", Euclidean n-space. In the case of curves in R2 we introduce an important invariant, curvature, whose generalizations guide later developments.

Definition 6.1. A parametrized, differentiable curve is a mapping, a: (a. b) - R". for some -oo < a < b < oo. such that a(1) has derivatives of all orders. The curve a is regular if da 0 O for all t in (a. b). Such a curve can be written in terms of its coordinate functions:

a(t)=(xi(I),x2(1).... .e (t)) and so da =(x1(1),x2(t)....x;,(r)). We refer to the image of a as its trace, which is the geometric object of interest. More generally we can speak of curves of class k, that is, the mapping a : (a, b) --+ R" has continuous derivatives up to order k. One then requires the appropriate order of differentiability in each definition and theorem. To focus more on the geometry than the analysis we have ignored this subtlety by assuming curves to be smooth, that is. differentiable of all orders. The careful reader can discover the most general formulation of what follows by providing the least order of differentiability needed for each proof.

Since the trace of a curve lies in R", we can apply the extra structure enjoyed by R", which is a vector space with an inner product. Recall that the inner product or dot product on R" of two vectors, v = (vi, .... v") and iv = (wi, ... , w,), is given by the formula

=viwi+ +v"w". The derivative of the dot product of two differentiable curves, a, f: (a, h) -+ R". satisfies the Leibniz rule: d r

fa(r) ,B(r)] = a'(r) ,B(t) + a(r)

14'(r).

63

Development: Differential geometry

64

Definition 6.2. Given a parametrized, differentiable curve, a : (a. b) - R". the tangent vector to a at a(t) is given by

a'(t) = j (t). The norm of a'(t), IIa'(t )II =

« (t) a (t). is called the speed of a at a(t).

Notice that a regular curve has speed always greater than zero. EXAMPLES.

(1) Let a: iR - R" be given by a(t) = p + tq. for jt, q in R", q 36 0. This

gives a parametrization of the straight line through p and q of speed 11411.

(2) Let a: R - R2 be given by a(t) = (rcos(t). r sin(s)). The trace of this curve is the circle of radius r centered at the origin: a has speed r. Let 6: R -+ R2 be given by $(t) = (r cos mt, r sin mt). where m is a positive integer. Both curves have the same trace but fl travels the trace m times as fast: IIB'(t )II = Jm2r2 sine ml + m2r2 cost ..t = mr = m lla'(t)11.

(3) Let a: R - R3 be given by a(t) = (a cost, a sin t. bt), a. h 96 0. This is called the right helix on the cylinder of radius a of pitch 2Jrb.

(4) Let a: R - R2 be given by a(t) = (t3. t2). This curve is not

regular because a'(0) _

(3t2. 2t)I: o _ (0, 0). The trace has a cusp at the origin.

(5) If f : (a, h) -+ R is a smooth function, then the graph of J' is the regular curve given

by a(t) = (t. f(t)).

A helix

A graph

A given trace has many parametrizations. For example, once a parametrization a: (a. b) -+ R" is chosen, it may be altered as follows:

Definition 6.3. Let a: (a. b) -+ R" he a regular, parametrized, differentiable curve and g: (c, d) - (a, b) a real-valued function. If g is smooth and g has a smooth inverse, then we say that g is a reparametrizatlon of a. the reparametrized curve being 6 = a o g. It is convenient to find some property that makes the choice of parametrization unique. We can standardize the choice of parametrization as follows:

6. Curves

65

Definition 6.4. Suppose a: (a, b) -- R" is a regular, parametrized, differentiable curve and a < ao < b. The arc-length function s: (a, b) -a R is defined by

s(t) = f Ila'(r)Ildr

=f

.'(r) a'(r)dr.

o

0o

From elementary calculus, s(t) equals the distance along the curve a from a(ao) to a(t), oriented by the direction a goes to b and the choice of ao.

Proposition 6.5. The arc-length function s(t) is independent of reparametrization. PROOF. Suppose g: (c, d) -+ (a, b) is a reparametrization and g takes co to ao. We write

t = g(u) and fl(u) = a(g(u)). Then u

s(u) = f IIf'(» )lldv = fc. Ila'(r)II ca

since I dg

I

dg dv du

Ila'(r)Ildr = s(t), as

d v = ±dr. In the case that g is decreasing (for example, (0, I) 8 ' (0, 1) given

by x H I - x), the sign ± is negative. However, in that case we are integrating in the opposite direction of the integral determined by a and so the negatives cancel.

When a is a regular, parametrized, differentiable curve, we use arc length to obtain a standard parametrization of the curve. Let (c, d) denote the image of s: (a, b) -, R. Then ds

d r'

dt

dt JQO

III (r)Ildr = Ila'(t)II # 0.

By the familiar theorems of elementary calculus, s(t) is one-to-one, onto, and smooth.

Let g: (c, d) -+ (a, b) denote the inverse function for s(t), that is, t = g(s(t)). Since s(g(s)) = s, we have dg = and we see that g is smooth. Let (3 = a o g. It follows

ds/dt'

that s

f IIf'(w)lldw = 0

fs

Ila'(r)II 0

I aw

I dw

=

f

g(s)

Ila'(r)lldr = s(g(s)) = s.

Thus 0 has its arc length as parameter. We call f(s) the arc-length parametrization of a. Observe that a curve parametrized by arc length has speed given by: Ila'(t)II

dg ds

Ila'(t)II = Ila'(t)II = 1 Ila'(t)II

Ids/dtl

that is, the tangent vector to f has length I everywhere. We speak of a curve parametrized by arc length as a unit-speed curve. We reserve the variable s for the arc-length parameter when it is convenient and t for an arbitrary parameter.


66

EXAMPLES. (I) If a(t) = p+tqq, then fl(s) = p+s-911 (2) If a(t) = (r cos t, r sin t), then Ila'(t)11 = r, and s(t) = r(t - ao). This yields g(s) _ (s/r) + ao, and if we take ao = 0, then

,6(s) = (rcos(s/r), r sin(s/r)). (3) If a(t) =(a cost, a sin t, bt) is a right helix. then IIa'(t)II =

(s)acosl

a +b I'asin(

a +b

)b

a + b and

a +b-/

(4) The curve a (t) = (a cost. b sin t), for b 54 a, gives an ellipse in the plane. Furthermore,

cr'(t) = (-a sin t, b cos t) and 11a'(011 =

a2 sin2 t + b2 cost r =(b2 4-22) - a2) cos2 t.

The resulting length function,

s(t) =

J0

/

a2 + (b2 - a2) cos2 rdr.

is not generally expressible in terms of elementary functions; it is an example of an elliptic integral.

Early work on plane curves (Huygens, Leibniz, Newton, Euler) Now that we have introduced curves, their parametrizations, and their tangent vectors, we consider the particular case of curves in the Euclidean plane R2. Our next goal is to associate a function to each regular curve that measures how much

the curve is "bending" at each point. The classical (Greek) curves of the straight line and the circle are understood to "bend uniformly," that is, the straight line "bends" not at all and circles "bend" the same at every point. To define this function, the (unsigned) plane curvature Kj(t), we assign its value on lines and circles: For a straight line we want

K±(t) - 0. For a circle we want Kt(t) = 11r, where r is the radius of the circle, that is, small circles bend more than large circles. The problem of generalizing this function to an arbitrary curve is implicit in work of Appolonius (third century n.c.) and in work of Johannes Kepler (1571-1630). It is taken up by Gottfried Leibniz (1646-1716) in a paper of 1686 and developed further by Jakob Bernoulli (1654-1705) and Johann Bernoulli (1667-1748). The most significant treatment of curvature for curves and for surfaces (see Chapter 9) is due to Leonhard Euler (1707-83). Let a: (a, b) - 1R2 be a regular plane curve, parametrized by arc length. In the Euclidean plane, three noncollinear points lie on a unique circle. For s in (a, b), take si, s2, and s3 near s so that a(si ), a(s2), and a(s3) are noncollinear. This is possible as long as a is not part

6. Curves

67

of a line near a(s). Let C(si. s2, s3) denote the center of the circle determined by the three points. We use the dot product to form a function approximating the square of the radius of this circle,

P(s) = (a(s) - C(sI, s2, s3)) (a(s) - C(si, s2, s3))

If a(s) is smooth, then so is p(s). Since p(si) = p(52) = p(s3), by Rolle's theorem there are points tl E (St, s2) and 12 E (s2, s3) with p'(tt) = p'(12) = 0. Applying Rolle's theorem again we get u E (ti, t2) with p"(u) = 0. Now p'(s) = 2a'(s) (a(s) - C(s,, s2, s3)) and p"(s) = 21a"(s) (ar(s) - C(st, s2, s3)) +a'(s) a'(s)I Thus, since p"(u)

= 0,

a"(u) (a(u) - C(si, s2, s3)) = -a'(u) a'(u) Take the limit as si . s2, s3 go to s; then C(si, s2, s3) converges to a value C (s), and 11, r2 go

to s, so p'(s) = 0 and a'(s) (a(s) - Ca (s)) = 0. Furthermore, a"(s) (a (s) -Ca (s)) = -1. The circle centered at Ca(s) with radius a(s) - Ca (s) shares the point a(s) with the curve a and furthermore,

the tangent to the circle at a(s) is a multiple of a'(s). The circle, called the osculating circle (Leibniz 1686; osculare, to kiss), is tangent to a(s) in the sense that it is a limit of the circle based on three nearby points on the a(s) a(s) curve. The point Ca(s) is called the center of curvature of a at s, and the curve given by C. (s) is called the curve of centers of curvature. We define the (unsigned) plane curvature of a at s by the reciprocal of the radius of the osculating circle, K± (S) _

Ha(s) - Ca(s)II

This geometric construction relates directly to the analytic (coordinate) expression for a(s) in the following result.

Theorem 6.6 (Euler 1736). K±(s) = Ila"(s)II Since a'(s) a'(s) = 1, by differentiating we have that a'(s) a"(s) = 0, that is, a"(s) is perpendicular to a'(s). Since a(s) - Ca(s) is perpendicular to a'(s) as well, it PROOF.

follows that

a(s) - Ca(s) = µa" (s), for some µ E R.

However, -l = a"(s) (a(s) - Ca(s)) = a"(s) . µa"(s) = µ11a"(s)112. and so Ha(s) - Ca(s)II = I/LI' IIa"(S)II =

Ilarr(s)II

=

Ila"(s)II


68

We observe immediately that the definition and theorem generalize our conventions for the straight line and the circle. Given the arc-length parametrization of the circle of radius

r, a(s) = (r cos(s/r), r sin(s/r)), then

a'(s) = (- sin(s/r), cos(s/r)) and a"(s) = (-(l /r) cos(s/r), -(I /r) sin(s/r)). so 11a"(s)11 =I /r as desired. For a line, a(s) =A+ s(4/11911). a"(s) = 0, so K±(s) = 0. As we mentioned in the example of an ellipse, it is not always easy to reparametrize a curve by arc length. It is convenient then to be able to compute the plane curvature of any regular curve a(t).

Proposition 6.7. The plane curvature of a regular plane curve a(t) = (x(0. y(t)) is given by

yx

xy,

K+(1) _

((x')2 + (y')2)'/2

Let s(t) denote the are length, and, risking confusion, denote its inverse by t(s). The reparametrization of a(t) by arc length is given by PROOF.

fl(s) = a(1(s)) = (x(t(s)), v(t(s))).

To compute K±(to), we compute K±(S; sp), which denotes the plane curvature of fl(s) at so, for so = s(to). Now

fi'(s) = a'U(s))ds and Since

ds

fl (s) = a"(t(s))

z

2

(ds) +a'(t (s))dsz

= x'(t)2 + y'(t)2, we can write dt

ds

x'(t)2+y'(1)2

x,x"

+ V'y" ds2 =-('(t)2+y'(1)2)2' d2t

I

and

By Euler's formula, we find o (I" Y11)

K±(10) = K±(fl; so) = I

x'(t)2 + Y'(t)2 X" t'' - v 'X' _ (x,(t)2 + y,(t)2)2 (Y'.

(x (1)2 + '(1)2)2

(x . Y )II

(X'(1)2 + V'(t)2)312

By way of example, consider the ellipse, a(t) = (a cost, h sin t). By Proposition 6.7 we have

Kf(t)

ab (a2 sin2 t + b2 cos21)3/2

I

=

ah I

(a2 + (b2 - a2) cos2 1)3

6. Curves

69

For b > a notice that the curvature achieves a maximum when t = fn/2 and a minimum

when t=0or n.

The tractrix A curve that will be important in later discussions was discovered in the seventeenth century. The Paris physician Claudius Perrault (1613-88) set many mathematicians the problem of

describing the curve followed by a weight being dragged on the end of a fixed straight length, the other end of which moves along a fixed straight line. Leibniz and Huygens understood that such a curve had the property that at every point the tangent line met the fixed line a fixed distance away. Huygens named such curves Traktorien (in a 1693 letter to Leibniz). The name that has stuck is the tractrix (Loria 1902). We suppose the v-axis to be the fixed line, the fixed length

to be a, and the curve to begin at (a, 0) on the x-axis. By the analytic characterization, the tractrix the equation dv adx X' X

(x (t), y(t)) satisfies

.r

Squaring both sides of the equation and manipulating the terms we find (a )2 (.r')2 + (v1)2 = (x')2.

-X

The tractrix.

Taking the derivative of both sides of the first equation yields

xyn - yx " _ (x')2

21 -ax .r2

a-X

that

is,x"v'-v"x'=

a2 (x). 3 x'- a - X

This leads to the following expression for the plane curvature as a function of x: K± (X, v) =

a2(x3

x

X3

x2 a- x a3(x')3 I= l a

a- x

Another parametrization of the tractrix results from the integration

Y(x) = 1

a-x X

d x.

Letting x = a sin 0 we get C-)(0) = (a sin 0, a In tan(0/2) + a cos 0).

Proposition 6.7 gives the plane curvature in 0 as Kj(6)

tan 6

70


Another consequence of the previous equations is an arc-length parametrization of the

tractrix. Let (x')2 + (y')2 = I, then

(a

l2

x/

(x')2 = I, which implies that x' =

I

a

x. For

convenience, let us take a = I and the case x' = x; then x(s) = es,

I-x dx-

dy

x

ds

I-e2$

ds

and a routine integration yields

y(s) =

1 - e2s - arccosh

(e-s)

.

Since this requires 0 < e- < 1, we take the opposite orientation of the curve by switching -s for s. This leads to the parametrization of the tractrix as

O(s) = (e-s,

I - e-25 - arccosh(es)),

for s > 0.

In the arc-length parametrization for a = I, the curve has plane curvature e -S

K±(s) = IIO (s)II = 1

The tractrix will be a key ingredient in constructing surfaces with particular properties in Chapter 13.

Directed curvature

At present, K±(t) is simply the reciprocal ofa length. We now introduce a sign which makes

plane curvature more sensitive. We want to be able to distinguish a curve a: (a, b) - R2 from its reverse &(t) = a(b - t), and to distinguish convexity on either side of a point where the curvature vanishes (for example, a point of inflection). To make these distinctions we introduce an orientation:

Definition 6.8. A ordered pair of nonzero vectors [a, v], u, v E

R2,

is said to be in

standard orientation if the matrix representing the linear transformation taking a to (1, 0) and v to (0, 1) has positive determinant.

For example, the matrix taking u = (1, 1) to (1, 0) and u = (- I.3) to (0, I) is given by 114

A = ( 31/4

1/4)withdeterminantl/4.Thus[(1,

1), (- 1, 3)) is in standard orientation.

Switching rows of a matrix changes the sign of the determinant. This corresponds to the fact that if (a, u] is in standard orientation, then [v, aJ is not. Notice that the set (a, u) is linearly independent when [a, v] is in standard orientation. Given any unit vector uo = (u01, u02) E 1R2 there is a unique unit vector vo = (-u02, u01) with u'o perpendicular to ao and standard orientation (ao, vo]. Henceforth we use the notation i o 1 u0 to denote that 40 60 = 0, that is, a0 is perpendicular to vo. If a(s) is a regular curve parametrized by arc length, then denote the (unit-length) tangent

vector a'(s) by T(s). Let N(s) denote the unique unit vector perpendicular to T(s) with standard orientation (T(s), N(s)). Since T(s) T(s) = 1, T'(s) 1 T(s) and so a"(s) _ T'(s) is a multiple of N(s).

6. Curves

71

Definition 6.9. The directed curvature K(s) of a unit-speed curve a at s is given by the identity

a"(s) = K(s)N(s).

It follows from Theorem 6.6 that IK(s)I = K±(s) and so the same formula for the directed curvature of a curve not parametrized by arc length given in Proposition 6.7 holds, if we can determine the sign. The example of the curve given by the graph of a function a(t) = (1, f(M illustrates the choice. Here Ila'(t)II = I + (f'(t ))2, and we can compute

I 1

N(t) -

-(

f(t)

I

(f(t))2

1

(f(t))2

-f (t)

1

I +(f(t))2

I +(f(t))2

-f(t)

(t) T'(t)- I +r'(f'(t))2 ( I + (f'(t))2

1

1 _+( f'(t))2

The sign of the curvature is determined by the second derivative f"(0. which is positive if f(t) is concave up, negative if f(t) is concave down. Since any curve in R2 is locally the graph of a function, we see that the directed curvature at a point is positive if the curve turns to the left of the tangent vector, negative if to the right. Just how much does the directed curvature of a curve determine a curve? The answer is given in the following theorem.

Theorem 6.10 (fundamental theorem for plane curves). Given any continuous function K: (a, b) - R, there is a curve a: (a, b) -. R2, which is parametrized by arc length, such that K (s) is the directed curvature of a at s for all s E (a, b). Furthermore, any

other curve a: (a. b) -, R2 satisfying these conditions differs from a by a rotation followed by a translation. PROOF. To prove the theorem we turn to the theory of differential equations. The conditions

of the theorem are summarized by the differential equation for a function f: (a. b) -. R2:

writing f(s) = (ft(s). f2(s)). (f (s), f24 (s)) = K(S)(-f2(S), fl (S)), subject to f(ate) = it and Ili;ii = 1. Notice that a solution to this differential equation gives a unit-speed curve:

d (fi2(s) + f2 (s)) = 2fi (s)fi (s) + 2Ji(s)f2(s) = 2(fi (s). f2(s))

(14, (s), f24 (s))

= 2K(s)(ft (s), f2 (S))

f2 (S). fi (s)) = 0.

Thus I f (s) 11 = I for all s E (a, b).

To solve the differential equation we call upon a standard result in the theory which we will use again.

72


Lemma 6.11. Ifg(t) is a continuous (n xn)-matrix-valued junction on an interval (a, b), then there exist solutions, F: (a, b) -. R", to the differential equation F(t) = g(t)F(t). A proof of Lemma 6.11 is outlined in the exercises. In our case. g(s) is the 2 x 2-matrix given by 0 -K(s) g(s) = K(s) 0

The equation T'(s) = K(s)N(s) becomes T'(s) = g(s)T(s) and so the lemma obtains T(s) for the curve a(s) with the correct curvature. To complete the proof we simply integrate T(s) to obtain a(s). Notice that we can choose a(ao) anywhere in R2, and that we can choose u' = T(ao) as any unit vector. Changing i at a(ao) involves a rotation and that rotation passes through the differential equation so that another solution would appear as T(s) = peT(s), where pe is the matrix pe

rcos0 sin0

-sin0 cos0

A translation resets the point a(ao) wherever one chooses. Thus a second solution a(s) satisfies a(s) = Pea(s) + tiwo,

for some angle 0 and some initial vector tuo in R2. This proves the theorem. EXAMPLE. Suppose that a regular unit-speed curve in R2 has constant positive curvature,

r2 > 0. Now T(s) T'(s) = 0 by the unit-speed assumption, and T'(s) = r2N(s). If we write T(s) = (x(s), y(s)), then we have the differential equations

xx' + yy' = 0. and (x'. y') =r2 (-Y. x). This implies that x" = -r4x, which has x = a cos(r2s + to) + b sin(r2s + to) as solution. Since we only need to find one solution (and move it about by rotations and translations), ,2 we can take x(s) = r2 cos(r2s). Then y(s) = sin(r2s) by the condition x'2 + y'2 = 1. This is a parametrization of the circle of radius

12

and so the condition of constant positive

plane curvature forces the curve to be part of a circle.

In the next chapter we will generalize the success in characterizing curves in the plane through their directed curvature to curves in three dimensions.

Digression: Involutes and evolutes Some of the earliest results about plane curves were motivated by the desire to build more accurate clocks. Practical designs were based on the motion of a pendulum, requiring careful study of motion due to gravity first carried out by Galileo, Descartes, and Mersenne.

6. Curves

73

The culmination of these studies was the work of Christian Huygens (1629-95) in his 1673 treatise Horologium oscillatorium sive de mote pendulorum ad horologia aptato demonstrationes geometricae. Some of the ideas introduced in Huygens's classic work, such as the involute and evolute of a curve, are part of our current geometric language. We next sketch the principal ideas of Huygens's pendulum clock and collect the main geometric ideas in the context of extending the theory of plane curves. For a thorough historical study of this part of Huygens's work, see Yoder (1989).

The first practical notion is related to the curve swept out by the end of an "ideal" pendulum. We say that a curve is a tautochrone (or that it is isochronal) if a pendulum, running along the curve from rest, reaches the bottom of its swing in the same time regardless of the initial point on the curve. If a circle were a tautochrone, then a rigid pendulum would

have the same period independently of the choice of initial angle. This is not true, as is shown in any elementary physics text, though Galileo observed that it is approximately true for small angles. If a pendulum were made to swing along the path of a tautochrone, then it would deliver a reliable period without worry about the initial angle. This is the basis of an accurate clock. Without a lot of motivating details, let us introduce the following curve. Definition 6.12. A cycloid is the curve swept out by a fixed point on a circular disk as it

rolls along a straight line.

Of particular interest is the cycloid for a circular wheel rolling "under" the line, and the point chosen on the circle. Referring to the figure, take -n < 9 < it and let O' be the point on the circle that began at the point 0 after rolling through 9 radians. We denote the radius of the circle by r. Consider the radius LC' perpendicular to the x-axis and the associated diameter L K. Denote the origin as U - it follows that UL has length r9. Since the arc O' K also has length r9, LO'C'K has measure 9. By projecting onto the diameter LK from the

point 0' to the point N, we obtain the equations for 0' = (x(9), y(9)), x(9) = r(9 + sin 9),

y(O) = -r (I + cos9).

We denote this particular cycloid by ('(9) = (x(9), y(9)). Taking the derivative we find

'(O) =r(1 +cosO.sin9), and IIC'(9)II =r 2+2cos9. Notice that (9) is not a unit-speed curve. Our interest in the cycloid is based on the following remarkable property.

74

Theorem


6.13. The cycloid is a tautochrone.

PROOF SKETCH. We will not justify the physics in what follows as it would take us too far afield (see Dugacs (1955) or Gindikin (1988)). Let us assume that a tautochrone is being followed by a pendulum and parametrize the curve by the angle with the vertical made by the pendulum, say 0 = 0 (1). Let a (r) be the point along the tautochrone at time t reached by the pendulum from an initial point. Resolving vectors into horizontal and vertical components we find the acceleration due to gravity determines the differential equation

d2a

d!2= -gsin0.

Here g is the constant of acceleration due to gravity. The property of being a tautochrone implies that the motion of the pendulum along this curve is simple harmonic and so it satisfies the differential equation for simple harmonic motion:

d2a

Z

d2 = -k a,

where k is a constant associated to Hooke's law. These two equations imply g sin 0 = kza. Differentiating we get the relation =Pdo.

gcos0dLo

Consider the following diagram. Since the tangent to the curve followed by the bob of the pendulum is perpendicular to the pendulum, 0 is also the angle that the tangent makes with the horizontal. Therefore, we have dx dt

da

=cos0dt.

dy do =sin0dt dt

Substituting the relations derived above, we get

dx = k2 cos20do.

dy=

kZ

sin4cos0do.

On integration the equations yield

x(0) - xo = 4kZ (20 + sin 20),

y(4') - yo = - 02

For 0 = 0, xo = 0, and yo =

-02

cycloid generated by a circle of radius

cos 20.

the origin is midway between the two cusps of a

02

Having determined the curve that we would like the "ideal" pendulum to follow, we next consider the problem of forcing the pendulum to follow this prescribed curve. Fix a curve

6. Curves

75

a: (a, b) -+ R2 and consider the curve fi: (a, b) -+ R2 given by unwinding a taut string along a in the direction tangent to a at each point. From the construction, we may describe the curve fl(t) by the equation

f(t) = a(t) +A(t)a'(t). Furthermore, the condition that the string be taut implies that f'(t) is perpendicular to a'(t) for all values of t.

Definition 6.14. Given a curve a: (a, b) - R2, a curve fi: (a, b) -> R2 such that f (t ) lies along the tangent line to a(t) and )4'(t) 1 a'(t) is called an involute of a, and a is called an evolute of Q.

A pendulum with a flexible string constrained to unwind along a fixed curve a follows a path given by an involute of a. The practical problem of building the "ideal" pendulum clock requires that we construct the evolute of a cycloid. We next develop some of the properties of involutes and evolutes in order to solve this problem.

Proposition 6.15. 0(s) is an involute of a unit-speed curve a: (a. b) -+ R2 if and only if, for some constant c, fl(s) = a(s) + (c - s)a'(s).

PROOF. Suppose fl(s) = a(s) + (c - s)a'(s). Then 6 lies along the tangent line at each point. Computing the tangent to 6 we obtain fl'(s) = a'(s) - a'(s) + (c - s)a"(s) = (c - s)a"(s).

Since a is unit speed, a' I a", and so we conclude 16' 1 a'. When 6 is an involute of a, we can write ,6 in the form

fl(s) = a(s) +A(s)a'(s). Taking the derivative and then the dot product with a'(s) we get

f'(s) - a'(s) = a'(s) a'(s) +.k'(s)a'(s) a'(s) + X(s)a"(s) . a'(s).

76


Since a is unit speed and fl' I a', we obtain the differential equation

A'(s)+I=0. This implies that x(s) = c - s.

For a regularcurve a(s) for which K(s) 54 0 for all s there is a natural evolute associated to a(s). Recall from earlier in the chapter that a unit-speed curve a(s) has an associated curve - the curve of centers of curvature, Ca(s).

Proposition 6.16. The curve of centers of curvature Ca(s) associated to a unit-speed curve a(s) is an evolute of a. PROOF. Recall some of the properties of Ca(s):

(1) K(s) =

I

11a(s) - Ca(s)II

(2) (a(s) - Ca(s)) a"(s) = - I. Without an explicit parametrization, we cannot reparametrize Ca(s) to be a unit-speed curve. However, the property of being an involute is characterized by the facts that the point of the curve lies on the tangent of the evolute and, at associated points, that the tangents are

orthogonal. To prove the proposition then, it suffices to show that Cc(s) 1 a'(s) and that

a(s) = Ca(s) + A(s)C., (s). Since a(s) is a unit-speed curve we associate to each point of the curve the orthonormal

pair (T(s), N(s)}, where T(s) = a '(s) and [T(s). N(s)] is in standard orientation. By Euler's formula (Theorem 6.6) we know that T'(s) = K(s)N(s). We take this relation one derivative further to prove the following relation.

Lemma 6.17. N'(s) = -K(s)T(s). PROOF. Since N(s) has unit length, N(s) I N'(s). Thus N'(s) = v(s)T(s). Since N(s) is perpendicular to T(s) we differentiate to obtain

0 = ds(N(s) T(s)) = N'(s) T(s) + N(s) T'(s)

= v(s)(T(s) T(s))+K(s)(N(s) N(s)) = v(s)+K(s). The lemma follows. Since (a(s) - Ca(s)) I a'(s) we have that a(s) - Ca(s) = µ(s)N(s). Since (a(s) -

Ca(s)) a"(s) = -1, it follows that

- I = (a(s) - Ca(s)) a"(s) = p(s)N(s) K(s)N(s), and so µ(s) _ -I/K(s). This calculation also implies that

Ca(s) = a(s) +

K I

N(s)

6. Curves

77

Taking the derivative we get K'(s) N(s). C (s) _ -(K(S))2 This implies that Ct,(s) 1 a'(s) and that

Thus Ca is an evolute of a.

Notice that Ca(s) is a regular curve when K'(s) # 0 for all s. To finish this digression, we prove the following theorem of Huygens (1673).

Theorem 6.18. The cycloid has a congruent cycloid as its evolute. PROOF. Though our parametrization of the cycloid is not unit speed, we can still determine the curvature and hence the curve of centers of curvature. We fix the radius of the generating circle to be I - the general case will be clear. Apply the formula for the plane curvature of

an arbitrary regular curve, a(t) = (x(t), y(t)),

Kj(t) =

y'/ y3 2 ((x')2 + (y')2) /

For the cycloid C (0) _ (0 + sin 0, - I - cos 0), we obtain

Kf(6) _

(1 +cos0)(cos0) - (-sin9)(sin0) [(I + cos 9)2 + (sin 0)2]3/2 4 cos(9/2)

To construct the curve of centers of curvature, we write

K(9) N(0), where N(O) is the unit normal to the curve at {(B). Now

'(6) = (I +coS0, sine). and so the unit normal will be given by

N(9) =

f

2+2cos9(-sin0.

(

1 +cos9)

sing

\

I +cos9

) l +cos9 _ f -2 sin(0/2) cos(0/2) f cos(0/2)l 2 '. f cos(9/2) / 2

_ (- sin(9/2), cos(8/2)).


78

Note that over -n < 8 < r, the vector N(O) points into the upper half-plane, making K(B) positive. Substituting into the expression for N(8), we get the curve of centers of curvature:

Cs (8) = (8 + sin 8, - I - cos 8) + 4 cos(8/2)(- sin(8/2), cos(8/2)) = (8 + sin 8 - 2 sin 6, -1 - cos 8 + 4 cos2(8/2) = (8 - sin 0, 1 + cos 8). The transformation 0 --, 8' + jr carries C, (8) onto

C{ (0) = (0' + sin 0', -1 - cos 8') + Or, 2), which proves the theorem.

Huygens applied his researches and had pendulum clocks built according to the preceding discussion, using a pair of plates curved to follow a cycloid between which the pendulum would swing. Thus the bob followed the path of a tautochrone. His proof of Theorem 6.18 used synthetic and infinitesimal ideas; he did not have the calculus at his disposal. This example of an applied problem leading to developments in differential geometry is repeated throughout the history of the subject.

Exercises 6.1

If a(t): R - R" is such that a"(t) = 0. what can be said of a(t)?

6.2

Show that f(t) = tan(7rt/2), f: (-I, I) -. (-oo, oo). is a reparametrization. Is g: (0, oo)

6.3

(0. 1) given by g(t) = t2 +

a reparametrization? 1

(a) Reparametrize a(t) = (e' cost, e' sin r, e') by arc length. (b) Determine if p(t) is a unit-speed curve:

fi(t) = 2 (t + t2 + I,

t+

t +l

2In(t +

t2 + 1) ///

79

6. Curves

6.4 6.5

Consider the curve in R2 given by the graph of the sine function t r-. (t, sin t). Determine the directed curvature at each point of this curve.

Let k: (a, b) -+ R be a function with k(t) > 0 for all t. Let 9(s) = f k(s)ds be any antiderivative of k(s). Show that

a(s)=(Jcoso(s)ds + a, f sin0(s)ds+b) is a regular curve with curvature the given function k(s).

6.6

Suppose that a plane curve is given in polar coordinates p = p(0), a < 0 < b. Show n dp

p2 + (02d9, where p' =

that the arc length along the curve is given by

+a

ae

. Also

show that the directed curvature is given by the formula K(0)

=

2(P )2 - PP" + p2 .

((P')2 + p21312 6.7

Determine the formula for the involute of a curve that is not unit speed.

6.8

Compute the arc length for the cycloid C(t) = (t + sin t. I - cost). One branch of the cycloid is determined by t E [-tr, jr]. Determine the length of this branch.

6.9'

Determine the involutes of the circle of radius 1 and the parabola y = x2 for - I < x < 1. Is the involute of a parabola asymptotic to a horizontal line?

6.10' Determine the curve of centers of curvature associated to the ellipse a(t) = (a cost, b sin t). This is an evolute of the ellipse. Show that this curve is not regular.

The basic result from the theory of differential equations from which we derive our analytic results is as follows: the differential equation of interest is given by

where f is a continuous function on a convex domain D C R"+1 satisfying the properties:

(1) There is a value MI such that 11f(t.. )II < M, on D.

(2) There is a value M2 such that Il f(t, yi) - f(t.Y2)II < M201 - y211 for all (1, y;) E D, i = 1, 2. This is the Lipschitz condition on f.

The initial conditions are given by a choice of point (to, yo) E D. Suppose there are

auxiliary values a,b satisfying b > aMi and the set Ra,b = ](t, y)

I

It - tol <

a. IL - yoll < bI C D. Then we have the following theorem.

Theorem 6.19. Under the conditions just given and initial values (to, yo) E D, there is a uniquely determined function y(t) with domain it - tol < a satisfying

dy dt

(t) = f(t, y(t)). Furthermore, the solution curve lies in the set Ra.b

6.11' Prove Lemma 6.11 using Theorem 6.19.

11

Curves in space Sur les courbes A double courbure.

Clairaut (1731), Frenet (1852), Serret (1851)

In this chapter we study regular curves in space, a: (a, b) - R3. Directed curvature was successful in characterizing plane curves and so we seek to generalize the idea to curves in space. In two dimensions the normal to the tangent is easy to describe. In three dimensions there is a continuum of choices of a normal vector. Furthermore, we must contend with a new phenomenon - a curve can rise out of the plane spanned by the tangent and normal. These complications are overcome by a special feature of R3 - the existence of the cross product of two vectors. Recall that the cross product of two vectors in R3, i = (111 - u2, 243) and 0 = (VI, v2, VD, is given by U x v = (142 V3 - u3u2, U3 V1 - u I V3, uI V2 - u2ul ).

The principal properties of the cross product are

(I) Forii4av",i J (i xi3)andul(i x0). (2) Ux0=-(iixii). (3) U x (av + w) = a(U x v') + (U x ru). (4) IIU xvI12= (5) If a, 0: (a, b) - R3 are differentiable curves, then

Wt (a

x fl)(t) = a'(t) x '6(t) +a(t) xf'(t).

That is, the Leibniz rule holds for the derivative of the cross product of differentiable functions.

Definition 7.1. Let T(s) = a'(s) be the tangent vector to a regular, unit-speed curve, a(s). Define K(s) = IIa"(s)II = IIT'(s)II, the curvature of a. and, when K(s) # 0, let N(s) = T'(s)/K(s) denote the unit normal vector to a. Finally, let B(s) = T(s) x N(s) denote the unit binormal vector to a.

Since T (s) has constant length, T(s) is perpendicular to T'(s), and so N(s) is perpendicular to T(s). This definition fixes the choice of unit normal by choosing the derivative of the tangent vector. This also generalizes the definition of curvature from the planar case. Notice that K(s) > 0 for curves in R3. 80

7. Curves in space

81

From the properties of the cross product, we find that the triple

(T(s), N(s), B(s)) forms a frame, that is, a basis for all directions in R3 through a(s).

In fact more is true; these vectors are orthonormal: Notice that by choice, IIN(s)II = 1, and further IIB(s)112 = I)T(s)II2IIN(s)II2 - (T(s) N(s))2 = 1, so 11 B(s) 11 = 1. Furthermore, the matrix representing the linear transformation that takes

T(s) H el = (1. 0, 0). N(s)

ez = (0, 1, 0), and B(s) r e3 = (0, 0, I) has positive

determinant. When this condition holds we say that the ordered basis (T(s), N(s). B(s)) is right-hand oriented. The right-hand orientation is a property of the cross product; in fact, (u, u, u x v) satisfies the right-hand rule as a basis of R3. We summarize these properties by saying that the triple (T(s), N(s), B(s)) forms a right-hand oriented, orthonormal frame. Since three noncollinear points in R3 determine a plane and a circle in that plane, the limiting process used to define the plane curvature determines an osculating circle and an osculating plane for space curves. The osculating plane is spanned by T (s) and N(s). Since the cross product is orthogonal to the tangent and normal vectors, changes in the binormal measure how "nonplanar" a curve is.

Proposition 7.2. Suppose a(s) is a regular unit-speed curve. Then B(s) is a constant vector if and only if a(s) is planar. PROOF.

If a(s) lies in a plane, then the methods of Chapter 6 show T(s) and N(s) are

contained in that plane for all s. Since the unit normal to that plane is determined up to sign, B(s) is a constant multiple of that vector. Suppose

B(s) = q, with 11411 = 1. Consider the function

f(s) = (a(s) - a(0)) B(s). The derivative satisfies

d= a'(s) Ys-

B(s) + (a(s) - a(0)) B'(s) = T(s) B(s) = 0.

Therefore f is a constant function and it follows that I =' a(s) satisfies the equation (x2 - a(0)) q = 0 since f(O) = 0. This implies that a(s) lies entirely in the plane perpendicular to q'.


82

For a general unit-speed curve in R3, we have the relations:

B(s) 1 T(s) and B(s) I N(s). Also IIB(s)II = I. and so B'(s) B(s) = 0 or B'(s) 1 B(s). Consider the following equation:

0 = dt (B(s) T(s)) = B'(s) T(s) + B(s) T'(s)

= B'(s) T(s) + B(s) K(s)N(s)

= B'(s) T (s), and so B'(s) 1 T(s). This shows that B'(s) is a multiple of N(s). Definition 7.3. B'(s) = -r(s)N(s) and the junction r(s) is called the torsion of a at t.

EXAMPLES. (1) The circle: Let a(s) = (r cos(s/r), r sin(s/r), 0); then the tangent vector

is given by T(s) = (- sin(s/r), cos(s/r), 0) and

T'(s) = (-(I/r)cos(s/r), -(I/r)sin(s/r),0), so K(s) = (I/r). It follows that N(s) _ (- cos(s/r), - sin(s/r), 0). To compute B(s), we take the cross product: B(s) = T(s) x N(s) = Oei + Oe2 + lea.

Thus B'(s) = 0, r(s) = 0. and a(s) is planar (as we already knew). (2) The helix: Consider the curve a(s) =

a +b I'asin( a +b )

Iacosl Let w denote

l

b

a +b /

Then

.1a2 +b T (s) = (-aw sin ws, aw cos ws, wb),

and T'(s) = (-awe cosws, -awe sin ws, 0), so K(s) = It follows that N(s) = (- cosws, - sin ws, 0) and the binormal is given by

aw2.

B(s) = T (s) x N(s) = (bw sin ws, -bw cos ws, aw). Thus B'(s) = (bw2 cosws, bw2 sin ws, 0) = -bio2 N(s), and r (s) = bw2. Note that the constant torsion of the helix describes how it lifts out of the osculating plane at a constant rate.

7. Curves in space

83

Definition 7.4. To a regular, unit-speed curve a(s), the associated collection (K(s), r(s), T(s), N(s), B(s))

is called the Frenet-Serret apparatus. The orthonormal basis (T (s), N(s), B(s)( is called a moving frame or moving trihedron along the curve. The moving trihedron with its curvature and torsion were introduced in the Toulouse thesis of F. Frenet (1816-68) in 1847. Independently, J. A. Serret (1819-92) published similar results in the Journal de Liouville (1851) before Frenet's results had received wide recognition. Frenet published his results in the same journal in 1852. When such an apparatus comes from a curve we have the following classical theorem.

Theorem 7.5 (the Frenet-Serret theorem). I fu (s) is a regular, unit-speed curve with nonzero curvature K(s), then

T'(s) = K(s)N(s), N'(s) = -K(s)T(s) + r(s)N(s), and B'(s) = -r(s)N(s). PROOF. By the definition of curvature and torsion we have already that T'(s) = K(s)N(s) and B'(s) = -r(s)N(s). Recall that if fib,, iu2, W3( is any orthonormal frame in R3, then an arbitrary vector u" E R3 can be written

'u = ( v " wi)wi + (v w2)iu2 + (i .

17V3)@3-

A p p l y this fact to the moving frame and N'(s) to get

N'(s) = (T(s) N'(s))T(s) + (N(s) N'(s))N(s) + (B(s) N'(s))B(s). -K(s). Ogivesusthat We already know that N(s) N'(s) = 0. Finally, N(s) B(s) = 0 implies that N'(s) B(s)+N(s) B'(s) = 0 and so N'(s) B(s) _ Differentiation

-(-r(s)N(s)) N(s) = r(s). Thus we have N'(s) = -K(s)T(s) + r(s)B(s). We record this information more succinctly in the matrix expression:

-K(s)

K(s) 0

r(s)

T(s) N(s)

0

-r(s)

0

B(s)

0

0

T1 S)

=

N'(s) B'(s)

The skew symmetry (that is, the condition A = -A') of this matrix plays an important role in 'the next result, which generalizes the fundamental theorem for plane curves to curves in space.


84

Theorem 7.6 (fundamental theorem for curves in R3). Let k, f: (a, b) -+ R3 be continuous functions with k > 0. Then there is a curve a: (a, b) -+ R3, parametrized by arc length, whose curvature and torsion are k and f, respectively. Furthermore, any two such curves differ by a proper rigid motion, that is, if & is any such curve, then there is a linear mapping A: R3 --> R3, which preserves arc length, and a vector u E IR3 such that

PROOF. Consider the matrix-valued function

g(s) _

0

k(s)

0

-k(s)

0

f(s)

0

-f(s)

0

_ (a;j(s)).

If we write v, = T, V2 = N, and V3 = B, then the Frenet-Serret equations (Theorem 7.5) give us the differential equation Vj

3

v; _ Fa;jvj

or

v2'

j=1

VI

= g(s)

v3

v2

(*)

V3

and, by Lemma 6.11, there is a solution V1 (S)- v2(s), u3(s), dependent upon the initial conditions. We may take (VI(so), V2 (SO), v3(so)I, to be a choice of an orthonormal basis for the moving frame at so. We next show that the solution through this given set of values (vI (s), v2(s), v3(s)) is a moving frame. Observe that d

3

ds(V;

vj) = LaikVj Vk +ajkvi V. k=1

Let /3ij = vi vj. Our data then satisfy the differential equation 3

ij = L aik,6jk +ajkflik k=1

with initial conditions fli j(so) = dij, the Kronecker delta junction, which is given by the matrix

S;j =

1,

i=j,

10, i# j. In order to have a moving frame we need to show that 6ij(s) = d;j holds for all s. But notice 3

a;j=O=aij+aji=

aikSjk+ajksik, k=1

which holds by the skew symmetry of the matrix (aij (s)). Thus bi j satisfies the differential equation (**) and so, by uniqueness of solutions to differential equations, we have a moving frame.

7. Curves in space

85

To define the curve a(s) we integrate

a(s) =

s

JSO

vt(t)dt.

We leave it to the reader to verify that a(s) has the correct curvature and torsion. The assumption that K (s) i4 0 is necessary in order to obtain uniqueness in the sense of the theorem. If K(s) = 0 along a subinterval (at, bt) of (a, b), then a line segment in 1R3 realizes this part of the curve. To see how uniqueness fails, suppose M is the midpoint of the line segment between a(at) and a(bt) of a solution curve with a given K(s) and r(s). Suppose e > 0 is some small value such that a(s) lies on one side of the plane through M perpendicular to the line segment for at - e < s < at and on the other side of this plane for bt < s < bt + e. Rotate the solution around the axis given by the line segment in the half-space containing a(bt) through some nonzero angle while leaving the other half-space fixed. This will preserve the curvature and torsion but there is no way to superimpose this new curve on the one given by a, and the fundamental theorem fails to hold.

As with plane curves, we get all the relevant geometric data about the curve from its curvature (when nonzero) and torsion. Here are some examples of this idea. EXAMPLES. (1) As we have already seen for a regular, unit-speed curve a(s) in R3, a(s) is a plane curve if and only if r(s) = 0 for all t.

(2) If a(s) is unit speed and regular, K(s) is constant, and r(s) = 0, then a(s) is part of a circle of radius I /K.

The power of the Frenet-Serret apparatus lies in the manner in which it organizes the principal analytic properties of a curve. Applications of this viewpoint are found in the exercises. As a final example, we introduce a class of curves that generalizes the helix.

Definition 7.7. A curve a(s) is a general helix if there is-some vector v' such that the tangent T (s) and u maintain a constant angle with respect to each other. The vector u is called the axis of the helix. We can rake 0 to be unit length.


86

For example, u = (0, 0, I) for the helix a(s) = (toss, sins, s). Observe that if a(s) is a general helix, then T (s) u =a constant given by cos 9, where 9 is the angle between T (s) and v. From the definition we also see that all plane curves are general helices with the binormal B(s) playing the role of the constant vector u. The Frenet-Serret apparatus allows us to characterize general helices.

Proposition 7.8. A unit-speed, regular curve a(s) is a general helix if and only if there is a constant c such that r(s) = cK(s). If i is the axis of a general helix a(s), then T(s) v = cos9, a constant, and this implies T'(s) v = 0, orK(s)N(s) v = 0, that is, u 1 N(s). Apply the facts that we have PROOF.

an orthonormal frame, that 11 611 = 1. and that i is in the plane of T(s) and B(s), to get

v = (T(v) u)T(s) + (N(s) v)N(s) + (B(s) v)B(s) = (cos9)T(s) + (B(s) v)B(s) = (cos9)T(s) + (sin 0)B(s). This implies further that

0 = ds(0) = (cos9)T'(s) + (sin9)B'(s) = ((cos9)K(s) - (sin9)r(s))N(s). Since N(s) 54 0, we obtain (cos9)K(s) = (sin9)r(s) or r(s) = (cot9)K(s).

Conversely, when r(s) = cK(s), then c = cot 9 for some angle 9. Now let V (.s) _ (cos9)T(s) + (sin 9)B(s). Notice that II V(s)II = 1. Differentiating we get d

d-V(s) = (cos9)T'(s) + (sin 0)B'(s) = ((cos0)K(s) - (sin 0)r(s))N(s) and (cos 9)K (s) - (sin 9)r(s) = 0 by assumption. Thus V (s) is a constant vector 'v. Furthermore T(s) v' = cos9 and we have proved that a(s) is a general helix.

As usual, it would be helpful to be able to compute the curvature and torsion when the curve is regular but not unit speed. The relevant formulas are given in the following proposition.

87

7. Curves in space

Proposition 7.9. For a regular curve a: (a, b) - R3, not necessarily unit speed, the curvature and torsion are given by

Ila'(t) x a"(t)II

K(t) =

IIa'(t)II3

(a'(1) x a"(t)) a,,,(t)

and r(t) =

IIa'(t) x a"(1)112

PROOF. As in the planar case, we use the arc-length parametrization #(s) =a (t (s)), where

= x'(t)2+y'(t)2+z'(1)2 =

s(t) is the arc-length function andt(s)isitsinverse. Here d` Ila'(t)II and so Ws =

Ila'(r)p.

Furthermore,

d2t

12x'x" + 2y'y" + 2z'z" dt

ds2 -

2

(x12

a'(t) a"(t)

ds -

+ yi2 + 2'2)3/2

11a'(t)114

Suppose so = s(to) and we want to compute the curvature and torsion at a(to). Then the chain rule gives 2

K(to)2 = IIfl"(so)112 =

a"(to) II

_

a (to) (ds)2 +a(to)d-Z (a'(to) . a"(to)

Ila'(to)112 - \ IIa,,(to)112

2

)a

IIa'(t)II4

- 2 (a'(to) a"(to))2

Ila'(to)114

IIa'(to)116

(1)II

(a'(to) . a"(to))2

+

Ila'(to)116

Ila'(to)11211a"(to)112 - (a'(to) a"(to))2

IIa'(to) x a"(to)112

Ila'(to)116

IIa'(to)116

and the expression for K(t) is established. By the Frenet-Serret theorem we can compute r (to) as B(s(to)) N'(s(to)) = (T (so) x N(so)) N'(so). For the arc-length parametrized curve 6(s) we have B(so) =

fl'(so)

x

B"(so)

K(SO)

and N'(so) =

fl

(so)

- fl"(so) dK

K(SO)

K(SO)2 dS

From these expressions we find r111

Since fl'(so) = T(so) =

11 _

Ila'(to)II

(6'(so) x fl"(so))

fl,,,(so)

K (SO) 2

and 0"(so) =

Ila'(ta)IIZ

fl'(so) x fl"(so) - a,(to) x

+a'(to)ds2 we obtain

a"(tO)

Ila'(to)113


88

Similarly, we have fl"'(so) = Ila",(to)3 + C2a"(to) + Cia'(to) for some constants C1 and

C2 and so

r(to)--

0'"(so)

(0' (so) x 0"(so)) 2

K (s0)

(a'(to) x a"(to)) -

a,"

(10)

Ila'(to)1l6 a"(to)112,

Ila'(to) x

Ila'(to)116

and the proposition is proved. For example, consider the curve

a (t) = (a cos t, ac cos t - bd sin t, ad cost + be sin t), where c2 + d2 = 1. Then

a'(t) = (-a sin i, -ac sin t - bd cos i. -ad sin t + be cost ).

a"(t) = (-a cos t, -ac cost + bd sin t, -ad cos t - be sin t), a"'(i) = (a sin s, ac sin t + bd cos t, ad sin t - be cost )

It follows immediately that (a'(t) x a"(t)) al"(t) = 0. Thus r(t) = 0 for all t and a(t) is planar. After a tedious computation, we also find K(r)

_ Ilet'(t) x a"(t)II _

fab

Ila'(t)113

(2a2 sin2 t + b2 cos2 t)3/2

0.

This curve has the same curvature as an ellipse in the standard plane. By the fundamental theorem for curves in R3, a(t) is an ellipse lying in some plane determined by the constants c and d. Much more can done in developing special properties of curves from the Frenet-Serret apparatus. We include some of these ideas in the exercises. The successful classification of the smooth curves in R3 via curvature (K # 0) and torsion establishes a paradigm that one could hope to generalize to other geometric objects, for example, two-dimensional objects. We turn to this in the chapters that follow.

Appendix On Euclidean rigid motions

A missing ingredient in our discussion of curves in R" is a notion of equivalence - given two curves a(t), #(t), when should we consider them geometrically equivalent? Reparametrization is one reduction; certainly a curve and any oriented reparametrization (one for which the derivative is positive) ought to be considered equivalent. This focuses our attention on the trace of the curve with a direction along it. Consider the directed trace of a curve as a rigid object in R" and move it about in space by a motion of R" preserving rigid bodies. For

example, translation of R3 by the vector (I, 0, 0) corresponds to "moving one unit to the right;' and we can expect such a motion to preserve a rigid body. In this appendix we determine those mappings of R" to itself that are the rigid motions. Those properties of curves that remain unchanged by such motions are the properties rightfully called geometric. To clarify the notion of a rigid motion, we make the following definition, which translates rigidity into the idea of distance preservation.

Definition 7.10. A mapping f: R" -+ R" is an isometry if, for all v, w E R",

IIf(v) - f(')II = 110 - w11 The most obvious examples are translations, t,;,(v) = u + w. In R2 the rotations given by

PO(v)

(sing

cos9

) (v2

in the standard basis, are isometrics. To see this one simply computes Ilpe(v)112 =11(cosevi - sinev2, sinevi +cos0u2)112

=cos29ui - 2cosesineviu2 +sin29u2 + sin2ev2 +2cosesinev1v2+cos28v2

=v1+v2=IIVI12. Given an isometry f: R" -+ R", we can alter it by a translation to obtain an isometry that fixes the origin: F(v) = f(v) - f(0). We now consider such isometrics. Proposition 7.11. An isometry F: R" -)- R" that fixes the origin is a linear mapping. PROOF. We want to show that F(av + biu) = a F(v) + bF(t1) for all a, b in R and v, ivin R. First observe that F(-0) = -F(v): Since IIF(v)II = 11 V11 = II - 011 = IIF(-0)11. both F(v) and F(-v) lie on the sphere of radius 11011 in R". Now 11 FO) - F(-iu)II = 110 - (-v')II = 211011,

and so F(v) and F(-u") are antipodal. Therefore F(-v') = -F(v). 89

90


Next we recall the polarization identity:

Applying this identity we find

F(i) F(-iii) = (IIF(i) + F(-w)112 - IIF(O)112 - IIF(-w)112) = 2(IIF(v) - F(w)112 - IIF(i)II2 - II F(-w)112)

= 2(IIi--II2-IIi112-II-w112)

Thus F(v) F(w) = u w for all v, w in R", that is. F preserves the inner product on R". In particular, we find that an orthonormal basis (i1, ... , 6,1 is carried by F to another

orthonormal set ( F(u1), ... , F(in)}. Notice that this is also a basis: If 0 = a1 F(i1) + + an F(in ), then, for all i.

0=0

F(ii)

= ar F(ir) F(or) = ai ii ii = ar. Thus the set is linearly independent and hence a basis.

To finish the proof of linearity it suffices to show that

F(a1i1 +...+anin) =al Write F(a1 i1 +

F(v1)+...+a.F(Vn)

+anin) in the orthonormal basis (F(11).... , F(on)): n

i=

By the properties of the isometry already established we have

F(ii) F(a1i1 +...+anin) = ii . (a1i1 +... +anin) = ai. Thus we have proved the proposition.

By the defining property for isometries, it is clear that an isometry is a one-to-one mapping. An isometry fixing the origin, a linear isometry, has a like inverse. We denote by O(n) the orthogonal group of all linear isometries of R". Choosing a basis for R". we can write every element of O(n) as a matrix.

Appendix. On Euclidean rigid motions

91

Proposition 7.12. 0(n) consists of all n x n matrices A satisfying the equation A - = A'. PROOF. Fix the canonical basis for R", lei, ... , e,,), where

e; = (0, .... 0, 1, 0, .... 0). 1 in the ith place. In this orthonormal basis we can compute the entries in a matrix A representing a linear transformation by Since the inverse of an isometry is an isometry we obtain the relation

Ae, ej=A-'Ae; A-let=e; A- tel. Thus the matrix representing A -is the transpose of A. The proposition follows. It is a property of the determinant that det(A) = det(A') and so we find

I = det I = det(AA-t) = det(AA') = det(A) det(A') = (det(A))2. Thus the elements of 0(n) have determinant ±1.

Suppose that n = 3 and consider the effect of applying A E 0(3) to a unit-speed curve a(s). Let P(s) = A o a(s). Because A is a linear mapping we have f'(s) = A o a'(s) and similarly for all higher derivatives of f(s). This implies that

Kp(s) = IIf"(s)II = IIA oa"(s)II = Ila"(s)II that is, a(s) and f(s) have the same curvature. Proposition 7.9 implies that a(s) and fl(s) have the same torsion. This proves that the curvature and torsion of a space curve are invariants in the sense of equivalence under rigid motion. One should compare this fact to the fundamental theorem for curves in R3.

Exercises 7.1

Establish the properties of the cross product on R3 listed at the beginning of the chapter. Also prove the Lagrange formula

7.2

Show that the following characterize the straight lines as curves in 1R3:

(a) All tangent lines to the curve pass through a fixed point. (b) All tangent lines are parallel to a given line.

(C)K=0. (d) The curve a(t) satisfies a'(t) and a"(t) being linearly dependent. 7.3

Consider the curve f(s) = ((4/5) cos s, I - sins. (-3/5) coss). Determine its FrenetSerret apparatus. What is the nature of the curve?

92

7.4


Let a(s) be a unit-speed curve with domain (-e, e), and with K > 0 and r > 0. Let

fl(s) =

B(u)du.

J0

(a) Prove that fl(s) is a unit-speed curve.

(b) Show that the Frenet-Serret apparatus for fl(s) is (k, f, T, 1V, ill. where k = r, T, and (K, r, T. N, B) is the Frenet-Serret f = K, T = B, N = -N, and apparatus for a. 7.5

If a(s) is a unit-speed curve with nonzero curvature, find a vector w(s) such that

T'=wxT,

N'=wxN,

B'=wxB.

This vector w(s) is called the Darboux vector. Determine its length.

7.6`

Show that knowledge of B(s) for a curve a(s) with nonzero torsion everywhere determines the curvature K(s) and the absolute value of the torsion Ir(s)I throughout.

7.7'

For a sufficiently differentiable unit-speed curve a(s) show that the Taylor series of the curve at a point a(so) can be given by

\ - 6 (As)2 1 + N(s)(As)2 (2 + 6 As ) 2

a(so + As) = a(so) + T(s)As (1

\+

+ B(s)(As)3 6 where the remainder R4 satisfies lim

7.8

R4, R4

= 0. Projecting into the various planes

o.-.o (As)-' {T, N). (T. B), { N, B). determine the algebraic curves on which the curvature and torsion of a curve must lie. The tangent vector T(s) to a unit-speed curve a(s) determines a curve on the unit sphere in R3. Suppose that its arc length is given by the function.-s(s). Make sense of the equation

K(s) = as, where s denotes the arc-length function of the curve a. This equation has a generalization that we will meet in Chapter 9. What does one get by doing the analogous procedure for N(s) and B(s)?

7.9' The intrinsic equations of a curve are given by K(s) = ft (s),

T(s) = 12(5)

for two real-valued functions ftt, j2. By the fundamental theorem for curves in R3, these functions determine the curve up to placement. Furthermore, relations between the functions j and f2 may also characterize curves. Prove that a nonplanar curve lies entirely on a sphere if and only if the following equation holds between the curvature and torsion:

Appendix. On Euclidean rigid motions

93

7.10 Given a unit-speed curve a(s), determine the equation satisfied by the osculating plane to the curve as a function of s. Show that a curve is planar if and only if the osculating plane contains some nonzero point for all s. 7.11

Let s(t) be the arc-length function on a regular curve a(l) = (x(t), y(t), z(t)), not necessarily unit speed. Prove the following formula for the torsion of a:

7.12

x"

I

I

r

KZ(S')6

det

I x"ix

y

z

y"

z"

yn,

Z",

Take the theory of the previous chapter of involutes and evolutes and generalize it to curves in R3. In particular, show that an evolute /3(s) of a curve a(s) with Frenet-Serret apparatus (K, r, T, N. B) satisfies the equation

fl(s)=a(s)+IN +µB, where I =I /K and s =(I /K) cot(f rdt + C) for some constant C. 7.13

Suppose a(s) is a unit-speed curve with K 34 0, r i4 0. If aK + br = I for some fixed real numbers a, b, then show that there is another curve (the Bertrand mate) fl(s) such that there is a bijection f from the trace of a to the trace of /3 and at corresponding points, a and Q have the same unit normal.

7.14

Determine the analog of the Frenet-Serret equations for a regular curve that is not unit speed.

7.15

Suppose L: R" - R" is a linear transformation. Define the adjoint L' of L to be the transformation of R" determined by L(v) w = v L' (w). Show that for a particular orthonormal basis, V is represented by a matrix that is the transpose of the matrix representing L. Define the Rayleigh quotient of a nonzero vector "u as R(v) _

L(v)-v .

v-v

Show that R attains a maximum and minimum on R" - (0). Show that R is differentiable with gradient given by

VR(v) =

L(v) - L *(v) - 2R(00 0-v

Finally, use these properties to show that any linear isometry F: R" -). R" has a subspace K C R" such that (a) dim K = I or 2. and (b) L(u) E K for all u" E K, that is. F has an invariant subspace K of dimension I or 2. 7.16

Following the ideas of the previous exercise, suppose K is a linear subspace of W. Let

K' be the subspace of R" satisfying K' = (v E R" 10 w for all w E K). Show that if K is an invariant subspace of R" with respect to a linear isometry F, then Kl is also an invariant subspace. Show also that F 1K is an isometry. If K has dimension 2 and F 1K

94


is a rotation, then K is called a rotation plane. Prove that for n > 3 a linear isometry has a rotation plane. It is also possible that F I K is a reflection across a line contained in K or across the origin. Prove that any linear isometry of R" can be expressed as a composition of n or fewer reflections in hyperplanes. (See the article by Lee Rudolph, The structure of orthogonal transformations. American Mathematical Monthly 98(1991), 349-52.)

7.17' Suppose that A is an element of 0(3) with determinant -I, and a(s) is a unit-speed curve in R3. Compute the curvature and torsion of A o a in terms of the curvature and torsion of a(s).

Surfaces Et quia per naturam superficiemm quaelibet coordinate debet esse functio binanim

variabilium. Euler. Opera posthuma, vol. 1. p. 494

To build a foundation for geometry we first define the structures where geometry can take place. Euclid tells us in Definition 5 of the Elements that "a surface is that which has length and breadth only." A plane is a special kind of surface and taking this point of view a little further, we can expect that the non-Euclidean "plane" will not be like a Euclidean plane and non-Euclidean "lines" will not be the familiar lines y = mx + b of the Cartesian plane. The insight of Euclid's definition is echoed in the preceding quote of Euler - the essence of a surface is its two-dimensionality, that it is describable by two variables, length and breadth. Our goal in developing the classical topics of differential geometry is to discover the surfaces where non-Euclidean geometry holds. On the way to this goal we examine the major themes of differential geometry which include the notions of intrinsic properties. curvature, geodesics. and abstract surfaces. We begin with the theory of surfaces in R3. This theory may be said to have been founded by Euler in his 1760 paper Recherches sur la courbure des surfaces (see Struik (1933)). It was developed extensively by the French school of geometry led by Gaspard Monge (1746-1818). We will consider the early contributions to the subject in Chapter 9. We first introduce the relevant objects, structures, and examples.

U

.l ul)-ol

95

96


Definition 8.1. If S is a subset of R3, then S is a (regular) surface if for each point pin S. there is an open set V of R3 and a function x: (U C R2) -> V fl S, where U is an open set in R2, satisfying the following conditions:

(1) x is differentiable, that is, writing x(u, v) = (fi (u, v), f2(u, v), f3(u, v)), thefunctions j have partial derivatives of all orders. (2) x is a homeomorphism, that is, x is one-to-one and onto V fl S and has an inverse x - i : V fl S -+ U that is continuous. This means that x - I is the restriction to V fl S of a continuous function W -+ R2, for some open set W C R3 that contains V fl S. (3) The Jacobian matrix

aft

aft (u, v)

a (a' v)

J(x)(u, v) =

f2s (u, v) au

au (u, v)

22 (u, v)

/

av

(u, v)

at,

has rank 2. That is, for each value of (u, v) E U, J(x)(u, v) : R2 -. R3, as a linear mapping, is one-to-one, or equivalently, ker J(x)(u, v) _ {0). This condition on x is called regularity. The function x: U -> S is called a coordinate chart (or patch). Coordinate charts allow us to define coordinates for points on a surface in a neighborhood of a given point. This makes precise the idea that a surface is locally two-dimensional. We

can think of a chart x: (U C R2) - S as the correspondence that a cartographic map makes with the surface it describes. EXAMPLES. (1) For i and q" 54 0 E R3, let fl;,,y denote the plane in R3 normal to the vector

q" and containing the point u. In coordinates write u = (ui, u2, u3) and = (qi, q2, q3), and suppose that 93 54 0. A coordinate chart for r1g,4 is given by x : R2 -+ R3, x(r s) = I

r,

(ui - r)ql + (u2 - s)q2 + s

U3 q3

The mapping x satisfies the conditions x(ui, 142) = i and (x(r, s) - i) q' = 0. Thus x maps R2 to 1714.4. Since x is an affine mapping, it is differentiable. The inverse of x is easily seen to be given by x'1 (ui, 142- u3) = (U 1, 112), which is continuous. It follows that x is

one-to-one and onto lg,y. Finally, the Jacobian is given by

afi

ar aft

afi as

1

0

ah

0

1

ar

as

qI

q2

afi

afi

q3

q3

ar

as

8. Surfaces

97

which has rank 2. This shows that the planes in R3 are regular surfaces. (For planes that are perpendicular to a vector q' 96 0 with 93 = 0, a change of names of the axes allows us to apply the same argument.)

(2) Let S2 = ((xi, x2, x3) E R3 I (XI )2 + (x2)2 + (x3)2 = I ( denote the unit sphere in R3, and let D = ((u, v) E R2 I u2 + v2 < I ( denote the open unit disk in R2. We define the mapping x : D -+ S2 by

l - u2 - v2).

x(u, v) = (u, v,

The image of this patch covers only the upper hemisphere of S2. The mapping is differentiable and has inverse x- I (u i , u2, u3) = (u 1, 142). The Jacobian is of rank 2 and so the patch is regular. In order to cover all of S2 with similar hemispheric charts, we need five more patches like this one corresponding to pairs of hemispheres "upper-lower," "back-front;' and "east-west" The sphere is a model of the surface of the Earth and so it is a well-studied surface. We will discuss other coordinate charts on S2 in the next chapter. (3) Both of the previous examples are special cases of the following construction: Suppose f: (U C R2) -+ R is a smooth function, where U is an open set in R2. Consider the graph of f, the set ((u, v, f(u, v)) (u, v) E U}, as a surface in R3. Notice that derivatives of I

all orders exist for the function x(u, v) = (u, v, f(u, v)) which is injective and onto the graph. Finally, the Jacobian 0

I

J(x) =

0

o

of

au

av

has rank 2.

Thus smooth, real-valued functions of two variables provide us with many examples of surfaces. Curves in the plane give another way to construct surfaces.

(4) Suppose that g: (a, b) -+ R is a smooth function satisfying g(l) > 0 for all t E (a, b). We obtain a surface by rotating the graph of g around the x-axis. Define a coordinate chart for this surface, x : (a, b) x (0, 2n) -. R3, by

x(u. v) = (u, g(u) cos v, g(u) sin v). Since g is smooth, x is differentiable. The Jacobian 0

1

a

J(z)(u, u) _

(cos u) au

-g(u) sin v

(sin v) au

g(u) cos v

contains square submatrices with determinants g(u) fig, -g(u) sin v, and g(u) cos v. Between these three expressions, for each (u, v), at least one is non-zero so J(x)(u, v) has rank 2. Since g(u) > 0 for all u E (a, b), (g(u) cos v, g(u) sin v) is injective as a function


98

of v. Since we have rotated the graph of a function, x is injective. To construct the inverse function to x, suppose (a, b, c) is in the image of the chart x and solve for v: arccos(b/(b2 + c2)),

V

- { 27r - arccos(b/(b2 + c2)),

c > 0, c < 0.

We check that this expression is continuous at the only problematic point, that is, c = 0. In this case, sin v = 0 and hence cos v = -1, and v = rr. Thus the assignment is continuous. More generally, if a: (a, b) -+ R2 is a regular, parametrized, differentiable curve that does not intersect the x-axis and does not intersect itself, then we obtain a surface by rotating the trace of a(t) around the x-axis. In fact, any line in the plane may be chosen as the axis of rotation generating a surface from a curve that does not intersect itself or the chosen line.

Examples 3 and 4 show that the class of regular surfaces in R3 is large. To generate another class of surfaces, we recall two important theorems from the calculus of several variables (see Spivak (1965) for details).

The inverse function theorem. Let F: (U c R") R" be a differentiable mapping (aFj and suppose, at u E U, the Jacobian matrix J(F)(i) = R" - R" is a linear axe ///

isomorphism. Then there is an open set V C U, containing i , and an open set W C R" containing F(i) such that F: V -, W has a differentiable inverse F-1: W -+ V.

The implicit function theorem. Suppose f: R" x R'

R' is differentiable in an open set around (i , v) and X. v) = 0. Let M be them x m matrix given by

M=

of (u) ax"+;

1
If det M 54 0, there is an open set U C R" containing u and an open set V containing v so that, for each i E U. there is a unique s' E V such that f (i. s) = 0. If we define g: U -+ V by g(F) = s, then g is differentiable. These theorems are equivalent and they provide us with another method for generating surfaces.

Definition 8.2. Given a differentiable function f: R' -+ R' with n > m and its Jacobian

J(f) = (a j/ax;), a point u E R" is a critical point off if J(f)(i). as a linear mapping R" - R'. is not onto. The point f (u) E R' is called a critical value of f. If v E R' is not a critical value of f, then v is called a regular value of f. Theorem 8.3. If f : (U C R3) -+ R is a differentiable function and r E f (U) is a regular value, then f-1((r)) = (u E U I f(u) = r) isa regular surface in R3. PROOF. Let p = (xo, yo, zo) E f-1 ((r)). Since r 'is a regular value of f, we can assume

that az 54 0 at p. Define F: (U C R3) -+ R3 by F(x. y, z) = (x, y, f(x, y, z)). The

8. Surfaces

99

Jacobian of F has the form

J(F) =

I

0

0 aj

aj

0 aj

ax

ay

az

1

0

and so det J(F)(p) 54 0. Apply the inverse function theorem to get neighborhoods of p and F(P), say V and W, respectively, such that F: V - W is invertible and F-1: W --- V is

differentiable. To construct our coordinate charts take U' = W n (z = r) and let x: U' V n f-t ((r)) be given by x(u, v) = F-1 (u, u, r). This determines a regular surface. Corollary 8.4. For a, b, and c nonzero, the following functions have 0 as a regular value and so determine a regular surface in R3. namely, f-1 ({0)): ellipsoids,

x2

j(x, y, z) = a2 + x2

t

y2

b2

y2

+

z2

c2

-I;

- 2z2 - I ;

hyperboloids of l or 2 sheets,

f (x, y, z) =

andfor 0 < r < a, the torus,

f(x.y,z)=z2+( x2+y2-a)

a2

62

2

-

r2.

An instructive nonexamplc of a surface is the cone

{(u, v, w) I u2 + v2 = w2 and w > 0). There is no way to introduce coordinates to a neighborhood of (0. 0. 0) in a smooth manner, and so the cone is not a regular surface. If we weaken the requirements on a surface and substitute continuity for differentiability, then we get the definition of a topological surface. The cone is an example of such a topological surface. We get another nonexample by considering how the pages of a book come together. Away from the spine the pages are locally two-dimensional, but if three pages come together in a line segment, no point on the join has a neighborhood which is locally two-dimensional. We now develop the calculus on a regular surface. In particular, we define what it means for a real-valued function, f : S --+ R, or a function between surfaces, m : St -* SS, to be


100

differentiable. A first guess at a definition of differentiability of a function f : S -- R

is if for each coordinate chart x: (U C 112) - S, the composite U X S f t is differentiable. However, this presents a problem - there may be infinitely many different coordinate patches containing a given point to check for differentiability at the point. This is rendered unimportant by the following proposition.

Proposition 8.5 (change of coordinates). Suppose S is a regular surface and x : (U C R2) -+ R3 and y: (V C R2) -+ R3 are coordinate charts with W = x(U) fl y(V) C S nonempty. Then the change of coordinates

h = y-I ox:x-'(W) -- y-I(W) is differentiable, one-to-one, and onto, and its inverse h-1 = x-1 o y is differentiable.

PROOF. We know that h is one-to-one, onto, and continuous since it is a composite of homeomorphisms. Since h-I and h are symmetric in x and y, we only need to show that h is differentiable. Let (ro, so) E x-t (W) and write p = x(ro. so). Consider the following trick: y is a coordinate chart which we write as Y(u, v) = (fi (u, v), f2(u, v), f3(u, v)), and so

af,

J(y) _

aflaf2_af,af2 W e assume that function

au av

au au

af,

au

av

aft

aft

au

av

af3

af3

au

av

has rank 2.

54 0 at (uo, vo) = h(r0, so) = y- I (p). Extend y to the

F: V x R - R3,

F(u, v, t) = y(u. v) + (0, 0. t).

The function F is differentiable and F(u. v. 0) = y(u, v). Consider the Jacobian of F:

af,

J(F) =

au aft

af, av

aft

au

dv

af3

af3

au

av

At (uo, vo, 0) this matrix has nonzero determinant and so, by the inverse function theorem,

there is a neighborhood of F(uo. vo, 0) = x(ro, so) such that F-1 exists and is differentiable. Now we can write (h (r0, so), 0) = (uo. vo, 0) = F-1 o x(ro, so),

8. Surfaces

101

that is, h is obtained from the composite of F-1 and x, and by ignoring the zero third coordinate. Since these mappings are differentiable. so is h.

Definition 8.6. If f: S - R and p is a point in S, then f is said to be differentiable at p if for any coordinate chart x: (U C R2) -+ S with p E x(U), the mapping fox: (U C R2) -+ R is differentiable at x-gy(p). f is differentiable, if it is differentiable at every point of S.

Proposition 8.5 implies that if f: S --+ R is differentiable at a point in one coordinate chart, then it is differentiable at that point for any other chart containing the point. The proposition also allows us to define easily what it means for a function between surfaces 0: S, --+ S2 to be differentiable. Definition 8.7. A function q5: S,

--+

S2 between surfaces S, and S2 is differentiable

at p in S, if there are coordinate charts x: (U C R2) - Si and y: (V C R2) -+ S2 with p E x(U), O(p) E y(V). such that the composite function y 1 o 0 o x: U -. V is differentiable at x-1(p). 0: S, -- S2 is differentiable if it is differentiable at each point in Si. A function 0: Si - S2 is a diffeomorphism if it is differentiable, one-to-one, and onto, and has a differentiable inverse function. Two surfaces S, and S2 are said to be diffeomorphic if there is a diffeomorphism 0: Si -+ S2. 2

EXAMPLE. The ellipsoid I (x, y, z)

I

a2 +

b2

+ c2 = I and the sphere S2 are diffeo111

morphic via the diffeomorphism ¢: S2 -+ the ellipsoid,

O(x, y, z) = (ax. by, cz).

The notion of diffeomorphism yields an equivalence relation on the set of all regular surfaces in R3. 71vo diffeomorphic surfaces are equivalent in the sense that they share equivalent sets of differentiable functions and any construction involving the calculus on one carries over to the other. This equivalence relation is the basis for the differential topology of surfaces. However, it is too weak an equivalence to be useful in distinguishing geometric objects.

The tangent plane The derivative of a function is the best linear approximation of the function at a point. A geometric realization of linear approximation is the tangent plane - a linear subspace of R3 that is the best approximation to a surface at a point. To define the tangent plane we restrict our attention to a single coordinate patch, x: (U C R2) -+ S. Suppose p E x(U) and p = x(uo, vo). Now restrict the mapping x to x (u, vo) for uo - e 0 and q > 0 with (uo - e, uo + e) x (vo - q, vo + 17) C U. These restrictions determine two curves,

u r-+ x(u, vo) and v r+ x(uo, v),


102

called the coordinate curves on S through p. Define

xu =

and x =

a

au

ax

(X (u, vo))I (uo vo)

a

av

= au (uo. vo) =

(X (uo, u)) I (U0. L-0)

ax av

(uo, uo);

x and xu denote the tangent vectors in R3 to the coordinate curves at p.

Proposition 8.8. A mapping x: (U C R2) -+ S satisfies the regularity condition if and only if the cross product x x xu 54 0. PROOF. The coordinate expressions forxv and x are the columns of J(x). Since x x xv = 0

if and only if x is a scalar multiple of xu if and only if J(x) has rank less than two, we have proved the proposition.

It follows that for each coordinate patch on a regular surface in R3 we can define a normal direction, that is, x x xv. We define the unit normal vector to the surface Sat a point p as

N(p) _

x x xv

IIXu X

In order for this notion to be well defined, we need to show that it is independent of the choice of patch containing p. As it turns out, this may be too much to ask for.

Proposition 8.9. If p is a point in a regular surface S and x: (U C R2) -, S and y: (V C R2) -r S are two coordinate charts containing p. then at p we have XU xxv IIXuXXuII

YuXYo IIYuxY6II

8. Surfaces

103

PROOF. Let W = x(U) fl y(V ). The subsets x-' (W) C U and y-' (W) C V are related by the diffeomorphism

y- 'o x: x-' (W) - y' (W ),

y ' o x(u, u) = (u(u, u), v(u, u)).

The coordinate curves in the chart x are x(u, vo) and x(uo, v). They are expressed in the chart y by using x(u, v) = y(u(u, u), v(u, v)). The tangent vectors become

au

x°=Yu-+y

au

au

av

+yeav.

We can compute x x x directly to get

au av XU X xa = Yaau +Yaau au as

a;, au

au av - au av

au av Yaav +yuav

X

Y a x Yo = det J(y ' o x)ya x

It follows that

XU xx _ detJ(y'ox) Ilxu x x,,ll

y;,xyD

IdetJ(y- ' o x)111 Y5 X YtIl

On a given patch, the normal is simply

xu x x

yaxya =f IIYy X Yoll

We can "translate" the normal over

Ilxu X

the surface by moving patch to patch, where we choose the variables u and v so that the unit normal vectors agree on the overlaps. For surfaces such as the Mobius band, however, this cannot be done. To construct the Mobius band imagine a rectangular band of paper, make a half-twist, and glue the ends of the band together. The mathematical version of this construction gives a surface when we ignore the boundary (see the exercises). If we move the normal at a point around the surface, when we arrive back at the first chart from which the translation began, the normal has changed sign. We distinguish the class of surfaces for which a coherent normal can be chosen.

Definition 8.10. We say that a surface S is orientable if there is a collection of charts (xa : (Ua C 1R2) - S I a E A) that cover S, that is, S = UaEA xa, (Ua), and if p E S lies in the overlap of two charts, p E x# (Up) (1 xy(Uy), then the Jacobian J(xy' o xf) at p has positive determinant.

Since the sign of the determinant of the Jacobian determines whether the normals at a point in the overlap of coordinate charts agree, the definition of an orientable surface lets us translate a normal at any given point over the rest of the surface. Even without assuming orientability of a regular surface, ±N(p) does determine a plane.


104

Definition 8.11. Let p be a point in a regular surface S. and x: (U C R2) -+ S any coordinate chart containing p. The vector subspace of R3 of vectors normal to N(p) _ XU X xv II

x x Xv

is called the tangent plane to Sat p and is denoted by Tp(S). II

Notice that Tp(S) is a plane through the origin and so does not necessarily contain the point p. By translating this plane by adding the vector determined by p we get the "plane tangent to the surface S at p." It is visually satisfying to think of Tp(S) in this fashion. However, it is more important to think of Tp(S) as a set of directions, admitting the operations of addition and multiplication by a scalar. An immediate consequence of the definition is that x,,, xv is a basis for Tp(S). We next see that Tp(S) contains the directions of the tangent vectors to any regular curve on the surface S through p.

Proposition 8.12. A vector v E R3 is an element of Tp(S) if and only if there is a differentiable curve a: (-e, e) -+ S with a(0) = p and v = a'(0).

PROOF. Suppose that a: (-e, e) - S is a curve lying on S with a(0) = p. Choosing e > 0 small enough we can take the image of a to lie in the image of a patch x: (U C R2) -+ Scontaining p. Considerx-1 oa: (-e, e) -+ U. Thiscan be written (x-1 oa)(t) _ (u(t), v(t)) and so a(t) = x(u(t), v(t)). By the Chain Rule,

a'(t)=

ax du

ax dv

du

dv

aud +avd =x"dt +x°dt'

that is, a'(t) is a linear combination of x and x,,. Thus a'(0) lies in Tp(S).

If v E Tp(S), then we can write v = ax + bx,,. Suppose p = x(uo, vo). Consider the

curvef(t) = x(uo+at, vo+bt) for-c < t < esmallenoughsothat(uo+at, vo+bt) E U. Then #'(0) = v. Thus every vector in Tp(S) is realized as a tangent vector to some curve on the surface. There is another interpretation of the tangent plane that frees us of the dependence on coordinate charts (though the proofs of the relevant properties go through charts). To discuss this we introduce an important object associated to a regular surface S. Define the set of smooth functions on S. COO(S) = (f: S -i R I f is differentiable). A more local version of COO(S) is the set of smooth functions defined on some neighborhood

of a given point p E S:

COO(p) = (f: (V C S) -+ R I V an open subset of S. P E V; f differentiable). Both COO(S) and Cm(p) are vector spaces endowed with a multiplication: (fg)(q) _ f(q)g(q), for q in a neighborhood of p. Alternatively, we can define Tp(S) as follows: Consider the collection of all linear mappings v: C°O(p) -+ R satisfying the Leibniz condition:

v(fg) = f(p)v(g) +g(p)v(f).

8. Surfaces

105

The correspondence between such mappings and tangent vectors as we have defined them

is obtained by considering a curve through p in S. say, a: (-E, E) - S with a(O) = p. The curve defines a mapping a'(0): C°O(p) --), U.

a'(0)(f) = d (f o The standard results of the calculus imply that a'(0) is a linear mapping and satisfies the Leibniz condition. With a little more work one can show that for any v E Tp(S), u(f) = V f v. We will return to this idea in Chapter 14 where we will not have the use of an ambient space like R3 in which to define tangent vectors. A differentiable mapping between surfaces induces a mapping of tangent planes at corresponding points. To define the mapping suppose F: S1 - S2 is a differentiable function and p E S1. A tangent vector i in TT(Si) corresponds to a'(0) for some curve a: (-E, E) -- S1. Composition with F determines another curve F o a: (-E, E) -* S2: Define

dFp(a'(0))= dt(Foa) r=0

Proposition 8.13. The mapping dFp: Tp(Sj)

TF(p)(S2) is linear.

PROOF. Suppose p and F(p) lie in the images of coordinate charts x and z on Si and S2, respectively. Consider the composite

(UCIIt2)-$I

i-I

--- (V C:

R2).

F F o x(u, v) = .(P1 (u, v), P2(u, v)). If a curve a: (-E, E) -> Si is given by a(t) = x(u(t), v(t)), then we can write F o a(t) = X(F) (u(1), v(t)), F2(u(t ), v(t))).


106

Suppose 'v = a'(0) E Tp(S); with respect to the basis {x,,, x,,) we can write v = x

du + dt

x dv . With all this in place we compute

d

dFp(u) = dt(Foa)l,_o =

(aidu

aFidu dE2du aFZdv au dt + av dr' au dt + av di apt

du

au

av

a IP2

a P2

r dt 11 1 du

aF,

1=0

r=0

This identifies dFp with the linear mapping given by multiplication by the matrix determined by the Jacobian of r- 1 o F o x. Notice that the previous argument applies to the identity mapping id: S -+ S and two different coordinate charts containing a given point. The change of basis between (x,,, x,,) and (yu, yp) is given by the Jacobian of v r o x. This fact is useful when we check certain constructions for invariance under change of coordinates.

The first fundamental form We are now in a position to introduce some geometry. For the most basic structure, we turn to linear algebra and an inner product with which many geometric notions may be defined.

Definition 8.14. For a point p E S, a regular surface, the first fundamental form Ip: Tp(S) X Tp(S)

R

is the inner product on Tp(S) induced by the dot product on R3. Thus Ip(v, u,) = v iu. We also denote Ip(v, w) by (v. w)p.

Notice that the definition is dependent on the fact that S is a subset of R3; the inner product (u', v)p = u v is inherited from R3. This appears undesirable on first consideration as it seems not to be a feature of the surface but an artifact of the embedding. Later we will show that important geometric properties of surfaces can be developed from the first fundamental form that are independent of the embedding. In order to calculate with the first fundamental form it is useful to have a local expression, that is. a formula in terms of a coordinate chart. For p E X (U) c S. the tangent plane Tp(S) is spanned by x and x5. Tangent vectors ii = a'(0) and w = r6'(0) in Tp(S) are associated

to curves a: (-e, c) - S and P: (-q,,) - S. If we write a(t) = x(u(t), v(t)) and

8. Surfaces

107

fl (t) = x(u(r). u(t)), then we can compute 1p(u, w) by

Ip(v, w) = lp(a'(0), 0'(0))

_

Ip

du

x°dr

= I p(xu, xu)

dv

+xvdr

du

- x°dr

dv xvdr

dude dude dvdv dude + Ip(xt xt,) + Ip(xu, xv) ( dr dr dr dr + dt dr dr dr

We introduce the functions

E(u, V) = I p(x,,,

F(u, v) = Ip(x,,, xv), and G(u, v) = I p(xv, xe).

These functions are defined on U C R2. and are differentiable. They are called the component functions of the metric on S. If we write tangent vectors as column vectors with respect to the basis (x,,, xv ), then v = I

dv/dt ) and w = 1 dU/dl

and we can express

\\\

the first fundamental form by the matrix equation GE

F

where 0' is the transpose of u, that is,v as a row vector.

EXAMPLES. (I) The uv-plane, ((u, v. 0) E R31. has coordinates given by the single chart x : R2 -+ 1R3, x(u, v) = (u, v, 0). For this chart x = (1, 0, 0) and xv = (0, 1, 0). The components of the metric are given by the functions

E(u, v) = I = G(u, v) and F(u, v) = 0. If we remove the origin and the positive u-axis, there is another chart given by polar coordinates (r. 0). The chart takes the form

y: (0, oo) x (0, 27r) - R3,

(r cos 0, r sin 0, 0).

Then yr = (cos 0, sin 0, 0) and ys = (-r sin 0, r cos 0, 0). The components of the metric are given by

E(r,0)= I, F(r,0)=0, G(r,0)=r2. (2) Standard spherical coordinates on S2 (see Chapter I) are given by

x: (0, 27r) x (0, 7r) -. S2. x(0,,0) = (cos>isin0,sin>
x, =(-sin >/isin 0,cos>Gsin 0.0) and xe=(cos*cos0.sin Vi cos0.-sin 0)

Development: Differential geometrn

108

and so the components of the metric are given by

E(>Ji.0) = sin2O. F(*. 0) = 0. G(>/r, 6) = I.

If we compute the determinant of thematrix ) E G ), then the regularity of a coordinate chart implies

EG - F2 =

llx,.Il2 - (x

11.,C x .r,,11- > 0.

As an abstract inner product on T,,(S), we can infer from this calculation that the first fundamental form is positive definite, that is, Iv(v, u) > 0 and zero if and only if u = 0. Furthermore, I, is nonsingular, that is, It,(vtt, w) = 0 for all ut if and only if vtt = 0. These properties, in fact, are inherited from the dot product on llt3, but it is reassuring to find that they may be established from the local expression. This will be useful in generalizations to come later (Chapter 14). With the first fundamental form we can calculate the length of curves on a surface. If a: (a, b) -+ S is a curve on S. then the length of the curve between a(p) and a(t) along a is given by

s(t) =

a'(r)) dr.

Ila'(r)lldr = f

J

s

1U

In local variables, s(t) can be expressed as the integral

/ du

2

/ du

ddu dtr'

E(u(r). v(r)) ( dr l + 2F(u(r). v(r)) +G(u(r). Or)) I dr rd 1i

I

dr.

It is customary to abbreviate this equation by the expression

ds2 = Edu2 + 2Fdudv + Gd v2. and call ds the element of arc length or the line element on S. For example, if E = G = I and F = 0. then ds2 = du' + dv2 expresses an "infinitesimal" form of the Pythagorean Theorem. Of course, this is shorthand for ds dt

_

E

(du dt

+2F

drrdv

dt dt

+G

(dv dt

The rules for transforming the local expressions for a line element follow from the calculus. and conveniently they are what one would expect algebraically of such an expression.

EXAMPLES. (I) We have computed already the line element for the uv-plane. I(n. v. 0) E 123}, in the usual coordinate chart and polar coordinates:

ds2 = did + d u`' = dr2 + r'd02.

8. Surfaces

109

(2) More generally, consider the plane in R3 that is parallel to the plane spanned by the basis (wi, w2) containing the given point p"o. We can parametrize this plane by the coordinate chart x : R2 -. R3, x(u, u) = PO + ui 1 + uw2. This implies that x = wi and x = w2. The associated line element is

ds2 = (wi wi )du2 + 2(t

w2)dudv + (w2 w2)du2.

Taking wi and w2 to be unit vectors with wi -L w2, we get ds2 = du2 + du2 or the "Euclidean" line element.

(3) Consider the cylinder over the unit circle with coordinate chart x: (0, 27r) x R -. R3

givenbyx(O, v) = (cos9,sin8, v).Herexe = (-sinO,cos9,0)andx = (0,0. 1). Hence E(6, v) = sine 0 +cos2 0 = I, F = 0, and G = 1. Once again we have ds2 = doe + du2. To understand why this line element is the same as the Euclidean line element, consider the cylindrical rollers used to print a newspaper - no distortion of the print takes place as they transfer print onto a planar page. The cylinder is a piece of the plane rolled up without altering relations between points.

------- -------

(4) Let f: (U C R2) -+ R be differentiable and consider the surface given by the graph of f with the coordinate patch x(u, v) = (u, v, f(u, v)). Then

x = (I, 0. of/au) and x = (0, 1, of/av). It follows that the line element is given by

ds2 = (I + (af/au)2)du2 + 2(af/au)(af/av)dudv +(I + (af/av)2)du2. (5) Consider the coordinate chart for the lower hemisphere of the unit sphere, x: ((u. v)

I

U2 + v2 < 1) -+ S2.

x(u. V) = (u. V, - 1 - u2 - v2). Here the tangent vectors to the coordinate curves are xu = (I, 0,

x = (0, I, line element

u

I -u -u ds2 =

1-

u

1 -u -v

) and

). When we compute the dot products and simplify we get the

u2

-u2-u2

2uv

du2 + T--;T--V2 dudu +

I - u2 I -u2-u2

dv2


110

This line element will make an important appearance in Chapter 15. The expression for the are length of a curve in a surface, given locally by the line element, is something that should depend only on the curve and not on the choice of coordinates. We

now see how a line element transforms under a change of coordinates to show the invariance of this local expression. Suppose a: (-e. e) - S is a curve on S lying in the intersection of two coordinate charts:

a(t) = x(u(t), v(t)) = y(u(t). D(t)). Notice that (u(t), P(t)) = yt o x(u(l). v(t)). Expressing the tangent vector to the curve a(t) in these different coordinates we get

a' (t) = .r

du

dv

_

du

dD

dt + x`. dt = y" dt + 1, dt

.

Write the component functions for the metric associated to x as E, F, and G and those associated to y as E, P. and G. The change of coordinates y- 1 o x implies an

X. = y" au +

aD x`,

yp as .

au

au

Ju G=.rf x,.=E(av)

E F) (F

_

G

F

an

aD )2

aD

- -)V I+G.

auaD\ av Ju/

auaD au av ail aD

2

i)DdF.

au

+2Favav+(an)2

av

au

aD

all aif av

all

=J(v

-.

+2Fauau+G(all

an an au av

F

au

,5 - + V

aD

av

cx'

au

EF

(F E

)

i)u au

an av au

all

F)J(''ox).at,

F GG

It follows that the change of component functions of the metric is made by multiplying the given matrix of component functions on the right by the Jacobian of the transformation (u, 0) = y 1 o x(u. v) and multiplying again by its transpose on the left. Line elements may be compared by noticing from these computations that

duldt

(d v/dt) = J(yI ox)

(ddFildt

v/dt

8. Surfaces

111

This leads to the matrix equation

ds2 = (du

dv)

EF

CE F)

du dv

= CJ(y I o.r)Cdu))f CF G) CJO'

\

E'

F

ox)Cdv

du

=(du du) CF G)Cdo

Thus the line clement is preserved under change of coordinates. The transformational properties of the matrix of component functions make it significant as a local object defined on the surface. We will consider similar expressions throughout the text. We can think of the matrix of component functions as "invariant under change of coordinates" in an unusual

way - the change does not preserve the form exactly, but the correction term, here the Jacobian, preserves the relevant form. This kind of invariance is made precise through the use of the tensor calculus, which is developed in Chapter 16. In practice, the transformation of the line element is carried out like a change of variable in integration. Consider the surface S given by the graph of u2 + v2. One coordinate chart is given by EXAMPLE.

x: 1R

-> S.

x(u, v) = (u, v, u2 + v`).

This paraboloid also has polar coordinates:

y: (0, oo) x (0, 2;r) -+ S,

y(r, 0) = (r cos0, r sin d, r2).

Being a graph, the surface has line element given by

ds2 = (I +4u2)du2 + (8uv)dudv + (I + 4v2)dv2. Changing to polar coordinates we can write

du = cosOdr - rsinOd8.

is = r cos 0, v = r sin 8. which implies

dv = sinBdr +rcos8d8.

Substitute these expressions for it. v, du. and dv into the given line element to obtain the

following:

ds2 = (I + 4r2 cost 0)(cos2 Bdr2 - 2r sin 0 cos 6drd8 + r2 sin2 6d02) + 8r2 cos 0 sin 0(cos 0 sin 0dr2 + r(cos2 B - sine 0)drd8 - r2 sin 0 cos 8d92) + (I + 4r2 sine B)(sin2 0dr2 + 2r sin 8 cos BdrdO + r2 cos2 0dO2) _ (I + 4r2)dr2 + r2d82.

The tangent plane, together with its inner product, allows us to make geometric constructions and define certain quantities. For example, suppose two curves on a surface S


112

intersect at a point p. The first fundamental form may be used to determine the angle between the curves - we simply apply the definition of the angle between vectors in R3 from linear algebra. If a: (-E, e) -+ S and f : (-e, e) -- S are differentiable curves that satisfy a(0) = f(0) = p, then the angle 0 between a and O at p is given by the angle between the tangent vectors to a and 6 at p: Ip(a'(0), )3'(0))

cost =

1Io(a'(0), a'(0))lp(f'(0). $'(0))

For example, consider the tangent vectors to the coordinate curves, x,,, xi,: the angle between the coordinate directions is given by IP(x,,, xv)

0 = arccos

arccos

(Fl

(JP(T-'X-)IP(X,1'

x and x, are orthogonal when F = 0. We call such a coordinate chart an orthogonal parametrization. EXAMPLE. In the earlier computation for the paraboloid ((u, v, u2 + v22)) the rectangular coordinates give

F

EG

_

4uv (1 + 4u2)(1 + 4u2)

These coordinate curves are seen to be the intersection with the paraboloid of vertical planes

that are parallel to the u- or v-axis. The resulting curves on the surface meet in an angle given by arccos(F/ EG). The polar coordinates, on the other hand, give an orthogonal parametrization.

Area Through the first fundamental form we have defined the are length of a curve and the angle between curves. We next define area, another fundamental geometric quantity.

Definition 8.15. By a bounded region R in a surface Sin R3 we mean a subset of S that is contained in some ball of finite radius in R3.

The following theorem is a standard result in multivariable calculus.

8. Surfaces

113

Theorem 8.16. Suppose S is a regular surface and x : (U C R2) - R3 is a coordinate chart. If R C S is a bounded region and R C x(U), then the area of the region R is given by

area(R) = f 1 dA =

ff. -'(R) Ilxv x xvlldudv.

Area is obtained as the limit of a Ricmann sum associated to a system of partitions. In this case, we consider a transformation U -. S -+ Tp(S) for p a point in a small region lying in U. The tangent plane Tp(S) is a linear approximation to S near the point p. To prove the theorem one shows that the infinitesimal contribution to the distortion by this transformation is given by Ilxu x xull. For details, see your favorite multivariable calculus book (Spivak (1965)). Recall that Ilx x xv 112 = Ilx. II21Ixv 112 - (x xv)2, so a local expression for area is given by the integral

area(R) =

rf

J x '(R)

E(u, v)G(u, v) - (F(u, v))2dudv.

The area of a bounded region is a measure that should be independent of the choice of coordinate chart. In order to see how our local expression for an area transforms with respect to a change of coordinates we prove the following proposition. Proposition 8.17. Let h: V -. U be a diffeomorphism of open sets of R2 and let y = x oh, then

JJr'(R) Ilye x ylldud= IJx(R) Ilxx xlldudv. PROOF.

If the variables u, v are transformed to variables u, D. then the expression dude

transforms to I det J(h)ldudv. Thus

ff

A-

IIyr, x yolldudu = (R)

ff

y (R)

Ax

Ilxv x xv11 I det

J(h)ldudv

Ilxv x xvlldudv. '(R)

For a change of coordinates y ) o x, the component functions of the metric transform in such a way that det

and so

E F

FG

= det

E F

pG

(det J(y ) o x))2,

EG - F = I det J(y ( o x)l E( -P2.

As an example we compute the area of the lower hemisphere of S2. We parametrize the lower hemisphere from the unit disk with polar coordinates by

x: (0, 1) x (0, 2n) -. R3,

x(r, 0) = (rcos0, rsin0. - 1 - r2).


114

In this system of coordinates the tangent vectors to the coordinate curves are given by Xr = (cos 0, sin 0, (r/ I - r )) and xB = (-r sin 0, r cos 0, 0). It follows that E = I /(I - r2), F = 0, and G = r2. The area of the hemisphere is given by 2a

1Jf

EG - F2drd9 = R

j

t

r

r

JO

1

r

drd9

= 2n. As another example, suppose R is a bounded region on a cylinder and x : (0, 2rr) x R -1R3 is the coordinate chart x(0. v) = (cos 0, sin 0, v). Then area(R) = area(x-t (R)) since

E = G = I and F = 0. The expression EG - F2 is a measure of how much the surface differs from its parametrizing plane. Exercises 8.1

Determine if the following functions qualify as coordinate charts for a surface in R3. Take

x:(-1, I)x(-1,1)-+ R3: (1) x(u, v) = (u2, u - v, u2). (2) x(u, v) = (sin rr u, cos 2r u, v). 8.2

83'

(3) x(u, v) = (u, u + v, v3/2). Let x: R2 --> R3 be given by x (u, v) = (u - v, u + v, u2 - v2). Show that the image of x is a regular surface S, and describe S. Leta: (a. b) -+ R3 be a unit-speed curve with nonvanishing curvature. Let U = (a, b) x (R - (0)) c R2 be the domain of a mapping x: U -+ R3 given by

x(u, v) = a(u) + ua'(u). Show that x is a coordinate chart when x is one-to-one. The resulting surface is called a tangent developable surface. Determine the component functions of the metric for such a surface. 8.4'

Consider the coordinate charts x_ x : (0, 27r) x (- 1, I) -. R3 defined by:

x(u,u)_(((/2-using)sinu.((2-vsinZ))cosu,vcoosZ),

\\

z(u,i7)=112-tisin(4+2 ))cosu. - l2-usinl 4+2 I lsini. vcos(4+2 Show that these charts determine the Mobius band as a regular surface in R3. Determine the change of coordinates between patches and show that the Mobius band is not orientable.

8.5

Suppose that f: U -+ V is a coordinate transformation on regions in R2. Suppose that x: V -+ R3 is a coordinate chart for a surface in R3. Determine the effect of the coordinate

transformation on the metric coefficients associated to x and to y = x o f.

8. Surfaces

8.6

115

Show that the tangent plane Ta(S) at a point p to a surface S is given by

Tn(S) = (x)(p)(R2) C R3. where J(x)(p) is the linear mapping determined by the Jacobian evaluated at p. 8.7

Consider the following coordinate chart for fixed values a, b, c > 0 and u, v E Ht:

x(u, v) = (a sinh u sinh v, b sinh u cosh v. c sinh u). Determine the first fundamental form and metric coefficients for the chart.

8.8

Determine the component functions of the metric for a surface of revolution generated

by the graph of a function f : (a, b) --. R, that is, the set ((u, v) I v = f(u)r.

8b

Map projections You are quite right that the essential condition in every map projection is the infinitesimal similarity; a condition which should be neglected only in veryspecial cases of need.

C. F. Gauss (to Hansen, 11 December 1825)

"The world is round: maps are flat." So begins an introduction to mathematical cartography (McDonnell 1979) and so also do we find an interesting problem for geometers. Coordinate

charts for surfaces are named after cartographic maps. In this chapter we consider the motivating examples forcoordinate charts, the classical representations of the Earth, thought of as a sphere. The subject has a history almost as old as geometry itself. The basic problem is to represent a portion R of the globe (idealized as the unit sphere

S2) on a flat surface, that is, to give a mapping Y: (R C S2) - R2 that is injective and differentiable, and has a differentiable inverse. Such a mapping Y is called a map projection. A coordinate chart for the sphere, x: (U C R2) -> S2, with a differentiable inverse, determines a map projection by taking Y = x-t, and conversely, a map projection determines a coordinate chart. The purposes of cartography, such as navigation or government, determine the properties

of interest of a map projection. An Ideal map projection is one for which all relevant geometric features of the sphere are preserved in the image. Lengths, angles, and areas on S2 are carried by an ideal map projection to identical lengths, angles, and areas for an appropriate choice of unit. More generally, such a map between surfaces is called an isometry. Many geographical features of interest could be found on an ideal map - shortest distances between points would be represented by line segments. and so navigation would be as simple as using a ruler and a protractor. The relative sizes of land areas would be apparent. One of the first results in the study of mathematical cartography is due to Euler in his paper De repraesentatione superficiei sphaericae super piano (1778).

Theorem 8H ,1. There are no ideal map projections. PROOF. Euler's proof is based on certain differential equations that describe "infinitesimal similarity." Here a synthetic argument suffices. Let AABC be a triangle on S2 formed of segments of great circles. From Chapter I we know (informally) that great circle segments minimize lengths, and so the ideal map projection takes A B, BC, and AC to line segments in the plane. However, the angle sum of the spherical triangle exceeds r while the triangle in 116

8"". Map projections

117

the plane has angle sum it. Thus, such an ideal map projection cannot be angle preserving, a contradiction.

The theorem tells us that particular properties of a map projection must be chosen to suit to the purpose of the map and we must settle for the failure of other properties. T\vo properties serve most purposes:

Definition 81.2. A diffeomorphism f : S1 -+ S2 of surfaces (in particular, a map projection) is conformal if it preserves angles, that is, given two curves, a: (-E, e) -+ S1 and f : (-n, q) -+ S,, with a(O) = p = P (O), then the angle between a'(0) and f'(0) in TT(S,) is the same as the angle between d f p (a' (0)) and d fp(4' (0)) in T!(p((S2 ). A mapping

is equiareal if areas are preserved. Navigation requires conformal map projections, while government applies equiareal ones. Before discussing the classical examples. we develop general criteria for a diffeomorphism to be conformal and to be equiareal.

Proposition 81.3. Suppose f : Si -+ S2 is a diffeomorphism. Then f is conformal if and only if there is a nonzero function p : Si -+ R such that, for all p E S, and v, 9) E Tp(S, ),

ip(v, u) = P2(p)1!(p)(dfp(0). dfp(w)). PROOF. Suppose f : St -+ S2 is a diffeomorphism and there is a nonzero function p : S1

-

Iii satisfying the stated equation. We compare the angle 0 between two curves on S, and the angle 01 between the corresponding curves on S2:

cos0 =

rp(u, u')

Jlp(iO) Ip(ri,, w) P2(p) if(p)(dfp(v). dfp(u,)) VPZ(P) if (P) (dfP (i). dfp(u)) p2(p) lf(p)(dfp(w). d lp(r7,))

=

l f(p)(dfp(v), d fp(m)) VIf(p)(dfp(v), dfp(v)) i f(p)(dfp(W),

= cos0,.

dfp(m))

Thus, if ip(u, w) = p2(p)1 f(p)(dfp(u), djp(m)), the mapping is conformal.

If we have a conformal mapping f: S( -+ S2. since f is a diffeomorphism, a chart x : (U C K2) -+ Si determines a chart on S2, namely, v = fox : (U C KF2) -+ S2. Thus we can apply the same u, v coordinates on each surface. Furthermore, the local expression

for a curve on S,, a(t) = x(u(t), v(t)), carries over to S2 via the mapping f as f o a(t) _ y(u(t), v(t)). Thus du dv du dip dfp (XU +x`'dt) = ti"dt +}°dt' dt

The component functions of the metric at corresponding points, p = x(u. v) and f(p) _ v(u. v). may be written as E(u, v) and E(u. v), F(u. v) and P(u. v). and G(u, u) and G(u, v).


118

Now suppose that f is conformal, and that ti and w are orthogonal unit vectors in Tp(S1). Then Ip(u, w) = 0 implies that I f(p)(d fp(v). d fo(w)) = 0. Write V = d f'(5) and W = d fp(iu), and suppose that II V II = c, and IIW 11 = c2. By the linearity of the inner product we find 1

Ip(u,u+w)

_

_

lp(u, u) Ip(v + w, u + w)

Ip(V,V+W) Ip(V, V) Ip(V + W, V + W)

c2

This equation implies that c, = c2. We can now define a function p: S, -+ IR, given by P(P) = Cl. At p e S1 write x = ati + bib. Then aV + bW and

a2V V + 2abV W +b2W W

E=

= P2(p)(a2 +b2) = P2(p)E.

Similarly, P = p2(p)F and G = p2(p)G. Varying the orthonormal basis smoothly in a small neighborhood around p, we obtain this relationship between component functions for all points near p. We now fix a representation of S22, a coordinate chart, against which we can compare other projections. The representation of choice dates back to the second century 13.c. and is standard in geography. A point on S2 is determined by its longitude A, measured from the Greenwich meridian, and its latitude 0, measured from the equatorial plane. These coordinates are like spherical coordinates for S2 in R3 (see Chapter 1). but differ in the choice of reference plane for the latitude. In the language of the Chapter 8. we have the coordinate chart

x: (-jr, r) x (- 2 , 2)

52,

x(.k, t) = (cos ). cos 0, sin A cos 0, sin 0).

.r' applied to the surface of the Earth.

8bu. Map projections

119

The coordinate curves x(Ao, 0) and x(A, 4,) are called the meridians and parallels, respectively. This chart covers most of S2, leaving out the poles and the "International Date Line:" By rotating the chart, we could obtain other charts to cover S2 completely. For our purposes, this will not be necessary. We compute the first fundamental form directly from the expression for the chart: xt, = (-sin A cos 0. cos A cos 0, 0), x o = (- cos A sin m, - sin A sin m. cos 0).

Then E = xa xa = cost 0. F = xa x,, = 0, and G = x0 x0 = I. We can write the line element as ds2 = (cost 4,)dA2 + d42.

With this explicit metric, Proposition W'.3 implies

Corollary 8".4. If Y: (R C S2) -+ R2isamapprojectionandds2 = Edu2+2Fdudv+ Gdv2 is the line element associated to the composite coordinate patch

Y o x : (U C (->r. rr) x (-

2

2)) -

then Y is a conformal map projection if F = 0, E = p2(A, 0) cost 0, and 0 = p2(A, 0) for some nonzerofunction p on U. To derive a criterion for equiareal projections, we use the area formula in Theorem 8.16. The trick of applying the formula is to think of R2 C IR3 as the subset of points (u, v, 0). Then 1R2 is a surface and we can define different coordinate charts for it. For example, if Y is a map projection, then the composite

(-7r, ;r) x (-2, 2

___X

+ SZ

>RZ

gives a coordinate chart for a subset of JR2 C IR3. Suppose S is a bounded region in S2 (away from the poles and the International Date Line). Write S = x(T). Then the area of S is given by

EG-F2dAdO= rrcos0dAd4,.

area(S)=JJ

r

T

Let E, F, and G be the component functions for the metric associated to the chart Y o x on R2. Expressed in this composite chart the area of Y(S) is given by

area(Y(S)) _ (J .1

r

This leads to the criterion for being equiareal:

EG - F2 dAdm.


120

Proposition 8"`.5. A map projection is equiareal if the induced coordinate chart on R2 C R3 satisfies to - P2 = cost 0. More generally, a diffcomorphism of surfaces is equiareal if the induced metric satisfies El; - P2. Notice how these criteria are fashioned out of the basic stuff of the differential geometry of surfaces, the first fundamental form and the component functions of the metric associated to a coordinate chart. We now turn to the examples.

EG - F2

Stereographic projection This map projection, applied to star charts, was known to Hipparchus of Nicae (190125 n.c.) who is thought to have introduced the idea of longitude and latitude. The term "stereographic projection" was first used by F. D'Aiguillon (1566-1617). To describe the projection in coordinates, we center a sphere of radius I at (0, 0. I) so that the plane tangent to the south pole is the plane z = 0. A point P on this sphere satisfies P = (x, y, I + z) with x2 + y2 + z2 = 1. Let T = (0, 0, 2) denote the north pole. The ray T A meets the plane tangent to the south pole in a point Q = Y(P) = (u, v, 0). As vectors

in R3. P - T and Q - T are linearly dependent and so (P - T) x (Q - T) = 0, where x denotes the cross product.

This leads to the relations

(x, y, z -1) x (u, v. -2) = (-2y - v(z - 1), 2x + u(z - 1). xv - yu) = (0, 0.0). 2x

2Y

I-z , v = I-z . Introducing the longitude and latitude coordinates we have (x, y, z) = (cos x cos 0, sin ), cos 0. sin 0), and so the formula for stereographic projection

Thus u = is

Y o

x(J1,0)

_

( 2cosx cos0 2sinl cos0 I -sin0 ' I - sinm

Proposition 81.6. Stereographic projection is conformal.

0' J

8bit. Map projections

121

PROOF. We view R2 C R3 as a surface with a coordinate chart given by Y ox. The tangent vectors to the coordinate curves are

2sinA cosh 2cosA cosm

(Yo.r)a = (Y o x)m =

I -sinm

'

I -sin4,

,0

2 cos A 2 sin L' 0 I - sin 0' 1 - sin 0

The component functions of the metric are z

E(A,0)= (I - sin 0)2 The function p(A, 0) =

(1 -sin 0)2 is nonzero for

I - sin

2 (the north pole), and so, by

Corollary 8".4, Y is conformal. There are several other proofs of the proposition, but they do not stress the methods we are developing. Notice that the previous computations also show that stereographic projection is not equiareal. The metric coefficients associated to Y o x satisfy

EG

p2

=

16 cost tb

(1 -sin O)°

# cost 0.

More directly, the image of stereographic projection is all of R2; such a mapping simply cannot be equiareal. Stereographic projection comes up often in other parts of mathematics, especially complex analysis. Euler. Lagrange, and others introduced complex variables into the study of map projections where the new point of view proved useful. We next record a remarkable geometric property of stereographic projection that will be used in Chapter 15.

Proposition 8°".7. If the circle on S2 does not contain the north pole, its image under stereographic projection is a circle in the plane. If a circle on S2 passes through the north pole, then its image is a straight line in the plane. PROOF. Any circle on Sz is the intersection of a plane in R3 with the sphere. We consider three cases: (1) a circle passing through the north pole. (2) a circle that does not pass through the north pole and is not a great circle, and (3) a great circle that does not pass through the north pole. In the first case, the image of the circle under stereographic projection is the intersection of the plane giving the circle with the plane tangent to the south pole, that is, a line. In the second case, the planes tangent to the sphere at points on the circle C meet at a point S that is the vertex of a cone tangent to the sphere and containing C. The ray TS is not perpendicular to the z-axis and so it meets the plane z = 0 in a point M.

122


Suppose P is a point on C and Q = Y(P) is the image of P under stereographic projection. In the plane determined by T, Q, and M consider LT QM and LQPS. Extend ST' to meet the line tangent to the sphere through the north pole in the plane of T. Q, and M at a point R. Since the lines QM and P1 lie in parallel planes they are parallel. By Proposition 1.29 the alternate interior angles L RT Q and L MQT are congruent. Now T R - PR since the line segments are both tangent to the sphere. Thus L RT P = L RPT. Furthermore, LRPT = LSPQ being vertical angles. Suppose that P' is the point on T Q with LSP'T = LMQT. Then LET P'S is similar to ATQM. By the previous discussion, L SPP' has congruent base angles, so SP = SP'. Similarity of LET P'S and AT QM implies the following ratios

TS TM

SP'

SP

MQ

MQ-

TS . Now SP has the same length for all points P This can be rewritten as MQ = on the circle C, and so the length MQ is the same for all points on C, that is, the image of the circle under stereographic projection is a circle with center M determined by the cone point S.

In the case of a great circle not passing through the north pole the argument just given fails - there is no tangent cone and hence no cone point S. To remedy this consider the unique line through T that is perpendicular to the plane that determines the great circle. This line meets the plane z = 0 at a point M. Let P be on the great circle and Q = Y(P) and consider the intersection of the plane R (T, Q. M) determined by T, Q, and M with the sphere. Since TM is perpendicular to the plane in which the great circle lies, TM is parallel to the tangent plane at P and so parallel to the line I lying in the plane f> (T, Q, M) tangent to the sphere at P. The line I meets the line through T tangent to the sphere at the north pole and lying in fl (T, Q, M) at a point R. The pairs of alternate interior angles L P RT and LTMQ and L PT R and LT QM are congruent. Since both segments are tangent to a circle, PR - T R, and so L PT R = L T P R. Angle sums now imply that L T QM = L QT M and

so QM - TM. Thus the image of a great circle is a circle of radius congruent to TM.

8bps. Map projections

123

Central projection Stereographic projection takes most great circles to circles in the plane. For navigation it would be more convenient to represent great-circle routes as straight lines on a map. This is the attractive and unique feature of central projection. To construct the projection, fix the tangent plane at a point R on the sphere and join a point P in the adjacent hemisphere to the center of the sphere. Extend this segment to the tangent plane and this is the image of the point P. Central projection is defined on the open hemisphere with R as center. The analytic expression for this map projection is easily derived when R is taken to be the south pole:

Y(X,¢) = (-cosxcot0, -sinAcot0. -1). The main property, great-circle segments map to line segments in the plane, will play an important role in the discovery of a model of non-Euclidean geometry (Chapter 15). Proposition 8°1t.8. Under central projection great-circle segments map to line segments. PROOF. A great circle is determined by the intersection of the sphere and a plane through the center of the sphere. The image of the great circle under central projection is the intersection of this plane with the tangent plane, and so determines a line.

In both examples discussed so far, we have projected the sphere to a tangent plane. There are other ways to obtain a map projection to the plane; in particular, the sphere can be projected onto another surface that maps to the plane without distortion. For example, take a rectangle and roll it up into a cylinder that fits around the sphere. Projections onto the cylinder can be unrolled to give a planar map - such projections are called cylindrical. An appropriate piece of the plane can also be fashioned into a cone that may be placed on the sphere as the target for projection - such projections are called conical (see the exercises for examples). We next consider some examples of cylindrical projections.


124

Mercator projection This map projection is the best known. Its appearance in 1569 marks the beginning of a new era of cartography. Gerardus Kriimer (Mercator) (1512-94), a Flemish cartographer, flourished in the Age of Discovery when new demands were made of map makers for a world of commerce that was growing to include the entire globe. To understand the problem solved by the Mercator projection, we introduce the following idea.

Definition 8m.9. A curve on the sphere S2 is called a loxodrome if it maintains a constant angle with respect to the family of meridians. Since the meridians determine the North-South directions, the course set by a navigator on a constant compass heading follows a loxodrome. In general a loxodrome is not the shortest path between two points on the sphere, but it is simplest to navigate. Extending most loxodromes determines a spiral path to one of the poles.

The problem solved by Mercator is to find a map projection such that: (I) Meridians and parallels project to orthogonal lines. (2) Loxodromes project to line segments. To construct the Mercator projection, we send parallels to themselves, that is, the composite

Yox:(-7r.n) x (-Z. Z) -+ S2 -+ R2 C R3 satisfies Y o x(A, 0) = (A, v(A, 0), 0). The condition v(A, m) = v(4,) guarantees the rest of the first desired A loxodrome.

property. For the second, we characterize loxodromes.

Proposition 8".10. A unit-speed curve a: (-e, e) - S2 is a loxodrome if and only if Cl a(s) = x(A(s), 0(s)), where A'(s) = and 4,'(s) = c2 for some constants c1, c2. cos 0 (s)

PROOF. Suppose a : (-r, c) - S2 is a unit-speed curve and a(s) = x(A(s), 4,(s)) in the longitude-latitude coordinate chart. Then we can write a'(s) = A'(s)xA + O'(s)xm. The condition that the angle made by a with all meridians be constant is equivalent to

a'(s) xx Ila'(s)II Ilxxll =

ci, a constant,

a'(s) x0

= c2, a constant.

Ila'(s)II II-Toll -

Now a' (s) xx = A' (s )xx xx = A' (s) cost 0 and II xx II = cos 0, so we get A' (s) =

Cl

cos 0(s) Also a'(s) xm = O'(s)xm x0 = 4,'(s) and IIx#lI = I, so 4,'(s) = c2. Since meridians and parallels are perpendicular, the constants are the cosines of complementary angles so

ci+c2=l.

8bu. Map projections

125

Suppose a map projection Y satisfies conditions (I) and (2). Then, if (r (s) = x (I (s), m (s))

is a unit-speed loxodrome. (Y o a)(s) = (x(s), v(v(s)).0) parametrizes a line in R2. We can write AX(s) + Bv(O(s)) + C = 0 for some A, B, and C. This equation leads to the condition v'(O(s))O'(s) = a, a constant. x'(s)

By the proof of the previous proposition. c2d v/d4

ci/cos0(s)

=a or

dv

do

=

b

cos0

Integrate this differential equation by separating variables. We are free to choose the unknown constant b. However, the following observation leads to an auspicious choice.

Proposition 8.11. A map projection Y o x(,1, 0) = (A, v(0), 0), satisfying dv/do _ b/cos 0, is confotmal if and only if b = ± I. PROOF. Since Y o x(1A, m) = (A, v(O), 0) we have

(Y o x)a = (1, 0, 0), and (Y o x)o = (0, d v/do, 0).

Thus the components of the metric induced by Y o x are given by E = 1, F = 0, and 0 = (d v/do)2 = b2/cos2 0. If cost O = p2 E, we have p2 = cost 0. Then I = p2 c = P.

andb=fl.

We finally arrive at the Mercator projection by integrating dv/do = I/cos0 with the initial condition v(0) = 0: Y(x, 0) = (A. In tan

(4

+Z

I ,

OI

.

Mercator's achievement is considerable; the calculus had not yet been developed in 1569.

Mercator projection is one of the family of cylindrical projections of the sphere. Consider the sphere S2 inside an infinite cylinder tangent to the equator. Central projection of the sphere to this cylinder is not conformal, but adjusting it to "make the cylinder infinitesimally close" leads to the Mercator projection. Arguing with infinitesimal triangles, as Mercator might have, one arrives at the same differential equation

dv/do = I/cos0. Mercator was one of the first compilers of logarithm tables and so the infinitesimal corrections needed to discover his projection were at hand.

126


Lambert's cylindrical projection All of the examples presented so far have not been equiareal.

Furthermore, they have all been constructed via projection from a point. In the next classical example, due to J. H. Lambert,

we consider an orthographic projection, that is, a mapping for which all lines of projection are orthogonal to the plane of projection. Instead of a plane we substitute a cylinder of height two, exactly fitting around the sphere. The lines of projection are required to be perpendicular to the cylinder at each point. From this description, the Lambert projection is expressed by Y o x(x, 0) = (>`, sin 0, 0). Rolling out the cylinder on the plane we get the image of the Lambert projection, the open

rectangle (-n, n) x (-1, 1). Proposition 81.12. The map projection Y o x(x, m) = (x, sin 0, 0) is equiareal. PROOF. Following our criterion for equiareal maps, we compute the component functions of the metric associated to Y o x as a surface chart. (Y o x)A = (1, 0, 0).

(Y o x)o = (0, cos 0, 0).

Thus E = 1, P = 0, and 0 = cost m. It follows immediately that to - P2 = cost 0, so, by Proposition 8".5, the map projection is equiareal. Notice that this projection has some other nice properties: It takes meridians to vertical lines and parallels to horizontal lines. In fact, it is unique as an equiareal projection with

Y o x(x, 0) = (x, u(0), 0). The possibility of such a cylindrical equiareal projection may have been known to Archimedes, who had computed the area of a sphere and the area of a cylinder of the same height. Lambert's work appeared in his Beytrage zum Gebrauche der Mathematik and deren Anwendung (1772) in which Chapter VI is dedicated to finding the general form of a conformal map projection. This is one of the first mathematical treatises on the subject of cartography. Lambert gives several new projections which bear his name and are in use today.

Azimuthal projection Map projections are a particular use of change of coordinates. From our experience of plane geometry we can try to set up polar coordinates on the sphere. This leads to the so-called azimuthal projection: Fix a point P on the sphere and a point Q in the plane. Send points on a circle of radius p centered at P on the sphere to the points on a circle of radius p centered at Q in the plane. To obtain an analytic expression, we choose a convenient point as P. the north pole, and one as Q, the origin in the plane. This allows us to map the longitude as the angle for polar

Pis. Map projections

127

coordinates in the plane. Distance on the sphere between the north pole and an arbitrary point (A, ¢) is measured along a meridian and is given by (n/2) -0. We now simply change the coordinates on the sphere from (A, 0) to (A, p), where p = n/2 - ¢. The change of coordinates that follows is encoded by the line element:

ds2 = cost 0 dA2 + dO2 = sin2(,r/2 - O)dA2 + (-d(,r/2 - 0))2 = sine p dAZ + dp2. This expression tells us that along meridians, that is, along a curve with fixed A, ds = dp and so distances are preserved. If we rotate the sphere to center the projection at an important

point, then the distances from the center of the coordinates to other points on the globe are compared easily by following the new set of meridians. Notice that in these rotated coordinates, the line element remains ds2 = sin2 p dA2 + dp2, the polar line element on s2. Many other map projections exist, suited to the specific needs of map makers. The mathematical problem of characterizing all conformal and equiareal map projections was solved by Lagrange in his 1779 paper "Sur la construction des Cartes gEographiques" (Oeuvres, volume 4, pp. 637-92). We have omitted a discussion of some later developments

of mathematical cartography which include the work of N. A. Tissot (1824-1904), who gave a uniform theory of distortion for map projections (see Robinson (1960)). In later chapters we will consider the more general problem of developing surfaces, that is, finding isometrics between surfaces, mappings that preserve lengths, angles, and areas.

Sample map projections The following maps are examples of the projections discussed in the chapter. We restrict our attention to Australia and its neighbor New Zealand. Since these countries are in the Southern Hemisphere, projections such as stereographic or central projection do not produce completely unrecognizable maps. The grid lines are 10° apart. The maps were produced using Mathematica and John Novak's package called WorldPlot.

Stereographk projection.

128


Mercator projection.

Lambert cylindrical projection.

8bit. Map projections

129

r "

Azimuthal projection with center the South Pole.

Exercises 8#",.1

Complete the set of charts based on longitude-latitude coordinates to cover all of S2. Do the same with the coordinates determined by the inverse of stereographic projection.

8'.2'

Suppose f: S, -+ S2 is a diffeomorphism between surfaces that is both conformal and equiareal. Show that f is an isometry, that is, there are coordinate patches around corresponding points on the surfaces with ds, = ds2.

S'63

Apply the criterion forconformality to Lambert's cylindrical projection anddetermine the regions of maximal failure of this property.

8".4

From a satellite, the projection of Earth onto a plane of a piece of film in a camera is orthographic. This is the essentially the map projection given by

(x, y, v/1 - x2 - y2) - (x, y, 0). Determine a complete set of charts for S2 of this sort, and determine if this map projection is equiareal or conformal.

g"'.5

Project a loxodrome on the sphere orthographically to the plane tangent to one of the poles. Show that it determines a logarithmic spiral in the plane.

8"',6

Find the length of a loxodrome on a sphere that starts at the equator at an angle 6 with the meridian and ends up by winding around the pole. Contrast this with the length of the great-circle route.

8'.7'

The plane and the cylinder were used in map projections because there is an ideal map projection between them. Another so-called developable surface (to the plane) is the cone. Give coordinates to a cone as a surface of revolution, and give coordinate charts for it. Show that the coordinate charts you have chosen provide an ideal map projection of the cone to the plane.

130

Development: Differential geometn

8"'.8

In the previous exercise you showed that the cone was developable on the plane. Thus one could fashion a map projection by mapping the sphere to a cone and then to the plane. Choose a parallel 4$o in the northern hemisphere and construct the cone tangent to the sphere at this latitude. Project to the cone from the center of the sphere. Give an analytic expression for the resulting map projection and determine if it is conformal or equiareal as a function of the latitude.

8/e.9

Consider the following mapping to the cone tangent to the parallel at 00: Send a point x(x, 4') to the point on the cone with angle ). for its first spherical coordinate

and j(4,) = f 2(C - sin 0), where j(4,) determines the distance of the mapped point from the vertex of the cone, and k = cos mo and C = 12kk of the previous exercise, show that this projection is equiareal.

.

Using the results

9 Curvature for surfaces Investigations, in which the directions of various straight lines in space are to be considered, attain a high degree of clearness and simplicity if we employ, as an auxiliary, a sphere of unit radius described about an arbitrary center, and suppose the different points of the sphere to represent the directions of straight lines parallel to the radii ending at these points. C. F. Gauss (8 October 1827)

In Chapter 8, surfaces in R3 were introduced and some of their basic structure, such as the first fundamental form, was defined. We now define analogs of curvature and torsion for surfaces in R3. The goal is to describe how a surface "curves" at a point. The first measure we introduce is naive - it will depend on curves in the surface. Later in the chapter, we associate a more appropriate, two-dimensional measure to the same task. Based on these ideas, the generalization for surfaces of the fundamental theorems for curves is realized in Chapter 10. Following the historical path of the subject, we begin with Euler's work.

Euler's work on surfaces Suppose p is a point in a surface S and N(p) is a choice of unit normal. Let i be a unit tangent vector in Ta(S). In IR3 take the right-handed frame [u, N(p)] to define a plane. Translate the plane so that the origin is at p and consider the intersection of this plane with S. This gives a curve on S near p.

Suppose co(s) is a unit-speed parametrization of the curve so that We define the normal curvature of S in the u direction at p, k (v), to be the directed curvature K (s) 131


132

of cr(s) at s = 0 as a curve in the plane determined by I. N(p)]. At a fixed point p, k"(0) is a function of the directions in the tangent plane. In a classic paper Recherches sur la courbure des surfaces (Omnia Opera, volume 28, pp. 1-22) Euler published the following useful theorem.

Theorem 9.1 (Euler 1760). If the normal curvature k,, (u) is not a constant function of v, then there are precisely two unit tangent vectors X) and X2 such that k (XI ) = k) is maxima! and k. (X2) = k2 is minimal. Furthermore, X1 is perpendicular to X2. PROOF. Since is continuous in v', and v lies on the unit circle in Ta(S), which is compact, the directions X1 and X2 exist. We prove that if X( is the direction in which the maximal normal curvature occurs and v makes an angle 0 with X), then

k(cos2 0 + k2 sin2 0. Choose coordinates for 123 so that p = (0, 0, 0) in R; and that Tn(S) is the x y-plane. By the implicit function theorem (Chapter 8). we can take a neighborhood of p in S to be the graph of a function, ((x, y, z) I z = f (x, y)}. A particular choice of a coordinate chart for S near p is

x(u, v) = (u, v, f(u, v)) which we fix up so that x

o) = (1.0.0), x I(0 0) = (0. 1, 0), that is,

j(0,0) = 0,

a/ I(0.0)

af I

= 0, and

ax

Another trick is needed - we rotate the xy-plane until by

aY (o.o)

aÌf

axay

= 0.

(*)

= 0. Rotations are given

I

(0.0)

po(x, y) = (x cos 0 - y sin 0. x sin 0 + y cos 0). and they do not alter the conditions fixed in (*). Let fe = f o pg. Then

afe

av

= - sin 0

cos B of , and

ay

a2le axay

-sin0cos0a-jj+(cosz0-sin20) ax ay

8x2

a2 f

(sin 29 (a2 j =cos2B-ay +2 \ ax / / \ ay To obtain axay = 0 when axe

a2 fI .

ax2

, solve for 0 in the equation a

tan 20 =

+sinBcosBa?j

2azj axay

ja2f

axe - are

(0.0)

ay2

of ax

+

9. Curvaturefor surfaces

133

Let 0 = r/4, if a2 f/8x2 = a2 f/a y2. Now rotate the xy-plane through 0 as chosen and set f to be fo as constructed. When the surface S intersects the xz-plane, the resulting curve is parametrized (not necessarily unit speed) by c(t) = (t, f(t, 0)). Since x is in the t-direction at the origin, x' = I and y' = of/ax = 0. The directed curvature is given by

k

[(x')2'+ (y') )3t 2

axe

(0, 0).

Similarly, the intersection of S with yz-plane is the curvee(t) = (t, f (0, t)) with directed 2

curvature ay (0, 0) = k2. If v is a unit-length vector in some direction in the xy-plane, then v = (cos0, sin 4,) and the curve defined by the intersection of the plane [u, e31 with S can be parametrized by c;,(t) = (t. f (t cos 0, t sin 4,)). The directed curvature is given by

k (v) = dt f2 (t cos 0, t sin 0)

= cost 4 axf (0, 0) + sine 4 11=0

2

a-f (0, 0) Y

= (cost m)ki + (sin2 O)k2.

As a function of 0, 0 < m < 2tr,

achieves its minimal value at k2 and its maximal

value at ki. This proves the theorem. The values k i and k2 are called the principal curvatures of S at p and the directions X1, X2 the principal directions. Together the principal curvatures give a qualitative picture of the surface near a point. From the previous proof we can write the equation of the surface S near p as a graph z = f (x, y) with R(x, z = kix2 + k2y2 + R(x, y), where lim x2 + y2 y) = 0.

Thus, near p, the surface is approximated, up to order 2, by the quadratic surface z = kjx2 + k2y2. We next consider what the signs of the principal curvatures tell us.

Case 1. ki, k2 > 0 or ki, k2 < 0 at p. Such a point is called elliptic and the quadratic surface approximating S at p is a paraboloid. The level sets given by a constant z value are ellipses. From the defining equation for S it follows that for x and y sufficiently small, z = f(x, y) is nonzero and always positive or always negative. Therefore, near p, S lies on one side of the tangent plane.

134

Development: Differential geometnv

Case 2. ki > 0, k2 < 0. Such a point is called hyperbolic and the quadratic surface approximating Sat p is a hyperbolic paraboloid. The level sets given by a constant z value are hyperbolas. In this case, near p, there are always points of S on either side of the tangent plane.

Case 3. ki > 0, k2 = 0 or ki = 0, k2 < 0. Such a point is called parabolic and the quadratic surface approximating Sat p is a parabolic cylinder, z = kix2 or z = k2y2. Near p, all points lie on the same side of the tangent plane.

Case 4. k1 = k2 = 0. Such a point is called planar. Examples such as f(x, y) = x3 or f (X, y) = ±xa show that the behavior of S with respect to the tangent plane near p may vary.

When k1 = k2 i4 0, then every direction is a principal direction. Such a point is called an umbilic point and we may take any pair of orthogonal directions at such a point as principal directions. For example, all points on a sphere are umbilic. When we have developed further tools, we will prove this result in a manner that does not use a special choice of coordinates. We now jump forward 68 years to the pioneering work of Gauss on surfaces.

The Gauss map For curves, a measure of curvature is given by changes in the normal vector. This notion was

generalized to surfaces by Carl Friedrich Gauss (1777-1855) in his seminal 1827 paper, Disquisitiones generales circa superficies curvas (Werke, volume 4, pp. 217-58). Gauss took his inspiration from astronomy and the idea of the "celestial sphere" Like many other contributions of Gauss, Disquisitiones had a profound effect on subsequent work on this subject. The mapping, now called the Gauss map, was introduced earlier by 0. Rodrigues (1794-1851) in two papers that appeared in 1815.

Definition 9.2. Let S be an orientable surface. Define the Gauss map to be the mapping N: S -. S2, the unit sphere in R3 centered at the origin, given by associating to each point in S the unit normal at the point. Locally we have the formula

N(p)_ xuXxt, Ilxu X XV11

9. Curvature for surfaces

135

Notice that this expression depends crucially on the choice of coordinates. For example, switching u and v changes the normal N(p) to -N(p). In what follows we need to pay attention that results depending on the Gauss map in fact do not depend on the choice of coordinates.

For a regular surface, we leave it to the reader to show that N is differentiable and so it has a differential map dNp: Tp(S) - T,u(p)(S2). What makes the Gauss map useful is the fact that the tangent plane to S at p is the plane normal to N(p) and it is a property of S2 that Tq(S2) is the plane normal to q. Thus Tp(S) = TN(p)(S2), and dNp can be taken to be a linear mapping dNp: Tp(S) -e Tp(S). To compute dNp we may use curves on S through p. Consider a tangent vector at p as

the tangent to a curve, that is, if v is in Tp(S), then there is a curve a: (-e, e) - S with a'(0) = ti. It follows that dNp(u') = (N o a)'(0). This is the rate of change of the normal vector at p when restricted to a(t). EXAMPLES.

(I) Let P be the plane ax + by + cz + d = 0. The unit normal is N =

(a, b, c)

a +b +c

a constant vector, and so d N = 0.

(2) Consider the unit sphere S2 in R3. For a curve a(t) on the sphere we can write a(t) =

(x(t), y(t),z(t)),where (x(t))2+(y(t))2+(z(t))2 = I. It follows that 2xx'+2vv'+2zz' = 0 or a'(t) 1 a(t). Thus N(x, y, z) = (x. y, z) is a choice for the unit normal vector to S2. It follows immediately that dNp = Id.

(3) Let us now consider a saddle-type surface, for example, the graph of x2 - y2. Through (0, 0, 0) the surface includes two parabolas with different orientations, hence the saddle. Let x(u, v) = (u, v, u2 - v2) be a coordinate chart. Then

x = (1, 0, 2u),

x,, = (0. 1, -2v).

136


It follows that

N(p) = N(x(uo, vo)) =

(-2uo, 2vo. 1)

4u0+4vO2+1

To obtain an expression for d Np at p = (0, 0, 0), let a: (-e, e) -* S be a unit-speed curve with a(0) = (0, 0, 0);

a(l) = (u(t), VOL u2(t) - v2(t)), so N o a(t) _

(-2u(t), 2v(t), 1)

4u2(t)+4v2(t)+ I

Finally, evaluate the derivative at t = 0 to get

(N o a)'(0) = (-2u'(0), 2v'(0). 0). Now a'(0) = (u'(0), v'(0), 0) and we have that dN(o.o.o)(v) = dN(o.o.o)(u'(0), v'(0). 0) = (-2u'(0). 2v'(0).0). If we write dNp in matrix form with respect to the basis x,,, xv, we get

-2 2)

dN(o.o,o) = C 0

2

.

The mapping d Np has a remarkable property with respect to the first fundamental form. To describe this property we introduce the following idea from linear algebra.

Definition 9.3. Let V be a real vector .space with an inner product ( , ) : V x V -, R. If A: V -* V is a linear mapping, then A is said to be self-adjoint if, for all v. W E V,

(A(v), w) = (v, A(w)). Proposition 9.4. The differential dNp: Tp(S) -+ Tp(S) is self-adjoint with respect to the first fundamental form. PROOF.

Because Ip is an inner product, it is linear in each variable and so it suffices to lp(xu, dNp(x,,)). To

check the proposition for the basis vectors, that is, we compute determine d Np

(N o x(u, vo))'(uo) =

IN J au

= N,

(N o x(uo. v))'(vo) =

(Uo.un)

INJ av

= N,,, (U0.vo)

thus we need to check that l p(N,,, x,,) = Ip(xu, N,.). We know that l p(N, lp(x,,, N). Taking partial derivatives, we obtain Ip(N0,

Ip(N, x,,,,) = 0 = l p(N,,.

Ip(N. x,,,,).

0=


By the differentiability of the coordinate patch,

137

x,,,,. Thus,

Ip(x.,dNp(xo)) What about self-adjoint linear mappings? We record certain useful properties of these mappings. Fix a two-dimensional vector space V with inner product ( , ).

Property 1. Suppose A: V - V is a self-adjoint operator with respect to

Then A

has two real, not necessarily distinct, eigenvalues.

PROOF. Apply the defining equation (A(v), w) = (u, A(w)) to the elements of any basis for V and the resulting matrix representation for A is given by A- _

(a b

dl. b

To compute the eigenvalues of A we solve for the real roots of the characteristic polynomial det(t I - A) = t2 - (a +d )t + (ad - b2). When we substitute these values into the quadratic formula, we get two real roots, and so A has two real eigenvalues.

Property 2. Suppose A: V - V is a self-adjoint operator. Suppose A has eigenvectors el, e2 of unit length. If the eigenvalues for A are not equal, then et is perpendicular to e2. PROOF. As eigenvectors we have Aei = .Lies and Ae2 = A2e2 for some real numbers A1, A2. Now

At (el, e2) = (Atei, e2) = (Aei. e2) = (et, Ae2) = (ei, A2e2) = A2(ei, e2)-

If), l 0 12, then we must have (el, e2) = 0, that is, el is perpendicular to e2.

Property 3. Suppose A : V -o- V is a self-adjoint operator with eigenvalues A i > A2 and associated eigenvectors et and e2. Take ei and e2 to be of unit length. Suppose v E V is also of unit length. We may write v = cos(9)et + sin(9)e2, and (A(u), v) = AI cos2(9) + A2 sin2(9). Furthermore, the expression (A(v), u) achieves its maximum Ai when 9 = 0 or it and its minimum A2 when 9 = ±,r/2. PROOF. Since et -L e2, lei, e2) is an orthonormal basis for V. Now (A M, v) = (A(cos(9)ei + sin(9)e2),cos(9)ei + sin(9)e2) = (cos(9)Aiei + sin(9)A2e2,cos(9)ei + sin(9)e2)

= Al cos2(9) + 12 sin 2(9). The rest of the statement follows as in the proof of Theorem 9.1.

This linear algebra produces results that resemble the formula of Euler discussed earlier. We can apply these properties to give a second derivation of Euler's result once we have


138

the appropriate setting for the Gauss map. We now introduce the function that plays for surfaces the role that curvature and torsion play for curves. Recall that a quadratic form

on a real vector space V is a function B: V - R that satisfies 8(r v) = r2 B(V) for all V E V and r E R and such that the associated mapping bB: V x V - R given by 1

bB(v, w) = 2(B(v + w) - B(v) - B(w)) is an inner product on V.

Definition 9.5. At a point p in a surface S the second fundamental form on S at p is the quadratic form llp : Tp(S) - R defined by llp(v) = - Ip(dNp(v), v). There is a geometric interpretation of Ilp(u) when v' is a unit tangent vector. The pair [v, N(p)1 is an orthonormal basis for a plane in R3. The intersection of this plane with taken to be unit speed with u. Since S is a curve c;,(s) on S near p = ca(s) is also a plane curve, we can talk about its directed curvature K(s), which is given by

c"(s) = K(s)N(s). By the definition of the normal curvature in the direction v, we have

c;;(0) = k,,(v)N(p) Theorem 9.6. For 'v E Tp(S) with Ilvll = 1. Ilp(i)) = PROOF. Since lcrt,,)(N(c6(s)), c,(s)) = 0, by taking derivatives we have

lcr(s)((N o ct)'(s). c'6 (s)) + lc;.ts)(N(c-v(s)), c "(s)) = 0, or C.

Ip(dNp(v),v)+Ip(N(p),c'"(0))=0, and so -1 p(dNp(v), v') = Ip(N(p), K(0)N(p)) = Since the second fundamental form is defined with the self-adjoint operator dNp, the properties of such operators imply properties of Iip, In particular, Property 3 together with Theorem 9.6 give another proof of Theorem 9.1. There are many other planar curves through p on S with vas tangent vector. The following generalization of a theorem of Euler (Theorem 9.1) is due to Jean-Baptiste Meusnier (1754-

93), a student of the leading French geometer of the time. Gaspard Monge. Meusnier published only one paper in mathematics and went on to publish important papers in other scientific fields. He died a hero's death in the siege of Mayence.

Theorem 9.7 (Meusnier 1776). If w E R3 is any unit vector that is perpendicular to v E Tp(S) and w makes an angle f with N(p), 0 < rq < ,r/2, then the unit-speed curve c,;, (s) through p with S has directed curvature

v given by intersecting the plane determined by [v, iu1 with Kp, _

IIp(i ) cos 0


139

PROOF. Since v = cw(0) is perpendicular to N(p), 0 = lp(cw(0), N(p)). More generally. we can write lcr,(s)(cW (s), N(cr.(s))) = 0. Taking the derivative of this expression we find 0

ds loj,(5)(c'W(s), N(cw(s)))

= lci.(s)(cl (s), N(c,;,(s))) +

ds N(c,;,(s))).

However, the expression

dsN(c,i,(s))I

T=0

so the previous equation implies

lp(cu,(0). N(p))

Ip(u, dNp(u)) = Ilp(ii).

By construction. C@' (0) = KW w. Furthermore,

cos fi = (tZ. N(p)). Putting these facts together we get

Ilp(i) = Ip(cW(0), N(p)) = lp(K,,,w, N(p)) = K@ COS

We now apply more linear algebra to the study of the Gauss map. Write d Np as a matrix in the orthonormal basis (Xi, X21 given by the principal directions at p:

dNp =

(

k)

0

0

-k2

Definition 9.8. TheGaussian curvature of S at p is defined by K(p) = det(dNp) = kik2.

Observe that K(p) > 0 atelliptic points (convexo-convex points in Gauss's terminology),

K(p) < 0 at hyperbolic points (concavo-convex points), and K(p) = 0 at parabolic and planar points. The nature of the approximating quadratic surface is determined by the sign of the Gaussian curvature. Another important observation is that the Gaussian curvature does not depend on the choice of coordinates around p; any change of coordinates leads to a change of basis for the tangent plane at p, but K(p) remains the same because it is a determinant.


140

Our first step in developing Gaussian curvature is to get a local (hence calculable) expres-

sion for it. Let P E S be in a patch x: (U C R2) -> S. Then N(p) = (xu x xv)/Ilxr, x xV and if we let w = a'(0) E Tp(S), for some a: (-e, e) -+ S, then

t'u = a'(0) = dt (x(u(t), u(t)))

II

= u'(0)xu + v'(0)xv, r=0

dN.(iu) = dt N ox(u(t). v(t))11=0

u'(0)Nu + v'(0)N,..

Since N and N lie in Ta(S)

Nu=allxu+a2lxv, Nv = al2xu + a22xv, and so we can write d Np: Tp(S) -. Tp(S) in matrix form in terms of the basis x,,, x, as

al

alz)

(u'(0)

a22/

V(0)

We now express the aijs through the second fundamental form. For convenience we write Ip(w, v) = (w, v'). IIp(w)

-(dNp(w). w) -(N.. xr,)(u')2 - ((Ne, xv) + (Nu, x,.))u v' - (Nv. xv)(v')2

Define the following functions on U: e

f g

(N., xu)

(N, xuu).

(Nv, xu)

(N. xuv) = (N, xvu)

(Nv, xv)

(N. xvv)

(N., x,,),

We can now write Ilp(w) = e(u')2 + 2 fu'v' + g(v')2. In terms of the functions (ark ) we have

11P60 = -

all 021

a 12

11'

a22v,/ \xvVl

\xv

- [(a11u'+al2v)(u')(x,,.xu) + ((a2l u' +a22v')(u') + (a1I u' +a12V')(V'))(XU, xv) + (a21 u' + a22

(xv, xv) ]

-[(u')2(aIIE+a21F)+(u'v')(a12E+a22F+aIIF+a21G) + (u')2(a12 F + a22G)]. Comparing this with the new functions e, f, and g we get the expressions

alIE+a21F= -e and a12F + a22G = -g.


and since N. = a 11 xu + a21 xv and N = a 12x. + a22xv we have

at1F+a21G=-f= a12E+a22F. All of this is expressed compactly in the matrix equation

re f1 =( a :2 fg a

a21

a22

_

-1 EG - F2

atz

FG

a22

from which it follows that

all

E F

a21

,

e

(f

f

g

G

-F

F E

Theorem 9.9. The following equations hold:

_ all

fF-eG EG-F2

_ gF- fG 0212

EG-F2'

_ eF- fE _ EG- 'azz

a21

Corollary 9.10. K(p) = det d Np = det(aj j) =

fF-gE EG-F2.

",g

EG - fFz2'

As an example, we calculate the Gaussian curvature for a surface of revolution (Example (4) in Chapter 8). Let

c(t) = (x(t). µ(t)): (-r, r) -+ R2 be a regular curve that does not cross itself or the taxis. A coordinate chart for the surface of revolution obtained by revolving the curve c(t) around the taxis is given by

x(u, v) = (X(u) cos v, a(u) sin v, µ(u)),

for-r
xv = (-A(u) sin v, x(u) cos v, 0).

It follows then that E = (x')2 + (µ')2, F = 0, and G = x2. The normal determined by

EG -

the coordinate chart has length Ilxu x xvll =

=

.l

auxiliary functions are given by

e = (N. x,,)

xu x xv

Ilxu x xvll'

')2 + (µ')z (A')2 +

.

(l')2

xuu)

(xp x xv, xuu)

(A')2 + (A17). The other


142

p

By similar calculations we obtain f = 0 and g =

In the basis {x,,,

(p')2 dNx(,,,,,) takes the form

X"pr - A'p" d Nx(u.,,) _

0

((A')2 + (l2')2)3I2

-p

0

(A')2 + (p')2

x

In this form it is evident that x and x are the principal directions with principal curvatures

zp-zp

-p

"

and This implies that meridians and parallels on a ((m')2 + l (x')2 + (p')2. surface of revolution are the curves of least and greatest normal curvature. The Gaussian curvature is given by the determinant of dNp:

K(p)

,L"(p')2

- A((x')2 + (g1)2)2-*

When (x')2 + (p')2 = I, it follows that x'x" = -p'p" and so we get the simpler formula

For example, suppose that c(t) = (cost, sin t). The surface of revolution is S2. The formula gives another proof that K(p) = I for all p E S2. A surface that will be of interest in later chapters is the surface of revolution of the tractrix (see Chapter 6). A parametrization for this curve is given by

6- (t) = (sin(s), In tan(t/2) + cos(t)). By the discussion so far we find

K(P) -

A'p'p" - x"(p')2 x((A')2 + (p')2)2 - COS u

-

cost u cos u(sin2 u + 1) sin u

+ sin u

sing a

( sin u t cost u +

cos4

U

sing U 2

COS4 u

sing u

= -1

-

/

Thus revolving the tractrix leads to a surface of constant Gaussian curvature, like the sphere, but negative everywhere. Besides the interpretation of Gaussian curvature in terms of principal curvatures, there is another geometric interpretation, in fact, the one that guided Gauss.


143

Proposition 9.11. Let p be a point in a surface S with K(p) # 0. Then

K(p) = lim

area N(A)

A- p

area A

where A is a connected region around p and the limit is taken through a sequence of such regions that converge to p. PROOF. We first record another appearance of Gaussian curvature in the formula

To see this we write N. = a) ix + a2)xv and N = a)2xu +a22Xv and apply the linearity of the cross product: N. x N = (a))x + a2IXv) x (a(2xu + a22xv) = (at)a22 - a(2a2I)xu x xt, = K(xu x xv)

In a small enough region to be contained in a coordinate patch, the area of a region A around p is given by the formula

fr area(A) =

JJX

Ilxu x (A)

Mapping the region to S2 by the Gauss map we get

area(N(A)) = jf.

x '(A)

x '(A)

IINk x Nvlldudv

Kllxu x xvlldudu.

Here we should write IKI Ilxu x xvlldudv. For pan elliptic point this makes no difference. At a hyperbolic point, the Gauss map reverses orientation and so, to obtain the area of N(A),

we must multiply by -1 giving -IKI = K. We can now apply the mean value theorem for double integrals (see Lang (1987)) which yields

ffx-'(A) Kllxu x xuIldudv

p f f - i (A) Ilxu x xvlldudu t

A

x

(I/area(A)) ffx-'(A) Kllxu x xvlldudu lim areaA-.0 (I/area(A))ffx_'(A) Ilxu x xvlldudu lim

area A-.O

Kllxu x xv11 I llxu X X, 11

= K(p). some point in A

Another sense of how the surface curves is caught up in this ratio of areas. For example

on a sphere of radius r this formula readily yields K(p) = l/r2, and once again we see that the bigger the sphere, the smaller its curvature and the better it approximates the plane.


144

Exercises 9.1

Show that, for a regular surface, the Gauss map is differentiable.

9.2'

Determine a formula for the Gaussian curvature K for a surface given by the graph of a smooth function z = f (x. y) in terms of the function f.

9.3

Suppose that S is a surface in 1R3 and suppose that the principal curvatures ki and k2 satisfy Iki l < 1 and Jk21 < 1 at every point in S. If a(t) is a regular curve in S, does it follow that IK(t)I < I for every t in the domain of a?

9.4

Determine the principal curvatures of a cylinder, an ellipsoid, and a saddle.

9.5

Define the Dupin indicatrix Do of a surface Sat a point p to be the subset of T, (S).

Dp=(vETn(S)IIlp(u)=Ior-1). Prove that if K(p) > 0. Dp is an ellipse, if K(p) < 0, Dp is a pair of conjugate hyperbolas, and if K(p) = 0. then Dp is a pair of parallel lines, or is empty. 9.6

A surface is said to be convex if it lies on one side of the tangent plane at some point of the surface. Interpret the property of convexity in terms of the principal curvatures and Gaussian curvature.

9.7

Define the mean curvature of a surface to be H(p)

2 trace(dNp) = 2 (ki + k2).

Prove that H2 > K. When does equality hold? 9.8

We say that a surface is minimal if H - 0 on the surface. Suppose on a coordinate chart

x(u, v) = (f,(u. v), f2(u, v), f3(u. v)). that ds2 = Odu2 + 4dv2, for some function m. Show that the image of this coordinate chart is minimal if and only if the functions j are harmonic, that is,

a2f a2j au2 + av2 = 9.9

0.

Let M be the surface determined by the coordinate chart x(u, v) = (a cosh v cos u, a cosh v sin u, av)

for 0 < u < 27r and -oo < v < oo. This is the surface of revolution generated by rotating the catenary curve y = acosh(z/a) around the z-axis. Prove that M is a minimal surface. 9.10' A curve in a surface S. a(t), is a line of curvature if a'(t) is in the direction of one of the principal directions of curvature for each t. Prove Rodrigues's formula: a(t) is a line of curvature if and only if

N(a(t)) = X(I)a'(1) for some differentiable function a(t). Prove further that the principal curvature in this

case is -x(t).

10 Metric equivalence of surfaces It is evident that any finite part whatever of the curved surface will retain the same integral curvature after development upon another surface. C. F. Gauss (8 October 1827)

The problem of constructing useful flat maps of the Earth led to the mathematical question of the existence of an ideal map projection (Chapter 8"'), that is. a mapping Y: (R C S2) - R2 preserving lengths, angles, and area. The archaic terminology for such a map projection is a development of the sphere onto the plane. Euler showed that no such development can exist

(Theorem 8'.1). A natural generalization of this problem is to determine whether there is a development of a given surface onto another. A diffeomorphism between two surfaces, 0: S( - S2, identifies the analytic structure of the surfaces, such as coordinate charts, at corresponding points. However, in order to identify the geometric structures of two surfaces, concepts such as arc length and area must also correspond. These data are encoded by the first fundamental form.

Definition 10.1. A diffeomorphism 0: S1 - S2 is an isometry if for all p E Si and iui, t'u2 E Tp(Si) we have

Io(wi, w2) = l (p)(dmp(wi), dOp(w2)) Two surfaces are isometric if there is an isometry between them.

The simplest examples of isometries are provided by the rigid motions of R3. In the appendix to Chapter 7 we showed that any rigid motion F: R3 -+ R3 is given by an invertible linearmapping A: R3 - R3 satisfying A-1 = A' plus a translation, F = A+tio. If S C R3 is a regular surface, then so is F(S) = S' and FIS : S -- S' is easily seen to be an isometry of surfaces.

If we further restrict our attention to isometrics of R3 that fix the origin, the linear isometrics, then restriction of any such mapping to the sphere S2 = (v" E R3 I v . u' = 1) is an isometry A : S2 -+ S2. In fact, every isometry of S2 to itself arises in this manner. This class of mappings includes rotations of the sphere around a line through its center and reflection across a plane containing the center of the sphere. Not every isometry between surfaces is the restriction of a rigid motion of R3. For

example, the cylinder (cos 8, sin 8. z) with 0 < 8 < 21r and -oo < z < oo may be "opened up" to an infinite open strip (9, z, 0) by the isometry. 0(cos 0, sin 9, z) = (9, z, 0). This mapping is not the restriction of a linear mapping. The isometrics that do not arise from rigid motions of R3 are the most interesting. 145

146


Furthermore, diffeomorphic surfaces need not be isometric. For example, two spheres of different radii are diffeomorphic, but not isometric - the arc length of a complete great circle depends on the radius of the sphere. The notion of isometry determines an equivalence relation on surfaces. The notion of isometry makes precise the ideas of rigid motions and congruence. We define a congruence to be a self-isometry of a surface. 0: S -> S. Two figures, that is, subsets of S. are congruent if there is an isometry with i (frgurei) = figure2. A figure made up of segments of curves on a surface may be thought of as rods in a configuration and the term rigid motion is synonymous with congruence. It follows from the definition that the inverse of an isometry is also an isometry and so the set of congruences of a surface forms a group. The importance of this observation cannot be overestimated. It is the basis of another approach to geometry via so-called transformation groups, initiated by Felix Klein (1849-1925) and Sophus Lie (1847-99). (An introduction to this approach is found in Ryan (1986).) Properties that are preserved under isometrics are the most important to the geometry of surfaces. We call such properties intrinsic. Properties that depend on the particular description of a surface are called extrinsic. For example, the fact that the z-axis is asymptotically close to the surface of revolution of the tractrix is an extrinsic property of the surface. We will see later that this surface intrinsically looks the same from almost every point on it. To investigate intrinsic properties, we often argue locally with the apparatus of functions associated to a coordinate patch.

Proposition 10.2. If 0: Si -+ S2 is an isometry, and p E S1, then there are coordinate charts x: (U C R2) - Si around p and x: (U C R2) - S2 around 0(p) such that the component functions of the metric associated to x and .t. respectively, satisfy E = E,

F= P,andG=G. PROOF. Let x: (U C R2) -+ S1 be any coordinate chart around p, and define x: U - S2 to be the coordinate chart given by x = Oox. This satisfies the properties of a coordinate patch and zv = by virtue of the properties of a diffeomorphism. Now i, =

Since 0 is an isometry, E(u, v) = E(u, v), F(u, v) = F(u, v), and G(u, v) = ((u, v) by direct calculation.

The proposition gives an important pointwise property of an isometry. An immediate corollary is that length, angle, and area are preserved by an isometry. For properties sufficiently local, it is enough to have a local version of isometry to compare surfaces geometrically.

Definition 10.3. A mapping 0: (V C Si) -> S2 of a neighborhood V of a point p in Si is a local isometry at p if there are neighborhoods W C V of p and 1v of 0(p) with 01 w : iV - W an isometry. Two surfaces are locally isometric if there is a local isometry at every point for each of the surfaces.

A local isometry may fail to be an isometry. Forexample, a cylinder is locally isometric to the plane (roll it out). However, because the plane and the cylinder have different topological properties, no single isometry identifies the cylinder with part of the plane.

10. Metric equivalence of surfaces

147

The next theorem is a partial converse to Proposition 10.2.

Theorem 10.4. If there are coordinate patches x: U - SI and z: U - S2 such that E = E, F = F, and G = c on U, then m = . o x-I : (x(U) CSI) -s S2 is a local isometry.

PROOF. Suppose a(t) = x(u(t), v(t)) is a curve in SI with a(0) = p, and v' = a'(0) = xuu' + X,, V' E To(SI). Now 0 o a(t) = . (u (t). v(t)) and so dop(a'(0)) = zuu' + i,v'. Since I p(a'(0), a'(0)) = E(u')2 +2Fu'v' + G(u')2 and I0(p)(dOp(a'(0)), d0p(a'(0))) = E(u')2 + 2Pu'v' + G(v')2, we get Ip(v, v) = 1m(p)(d4p(vv), d0p(i )). Applying the polarization identity

2Ip(v, w) = Ip(v + w, v + w) - lp(v, v) - Ip(iv, iv), we obtain that 0 is an isometry.

When are two surfaces isometric? If they share the same geometry at corresponding points, then we expect them to share all other intrinsic properties as well. We now come to the central role of curvature in geometry. At any point of a regular surface S C 1R3 there is a (not necessarily orthogonal) basis for the R3 directions from the point that can be taken conveniently to be {xu, xu, N) with the choice of a coordinate chart. Consider the derivatives with respect to u and v of this trihedron:

xuu = 11l lx,+rilxv+LIN, xuv = r12xu + r12xv + L2N,

xvv = 122x, + r22xv + L3N,

Nu =allxu+a2lxu, Nv = a l2xu + a22xv.

The functions f'., defined by the previous equations, are called the Christofel symbols of S in the parametrization x. The indices i, j, and k vary over I and 2 where I corresponds to the variable u and 2 to the variable v. The lower indices ij refer to the second partials of the chart x, and the upper index k to the component in the basis {xu, xv, N). Since Ip(xu, N) = Ip(xv, N) = 0, we find

LI = lp(LIN. N) = Ip(xuu, N) = e, one of the component functions associated to the second fundamental form. Similarly,

L2 = f, and L3 = g. Writing (,) for Ip( , ), and fu for (3f/au), fv for (aj/av) for a function f (u, v), we also have Eu = 2(xuu.xu). F. = (x,u,xv) + (xu.xuv),

Gu = 2(xuv, xv),

Ev = 2(xuv,xu). Fu = (xuv,xv) + (X,,Xuv),

Gv = 2(xuv, xv).


148

Expanding these inner products in terms of the Christoffel symbols we obtain the following relations, here written as equations and in matrix form:

f,', E+rIIF =2'E,,.

(rl

E\F

G/ r?i / -

I'i,F+r11G=Fu-7' E,,,

E'2E+f12F=2E1 it r12F+ri2G='Gu,

\F GJ \riz)

r'22 E + r22 F = FL, - I' Gu. {

2

\F G)(r22)

r'22F+ r222 G = 2G

\F.EEJ "GU)

).

= (F2' G,

Thus the functions T are expressible in terms of the metric coefficients E, F. and G. and their derivatives. Proposition 10.2 implies that at corresponding points in isometric surfaces the Christoffel symbols agree. Any expression given in terms of Christoffel symbols that is independent of the choice of coordinates is then intrinsic. We now record the most important theorem of Gauss's Disquisitiones (§ 12). The result makes curvature and the intrinsic properties of surfaces the principal objects of study for differential geometry.

Theorema egregium. Gaussian curvature is intrinsic. PROOF. We give two proofs. In each we show that the Gaussian curvature K can be expressed entirely in terms of the component functions of the metric and their derivatives. We begin with the formula for Gaussian curvature (see the proof of Proposition 9.11):

N. x N. = K(xu x x1,). Take the dot product of both sides with xu x xu and then apply a formula of Lagrange (exercise 7. l) to get

K(EG - F2) = (Nu xu)(Ni, xv) - (N,,

xu)(N1, xu).

By the definition of the normal N we get

N,, = NV =

xuu x xo + xu x xu

EG - F

,I-

xuv X Xt. + Xu X Xi.u

+

a

au

I

EG - F

a

+ av av

EG-F

(XU X Xv).

(xu x xu).

Substituting these expressions in the equation for curvature and simplifying, we obtain the following expression, where the column vectors shown are in fact 3 x 3 matrices and so we have a determinant:

K(EG - F2)2 = ((xuu x

x xvo) xv) - ((x,, x xuo) xo)((xuv x xu) xu)

/0. Metric equivalence of surfaces

Xuu

Xu

Xv XU

) det ( 1,

= del

Xu l - det Xuv ) del

(XvV)11

x

Xuu

xv XU

xv I

v

Xuu

149

Xuu

x

x

(XUV)']

x

Xv Xvxv

= det

- del

Xuu Xv

xuu XVV

xuu . XU

XU xvv

E

F

Xv xuu

F

G

= det

Xuo Xuv

- det

XU xuv xv Xuo

xv

Xuv Xu E

xuv Xv

F

G

F

We can express the second partials, xuu, xvv, and xvv, in terms of the component functions of the metric and the Christoffel symbols. Furthermore, we can rewrite xuu xuu - xuv xuv

as follows: Since x xvv = F - 21 Gu we find x,,,, xvv + xu xvvu = Fvu - 2Guu. Also XU xuv = i Ev implies that xuv xuv + x xuuu = 11 E. Since the coordinate patch is assumed to be differentiable to all orders, xuvv = xuvv and Fvu = F. Therefore, Xuu Xuv - Xuv . Xuu = Xuu Xvv + XU Xuvv - XU Xuvv - xuv Xuu

=Fuv - I2 Guu- I2Evv Putting this all together, we get the formula for Gaussian curvature

K=

I

(EG - F2)2

70 - det

- 'Evv + Fuv - 2Guu

IEu

Fu - I' Gu

E

F. - 2' E, F

F

G

det

(

I

Ev

i Ev

E

Gu

F

2'

Gv

2' Gu

F G

Corollary 10.5. Isometric surfaces have the same Gaussian curvature at corresponding points.

This proves that Gaussian curvature is truly a geometric feature of a surface. The definition of the Gaussian curvature relies on many features of the embedding of the surface in 1t3 such as the Gauss map. It is remarkable (egregium) that this function is independent of these trappings. Euler's theorem on the nonexistence of an ideal map projection follows directly from theorema egregium since the sphere and the plane have different curvatures. There are further relations between the Christoffel symbols that follow from the differentiability of the parametrization. They also lead to our second proof of theorema egregium which is based on a calculation similar to the one at the heart of the Frenet-Serret theorem.


150

The following equations hold by virtue of the differentiability of a coordinate chart: (xuu)v - (xuv)u = 0.

(1)

(xuu)u - (xvv)u = 0,

(2)

Nuv - Niu = 0.

(3)

The first equation produces the following flurry of subscripts and superscripts: a -xuu =

(r,,, )uxu + r;,xuu + (ri, )ux,, + ri,XV, + eN + e,,N

_ (ri, )oxu + r (r1 Vxu + r12x + fN) + (r21), xu + ri, (r22xu + r22x0 + gN) + e(a12xu + a22x,,) +

((r )v + ri t rig + r1, r22 + ea12)xu + ((ri, ) + r;, r12 + ri, r22 + ea22)xu + (r;, f + r;, g + eu)N.

a xuv = (rj2)uxu + r12(r;,xu + ri,x + eN)

+(r212 )ux,, +r2 12 xo+fN)+ fuN+ f(aiixu+a2lxu) 12 (r,2xa+r2 _ ((r,2)- + rizri, +ri2ri2 + fai,)xu 1

+

rl2r, + ri2ri2 + fa21)x, + (r12e + r12.f + fu)N.

Since xu, x,,, and N are linearly independent, equation (I) leads to the relations

(r:1)V+riir12+riir22+ea,2 -(ri2)u - ri2rii - ri2ri2 - fail =0, (ril)e+rllr,2+ri,r22+ea22-(r,2)u-r12r11 -ri2ri2-fa21 =0,

riif+ri,g+ev -r;2e-r12f-fu=0. From Theorem 9.9 we know that a22 = second line we obtain the relation

fF-gE and a21 = E G - F2

EG

- F2

and so from the

+r2 l r22

- r12r11 - ri2ri2 = fa21 - ea22 fE fF-gE -fEG-F2eEG-F2

(ril) , - (ri2)u+rl l r12 _ eF-

=EK. =Eeg-f2 EG - F2 Once again we have proved that K is expressible in terms of Christoffel symbols and the component functions of the metric. How well does the apparatus of functions (E, F, G, e. f, g) and their associated functions, the Christoffel symbols, characterize a surface? This natural question extends the relation of the Frenet-Serret apparatus to a curve in R3, that is, the fundamental theorem for


151

curves (Theorem 7.6). Let the Gauss equations refer to the relations among the Christoffel symbols obtained from (x,,,,) - (xuv)u = 0 and (xuv)v - (xvv)u = 0: (a)

(b) (c)

(d)

(rii)v - (r1 2)u + rii rig +p2 rz2 - ri2ri i - ri2ri2 = EK. (ri2)u - (rii)v + ri2ri2 - r rz2 = FK.

i

i

i

(r22)u - (1-124 + ri2rii + rz2ri2 - ri2ri2 - ri2rz2 = GK, (ri2)u - (ri2)u + & 12 - r22ri i = FK. 12'

Let the Mainardi-Codazzi equations refer to the relations coming from the differentiability

of N, that is, Nu, - Nvu = 0. They are

ev - fu = er;2 + f(r12 - l' ) - gri m fv - 91, = er22 + f(ri2 - rig) - gri2 11

G. Mainardi (1800-79) and D. Codazzi (1824-75) were colleagues at the University of Pavia during a particularly active period in the development of differential geometry. Mainardi published his paper on the relations among the six functions E, F, G, e, f, and g in 1856. Ten years later. Codazzi published his account of these relations in terms of directional cosines. In this equivalent form, the equations became widely known as part of his second-prize winning memoir submitted to the Paris Academy in 1860. In fact, the equations were known in 1853 by K. Peterson (1828-81), a Russian geometer who studied in Dorpat under Senff and Minding (Reich 1973). The Gauss equations and the Mainardi-Codazzi equations together are called the compatibility equations for surfaces. Ossian Bonnet (1819-92) proved the following version of the fundamental theorems we have proven for curves.

Theorem 10.6 (fundamental theorem for regular surfaces). Let E, F, G, e, f, g be differentiable functions defined on an open set V C R2 with E > 0 and G > 0. Suppose that these functions satisfy the compatibility equations, and EG - F2 > 0. Then for each q E V

there is a neighborhood U of q, U C V, and a diffeomorphism x: U - x(U) C R3 such that the regular surface x(U) has E, F, G, e, f, and gas component functions of the first and second fundamental forms. Furthermore, if U is connected and z: U -+ . (U) C R3 is another such diffeomorphism, then there is a vector v E R3 and a matrix A in 0(3) such

that g=Aox+v. The proof, like that of the previous fundamental theorems, is a side trip into the theory of differential equations and would take us too far afield (see Spivak 1975, volume 3, pp. 7985). It follows from the theorem that the six functions E, F, G, e, f, and g determine the local theory of a surface.

Special coordinates The simplifying feature of curves is the arc-length parametrization. There is no such canonical choice of coordinates for surfaces. This fact is a source of both difficulty and freedom.

152


We use it to our advantage and try to construct coordinate charts whose properties are closer to the geometric features of the surfaces. This will simplify the computation of important measures such as the Gaussian curvature. Artful choices of coordinates play a crucial role in certain arguments in later chapters. Here we introduce two useful constructions to give the flavor of these ideas. A familiar feature of Cartesian coordinates in elementary geometry is that coordinate lines meet at right angles. On a surface, this condition implies that the second component

function of the metric, F, is identically zero. There are many ways to construct such a coordinate patch for a surface. We now consider a particular choice that arises from the surface's geometry.

Definition 10.7. A curve a: (a, b) -, S in a regular surface S is a line of curvature if, for all t E (a. b), a'(t) is a nonzero multiple of a principal direction at all ). Euler's Theorem 9.1 shows that when the principal curvatures differ, the principal directions are orthogonal. A patch for which the coordinate curves are lines of curvature is orthogonal, that is, F = 0. Another set of curves on a surface is related to the second fundamental form:

Definition 10.8. A curve a: (a, b) -- Sin a regular surface S is an asymptotic line if, 0. for all t E (a, b), Asymptotic lines do not always exist at each point in a surface. To see this write the differential of the Gauss map in the orthonormal basis of principal directions, {X1, X21:

dNp= I 0'

k2

IIp(v) = 0 has no nonzero solutions if ki\> k2 > 0. Furthermore, if ki > 0 and k2 < 0 at each point in a neighborhood of a point in a surface, then there are two asymptotic directions at each point. Thus if K(q) < 0 for all q in a neighborhood of a point in a surface, then we can try then to construct a coordinate chart with asymptotic lines for coordinate curves. There are appropriate differential equations for lines of curvature and asymptotic lines whose solutions lead to the coordinate charts we seek. However, we present a more general method for constructing special coordinate patches. If W is an open subset of a surface S and V (q) is a smooth assignment, q r-> V (q) E Tq (S), for each q E W. then we call V a vector field on W. In a coordinate patch x: (U C R2) -+ S this can be expressed by a pair of smooth

functions a(u, v), b(u, v) such that V (x(u, v)) = a(u, v)x,, +b(u, v)x,,. An Integral curve for a vector field V (q) is a curve a(t) satisfying the equation a'(t) = V(a(t)). Note that in general we cannot construct a patch whose coordinate curves are integral curves for the fields of principal directions - these vector fields (X1, X2) are orthogonal and of unit length and so F = 0 and E = G = 1. Thus the line element takes the form ds2 = d u2 +d v2 from which we can deduce that the surface has Gaussian curvature K =_ 0. We can get around this issue by looking for lines of curvature or asymptotic lines as the coordinate curves for our charts. More generally, given a vector field V (q) on an open set


153

W C S, we say that a curve a (t) is a line in the direction of V (q) if a'(t) = x(a (t ))V (a (t )) for some nonzero, differentiable function x : W - R. In this terminology, lines of curvature are lines in the direction of a principal vector and asymptotic lines are lines in asymptotic directions.

Theorem 10.9. Suppose p is a point in a regularsurface S, and in a neighborhood W C S of p two linearly independent vector fields X (q) and Y(q) are defined. Then there is a patch around p whose coordinate curves are lines in the directions of X and Y. PROOF. Fix a coordinate patch x: (U C R2) -+ S around p such that (0, 0) E U and x(0, 0) = p. An integral curve a(t) = x(u(t), v(t)) of the vector field X(u, v) _

a(u, v)x + b(u, v)x would satisfy the differential equations u' = a(u, v) and v' = b(u, v).

Since X(q) does not vanish in the neighborhood of p, one of the quotients u'/u' or v'/u' represents a well-defined smooth function in the neighborhood of (0, 0) in U. Say it is u'/u'; this gives the ordinary differential equation

du a(u, v) = .fx(u, v) = b(u, v) dv By Theorem 6.19 on the existence of solutions of such differential equations, there is a solution ux(v; uo, vo), for initial conditions ux(vo; uo, vo) = uo, in a neighborhood around (0, 0). Similarly, there is a solution uy(v; uo, vo) for the analogous differential equations associated to integral curves for the vector field Y(q). Now each point (r, s) E U, sufficiently close to the origin, lies on a pair of curves ux(v; 0, vo), and u y (v; uo, 0). Consider the function sending (r, s) to (uo, vo). By the linear independence of the vector fields, and the uniqueness of solutions of differential equations, this change of coordinates is well defined and smooth, and determines a diffeomorphism

of a neighborhood of the origin in U to some open set in R2. Let 0 be an inverse to this function. The composite y = x o m is a coordinate chart for a neighborhood of p in S. The coordinate curves of this patch run along curves that satisfy the differential equations given earlier. Thus they are lines in the directions of X(q) and Y(q). Notice that we have not necessarily constructed integral curves for either vector field; we have determined that the ratio u'/v' is correct, but we have not solved u' = a(u, v) and v' = b(u, v). By quadratures we have in fact solved equations u' = l (u, v)a(u, v) and v' = l (u, v)b(u. v) for some integrating function l (u, v). Corollary 10.10. Suppose p is a point in a regular surface S. If p is not an umbilic point, then there is a patch around p with coordinate curves that are lines of curvature. If K < 0 in a neighborhood around p, then there is a patch around p whose coordinate curves are asymptotic lines. PROOF. We leave it to the reader to show that the set of umbilic points in a regular surface is closed. Around a nonumbilic point p, then, there is a neighborhood of nonumbilic


154

points. The principal directions determine a pair of linearly independent vector fields on that neighborhood to which Theorem 10.9 applies. If the curvature in a neighborhood around p is negative, then there are two linearly

independent unit vectors satisfying IIq(ui) = Ilq(i2) = 0 for q in this neighborhood. These may be written as smooth linear combinations of the principal directions. We may apply Theorem 10.9 again to obtain a patch with coordinate curves that are asymptotic lines. If a patch has lines of curvature as coordinate curves, then the analytic data attached to the surface are simplified. As we have already said, the metric component function F = 0

by Euler's theorem. We can also deduce that the component function f of the second fundamental form vanishes. This follows easily from the matrix representation for dNp. Since the principal directions are eigenvectors for this operator, it is a diagonal matrix. This yields

( aOr

au) (./ g/ 0 E/ EG G on a regular surface, f = 0. This matrix equation also implies the

and since E i4 0 formula K = eg/EG.

If the coordinate curves are asymptotic lines, then Il ll p(x,.) = 0. Expressing Hp in terms of the basis (xv, xv,) this implies that e = 0 = g. This gives the simpler expression

_fz

for Gaussian curvature

K=

EG - F2

The simplification of the components of the first and second fundamental forms leads to a simplification of the associated Mainardi-Codazzi equations. We record the results here for use later in Chapter 14. If the coordinate curves are lines of curvature, the discussion above implies that the Mainardi-Codazzi equations take the form

e = eI i2 - gri r

-gu = erzz - grit.

Rewriting the Christoffel symbols in terms of the components of thee metric we get

2 lE+

)G,

g°

2\E+G/.

If the coordinate curves are asymptotic lines, then the Mainardi-Codazzi equations are

fv = pr ,I - r12),

(EGf

F2)

(!GEU

fo = f(r222 - r;z).

- FF + z FEv + Z FED, - 2

(EGf Similarly, the form of fv is given by

fv

(EGf

FZ)

(' a

These expressions play an important role in later proofs.

.


155

Exercises 10.1

Show that a surface S is isometric to a surface of revolution if it has a coordinate chart around each point such that the line element can be expressed

ds2 = A(du2 + dv2),

for A some function of u or v alone. Show that this chart determines a conformal mapping of the surface to the plane. 10.2

Determine the Christoffel symbols r, f for the polar-coordinate parametrization of the plane.

103'

Suppose that F = 0 for some coordinate chart. Prove that, on this chart.

K_2

TV

au

EG

10.4

aG

aE a

I

au

a

au

EG

EG

Looking forward to the generalization of the differential geometry of surfaces to their higher-dimensional analogs (Riemannian manifolds), we present another version of theorema egregium, notation-wise. All indices shown vary over the set (1, 2) corresponding to coordinates (u, v). Let u = u1 and v = u2. Consider the matrices

(gil) =

EF

(F

(g'1) = (go -I.

G

Show that the Christoffel symbols are defined by

ri

gi`

(at1 +

2

au;

- agk auk

aul

Define the Rlemann curvature tensor to be the function Risk

auk rik

aau k

r'; +

rkr' - r, rk, 0oi.2

along with the transformed version

R'jki =

gi.Rikr m=i.2

Show that theorema egregium now becomes

K

_

R1212

det((g,j))

156

10.5*


A surface S in R3 is ruled if through each point p E S there is a line in R3 entirely contained in S. Show that the line through p lies along an asymptotic direction. Prove that if a surface is ruled, then K < 0 at each point. Give an example of a ruled surface that is not a cylinder.

10.6*

Show that a surface is minimal if and only if the asymptotic directions are perpendicular.

10.7

Consider the surfaces determined by the coordinates (-a , the catenoid: a v = ur sin vr, z = av,. the right helicoid.

v = u sin v,

Show that the mappings v = yr and u = F_2 + a2 0r ill

=

u - a determine an

isometry between these two surfaces. 10.8

Suppose that the functions e, f, and g are all zero for a surface S. Show that the surface is a portion of the plane.

10.9

Does there exist a surface with coordinate chart x = x(u, v) and associated functions

F,=I,F=O,G=e", e=e". f=0,g=1? 10.10

Show that the umbilic points form a closed set in a regular surface.

10.11

Suppose that K(p) = 0 for all p in a neighborhood W of a point in a surface S, and that the mean curvature H(p) 96 0 for all p in W. Show that the unique asymptotic curve through each point p E W is a straight line in 1R3, and that the tangent plane is constant for all points along any such line.

11 Geodesics But if the particle is not forced to move upon a determinate curve, the curve which it describes possesses a singular property, which had been discovered by metaphysical considerations; but which is in fact nothing more than a remarkable result of the preceding differential equations. It consists in this, that the integral f v ds, comprised between the two extreme points of the described curve, is less than on every other curve. P. S. Laplace, Mfcanique CEleste (1799)

In the last few chapters we developed some of the analytic and geometric properties of surfaces in ]R3 that are consequences of the properties of the first and second fundamental forms. The important geometric properties are intrinsic, that is, preserved by an isometry of the surface. To develop further elementary geometric notions we next introduce a notion of "line" on a surface. A "line" on a surface should enjoy some of the familiar properties of lines in the plane: For example, lines are (I) The curves of shortest length joining two points (Archimedes). (2) The curves of plane curvature identically zero (Huygens, Leibniz, Newton). (3) The curves whose tangent and its derivative are linearly dependent.

Furthermore, to have geometric significance, the defining notion for a "line" on a surface should be intrinsic. We first try to adapt the condition of zero plane curvature. Let a: (-e, e) -+ S be a curve on S, parametrized by arc length. We denote the tangent vector, as usual, by T(s) = a'(s) and consider a"(s) = T'(s). Since a(s) is a unit-speed curve, T'(s) 1 T(s); since T'(s) is a vector in J3 and the set (x,,, x, N) is a basis for JR3 at the point a(s), T'(s) has a component pointing in the normal direction and a component lying in the tangent plane. The intrinsic part of this vector will be seen to be the component lying in the tangent plane.

Definition 11.1. The intrinsic normal to a(s), a unit-speed curve on S, is the vector

na(s) = N(a(s)) x T(s). The intrinsic normal, na(s), lies in Tat,5l(S), the tangent plane to Sat a(s). Since T'(s) is perpendicular to T(s), we have

a"(s) = T'(s) = (T'(s), N)N + (T'(s), na(s))na(s). 157

158


Theorem 9.6 tells us that the magnitude of the component in the direction of N. (T'(s), N), is the plane curvature of the curve given by the intersection of the plane spanned by (T(s), NI

with the surface S. This is the normal curvature of the surface at a(s) in the direction lla(s)(T(s)) = (T'(s), N). The other component, lying in the T(s), given by tangent plane and normal to the tangent vector, is the subject of the following definition, introduced by Ferdinand Minding (1806-85) in 1830. Minding was a student of Gauss and later professor of Mathematics at Dorpat (now Tartu in Estonia). This university was far removed from the research centers of mathematics, but it was an outpost for work in differential geometry at the time, mostly through the efforts of a teacher of Gauss, M. Bartels (1769-1836), who came to Dorpat in 1821. He was a correspondent with Gauss during his lifetime and he was keenly interested in differential geometry.

Definition 11.2. The geodesic curvature of a unit speed curve a(s) on a surface S is the function kg(s) = (a"(s), na(s)) = latsr(a"(s). na(s))

The definition identifies the amount that a"(s) is "perceivable" in the tangent plane to S at a(s). Extrinsically, notice that a unit-speed curve a(s) in R3 has a"(s) = K(s)1V(s),

where IV(s) =

a surface S in R3, K2(s) = k2(a'(s)) + k2(s) since a"(s) _

II« (s)II k (a'(s))N(s)+kg(s)na (s). We now prove that the geodesic curvature is an intrinsic feature of a curve, that is. it is dependent only on the component functions of the metric, and so it is preserved by isometries.

Theorem 11.3 (Minding 1830). Geodesic curvature is intrinsic.

PROOF. We calculate kg for a unit-speed curve a(s) = x(u(s). v(s)) lying in a coordinate

ll. Geodesics

1S9

patch x : (U C R2) -> S. From the proof of theorema egregium we can write

T (s) = a'(s) = u'x + v'X, T'(s) = a"(s) = u"xu + v"Xp + (u')2Xuu + 2u'v'Xuu + (li )2X'.

U "xu + v"x0 + (u')2(r1,X, + rl,xu + eN)

+ 2u'u'(r12x + r12x + fN) + (U)2(r22X,, + r22xv + gN). By the unit-speed assumption, the part of T'(s) that lies in the tangent plane points in the na(s) direction and so it is perpendicular to N. Ignoring the component in the normal direction we have kg(s)na(s) = (u" + (u')2r1, + 2u'v'r12 + (VI)2 r22)xu

+ (u" + (u')2r11 + 2u'v'rl2 + (v')2r22)xV. Since

na(s) has unit length,

kg(s) = (kg(s)na(s).na(s)) = (kg(s)na(s), N x T(s)). The cross product satisfies (u", v x i))) = (w x i, v) and this leads to the formula:

kg(s) = (T(s) x kg(s)na(s), N) = ((X" U,

+

x (U" + (u')2rl, + 2u'v'rl2 + (v )2r 2)X,,, N) x (v" + (u')2r2 + 2u'v'rziz+ (v')2rzz2)x,,, N).

To To simplify this expression we use the relations x x xv = EG - F N and (N, N) = 1:

kg(s) = EG - F2(-u"u' + u'v" + rl, (u')3 + (2r12 - r,1)(u')zv + (r22 - 2r12)u'(u2 - r22(V )3)

EG-F2det( u'V/

u"+(u')zrli+2u'v'rl2+(v)2r22

v + (u ,)z r z+ 2u v r12z + (VI)2rz22 )

If 0: S -> S' is an isometry, then, at O(x(uo, vo)), the expression for kg is the same for the curve $ (x(u(s). v(s))) since it depends only on the components of the metric, the Christoffel symbols, and (u(s), v(s)). It is useful to have an expression for geodesic curvature when a curve fl(t) is not unit speed. This is made easy by the properties of the cross product in llt3. Write the arc-length parametrized curve associated to ,8(t) as a(s) = 6(t (s)), where t(s) is the inverse function of the arc length. Then z

T (s) = a'(s) = 0'(t (s)) d and T'(s) = a" (s) = P"(t (s)) (T) +

z

(t (s)) dsz


160

()3.

It follows that T(s) x T'(s) = [#'(1 (s)) x 6"(t (s))] ds

Expressing everything in sight

in terms of to = t(so) we find

kglBuor = (T(so) x T'(so). N) =

(f'(to) x fl"(to), N) = (#'(Io) x #"(to). N) (ds/dt)3 (f'(to), f'(to))3/2

Finally, if we write fi(t) = x(u(t), v(1)) in a coordinate chart for the surface, we find

P'(t) = u'xu + v'x U"xu + v"Xv + (u')2Xuu + 2u'v Xuv + (v')2xvv

=u"xu +v"x1,+(u')2(I jrxu+r21XV+eN) + 2u'u'([ 12xu + 112x + JN) + (v')2(r22.ru + r 222x1, + gN). This gives the formula

EG - F kglBt1or

(#'(to), fi'(to))312

det

u'

u" + (u')2I'i r + 2u'v'f 12 + (v')21'22

V,

U' + (u')2rlr + 2u'v'r12 + (U)2('22

If T'(t) points entirely in the direction of the normal to the surface, then the intrinsic part of T'(t) vanishes. An inhabitant of the surface would see T'(t) = 0, since such an observer could only perceive acceleration in the tangent plane. This leads to the following class of curves on a surface.

Definition 11.4. A geodesic on a surface S is a curve on S whose geodesic curvature is identically zero.

For example, on the unit sphere S2 c 1R3, consider a curve of constant latitude. Since a latitude is the intersection of a plane fl with the sphere, it is a planar curve and T'(s) lies in 11. If the normal at a point at this latitude does not lie in fl, then T'(s) has a component in the normal direction and kg 54 0. It follows, by rotating the sphere if necessary, that planar sections of the sphere are geodesics if and only if they are great circles.

By a similar argument on a surface of revolution, a meridian through a point of a graph is a geodesic if and only if it is over a critical point of the graph.

11. Geodesics

161

From the local formula for geodesic curvature we see that kg is identically zero whenever the following differential equations hold

u"+(u')2r11 +2u'v'r12+(v')2ri2 =0, 11f

v" + (u')2r

i

I

+ 2u''ri2 + (v')2ri2

(*)

= 0.

From this analytic expression we can immediately determine the geodesics on the plane. The line element is given by ds2 = du2 +dv2 and so all of the Christoffel symbols vanish. The geodesic equations (*) become

u"=0, v0. which determine a line in the plane, as expected.

Some important properties of geodesics are consequences of the differential equations given by (*).

Proposition 11.5. Suppose fl: (-e, e) - S is a curve on a surface S and P(t) _ x(u(t), v(r)) in a coordinate patch an S. If f satisfies differential equations (*), then

(I)

fl is a geodesic.

(2) fl is parametrized by a multiple of arc length.

(3) For r > 0 the curve 4: (-e/r, a/r) - S given by 4(t) _ (rt) is also a geodesic. PROOF. When we write fi"(t) in terms of

(xe, x, N), the equations (*) imply f"(1) =IN

and so

dr (p'(r), '(r)) = 2(f"(t). #'(t)) = 2(IN, )9'(t)) = 0. Thus 0'(t)) is constant and f(r) is parametrized by a multiple of arc length. Finally, (*) is a system of degree two homogeneous differential equations and so the substitution t H rt in x (u(t ). v(t )) does not affect the conditions (*). Thus fi(t) as defined in the proposition is also a geodesic.

We next consider a deeper application of the local formulas and show that geodesics are the curves that locally minimize distance. Thus they satisfy the condition of Archimedes for being lines.

Theorem 11.6. If a: (-e. F) - S is a unit-speed curve and for any a, bin (-e. e) with a(a) = p, a(b) = q, a is the curve of shortest distance joining p and q in S, then a is a geodesic.

PROOF. The proof is a page out of the calculus of variations, a subject that finds many applications in differential geometry. The goal is to show that kg(s) = 0 for a < s < b. Suppose this fails and let so E (a, b) with kg(so) 96 0. By continuity there are values j. k. a < j < so < k < b, with kg(s) 96 0 on (j. k], and a([ j, k]) C x(U) for some coordinate patch x. Let A: (j. k] --). R be a smooth function satisfying (I) A(j) = A(k) = 0.

(2) A(s)kg(s) > 0 for all s E (j, k).


162

Let na (s) = N x a'(s) be the intrinsic normal to a. Define u, v: If. k] -# R by A(s)na (s) = u(s)x +D(s)x0. Extend u and v to [a, b] by u(s) = 0 = D(s) fora < s < j and k < s < b. Write a(s) = x(u(s), u(s)), and define

C(r, s) = x(u(s) + ru(s), v(s) + ri(s)) with Iri small enough for this expression to make sense. By the properties demanded of x,

it follows that C(r, a) = p, C(r, b) = q, and C(0. s) = a(s); r parametrizes a family of curves between p and q. For a fixed value of r, we write ar(s) = Or. s). q

Consider the length of ar(s) as it varies over this family

L(r) = f

P

da

das

(ds ' d

)ds.

By assumption L(r) has a minimum at r = 0. Consider L'(r).

L'(r) =

d

f

dr

=

n

ac

=

a a

dar dar

(ds

la

ar

as

)) ds.

' ds

ds

((asa2C

c

aas

as '

ds)ds

ds '

ar'

Ln

For r = 0,

J(dalr dar

r_o

0 = L'(0) =

as) fas'

= I. and so applying integration by pans we obtain

fh(a2c

ac

asar' as)

r_o ds

-f b[ds(aC' as)Ir=v(aC' j.b (8C

b

1a'

as

)Ir-ola

-

8sC)r-olds

a2C

ar a2

)Ir=oas.

By the definition of C(r. s) we have, on the interval [j, kJ,

ac

= u(s)x + D(s)x1, = A(s)na(s).

ar

s)

Since na(s) is perpendicular to

0

= L'(0) =

=

-

fb

_f

= a'(s), we know

ac a2c

(ar

.

2

c

aas

T

,

k

)I

ds=-

k

,l(s)kg(s)ds < 0.

(.k(s)na(s),

= 0. Hence =o

11. Geodesics

This contradiction follows from the assumption that kg(so) a is a geodesic.

163

0. Thus we have shown that

The term geodesic comes from the science of geodesy, which is concerned with measurements of the Earth's surface. Several mathematicians have been involved with geodesy, including Gauss. The Konigsberg mathematician F. W. Bessel (1784-1846) published an analysis of several geodetic surveys to determine more precisely the shape of the Earth as an ellipsoid of rotation. C. G. Jacobi (1804-51) made a study of the "shortest curves" on an ellipsoid of rotation which he referred to as "geodesic curves:' motivated by such curves on the Earth's surface. Earlier writers such as Johann Bernoulli and Gauss had used the term "shortest curves" (linea brevissima) without worrying about ambiguities such as the choice of direction along a great circle joining two points on a sphere. In 1844 the term "geodesic curve" replaced "shortest curve" in the influential work of J. Liouville (1809-82) who took the term from the work of Jacobi.

The equations (*) associated to a geodesic curve are grist for the mill in the theory of differential equations. The following general theorem leads to many useful geometric results. We refer the reader to Spivak (1970, vol. 1) for a proof. Theorem 11.7. Let F: R" x R" -+ R" be a smooth function and consider the differential equation

dz dtz

/

F (c

dt)

Then, for all (io, uo) in R" x R', there is a neighborhood U x V of (xo, vo) and an e > 0

such that, for any (i. v) E U x V, the equation has a unique solution c : (-e, e) R" satisfying the initial conditions cj (0) = x and c,(0) = u. Moreover, the mapping f: U x V x (-e, E) -+ R" given by f(x, u, t) = c6(t) is smooth. The system of differential equations (*) satisfies the conditions of the theorem, and so we can apply it to determine the existence of geodesics. The restriction to a neighborhood in the theorem implies a restriction on the length of tangent vectors to a surface. This is made precise by defining the associated norm on the tangent space Tp(S) determined by the inner product: For u E Tp(S), 11i311 = Ip(u, u).

Corollary 11.8. If p is a point in a surface S. then there is a neighborhood U of p and an e > 0 such that if q E U and i E TqS with 116 11 < e, then there is a unique geodesic y;,: (-1, 1) -> S with y5(0) = q and 4(0) = U. Moreover the mapping

C:((q,15)IgEU.vETq(S), and 11i)11
By applying the homogeneity property of the equations (*) we can reparametrize the results of the corollary as follows:

164


If p is a point in a surface S, then there is a neighborhood U of p and an e > 0 such that if q E U and i E Tq S with 11011 = 1, then there is a unique geodesic y;,: (-e. e) -+ S with y;,(0) = q and yo'(0) = u. Moreover the mapping

C:((q,u)IgEU,0ETq(S),and IIUll=I)x(-e,e)-+S given by C(q. v, t) = y;,(t) is smooth.

The uniqueness part of the theorem allows us to finish a (long-awaited) proof that great circle segments are all of the geodesics on the sphere. To wit, if y (s) is a unit-speed geodesic through a point p = y (O) with v = y'(0), then by an isometry of S2 we can take this point to be on the equator and v pointing due north. The known geodesic satisfying these conditions is the meridian through this point on the equator, a great circle. Thus the transformed y lies

along a great circle. The isometry may be given by a rotation or two so its inverse takes great circles to great circles. Hence, y lies along a great circle. We now obtain some useful consequences of the existence of geodesics. In the normed linear space Tp(S) we can define the (open) ball of radius S, and the sphere of radius S, respectively: Bs(Op) = (iu E Tp(S) 111@11 < S) and St(Op) = (w E Tp(S) I IIw11 = S).

With this notation we can define the following important mapping.

Definition 11.9. The exponential map expp: BE (0p) -+ S is defined by expp(w) = Yu,/IIWII(Ilwll):

that is, expp(w) is the point in S gotten by traveling along the unique geodesic through p determined by the direction @111 w II for length 11 w II.

The key properties of the exponential map are listed in the next theorem. A complete proof of these properties would require a deeper study of the differential equations involved. We refer the reader, once again, to Spivak (1970, vol. I) for details.

l 1. Geodesics

165

Theorem 11.10. To each p E S there is a value 0 < E p < oc such that (1) The finectiot expp: BP(Op) -, S maps BEp(Op) diJfeomorphically onto a neighborhood of p. (Such a neighborhood of a point p is called a normal neighborhood.) (2) Any two points in expp(B(,,(0p)) are joined by a unique geodesic of length less than 2ev. The exponential map gets its name from its form in a special higher-dimensional context where it is realized by exponentiation. Historically, it is a little out of place in this discussion.

However, it greatly simplifies the arguments to follow, and it organizes several notions from the classical theory of surfaces. In Chapter 13 we will consider useful coordinate charts constructed from the exponential map. These coordinates appear in the work of Gauss (1828) and Riemann (1854) and so the exponential map may be traced back to GSttingen. We now turn to some synthetic notions that can be expressed analytically via the exponential map.

Euclid revisited I: The Hopf-Rinow theorem The value Et, in Theorem 11.10 tells us how far from the point p we can travel along any unit-speed geodesic emanating from p. Euclid's Postulate 11 requires that any straight line segment be extendable in a straight line any given length. In the language of surfaces we interpret this postulate as the following property of surfaces.

Definition 11.11. A surface S is said to be geodesically complete if even. geodesic y: [a. b] -* S can be extended to a geodesic y: R - S. Given a geodesic y : [a, bj - S, let p = y(a). and consider the exponential mapping expp. Geodesic completeness implies that, in any direction from p, expp may be applied to a vector of any length. This leads to a reformulation of the definition.

Proposition 11.12. A surface S is geodesicall r complete if and on/v if for even. p E S, the domain of expp is all of Tp(S). Geodesic completeness is a global property of the surface S. Naively, to verify if it holds. we would need to check every geodesic emanating from each point. It is easy to construct examples where geodesic completeness fails: For instance, let S = S` - Isouth pole); then from any point in S the direction "due south" yields a geodesic of limited length. We may interpret Euclid's Postulate I as another global property of a surface S. For any pair of points p, q r= S, there exists a unique length-minimizing geodesic joining p and q. The choice of the term length-minimizing is important: contrast the two geodesics joining New York City and Chicago on the great circle (kg = 0) through these cities. On the sphere, if the two points are antipodes, there are infinitely many length-minimizing geodesics joining them.

166


Counterexamples to the existence part of Postulate I are also simple to construct. Consider S = 1R2 - ((0, 0)), the Euclidean plane with the origin removed. Then there is no geodesic

joining points (-1, 0) and (1, 0). In what follows, we examine the question of existence of geodesics joining points (the uniqueness question is considered in the next chapter). Our discussion leads to a condition that a surface must satisfy to obtain the analogs of Euclid's Postulates I and H. A minimal condition to realize Postulate I is that the surface S be connected. In fact, from the definition of a surface, connectedness implies that for any pair of points in S, there is some piecewise regular curve joining the points. To a regular curve a : [to. tj J --> S, denote

the length of a by L(a) = fro

(a (t), a (t)) dt.

Definition 11.13. The distance between two points p and q in a connected surface S is given by

d (p, q) = inf ( L (a) I a is a piecewise regular curve in S joining p to q ).

Proposition 11.14. d: S x S -- R makes S into a metric space. PROOF. The triangle inequality d (p, r) + d (r, q) > d (p, q) follows by joining curves and the definition of distance as an infimum. Certainly d (p, q) = d (q. p), so it suffices to show

that d(p, q) = 0 implies that p = q. Suppose that d (p. q) = 0 and p 54 q. Let ep > 0 be such that the map expp: B,, (Op) S is a local diffeomorphism. Notice that for ep small enough, all points of distance less than ep from p lie in expp(BEP(0p)), that is, for r E exp(BE0(0p)) there is a u E BP(Op) such

that d(p, r) = 1Ii II < Ep and r = expp(u). Since d(p, q) = 0 < ep. q E expp(BE0(0p)) and so q = expp(w) for some w E BEO(Op). Since p # q and exp is a one-to-one mapping,

w # 0. However, it then follows that 1ItI1 # 0. The fact that d(p, q) _ IIwII yields the contradiction. Thus p = q. The surface R2 - ((0, 0)) lacks a property that is related to geodesic completeness it fails to be a complete metric space. Recall that a sequence of points in a metric space (M, d) is a Cauchy sequence if, for any E > 0, there is an index N = N(E) with d(ak, at) < E whenever k,1 > N. A metric space M is complete if every Cauchy sequence of points in M converges to a point in M. To see that R2 - ((0, 0)) is not complete, consider

the Cauchy sequence ((I/n, 0)). In fact, metric completeness of a surface as a metric space (S, d) is the key to Postulates I and II.

Theorem 11.15 (the Hopf-Rinow theorem). Let S be a connected surface. Then the following conditions are equivalent.

(I) (S, d) is a complete metric space. (2) S is geodesically complete.

ll. Geodesics

167

Furthermore, either condition impliesthatforanyp.q E Sthere isa geodesic y: [a, b] -. S

with y(a) = P. Y(b) = q, and L(y) = d(p, q). PROOF. We first show that (2) implies the last statement of the theorem. Suppose that

S is geodesically complete. Let p = d(p.q). Consider ep > 0 small enough so that expp: B.,(Op) -. S is a local diffeomorphism and between any pair of points in expp(B.P(0p)) there is a unique unit-speed geodesic. For some 0 < 8 < ep, let £ = expp(S6(Op)); notice that S6(0p) is compact and hence so is E. Let po be the point in £ such that d(po, q) < d (s. q) for all s E E.

Such a point exists by the continuity of d( , q) (a property of distance functions) and the compactness of E. By the choice of ep, po = expp(8uo) for some unit vector vo. We claim

that expp(piio) = q. To prove this claim we show that the geodesic expp(tio) = y(t), 0 < t < p, satisfies

d(Y(t),q)=p-t. By connectedness there are piecewise-regular curves joining p to q. Any such curve must pass through E. Since distance is defined in terms of piecewise-regular curves and distance is minimized by geodesics to points on E, at the point po we can write

p=d(p,q)=min(d(p,s)+d(s.q)}=d(Po,q)+8. SEE Thusd(y(8),q)=p-8. Let ro = supir E [8. p] d(y(r), q) = p - r). By the continuity of the metric, d(y(ro).q) = p - ro. Suppose that ro < p. Around y(ro) there is a normal neighI

borhood given by expy(ro)(B.-(0ytrol)). Let 0 < 8' < e'. This gives another sphere £' = expy(,)(S6'(0y(,e))) and a point po E £' such that

d(po,q)
d (Y(ro), q) = min (d(y(ro), s') + d(s', q)) = 8' + d (p , q), s'E E'


168

and so d(po, q) = (p - ro) - 8'. This implies that

d(p, po) ? d(p. q) - d(p , q) = ro + 8'.

The piecewise-geodesic curve gotten by going along y to y(ro) and then along the geodesic from y(ro) to po' has length ro +8'. This curve achieves the minimal length. We show in a lemma and corollary to follow that a piecewise-geodesic curve of minimal length must be, in fact, part of a single geodesic. That geodesic must be y by uniqueness and so y (ro+8') = po.

But this violates the definition of ro since d (y(rb + 8'), q) = p - (ro + 8'). Thus ro must be p and q is joined top by a length-minimizing geodesic.

A consequence of this proof is that if A C S is a subset of diameter p and p e A, then expp: Tp(S) -- S maps the closed disk Bp(Op) onto a compact set containing A. This implies that the closure of A is compact, and so, by an elementary result in point set topology, any Cauchy sequence of points in S converges to a point in S. Hence (S. d) is a complete metric space and (2) implies (1). Suppose that (S, d) is a complete metric space, and y: (a, b) -+ S a unit-speed geodesic. Then considera Cauchy sequence in (a, b), (t (, converging to b. The unit-speed assumption is a Cauchy sequence in S and so it has a limit, which implies that the sequence we write as y(b). By Corollary 11.8 (existence and uniqueness of geodesics), since y(b) is defined and y'(b) can be computed as a left limit, we can extend y beyond b. Continuing the argument along in both directions we can extend the domain of y to all of R and S is geodesically complete.

Lemma 11.16. Ifa: [a, b(-. S is a piecewise-regular curve lying in the image ofBf,(Or) under the exponential map and

a(t) = expp(u(t)X(t)) for0 < u(t) < ep and IIX(t)II = 1, then

b

(«'(t), a'(t)) dt > la(b) - u(a)I,

-

Q

where equality holds if and only if u(t) is monotonic and X(t) is constant, that is, a(t) is a segment of a radial geodesic from p.

PROOF. Let A(u, t) = expp(uX(t)) so that a(t) = A(u(t). t). It follows that

a'(t) =

aA du

aA

au dt + at

Since IIX(t)II = I for all t, IIaA/auII = I by the construction of the exponential map.

(aA 1A)

= 0 by an argument similar to the proof of Theorem 11.6 (see au at also the proof of Lemma 13.2). It follows that Furthermore,

11

(a'(t). «'(t)) _

12

(dA

dt

Z

+ Iat

'

)

atdt

ll. Geodesics

and so

f

j

a

(a'(t),a'(t))dt

a

ad

169

di

Iu(b) -

To obtain equality we need 8A/et = 0, that is, X'(t) = 0 or X(t) is constant. We also need that u(t) is monotone.

Corollary 11.17. Ifq, q' E expo(BEO(0p))arejoinedbyaunit-speedgeodesicy: [0, r] S and by a piecewise-geodesic curve a : [0, r] -' S, then

f

(y'(t), y(1)) d, < J r

0

(a'(t).a'(t))dt

0

and equality occurs if and only if a (t) is a reparametrization of y (1).

The metric conditions of the Hopf-Rinow theorem imply the analogs of Euclid's Postulates I and 11 from the point set topology of the surface S. For example, if S is compact and connected, then it is a complete metric space and so it is geodesically complete. Global properties such as metric completeness or compactness cannot be determined merely from local information. However, they are important ingredients in the construction of models for geometries as we shall see in later chapters. Exercises 11.1

Suppose Si and S2 are surfaces and 0: S1 - S2 is an isometry. Suppose that y: (-e, e) Si is a geodesic. Prove in detail that 0 o y: (-e, e) - S2 is also a geodesic.

11.2'

Let S be a surface in R3 and fl a plane that intersects Sin a curve a. Show that a is a geodesic if Il is a plane of symmetry of S. that is, the two sides of S are mirror images across n. Apply this to the sphere and an ellipsoid of rotation.

11.3

Let y be a straight line segment in R3 contained in a surface S. Prove that y is a geodesic.

11.4'

Let a: (0, 11 - S2 be the curve a(t) = (cos e' , sine, 0). Show that a is a geodesic on S2, but in the latitude-longitude parametrization of S2, a does not satisfy the differential

equations (*). to be satisfied by a geodesic, not necessarily

11.5

Determine the analog of equations unit speed.

11.6

Let 11(0, w) = 3'(11(0 + ti) - 11(0) - 11(w)) denote the inner product determined by the second fundamental form. Prove the analog of the Frenet-Serret theorem for curves a in S. Use the frame (T. na, N] and show

T' = kxna + II(T)N, nQ = -ksT + II(T, na)N,

N' = - II(T)T - lI(T, na)na. The function r8(t) = lla(j)(T(t), na(t)) is called the geodesic torsion of a.

170


11.7

Suppose that S is geodesically complete. Show that for any point p E S, the mapping exp, has domain all of Ta(S).

11.8

For the reader with some knowledge of topology: Show that the topologies on a surface S as a subset in R3 and as a metric space (S, d) are the same topology.

11.9

Suppose that a surface is geodesically complete and S satisfies the property that there is some r > 0 such that, for all p, q E S, d (p, q) < r. Show that S is compact.

11.10

Suppose that St is connected and geodesically complete and S2 is connected and satisfies the property that every pair of points is joined by a unique geodesic. If ¢: S, - S2 is a local isometry. show that it is a global isometry.

12 The Gauss-Bonnet Theorem This theorem, if we mistake not, ought to be counted among the most elegant in the theory of curved surfaces. C. F. Gauss, 27 October 1827

A triangle in the plane is determined by three line segments and the region they enclose: a triangle on a surface is determined by three geodesic segments that enclose a region. In Disquisitiones generales circa superficies cun'as (Gauss 1828). Gauss deduced some of the basic properties of geodesic triangles. In particular he proved a general relation between area and angle sum. which extends the known cases of the sphere (Proposition 1.2) and the plane. In his paper of 1848, Stir la Theorie generale des Surfaces. Bonnet introduced the term for geodesic curvature kg and applied Green's theorem to prove a far-reaching generalization of Gauss's theorem for polygons whose sides are not necessarily geodesic segments. For compact, orientable surfaces the theorem of Gauss and Bonnet has remarkable topological implications. In order to state the main theorem of the chapter we need to set out a handful of assumptions. Fix an orientable surface S C R2 and suppose R C S is a region in S satisfying the following properties:

(1) The region R is simply connected in S. That is, any closed curve in R can be contracted to a point without leaving the region. Thus R is like a convex subset of S. Furthermore, the interior of R in S is connected. This condition excludes cases like the region enclosed by a figure 8. 171

172


(2) The boundary of R is given by a curve a : [0. s ] -+ S. Furthermore, a is piecewise differentiable and closed, that is, it may be written

a = (ai.a2,....an).

a,: [s,_i.s,] -+ S,

where each ai is unit speed and differentiable on its domain. Also ai (s,) =a,+ I (s') for i = 1, ... , n -1, and a i (0) = a (s,,). The curves a; (s) are the sides of a general sort of polygon lying on S. The points a, (s,) are the vertices of the polygon. At each vertex the tangent vectors a, (s,) and a;+, (s,) (along with al (0) and a;, exist and the angle between them is well defined. We call this angle the exterior angle at the ith vertex and it satisfies

cose, = Ia,c.s,t(a;(s'),a,(s,

(and cos e = (0), a;, If we assume 0 < Iei1 < ir, then the previous equation determines e, up to sign. A cusp is a vertex where a! (s,) _ -a, +i (s,) and so the exterior angle is ±7r. If a vertex is not a cusp, then we can fix the sign of e, by requiring the angle to lie between 0 and n and checking the handedness of the ordered basis [a, (si). a' t t (si), N(a, (s, ))1. If the basis is right-handed, the sign is positive; if the ordered basis is left-handed, it is negative. To fix the sign at a cusp, choose a nearby

point on ai, say a, (s, - >)) for n > 0 and close to 0. choose a nearby point on a,+1, say a,+i (s, + r)), and consider the ordered

basis[a,'(s, -n),a,+,(s,+r1),N(a,(s,))1. Since the curves are one-to-one and the sur-

face is regular, this ordered basis is well defined near the vertex and its handedness remains the same as q goes to zero. This determines the sign of n at a cusp. (3) The region R and its boundary are contained in a coordinate patch x: (U C R2) - S. Furthermore, we can take the component functions of the metric for x to satisfy F = 0, that is, the coordinate curves are orthogonal (see Corollary 10.10).

At the heart of the proof of the Gauss-Bonnet theorem is a result about piecewise differentiable curves on a surface like the boundary of the region R. Though out of place historically, this theorem due to H. Hopf (1894-1971) has a natural place in our discussion and it was quietly assumed by Gauss and Bonnet. To state the result we introduce functions ¢, : [s, _ s, ] - R along the curves a, . These functions measure how the tangent a' (s) turns around the region R. Let 0, (s) measure the angle from x,, at ai (s) to a,(s). Such a function is determined up to multiples of 27r and it satisfies the formula cosOi(s) = Ia,tsi (a;(s).

To determine the values i, (s), begin by choosing an angle to represent Ot (0). Then extend ¢t along 10. st ] by continuity. At s = st we make a jump in order to measure the

12. The Gauss-Bonnet Theorem

173

angle via the curve a2. The size of the jump, and hence the value of 02(si), is determined by the equation 02(sl) = EI + 01(sl)

Now extend h2(s) along a2(s) by continuity, and so on. In this fashion we can construct all of the functions 0i exactly. We now state the key topological result for such curves on a surface (Hopf 1983). Theorem 12.1 (Hopf's Umlaufsatz). The tangent along a closed piecewise d i ff e r e n t i a b l e curve enclosing a simply connected r e g i o n turns through 2tr, that is, 0

e_

O1(0) + 2tr. The proof of this innocent-sounding result is lengthy and involves ideas that really belong in a course on topology. In order to avoid such a long detour we postpone a sketch of the proof to the end of the chapter. For now we conclude a useful corollary that follows from the construction of the functions 0i.

Corollary 12.2. > r 0'(s) ds = 2n - E EJ. J=1

4

J=1

Under the assumption on the coordinate chart that F = 0 the Gaussian curvature takes the form (Exercise 10.3) K

2 EG (a,v

\ EG/ + 8u ( EG)/

We next obtain a local expression for the geodesic curvature along each curve ai using the functions Oi and the fact that F = 0. This convenient expression is related to the Gaussian curvature via Green's theorem.

Lemma 12.3. Along the curve ai (s),

Gudv ds

I

kg(s) = PROOF.

Let e _

2 EG

and e =

- E,Vdu ds

+

dOi

ds

be the unit-length coordinate vectors associated to

the patch x. We have assumed already that e 1 e,,. Since each ai is unit speed we obtain the following equations for the tangent and intrinsic normal: a; (s) = cos Oi We, + sin 0i (s)e,,,

na, (s) = - sin 0i (s)eu + cos 0i (s)e,,.

Taking the derivative of the tangent vector we find

a;'(s) _ -sinOi(s)

=

ieu +cos0i(s)e;, +cosOi(s) d' e +sinOis)e;,

d dOi na;(s)+cosOi(s)e;,+sinOi(s)e'v.


174

Recall the formula for geodesic curvature: kg(s) = 1a,(S)(a:'(s), na, (s)). This gives

kg(s) _ - sin 0;(s) cos0;(s)(e,,, e',) - (sing O;(s))(eu, (cost O; (s)) (e, e;) + sin O; (s) cosO, (s)(ev, e',) + d !!

.

Since e and e are unit speed and perpendicular, (e,,, e;) = 0 = (ev, e;,) and (ev. ev) -(e;,, en). Therefore, the formula simplifies to

kg(s)=(e;,,e,) + d To compute (e;,, ev) we first consider e;,:

_de,, _ d f x,, ds

e°

_

fE

ds

E du E. dv E+ 2E3/2 TS -xi' (i:5Th;

dv ( du +xuvds (x°Ods

I

xv) _

Since F = 0, F. = 0 and so

E. It follows that

du

I

(e°, ev) =

EG

(xuu- xv)

I

ds + 17M

(xuv, xv)

dv ds

dv du Gdv-Evds Tds s

I

2 EG This proves the lemma.

J kg(s)ds, the total geodesic curvature along a, the boundary

We now compute

9

J=1

of the region R. By Lemma 12.3 we have

EJ n

J

kg(s)ds=

i=) aj

i=)

f

IG

I

, 2 EG L

Ts

7,

j_) a

dp1.

(- )

Recall Green's theorem for a region W C R2 with boundary \\a simple closed curve aW:

J Pdu + Qdv w

=

jj

w

au

av /

Substituting the integral of interest, we get

G dv -

I

EG

2 Jax-'(R) 2

JJ'1'(R

= -jj

jJ.r

>

Ev du EG

(a,V ( EG)+

(R> 2 EG

u

( EG)/ dude

(av ( EG) + au

(:))

KdA. R

This observation with Lemma 12.3 proves the main theorem.

,.T

dude


175

Theorem 12.4 (the Gauss-Bonnet theorem). If R is a simply connected region in a regular surface S bounded by a piecewise differentiable curve a making exterior angles E i.

EZ, ... , e at the vertices of a, then

r kg(s)ds+Jr KdA=2n-EEi.

J=1

J°j

JR

j=1

The previous formula, due to Bonnet (1848), generalizes a formula in Gauss's Disquisitiones (1828). It contains geodesic curvature, Gaussian curvature, and exterior angles, mixing up curves, angles, and areas into a remarkable relation. We next turn to applications.

Euclid revisited II: Uniqueness of lines Gauss's version of the Gauss-Bonnet theorem is concerned with geodesic triangles. Let R = 6ABC be such a triangle with sides given by geodesic segments. The interior angles

are given by LA=n-EA,LB=n-EB,and LC=ir-EC.Thus27r-EA-EB-EC=

LA+LB+LC-tr.

Since geodesic curvature vanishes on geodesics, we have proved the following result from Disquisitiones (§20):

Corollary 12.5 (Gauss). For a geodesic triangle An, A BC on a surface,

KdA= LA+LB+LC-n. on BC

Corollary 12.6. (a) The interior angle sum of a triangle in the Euclidean plane is r. (b) On a surface of constant positive curvature, the area of a geodesic triangle is proportional to the angle excess. (c) On a surface of constant negative curvature, the area of a triangle is proportional to the angle defect.

Statement (a) follows from the vanishing of Gaussian curvature on the plane. Statements (b) and (c) follow from Corollary 12.5 and the equation for constant Gaussian curvature PROOF.

AR

KdA=KARdA=Karea(R).

176


This corollary reproves an equivalent of Postulate V for the plane (Theorem 3.6), and it recovers the formula for the area of spherical triangles (Proposition 1.2). It also leads to a condition on a surface that models non-Euclidean geometry: In order to obtain the formula for the area of a non-Euclidean triangle (Exercise 5. 10), the surface must have constant negative curvature. In fact, more can be said. In Chapter II we discussed Euclid's Postulate I and the existence of geodesics joining two distinct points in a surface. This left open the question of the uniqueness of geodesic segments joining two points. There are infinitely many geodesics joining the north and south poles of a sphere. However, this cannot happen on surfaces of nonpositive curvature.

Corollary 12.7. On a surface of nonpositive curvature, two geodesic segments yi, y2 meeting at a point p cannot meet at another point q so that the curves form the boundary of a simply connected region.

PROOF. Suppose such a figure could be formed. By the Gauss-Bonnet theorem.

11.

KdA+El +E2=2n.

Since geodesics in a given direction are unique, yj and y2 cannot be mutually tangent, so c 1 and E2 are both less than n. Since K < 0, the equation cannot be satisfied.

Putting this corollary together with the Hopf-Rinow theorem we have proved that a geodesically complete surface of nonpositive curvature satisfies Euclid's Postulate I. that is, given two points in such a surface, there is a unique geodesic joining them.

Compact surfaces Suppose that a surface S is compact, without boundary, and orientable. Compactness implies that integrals of the form f fs f d A are finite for any continuous function f: S -+ R. In particular, we can integrate the Gaussian curvature over the entire surface S. The Gauss-

Bonnet theorem tells us the answer for a reasonable subset of S. In particular, we can interpret the formula of Gauss and Bonnet as an expression for the total geodesic curvature along the boundary of such a region. We suppose further that the surface S has a triangulation,

P=(A,CSIi=1,...,m),


177

where each A; is a generalized triangle on S. that is, A, = x(DA;B;C;) and AA;B;C; is a triangle in the domain U C R2 of a coordinate chart x. Furthermore, we assume that S = LJ1 A;, and 6i fl Lj is either the empty set, a vertex, or a shared side of each triangle. The existence of triangulations of surfaces was taken for

granted in the nineteenth century. Gauss, as Director of the Gottingen Astronomical Observatory, had considerable experience doing geodetic surveys in which large geographical regions are broken up into geodesic triangles (see volume 9 of his collected works). For small enough regions of a surface, the geodetic approach of laying out geodesic triangles can lead to a triangulation. To prove that all compact surfaces have a triangulation requires the use of topological results such as the Jordan curve theorem. The first complete proof was due to Tibor Radb in 1925. The interested reader may find a proof in Moise (1977) and Doyle and Moran (1968). Using the orientation we can take each triangle to be oriented as follows: Each boundary 2A; = ai is parametrizable as a unit-speed, piecewise differentiable curve. Choose the direction of the parameter so that (a' (s), nQ, (s), N(a; (s)) is a right-handed frame at each regular point on the curve.

Associate to a triangulation P the numbers

F = the number of faces (triangles) in the collection P.

E = the number of edges in the collection P, V = the number of vertices in the collection P.

Define the Euler-Poincarc characteristic of the triangulation P as the integer Xp(S) _

V - E + F.

Proposition 12.8. if S is a compact, orientable surface without boundary, then (J KdA = 27r XP(S). J s

PROOF. From the definition of a triangulation, each triangle contributes three edges to E but each edge is shared by two triangles. Thus 3F = 2E.


178

By the Gauss-Bonnet theorem, we have the following:

fj KdAIL KdA=(iri+irz+ir3-n)+j kg(s)ds, r=i

r-i

r=i

ar

where the irj are the interior angles in each generalized triangle. Because each triangle is oriented, contributions by a given edge to EF I faj kg(s) ds cancel for adjacent triangles. It follows that the integral Er f,,,, kg(s) ds vanishes. This implies that F

AS

K d A = F_(ir i + ire + i13) - tr F. t=1

At each vertex, the interior angles sum to 27r, so the sum of all the interior angles is 27rV. Also, since 3F = 2E, we have 7r(3F - 2E) = 0, and so

AS

KdA =2rrV-nF=2trV-nF+JrOF -2E) = 2rr(V - E + F) = 27rXp(S).

Gauss referred to the integral f fs K d A as the total curvature of the surface S. Since the total curvature does not depend on the triangulation, we have proved the independence of the Euler-Poincare characteristic.

Corollary 12.9. Given two triangulations P and P' of a compact, oriented surface S without boundary, Xp(S) = Xp-(S). We can write X (S) for the value 2L

JJ AS

K d A or for V - E + F associated to any

triangulation of S. Euler introduced his formula, V - E + F = 2, in 1758 for polyhedra that are homeomorphic to a sphere. This led to a combinatorial proof that there are only five Platonic solids. In 1639, however, Rend Descartes (1596-1650) had proved a result that anticipated Euler's formula and the Gauss-Bonnet theorem. To state his result, consider a polyhedron that is homeomorphic to a sphere, such as the icosahedron or the cube. At each vertex the sum of the interior angles of the faces that meet at that vertex exceeds 27r by an amount we call the excess of the vertex. If one sums the excess of all the vertices, then the result is 4n. Descartes's proof uses spherical trigonometry. The paper in which Descartes proved this result appeared in print in 1860 and so it had no direct influence on the developments just given or on Euler's work. The correct context for Euler's formula was found by Jules Henri Poincare (1854-1912) in 1893. Using his newly defined homology theory. Poincare showed that the alternating sum V - E + F is a topological invariant of a surface. This implies its independence of the choice of triangulation. In this context the Gauss-Bonnet theorem implies the topological invariance of the total curvature. This result


179

marks the beginning of the modem study of the interplay between Geometry, Topology, and Analysis. If we deform a surface without altering its topological type, then Poincare's theorem implies that there is no change in the Euler-Poincar6 characteristic. The invariance may be interpreted as a kind of rigidity. For example, if you poke your finger into a balloon (idealized as S2), it swells on the side to decrease the Gaussian curvature away from your finger where

the curvature has increased. However, if you require that the deformation maintain the condition of constant curvature, the class of deformations is severely restricted. In the case of S2, it contains only the translations, rotations, and reflections. The Euler-Poincarr characteristic also restricts the ways a surface may lie in R3. For example, the torus T can be shown to have Euler-Poincare characteristic X (T) = 0. Since T is compact, it has a point that is furthest away from the origin. At that point, the Gaussian curvature is positive. This implies that there are points on the torus where the Gaussian curvature is negative since f fTZ K d A = 0. Thus there is no way for the torus to lie in R3 with constant curvature. We now turn to a surprising application of the Gauss-Bonnet theorem to the theory of curves in R3.

Theorem 12.10 (Jacobi 1842). Suppose a: (0, rJ -+ R3 is a unit-speed, differentiable, closed curve with a'(s) anda"(s) linearly independent for all s E [0, r]. If (T, N, B) is the Frenet-Serret frame associated to a. and the image of N on S2 is without self-intersection, then the curve N(s) divides S2 into two sets of equal area.

JJ=

PROOF. Let R denote one of the regions into which N(s) divides S2. Suppose a is oriented so that N(s) is the boundary of R with R lying to the left of N(s). By the Gauss-Bonnet theorem we have R

fg R

Here kg(s) is the geodesic curvature of the curve N(s) on S2. From this formula the theorem follows by showing faR kg(s) ds = 0. Though a(s) is parametrized by arc length, N(s) need not be. The formula in Chapter I I for the geodesic curvature of N(s) is

kg(S) = IN(s)(N'(s) x N"(s), Ns2(N(s)))/IIN'(s)113,

where Nsx (p) = p is the normal to the surface S2. By the Frenet-Serret formula (Theorem 7.5), N'(s) = -K(s)T(s) + r(s)B(s) and so IIN'(s)II = K2(s) + r2(s). Also

N"(S) = -K'(s)T(s) - K(s)T'(s) + r'(s)B(s) + r(s)B'(s) = -K'(s)T(s) - (K2(S) + r2(s))N(s) + r'(s)B(s). Putting this into our expression for geodesic curvature we get

kg(s) = (K2 +Ir2)3/2 ((-KT + rB) X (-K'T - (K2 + r2)N + r'B). N) K(S)r'(S) - K'(S)r(S) (K2(S) + r2(S))3/2


180

To compute the integral f o kg (s) d s we effect the following change of variables. The function Kr

l

arctan (-I has as derivative d Tl KT' - K'T darctan (- l = KZ +r 2 s K

Denote the arc length along N(s) by a. Then

ds = do

I

K +r

.Thus

T KT' - K'T d arctan (-> = (K2 + T2)3/2 da K

Since a(s) is a closed curve,

Jr kg(s)ds = J0

o r)

d arctan (T) da = 0.

L(o)

da

K

This proves the theorem.

We now turn to a discussion of curves and a sketch of the proof of the Hopf Umlaufsatz.

A digression on curves Closed curves play an important role in the study of geometry and topology. In this digression

we record the key properties that are needed to prove the Gauss-Bonnet theorem. These properties rely upon certain topological features of curves. Suppose a : 10, r] R2 is a closed, differentiable, unit-speed curve in the plane. Suppose further that a(s) is one-to-one and differentiable, and that a'(0) = a'(r). The Jordan curve theorem (Henle 1979) implies that a(s) encloses a bounded region. Assume that a(s) is oriented so that the enclosed region is on the left, that is, na(s) = N(a(s)) x a'(s) points inward.

The principal example of a closed curve is the unit circle S' C R2. The circle comes equipped with a mapping p: R -+ St, given by p(r) = (cos r, sin r). This mapping "winds" the line onto the circle and it has the property that around each point 9 in St. there is an open set 9 c_ U C St. called an elementary neighborhood of 9, such that p t (U) C R is a union of countably many identical (that is, diffeomorphic) copies of U. Each copy of U in p-I (U) is identified with U via the mapping p. For example, the

open set around the point (I.0) given by the angles between -n/6 and n/6 has inverse image oc

p-t ((cos0, sin 9)

1 -7r/6 < 0 < n/6(= U (2nk - (tr/6), 27rk + (,r/6)). k=-oo


p-1(U)

(=

181

A special property of this mapping is that it allows us to work on the circle or on the line when we work in the small. This leads to a factorization of a curve on S' through the mapping p. If we are given a mapping St. we can construct a lift of f, that is, a f : [a, bJ mapping f : [a, b] -* ][$ such that p o f = J'. To construct the lift, choose a point a in in with p(a) = (a). Take an elementary neighborhood U around .f (a) and consider its inverse image under f. This is an open set around a in [a. b]. By assumption there is a diffeomorphism that identifies U with the piece of p-I (U) containing a. Consider the composite of f with the diffeomorphism. The composite is defined on f-t (U) and

gives a mapping J: f-t (U) -> ]R with f = p o f. We now move along [a. bJ toward the edge of f-t (U) and extend f a bit further by the same argument. Because [a, b] is compact, the iterated extension eventually gives us the lift we want.

Notice that the tangent vector to the closed curve a(s), denoted by T(s) = a'(s). determines a differentiable mapping T : [0, rl -+ S1. Furthermore, this mapping satisfies TO = T (r). Construct a lift of T. say T : [0. r] -> R. Then p o T(0) = p o T (r) and so T(r) T(0) is a multiple of 27r.

-

Theorem 12.11. If a: [0, r]

R2 is a closed. unit-speed curve in the plane, such that a(s) is one-to-one and differentiable, and a'(0) = a'(r), and if a(s) is oriented so that the enclosed region is on the left, then ?(r) - T(0) = 27r.

(r,r)

A SKETCH OF A PROOF.

auxiliary function H : Sr b < r}. Define H by

To prove the theorem we introduce an R2. where Sr = ((a, b) I 0 < a <

of (Y) - a(x)

[la(y) - a(x)II '

H(x. y) =

T(s)

-T(0).

ifx < y and (x, y) 54 (0, r).

ifx =y=s, ifx=Dandy=r.

The mapping H is continuous on Sr. and H(s, s) = T (s). By taking the family of piecewise linear curves pictured and composing with H, we obtain a continuous deformation of the

mapping T to the mapping given by H restricted to the edges (0) x [0, r] U [0. r] x (r}. Two facts finish the proof:

Fact 1. The difference of the values at the endpoints of a lift of a closed curve is unchanged under deformation.


182

To see this define another auxiliary function

A,(s)=

H(0, 2s),

if t = 0 and 0 < s < r/2,

H(2s, r/),

if t = 0 and r/2 < s < r,

\\ Hls,l? -Ilsl, Hls \\

s+2r(

2

t -t

2

if0
l-t)1 if0
By the definition xo(s) is H restricted to the edges (0) x 10, r) U [0, r] x (r) and XI (s) is T(s). A similar argument to the one given earlier to construct a lift of a curve can be generalized to lift the entire family x,(s) to a mapping ;,(s). The difference of the values at the endpoints of the lifts, that is, A, (r) - A, (0), varies continuously overt. However, this difference is also a multiple of 2rr. Since the set of multiples of 2n is discrete, the value A,,(r) - x,(0) is constant as a function of t. This proves Fact 1.

Fact 2. The difference of the values at the endpoints of a lift of the curve gotten by restricting H to (0) x 10, r] U (0, r] x jr) is 2rr.

We prove the fact for curves a(s) with a(0) = (0, 0) and a'(0) _ (I.0) and a(s) _ (x(s), y(s)) satisfying y(s) > 0. For more general curves we can always find a value 0 < so < r for which the trace of a lies entirely on one side of the tangent line to a through a(so). Extend the domain of a to all of IR by a(s+r) = a(s). Now work with the one-to-one curve a: (so, so + r] - R2 and reparametrize it to have domain [0, r]. By a rotation and translation we can arrange the particular conditions desired. The parameter for the particular case may now be rotated back to the original curve a by a family of one-to-one curves all with the same trace as a, and the argument given in Fact I can be applied to this family to show that the difference of the values at the endpoints of the lifts remains constant.

With these assumptions, H(0, t) lies above the x-axis for 0 < t < r. Furthermore,

H(0, 0) = T(0) and H(0, r) = -T(0) and so

H1lO1xl0.rl(0) = n. A

similar argument for HIlO,rixlri gives Hllo.rlxlr)(r) - HII0.rlxirl(0) = n. Putting the contributions of each part of the curve together we get H llo.rlxlrl (r) - HlfO} xfO.ri(0) = 2n.

The general technical setting for results of this kind is the homotopy theory of closed curves. There we would speak in terms of winding numbers, covering spaces, and homotopy classes of curves. Since we will not need these ideas later, we refer the interested reader to Massey (1991) for a more thorough discussion.

If we write T(s) = (cos T(s), sin T(s)), then

T"(s) = (-T'(s) sin ?(s), T'(s) cos f (s)) = T'(s)N(s), and so the directed curvature of a is given by K(s) = T'(s). The difference in values of the lift at the endpoints can be written as the integral

T(r) - T(0) =

J0

r T'(s)ds

= I0

This value is called the total curvature of the curve a.

K(s)ds.


183

Our next task is to generalize Theorem 12.11 to a case more suitable to generalized polygons. Suppose that at is a closed, piecewise differentiable curve in the plane. That is,

where 0 = so < s, < .. < s and each a; : [s;_,, s;] -+ R2 is a one-to-one, unitspeed, differentiable curve. Then a; (s;) = a;+, (s;) and a a i (0). Assume that a is one-to-one, that is, a does not cross itself. We call the trace of a a generalized polygon in the plane, that is, a polygon made up of curved sides. At each vertex, we define an external angle as we did for surfaces at the beginning of the chapter: Suppose 0 < lei I < r and tE; = arccos(a' (s;)

(s;))

(±E =

a sign to the angle, measure from a' (s,) in a counter-clockwise direction to a'+l (s;): if the answer lies between 0 and n, the sign is positive. If the answer exceeds n by any amount, say q, then E; is given by -(i) + n/2). There is an ambiguous case that needs discussion - suppose that the vertex is a cusp, that is, a'(s;) and a,+, (s;) point in opposite directions. To settle this case as n or -n, choose points near the vertex along a; and a;+i . respectively. The tangent vectors at these points, as vectors at the origin, determine an angle to which the aforementioned convention applies. This sign does not change as we let these points converge to the vertex, and so we take that sign for the cusp. We can associate a lift of the tangent directions along this piecewise differentiable curve

as before: Choose an angle that represents Ti (0) = a, (0), say a, and let T, (0) = a. By the same argument as given before, we can extend this lift of T, (s) to T, (s, ). To cross over to a2, the angle of the tangent vector turns through the exterior angle el. Define T2(sl) = T, (s,) + El. This determines the representative for T2(s,) and so we can continue to lift along a2. Continuing in this manner, we can lift each T (s) ending with T

Proposition 12.12.

+ E = T, (0) + 27r. Construct a circle centered at each vertex V; = a; (s;) = a;+, (s;), small enough so that only portions of a; and a;+, lie in it. Denote the first point of entry into the circle along a; by Pt = a; (ti) PROOF.

and the point of departure along a;+, by Q; = a;+, (u;). As the radius of the circle goes to zero, the angle L P; V; Q; converges to e; .

Construct a curve flr within the circle at V; such that fl;: [ti, u; ] R2 is differentiable and unit speed, and has f'(t;) = a'(t;) and $'(u;) = a'+I (u;). Lifting Tp, (s) so that Tp, (t;) = T, (ti), we get that the total change along fi; of the tangent vector from t; to u, is close to e,. By construction, the curve made up of the portions of the a; outside the circles and f; in the circles is sufficiently differentiable to apply Theorem 12.11 and get a difference in the lift of the endpoints to be 27r. Since the contributions of the f; is close to the sum of the exterior angles, the lift we constructed for the piecewise differentiable curve has a difference at the endpoints close to 27r minus the contribution of fi,,. This proves the theorem.


184

Finally, we complete the proof of the Hopf Umlaufsatz (Theorem 12.1). One of the assumptions set out in the beginning of the chapter is that the region R, a subset of a regular orientable surface S, the boundary of which is a generalized polygon on S, is entirely contained in the image of a coordinate chart for S. Thus we can express R as x (W ), where W is a subset of the domain U C 1R2 of the coordinate chart x satisfying all of the assumptions that R does. The coordinate chart x induces an inner product on the tangent space at each point of U that is different from the Euclidean inner product as follows: If rri (t) = (ui (t), vi (t)) and 172(t) = (u2(t), V2 (t)) are curves in U that meet at Y11 (0) = 112(0), then the inner product between the tangents to the curves may be taken to be

(0),xuu2(0)+xv v2(0))

(i (0),

For example, the cosine of the angle between the line parallel to u-axis and the line parallel to the v-axis through a point in U is given by F(uo, vo)/ E(uo, vo)G(uo, uo). Denote the usual Euclidean inner product (the dot product) on R2 by (, )E. We can interpolate between the surface inner product and the Euclidean inner product by forming the sum (0). r12(0)), = (I - t)(ni (0), n2(0))E + r(qi (0), 172(0))s

The reader can check that, for all t E (0, 1), the function (, ), is an inner product. Suppose R2. We that the boundary of W is a piecewise differentiable and closed curve 6: (0, can form the functions Oj'(s) that lift the tangents along the curve with respect to the inner product ( , ),. Notice that the angles between xu and a '(s) may differ as t varies. When t = 0 the inner product is Euclidean and so our argument above gives the Hopf Umlaufsatz for the surface R2. As the inner product changes continuously, the functions 0,(s) change continuously, and so the difference of the values of the lift at the beginning and endpoint of the curve plus the last angle gives a continuous mapping to multiples of 2,r. Since this set is discrete, the value is constant. Thus, on the surface, the equation On(sh) + e = 0, (0) + 27r

holds, that is, the tangent along a closed, piecewise differentiable curve enclosing a simply connected region in S turns through 2,r. This last part of the proof anticipates a more general notion of surface. For some values oft, the inner product ( , ), may not correspond to the inner product at a point in some surface lying in IR3. The pair (U, (, ),) for a fixed value oft is a coordinate chart for an abstract surface, a topic we will take up in Chapter 14.

Exercises 12.1

Let T2 be the torus in IR3. Triangulate it and compute its Euler-Poincare characteristic.

12.2

Show that on the infinite cylinder S' x R C JR3, there are infinitely many geodesics joining

two points (x. y, z), (x', y', z') with z 54 z'. We know that K - 0 on the cylinder. Is this a contradiction to Corollary 12.7?


18S

123' Let R' C S be a subset of a surface with boundary 8R' a piecewise differentiable curve on S. Suppose further that R' can be triangulated so that each triangle is simply connected and lies in a coordinate chart. Show that the Gauss-Bonnet theorem for R' takes the form

> ej+ J aR i 12.4

kg(s)ds+ I

f

!R

KdA=2trX(R').

On a surface S with Gaussian curvature everywhere nonpositive, show that a closed geodesic, if one exists, cannot be the boundary of a bounded region in S.

12.5' Begin with a sphere and add g handles to it. A handle is a cylinder that is joined to the sphere by removing two disks from the sphere and glueing the cylinder on at the holes one end along one edge, the other end at the other edge. For example, when g = I one gets a surface that is homeomorphic to the torus. Show that this gluing procedure gives an orientable surface in R3 and that such a surface has Euler-Poincarf characteristic 2 - 2g. 12.6

From Euler's formula prove that there are only five Platonic solids.

12.7

The sphere S2 can be triangulated by geodesic triangles, all of them congruent to each other. Is it possible to construct an analogous tiling of the sphere using congruent geodesic quadrangles so that four such quadrangles meet at each vertex?

13 Constant-curvature surfaces I remark only that I have developed Astral Geometry so far that I can completely solve all its problems as soon as the constant k is given. C. F.

Gauss (to Gerling. 16 March 1819)

We finish the task of building a foundation for the geometry of Part I by finding the conditions a surface must satisfy in order to provide a model of the geometries of interest. We take our

cue from the metric-space properties of surfaces found in Chapter I I. In particular, it now makes sense to talk of a circle in a surface S of radius r centered at a given point p e S.

that is. the set of points q E S such that d(p, q) = r. Of course it is possible that this is an empty set for certain radii. When such a circle determines a curve on S, then we may compute its circumference as a function of the radius. Recall that the circumference of a circle of radius r is 27rr in the Euclidean plane and it takes the form 1

27r R sin (R) = 27rr - R= + 2 rk sinh

(k)

YTr1

= 2irr + -Sky +

, on a sphere of radius R (for r

< n R),

and

. in hyperbolic geometry.

The Taylor series for the non-Euclidean cases illustrates the deviation from the familiar Euclidean formula. In synthetic and Cartesian geometry circles enjoy many useful properties. We examine some of these properties and interpret them on a general surface in what follows. In particular, we can compare surfaces by computing the circumference of circles. In Chapter 10 we developed particular systems of coordinates related to the geometry of a surface. We now introduce another pair of systems that generalize rectangular and polar coordinates on the Euclidean plane. In the plane the pencil of lines through the origin leads to polar coordinates. The pencil of lines perpendicular to a given line leads to rectangular coordinates. On an arbitrary surface we consider the pencil of geodesics through a given point. The exponential map may be used to define a coordinate chart.

Suppose p is a point in a surface S. Fix a vector w of unit length in Tp(S). Let

a: (-tr, n)

Tn(S) be a smooth mapping that satisfies the property that 11a(0) 11 = I

and Ip(m, a(8)) = cos8 for all 0 E (-rr, n). The image of (-rr, rr) under the mapping a(0) is the unit circle in Tp(S) (minus one point) and a(0) parametrizes the circle according to the angle made with the fixed vector w. For some ep > 0 the exponential map is a diffcomorphism of BP(Op) C Tp(S) onto a neighborhood of p E S (Theorem 11.10). This leads to the following coordinate chart. 186

13. Constant-curvature surfaces

187

Definition 13.1. Geodesic polar coordinates are given by the coordinate chart x: (0, EP)X

(-a, n') - S. x(r, 0) = expp(ra(0)).

A geodesic circle of radius r < EP centered at p is the locus of points in S of distance r from p. Alternatively, it is the image of the set (u E Ta(S) I Ili II = r} under the exponential

map. For a fixed angle 9o, the geodesic given by y: (0, r] - S, y(s) = expp(sa(Bo)) is a geodesic radius of the geodesic circle of radius r. This definition contains the obvious generaliza-

tions of the familiar notions of a circle, a radius, and polar coordinates in the Euclidean plane. The restriction to radii of length less than ep allows us to do some analysis through the exp map. We first record an important geometric result generalizing a property of planar circles to geodesic circles. Gauss gave two different proofs of this lemma in § 15 of Disquisitiones.

Lemma 13.2 (Gauss lemma). Geodesic circles are perpendicular to their geodesic radii.

PROOF. Fix a point p in the surface S and suppose that the exponential map is a diffeo-

morphism on the open ball BEO(0p) C Tp(S). Suppose that a: (1 C R) - Tp(S) is a curve in the tangent plane satisfying Ila(t)II = r < E for all t E 1, an interval. The curve fl(u) = expp(a(t)) is a portion of a geodesic circle of radius r. Define the family of curves

0 < s < I.

a,(s) = exp,(sa(t)).

Then a,(I) = [3(t), and, for a fixed to E 1, a,o(s) is a geodesic. Our goal is to prove that for each to E I d

( dt a,0) r=rp

d

a,a(s)

ds

L) 0

.

The idea of the proof is to consider the family of curves a, (s) as a variation of a fixed geodesic a,o(s). In the proof of the fact that a shortest curve in a surface is a geodesic (Theorem 11.6), we considered a variation of the length function. In this proof we vary an energy function defined by

E(t) =

dd dcrldj(L.

This function is constant because it is related to the length of the geodesic radius from p to


188

)S(t). Taking the derivative with respect tot we obtain

0 = E'(ro) =

i

d

da, dor,

) fo (ds ' _Ws )ds

d1 1

= fo

a

at

2J

do, do,

`ds

ds

=10

2

aa,

at

l aa, Loll)

1 at ' as

as 11=10

=10

- 2f' '

s

ao, (1)

aa,

ds )ds11.10

a2o,

dt' ds)I,=rO-cat' as=) r=10 J ds s=1

=2

àlas'

L_,

d (do: dor

dor

a2a,

I

)dsl f_1p =

(aa,

o

aaro

as

S=0

at '

a20,,

asz

)1'=10 ds.

(801(0)

Hat

aa,0 (to) I

as

Since a,(0) is just the point p, the second term is 0. The first term is the inner product we wish to compute. It suffices then to show that the integral that is the last term in the penultimate computation vanishes. Now a,0(s), being a geodesic, and furthermore being a constant-speed geodesic (Propo-

sition 1 1.5), satisfies the property that d.a2 = IN, that is, the acceleration of the curve at,

points in the direction normal to the surface. Since dt1 is the tangent vector to the curve a,(so), it lies in the tangent plane to the surface and so we conclude

at1

as2

)I

=to

=0.

This proves the lemma.

Let us now consider the line element for geodesic polar coordinates. The coordinate curves are given by geodesic circles when r is constant, and geodesic rays emanating from

p when 0 is constant. By the definition of the exponential map, the curves with 0 held constant are unit-speed geodesics and so E = I. By the Gauss lemma, the coordinate curves meet orthogonally and so F = 0. Thus ds` = dr2 + G(r, 0)d02.

This is a particularly simple expression for the line element from which we can easily compute the associated apparatus of intrinsically interesting functions. It is also a general result, so it holds in a neighborhood of any point in a given surface.

/3. Constant-curvature surfaces

189

The lone function G(r, 8) left unspecified is not without restriction. In order to get at the conditions on G, recall a strange property of polar coordinates for R2; the origin is a singular point for this coordinatization of the plane. It is, however, a kind of false singularity as the change to rectangular coordinates shows. If we consider the set (0, ep) x (-Yr. n) as a subset of R2 in polar coordinates, then we can effect the change of variables from rectangular coordinates:

X(x, y) =

(/x2 + y2. arctan (y)\ . I

x

This is the usual inverse to (r, 8) r-. (r cos 8, r sin 8). However, the origin is not a singular point for rectangular coordinates. We next study what happens to geodesic polar coordinates at p if we pass to the limit as r goes to 0 and use rectangular coordinates to smooth out the singularity. Since the change of coordinates given by X is an isometry, we can recompute ds2 in x

and y: Recall that dr

-xdy -ydx and so =xdx + ydy d8 r r2

ds2 = Edx2 +2Pdxdy+Gdy2 = dr2 + G(r, 8)d82

(XdX+YdY)2+G(rO)(XdY_YdX)2 r r2 2 + G 42 dx2 + 2 ( 1 - Z ) Z dxd y +

=1

(

2 + G 42 ) d y2.

It follows from this expression and x2 + y 2 = r2 that

E- I =

2

r2

(G r28)

1) , and

r

d - I = x2

(G(r, 8)

I

)

(*)

and so x2(E - 1) = y2(G - I). Since this equation holds for any x, yin the region of definition, it follows that t - I = my2 + and C; - I = mx2 + . Substituting this into the equations (*) we find that

G(r,8) r2

= I+mr2 +

and so

G(r,8)=r+m 32r +

This implies the following conditions on the function G:

Proposition 13.3. For geodesic polar coordinates with ds2 = dr2 + G(r. 8)d82, and for any fixed angle 8o, -n < Bo < rr, we have

lim G(r, So) = 0, and lr-O im

a

ar

1


190

In finding the analytic conditions on G(r, 9) we computed a bit of the Taylor series of

(r, 9) around r = 0. The particular form of the line element leads to the following simple form of the Gaussian curvature in this coordinate patch:

-I

K=

aZ V'U(-r-,O)

(r, 9)

art

Substituting our results for the series associated to G we get the equation

-K

m; +...)=-KI-G = aZ.

=3mr+...,

arz

and so we conclude, near r = 0, that 3m = - K and r3

where K = K(p) is the Gaussian curvature at the point p. Notice that another choice of the unit vector iii which determines 9 = 0 gives another form of the polar coordinates for which the variable r remains the same. Thus these formulas do not depend on a choice of 0 along which r goes to 0. We can now compute the circumference, denoted by circump(r), of a geodesic circle of radius r centered at a point p. In geodesic polar coordinates, for fixed r < e p, the geodesic circle of radius r is the coordinate curve x(r, 9), -n < 9 < n, and its circumference is given by

G(r.9)d9

circump(r) = IF

r3

(r-K(p)6 rK(p)

+...

d9

3

In 1848, J. Bertrand (1822-1900) and V. Puiseux (1820-83) gave the following very geometric interpretation of the Gaussian curvature.

Proposition 13.4. if circump(r) is the circumference of a geodesic circle of radius r centered at the point p in a surface S, then

K(p) = lim

r-.O

6trr - 3circump(r) 7rr3

The proof follows from the formula for the circumference of a circle just given. The proof of Lemma 13.2 generalizes to other families of curves. For example, if we fix a unit-speed curve a: (-q, ill -)- S, then, in a neighborhood of a(0), we can define a family of curves Ye(s) = export, (sr(N x ap(t))) = expa(,)((sr)na(t))


191

for r < e. the bound associated with a(O) by exp in this neighborhood. The curves i H y, (so) are called geodesic parallels to the curve 01) and they are a fixed distance from a(t). The family y,,,(s) consists of geodesics that are all perpendicular to a(t). and following the proof of Lemma 13.2 with little change, they can be shown to be perpendicular to the geodesic parallels. This leads to a version of rectangular coordinates in a neighborhood of p = a(0) with the curve att) acting as the x-axis and the normal geodesics acting as the lines x = xo. The associated chart gives coordinates called geodesic parallel coordinates and the line element in this chart also takes the form ds2 = dx2 + G(x, v)d y'. When a(t ) is a geodesic, this procedure generalizes the pencil of lines perpendicular to a given line.

Euclid revisited III: Congruences While discussing geodesic circles we can interpret Euclid's Postulate III in our language of surfaces. On a surface S, suppose p e S is taken to be the center and q E S is joined to p by a geodesic segment taken to be the radius of a circle in S. When the exponential map is C Tp(S), where r > d(p. q). we can construct the circle centered at p defined on with radius pq as a geodesic circle. In general. however, this construction is restricted to small radii by the general theory of differential equations on the surface (Corollary 11.8). This restriction can be overcome by assuming that the surface is geodesically complete. Then the exponential map has all of the tangent plane T1(S) as its domain. When exp is one-to-one, the image of the set of points a fixed distance from Op in Tp(S) maps via exp to a geodesic circle of that radius. In synthetic geometry. Euclid. and later Archimedes, Gauss, Lobachevskii, and J. Bolyai, considered all circles with a given radius to be congruent. As Proposition 13.4 shows, on a general surface the circumference of a circle of radius r may depend on the Gaussian curvature at its center. This leads us to a further assumption about the surfaces we will consider as models of non-Euclidean geometry - such a surface must have constant Gaussian curvature. From the formula for the circumference of a circle in non-Euclidean geometry (Theorem 5.21) and the theorem of Bertrand and Puiseux (Proposition 13.4). we can give finally an interpretation for the mysterious constant k of non-Euclidean geometry found in the work of Gauss, Lobachevskii, and J. Bolyai. To wit, comparing Taylor series for the circumference of a circle of radius r we get

2rrksinh(r)=2rrr+3A-,

+...=2nr-n 6(p)r;+....

k

Thus 2/k2 -K(p) _ -K, a constant. This sets another condition on a surface modeling non-Euclidean geometry - it must have constant negative curvature. This is also apparent from the formulas for area given by Corollary 12.5 and the relationship between area and angle defect in non-Euclidean geometry. Using geodesic polar coordinates and the assumption of constant Gaussian curvature we can also make precise Euclid's Postulate IV. Given two right angles at different points in a surface, then near the vertices of the angles we can construct a local isometry by composing


192

geodesic polar coordinates and their inverses carrying one point to the other and the right angles onto one another by making the right angles correspond to the geodesic curves 0 = 0

and 9 = n/2. To say that the angles are congruent, however, we require that this local isometry extend to an isometry of the whole surface to itself. Thus Euclid's Postulate IV is a statement about the existence of many isometrics of the surface to itself. If for each pair of points p, q E S, and pair of unit tangent vectors i E Tp(S), w E Tq (S), there is an isometry taking p to q and ii to iuu, then we call the surface S point transitive. The development of these ideas is part of the theory of transformation groups; we will not develop these issues further, except to observe that certain surfaces important for non-Euclidean geometry have this property (see Chapter 15). Notice that a point-transitive surface is a surface of constant curvature.

The work of Minding A general question discussed in Chapter 10 is whether two surfaces can be compared geometrically; that is, given surfaces S1 and S2, when is there an isometry between them? This problem has a local version called the problem of Minding - given two surfaces Si and $z, when does dsl = dsz, where dsi is the line element on S,? A necessary condition for a solution is equality of the Gaussian curvature at corresponding points. The solution to this problem in the special case of constant-curvature surfaces is due to Minding, who also introduced geodesic curvature.

Theorem 13.5 (Minding 1839). 7ivo surfaces S1 and Sz with constant Gaussian curvatures Ki and K2 are locally isometric if and only if Ki = K2. PROOF. We have already shown (Corollary 10.5) that locally isometric surfaces have the same Gaussian curvature at corresponding points. To prove the converse, choose p E Si and q e S2 and introduce geodesic polar coordinates at each point with the same coordinates (r, 9) in (0, e) x (-it, zr). The line elements take the form

dsi = dr 2 + G 1 (r. 8)d02 andds; =dr2+G2(r,8)d02. Furthermore, the functions G i and G2 satisfy l

32,/C _ -K' = -K2 art

-

a2,/rC2

I

,2

art

a2

This is a special case of the differential equation are + K f = 0 with the initial conditions

lim,,o f = 0 and limr-.o

a

ar

= 1. where K = Ki = K2, a constant. The uniqueness

of solutions to differential equations with initial conditions implies GI = G2. The desired local isometry is the composition of expp I with the coordinates around q.

To extend our understanding of constant-curvature surfaces, let us consider particular surfaces in R3 where K = 0, K > 0, and K < 0. Minding's theorem tells us that we can


193

compare locally any two surfaces sharing the same constant curvature. A given constantcurvature surface then is a local model for all surfaces of that curvature. One route to constructing model surfaces is the differential equation to be satisfied by ,V,

art+Kf=0 such that of =0 and limo

I.

Surfaces with K = 0

G

a2

In this case

8r

=

0 and so

G(r, 0) = rfl (9) + f2(9).

The initial conditions which are independent of 9 imply that ft(0) = I and f2(0) = 0; thus G(r, 0) = r2. This is the line element for polar coordinates on the Euclidean plane, which we take as our model surface of constant zero curvature.

Surfaces with K = I/ R2 a2

The differential equation art + R2 f = 0 has the general solution f (r. 0) = A(9) cos(r/R) + B(9) sin(r/R). The initial conditions imply that

f(r,9) = G(r,9) = Rsin(r/R), and the line element becomes ds2 = dr2 + R2 sin2(r/R)d92. In Chapter 8'° we met such a line element associated to the coordinates given by

x(r, 0) = (R sin(r/R) cos 0. R sin(r/R) sin 0, Rcos(r/R)). These are polar coordinates on the sphere of radius R centered at the origin in R3, given by longitude and distance from the north pole, that is, the azimuthal projection. We already know that such a sphere is a surface of constant curvature I/ R2. By Minding's theorem we take it as the canonical local model of such a surface. A particular feature of the spheres in R3 is that they are compact surfaces and that geodesics are closed curves. If we look for those surfaces in R3 of constant Gaussian curvature that are also compact, then we find that spheres are the only examples.

Theorem 13.6 (Liebmann 1899). A compact, connected, regular surface S in R3 with constant Gaussian curvature and without singularities or boundary is a sphere. PROOF. We first observe that a compact surface has a point on it with positive curvature. To see this, consider the z-coordinate of points on the surface. This is a continuous function


194

on a compact space and so it achieves a maximum at some point p E S. At such a point of maximal height the tangent plane Tp(S) bounds the surface, that is, S lies entirely on one side of this tangent plane. Thus K(p) must be nonnegative by the characterization of the Gaussian curvature as the product of the principal curvatures. If the point were planar, then the surface would contain a line segment, and so, by completeness, a line. This would violate compactness (S is closed and bounded in R3). Thus the curvature at p is positive. We next prove that every point on our surface S is umbilic. If this were not so. then there would be points p where the principal curvatures k1 > k2 were locally a maximum and a minimum, respectively. The following lemma, due to David Hilbert. shows that this cannot happen.

Lemma 13.7 (Hllbert 1901). Suppose p is a nonumbilic point in a regular surface S such that k 1 > k2 are the principal curvatures on S. Suppose k 1 achieves a local maximum

at p and k2 achieves a local minimum at p. Then K(p) < 0.

Since p is a nonumbilic point of S. there is a coordinate patch around p with coordinate lines given by lines of curvature (Corollary 10.10). For such coordinates we PROOF.

have

k1 =

Ee

g

and k2 = G .

Furthermore, the Mainardi-Codazzi equations take the form e,,

Eve+g)= E G Ev(k1+k2). 2

- 2

Gv(e gu

2

+ G

F_

gGu

(k1 + k2)

2

Writing e = k1 F we take the partial derivative of both sides with respect to v, and get

ev = a E + k, E. Using this to simplify the Mainardi-Codazzi equations we get

2E02 -kt)

8v

Similarly, writing g = k2G, we obtain the equation aa2 = 2G (k1 - k2). Since k1 is a

local maximum and k2 a local minimum at p, we have av = 0 = as , and so, at p,

E,=Gu=0. Since the coordinate curves are perpendicular, the Gaussian curvature can be computed from the formula

K

2 EG \

(

EG ) +

a ( '))

Expanding the derivatives and simplifying we get the expression

_ K

Gu

2EG(Evv+Guu)+4EG (E + EG

+ G + EE )

Thus, at p, the Gaussian curvature takes the form K = 2EI (Evv + Guu).


195

Since k, is a local maximum and k2 a local minimum at p, the second partials of ki and k, at p satisfy a2k2

a'k-

< 0 and

av2

au

>

0.

We relate E,,,. and G,,,, at p to the principal curvatures by

E,,,, _

a2ki

2E

>0, av2 (k2-ki) -

G

_

a'-k2

2G

>0.

au2 (kj-k-) -

",

Substituting into our expression for Gaussian curvature at p, we see that K(p) < 0. To finish the proof of Liebmann's theorem, we show that if every point on S is an umbilic

point, then S is a sphere. Suppose x : (U C R2) -> S is a coordinate patch around a point p. Since every point is an umbilic point, every curve is a line of curvature, and from the definition of lines of curvature we can apply Rodrigues's formula (Exercise 9.10), which gives

d (N(a(t))) = A(a(r))a'(t) for any curve a(t) on S. In particular, for the coordinate curves we find

Iu

(N(x(u, 0))) = ,l(u.0)x,,.

dv(N(x(0, v))) = x(0. v)x,.,

that is, N = hx and N, _ Ax,.. Taking a second derivative we get the expressions N,,,, = obtain

+Ax,,,. and N,,,, _

Ax,,,,. Now x,,,, = x. and N,,,. = N,,,,, and so we

--uxr + A,,x = 0. Since x,, and x, are linearly independent, A = A, = 0 and so x is a constant. Integrating Rodrigucs's formula we get, for any curve a(t) on S, a (t)

+ zN(a(r)).

and this equation characterizes curves on a sphere of radius I /X. By connectedness, all of S lies on a sphere. Since S is without boundary, S is a sphere.

Surfaces with K = - I/ R2 Minding's theorem tells us that any model of a surface of constant negative curvature is locally isometric to any other, so we need only one model to study the geometry of such a surface near a point. For convenience fix R = I. In Chapter 9 we showed that the surface of revolution of the tractrix 0(t) = (sin(s). Intan(r/2) +cos(l)) is a surface of Gaussian

curvature K = -1.


196

The differential equation that determines geodesic polar coordinates on a surface of constant negative curvature is 3rf

R2

f = 0. This has a general solution

f(r, 9) = A(©)cosh(r/R) + B(8) sinh(r/R). The initial conditions imply that

f(r, 9) =

G(r, 0) = R sinh(r/R),

and the line element becomes ds2 = dr2 + sinh2(r)d92 when R = I. Thus, without constructing geodesic polar coordinates on the surface of revolution of the tractrix, we have obtained the line element that would result. The surface of revolution of the tractrix is called the pseudosphere in the literature of

the late nineteenth century. As we will see in later chapters, the local geometry of this surface is non-Euclidean. Trigonometry on the surface is hyperbolic trigonometry, that is, trigonometry of a sphere of radius It is also like the sphere of radius -I because its surface area is 4n.

Is this the surface that we have been seeking as a

The pseudosphere.

model of non-Euclidean geometry? In fact, the pseudosphere falls short of ourgoal - it is not complete. Above the x)-plane. we could continue the pseudosphere by reflection, but on the circle of points where z = 0, the surface is manifestly singular. Geodesics starting near this "equator' are unable to continue past it smoothly and so the surface is incomplete. We begin the next chapter by trying to correct this failure of the pseudosphere. That is, we consider complete surfaces of constant negative curvature in 1R3. The general problem of finding such a surface has a surprise ending due to David Hilbert.

Exercises 13.1

Suppose that a coordinate chart can be chosen for a neighborhood of a point in a surface

such that the coordinate lines u = constant or v = constant are geodesics. Prove that the line element can be brought into the form ds2 = due + duZ. that is, the surface is developable on the plane near the point. 13.2

Prove the theorem of Diguet (1848) that

K(p) = lim

r-O

12(,rr2 - areap(r)) 7r r4

where areap(r) is the area enclosed by a geodesic circle centered at p of radius r. (Hint: Use the formula for area given in Theorem 8.16 with geodesic polar coordinates.)


13.3

197

Prove that the tangent developable surfaces (x(s, 1) = a(s)+ta'(s))defined in Chapter 8 are locally isometric to the plane.

13.4' Prove one of the important properties of the exp mapping as follows. Suppose that p and

q are such that q = expo(w), Ilwll < e. and exp: B,(0p) --. S is a diffeomorphism. Show that the geodesic radius joining p to q is the shortest path in S. (Hint: Use the particular form of the metric in geodesic polar coordinates to show that a variation of the geodesic radius has longer length.)

13.5 A surface of revolution, x(u, v) _ (A(u) cos v, A(u) sin v, A (v)), has constant curvature K if A'2

+A'2

= I and K

For K = I the solutions to k" = -A take the form

?(u) = rt cos u + r2 sin u = r cos(u + b). If we take b = 0, solve for p (u) (you should get an elliptic integral). Describe the surfaces

that arise if r I (Wilsttp). 13.6` Prove the analog of the Gauss lemma for geodesic parallel coordinates, and obtain the associated line element ds2 = dx2 + G(x. ))d v2. What form does G(x, y) take on the sphere of radius R?

13.7' Show that the principal curvatures on a surface are differentiable as follows: Let H denote the mean curvature and K the Gaussian curvature of a regular surface. Prove the formula for H:

H- Eg-2Ff+Ge 2(EG - F2)

Show that the principal curvatures ki are computed by

k;=Hf VH2 - K 13.8

Determine the surface area of the pseudosphere. More generally, determine the surface area of the surface of revolution of a tractrix determined by dragging a weight on the end of a fixed length, say a.

PART C

Recapitulation and coda

14 11

Abstract surfaces The concept "two-dimensional manifold" or "surface" will not be associated with points in three-dimensional space; rather it will be a much more general abstract idea. Herman Weyl(1913)

In the previous chapters we showed that a surface serving as a model of non-Euclidean geometry must be geodesically complete and have constant negative curvature. We next search for surfaces in R3 satisfying these conditions. One candidate, the surface of rotation of the tractrix, has constant negative curvature, but it fails to be complete. In fact, we will find that these conditions cannot be met by any surface in R3! This leads us to consider a more abstract notion of a surface, that is, an object that need not be a subset of some Euclidean space. The material in this chapter is a bit out of the historical sequence. The development of the definition of an abstract surface between Riemann and Poincare is another story (told

in detail in Scholz (1980)); we give the modem version here. Theorem 14.4 frames our story, which picks up the historical thread again in Chapter 15. To begin, let us suppose we have a surface S C R3 of constant Gaussian curvature, K == -C2. Recall from Chapter 10 that for each point p in S there is a neighborhood of p with asymptotic lines as coordinate curves. We first show that these coordinates have some even nicer properties.

Proposition 14.1. if p is a point in a surface S C R3 of constant Gaussian curvature K- -C2, then there is a neighborhood of p with all coordinate curves unit speed and asymptotic.

PROOF. The tool of choice is the Mainardi-Codazzi equations. Suppose we consider a coor-

dinate patch, x : (-e, e) x (-q, q) - S, around p = x(0, 0) with asymptotic coordinates. At the end of Chapter 10 we showed that

fu = (EGf

F2)

(EG f

F2)

(EG - F2) + FE (EG - F2 ) + FG - G E 1

.

Furthermore, e = 0 = g and so the value of the Gaussian curvature is given by

-f2 K

EG-F2. 201

202


From K . -C2 it follows that f2 = C2(EG - F2), and so 2ff = C2(EG - F2),,. Substituting the expression for f, given above we get z

(EG - F2) + 2FEV - 2EG,,, which implies that FE = EG,,. Similarly, we obtain FG = GE from the other Mainardi-Codazzi equation. Multiplying these equations by the appropriate functions we get

F2E,. = EFG = EGE, and F2G = FGE = EGG, from which it follows that (EG - F2) E = 0 = (EG - F2)G,,. Since the surface is regular,

EG - F2 54 0 and so E = 0 = G. It follows that E = E(u) and G = G(v). We now make the change of coordinates given by

u Fa f E(t)'12dt and v i--+ 0

J0

`G(r)1j2dr.

These are just the arc-length functions of the coordinate curves and the inverse of this coordinate change composed with x has unit-speed coordinate curves in asymptotic directions.

An immediate consequence of the proposition is a neighborhood of the point p in S with line element given by

ds2 = due + 2Fdudv +dv2. Any such system of coordinates is called a Tchebychev net and it solves the problem

of finding a parametrization x: (a, b) x (c, d) -+ S of a neighborhood of a point in S with arc length as parameter along the coordinate curves. This problem was studied by P. L. Tchebychev (trèbylev 1821-94) in a paperof 1878. If one views the patch (a, b) x (c, d) as a piece of cloth with fibers parallel to the u- and v-axes, then x puts the cloth onto a patch of S without stretching the fibers. In a coordinate patch given by a Tchebychev net the Gaussian curvature takes on a simple form:

Corollary 14.2. Let S be a surface in JR3 and x : (a, b) x (c, d) -+ S a Tchebychev net on S. Let w(u, v) denote the angle between the coordinate curves x,, and x,, at (u, v). Then the Gaussian curvature satisfies

-1

"dew

sin w auav

PROOF First notice that F = cos w by the basic properties of the dot product. Since E _

G = 1, EG - F2 = I - cost w = sinew.

14. Abstract surfaces

203

We apply the formula for curvature given in the proof of theorema egregium:

a2coSw

0

auav K

--F2)2detl acosw

3Cos(0

au

I

TEG

0 1

L

sin4 w

(I - cos2 w)

a2cosw

auav

Cos w

cosw

1

+ cosw

/

acoswacoswl au

av

j

a2w aw aw aw aw sin w--sin WCOS(--+sin wCosw-auav au av au av 2

3

1

sin° w L

_ -I

I

av

2

a2w

sinw auav We next use the hypothesis that S is geodesically complete to extend a single special coordinate patch over the entire surface by a kind of analytic continuation. Proposition 14.3. Let S be a complete surface of constant negative curvature in R3. Then there is a Tchebychev net x : R2 - S. PROOF. Let po be a point in S. Suppose ! E R is the least upper bound of the set (r > 0 1

there is an asymptotic Tchebychev net x : (-r, r) x (-r, r) - S with x(0. 0) = po}. Since the Tchebychev net has coordinate curves parametrized by arc length, the images of x are open squares on S. Consider the union of all of these patches, which gives a coordinate

patch x : (-1. l) x (-!,1) -+ S. Let R denote the image of x. Consider the boundary of the subset R, denoted bdy R. It is closed, bounded, and hence compact, and by the metric

completeness of S, bdy R C S. Around each point q E bdy R there is an asymptotic Tchebychev net yq : (-Eq, Eq) X (-Eq, Eq) -+ S. This provides an open cover of bdy R. and by compactness, only finitely many of the patches, yq,, i = I, ... , n, are required to cover the boundary. Taking E to be the minimum value of the Eq,, we can extend x to a square of side I + e, which contradicts the least upper bound property of 1. Since we know at least one asymptotic Tchebychev net exists around each po. we find that the largest such coordinate chart has domain all of R2.

Hilbert's theorem We now come to the remarkable theorem due to Hilbert. Hilbert's 1901 proof utilizes the machinery of covering spaces - a worthy but lengthy detour to our story. We present a proof due to E. Holmgren from a paper that appeared in 1902.

Theorem 14.4. There is no surface in R3 of constant negative curvature. PROOF Combining Corollary 14.2 and Proposition 14.3 we know that a surface S C R3 of

constant negative curvature K = -C2 is equipped with a function w: R2 - R such that

0
a2w

= C2 sin w. FU -8V

(O)


204

Since the sine is positive over the values of w, the differential equation satisfied by w implies

that aw/au is an increasing function of v, and so for all u we have aw aw (u. 0) for v > 0. (u, v) > au all

Integration over the interval [a, b] with respect to u gives the inequality

J

n a'(u, v)du >

J

n u (u, Wit,

that is,

w(b, v) - w(a, v) > w(b, 0) - w(a, 0), for v > 0 and a < b.

(t)

Integrating the differential equation (O) with respect to v implies au 76 0 everywhere in the plane, and, in particular,

aw(0. 0) # 0. The differential equation also enjoys a certain

symmetry: If w(u, v) is a solution, then so is w(-u. -v). Switching solutions if necessary, we can assume that d u (0, 0) > 0.

Now choose three values 0 < u, < U2 < u3 such that -u (u, 0) > 0 for 0 0 we have that

w(u3, v) - w(u2, v) > e and w(uI. v) - w(0. v) > E. Suppose u, 0. It follows that

w(u,. v) < w(u, v) < w(u2, v). Now w(u,, v) > w(u,, v) - w(0. v) > E. Since w(u3. V) - w(u2, v) > E, we have w(U2, v) < W(u3, V) - E < 7r - E. Thus

E < w(u, v) < jr - E. It follows that sinw(u, v) > sine in the strip [u,, u21 x [0, 00). With these estimates we now integrate C2 sin w(u, v) over a box 1111, U21 x 10, T1. fT f'12

fT

fu2 8w

C2 sinw(u, v)dudv =

,

,

auav

dudv

= w(u2, T) - w(u,, T) - w(U2, 0) + w(u,, 0). Rearranging terms we get

n > w(u2, T) - w(ul, T) = w(u2, 0) - w(u,, 0) + v

T fu_ J 2 C2 sinw(u, v)dudv u,

> w(u2, 0) - w(u,, 0) + C2T(u2 - ut) sine.


205

However, this cannot be true for large T, so there is no function w defined on all of R2 satisfying the differential equation (O). This result puts the search for a model of non-Euclidean geometry into perspective. Our normal powers of visualization are restricted by the experience of our natural "space;' R3. It was not for lack of effort that a model of non-Euclidean geometry was not discovered it simply cannot be constructed in R3, except in a local manner. One of the tools in Hilbert's proof of Theorem 14.4 is the following interesting consequence of the existence of an asymptotic Tchebychev net.

Proposition 14.5 (J. N. Hazzidakis 1880). if S is a surface of constant negative curvature and x : (a, b) x (c, d) -a S is an asymptotic Tchebychev net, then any quadrilateral formed by coordinate curves in the image of x has area satisfying

-

area =

K

> a; - 2rr i=1

<

27r

-K

where the ai denote the interior angles of the quadrilateral.

PROOF.

Let the curvature be given by K = -C2. A quadrilateral is determined by the

sides x(ul, vj) to x(u2, v0, x(u2, VI) to X(U2. V2), x(u2, 112) to x(ui, V2), and x(ui, V2) to x(uI, vi). The formula is proved by considering the Gaussian curvature in such a patch, K = -w,,,,/ sin w, or equivalently, C2 sin w = w,,,,. Since we have a Tchebychev net,

E = G = I and F = cosw. It follows that dA = EG - F dudv = sin wdudv. rV2

C2area=J

v, V2

u2

Ju,

C2dA=J

V2

U u2

C2sinwdudv v,

u,

U

= L,ui L,u,

wuvd ud v

= w(u2, V2) -w(u2, u,) -w(ui, v2)+w(ui, vi) 4 ON

i=1

Since the interior angles ai are all less than 2r, it follows that C2area < 2n.

206


The formula of Hazzidakis may be applied to obtain another proof of Hilbert's Theorem by showing that the area of a complete surface of constant negative curvature is infinite. The formula puts the upper bound of 2n/C2 on the area of any quadrilateral of asymptotic sides, but this contradicts Proposition 14.3. In order to find a model of non-Euclidean geometry we next turn to a generalization of surfaces in R3.

Abstract surfaces The concept of a surface in 1R3 described in Chapter 8 is concrete; the definition refers to a subset of R3. The important geometric tools, such as the first and second fundamental forms, are derived from the Euclidean geometry of R3. The real workhorse for the geometry of surfaces in R3, however, is the coordinate chart. To generalize the concept of surface we focus on these constructs. Definition 14.6. A smooth abstract surface (or smooth two-dimensional manifold) is a set S equipped with a countable collection of one-to-one functions called coordinate charts or patches

A=(xa: (Ua CR2)- S;aE Al such that

(I) Ua is an

open subset of R2.

(2) Uaxa(Ua) = S. (3) If a and ,6 are in A and xa(U0) fl x# (U,6) = Vao 0 0, then the composite xah ox

:

X.6 h(Va8) - xa h(Vati)

is a smooth mapping (the transition function) between open sets of R2. The collection

A generates a maximal such set called an atlas of charts on S. That is, if x : U -, S is another chart such that xa h o x and x - o xa are smooth for all Of E A, then x is in the collection generated by A. The atlas generated by A is called a differentiable structure on S. There is a further more technical assumption about the set S with atlas generated

by A: (4) The collection of subsets determined by the x (U) C S for U in the atlas is a subbasis for a topology on the set S. which is required to satisfy the Hausdorfcondition. That

is, if p and q are in S. then there area and ,8 such that p E xa(U.). q E xq(UR). and xa(U0) fl x,6(UB) = 0. Furthermore, this topology is required to be second countable, that is, a countable collection of charts from the atlas determine the open

sets in S such that any other open set is a union of open sets from the countable collection.


207

The reader who is unacquainted with point set topology can find a discussion of the Hausdorff condition, second countability, and subbases for topologies in any good introductory topology book (e.g., Munkres (1975)). An atlas for a surface may be likened to a cartographic atlas - each coordinate chart is a page in the atlas, and on intersections of charts, there is a differentiable correspondence between pages. The abstract surface, however, need not float in space like the Earth. It is not required to be a subspace of some

particular R. ExAMPI.ES.

(1) The simplest surface is R2 with the identity chart.

(2) All of the regular surfaces in R3 we have met so far are two-dimensional manifolds.

(3) Let RIP2 denote the set of lines through the origin in R3. A set of algebraic coordinates may be defined for RP2 by tak(rx, ry, rz) ing the equivalence classes of 3-tuples, (x, y, z) whenever r 96 0. In each equivalence class there are two representatives satisfying x2 + y2 + z2 = I. If we take a coordinate chart for S2 such that xa (Ua) n -xa (Ua) = 0, then this defines a coordinate chart on RIP2 by identifying lines with their representatives on S2. The surface RIP2 with the atlas generated by these charts is called the real projective plane.

Having defined the objects of interest, we next identify the important mappings between them.

Definition 14.7. Given two surfaces S and S', a function 0: S - S' is differentiable at a point p E S if, for any coordinate charts xa : (Ua C R2) -* S around p and yy : (Vp C R2) -+ S' around f (p), the composite yi I o ¢ o xa is a differentiable mapping Ua -+ Vp, that is, y9 I o 0 o xa has continuous partial derivatives of all orders. A function 0: S - S' is differentiable if it is differentiable at everypoint p E S. A function 0: S -> S' is a diffeomorphism if 0 is differentiable, one-to-one, and onto, and has a differentiable inverse function.

Two particular collections of differentiable functions are the class of smooth curves,

A: (-e, e) -+ S, and the class of smooth, real-valued functions on S, f : S -+ R. The meaning of differentiability here is determined by composing each function with a chart or its inverse and applying the notion of differentiability between portions of R and R2. We denote the set of all smooth, real-valued functions on S by C°'°(S). More locally, if p E S, then let C°O(p) denote the smooth, real-valued functions defined on a neighborhood of p in S. Both COD(S) and COO(p) are vector spaces with a multiplication: For f, g E C°O(p), r E R, and q in a neighborhood of p where f and g are defined,

(f + g)(q) = f (q) + g(q).

(rf)(q) = r(f (q)).

(fg)(q) = f(q)g(q).

Such a structure is called an algebra over R. The relationship between curves and smooth functions allows us to define what we mean by tangent vectors for an abstract surface. Because an abstract surface may not lie in a


208

Euclidean space, the idea of a tangent vector to a curve must be based on some notion other than the "direction" of the derivative of the curve. As we saw in Chapter 8, the directional derivative also characterizes tangent vectors; this leads to the following definition.

Definition 14.8. Given a curve a: (-E, e) - S through a point p = x(0) in S, define the tangent vector to x at t = 0 as the linear mapping

x'(0): C0°(p) - R,

x'(o)(f) = d (f 01(t)) r=0

The collection of all such linear mappings for all smooth curves through p is denoted by Tp(S). the tangent space of S at p.

This generalizes the ideas from Chapter 8. However, it is difficult to make geometric sense of such a tangent space. In order to make some connection with our intuition, we prove the following result. Proposition 14.9. Tp(S) is a two-dimensional vector space. If X is a tangent vector, then

X(fg) = f(p)X(g) + g(p)X(f). that is, X satisfies the Leibniz rule for the product of functions in CO°(p). In order to proceed, we introduce coordinates, prove our results for a single chart, and then show that our results are independent of the choice of chart. Suppose x : (U C R2) -* S is a coordinate chart around p = x(0, 0). We can write a curve x: (-e, E) -+ S with a(0) = p by passing through the chart as A(t) = x(u(t). v(t)), where (u(t), v(t)) = X-1 (A(t)). In this notation, the tangent vector determined by X = x'(0) acts as follows on PROOF.

a smooth function:

x'(0)(f) =

d

d f(x(u(t), v(r)))I

= u (0) au + v'(0)

o

afdu

of dv

au dt

r-o + av di

=0

av .

We now introduce the analog of the coordinate tangent vectors xu and.rv to a surface in R3. The coordinate curves x(u. vo) and x(uo, v) are particular examples of smooth curves in S and their tangent vectors satisfy

x(u, v(,)'(f) = a-u and.x(uo, 0'(J) = av We name these particular tangent vectors as

a

au

and

a

av In the case of a more general curve a(t), we can write a

al

and so tangent vectors are given as linear combinations of the basis tangent vectors su and

av at a point p. Furthermore, any linear combination of att and a is realized as ,1'(0)

209


for a curve X by simply taking the line through (0, 0) E U of the required slope composed with x. The sum and scalar products of tangent vectors are clear from this representation and Tp(S) is a two-dimensional vector space. The Leibniz rule follows directly from this local description as it is a property of partial derivatives. Finally, if y: (V C R2) -+ S is some other coordinate patch around p, then the transition

function y-1 o x : U fl x- I (y(V )) -+ V fly 1 (x(U)) determines a change of basis for Tp(S). To see this, write x(u, v) and y(r, s) as coordinates. Then the coordinate curves for x become smooth curves in the patch determined by y and so a

au

=a

a

ar

+ba andav a =ca +a. as ar as

By the definition of a surface, this linear transformation (

a

d ) is invertible and so gives

an isomorphism of T,(/S) to itself.

In fact, the matrix I a

b I is none other than the Jacobian of the transition function

y-) o x. To see this, write the curve x(u, 0) = y(r(u), s(u)) where (r(u), s(u)) = y-1 o x(u, 0). Given a smooth function f on S, we have

a(fox) au

a(fox) _ a(fo y) ar + a(fo y) as

a(fo y) ar + a(fo y) as ar au as au'

av

ar

av

as

av'

We denote this transformation in the notation of the tangent space by a

ar

as

a

Here J(y 1 o x) denotes the Jacobian. Since x-1 o y is also differentiable, we obtain the

inverse of J(y I o x), namely, J(y'I o x)-I = J(x-) o y). Next we can describe the differential mapping associated to a differentiable function between abstract surfaces.

Definition 14.10. Given a differentiable mapping 0: S -+ S' and a point p in S, the differential of ¢ at p, d4p: Tp(S) --j- TO(p)(S), d is given by dOp(x'(0)) = Wt (0 0 X(t))I r=0

We leave it to the reader to prove that d4p is linear, and has the following local form: If x: (U C R2) -+ S is a coordinate patch around p, a(t) = x(u(t), v(t)), and


210

y: (V C R2) basis

a

a ar' as 111

S' is a coordinate patch around q(p) with coordinates y(r. s), then, in the

for T#tpt(S') we have

ara

d¢p(1'(0)) =u'(0) (au

ara

asa

asa

ar + au as + v'(U) (av ar + av as) ar ar au

=

as aal

av as av

u(o)) _ J(v o 0 o x)(a'(Op. v'(O)

The Chain Rule follows easily from this formula, that is, if 0: S -+ S' and 0: S' -+ S" are differentiable mappings, then d(* o 0)p = d 1Gm(p) o dOp

Notice that the Jacobian appears locally when we simply consider the identity mapping id: S - S. In different coordinate charts around a point, the previous computation shows that d(id),, = Ay- t o x) in local coordinates. What remains lacking in this course of generalization is the line element. For surfaces in R3. ds was induced by the usual dot product for R3. Since tangent vectors were, in fact, vectors in R3. it was easy to see that the first fundamental form was well defined and independent of the choice of coordinate chart. On an abstract surface we have considerable freedom in choosing a line element, as long as the choice is made coherently with respect to the coordinate charts. We next define the analog of the first fundamental form for abstract surfaces, a Riemannian metric, and then discuss what properties such a thing must have to be well defined.

Definition 14.11. A Riemannian metric on an abstract surface S is a choice of positivedefinite inner product (. ) p on each tangent plane, Tp(S) for P E S, such that the choice varies smoothly from point to point. In detail, we require (. )p to satisfy, for X. Y. and Z in Tp(S), and r E R.

(1) (rX + Y, Z)p = r(X, Z)p + (Y, Z)p. (2) (X. Y)p = (Y. X)p. (3) (X, X)p ? O and (X. X)p = 0 if and only if X = 0. To make precise what we mean by "varies smoothly" we examine everything locally. In a

coordinate chart x : (U C R2) - S, we have a basis for the tangent space at each point, a

a

au' av functions

The inner product is determined by linearity in terms of this basis and by the

Ea

at

au' au fp

.

F=(a a) auavp

.

G= a a )P.

avav

A Riemannian metric requires that these functions E, F, and G be smooth functions of u and v. Another requirement is that they "match" on the overlap of two patches. This is


211

made precise by showing how these component functions of the metric transform under transition functions.

Lemma 14.12. If x : (U C R2) - S and y: (V C R2) - S are two coordinate charts for S and p is a point in x(U) fl ),(V), then, at p.

F G) _ (Y-t o x)t C p P) (Y

t

o x).

where { E, F, G) are the component functions of the metric associated to x. It, F, (; ) those associated to y, and J(v t o x) is the Jacobian matrix associated to the mapping tox:x-t(x(U)fly(V))->V.

y

The proof is left to the reader. The transformational properties of the local description of a metric make it an instance of a "tensor field" on S. In Chapter 16 we will consider more fully the analysis and algebra of tensors, which form the basis of an approach to the local aspects of differential geometry for higher-dimensional manifolds. Having defined a Riemannian metric, it is not immediate that such a construct exists in general. The Riemannian metric on a surface in R3 obtains the correct properties by inheriting them from the ambient space. In order to establish the existence of Riemannian metrics for abstract surfaces, one must utilize all the topological assumptions we made in the definition. We refer the interested reader to Warner (1971) for details. EXAMPLES. (I) The flat torus: Let I = (0, 11 denote the closed unit interval in R and consider the equivalence relation on I x given by (0, v) (1, v) and (u. 0) (u. 1). The quotient space T of equivalence classes is a space homeomorphic to the usual torus T St x St. We now endow T2 = I x // with a Riemannian metric given by the geometry of the unit square as a subset of R2.

In particular, E = G = I and F = 0. This model of the torus will be shown to have constant zero curvature and so it differs significantly from the usual model of the torus in R3. which has points of positive and negative curvature. (2) Unrolling a surface of revolution: Suppose we take the graph of a single-variable function

y = f(x) > 0 over an open interval (a. b). Rotating this curve around the x-axis in R3 gives us a surface with coordinate patches

yt : (a, b) x (0,2tr) -+ S.

y2: (a, b) x (-n, tr) - S,

yt (u, v) = Y2(u, v) = (u, f(u) cos v, f (u) sin v).

The abstract surface we are interested in is given by S' = (a. b) x R with a single chart given by the identity. The Riemannian metric is given by taking any curve in S' and composing it with yt or y2 as extended by periodicity to S' and computing E. F, and G accordingly.

The result is determined by f(x):

E(u, v) = I + (J4 u))2,

F(u, v) = 0.

G(u, v) = (.f(u))2.

212


Thus we have "unrolled" the geometry of the surface of revolution onto the stripe (a. b) x R in R2. This construction has the property that it preserves the local geometry of the surface of revolution, but it changes the topology; the abstract surface gotten from unrolling the surface of revolution is simply connected and the surface in R3 is not. (3) The basis for our discussion of map projections in Chapter 811 was the longitude-latitude

coordinates on the sphere. In fact, we can view the subset (0, 2n) x (-n/2, n/2) C R2 as an abstract surface modeling the sphere by assigning the Riemannian metric ds2 = cost Odil2 + d02.

In the presence of an inner product and a notion of tangent vector to a curve, we can define the are length of a curve a : (a, b) -- S by

s(t) =

(a'(r), a'(r))a(T) dr.

1

J

and the angle 0 between curves a1 (t) and a2(t) is determined by {a I (tP), a' (t p)) p (aj UP),

(t11))p(a2(tp). c4(tp)) p .011

where ai (t,) = a2(tp) = p. The notion of area generalizes to abstract surfaces as well. Suppose R C S is a regio in S lying entirely in a coordinate chart x : (U C R2) -+ S. Then the area of R is given by the integral

f

f dA =

f f -1(R)

EG - F2dudu.

The geometric features of a surface in R3 discussed in Chapters 8 through 13 were deduced from the associated first and second fundamental forms. In the case of an abstract surface, however, there is no normal direction defined on the surface and so the second fundamental form does not exist. The work of Gauss and, in particular, his theorenw egregium free us to work without a normal direction.

Definition 14.13. Given a coordinate chart x: (U C R2) -> Sfor an abstract surface S, define the Christoffel symbols as the functions r*: : (U C R2) --> R satisfying the systems of equations

(E F ) ( I'l F

I

*F

21 E

)

( Fu - I

" E,

\ F G)

r'2 2 )

-

(FoGiGI-

r,2 )

'E

213


The Gauss-Riemann curvature is defined on the image of x as the function

-2' Ec,,,+ F,,,, - 2'G,,.

K=

G

F

den

I

(EG - F2)2

'

G

- det

0

21 E"

11 G

2 E1,

E

F

G,,

F

G

E

F. - 2' E F

F

G

2' E.

From our second proof of theorema egregium we could also have defined K as the expression 21)v

K= E

-

22)u

22

+ rl

21

+

r22 - r12r1t - r12r12)

Example (1). the flat torus, has constant zero curvature by the previous formulas. This is very different from its diffeomorphic and familiar image in R3. We next define the "lines" on an abstract surface. The way to proceed is by asking for the curves that locally minimize distance, here determined by arc length, along the curve. As in Chapter 11, this is a problem in the calculus of variations. We observe that the answer is local and so we can argue in a single coordinate chart. The argument carries over from the case of surfaces in R3 (Theorem 11.6) and we can make the following generalization.

Definition 14.14. A curve y : (-r, r) -. S is a geodesic if, for each -r < t < r, in a coordinate chart around y (t ), the following differential equation is satisfied:

u" + (U,)21-', + 2u'v'r112 + (t")2r2, = 0,

I v"+(u')2F11 +2u'dr12+(v')2r22 =0. EXAMPLES. Let us consider the flat torus. In this case, the Christof-

U

fel symbols vanish and so the geodesic equations become u" = 0 = v". In the chart for the torus, this is simply a straight Euclidean line along the torus. When the line has a rational slope, notice that the geodesic eventually rejoins itself and is closed. When the slope is irrational, the line has an image dense in the torus, that is, every point in the torus is arbitrarily close to the trajectory of the line.

When we unroll a surface of revolution onto a stripe in the plane, the Christoffel symbols become

f"(u) f(tt)f"(u) r12= 2 I

+ U (u)) 2

I(rn '

122= 1

-f'(tr)f(u) I

,

+ (f (u))

The geodesic equations become u", + (u')2

f (u)f'(u) I + (f (u)) 2

I

(u) v"+2rr''fj(u)

=0.

- 0, - (v')z 11'(u)f(r1) + (f'(u))2

I

2

214


Vertical lines, t i-+ (uo, t), are geodesics whenever uo is a critical point of f(u). By parametrizing the horizontal lines

t H (r(t), vo) such that r" = -(r')2

J'(r) f ma(r)

(I +(f'(r))2/)

we also obtain a geodesic on this abstract surface.

Because a differential equation is the basis for the definition of a geodesic, we obtain all the familiar local results about them. For example, there is a unique geodesic in each direction from a point. The exponential mapping can be defined as in Chapter 11, and geodesic polar coordinates exist. The arc length provides the structure of a metric space on an abstract surface and the topology induced by this metric space structure coincides with the topology defined by the charts (Hicks 1965, p. 80). Finally, the Hopf-Rinow theorem (Theorem 11.15) holds for abstract surfaces. To compare two abstract surfaces, each with a Riemannian metric, we define what we mean by an isometry. Definition 14.15. A one-to-one, onto, differentiable function m : S -+ S' between abstract surfaces, each with a Riemannian metric, is an isometry ifforall p e S, and V, W E Tp(S). (dOp(V ). dOp(W))m(p) = (Y. W) p.

Since an isometry induces an isomorphism of tangent planes at each point, a generalized version of the inverse function theorem for abstract manifolds tells us that an isometry is a diffeomorphism. It follows that we can use the charts from S to give charts on S' and so at a point, there are charts on each surface such that the isometry sends the associated functions E, F. and G on S to identical functions on S'. It follows that the Gauss-Riemann curvature is preserved by isometries.

For example, consider T2. the flat torus, and ask if there is a surface in R3 that is isometric to it. Since the flat torus is a quotient of a compact space, it is compact, and since the curvature is constant on it, a surface in R3 that is isometric to the flat torus would be a compact surface with constant curvature. In Chapter 13 we proved that the compact, constant curvature surfaces in R3 are spheres and so there is no surface in R3 isometric to

the flat torus. However, there is a mapping *: T2 - R4 given by (r, s) r-- (cos r, sin r, coss, sin s).

We claim that this is an isometry from T2 to the image of >y contained in S' x S1. The coordinate charts on S1 x S' are images of open intervals (a, b) x (c, d) under the same mapping >' and so >' is certainly a diffeomorphism. The natural choice of metric on St x S' is the one that is induced by the dot product on R4. We check now the first of the component functions of the metric:

E=(aau' aua) =

at

(uo.ib)

(cost, sin t, cos vo, sin vo)

dt

(cost. sin t, cos vo, sin vo)

_ (- sin r, cost, 0, 0) (- sin i, cost, 0, 0) = I = E. Similarly, F = E, and G = 6. and so

is an isometry.


215

This example identifies a condition satisfied by the surfaces of our classical discussion.

When an abstract surface S is a subset of R" for some n, then it inherits a Riemannian metric from the dot product on R". If this occurs, we say that the surface S with the inherited

Riemannian metric is embedded in R". When an abstract surface S' is diffeomorphic to a surface S embedded in R", then we say that S' has an embedding in R". Finally, if the diffeomorphism S' -- S is an isometry, we say that S' has an isometric embedding in R". Having an embedding identifies the topology of the surface S' with a surface embedded in R". Having an isometric embedding identifies the geometry of the surface S' with the geometry of a surface in R", in particular, the Riemannian structure on S' is identified via the isometry with the canonical Riemannian structure on subsets of R". By definition all of the classical surfaces in R3, if taken abstractly, have isometric embeddings in R3. The remarks above show that there is no isometric embedding of the flat torus T2 in R3. Example (3). ((0, 21r) x (-rr/2, 7r/2), ds2 = cost Odx2 +dm2)

has an isometric embedding in R3 given by the unit sphere S2. Hilbert's Theorem (14.4) shows that there is no isometric embedding of a complete, constant-negative-curvature surface in R3. Getting a little ahead of ourselves, we mention that D. Blanu§a (1955) exhibited an isometric embedding of a complete, constant-negative-curvature surface in R6. It is not known if it is possible to embed isometrically such a surface in R5. In the next chapter we reach the end of our search for a model of non-Euclidean geometry by constructing an abstract surface that is complete and has constant negative curvature.

Exercises 14.1

Suppose that A and A' are differentiable structures on an abstract surface S. Define the relation on differentiable structures by A A' if the union AtJ.A' defines a differentiable structure. Show that is an equivalence relation. Show that an atlas generated by A is the largest equivalence class under containing A.

14.2' Let V, = Ix = (XI, x2, x3) E R3 I x, 54 0) for i = 1. 2. 3. Consider the mappings

-

X2 x3

OI(x) _x1'- X1

r1

.

' 2(X) =X2-

..I r3 r2

11

.

x,

3(X) _\r3 -. XI, .

-

R2. The mappings 4'; satisfy the property 0, (.x. v. z) = defined for 4, : V1 4';(rx. rv, rz) for r E R. r ¢ 0, and so they pass to mappings defined on RIP2 - R2. The inverses of the 4'; take the form

mi

021(u,v)=lu, 1,v1,

031(u,t')=1u,v, 11.

Show that these mappings determine a differentiable structure on 14.3

Define what it means for an abstract surface to be orientable.

14.4

Prove Lemma 14.12.

RIP2.

216


14.5 Given a differentiable mapping 0: S -+ S'. prove that do, is linear. Also, prove the

chain rule, that is, if 0: S - S and * : S' - S" are differentiable mappings, then d(* a O)p = d*m(p) ° dOp 14.6

Suppose that ( , ) ( and (, )2 are two Riemannian metrics on an abstract surface S. Show that a linear combination with positive coefficients of these inner products is also a Riemannian metric on S.

15 Modeling the non-Euclidean plane In recent times the mathematical public has begun to occupy itself with some new concepts which seem to be destined, in the case they prevail, to profoundly change the entire order of classical geometry.

E. Beltrami (1868)

The notion of an abstract surface frees us to seek models of non-Euclidean geometry without the restriction of finding a subset of a Euclidean space. A set, not necessarily a subset of some 1lt", with coordinate charts and a Riemannian metric determines a geometric surface. With this new freedom we can achieve our goal of constructing a realization of the geometry of Lobachevskii, Bolyai, and Gauss. In this chapter we present the well-known models of non-Euclidean geometry due to E. Beltrami (1835-1906) and J. Henri Poincard. This chapter contains many computational details, like a lot of nineteenth-century mathematics. The foundations for these calculations lie in the previous chapters. It will be the small details that open up new vistas. We begin by considering a problem Beltrami posed and partially solved in a paper of 1865. He asked for local conditions on a pair of surfaces, Sr and S2, that guarantee that there

is a local diffeomorphism of Si - S2 such that geodesics on Si are taken to geodesics on S2. Such a mapping is called a geodesic mapping. Beltrami solved the problem when one of the surfaces is the Euclidean plane. He gave conditions for the existence of a mapping taking geodesics on a surface S to straight lines in the plane.

Theorem 15.1 (Beltrami 1865). If there is a geodesic mapping from a surface S to the Euclidean plane, then the Gaussian curvature of the surface S is constant. PROOF.

Suppose f : (W C S) -+ lR2 is a geodesic mapping, a local diffeomorphism

defined from an open set W C S to the plane. Let U C R2 be an open set lying in the image

off and take x: (U C R2) - S to be the chart given by x = f-r. Since f is a geodesic mapping, straight lines in U go to geodesics in S. Write ds2 = Edu2 + 2Fdudv + Gd v2 for the associated metric. Suppose y(t) _ (u(t), v(t)) maps to a geodesic x o y(t) on S. By assumption y(t) is part of a Euclidean line, that is,

au(t) + bv(t) + c = 0, for a and b not both zero. Since au' + bv' = 0 and au" + by" = 0 we have

(u' v'l((a u"

0) and so u'v" - u"v' = 0. 217


218

The general condition for a curve to be a geodesic is given by the differential equation (Proposition 11.5) U,

0 = det

v'

u"" + (u')2r11 + 2uv'r;2 +

)-I"12)

u" + (u')2r11 + 2u'v r12 + (v')21 , /

(2r12 - r11)(u)2v

=

+ (r22 - 2r12)u'(v')2 - r22(v')3. By choosing particular lines of various slopes in U for y(t), we obtain the relations

r11 = 0 =122.

2r12 = rl,, and 2r12 = rig

From Chapter 10 recall the Gauss equations which relate the Christoffel symbols and Gaussian curvature:

-22 = EK.

(a)

(r?) - (ri2)u + ri l riz + ril rz`z - rizri

(b)

(ri2)u - (ri 1)v + r12r12 - r11 r22 = FK.

(c)

(1722)- - (r12)v + r22r11 + ri2r;2 - r;2r;2 - r12r22 -GK.

(d)

(r12)v - (r22)u + r12 r12 - r22r1, = FK.

l

The relations associated to a geodesic mapping to the plane imply some simplifications (a)

EK =ri2ri2 - (r2l2)u

(b)

FK = (ri2)u - (2r12)v + ri2r12

(c)

GK = rizriz - (r12)1

(d)

FK = (1 2)v - (2r,12),, + r12ri2.

Subtracting (d) from (b) we get (riz). = (ri2)L,. Thus we can rewrite (b) and (d) as

FK = rizr12 - (ri2)v

(b)

(d)

FK = ri2ri2 - (r:2)u

We now use the equations (r12)u0 = (r12), and (rig),,v = (ri2)vu. Taking the partial derivative of line (a) with respect to v, and of line (b) with respect to u, we get

a(EK) av

a(FK) au and so

= E K + EKv = 2r122 (r122)u - (r122 )uu, °

= FK

r12(r12)u + r12(r12)u - (r12)uu.

EKv - FK + K(E, - Fu) = 2r12(r12)v - r12(r12)u - r12(r12),,. Substituting

from (a), (b), and (d), we find

EK - FK

K(EL, -

2r12(r12r12

- FK)

- ri2(r12r12 - EK) - r12(r12r12 - FK) K(r12E - r12F). K(E -

1 5. Modeling the non-Euclidean plane

219

Since I'11 = 0 and I'i i = 2r12, we can write (see Chapter 10)

r22F-2r 2

r 1'2E - I'2F = I'i2E + = r 12E+f 1

2

12F - ri i G

1

1G)

= Thus

EKv - FKu = 0. By computing a(au)

and obtain

FK, - G K

a(GK)

we carry out a similar derivation

= 0. In matrix form this is

E

(F G)(

KK

u)-(0)

Since EG - F2 # 0, it follows that Ku = K = 0 and so K is constant. In fact, one can show that locally there is a geodesic mapping of each of our models of surfaces of constant curvature to the plane. The case of constant zero curvature is trivial. We leave it to the reader to study the pseudosphere, the surface of revolution of the tractrix. In the case of the sphere we have already seen in Chapter 8' that central projection takes great circles to straight lines. Let us consider this mapping more analytically.

Proposition 15.2. The inverse of central projection of the lower hemisphere of a sphere of radius R centered at the origin to the plane tangent to the south pole (0, 0, - R) has the form x : IIt2 - S2,

x(u, v) =

PROOF.

R

/R2 +u +v

(u, v, -R).

Let Q = (u, v, -R) denote a point in the plane tangent to the south pole and

consider the line segment in R3 joining Q to the origin. This passes through a point P on the sphere. Write the coordinates of the pointx(u, v) = P = (r, s, t). The linear dependence of OP and OQ implies

(0,0,0) = OP x OQ = (-Rs - tv, Rr + tu, rv - su).


220

From this equation it follows that r = t implies that t =

- R2

R +u +v

u

v

and s = t . The condition r2 + s2 + t2 = R2

and the proposition follows.

From the inverse of central projection we can endow the plane with the geometry of the sphere by inducing a Riemannian metric on R2 via the mapping x : R2 -+ S2. Since the sphere is a surface in R3 we compute directly:

xy =

R

(R2

(R2 + u2 + v2)3/2 R

+ V2, -uv, Ru),

(-uv, R2 + u2, Rv).

(R2 + u2 + uz)3/z

These coordinate vectors determine the following line element on the sphere and hence on the plane in the induced metric:

ds2 = R2

(R2 + v2)du2 - 2uvdudv + (R2 + u2)dv2 (R2 + u2 + v2)2

The reader will want to show that the plane with this metric is an abstract surface of constant Gauss-Riemann curvature I / R2. Notice that as R goes to infinity, this metric converges to

du2 +

dv2.

The Beltrami disk Computing the curvature associated to the metric induced by central projection generates a morass of algebra that arrives at K(p) = I/ R2 for all p. The key feature of the calculation is that it depends only on R2 and not on R. From this observation, Beltrami made the leap R to that brings us to the first model of non-Euclidean geometry. He replaced R with obtain the line element

(v2 - R2)du2 - 2uvdudv + (u2 - R2)dv2 (-R2 + u2 + u2)2 (R2 - v2)du2 +2uvdudv + (R2 - u2)du2 = R2 (R2 - u2 - v2)2

ds2 = -R2

This formula determines a Riemannian metric on the abstract surface given by the interior

of the disk of radius R in R2 centered at (0, 0). The curvature is constant and equal to - 11R2. Furthermore, since we have only changed the constant R2, the differential equations

r;, = 0 = r22, 2r12 = r;,, and 2r12 = rig continue to hold. Thus the geodesics on this abstract surface are Euclidean line segments. Recall that the desired surface to model non-Euclidean geometry must have constant negative curvature and be geodesically complete. In fact, the surface we have just constructed is geodesically complete. However, we prove this property by a roundabout route. What

15. Modeling the non-Euclidean plane

221

follows up to the proof does not depend on geodesic completeness, so the reader may proceed with a promise that our goal has been achieved already. We now revisit the geometry of Lobachevskil and Bolyai to see that we have a concrete realization of their non-Euclidean plane. Fix the value of R as 1. Our abstract surface is the interior of the unit disk in the plane, which we denote by

YD= {(zi.v) ER2 1 u2+v2 < I). We decorate the metric with a B (for Beltrami):

ds8 -

(I - v2)due+2uvdudv+(I -u2)dv2 (1 - U2 - v2)2

Let IIDe = (11), dsB) denote the Beltrami model, that is, the abstract surface 1D with the Beltrami metric. It is immediate that Euclid's Postulate V does not hold on DB; consider the accompanying diagram, where we have pictured a point not on a given line and an infinite family of lines (that is. geodesics) through the point, each of which does not intersect the given line. To derive results of a more analytic but local nature we choose a convenient point in lB. Later argument.,, will show that these particular computations hold everywhere in 1DB. When u = 0 or

v = 0 the middle term of dsB vanishes and so the geodesics u H (u, vp) and v H (uv, v) are perpendicular to the v-axis or u-axis, respectively. At the center of the disk the axes themselves are perpendicular geodesics. Our convenient point is the origin (0, 0). To discuss distance from (0, 0) and circles centered at (0, 0) polar coordinates on D are convenient: Let

it=rcos9,v=rsin9. To transform the metric we use du = cos Odr - r sin 6d0 and d v = sin Odr + r cos 9d9. The Beltrami metric can be rewritten in the following convenient form in which to substitute the new coordinates dsB =

(I - u2 - v2)(drt22 + dv2) + (udu + vdv)2

(I - u2 - v2)2 (I -r2)(dr2+r2d92)+r'-dr'(1 r2)2

dr2

)'-d9,``

(1 -r2)2 + I -r,


222

A line through the origin in IIDB has polar equation 0 = O. a constant. If we write 0 = (0,0) and P = (r cos 00, r sin 00), then the distance in IDB, here denoted de(O, P), is given by

de(O, P) =

dt12

J

,I

=

2 In

( I ±r)

Notice that as r approaches I the distance goes to infinity. Thus lines through the origin are infinite in length. Later we will construct all of the isometrics of another model that is isometric to IDB and the previous computation will prove the geodesic completeness of both models.

By fixing r and varying 0 we get a circle of radius p = 2 In (._±L) in D. Its circumference is given by the integral IT

circum(p) = 0

rd0 1-r

-

tar 1-r

We next solve for r in terms of p:

p = 2 In

r

I

andr=

implies + r = e2p, so I + r = e2° - re2o e2p - 1

e2p + I

_

eP - e-P

eP + e-P

=tanhp.

The circumference becomes

circum(p) _

2a tanh p

= 2a sinh p.

1 - tanh2 p

This establishes the formula of Gauss (Theorem 5.21 with k = 1).

U = (1,0)

We next prove the Lobachcvskil-Bolyai theorem (Theorem 5.16) on the angle of parallelism. Choose (0. 0) as the point of intersection of the perpendicular geodesics given by


223

the u- and u-axes. Let A = (0, r) be a point a distance p = arctanh(r) from the origin. Following Saccheri and Gauss (Definition 4.6), the parallel through A to the right of the v-axis is the first nonintersecting line. This line passes through the point on the limit circle S2 = (1.0). The parallel line is determined by the line segment joining A to 12, which can be parametrized by

C1(t) = (ui(t). ri(t)) = U. r - rt). for 0 < t < 1.

This gives ui = I and v, = -r. The line from A to the origin along the v-axis is parametrized by C2(t) = (u2(t). v2(t)) = (0, r - t) with tt2 = 0 and v2 = -1. We put these data into the Beltrami metric to compute the cosine of the angle between the curves C( and C2 at A, that is, the angle of parallelism: cos

fl (p)

E(O,r)uju',+F(0.r)(uivZ+u,vi)+G(0.r)v'u;

_

1(O.r)(C,(0), C,(0))

4O.rI(Cj(0). Cj(0))

r/(I - r22)22

V(11(1 - r2)) + (r2/() - r2)2) 1/(1 - r2)2

r = r = tanh p. I -,.2+,.2 This implies the Lobachevskii-Bolyai theorem: tan

11(p)

I - cos l"I(p)

2

sin 11(p)

I - tanh(p) 1 - tanh-(p) 2e-P

= e-P (eP + e-P)2 - (eP - e-p)2

We close our discussion of the Beltrami model by establishing the theorem of Saccheri (Theorem 4.4) that non-Euclidean parallels are asymptotic. Con-

sider the point A = (0. r) and the parallel to the u-axis through A given by the line segment A S2 with

S2 = (1.0). Choose a point P = (a. 0) on the uaxis and consider the line segment joining P to the line AS2, meeting at the point Q = (a. r - ra). The parametrization of this segment given by C(1) = (a, r(1 - a)!), 0 < t < 1, allows us to compute the

length of PQ. Since u' = 0 and v' = r(1 - a) for the curve C. the arc length of PQ is given by the integral

J

G(C(!))(u')2dt =

U

=r

I

- a-

'

J0

(r(I

-

1 - a2

-a - - r2(1 - a)2!2)2

dt (

0

(I +a)-r2(l -a)!22

=

I

2

In

di

I+a+r 1 -a 1 +a - r1 -a'


224

As a goes to I, the limit of this expression is zero, that is, the length of PQ goes to zero and parallels are asymptotic. We postpone a discussion of horocycles and distance in the Beltrami disk until we have developed the isometric models of non-Euclidean geometry that bear Poincard's name.

The Poincare disk One drawback of the Beltrami disk is the representation of angles -the rays may be Euclidean

line segments, but the angles can be far from their Euclidean appearance in measure. Fix a Euclidean angle between two line segments, say n/2, and place this Euclidean figure at different points in IDa. The angle measure depends on the position of the vertex. For example, away from the u- and v-axes, the line segments u " (u, vo) and v r. (uo, v) cross at (uo, vo) in an angle determined by the function F(uo, vo). We now apply some remarkable transformations to obtain a second model of nonEuclidean geometry that has the interior of a Euclidean disk as underlying abstract surface. In this model, however, angle measurement agrees with its Euclidean representation. Of course, geodesics will fail to remain Euclidean straight lines. However, they are transformed into reasonable classical curves. We begin with another visit to the sphere. Recall from Chapter 8" that orthographic projection is a mapping from 11D C R2 to the lower hemisphere of S2 for which all lines of projection are perpendicular to the plane containing D.

Proposition 15.3. Orthographic projection of the Beltrami model IDS to the lower hemisphere of the sphere of radius one centered at (0, 0, I) is a conformal mapping. PROOF. Orthographic projection is given analytically by the mapping IIDB -. S2 + (0, 0, 1),

I-u2-v2).

(u, V) i-+ (u, v, I -

In Chapter 8 we computed the line element for the lower hemisphere of S2 C R3 with the coordinate chart associated to orthographic projection:

ds2=

(I -v2)du2+2uvdudv+(1 -u2)dy2 1-u2_v2

Comparing this with the line element on the Beltrami disk we see ds2

= (I - u2 -

By Proposition 8"'.3, the mapping is conformal.

v2)ds2B.


225

Orthographic projection takes the geodesics in De, Euclidean line segments, to semicircles on the sphere that meet the equator at right angles. Follow this mapping by stereographic projection from the north pole to the plane that is tangent to the south pole. The lower hemisphere maps to the disk of radius 2, centered at the origin,

D2=((x.v) Ix2+y' <4}. We induce a Riemannian metric on I1D2 by transfering dsB from 11 B via the diffeomorphism given by the composite of orthographic and stereographic projection. Parametrize

the lower hemisphere of the unit sphere with center at (0, 0. 1) as (u, v) r> (u, v. I 1 - u- - v ). Following Chapter 8", we find that such a point is sent by stereographic projection to (x. y. 0) given by 2u

1-rr -u-

I+ If we write w = -

2v

I+ I - u-- v-

I - u22 -_P', then it follows that

(udu+vdu)2 dw= udu+vdv anddu12 _ w'(I -rr2-v2)2 1-u--vThis implies that we can write

- rr2 - u2)(du2 +dv`) + (udu + vdv)2 ds2 (1 - u2 - v2)2 e - (I (du- + d v- + du12 ). u, Stereographic projection is given analytically by x = 12u and Y = 1

dx =

2(1 - w)du + 2udw (I - w)2

2v. w This gives

dv - 2(I - w)dv + 2vdw (l - w)2

226


It follows that

dx2+dy2 =

4

I(I -w)2(du2+dv2)+2(1-w)(udu+vdv)dw+(u22+v2)dw2].

(I -

Since w2 = I - u2 - v2 we can substitute I - w2 for u2 + v2 and -wdw for udu + vdv. 4

dx2 +dy2 = (I - w)4I(I - w)2(du2 + d v2) - 2w(l - w)dw2 + (I - w2)dw2] z

_ (I 4w)2(du2+dv2+dw2)= (14ww)2dspp. Now let ds2 = (1

R-

Since

_

x2 + y2 4

-1

- w)2(dx2 + d Y2).

4w2

_ 2w(l - w)

u2 + v2

(I-w)2-1

(1-w)2

2w

1-w,

the metric induced by the mapping on 1)2 takes the form

dx2 + dy2 x2 + y2 )2.

-

ds2R =

\I

4

The subscript R refers to Riemann; a metric of this form appears in the middle of his famous Habilitation lecture as part of a very general discussion of higher-dimensional manifolds with constant curvature. We will discuss these ideas in the next chapter. We write Dp to denote the abstract surface with Riemannian metric (D2. dsR). which is called the Poincare disk. As Dp is isometric to HDB, it too is a model of non-Euclidean geometry. Since stereographic projection is conformal and takes circles on the sphere to circles in the

plane (see Chapter 8'). the geodesics in 1,' are the images of the geodesics in D. that is, circle segments which intersect the boundary circle in right angles. The advantage of Dp over Ds is the conformal representation of angles. The metric is a multiple of the Euclidean metric and so angles between curves have the same measure as their Euclidean representation. This allows us to visualize properties of angles more directly. For example, we prove That there is a regular octagon in Dp with all right angles. From the center of the disk IiD2 send out eight rays each n/4 apart. At infinity, the angle between the line parallel to an adjacent pair of rays and each ray is zero. Consider the family of (Euclidean) circles concentric to

the limit circle. For each such circle construct the regular octagon with the intersection of the circle with the eight rays as vertices. As the circles decrease in radius, we obtain a regular octagon with increasing angles. The metric converges to the Euclidean metric at the origin, and so near the origin, the octagons are


227

nearly Euclidean, that is, the angles are close to 37r/4. By continuity there is a regular octagon for some concentric circle with all right angles. Poincare arrived at this model of non-Euclidean geometry from an entirely different direction than Beltrami. His investigations of complex differential equations, based on the work of L. I. Fuchs (1833-1902), led him to consider figures in a disk made up of arcs of circles meeting the boundary at right angles. The trick of sending such circles to straight line segments in the plane by composing stereographic projection with orthographic projection led him to a result he needed. He did not make the connection with the Beltrami disk until later in one of the most well-known accounts of mathematical discovery:

"At that moment I left Caen where I then lived, to take part in a geological expedition organized by the Ecole des Mines. The circumstances of the journey made me forget my mathematical work; arrived at Coutances we boarded an omnibus for I don't know what journey. At the moment when I put my foot on the step the idea came to me, without anything in my previous thoughts having prepared me for it; that the transformations I had made use of to define the Fuchsian functions were identical with those of non-Euclidean geometry. I did not verify this, I did not have time for it, since scarcely had I sat down in the bus than I resumed the conversation already begun, but I was entirely certain at once. On returning to Caen I verified the result at leisure to salve my conscience."

This account is often quoted in discussions of mathematical discovery and the psychology of mathematical invention. For a detailed discussion of the related mathematics see Gray (1986). In fact, Beltrami had presented this model in his 1869 paper Teoriafundamentale degli spazii di curvatura constante (Beltrami 1869). That paper was concerned with extending the results of the Saggio (Beltrami 1868) to higher dimensions in the context of Riemann's theory of manifolds. Poincare established another context, complex function theory, where the non-Euclidean plane appeared naturally and fruitfully. The popular name of the Poincare disk model is due to Poincare's efforts to bring the new geometry into the mainstream of nineteenth-century mathematics. Poincare's inspiration, around the year 1880, continued after the disk model. Further studies in complex analysis led him to another realization of the non-Euclidean plane.

The Poincare half-plane In a ground-breaking paper for many areas of mathematics (Poincare 1882), Poincare (then 28 years old) applied the methods of stcreographic and orthographic projection to provide another conformal model of the non-Euclidean plane. To construct the model, project the Beltrami disk IDB C R2 x 10) C R3 orthographically to the lower hemisphere of the sphere of radius one centered at (0, 0. I) as we did to construct I)p. Now rotate the sphere around the axis through the center parallel to the x-axis through a right angle to put the lower hemisphere in the half-space y > 0. This is accomplished by the mapping

(u, v, I -

1 - u2 - v2) H (u, 11 - u2 - v2, v + 1).


228

Finally stereographically project from the north pole. This takes the "right hemisphere" of the sphere to the upper half-plane in R2 given by H = ((u. v) v > 0). The composite mapping IIDa -> H is seen to be I

2u

(u, U) H

2

-v,

1 - 'u-- uI-u

What is the metric on H induced by this mapping? Let w = -wdw = udu + vdv, and the Beltrami metric can be expressed as dse =

12

w

I-u-u

so that

(dug + d v2 + d w2 ).

Write x = 2u and y = 2w to get

I-u

1-v

dx =

2(1

- v)du+2udv (I - v)2

anddY =

2(1 - v)dw+2wdv (I - v)2

From these expressions we get dx2 + d y2

= (I

4 u)41(1 - v)2(du2 + d w2)

+ 2(1 - v)(udu + wdw)dv + (u2 + w2)dv2J 4

_ (I - u)4

J(1 - v)2(du2 +dw2) - 2v(l - v)dv2 + (1 - v2)dv2J 2

_ (1 4v)2(du2+dv2+dw2)= (14wv)2ds2=y2ds2. Thus we obtain the induced metric dsP

=

dx2 + dy 2 Y2

. Let H now denote the abstract

surface with Riemannian metric (H, dsP), which is called the Poincare half-plane. This


229

surface is isometric to DB and so models the geometry of Lobachevskii and Bolyai. The geodesics are the images of geodesics in De, that is, either vertical lines or semicircles which meet the x-axis at right angles. Like the Poincart disk, the Poincard half-plane is a conformal model of non-Euclidean geometry. The advantage of H over the other models is its connection with complex variables. As Poincard understood, by thinking of H as the upper half of the complex plane, wellknown methods of complex analysis can be brought to bear. We next take advantage of this serendipity. We do not assume that the reader has much familiarity with complex analysis - acquaintance with the elementary properties of the complex numbers will suffice. As a result of the development of the theory of elliptic functions and modular forms the following set of transformations are central in complex analysis. Definition 15.4. Given a 2 x 2 matrix A with complex entries and nonzero determinant, the linear fractional transformation determined by A is given by

A=1 a

db

=zr-*TA(z)=az+h

cz+d*

The most basic properties of linear fractional transformations are recorded next and their proofs left to the reader (see Ahlfors 1966). Proposition 15.5. (1) If A and B are two 2 x 2 complex matrices, then TAB = TA To TB. In particular. (TA)Ì = TA_i.

(2) The following matrices and their associated linear fractional transformations generate all linear fractional transformations. That is, every T,4 can be written as a composite of the following:

4---I- z i-- z + b translation,

4--' 2 r az, fora > 0, similarity, .-- z i-' az, for Jai = 1. rotation, inversion. z

The domain of a linear fractional transformation is the set of all complex numbers except z = -d/c. By moving our discussion to the Riemann sphere, we can extend the domain to all complex numbers. The Riemann sphere is the result of mapping the complex numbers to the sphere by stereographic projection and adding the north pole as infinity. We extend a linear fractional transformation to a mapping of the Riemann sphere to itself by letting

TA(-d/c) = no and TA(oo) = a/c. In this formulation a line in the complex plane is a circle through no on the Riemann sphere and so we can talk of both circles and lines as circles. This brings us to a useful proposition:


230

Proposition 15.6. Linear fractional transformations take circles to circles.

PROOF. The equation for a circle in the complex plane centered at p with radius r is

(z - p)(2 - p) = r2, where 2 denotes the complex conjugate of z, (a + bi) = a - bi. Multiplying this equation out we get a more useful form:

AIz12 + Bz + BZ + C = 0, where A and C are real and IB12 > AC. Notice that this form includes a line as the case

A=0. To prove the proposition we merely apply Proposition 15.5 and check the generators of the set of linear fractional transformations. Clearly translation takes circles to circles. If TA(z) = az for a = peiD, then we can multiply the equation for a circle by Ja12 to get 0 = AlZ12Ia12 + B21a12 + BzlaI2 + C1a12

= A IazI2 + Baaz + Ba(az) +

C1a12,

and IBa12 = IB121a12 > ACIa12. Thus rotations and similarities take circles to circles. Finally. we check inversion: Divide the equation of a circle by IZ12 to get Clll2

II

+B-z +B-z +A=0.

Thus the equation for a circle holds with A and C reversed for the image of inversion. Since every linear fractional transformation is a composite of the generators given before, the proposition is proved.

An important property of linear fractional transformations relates them to projective geometry and gives the foundation for another approach to non-Euclidean geometry that was developed by Felix Klein (1849-1925). To state the property we need the following notion.

Definition 15.7. Given distinct complex numbers zi, z2, z3, and z4, their cross ratio is given by (ZI, z2: z3, Z4) =

ZI - Z2

Z4

ZI - Z3

Z4 -

-3 Z2

.

The cross ratio depends on the ordering of the complex numbers zj. It is left to the reader to determine the effect on the value of the cross ratio of a permutation of the values z;. We extend the cross ratio to the Riemann sphere by allowing one of the zi to equal oo: (zI. z2; z3, oo) =

zI - z2 ZI - Z3

The relation between the cross ratio and linear fractional transformations is given in the following result.


231

Proposition 15.8. Given a linear fractional transformation TA and four distinct points on the Riemann sphere zr, z2, z3, and z4, (Zr, Z2: Z3, Z4) = (TA(ZI ), TA(Z2); TA(z3), TA(Z4)).

PROOF. The Proposition follows from the simple computation: azl + b

az2 + b

czr + d

cz2 + d

_

(ad - bc)(zr - z2)

TA(zr) - TA(z2) = (czl + d)(cz2 + d)

The cross ratio determines a linear fractional transformation as follows: Given three distinct points on the Riemann sphere z2, z3, and 24, define Z - Z2 7x2.23.24(2) = (z, z2; 23, 24) =

Z4 - Z3

Z-Z3 24-Z2

In this form we can see that the linear fractional transformation Tz2,Z,,Z4 takes Z2 to 0, Z3 to oo. and z4 to 1. From this fact we can take any circle (in the extended sense) to any other circle by choosing three points on each and applying the transformation (TW2.w,.w4)-I o T.2,:3,.4. This also allows us to determine when four points lie on a circle or line. If we take three of the points to be 22, z3, and Z4, then the transformation T22.2,.%4 takes these three points to 0, oo, and 1, respectively. These are points on the extended circle that is the real line in the complex plane. The inverse of this transformation takes the real line to the circle determined by z2, z3, and 24. If any other point zr lies on this circle, then T,,13,,' ,(z1) is real. That is,

four distinct points on the Riemann sphere lie on a circle if and only if their cross ratio is real. We now concentrate on the Poincare half-plane H. The main object of study in Poincare's

paper of 1882 is the set SL2(R) of real linear fractional transformations with determinant = I; this is the group of real linear fractional transformations, TA (Z)

az + b, "'here

A=Ia

b 1 , a, b, c, d E R, and ad

d

- be = I.

Notice that any real 2 x 2 matrix with determinant positive determines an element in SL2(IR) by rescaling the/entries by the/determinant. The group SL2(R) is generated by translations, similarities (I a

` ` 0 1 and I f I, I determine the same linear fractional

transformation), and inversion. The important property of these transformations is given by the following result.

Theorem 15.9. A real linear fractional transformation in SL2(R) is an isometry of the Poincarf half-plane K Furthermore, any isometry of H is given by such a real linear fractional transformation. PROOF. We first check that these real linear fractional transformations take H to itself. If TA is in SL2(R), then

TA (u + iv) _

au+aiv+b = (au+b)+aiv (cu+d) -civ cu+civ+d (cu+d)+civ (cu+d) -civ

- (au + b)(cu + d) + acv2 +

(ad - bc)vi

(cu + d)2 + (cv)2

(cu + d)2 + (cv)2'


232

Since ad - be = 1, the transformation takes H to itself. It also takes the u-axis (the real line), the boundary of H. to itself. Since the singular point of TA, z = -d/c, is a real number, and the image of oo is also real, TA is one-to-one and onto H. To show that T4 is an isometry, we write w = TA(z), and compare the metric dsp at corresponding points. This is made especially easy by changing coordinates from (x. y) to

(z,2)where z=x+iyand2=x-iy.Then dz=dx+idyandd2=dx-idvandit follows that dx2 + dy2 = dzd2. Solving for y we get y = z 2

dsP =

dx2+dy2 y2

2i

z and so

= 4 (Zdzd2 - 2)2.

At w = TA (z), the change of coordinates takes the form a(cz+d)dz-c(az+b)dz

dw =

- (ad -bc)dz -

dz

(cz + d)2

(cz + d)2

(cz + d)2

Since a, b. c, and d are real, w = TA(z) = TA(2) and so dzu = compute

dwdtu

(w - tp)2

_

d`

(c2 + d)2

We can now

dzd2

((az+b)(c2+d) - (a22+b)(cz+d)12 dzd2 [(ad - bc)z - (ad - bc)212

dzd2

(z - 2)2

.

Thus TA is an isometry, globally defined on H. To show that the set of all such transformations, SL2 (R), contains every congruence, that is, global isometry, of the Poincare half-plane, we show that any pair of points P and Q in H and any pair of unit-speed geodesics, yi passing through P, n through Q, can be mapped by a real linear fractional transformation taking P to Q and yi onto Y2. The existence and uniqueness of geodesics through a given point and the fact that isometrics take geodesics to geodesics imply that we have constructed all of the congruences. Geodesics in H may be vertical lines or semicircles with centers on the real axis. We treat the case of a pair of semicircles and leave the other cases to the reader.

Suppose that P lies on the semicircle of radius ri and center ci on the real line, and Q lies on the semicircle of radius r2 and center c2. Let al and bi denote the intersection of the


233

first semicircle with the real line and orient (al. P, bi) in the direction of the geodesic yi through P. Likewise, orient the geodesic y2 through Q as (a2. Q, b2). We may take both geodesics oriented counterclockwise. Apply real translations to bring the centers of both semicircles to the origin. Now apply real similarities to make the radius of each semicircle unity. The points P and Q become e'm and e'" on the unit semicircle. It remains to show that there is a real linear fractional transformation taking the unit semicircle to itself and e'9 to e'". To construct such a transformation, form an intermediate transformation taking - I to 0, e'0 to i, and I to oo. Using the cross ratio, this can be done by

z+I

er0-1

I

e 'a+l

Write era = cos 9 + i sin 0 and compute

ei9-I re'B+1

(cos9-t)+isin9 (cos9+1)+isin9 _ -sin9(cos0+ 1)+sin9(cos9 - I)+i(cos29 - I + sin' 0) (cos 9 + 02 + sine 9

_ -sing cos9+I ER This shows that the intermediate transformation has determinant

2os9sin+9

c

which is posiI

five forO < 9 < n, and so it determines an element of SL2 (R). Similarly, the transformation taking - I to 0, e"? to i, and I to oo is in SL2(R). The composite of the first with the inverse of the second takes the unit semicircle to itself and e'6 to e". To map the given geodesics to one another, we carry out the aforementioned translations, similarities, and transformations in the correct order. This proves the theorem. The consequences of this theorem are many and substantial. We remark on those most relevant to our goals.

(I) Earlier in the chapter we showed that a line through the Euclidean origin of the Beltrami disk is infinite. Geodesic completeness of the three models follows from the theorem. The image of a line through the origin in ID5 is a geodesic in H. Since the transformation is an isometry, there is at least one geodesic in H that is the diffeomorphic image of a onedimensional subspace of T(o.o)(®a) under the exp map. Applying isometrics we get that

every geodesic in H is infinite, and finally, reversing the isometry to DB and following through to ®p, we see that both these models are also geodesically complete. At this point, we have completed the story of Postulate V. The existence of a geodesically complete abstract surface of constant negative curvature implies that Postulate V cannot be proved from Postulates I through IV. Euclid is vindicated in his choice to call Postulate V an assumption, and a search stretching over two millennia is ended, not with a proof, but with a new geometric universe to study.


234

(2) The simplicity of the metric on IIl allows us to find an explicit formula for distance in the Poincart half-plane. We first derive the distance between two points on a vertical geodesic

P = (a, y j) and Q = (a, Y2). The path C(t) _ (a, y i + t (.v2 - YO), 0 < t < 1. provides a parametrization of the geodesic with C'(t) _ (0, v2 - yi). The distance between the points P and Q in H is denoted dH(P. Q) and is given by

dH(P, Q) _

dx2+dy2 cut

-

I

(y2 - vi)dt

yr+t(y2-YO

y2

= I ln(yl +I (Y2 - yl))

tI

I=(ln\yi/I

It is also clear from this formula that vertical geodesics are infinite in length. Given two points that do not lie on a vertical line, say P and Q. we can determine the

semicircle with its center on the real line that passes through P and Q by a Euclidean procedure; consider the perpendicular bisector of the Euclidean line segment PQ. The intersection of this perpendicular bisector and the real line is the center of the semicircle. Suppose that semicircle has center c and radius r so that it intersects the real line at points

c-r and c+r. Consider the following transformation that takes c-rto0, P = zi = xi+iyi to 1, and c + r to 00:

T: zH i

z-(c-r) zi -(c+r) z-(c+r) zi-(c-r)

We claim that this linear fractional transformation is in SL2 (R). To establish this we compute

izt -(c+r) zt -(c-r)

i(xi -c)2+yj -r2+i(2rvi) (xi -c+r)2+ vi

(xi - c)+ivi - r

(xi -c)+ivl +r

Since zt lies on the semicircle of radius r and center c. (xj - c)2 + v = r2. and we see that i

zi - (c+r)

zi - (c - r)

4r2yt

(xi -c+r)2+yi '

=

-2rvi

2. The determinant of the transformation is

(xl -c+r)_+yi

a positive real number.

If we apply the transformation to Q we get

.z2-(c-r) zl -(c+r) T(Q)= Z2 - (c + r) zl - (c - r)' and this is i times the cross ratio of four points that lie on a circle, that is Q. maps to a point of the form iy with y = (z2, c - r; c + r, zl ). By the earlier formula for vertical lines, we get

dH(P, Q) = dH(TP, T Q) = I ln((z2, c - r: c + r, Z1 ))j In

z2-(c-r) zl-(c+r) z2 - (c + r)

zt - (c - r)/

235


This form of the distance function relates directly to the cross ratio. By manipulating the

expressions a little, one can simplify this expression to prove, when P = (xi,yi) and Q = (x2, Y2), that

ri -c+r y2

dH(P,Q)= In

(XI-C+1. yl

Another simplification comes by the fact that the cross ratio (z2. C - r: c + r. zi) is a real number (since the points lie on a circle). This number times its conjugate is simply the cross ratio squared, but the complex expression becomes

I:2-(c-r)1'- Izi -(c+r)1? _ (dE(:2.(c-r)) dE(z1,(c+r))2 Iz2 - (c + r)12 1:1 - (c - r)1 dE(:2.(c + r)) dE(zi,(c - r))) where dE(:. W) is the Euclidean distance between the points z and w as elements in 1R2. Thus dE(z1.(c+r))\11 dE(:2.(c-r))

d)HI(P,Q)= In

dE(:1.(c-r))/

.

The other models of the non-Euclidean plane are isometric to H and so we can relate the distance function on H to the distance functions on D8 and ®p. We treat the case of ID8 here and leave Dp to the reader. Suppose P and Q are points in ®8. Let the distance function on 11D8 be denoted by d8(P. Q). Because we have an isometry ¢: D8 -+ IHI, we can compute distance by

dB(P, Q) = dgg(O(P), O(Q))-

Extend the line segment PQ in IlD8 until it meets the limit circle at points A and S2. The points A and n go to the intersection of the geodesic in H through q5 (P) and 0(Q) and the real line. The formula for distance in H in terms of Euclidean distances yields

d8(P, Q) =

In

dE(O(Q).O(A))

dE(O(P). c(c2))

(dE(O(Q).O(ff)) dE(O(P).O(A))

Recall that the isometry 0 is given by

2 1-u --v

2u

- u'

1-v

Write P = (U I. vi), Q = (u2, 112), A = (uo, vo), and S2 = (uOO, voo). To compute de(P, Q) we need to simplify the following horrendous expression involving the squares of the distances: z

2112

2uo

2

+

2

I-vu}

I - u; - 112

I-vi

1-v2

z

2u2

(2141

I-

2

1

tr i - vv2

I- vi

uOO)

?'

2

2u2

(1-v2

2u, 1 -vOC,

2

+

2

1 - uZ - v?

1-v2

(2uI

1-vi

-

2uo

1-uo,

2

+

(211 - u1 - vi

l-vi

236


2 Now we can use u2 + vo = I = u + v;o, and the fact that P. Q, A, and n lie on a line, say

v = mu + b, to significantly simplify this expression (a vertical line in D8 needs another simple argument). It becomes, in fact,

/u2-up

1

dB(P,Q)=-2 Inl

Ui - Uoo\ ui - up

U2 - Uoo

With the Euclidean metric on DB, notice that

dE(Q A) = %AU2 - uo)2 + (v2 - Vol =

(m2 + I)(u2 - Uo)2.

By similar expressions for dE (Q, S2), dE (P, A), and dE (P, 12) we can form the expression

dE(Q, A)

dE(P, S2)

dE(Q, S2)

dE(P, A)

_

U2 U2

rtp

ui - uoo

Uoc

III - Up

Substituting this into the formula for distances in DB we get

dB(PQ)

In

(dE(Q.A) dE(P, S2)l dE(Q.92) dE(P, A))

This generalizes the case of lines through the origin that we computed earlier. If we treat the four points P, Q, A, and S2 as complex numbers, then we can use the fact that they lie on a line to substitute the cross ratio (Q, A; 0, P) for the expression in Euclidean distances. Therefore we can write 1

d8(P, Q) = 2I ln((Q. A; Q. P))I The presence of the cross ratio and the representation of geodesics as lines strongly suggest connections to projective geometry. These connections were made by Felix Klein in his paper of 187 I, Uber die sogenannte Nicht-Euclidische Geometrie. He based his work on an idea of Sir Arthur Cayley (1821-91) whose paper of 1859, A sixth memoir upon quantics, contained a discussion of distance on subsets of projective space. Today the Beltrami model

is most often referred to as the Klein model, or the Klein-Beltrami model, or even the Cayley-Klein-Beltrami model. The modern term for non-Euclidean geometry, hyperbolic geometry, was coined by Klein in this paper. Our emphasis on the differential geometric approach, and the dates of publication of the relevant papers of Beltrami and Klein led to our choice of nomenclature. For an exposition of the projective viewpoint see Coxeter (1968) or Klein (1928).

(3) From the convenient center of the Beltrami model with polar coordinates we were able to determine the circumference of a circle of radius p and center (0. 0). Such a circle in DB is also a Euclidean circle (a special circumstance for this center) and furthermore, orthographic projection maps such circles up to circles on the sphere. One of the special properties of stereographic projection is that it takes circles on the sphere to circles (in the extended sense) in the plane (Proposition 8'.7). Thus these circles of any radius centered at (0, 0) in D8 map isometrically to non-Euclidean circles in H which have the form of


237

Euclidean circles but with different centers. By applying an isometry of H we can move the center of a given circle to any point of H. Since linear fractional transformations take Euclidean circles to Euclidean circles, we have shown that all non-Euclidean circles in H are curves given by Euclidean circles.

Let (x - a)2 + (y - b)2 = r2 (here r > b > 0) be the equation of a circle in H. The Euclidean center, (a, b), is not the non-Euclidean center. To determine the center in H we can choose a diameter and halve it. The most convenient diameter is the vertical one, the

line ((a, y) I y > 0). Let B = (a, b - r) and T = (a, b + r) denote the bottom and top, respectively, of the circle. The line segment BT is a diameter and so the midpoint, P = (a, p), is the center of the circle as a circle in H. To compute the radius we use the distance formula:

dH(P, B) = In

(bpr)

= In

(b + r)

= dH(P, T).

This implies that p = /b2 - r and the center is found. The radius becomes

p - In

bb

-r

-r

=1n(

b+r

b-r

These equations are easily reversed to determine the non-Euclidean circle with center P = (a, p) and radius p. We leave it to the reader to prove that it is the Euclidean circle of radius psinh(p) and center (a. pcosh(p)). Similar results hold for the circles in Dp. The Beltrami model is more difficult - circles with center away from (0, 0) are not Euclidean circles. In fact, non-Euclidean circles in HB are represented by Euclidean ellipses. This is made apparent by carefully studying orthographic projection of circles on the sphere to the disk. The appearance of conics in this setting is clarified in the projective formulation of Klein (1928). (4) Another non-Euclidean object is the horocycle. Recall that this can be described as a "circle with center at infinity." One of the beauties of the models we have considered is that "infinity" is a visible place - the limit circle for DDB or Dp, and the real axis (with oo) for H. In H the simplest pencil of parallel lines is the set of vertical lines that are all parallel to each other in the "up" direction. If we choose a point P in H and take any semicircular geodesic I through P, it forms an angle a with the vertical line through

P. To find the other point on 1 that corresponds to P (see Chapter 4), we look for the vertical line that makes an angle a on the opposite side of the vertical as at P. Since our model is conformal, we can solve this question with Euclidean tools. By reflecting H across the vertical line through the "uppermost" point of 1, we take I to itself and the vertical through P to the vertical through the corresponding point. This shows that the points that correspond to P all share the same y coordinate, and so a horocycle with respect to the pencil of vertical lines is a horizontal Euclidean line.

238


Every other horocycle can be constructed by applying the isometries of H. A real linear fractional transformation takes a horizontal line to a circle with one point on the real axis (the image of oo). The picture for Dp is the same - horocycles are Euclidean circles with one point on the limit circle. The case of D8 is similar to Dp but involves Euclidean ellipses instead of circles. (5) We close our discussion of the models of non-Euclidean geometry with area. In particular, we prove Gauss's theorem (Exercise 5.10 and Corollary 12.5 - the area of a non-Euclidean triangle is given by the angle defect. The easiest model in which to manipulate triangles is the Beltrami model. Since we know all the isometries of H l and, in principle, of HD8, we only need to consider a right triangle with one vertex at the convenient point (0, 0) E D8. Set out

two copies of a right triangle, AOPQ and E OP'Q' with 0 = (0, 0), P along the u-axis,

P along the v-axis. Suppose further that OP = P'Q', PQ - OP', and OQ - OQ' and that the right angles are at P and P. At (0, 0) in D8 the measure of non-Euclidean angles coincides with the Euclidean measure since dse = due +dv2 there. Let a denote the angle measure of angles LPOQ - LP'Q'O. and let fi denote the angle measure of

LPQO = LP'OQ'.

Let the Euclidean coordinates of P and Q be given by P = (u, 0) and Q = (u. v). By extending the sides of the triangle to meet the limit circle we determine the points with which to compute the non-Euclidean distance along the legs:

-u-v +

( a=de(P,Q)=ln 2 1-u -v I

b=d8(O,P)=

(j), v

I In+u I

u u2

I

andc=de(O,Q)= 2ln

I+

-1-

u +v


u +v = tanh c, and

This implies that u = tanh b,

1 - tanh2 b, we can write cosh b =

V

1-u

239

= tanh a. Since sech2 b =

and so tank a = v cosh b.

l

u

The point P _ (0, v') is determined by I

a=dB(O,P')= IInl +V,J, sov'=tanha=vcosh b. 2 Since the non-Euclidean and Euclidean measures of angles agree at O. we can compute sines and cosines from the Euclidean right triangles: sin a =

=

v'

v

cos 16 =

u +v

v cosh b

u +u

u +u

= cosh b sin a.

Away from the origin we may use Euclidean polar coordinates. The line element ds2

transforms into

dr2

2

r2d02

ds = (I - r2)2 + I - r2 Following Chapter 14, to compute the area of A0PQ, we integrate

area(AOPQ) = ff

EG - F2drdO OP Q unhbsec0

a 0

r (l- r 2)3/2drdO

0

a O

('ll_tanh2bsec2g_l)d0;ktt5 1sins

IS" _ -a + arcsin

dl (1 - tank b) - r2 Gsin a

1 - tanh2 b

= aresin(cosh b sin a) - a = aresin(cos 0) - a

=2- -a=n-(a+Q+2 and this is the angle defect of AOPQ. Since 1)B is isometric to Dp and H, the same result applies for triangles with a vertex at the image of (0. 0). In H we can apply the elements of SL2(R) to prove the result for all right triangles in H. and hence in DB and Dp. By dividing an arbitrary triangle into right triangles we obtain the result for any triangle. We can prove many other results of non-Euclidean geometry through the analytic methods developed here. The methods of differential geometry provide the foundation on which Euclidean and non-Euclidean geometry are special cases of a vastly richer set of geometric objects, each with its own story to tell. Though we have seen an end of the long-standing question of the Parallel Postulate, we have by no means seen the end of the development of differential geometry. In the next chapter we explore a few directions that we can follow from here.


240

Exercises 15.1

Suppose there is a geodesic mapping of a surface S to the plane. Taking the inverse of the mapping as coordinates on S, show that the induced metric on the plane satisfies the differential equations:

0.

15.2

By a change of scale on the coordinates u and v, show that a solution to the differential equations given in Exercise 15.1 is the following set of metric functions where R and a are constants:

_ E 15.3

R

2

a'- + u2

uu

2

2

v2)2. F = - R (a2

(a2 + u2)

v2)2. G = R (a2

+ u2 +

(a2 + u2 +

v2)2.

+ u2 +

From the rectangular parametrization of the lower hemisphere of the sphere of radius R centered at the origin in R3 by the inverse of central projection, compute the component

functions of the metric, E. F, and G. 15.4

Show directly that the line element on 1) = 1(u, v) E R2 1 u2 + v2 <

(1 -v2)du2+2uvdudv+(I -u2)dv2 ds2 a -(I - u2 - v2)2

gives a metric for a surface of constant Gauss-Riemann curvature -1.

15.5'

In the Beltrami disk D8. let A = (0, r) and 0 = (0.0). Take the u-axis as a line through 0 not containing A. Suppose I is a line through A that does not intersect the u-axis but is not one of the two parallels through A to the u-axis. Show that ! and the u-axis share a perpendicular.

15.6

Prove the assertion for abstract surfaces made by Riemann in his celebrated Habilitationsvortrag that the metric

due + dv2

ds2 =

u2)lz

1

I + . (u2 +

determines a surface of constant curvature given by the value a. 15.7

Show that the transformation of the complex plane

is an isometry of the abstract surfaces Dp -+ H.


15.8

241

Show that the set of complex numbers satisfying the equation

Aizi2+B2+Bz+C=0, where A and Care real and IB12 >A C, forms a circle if A 96 0 and a line if A= 0.

15.9'

Derive the formula for distance in H directly as follows: We can parametrize the geodesics that are not vertical lines conveniently. Recall that cosh2 t - sinh2 t = 1. Dividing by cosh2 t and rearranging terms we get tanh2 t + sech2 t = 1. Since tanh t > 0 for all real t, the following is a parametrization of the semicircle with center c on the real axis and radius r (the so-called Weierstrass coordinates): t r-s (r tanh t + c, r sech t).

Use this parametrization and the Poincart: metric to obtain the formula for distance between two points on a semicircular geodesic. 15.10

Let zi, z2, z3, and z4 be distinct complex numbers and let w = (zi. Z2; z3. z4) denote their cross ratio. Determine, in terms of w, all of the values taken on by the cross ratio when the tour given distinct values zi, z2. Z3, and z4 are permuted.

15.11

Show directly that a circle of radius 0 < r < I in IDB centered at (0, 0) maps via the isometry 0: Ds -. H to a Euclidean circle.

15.12

Show that the non-Euclidean circle in H with center P = (a, p) and radius p is the Euclidean circle of radius p sinh(p) and center (a, p cosh(p)).

15.13

Show that a non-Euclidean circle in Dp is a Euclidean circle, and determine the center

and radius from the Euclidean data, (x -a)2 + (y-b)2 =r2for0 < r < I - (a2

+b2).

15.14' Use the analytic methods developed so far to prove the following important results in non-Euclidean trigonometry - Bolyai's theorem (Theorem 5.20): sin A sinh a

_

sin B sinh b

_

sin C

sinhc'

and the hyperbolic Pythagorean theorem (Theorem 5.19):

cosh c = cosh a cosh b.

16 Epilogue: Where from here? Alongside Euclid, who still holds pride of place. Lobachevskii and Riemann have played a particularly important role in research on the foundation of geometry, but Riemann comes before the others.

Sophus Lie (1893)

On the tenth day of June 1854 G. F. Bernhard Riemann (1826-60) delivered a lecture to the Philosophical Faculty of the University in Gottingen to fulfill the requirements of promotion to Privatdozent. As was customary, Riemann offered three possible topics for his lecture. The first two dealt with parts of his Habilitationsschrift (a second thesis also required for promotion), and the third with the foundations of geometry. Against usual practice, Gauss chose the third topic and Riemann offered the lecture Ueber die Hypothesen, welche der Geomerrie zu Grunde liegen (On the hypotheses that lie at the foundations of geometry). In it he launched the next stage of development of differential geometry. A translation of Riemann's words follows this chapter. The text of the lecture did not appear in Riemann's lifetime. Its posthumous publication in 1868 (in the 1866/67 Abhandlungen der Koniglichen Gesellschaft der Wissenschaften zu Gottingen) brought a nearly immediate response from the community of mathematicians

working on differential geometry. The period of the subject driven by fundamental elementary questions, especially non-Euclidean geometry, ends coincidentally in 1868 with the publication of Beltrami's Saggio di interpretazione della geometria non-euclidea (Beltrami 1868). Beltrami did not finish his study of non-Euclidean geometry with this paper. He left unresolved the problem of realizing the three-dimensional non-Euclidean space of Lobachevskii and Bolyai, and, in fact, he thought that such a space was impossible. Shortly

after Riemann's lecture appeared, Beltrami published an analysis of spaces of constant curvature in many dimensions (Beltrami 1869) based on Riemann's ideas and including a detailed version of non-Euclidean three-dimensional space. The reader will find that the mathematical content of Riemann's lecture is sparse and condensed. However, there was enough detail present to give direction to subsequent generations of geometers. One of the key points is the separation of the point set properties of a higher-dimensional space from its possible metric properties as we have seen for an abstract surface with or without a Riemannian metric. Many metrics may be defined on a single point set. When the point set is the space of our universe, then we see that there is no "natural" choice of the geometry of the physical world. This insight dispels the philosophical confusion over the primacy of Euclidean or non-Euclidean geometry. Riemann had steeped himself in Gauss's theory of surfaces and fully appreciated the intrinsic viewpoint; it is the cornerstone of his approach. 242

16. Epilogue: Where from here?

243

We begin our discussion of Riemann's ideas with his "concept of space:' that is, the appropriate notion of a point set for higher dimensions.

Manifolds (differential topology) What is a many-dimensional space? Riemann proposed the concept of "multiply extended quantities;' which was refined for the next fifty years and is given today by the notion of a manifold.

Definition 16.1. An n-dimensional manifold is a set M (often denoted M") equipped with a countable collection of injective functions called coordinate charts or patches

A = {xa : Ua - M: a E A) such that

(1) Each Ua is an open subset of R".

(2) Uaxa(Ua) = M. (3) If a and f3 are in A and x, (Ua) fl xp (Ut) = VaH 34 0, then the composite

X..1oxp:xR1(Vap)-*xaI(Vap) is a smooth mapping (the transition function) between open sets of R". The collection A generates a maximal such set of charts with respect to this property called an atlas of charts on M. That is, if x : U --* M is another chart such that x. I o x and .x- f o xa are smooth for all a E A, then x is in the collection generated by A. The atlas generated by A is called a (smooth) differentiable structure on M. (4) The collection of subsets {x(U) } for x in the atlas determined by A is a subbasis for a topology on the set M. which is required to satisfy the Hausdorff condition. That M is, if p and q are in M, then there are coordinate charts in the atlas y: V

and =: W - M such that p E y(V), q E z(W), and y(V) fl :(W) = 0. Finally, this topology is required to be second countable.

Here we have simply generalized the definition of an abstract surface, which is a twodimensional manifold. If P E M" is a point in the image of a coordinate chart x: (U C R") -- Al, then

we denote x-1 (p) = (X1, ... , .r" ). This superscript convention is part of the algebraic formalism of tensor analysis on which the modern local theory of manifolds is based. We will describe tensor analysis in more detail later. EXAMPLES. (1) The simplest n-dimensional manifold is R" with the atlas generated by the identity chart. More generally. the atlas containing the identity chart makes any open subset of R" an n-dimensional manifold.

(2) All of the abstract surfaces of Chapter 14. and hence all of the classical surfaces in R3 of Chapter 8, are two-dimensional manifolds.

244


(3) The unit sphere centered at the origin in R" is the set

S"-1 = ((x'.....x") E R' We define coordinate charts for the sphere as follows: Consider the open subset of R"-

E =I (Y'. ... , vn-1) I E(yt)2 < I ).

i

For each i, I < i < n, define the mappings xt : E - S"-'.

x}(y'...... y"-') = (y' . .... y'-'.

I-

yn-1).

i

The set of 2n charts (x} I I < i < n) covers S"-' and it generates an atlas making the sphere an (n - l)-dimensional manifold. (4) Let RP" denote the set of lines through the origin in R"+1. One can define algebraic , rx") coordinates for RP" through the equivalence relation (xo..... x") (rxo, whenever r # 0. In each equivalence class there are exactly two representatives such that

Ei(xi)2 = I. Coordinate charts for S" that satisfy x(U) fl (-x(U)) = 0 determine coordinate charts on RP". The n-dimensional manifold RP" with the atlas generated by these charts is called the real projective n-space. The notion of differentiable mappings between manifolds is a direct generalization of the definition for mappings between abstract surfaces, with the added richness of possibly different dimensions.

Definition 16.2. Given two manifolds M and M', of dimensions n and n', respectively,

a function 0: M - M' is differentiable at a point p E M if, for any coordinate charts xa : (Ua C R") -+ M around p and yp : (Vg C R") -- M' around f (p). the composite ,vp t o4oxa is a differentiable mapping U. -+ Vp. A function 4) : M -- M' is differentiable if it is differentiable at every point p E M. A function 0: M -> M' is a diffeomorphism if 0 is differentiable, one-to-one, and onto, and has a differentiable inverse function. Denote the set of all smooth, real-valued functions on M by CO0(M). Locally, if p E M. then let CO' (p) denote the set of smooth, real-valued functions defined in a neighborhood of pin M. Both COO(M) and CO0 (p) are real algebras via the formulas: If q is in a neighborhood

of p on which f and g are defined, then (f + g)(q) = f(q) + g(q),

(rf)(q) = r(f(q)),

(fg)(q) = f(q)g(q)

The definition of tangent vectors and tangent space is the same as for abstract surfaces.

Definition 16.3. Given a curve A: (-e. e) - M through a point p = A(0) in M, define the tangent vector to A at t = 0 as the linear mapping A'(0): C00(p) -+ R,

A'(0)(f) =

dt (f o

11=0


245

The collection of all such linear mappings for all smooth curves through p is denoted by Tp(M), the tangent space of M at p.

Proposition 16.4. TT(M) is an n -dimensional vector space. Furthermore, if X E Tp(M)

and f, g E C°D(p), then X(fg) = f(p)X(g) +g(p)X(f), that is, X satisfies the Leibniz rule for the product of Junctions in C' (p). A basis for the tangent space Tp(M) is given by the coordinate curves associated to a chart whose tangent vectors can be expressed as the partial derivative operators a l

-, i = 1,...,n au'

Definition 16.5. Given a differentiable mapping 0: M -+ M' and a point p in M, the differential of 0 at p, dOp: Tp(M) --> Tmcpt(M'),

is given by d0p(A'(0)) =

dt

(0 o MO) i=0

As in the case of surfaces, dOp is linear, and has the following local form: If x : (U C

R") -. M is a coordinate patch around p, x(t) = x(x1(t),....x"(t)), and y: (V C R") --, M' is a coordinate patch around O(p) with coordinates y(y', .... Y"). then

d0p(A'(0)) _ E(x')'(0)a(y i./

o8x0ox)i

8a .

It follows from the linearity and invertibility of the differential associated to a diffeomor-

phism m: M -- M' that dim M = dim M'. (5) A source of manifolds with beautiful properties is the family of groups of matrices. In particular, we may consider the collection of all n x n matrices with real entries as the space R"2. The determinant mapping det : Rn2 - R is smooth and the inverse image of the regular value 0 is a closed subset of R"2. The complement of det- 1 ((0)) is an n2-dimensional manifold, GL" (R). The symmetric n x n matrices, that is, those that satisfy R'2 A = A'. form a subset of of dimension n(n + 1)/2 obtained by mapping Rn("+1)l2 to R"2 as the entries along the diagonal and above the diagonal of a symmetric matrix. This determines a subspace of the manifold GL, (R) of invertible symmetric matrices. Define the mapping F: GL"(R) -> GL"(R) by F(A) = AA'. This mapping is smooth, and it takes a matrix and makes it symmetric since (A')' = A and (AB)' = B'A'. In the discussion at the end of Chapter 7, we proved that the group O(n) of rigid motions of R' is made up of invertible n x n matrices satisfying A-1 = A' (Proposition 7.12). This condition allows us to express O(n) as F-1(1) where I is the identity matrix. To determine the dimension EXAMPLES.

of O(n) we look at the differential dFA: T4(GL"(R)) -> Tj(GL"(R)) for A E O(n). Since F is constant along the O(n) directions, TA(O(n)) lies in the kernel of d FA. Since F(GL"(R)) lies in the symmetric matrices, the image of the tangent space has dimension at most n(n + 1)/2. By applying a version of the implicit function theorem (Chapter 8) one

246


can prove that the dimension of O(n) is as large as it can be, that is, n(n - 1)/2, which is n2-n(n+ 1)/2, the dimension of the kernel ofd FA. Thus O(n) is an n(n- 1)/2-dimensional manifold.

(6) Given a manifold M" we can associate to it a 2n-dimensional manifold by "gluing together" all of the tangent spaces of M: Let TM = UPEM TT(M). A point in TM has the form (p, X) where p e M and X E Tp(M). If x : (U C R") -+ M is a coordinate chart containing p, then we can define a chart for TM containing (p, X) by letting i : (U x R" C

R2') - TM be given by

(x(xI

a a

n),

Z

.

ax'

To prove that we have a manifold, we check the transition functions between charts. If y: (V C R") -- M is another chart containing p, then we can write p = x(x 1 , ... , x") _ y(y1, ... , y") and x' = xi (y1.... , y"), so

a

n = J'+ L-j

a

J=1 ax' ayj

axi

The effect on the coordi-

nates for the tangent space is given by

x-1oY(Y...... y".Z1.....z")

X"....x"

Z'ax'

''ax" 3'

The Jacobian of this transformation is given by 0

J(.r-o},)=fJ(.r-toy)

H

J(x

O y) J

where H denotes the matrix

82xj [ice

Y

The determinant of the Jacobian J(.1 -1 o v) is (det J(x-1 o y))2, which is nonzero, and so the transition functions are smooth and nondegenerate. Thus we have shown that T U is a 2n-dimensional manifold.

The manifold TM is equipped with a smooth mapping A : TM -+ M given by rr (p. X) = p. We call this mapping n : TM - M the tangent bundle of M. The manifold TM is also called the tangent bundle with the associated mapping it understood. Let M = R" and consider the tangent bundle TR". Because R" has an atlas generated by a single chart, TR" may be expressed as R" x R", that is, a product. This simple form does not hold for most n-dimensional manifolds. The failure is often a consequence of the topology of the manifold. The study of the topological properties of manifolds is the thrust of differential topology, which has developed into an important area of mathematics in the century after Riemann's lecture appeared (see Dubrovin et al. (1984,1985,1990) or Guillemin and Pollack (1974)).


247

Vector and tensor fields With the tangent bundle, n : TM -+ M, we can define vector fields in a simple manner.

Definition 16.6. A differentiable vector field on a manifold M is a smooth mapping X : M --+ T M satisfying n o X = idM, that is, X (p) = (p, v) for some v E Tp(M). The economy of language is clear - a vector field is a differentiable mapping of manifolds with a certain property. In order to describe a vector field locally, we choose a coor-

dinate patch and write X(p) = (p, v(p)). Then v(p) _

i-

a'

ax,

for functions ai that

depend on p. Equivalently, if we choose a set of functions (at: U -+ 1t I < i < n) on each coordinate chart x: U -+ M and require on the overlap of charts that the sets of I

functions transform into one another coherently, then we have determined a vector field. The transformation rule for coherence is derived as follows: For two coordinate patches

x: U -+ M and y: V -+ M we may write

X(P) _

i

a1(p)ax, =

j

av(P)aaY;

where a''U : U -. IR and aiV : V -+ )R are differentiable functions. By the chain rule we have

a

ax i

=

aYj a

axi yaj' and so, for all j, a`' =

aY' > a' axi .

This is the desired transformation rule and it is the defining property for a (o)-tensor. The sums Fi a' (p)a/axi, one for each coordinate chart x: (U C IR") -+ M. form the (o)-tensor field (or vector field). Notice that it is the analytic structure, the chain rule, that determines the transformation rule, not the geometry. The coordinate charts provide enough analytic structure on a manifold to speak locally of derivatives, vector fields, and integrals. In a given coordinate chart, such expressions may be convenient to manipulate while being meaningless globally due to a lack of coherence on overlapping charts. The problem we consider next is how to decide when a collection of analytic expressions is not dependent on the choice of coordinates. We first consider some examples: (1) Dual to vector fields are the expressions that appear in line integrals, that is, locally, Oidx'. A change of variables leads to a transformation rule:

E9udx' = >Brdyt

implies

8' = L9vax Y

i

This follows from the rule dx' _ Ek ax, dyk, familiar from the calculus. A collection of sums >i BUdx', one for each coordinate chart, is called a (?)-tensor field

248


(or a 1-form) on M, if the coefficient functions on overlapping charts are related by the transformation rule.

This example is important to the calculus on a manifold. If f : M -+ R is a differentiable function, then we will consider it a (o)-tensor. The partial derivatives with respect to different charts transform according to the rule governing a (o)-tensor,

of

rof ax' ax ayj'

ayj i

where f (YI ..... Y") = BY I(x 1, ... , x"), ... Y"(x 1.... x')). To recover f from its derivatives, we apply Stokes's Theorem and integrate the I-form

df-àx'dx'ay dyj. i=I

j=1

We note that the "derivative" df of the (o)-tensor j is a (1o)-tensor.

(II) In Chapter 14 we considered the case of a Riemannian metric on an abstract surface S. To set the stage for generalization, we introduce new notation for the component functions of a Riemannian metric on a surface: g11 = E. 912 = 921 = F. and 922 = G.

The line element, as an aggregate of four functions (gij), is an example of a C2) tensor field on the surface S. To establish the transformation rule, write the line element on overlapping charts as

ds2 =>g,sdx'dxs = Lemma 14.12 implies that

ax' axs - . ayj ayj

gij = E gis r.s

We now give the general definition of tensor fields on an n-dimensional manifold.

Definition 16.7. For each coordinate chart x : (U C R") -+ M, an aggregate offunetions (Tj,:..j" U - , -

RI

1 < i1..... is. jl.... , jr :.-S n)

constitutes an (")-tensor field on M if on overlapping coordinate charts, we have the transformation rule k,...k,

TI,...t, =

i, ,....1,.ji.... ,j,

i,...i, ayki T;,...,, axi,

ayk, axi' axi, ayt,

axi, ayi, .

where T denotes the aggregate of functions associated to the coordinate chart y: (V C

R") - M.


249

In order to build the natural algebraic home for tensors we would need a sizeable detour

through the linear algebra of tensor products and the dual and tensor product bundles associated to the tangent bundle. Riemann and his immediate successors did not have these notions, though they manipulated the complex expressions associated to these constructions. The key feature of the definition of a tensor field is the transformation rule which implies that the tensor field does not depend on a choice of coordinates. The founder of this algebraic structure is G. Ricci-Curbastro (1853-1925). He and his student T. Levi-Civita (1873-1941) presented a unified treatment of tensors and their calculus in a paper titled Mdthodes de calcul difft rentiel absolu et leur applications that appeared in 1901 in the Mathematische Annalen at the invitation of Felix Klein. The absolute differential calculus of the title refers to the independence of choice of coordinates. The absolute differential calculus was renamed tensor analysis by Einstein in 1916. We will discuss the calculus of tensor fields in what follows.

Metrical relations (Riemannian manifolds) In the second part of his lecture, Riemann turned to the problem of making elementary geometric constructions possible on a manifold. In making such constructions, one requires lines and circles. To identify these special curves in this context we need a notion of length, that is, a line element. The definition from the theory of abstract surfaces generalizes easily for any dimension.

Definition 16.8. A Riemannian metric on a manifold M is a choice of positive-definite inner product ( , ) p on each tangent space Tp(M), for P E M, such that the choice varies smoothy from point to point. A manifold with a choice of Riemannian metric is called a Riemannian manifold. For a particular coordinate chart x : (U C R") -+ M, the inner product may be represented by a symmetric matrix of smooth functions (g,,(x' , ... , x")). If X and Y are tangent

vectors at p=x(u'..,..u"), then Xa`

8x'

,Yb'

and

b'

(X, Y)p = (a'..... a")(g,,(u'...., u")) b Independence of the metric on the choice of a coordinate chart requires that the functions g,, form a (2)-tensor.

Lemma 16.9. If x : (U C R") - M and y: (V C R") -i M are two coordinate charts for M and p is a point in x(U) fl y(V), then

(g,1) = J(y

1

oxY (8rs)J(y-t ox)

where (g,,) is the matrix of component functions of the metric associated to x. (k,,) is the matrix associated to v, and J(y -' o x) is the Jacobian matrix associated to the mapping v-' 0 x: x - ' (x (U) fl y(V )) -s y ' (x (U) n y(V )).


250

The proof is left to the reader. Component-wise, this matrix equation gives us

gii =F, L.rs r, s

ayr ays ax' ax7

that is, {gii) is a (0)-tensor field on M. The fact that the metric is positive definite implies that the matrix (gii) is invertible; we denote its inverse by (gki). The Jacobian is also an invertible matrix:

J(y lox)laxil and soJ(y-lox)-J(x-1oy)=(aYj). These relations imply the identities Ek

ay,r

k

axY' = SS and Ek gikgki = S where Sm is m, Sm = 0. We may apply these rules to

the Kronecker delta function, S, = 1, and. if n form new tensor fields.

Proposition 16.10. Suppose (T"""') is an

field on M, an n-dimensional Rie-

mannian manifold. Then the following expressions n

il...jp...it Uxi1-J'

M

vxil..i,

Til...i'_,nlip+,...1,

gxm il...i,

LmTi,...

JA...iq-ImJq+1... J

Jl... iq...i,

M=1

m=1

constitute an ("+-I,)- and an (r-+:) -tensor field, respectively. The operation of contraction of indices on a tensor field is defined by

-

wil...op ...(,

Ti, ...Ip-.I mlp+I...j, Jl...iq

il... iq...i,

Imiq+'...i,

m=1

-tensor field.

The contraction of an (')-tensor field is an Here

- means to omit the index; for example, ii i2 ... in = i2

in.

It suffices to demonstrate the transformation rules. We prove one case and leave the others to the reader. PROOF.

n

Tkl...kp-lmkp+l...k,

m=1

il...i, aykl

8Vkp-1 aVm aVkp11

ayk. a.YJ1

axi'

g,, a},x aym Til ...i, axi'

axip-1 axip axip+'

axi, aytl

ayt'

axu u.v.m.i1.....i,.

(1xu

A...-J, Til...i, ay k, gulp

11...i, axi'

Uil...jp...i, aVk1 uil..i, aXll

u.i,.....i,

aykp-I dVkp+1

ayk, axn axis

axip'

ax', aVx avtl

... axip-I

... axi' avt.

a)'k,, ' aykp41.

axip

ayk, axu axJ1 axip+I ... axi

ayx avtl

ax/' ayt'


Here we used

251

ax, av'n u = = 3' and the definition of U. avm ax'o 0 nr-1 n

These procedures of "raising;' "lowering," and contracting indices allow us to change the types of tensors, sometimes arriving at geometrically significant tensors from other such ones (see Proposition 16.14). In the presence of a Riemannian metric we define the arc length along a curve a(t) in a manifold M by

s(1)=

f

` p

and the angle 0 between curves a( (t) and A2(t) is determined by (); (1p), A'2(tp))p

cosh =

(A (tp), x'l (tp))p(A2(tp), A (

))p

where ,l,(tp) = p for i = 1, 2. The line element is given by the local expression

ds2 = r g,,dx'dx1. r.J

ay

The transformation rule (based on the chain rule) dyk =

r,l

ax ./

y'd y'.

g,,dx' dx'

ds2

-dx 1, gives

r.s

that is. arc length is independent of choice of coordinates. By separating the manifold from its metrical properties, Riemann made them independent aspects of a geometric object. For example, let M2 = 1R2 and contrast the Riemannian manifolds (IR22, ds2 = dx22 + dye) and (IR2, ds2 = dx2 + e2`dy2).

In the first case we have the Euclidean plane and in the second case another model for the geometry of Bolyai and Lobachevskii. Riemann attributed the confusion over Postulate V to the assumption that the plane, as a two-dimensional manifold, has a unique geometry. By separating the metric relations from the space itself, he sought to eliminate the confusion. For three-dimensional space R3, Riemann showed that there are many metrics and so geometry as a science of the space in which we live must depend on physical experience. One direction of generalization was prompted by "experience," that is, physics. In 1905. Albert Einstein (1879-1955) introduced his special theory of relativity. In 1908. Hermann Minkowski (1864-1909) lectured to the 80th Assembly of the German Scientific Union in Kiiln on Space and Time. The text of the lecture begins The views of space and time which I wish to lay before you have sprung from the soil of experimental physics, and therein lies their strength. They are radical. Henceforth space by itself and time by itself are doomed to fade away into mere shadows, and only a kind of union of the two will preserve an independent reality.

252


The union of space and time that lies at the basis of special relativity maybe described by a slight generalization of the Riemannian metric. We drop the condition that the inner product be positive definite and instead require the weaker condition that it be nondegenerate. Thus,

for p E M, for X, Y. and Z in Tp(M). and r E R.

(1) (rX + Y, Z)p = r(X. Z)p + (Y. Z)p. (2) (X, Y)p = (Y, X)p. (3) If (X, Y)p = 0 for all Y E Tp(M), then X = 0. If we represent such a metric locally by a matrix (g;j), then Sylvester's law of inertia (Lang 1971) implies that, at a particular point p E M, there is a change of coordinates in which the line clement takes the form ds2

= Eg;jdr'dx1 = >dvkdyk i

k=1

dymdym. m=/+I

In this case we say that the index of (g;1) is n - 1. A pseudo-Riemannian manifold is a manifold with a nondegenerate metric of constant index. A Lorentz manifold is a pseudo-Riemannian manifold with a metric of index 1. The simplest Lorentz manifold was introduced in Minkowski's lecture. Minkowski space is the manifold R4 endowed with the Lorentz metric

dsZ = dx2 + dy2 + dz2 - cdt2,

where c denotes the speed of light. Tangent vectors in Minkowski space fall into three classes spacelike,

if (X, X)p > 0 or X = 0,

null,

if (X. X)p = 0,

timelike,

if (X. X) p < 0.

This classification is based on causality and the physical assumption that the speed of light is constant in every frame of reference. A curve or world line of a physically meaningful object must have its tangent vector everywhere timelike or null in order to preserve this assumption. The study of differential geometry and, in particular. pseudo-Riemannian manifolds was spurred on by the development of relativistic physics. The interested reader may consult O'Neill (1983) for a comprehensive introduction to pseudo- Riemannian geometry.

Curvature When can two Riemannian metrics be transformed into one another? This is a natural question for Riemannian geometry called the local equivalence problem. Several constructions may be reduced to this problem, for example. on the overlap of two coordinate charts, we

16. Epilogue: Wherefrom here?

253

require a solution to the local equivalence problem in order to define a Riemannian metric. If two manifolds are to be isometric, then at corresponding points we must have a solution to the local equivalence problem. Certain analytic questions from the study of the heat equation can also be reduced to this problem. On this question Riemann offered a counting argument that was a key step in later research. A metric is determined by n(n + 1)/2 functions on each coordinate chart, namely, the component functions (gig I. A change of coordinates is given by the expression

x (x' , .... x") = y(y' W.... , X")'... , y"(x t , ... 'X")). This introduces n functions which change at most n of the functions (g 1), leaving n(n - 1)/2 choices. Riemann argued that there must be some set of n(n - 1)/2 functions that determine

the metric. When n = 2, there are three functions, E. F. and G. that make up the components of the metric. TWo may be fixed by a change of coordinates, leaving one function that fixes the metric completely. That one function is the Gauss-Riemann curvature. The work of Minding (Chapter 13) bears this out. The general case is answered by the Riemann curvature tensor which is determined by a set of n(n - 1)/2 functions (Proposition 16.15). The theorem that these functions fix a metric uniquely was proved in 1951 by the French

geometer E. Cartan (1869-1951). The special cases of n = 2 and n = 3 were solved by G. Darboux (1842-1917) and L. Bianchi (1856-1928), respectively. To introduce the Riemann curvature tensor, we consider the work of E. B. Christoffel (1829-1900), in particular, his 1869 paper in Crelle's Journal, Uberdie Transformation homogener Differenrialausdrucke zweiten Grades. Though it appeared after Riemann's death, this work is a natural extension of Riemann's ideas. Christoffel acknowledged his debt at the end of the paper, where he thanked Dedekind for making unpublished papers of Ricmann available. The Christoffel symbols (Chapter 10) were introduced in this paper. Christoffel's

computations form the basis for the invariant approach to Riemannian geometry which characterized its next stage of development. Suppose that a given metric has component functions (gj1) in one set of coordinates, and (g;f) in another set of coordinates. Recall the transformation rule for the metric,

gij =

aye a, r.s

grs Vi axe

This expression may be differentiated to give the equation ag.axle

- . r< < ay' ax' axe axt + ` s gas ( a'ye avs ages aye ays ay`

ayr a-v'

axi axle ax% + axi ax%axA

These derivatives may be permuted and summed with signs, to obtain the Christoffel symbols of the first kind associated to (g, j ):

[i k ! 1=

1

2

ag) ag,k _ agile axk + axi ax'

254


These functions are sometimes denoted f jk,i . Now let [St. r] denote the Christoffel symbols of the first kind associated to the metric Substituting the analogous expressions for

the [st, r] into the definition of the [jk, i] and simplifying we get agi;

I

[Jk, i 1 =

= rs.l

ag k

agjk

axk + ax) - ax'

2

[s, r]

-lay, ays ay'

+ E grs

ax' axj aXk

r.s

ayr a2xs az' axjaxk .

The Christoffel symbols of the second kind associated to (gjj ] are defined as n

g"ljk,l].

_

(Christoffel's original notation was [ jk. i ] _ ;k and Ilk = I k } . In Riemann's work we find pr;k = 2[ jk, i].) Substituting the expression for [ jk, i],We get

r,k r.s,r,l g

[Sr r]

ayr ays ay' aX1 axe aXk +

ax' ax' -'IS

ay ay"

_ _

", r.s.t.u

Siu

r"

r r

ayr

r-

a2ys

; 9' grs aXt aXjaXk

ays ay' ax' axr ayr 1,ayr r]--+ E k""krs--a.' aXi aXk ay" ay, axt axjaxk a2 ys

[Sr.1ays rJ--y' ax' + aXj aXk ay"

ays ay' ax' +

ax'

a2 ys

grsay" aXjaXk

r.s.u

a2 ys

ax'

aXj aXk ay" S ays axiaxk'

ay, where we have applied the relations

= Sr and

ay°

k"rkr, = S u. Fix an index

s = a; we can use the previous equation to isolate a second partial derivative. a2ya

a

aya

a2ys

axiaxk = Ea$ axiaxk -

axr

ax' (ays axiaxk

r" ay., ay' axr (lT'ik _ i

I' aya /k ax'

We have proved the following result.

siu

st

3.1

a2w'

xjs axle

ay")

ra ays ayl s' axe aXk

aya

ax'

)


255

Lemma 16.11 (Christoffel 1869). Let r;k denote the Christoffel symbols of the second kind associated to the metric (g j) and coordinates (x 1, .... x") and f sf those associated with (grs) and (y', ... , y"). With respect to a change of coordinates

(YI (x1,... ,x"),... ,y"(x1,... ,X")) taking the metric (g; j) to ($rs), we have, for all I < a. j, k < n, 82ya Jk ax'

axjaxk

s` axj aXk

s.f

.

This equation shows that the second mixed partials of the change of coordinates are expressible in terms of first partials and Christoffel symbols. It follows that this is true of all higher iterated partial derivatives. By his previous work on invariant theory. Christoffel reduced the local equivalence problem to an algebraic problem that he had already solved in 1864. To discover the Riemann curvature we take another derivative. a3 ya

(arjk aya

axjaxkaxl =

1: ( axl ax"'

a2 ya

+ rm

kax"'ax a2ys ayf

ataf ay" ays ayf

- s.f.u ayu axl axj aXk -

St

ay., a2yf

axjaxl aXk + aXj aXkaXl

Expanding the second partial derivatives according to Lemma 16.11, we obtain the (somewhat horrifying) expression

I_

a3yk

arm k 8ym

`3ya

r krMI

aX ax +

aXjax aX =

a rjkmrs'

ara ay" a c ay'

ayf ay' _ aX"f

x,

s

ra , ay" ay' av'

ra rr ays ay' +

-

rs.r

sf

)l aXr aXk

s`

ayf ays

rasf rrkf r.s.f

ax" Dxl

ayu Dxt Dxi aXk UV -,XI aXl aXk

s,f,u,u

+

su

Tar' SI

ayu

Dv" art

uu aXk axl aXj

The point of all this computation is to compare what we get when we equate the mixed 3 a 3 a partials (recall that the Gauss equations leading to theorema axk egregium arose this way). Set on the left everything that is made up of expressions related to the coordinate chart x, and on the right everything related to y; we obtain

[( !rj aXlk

aXkl

r are;

=sI

(ayu

/

+

ar

r

(rjkl'rl - r;ll'rk)j

u - (Iy,) +

ax"'

ays avf av a -", a (Fuf,rsf - HIV Su) I axj a.rk a.rl


256

If we multiply both sides by

ax m

and sum over all a, we obtain the equation

a

m aXIk

+> (r%krrt - r;,r,0

- aXk

_ ua

'i

r

arsr _ arsu ay

(ay°

a

o

a

+v' (rstruu - I'Surr.)

al'S aye ay° axe"

ax%axk axe aya

Definition 16.12. The Rlemann curvature tensor is the (j)-tensor field given by (RRtk I < m, j, k, l < n), where Rrk

r - [(axrk m

axkr) + r

`r', rrl - r; rrk

The previous derivation gives the transformation rule:

R,rkmaayàyàyuax'" RSul s.r.u.a

a% axk ax, ay,a

A necessary condition for a solution to the local equivalence problem emerges from the transformation rule - in order for two Riemannian metrics to be transformable, one to another by a change of coordinates, this equation must hold for all choices of m, j, k, and 1. Since the ingredients that go into computing the functions are derived from the component functions of the metric and their derivatives, we can test a pair of metrics for equivalence directly. A special case of the local equivalence problem was treated in one of Riemann's unpublished works, his Paris Academy paper of 1861. The problem set by the Academy in 1858 concerned heat conduction and systems of isothermal curves. In his paper Riemann recast the problem in geometric terms which required that a given metric be transformed into the standard Euclidean metric. His essay lacked certain details and the Academy decided not to award the prize until the essay was revised. Riemann's failing health prevented him from completing the essay and the prize was withdrawn in 1868. (y"

If a metric takes the form of the standard Euclidean metric in a coordinate system y"), then the Christoffel symbols P' = 0 for all a, s, and t. Thus a necessary

condition for a metric (gig) to be transformable to the Euclidean metric is the vanishing of the Riemann curvature tensor, that is, R,'i14 = 0 for all m, j, k, and 1. In fact, this condition is also sufficient. To see this we rewrite the problem following Riemann's formulation. If (x 1.... , x") give the coordinates for the given metric (gjj ), then let ()''..... y") be the desired coordinates for the standard Euclidean metric (kid = Sid). Riemann derived the particular case of Lemma 16.11, namely,

fir'av

r=I tk aXr

a

= aXJaxk = 3 5

ayJ3

axt


257

We can write this in a convenient matrix form as the equation

rik rik r2k

...

r2k

rik

aYg lax'

r2k

avR/ax2

aY /ax' a

aye/ax2

axk rnk

or Fk(x)(°y0) =

rnk

...

rnk

aya/ax"

aye/ax"

where Fk(x) denotes the matrix of Christoffel symbols

(r l "(x)) and V denotes the gradient. A necessary condition for the existence of the new coordinates v") is given by the equation of mixed partials which leads to the Riemann curvature tensor:

aat axk

which implies Rjmlk = 0 for all m, j, k, and

-azk aat (°)", 1.

To prove sufficiency one generalizes a familiar integrability result from two dimensions:

If f, g: R2

R satisfy of =

in an open set containing (0, 0), then, for ax zo E R, there is a function F: (U (: R2) - R with U an open set containing (0, 0) such FACT.

that F(0, 0) = zo, and VF(x, y) = (f (x, v), g(x, ),)). In n dimensions, the integrability conditions for the differential equations satisfied by

the y's are those that follow from azt Fk(x)(°

aak Ft(x")(°y-).

The effectiveness of Riemann's notion of curvature is shown by the following result. We refer the reader to Spivak (1970, vol. 2) for a choice of seven proofs.

Theorem 16.13 (Riemann 1861). An n-dimensional Riemannian manifold for which all the functions making up the Riemann curvature tensor vanish, that is, RRkt = O for all choices of i, j. k, and 1, is locally isometric to R" with the standard Riemannian metric

ds2 = E. dx'dx'. A Riemannian manifold that satisfies 0 for all i. j. k. and / is called flat. The Riemann curvature tensor measures the deviation from flatness of a manifold. Associated to the Riemann curvature tensor is a (4)-tensor field obtained by lowering an index R(jkt =

grr

r-1

Notice that if Rjkt = 0 for all choices of indices, then R;jkt = 0 as well. We will tease out a geometric interpretation of a subset of the R;jkt in what follows. The relation between the Riemann curvature tensor and the Gauss-Riemann curvature (Definition 14.13) is established by considering surfaces through a point that lie in a given


258

manifold. Suppose X and Y are linearly independent tangent vectors at a point p in an ndimensional Riemannian manifold M. We may choose the vectors X and Y as the coordinate

vectors as and T-2 to a surface SXy gotten by fixing the rest of the coordinates of a chart x: (U C R") -. M with x(0.... , 0) = p. The surface is the image of x applied to that part of the x'x2-plane in R" lying in U. By applying linear isomorphisms to R" we can li

always arrange for any given coordinate chart around p to have such properties. The rest of the charts for M transform smoothly when restricted to the appropriate subspaces of their domains and so we obtain an abstract surface SXy containing p. Since the tangent plane of SXy is a linear subspace of the tangent space of M at p, we may restrict the metric on M to a Riemannian metric on SXy. We compute the Gauss-Riemann curvature of this surface.

Proposition 16.14. Given linearly independent tangent vectors X and Y at a point p in an Riemannian manifold M, the Gauss-Riemann curvature at p of the surface SXy, denoted KXy (p), is given by 81212

XyP

-EG-F2'

where R1212 is a particular component of the lowered Riemann curvature tensor restricted

to SXy and E = g11 = (axi axl) P

a

and G = 922 =

' F = 912 = axl 'aXz) P

a )P.

1TX2' TX-2

PROOF. The proposition is a restatement of the formulas from Theorenra Egregium as they apply to abstract surfaces. To wit we have

_ R2,2

arzz _ ar21 1

1

ax1

aXz

ar2

are

1

1

1

2

1

2

1

+ (r22r1, - r12r12 + r22r12 - r12rz2) = GK

8212 = aXI -axe + (r22ri, -

-FK.

It follows that 81212 = ER212 + FR212 = (EG - F2)K. The quantities KXy are called the sectional curvature of the manifold at a point p in the directions X and Y. Not all of the component functions Rjkt or Rijkt of the forms of the Riemann curvature tensor are independent. In fact, there are many symmetries indicating that a deeper algebraic

structure must be enjoyed by this tensor. We record these symmetries in the following proposition. The proof is left to the reader. Proposition 16.15. The following equations hold: (1) Rjkt =

Rijkt = -Rjikt (2) Rjkt + Rktj + RI. Ijk = 0. (3) Rijkt = Rktij


259

By exploiting the symmetries of the curvature, one proves that the

n(n2

1)

=

(n)

functions R,,,, determine R,jtt and hence R" k1. These functions are the missing data that determine a metric conjectured to exist by Riemann. A Riemannian manifold is said to have constant curvature if the sectional curvatures KXy = c, a constant, for all linearly independent X and Y. Constant Gauss-Riemann curvature was a requirement of an abstract surface in order that it have enough congruences. In his lecture Riemann proposed the condition that "lines have a length independent of their configuration, so that every line can be measured by every other." The local equivalence problem and a version of the exponential mapping imply that such a manifold M has constant curvature. Riemann also proposed examples of manifolds of constant curvature.

Proposition 16.16. The subset ((x 1.... , xn)

1 + 4 E, (X,)2 > 0) of Rn with the

I

metric ds -

E; dx'dx'

=

z

tl+

4 r (x` )2

is an n-dimensional Riemannian manifold of constant curvature.

a,, PROOF The component functions of this metric may be written g,, = F,, , where F =

I+4

(x')2.

If we write f = In F, then we have ag,,

-

2aF F3 axk

3Xk

-28,, of

= F2 ark .

This leads to the following expression for the Christoffel symbols of the first kind: 1

[jk. i ]

F,2

S,

of axt

+ S,t

of

of 1 S,

axJ

ax'

)

The Christoffel symbols of the second kind are given by

Clt = Eg+m[jk nt] _ ,n=1

F`S'm[jk.ml = F2[jk, i]. nr=1

Similarly we can relate the lowered form of the Riemann curvature tensor with its usual form by n

n

n F,2 RJ,J = L'

gmRJ,t =

R,1,, =

n+

M=1

r

n,=1

I

F2

r R),1.

To compute the sectional curvature we use the fact that Kxy(p) is linear in X and Y and


260

where i

so it suffices to compute K a

Ri ..

iij

=

ari.. Ji- ar!. axi

axi

a2f + axiaxi

ii

M

j. M,

-r'"J, r'1-i

a2f axiaxi

n

+ E F°(Ijj, m]Imi, i) - [ii, m)Imj, i)) M=1

)2+ (af\21af2sI a2f = axiati + a xiaxi + (axi In our case f(x1, ... , x") = In l I + a 4

`

of _

axs

ax'

2F'

aY )

(Xs)2) from which it follows that s=I

a2f and

a.r'axs

=

, (2aF -

(ax"')2).

4F-

Substituting this into the preceding expression we obtain Riiij = a/F°, and since gii =

gjj = I/ F2, K

a

To '.

a =-

Ril1i

=a.

giigjj

When n = 3 and a = -I we obtain a model for Bolyai's and Lobachevskii's nonEuclidean space. In particular, the model is represented by the interior of the ball of radius 2 in R3 and so, topologically, it is simply connected. Furthermore, the boundary of the model lies an infinite distance from any point in the space and so it is complete. The arguments of Bolyai and Lobachevskii are realized in this manifold by restricting to planar sections of the ball; these are non-Euclidean planes that are isometric to the Poincart disk. These remarks were established by Beltrami in his paper on spaces of constant curvature (Beltrami 1869). He also presented new examples of constant-curvature manifolds by applying the higher-dimensional analogs of stereographic and orthographic projection. In particular, he identified the upper half-space model (H' = 1W.

x") ER" Ix">0).ds2=

(dx1)2+...+(dx")2/ .Y"

as an n-dimensional Riemannian manifold with constant negative curvature. In H3 the planes given by ((x1 , x2, c) I c a positive constant) are horospheres and it is immediate from the form of the line element that the induced geometry is Euclidean. Thus Beltrami showed analytically that the Euclidean plane lies inside a three-dimensional non-Euclidean space (Theorem 5.13), a result shown synthetically by Wachter, Bolyai, and Lobachevskii.

16. Epilogue: Where firm here:'

261

Covariant differentiation The quantities making up a tensor field on M are part of the analytic data associated to a manifold. To extend our understanding of these analytic expressions, we turn next to derivatives of the functions making up a tensor field. First consider a vector field given locally on coordinate charts x : (U C I[.') ---> M and r: (V C r) -+ M by X =

-

ai; aa, = > ai a I,] a coordinate direction ar

_

r r a1' at

Suppose we take the partial derivative of a'. with respect to This operation implies a transformation rule as follows: Since

ax,.

aa'`.

aa',, ax' a i

ai.9

axj ayy a.Y' +a ax'a.YI ati'y

,

a.x'

02 i,r

The presence of the second-order partial derivative shows us that we do not have a tensor. If the change of coordinates were linear, then the second derivative would vanish and the

expression aVq a would transform as a (:)-tensor. In the preliminary stages of his development of general relativity, Einstein struggled with this problem, and for a brief time he postulated that the only physically interesting changes of coordinates were linear. The restriction to linear changes of coordinates is unreasonable for geometry and physics and so another approach is needed here. Let us expand the previous expression with the help of Lemma 16.11.

a.Y, al?, 321.r ax'

aai

aa'L

a1 (!

ax, avy ax'

+RUa.Y'a.Y! al'g

aat; ax a

axl

I"* aY'

kv

, a v'

ax,

rs'ax' axj

a>y

,

a.Y!

a

ai.ivy

a'-. Rearranging the terms we find

ax,

, r 1 r; , 1 av" a.Yt +a f,y = EaaU axe +aL,i f ax' a}'9

J°1'

dry that is, the expression

r",

ax, aV

as' where we have used r,, al

+aL,

aar. a; qq

/

+ Ea'.I'"y transforms as a (:)-tensor. ,

Definition 16.17. The covariant derivative of a ((1))-tensor (a') is given by u

The set of functions (a' ') constitutes a (:) -tensor.


262

The case of a (1)-tensor (0j) is similar. The analogous calculation leads to the expression ae; F Br fr 9i:j=axjOil r

The covariant derivative of a (0)-tensor is a (Z)-tensor. Let us consider what we have discovered in these formulas. First of all, notice that for the flat Euclidean metric the covariant derivative reduces to the ordinary partial derivative.

Secondly, the new expression is a tensor again, that is, we have taken an aggregate of functions that behaves as a well-defined field (of tensors) on a manifold, and the covariant derivative once again determines a well-defined field on the manifold. Finally, we give the definition of the covariant derivative for a general tensor field. It is made up of parts depending on the upper and lower indices and behaves like a partial derivative following the Leibniz rule with respect to multiplication by a function on the manifold.

Definition 16.18. The covariant derivative with respect to xk of an (;)-tensor

::.

,I

is given by r

ar''...,

axk s

n

+

++

T'I...j.

rkm

1=1 m=1 It

A=1 µaI

The functions (T.''

''

constitute an (S )-tensor.

So far we have presented a formalism to generalize the partial derivative and stay within

the space of tensors. To put some geometry back into the discussion let us consider a special case of interest, geodesics on an n-dimensional manifold M. We begin by recalling the Euclidean case. We can characterize a geodesic in IR" in three equivalent ways: (1) It is the shortest curve joining two points along it. (2) The second derivatives of the components vanish.

(3) The tangent vectors along the curve are all parallel to each other. 2

The second property invites us to generalize the operator dt2 in the context of tensor analysis. This is straightforward to do. Suppose that a: (-r, r) -. M is a regular curve. Then the curve may be expressed in local coordinates by

a(t) =

x(x' (r)..... X"(1)) = Y(y' (t )..... y"(t ))

for overlapping charts x: (U C 1R") - M and y: (V C IR") -+ M. The tangent vector may be written a '(t)

dyj a

dx' a =

_ ,

1

dt axi

,

dt ayj'

263


dV'-

and the ordinary rules of differentiation imply dt

dC' ayr

=

dt axe .

that is, the tangent

vector along a(t) transforms as a (o)-tensor field. Taking another derivative of the transformation rule we have

d2t'r

d.-(' dxJ

d2 r' al .r

dt-'

r

E

dt'- ax' +

d'x' a'r dt' ax' +

r.J

(121-r

dt lit ax'a.Y'

dxt - dx' dt dt (F-

ay'' aV'

BV'

r" ax' axt

" a.Ck -

From this expression we obtain

d2 i" dt'-

r dy'' dy'

r°' d d =

d'xt

t dx' dx1 dt

'J dt

dt' +

aVr axi

.

This is the defining property for covariant differentiation with respect tot; it takes a particular

vector field, the field of tangents to the curve, and gives back a vector field of second derivatives. We denote this operation by

dt - ddt dt)

Da'

dx' dxt

d'-xt

a

ait

We generalize the covariant derivative along a curve a(t) to give the derivative of a vector field defined along a(t).

Definition 16.19. Given a regular curve on a manifold M with coordinates a(t) _

x(X1(t)......Y"(t))andavectorfeld X =

s

a' ar definedonaneighborhood contain-

ing the trace of a (t), the covariant derivative of X along a (t) is given by the expression

DX dt

-'

da'

d.Y't

a .

dt + I k' -;ka r dt

3.v'

The previous calculation generalizes to show that dX is also a vector field along a(t ). that is, it has the correct transformation rule. In order to appreciate the importance of this formula we prove a somewhat technical lemma. This result lies at the heart of the derivation of the equation satisfied by a geodesic.

Lemma 16.20. If M is an n-dhnensional Riemannian manifold with metric denoted be

(.

). then d

dt

(a (t). a'(t)),vl = 2

-a ,

a'(t)' lla(r)


264

`

PROOF. We simply compute

(dY

1

i, j

k.f

d2x' dx'

&j =

t2

dj

!

k

8j dl + L+ r kdd

dt

dr i

dxj dxk dx'

&jrkl dt dt dt

di +

T

'.1

d2x' dxj dxi dxk dx' F(kl, j'1 dr dt gii dt2 dt + j.k.t d2x' dxi gii dr2 dt +

=

18gjk dxi dxk dx' dt dr

E 2 axk dt

If we compute the arc length along a curve a(t) that satisfies

ct' = 0 we see from the

lemma that (a'(1). a'(t)) is a constant and so arc length is proportional to the parameter.

Da'

This property ties curves with = 0 to arc length. Analysis similar to the proof of Theorem 11.6 shows that such curves in fact provide the shortest distance between points along them. We are led to the following definition.

Definition 16.21. A curve y : (-r, r) -+ M is a geodesic if dry = 0, that is, for each -r < t < r, in a coordinate chart around y(t), the following differential equations are satisfied for i = 1..... n: dxi dxk

d2x'

rjk dr dt = 0. dt + F_ j.k All of the standard results about existence and uniqueness of geodesics follow from the theory of second-order differential equations. This gives us the exponential mapping and the special coordinate systems based on exp. The geodesic equations bear a formal identity with the equations arrived at through the geodesic curvature definition for geodesics on surfaces in R3. Is there a sense in which these equations and covariant differentiation are "geometric"? The last stage of the development

of differential geometry that we will discuss in this book begins with an answer to this question. For this we recall the situation of surfaces in R3. Consider a plane n in R3 through the origin. The tangent plane to n at each point is simply a copy of fl and we may identify every Tp(Il) as Il. This makes taking the derivative of a vector field simple. If X : fl - T Il is a vector field and x : (U C R2) -+ Il a coordinate chart, then X(x(a, b)) and X(x(a + t. b)) can be thought of as being in the same plane. Thus

X(x(a + t, b)) - X(x(a, b)) 1-o makes sense.

t

=

ax ax'(x(a.b))


265

When we consider a more general surface S C R3, the vectors X(x(a + t. b)) and X(x(a, h)) are in different tangent planes which are usually not parallel. Hence the difference quotient with limit definition of the derivative does not make sense. Every tangent plane to the surface S. however, is abstractly isomorphic to every other. The problem is to decide how to choose an isomorphism between T,,(S) and Tq(S) for each p and q in S. In 1917 T. Levi-Civita introduced a formalism for making these choices of isomorphisms. This solved the problem of giving a geometric interpretation of covariant differentiation by generalizing the parallelism enjoyed by Euclidean space. A similar approach was arrived at independently by J. A. Schouten (1883-1971) around the same time.

Definition 16.22. Given a cunve a: (-r, r) -, M, where M is an n-dimensional Riemaannian manifold, if X is a vector field defined along a, then X is said to be parallel along

aif i (a(t))=0forall

r.

In these terms, a curve is a geodesic if the tangent is parallel along the curve. Since the covariant derivative in a Euclidean space coincides with an ordinary derivative, we see that a vector field is parallel along a curve if it is constant along it. To see how this leads to an interpretation of covariant differentiation, we need a technical result:

Theorem 16.23. Given a regular curve a : (-r. r) - M and a vector field X defined in a neighborhood of the trace of a. there exists a unique vector field X defined along a(1)

with 1(a (0)) = X(a(0)) and k parallel along a. The proof relies upon the properties of the differential equations for parallelism (see Spivak (1970, vol. 2)). We note that the equations are linear and so if X = rxi + X2. then X = r.ri + X,. We use this property to define a mapping from Ta(o)(M) to Tacs,(M) for

-r < s < r: Let r,(X(a(0))) = X(a(s)). For any tangent vector to M. Y E Taion(M), there is a vector field Xy defined in a neighborhood of a(0) with Xy(a(0)) = Y. Thus our definition makes sense. The mapping r,,. is called parallel transport along a(t) and it is a linear isomorphism. The inverse of r,, is defined by reversing the curve and linearity follows from the differential equations. A geometric interpretation of covariant differentiation is contained in the following result.

Theorem 16.24. Let x : (U C R") -* M be a coordinate chart for an n-dimensional Riemannian manifold M around a point p E M with p = x(0.... , 0). Let ck : (-r, r) -. M be the regular curve on M given by the coordinate cunve

cR(1)=x(0,....0./,0,....0). Let r,: Tt,(Al) -> T,.A(,.)(M) for -r < s < r denote the isonorphisms given by parallel

transport along ck. Then, for any vector field X = >1 at a defined in a neighborhood rt

around p, we have Et

as

d

d.rt

r-0

rte (X((-t(h)))-X(p) h


266

PROOF. Denote the tangent vector r, I (X(ck(h))) by h(0). For each h near 0 we have a (possibly different) parallel vector field along ck(1) given locally by

kh(t) _

bt(h,t)aX1, atck(t),with bt(h.h) =at(ck(t)).

Since Xh is a parallel vector field, the following differential equations are satisfied:

dbt(h, t)

I = 0. + Ebi(h. 0L C' f'f(ck(tD

,./

Recall that the derivative with respect to t along ck (t) is the partial derivative with respect to xk so this differential equation becomes abt (h, 1) axk

+ 'V'bj(h , t)I'k1 (ck(t)) = 0 .

i

By the mean value theorem we may write

bt (h, h) = bt (h, 0) + h

db'(h, t) I

-

,

t-u,

for some ur E (0, h).

With this in mind, we compute

lim

rh'(X(ck(h))) - X(ck(0))

h-0 = limo

h

V" (br(h. h) - h

- h-0

(aal

-

(bt(h,0) -a(ck(0))) a

= lim

h-.o It

h

dbt(t

t) I

ax1

a tU

- at(ck(0))) aar

at(ck(h)) - at(ck(0))

dbt(h.1)

h

dt

+h-0

a I

,_

axr

bt(h,ur)l1j(ck(ur)) Xr

axk

r = ` axk + ua/r j ,

r

5-a

XI

i

a

r =Ea:kax

The parallelism introduced by Levi-Civita has some further properties that are explored in the exercises. Not only does it give an interpretation of the covariant derivative, but it


267

uncovers a new foundation for differential geometry. At the heart of our flurry of indices is Christoffel's lemma (16.11) which gives a kind of transformation rule for the rilk:

a2 a ax %ax

- L. rik aya ax

ayI aXj ay, ax

I < i. j, k < n) is a collection of n3 functions that satisfy the same Suppose (G'. transformation rule as the r, jk. Using the theory of differential equations as a tool, one can develop a notion of covariant differentiation and parallel transport by following the formal development given earlier based on the functions G'k' Such a choice of functions (G' k) is now called a (classical) connection for the notion of parallel transport it determines. In particular, this choice of structure is prior to the choice of Riemannian metric and belongs to the analytic machinery associated to the manifold. This level of generality was introduced by H. Weyl (1885-1955) in 1918, who was motivated by his attempt to unify relativity and electromagnetism. Among the possible connections associated to a Riemannian manifold, the one based on a choice of Riemannian metric and its associated Christoffel symbols is unique. This is the content of the following result: I

Theorem 16.25 (fundamental theorem of Riemannian geometry). Given a Riemannian manifold M there is one and only one connection on M such that parallel transport is an isometry and the connection is symmetric, that is, G'jk = G,'tj for all i, j, k. The connection singled out by this theorem is called the Levi-Civita connection or the canonical connection on (M, ( , )). The coordinate viewpoint that we have presented gave way in the twentieth century to coordinate-free constructions that do the work of the tensor calculus globally. The basic ideas

for these developments lie in Levi-Civita's work and parallelism. We close with Gauss's motto from his Copenhagen Preisschrift, 1822, a paper in which he solves "the problem of mapping the parts of a given surface onto another such that the image and the mapped part are similar in the smallest parts." Ab his via sternitur ad maiora

From here the way to greater (accomplishments) is smoothed. It is the author's hope that this book has smoothed the way for the reader to study the current formulation of differential geometry with a sense of where the subject has been and a deeper intuition about its inner workings and formalism.

Exercises 16.1

Prove that a connected I-manifold is diffeomorphic to either the real line or a circle.

16.2

Given two manifolds M' and N", construct the direct product manifold M x N of dimension m + n.


268

16.3

Show directly that the upper half-space,

(H" = ((x' .... , x") E R" I x" > 0), ds2 =

(dx

1)2

+ ... + (dx")2 .r"

/

is a Riemannian n-manifold of constant curvature.

16.4

Use the transformation rule for the Christoffel symbols to show that rjk is not a tensor of any order.

16.5

Show that the sectional curvature Kxy is linear in each argument, that is.

Kax+by.z = aKxz +bKYz, and Kx.ay+bz =a Kry +bKxz. 16.6

Prove Ricci's lemma, that is, for a Riemannian metric (g,j) on a manifold M we have

$im=0=B,:k

fort
Prove the Bianchi identity for the Riemann curvature tensor:

R jk:t+Rikt:J+R, :k =0. 16.8

Suppose that (a') denotes a (o)-tensor field on a Riemannian manifold M. Prove the following identity due to Ricci:

ai:I:k

-'

i - a':k:j = L a Risk.

!=1

16.9

Let ['(TM) denote the set (a vector space) of vector fields on a manifold M. Suppose that (Gj'k) is a choice of classical connection on M and suppose that the function V : ['(TM) x

r(TM) - 1'(T M) is defined by V

(52;7 .

a

ax)

k

Gi;

a

exk

and the fact that V is linear in both variables. Show that V further satisfies:

V(fX, Y) = fV(X, Y), for f e C'°(M), and

V(x, fY) = fv(x. Y) + X(J)Y. The properties of V define what is called a Koszul connection. Show further that, for the choice of the canonical connection (f j'k) on a Riemannian manifold, the covariant derivative of a vector field X along a curve a(t) is given by

dt

Riemann's Habilitationsvortrag: On the hypotheses which lie at the foundations of geometry* Plan of the investigation As is well known, geometry presupposes as given both the concept of space and the basic principles for constructions in space. It gives only nominal definitions of these things, while their essential specifications appear in the form of axioms. The relationship between these presuppositions is left in the dark; one does not sec whether, or to what extent, any connection between them is necessary, or a priori whether any connection between them is possible. From Euclid to Legendre, to name the most famous of the modern reformers of geometry. this darkness has been dispelled neither by the mathematicians nor by the philosophers who have concerned themselves with it. This is undoubtedly because the general concept of multiply extended quantities, which includes spatial quantities, remains completely unexplored. I have therefore first set myself the task of constructing the concept of a multiply extended quantity from general notions of quantity. It will be shown that a multiply extended quantity is susceptible of various metric relations, so that Space constitutes only a special case of a triply extended quantity. From this however it is a necessary consequence that the theorems of geometry cannot be deduced from general notions of quantity, but that those properties that distinguish Space from other conceivable triply extended quantities can only be inferred from experience. Thus arises the problem of seeking out the simplest data from which the metric relations of Space can be determined, a problem which by its very nature is not completely determined, for there may be several systems of simple data that suffice to determine the metric relations of Space; for the present purposes, the most important system is that laid down as a foundation of geometry by Euclid. These data are - like all data - not necessary. but only of empirical certainty, they are hypotheses; one can therefore investigate their likelihood, which is certainly very great within the bounds of observation, and afterwards judge the legitimacy of extending them beyond the bounds of observation. both in the direction of the immeasurably large. and in the direction of the immeasurably small.

1. Concept of an n-fold extended quantity In attempting to solve the first of these problems. the development of the concept of multiply extended quantity, I feel particularly entitled to request an indulgent criticism, as I am little trained in these tasks of a philosophical nature where the difficulties lie more in the concepts than in the construction, and because I could not make use of any previous studies, except

for some very brief hints that Privy Councillor Gauss has given in his second memoir on biquadratic residues, in the Gottingen Gelehrte Anzeige in his Jubilee-book, and some philosophical researches of Herbart. * Based on a translation by Michael Spivak used with permission. 269

270

Riemann's Habilitationsvortrag

1

Notions of quantity are possible only where there already exists a general concept which allows different realizations. Depending on whether or not a continuous transition of instances can be found between any two of them, these realizations form either a continuous or a discrete manifold; individual instances in the first case are called points and in the latter case elements of the manifold. Concepts whose instances form a discrete manifold are so numerous that some concept can always be found, at least in the more highly developed languages, under which any given collection of things can be comprehended (and in the study of discrete quantities, mathematicians could unhesitatingly proceed from the principle that given objects are to be regarded as all of one kind). On the other hand, reasons for creating concepts whose instances form a continuous manifold occur so seldom in everyday life that the position of sensible objects and colors are perhaps the only simple concepts whose instances form a multiply extended manifold. More numerous reasons for the generation and development of these concepts first occur in higher mathematics. Distinct portions of a manifold, distinguished by a mark or by a boundary, are called quanta. Their quantitative comparison is effected in the case of discrete quantities by counting, in the case of continuous quantities by measurement. Measuring involves the superposition of the quantities to be compared; it therefore requires a means of transporting one quantity to be used as a standard for the others. Otherwise, one can compare two quantities only when one is a part of the other, and then only to decide "more" or "less;' not "how much." The investigations about this which can be carried out in this case form a general division of the science of quantity, independent of measurement, where quantities are regarded, not as existing independent of position and not as expressible in terms of a unit, but as regions in a manifold. Such investigations have become a necessity for several parts of mathematics, for the treatment of many-valued analytic functions, and the dearth of such studies is one of the principal reasons why the celebrated Theorem of Abel and the contributions of Lagrange, Pfaff and Jacobi to the general theory of differential equations have remained unfruitful for so long. From this general part of the science of extended quantity which assumes nothing further than what is already contained in the same concept, it suffices for the present purposes to emphasize two points, the first of which will make clear the generation of the concept of a multiply extended manifold, the second reducing position fixing in a given manifold to numerical determinations and will make clear the essential character of an n-fold extension. 2

In a concept whose instances form a continuous manifold, if one passes from one instance to another in a well-determined way, the instances through which one has passed form

a simply extended manifold, whose essential characteristic is that from a point in it a continuous movement is possible in only two directions, forwards and backwards. If one now imagines that this manifold passes to another, completely different one, and once again in a well-determined way, that is, so that every point passes to a well-determined point of the other, then the instances form similarly a doubly extended manifold. In a similar way, one obtains a triply extended manifold if one imagines that a doubly extended one passes

Rientann's Habilitationsvortrag

271

in a well-determined way to a completely different one, and it is easy to see how one can continue this construction. If one considers the process as one in which the objects vary. instead of regarding the concept as fixed, then this construction can be characterized as a synthesis of a variability of n + I dimensions from a variability of n dimensions and a variability of one dimension. 3

I will now show, conversely, how one can break up a variability, whose boundary is given. into a variability of one dimension and a variability of lower dimension. To this end one considers an arbitrary piece of a manifold of one dimension - with a fixed origin, so that points of it may be compared with one another - varying so that for every point of the given manifold it has a definite value, continuously changing with this point, or in other words, one takes within the given manifold a continuous function of position, and, moreover, one such function that is not constant along any part of the manifold. Every system of points where the function has a constant value then forms a continuous manifold of fewer dimensions than the given one. These manifolds pass continuously from one to another as the function changes: one can therefore assume that they all emanate from one of them, and generally speaking this will occur in such a way that every point of the first passes to a definite point of any other: the exceptional cases, whose investigation is important, need not be considered here. In this way, the determination of position in the given manifold is reduced to a numerical determination and to the determination of position in a manifold of fewer dimensions. It is now easy to show that this manifold has n - I dimensions, if the given manifold is an n-fold extension. By an n-fold repetition of this process, the determination of position in an n-fold extended manifold is reduced to n numerical determinations, and therefore the determination of position in a given manifold is reduced, whenever this is possible, to a finite number of numerical determinations. There are, however, also manifolds in which the fixing of position requires not a finite number, but either an infinite sequence ora continuous manifold of numerical measurements. Such manifolds form, for example. the possibilities for a function in a given region, the possible shapes of a solid figure. and so forth.

II. Metric relations of which a manifold of ri dimensions is susceptible, on the assumption that lines have a length independent of their configuration, so that every line can be measured by every other Now that the concept of an n-fold extended manifold has been constructed, and its essential characteristic has been found in the fact that position fixing in the manifold can be reduced ton numerical determinations, there follows, as the second of the problems proposed above, an investigation of the metric relations of which such a manifold is susceptible, and of the conditions that suffice to determine them. These metric relations can be investigated only in abstract terms, and represented in context only through formulas: under certain assumptions, however, one can resolve them into relations which are individually capable of geometric

representation. and in this way it becomes possible to express the results of calculation geometrically. Thus, to put this work on solid ground, although an abstract investigation with formulas certainly cannot be avoided, the results can be presented in geometric garb.

272


The foundations of both parts of the question are contained in the celebrated treatise of Privy Councillor Gauss on curved surfaces.

1

Measurement requires an independence of quantity from position, which can occur in more than one way; the hypothesis that next presents itself, and which I shall develop here, is just this, that the length of lines be independent of their configuration, so that every line can be measured by every other. If fixing of position is reduced to numerical determinations, so that the position of a point in the given n-fold extended manifold is expressed by n varying

quantities xi, x2, x3, and so forth up to x,,, then specifying a line amounts to giving the quantities x as functions of one variable. The problem then is to set up a mathematical expression for the length of a line, for which purpose the quantities x must be thought of as expressible in units. I will treat this problem only under certain restrictions, and I first limit myself to lines in which the ratios of the quantities dx - the increments in the quantities x - vary continuously; one can then regard the lines as broken up into elements within which the ratios of the quantities dx may be considered to be constant and the problem then reduces to setting up at every point a general expression for the line element ds, which will therefore contain the quantities x and the quantities dx. I assume, secondly, that the length

of the line element remains unchanged, up to first order, when all the points of this line element suffer the same infinitesimal displacement, whereby I simply mean that if all the quantities dx increase in the same ratio, the line element changes by the same ratio. Under these assumptions, the line element can be an arbitrary homogeneous function of the first degree in the quantities dx that remains the same when all the quantities dx change sign, and in which the arbitrary constants are functions of the quantities x. To find the simplest cases, I next seek an expression for the (n -1)-fold extended manifolds that are everywhere equidistant from the origin of the line element, that is, I seek a continuous function of position that distinguishes them from one another. This must either decrease or increase in all directions from the origin; I want to assume that it increases in all directions and therefore has a minimum at the origin. Then if its first and second differential quotients are finite, the first-order differential must vanish and the second-order differential cannot be negative; I assume that it is always positive. This differential expression of the second order remains constant if ds remains constant and increases quadratically if the quantities dx, and thus also ds, all increase in the same ratio; it is therefore equal to a constant times ds2, and consequently ds equals the square root of an everywhere positive homogeneous function of the second degree in the quantities dx, in which the coefficients are continuous functions of the quantities x. In Space, if one expresses the location of a point by rectilinear

coordinates, then ds = JE(dx)2; Space is therefore included in this simplest case. The next simplest case would perhaps include the manifolds in which the line element may be expressed as the fourth root of a differential expression of the fourth degree. Investigation of this more general class would actually require no essentially different principles, but it would be rather time consuming and throw proportionally little new light on the study of Space, especially if the results cannot be expressed geometrically; I consequently restrict myself to those manifolds where the line element can be expressed by the square root of


273

a differential expression of the second degree. One can transform such an expression into another similar one by substituting for the n independent variables, functions of n new independent variables. However, one cannot transform any expression into any other in this way; for the expression contains n(n + 1)/2 coefficients which are arbitrary functions of the independent variables; by the introduction of new variables one can satisfy only n conditions, and can therefore make only n of the coefficients equal to given quantities. There remain n (n - 1)/2 others, already completely determined by the nature of the manifold to be represented, and consequently n (n - 1)/2 functions of position are required to determine its metric relations. Manifolds, such as the Plane and Space, in which the line element can be brought into the form E(dx)2 thus constitute only a special case of the manifolds to be investigated here; they clearly deserve a special name, and consequently, these manifolds,

in which the square of the line element can be expressed as the sum of the squares of independent differentials, I propose to call flat. In order to survey the essential differences

of all the manifolds representable in the assumed form, it is necessary to eliminate the features depending on the mode of presentation, which is accomplished by choosing the variable quantities according to a definite principle.

2

For this purpose, suppose one constructs the system of shortest lines emanating from a given point; the position of an arbitrary point can then be determined by the direction of the shortest line in which it lies, and by its distance along this line from the initial point, and it can therefore be expressed by the ratios of the quantities dx0, that is, the quantities

dx at the origin of this shortest line, and by the length s of this line. In place of the dx° one now introduces linear expressions At formed from them in such a way that the initial value of the square of the line element will be equal to the sum of the squares of these expressions, so that the independent variables arc: the quantity s and the ratio of the quantities At; finally, in place of the da choose quantities xi, x2, ... , .r proportional to them, such that the sum of their squares equals s2. If one introduces these quantities, then for infinitesimal values of x the square of the line element equals E(dx)2, but the next order term in its expansion equals a homogeneous expression of the second degree in the n (n - 1)/2 quantities (x i dx2 - x2dxi ). (x i dx3 - x3dx, ), .... and is consequently an infinitely small quantity of the fourth order, so that one obtains a finite quantity if one divides it by the square of the infinitely small triangle at whose vertices variables have the values (0, 0. 0, ... ). (x, , x2, X3, ... ), (dx,, dx2, dx3.... ). This quantity remains the same as long as the quantities x and dx are contained in the same binary linear forms, or as long as the two shortest lines from the initial point to x and from the initial point to dx remain in the same surface element, and therefore depends only on the position and direction of that element. It obviously equals zero if the manifold in question is flat, that is, if the square of the line element is reducible to E(dx)2, and can therefore be regarded as the measure of deviation from flatness in this surface direction at this point. When multiplied by -3/4 it becomes equal to the quantity that Privy Councillor Gauss has called the curvature of

a surface. Previously, n(n - 1)/2 functions of position were found necessary in order to determine the metric relations of an n-fold extended manifold representable in the assumed

274

Riemunn's Habilitationsvonrag

form. hence if the curvature is given in n(n - 1)/2 surface directions at every point, then the metric relations of the manifold may be determined, provided only that no identical relations can be found between these values, and indeed in general this does not occur. The metric relations of these manifolds, in which the line element can be represented as the square root of a differential expression of the second degree, can thus be expressed in a way completely independent of the choice of the varying quantities. A similar path to the same goal could also be taken in those manifolds in which the line element is expressed in a less simple way, for example, by the fourth root of a differential expression of the fourth degree. The line element in this more general case would not be reducible to the square root of a quadratic sum of differential expressions, and therefore in the expression for the square of the line element the deviation from flatness would be an infinitely small quantity of the second dimension, whereas for the other manifolds it was an infinitely small quantity of the fourth dimension. This peculiarity of the latter manifolds therefore might well be called flatness in the smallest parts. For present purposes, however, the most important peculiarity of these manifolds, on whose account alone they have been examined here, is this, that the metric relations of the doubly extended ones can be represented geometrically by surfaces and those of the multiply extended ones can be reduced to those of the surfaces contained within them, which still requires a brief discussion.

3

In the conception of surfaces the inner metric relations, which involve only the lengths of paths within them, are always bound up with the way the surfaces are situated with respect to points outside them. One can, however, abstract from external relations by considering deformations that leave the lengths of lines within the surfaces unaltered, that is, by considering arbitrary bendings - without stretching - of such surfaces, and by regarding all surfaces obtained from one another in this way as equivalent. For example, it follows that arbitrary cylindrical or conical surfaces are equivalent to a plane, since they can be formed from a plane by mere bending under which the inner metric relations remain the same, and all theorems about the plane - hence all of planimetry - retain their validity: on the other hand, they count as essentially different from the sphere, which cannot be transformed into the plane without stretching. According to the previous investigations, the inner metric relation at every point of a doubly extended quantity, if its line element can be expressed as the square root of a differential expression of the second degree, which is the case with surfaces, is characterized by the curvature. This quantity can be given a visual interpretation for surfaces as the product of the two curvatures of the surface at this point, or by the fact that its product with an infinitely small triangle formed from shortest lines is, in proportion to the radius, half the excess of the sum of its angles over two right angles. The first definition would presuppose the theorem that the product of the two radii of curvatures is unaltered by mere bendings of a surface, the second, that at each point the excess over two right angles of the sum of the angles of any infinitely small triangle is proportional to its area. To give a tangible meaning to the curvature of an n-fold extended manifold at a given point and in a given surface direction through it, one must proceed from the fact that a shortest line emanating from a point is completely determined if its initial direction is given.


275

Consequently one obtains a certain surface if one prolongs all the initial directions from the given point that lie in the given surface element into shortest lines, and this surface has a definite curvature at the given point, which is equal to the curvature of the n-fold extended manifold at the given point in the given surface direction. 4

Before applying these results to Space, it is still necessary to make some general considerations about flat manifolds, that is, about manifolds in which the square of the line element can be represented as the sum of squares of complete differentials. In a flat n-fold extended manifold the curvature in every direction, at every point, is zero; but according to the preceding investigation, in order to determine the metric relations it suffices to know that at each point the curvature is zero in n(n - 1)/2 independent surface directions. The manifolds whose curvature is everywhere zero can be considered as a special case of those manifolds whose curvature is everywhere constant. The common character of those manifolds whose curvature is constant can be expressed as follows: Figures can be moved in them without stretching. For obviously figures could not be freely shifted and rotated in them if the curvature were not the same in all directions, at all points. On the other hand, the metric properties of the manifold are completely determined by the curvature; they are therefore exactly the same in all the directions around any one point as in the directions around any other, and thus the same constructions can be effected starting from either, and consequently, in the manifolds with constant curvature figures can be given any arbitrary position. The metric relations of these manifolds depend only on the value of the curvature, until it may be mentioned, as regards the analytic presentation, that if one denotes this value by a, then the expression for the line element can be put in the form I

4

I+Or

2

E(dx)2

The consideration of surfaces with constant curvature may serve for a geometric illustration. It is easy to see that the surfaces whose curvature is positive can always be developed onto a sphere whose radius is the reciprocal of the curvature; but in order to survey the multiplicity of these surfaces, let one of them be given the shape of a sphere, and the others the shape of surfaces of rotation that touch it along the equator. The surfaces with greater curvature

than the sphere will then touch the sphere from inside and take a form like the portion of the surface of a ring, which is situated away from the axis; they could be developed upon zones of spheres with smaller radii, but would go round more than once. Surfaces with smaller positive curvature are obtained from spheres of larger radii by cutting out a portion bounded by two great semicircles, and bringing together the cut-lines. The surface of curvature zero will be a cylinder standing on the equator; the surfaces with negative curvature will touch this cylinder from outside and be formed like the part of the surface of a ring which is situated near the axis. If one regards these surfaces as possible positions for pieces of surface moving in them, as Space is for bodies, then pieces of surface can be moved

276


in all these surfaces without stretching. The surfaces with positive curvature can always be so formed that pieces of surface can even be moved arbitrarily without bending, namely on spherical surfaces, but those with negative curvature cannot. Aside from this independence of position for surface pieces, in surfaces with zero curvature an independence of position for directions also occurs, which does not hold in the other surfaces.

III. Applications to space I

Following these investigations on the determination of the metric relations of an n-fold extended quantity, the conditions may be given which are sufficient and necessary for the determination of the metric relations of Space, if we assume the independence of lines from configuration and the representability of the line element as the square root of a second-order differential expression, that is. flatness in the smallest parts.

First, these conditions may be expressed by saying that the curvature at every point equals zero in three surface directions, and thus the metric relations of Space are implied if the sum of the angles of a triangle always equals two right angles. But secondly, if one assumes with Euclid not only the existence of lines independently of configuration, but also of bodies, then it follows that the curvature is everywhere constant, and the angle sum in all triangles determined if it is known in one. In the third place, finally, instead of assuming the length of lines to be independent of place and direction, one might assume that their length and direction is independent of place. According to this conception, changes or differences in position are complex quantities expressible in three independent units. 2

In the course of the previous considerations, the relations of extension or regionality were first of all distinguished from the metric relations, and it was found that different metric relations were conceivable along with the same relations of extension: then systems of simple metric specifications were sought by means of which the metric relations of Space are completely determined, and from which all theorems about it are a necessary consequence;

it remains now to discuss the question how, to what degree, and to what extent these assumptions are borne out by experience. In this connection there is an essential difference between mere relations of extension and metric relations, in that among the first, where the possible cases form a discrete manifold, the declarations of experience are surely never completely certain, but they are not inexact, while for the second, where the possible cases form a continuous manifold, every determination from experience always remains inexact be the probability ever so great that it is nearly exact. This circumstance becomes important when these empirical determinations are extended beyond the limits of observation into the immeasurably large and the immeasurably small; for the latter may obviously become ever more inexact beyond the boundary of observation, but not so the former. When constructions in Space are extended into the immeasurably large. unboundedness and infinitude are to be distinguished; one belongs to relations of extension, the other to


277

metric relations. That Space is an unbounded triply extended manifold is an assumption which is employed for every apprehension of the external world, by which at every moment the domain of actual perception is supplemented, and the possible locations of a soughtafter object are constructed; and in these applications it is continually confirmed. The unboundedness of space consequently has a greater empirical certainty than any experience of the external world. But its infinitude does not in any way follow from this; quite to the contrary, Space would necessarily be finite if one assumed independence of bodies from position, and thus ascribed to it a constant curvature, as long as this curvature had ever so small a positive value. If one prolonged the initial directions lying in a surface direction into shortest lines, one would obtain an unbounded surface with constant positive curvature, and thus a surface which in a flat triply extended manifold would take the form of a sphere, and consequently be finite.

3

Questions about the immeasurably large are idle questions for the explanation of Nature. But the situation is quite different with questions about the immeasurably small. Upon the exactness with which we pursue phenomenon into the infinitely small, does our knowledge of their causal connections essentially depend. The progress of recent centuries in understanding the mechanisms of Nature depends almost entirely on the exactness of construction that has become possible through the invention of the analysis of the infinite and through the simple principles discovered by Archimedes, Galileo, and Newton, which modem physics makes use of. By contrast, in the natural sciences where the simple principles for such constructions are still lacking, to discover causal connections one pursues phenomenon into the spatially small, as far as the microscope permits. Questions about the metric relations of Space in the immeasurably small are thus not idle ones. If one assumes that bodies exist independently of position, then the curvature is everywhere constant, and it then follows from astronomical measurements that it cannot be different from zero; or at any rate its reciprocal must be an area in comparison with which the range of our telescopes can be neglected. But if such an independence of bodies from position does not exist, then one cannot draw conclusions about metric relations in the infinitely small from those in the large; at every point the curvature can have arbitrary values in three directions, provided only that the total curvature of every measurable portion of Space is not perceptibly different from zero; still more complicated relations can occur if the line element cannot be represented, as was presupposed, by the square root of a differential expression of the second degree. Now it seems that the empirical notions on which the metric determinations of Space are based, the concept of a solid body and that of a light ray, lose their validity in the infinitely small; it is therefore quite definitely conceivable that the metric relations of Space in the infinitely small do not conform to the hypotheses of geometry, and in fact one ought to assume this as soon as it permits a simpler way of explaining phenomena. The question of the validity of the hypotheses of geometry in the infinitely small is connected with the question of the basis for the metric relations of space. In connection with this question, which may indeed still be reckoned as part of the study of Space, the

278


above remark is applicable, that in a discrete manifold the principle of metric relations is already contained in the concept of the manifold, but in a continuous one it must come from something else. Therefore either the reality underlying Space must form a discrete manifold, or the basis for the metric relations must be sought outside it, in binding forces acting upon it. An answer to these questions can be found only by proceeding from the conception of phenomena which has hitherto been proven by experience, for which Newton laid the foundation, and gradually modifying it driven by facts that cannot be explained by it; investigations like the one just made, which begin from general concepts, can serve only to insure that this work is not hindered by the restriction of concepts, and that progress in comprehending the connection of things is not obstructed by traditional prejudices. This leads us away into the domain of another science, the realm of physics, into which the nature of the present occasion does not allow us to enter.

Appendix: Notes on selected exercises* I entreat you, leave the doctrine of parallel lines alone; you should fear it like a sensual passion; it will deprive you of health, leisure, and peace - it will destroy all joy in your life. These gloomy shadows can swallow up a thousand Newtonian towers and never will there be light on earth: never will the unhappy human race reach absolute truth - not even in geometry. R Bolvai (in a letter to J. Bolrai)

The following pages contain some hints and solutions to various exercises found in the book. In most cases, instructions for constructing solutions stand in for fully worked-out answers. In several cases, the reader is provided only with the necessary identities and a general strategy for completing the problem. I hope that the reader will turn to these pages not in despair but with the hope that the discussion presented here will lead to a deeper understanding of the subject. Several of the more complicated graphics in the text were produced using Mathentatica.

Chapter 1 1.2. Consider points on the sphere as vectors in R3. The standard formula for the angle between two vectors implies that the great-circle distance between A and B = arccos(A A. B). In spherical coordinates, the cosine rule (Theorem 1.3) implies that the great-circle distance

between A and B is arccos(cos(>Jri - >G2) sin 01 sin 0-2 +cos0i cos92).

1.3. If two planes are perpendicular according to our definition, then their normal vectors are perpendicular as well. Let N1, N2, and k3 be a choice of normal vectors to fl 1, fl2, and 113, respectively. Let I = NI x N3 and m = .N2 x N3. Then 1 is parallel to the line 11i n fl3 and m is parallel to the line f12 n 1713. From the properties of the cross product (see

the beginning of Chapter 7 or Banchoff and Werner, Linear Algebra through Geometry. Springer-Verlag, New York, 1983), we derive

l m(NI x

x N3)=(N3 x(N2 x N3))

N1

`Prepared by J. Cantarella. who would like to thank K. Richardson for her friendship, support, and intellectual conversation during the preparation of this manuscript, and Annie Campisi for inspiring him to complete the process This work is dedicated to the memory of his father. Robert Georgc Cantarella. 279

Appendix: Notes on selected exercises

280

1.7. On the diagram, let c be the center of the spherical circle, a be a point on the spherical

circle, and 0 be the center of the sphere. Note that LaOc is the angle subtended by the radius p at the center of the sphere. Use this fact to find a formula for the length of ab (the planar radius of the circle) in terms of p and R. This leads to the expression given in the exercise. If we rewrite 2n R sin

A

R

as 27r

sin(p/R) then L'Hopital's rule implies that the limit (1/R)

as R -+ oo is 27rp.

Chapter 2 2.9. Before attempting this problem, we recommend that the reader examine the first few pages of Chapter 4 to introduce the Saccheri quadrilateral. We refer to C the diagram on the left. Mark off D and E, the midpoints of sides AC and BC. Extend up, in both directions and drop perpendiculars to t7t from

F D11J\E G

A, B, and C meeting in F. G. and H. Now show tADF = ACDH and ABEG = L\CEH. This proves that PG = 2 DE and that-FAHC = GB. Theorem 4.3 shows that 77G:5 AB. Postulate V is not required. 2.13. To construct the reciprocal of a given line segment T -B, we refer to the diagram on the D left. Construct AC and BD, each unit length and each perpendicular F to AB. Then connect CD forming a rectangle. Connect the diagonal AD. Now mark off point E on AB such that AE AC.Construct

the perpendicular to AB at E. meeting Al) at G, and CD through

E

A

F. Show that GE is the reciprocal of AB. To do so, you must construct the perpendicular to GE at G, meeting AC in H and BD in 1, and demonstrate that the area of HI BA is equal to the area of unit square ACFE. Proposition 1.43 may be helpful. To construct the square root of a given line segment A B, see Euclid's Elements. Book II. Proposition 14.

Chapter 3 3.2. To prove (c), show that (c) implies Playfair's axiom. Suppose that Playfair's axiom does not hold. We refer to the diagram on the left. Fix a line I and a point a not on 1. Construct the perpendicular to / through a, meeting

P in a point b on I and draw the first parallel P to 1 through a. By P'

assumption, this line has a common perpendicular cd with 1. Suppose c is on P. Choose e on 1 such that d E be. and raise the perpendicular to I at e, meeting Pat f. Prove, by continuity, that some line through a, say P', is perpenh

d

e

1

dicular to of at some g E ef. Now use the Saccheri-Legendre theorem (Theorem 2.2)


281

to conclude that P' is parallel to 1. By construction, P was the first such parallel. This contradiction proves the result. 3.4. The error here is subtle. The key step in the proof is the construction through D of a line which cuts AC and A B. The assumption that such a line always exists is the assumption that LBAC is an angle such that every point in the interior of LBAC is on a line intersecting

both rays of the angle away from the vertex. By Theorem 3.9 (10), this is equivalent to Postulate V.

Chapter 4 4.2. Suppose 1, m have common perpendicular ab with a on 1. Follow the diagram on the left. Choose any con ! and drop the oernendicular tom from c. meeting

f

at d. Now acdb is a Lambert quadrilateral so by a preceding exercise, Td > ab. Choose eon ! so that c E ae and construct the perpendicular to in from e, meeting in at f. We must show of > cd. Extend cd and

drop the perpendicular to cd from e meeting at g. Now dgef is a Lambert quadrilateral, so of > gd. We can now argue by continuity that gd > cd. 4.5. To prove this result, a lemma is required. Suppose we have a quadrilateral A BCD such D E C that LBCD = LADC and AD = WE. We show that LDAB LCBA. Bisect CD at E and connect AE and BE. By Side-Angle-

Side, AADE = OBCE. Thus LEAD = LEBC. Also, AE B

A

p

S

R

T

BE, so AABE is isosceles. Thus LEAB = LEBA, so LDAB LCBA. Now P, Q, and R are all on the horocycle of points corresponding to Q. Fix Son the parallel through P, and T corresponding to Son the parallel through R. The horocycle of points corresponding to S is concentric with the horocycle of points corresponding to Q

so PS = BT (Lemma 4.11). Further, since S and T correspond,

L PST - L R TS. Using the lemma just proved on the quadrilateral PR TS we have proved that P and R correspond.

4.6. Let 1, m, n be parallel lines. Locate corresponding points a, b, c on !, in, and n respectively, and locate d, e, and f on 1, m, n away from a, b, and c. or n Suppose that HRA holds. Prove that ab, be are in a line by proving that a c

d

e

f

Labe and Lcbe are right angles. To do this, use the assumption that a, b, and c correspond to show Labe = Lbad - Lbcf - Lcbe.

(*)

Apply Euclid's Proposition 1.29 to prove that these are all right angles. Observe that this implies that the horocycle connecting a and c and the line connecting a and c coincide. To prove the other direction, use formula (*) to prove that ab is a common perpendicular to 1, in. Now use Theorem 3.9 to prove that this implies HRA.


282

Chapter 5 5.4. Construct the altitude BD of AABC. Consider the case shown in the figure, where BD is inside ABC. By the formulas leading up to the proof of B

tanh((b - x)/k) Further, tanh(c/k) . cos fl(b) = tanh(b/k) and cos fl (c) = tanh(c/k). Thus Theorem 5.19, we have that cos LCAB =

C

cosLCABcosfl(b)cosfi(c)=tanhl Further. forx E 1R, sin fl(x) = I/cosh(x/k). Thus sin 11(b) sin 11(c)

_

sin fl (a)

b k

x Itanhl k I.

(I)

\

cosh(a/k) cosh(b/k)cosh(c/k)

By the hyperbolic Pythagorean theorem (5.19). cosh(a/k) = cosh(p/k)cosh(x/k) and cosh(c/k) = cosh(p/k) cosh((b

- x)/k). Thus

_

cosh(a/k)

cosh(x/k) cosh((b - x)/ k) cosh(b/k)

cosh (b/k) cosh(c/k)

(2)

(__i - k I since cosh(- ) is an even function. Using

We observe that cosh(x/k) = cosh b k x

/

the addition formula for cosh, we have that

cosh

(_f - k) = cosh((b - x)/k)cosh(b/k) - sinh((b - x)/k) sinh(b/k).

Substituting this into the right-hand side of (2) and adding the resulting equation to (1) proves the result. The extension to the case where BD falls outside ABC is similar. 5.6. Since LOAE and 11(x) are both less than a/2, it suffices to show that sin 11(x) _ sin LOAE. From the proof of the Lobachevskil-Bolyai theorem (5.16)

we know that sin 11(x) = I/cosh(x/k). For the triangles ADAE, AOAB, and ADAB in the diagram, apply the hyperbolic Pythagorean theorem (5.19) to get the following identities: cosh

(DA)

= cosh

(k) cosh (k)

,

BB(v).

cosh(k)=cosh(2)cosh These identities imply

1

cosh(x/k)

_ cosh(OB/k) cosh(DB/k)

Suppose we have a right hyperbolic triangle DABC with sides opposite vertices given by a. b. and c and right angle at C. The formulas for cos(L ABC) and sin(LCAB) in Chapter 5 cos(LABC) Apply this to triangles LAOB and AADB to see that imply cosh(b/k) = sin(LCAB)

cosh(OB/k) _ cos(LOAB)sin(LADB) sin(LAOB)cos(LDAB)' cosh(DB/k)


283

Now apply Bolyai's theorem (5.20) to triangles AADB. AAOB, and L DA E to see

sin(LADB) sin(LAOB)cos(LDAB)

_

sinh(AB/k)/sinh(DA/k) =1. (sinh(AB/k)/sinh(c/k))sin(LDAE)

Finally. since LOAB + LOAE = LBAE, a right angle, we have that cos(LOAB) _ sin(LOAE). This leads to the desired identity.

Chapter 6 6.9. Since it is unit speed, according to Proposition 6.15, the circle has involute

,8 (s) = (cos s, sins) + (c - s) (- sins, cos s). The parabola may be parametrized as the non-unit-speed curve a (t) = (t, t2). It has involute determined by the general formula

p(t)=a(t)+ In this case, the result is

!

IIa()

)II)a'(t).

`1.r

I0(t)=(t,12)+ 1c-fOj \\

Involute of a circle.

I + 4x2dx 1 + 4t'-

/(1,2t).

Evolute of an Ellipse

6.10. Referring to the discussion following Theorem 6.18 for the general formula for the curve of centers of curvature, we compute or

(t) =

ab

(b2 cos2(t) +a' sin2(t))3/2

(notice that we have chosen the sign of the curvature positive) and Na (t) =

I

(b cos(t), a sin(t)).

b2 cos2(t) +a2 sin-(t)

Of course, Ca(t) = a(t) + K(t) Na(t). To prove that the curve is singular, we take the derivative of the curvature.

K'(t)=This is zero at t = 0,

rr, 37r/2.

3ab(a2 - b2) sin 2t 2(a2 sine t + b22 cost /)5/2

284


6.11. For any e > 0, define hE: (a + e, b - eJ x B"(0, R) --i. R", where B"(0. R) is the closed ball of radius R > 0 in R", to be hE(t, v) = g(t)v. Clearly, hE is continuous. We observe that the domain D of hE is compact and convex. Thus, by the extreme value theorem, there exists some Mi E R such that 11h, (t, v) 11 < Mr. We claim that there exists some value M2 such that IIhE Q. 'v r) - hE (t. v2) II < M21I v i -

vz11 for all (1, vi), (t,Yz) in D. SincehE is continuous, the function H: D x B"(0, R) -- R defined by

H(t. yr v2) =

IIhf (t, yr) - hE U, F2)II II3'I -x211

is continuous. But the domain of His compact, so by the extreme value theorem, there exists

some M2 such that H(t, vr, v2) < M2 for all (t, yi. Y2) in D x B"(0, R). Of course, this M2 satisfies the conditions of our claim.

We now apply Theorem 6.19. Choose A > 0 so that R > AMi. Clearly, A and R are legitimate auxiliary values as required by the theorem. Now choose initial conditions (to. )''o) = (a + E + A, 0). The theorem tells us that there is a uniquely defined function through 0, denoted vo(t): b(to, A) --# R", such that vo(t) = hE Q. "vo(t)). Here b(to, A) is the open ball around to of radius A in R. Now let ti = to + A/2 and apply the theorem for initial conditions (ti,. o(tr)). By uniqueness, the resulting function vr(t) extends the domain of definition of vo to (a + e, a + e + 3A/2). Repeating this process finitely many

times extends the definition of v0(t) to (a + e. b - e). Refer to this function as v((t). Repeating this process as e - 0 produces a final v(t) defined on (a. b). This completes the proof.

Chapter 7 7.6. The Frenet-Serret theorem implies that IIB'll = ITI. Note also that this implies -B'/118'11 = N. Thus N can be written in terms of B. Now show that T' can be written in terms of B. However, T = N x B. so T' = N' x B + N x B'. Since N is a unit vector, the Frenet-Serret theorem implies that K = T' N. 7.7. The expansion is the Taylor series for a(s), that is,

a(so+Os)=a(so)+a'(so)(As)+

a"(so)(As)2 2

+

of

(s0)(ts)3

+...

6

expressed in T, N. B coordinates. If we name the axes t. n. b, the projections of a obey the following expressions near a(so):

K2 n=2r.

2

b2 =

2r2 9K

3 n,

b=

KT 3 6t-.

7.9. Since the curve is spherical, there must exist some point c E R3 and a real number r such that (a(s) - c) (a(s) - c) = r2 for all s. Differentiate both sides three times to obtain

285


0 = (a(s) - c) (K'N +KN) = (a(s) - c) (K'N + K(-KT + T B)) = K'(&(S) - c) N - K'(a(s) - c) T + Kr(a(s) - c) B

_ -K +Kr(a(s) - cl . B. K

This tells us that (a(s) - c) B = K . The desired formula follows by differentiating. KZT

7.17. Since A is an isometry, A o a is a unit-speed curve. Thus I/IIA(a"(s))II

K.4ar(s) =

=

Ka(s)

Since A o a(s) is a unit-speed curve, we know that

(A(a'(s)) x A(a"(s))) A(a(s)) K.4c«(s)2

Using the definitions of matrix multiplication and the cross product, one finds that

(A(XI) x A(X2)) A(X3) = det(A)(XI x X2) X3. Using this identity, we can see that

det(A)(a'(s) x a"(s)) a"'(s)) r .4.a (s) = -

1

KA_a

(a'(s) x a"(s)) a,,,(s))

_ -r«(s).

Chapter 8 8.3. It is easy to show that x (u, v) is a coordinate chart. Since a is smooth, x is differentiable. By hypothesis, x is one-to-one. Now u

(a1

+ tal)

r(aI + ua,) a

va') au (a3 +

av

a

(vaj ())

v(va;(W))

a

at

=

aI + vaj ai(u) a; + vat a; (U) aj + va3 aj(u)

(vai(n))

Since K(u) 96 0, we know that a"(u) 96 0. Thus, since v # 0. this matrix has rank two. As for the component functions of the metric,

E = a'(u) + va"(u), a'(u) + va"(u)} = 1 + (vK)2. F = (a'(u) + va"(u). a'(u)) = 1. G = (a'(u), a'(u)) = 1.


286

8,4. We make two claims. First, we show that for 7r/2 < u < 2n, u = u - n/2, and u = v we have x(u, v) = i(u, v). Next, we prove that for 0 < u < rr/2,

u = u + 3a/2, and v = -v we have x(u. v) = i(u, u). We must restrict the domain of these assertions to prevent attempting to evaluate x(u, v) or i (u, v) outside their domains of definition. In

the first case, r/4 + u/2 = n/4 + (u - rr/2)/2 = u/2. Also, we have sin u The Mobius Strip

=sin (u

- 2) = sin u cos 2 - cos u sin 2rr = - cos u, IT

n It IT cos is = cos (u - 2) = cos u cos 2 + sin u sin 2 = sin u. Our claim follows trivially. The second case is similar. Using these identities, our claim can be verified by some algebra. This verifies that these charts determine the Mobius strip. To see this, consider the effect

of the following composite map, m: [7r/4, 27r + (r/4)] x (- 1. I) - R1. Let

x(u, v).

for

m(u, v) =

n < u < n, 4

i(u - 2, v). fore
m(27r + n/4, -v), as i(37r/2 + 7r/4, -v) = x(n/4, v). This is exactly the condition that defines the Mobius strip as a topological space. To see that the Mobius strip is nonorientable, we compute the determinant of J(i - o x)

at x(7r/4, 1/2) = i(37r/2+n/4, -1/2). Around this pointi-1 ox (u, v) = (3n/2+u, -v) so

detJ(i-'ox) =

--

au ai) au av

ai) au 5-u *TV

By Definition 8.10, this proves that the Mobius strip is not orientable.

Chapter 8'1 8°1.2. Suppose we have a conformal, equiareal map f: Si -+ S2. Further, suppose we use f to induce a coordinate chart on S2 from a coordinate chart x: U -+ S1. Let E. F. and G be the component functions of the metric on S1 with respect to x and E, F. and G be the components of the metric on S2 under fox. Now, by Proposition 8.3. there is a nonzero function p: S1 -+ R such that for all p E S1

E = p2(p)E,

F = p2(P)F. and G = p2(p)G.

Now the functions E, F, G and E, F, G are continuous, so the composite functions A (p)

E(p)G(p) - F (p), B(p) =

_

E(f(p))G(f(p)) - F2(f(p)) are continuous as well.

Suppose there exists some p E Si such that A(p)

B(p). Without loss of generality,


287

assume that A(p) is greater. Since these functions are continuous, there is some bounded neighborhood N of p where A(q) > B(q). Suppose x - ' (N) = V and compute the area of

N and the area of f(N). By our assumption that f is equiareal, area(N) = area(f(N)). But

area(N) = j'J A (x(u, v))dudu

> Jj B (x(u, v))dudv = area(f(N)). Thus no such p exists. Use this equality and our previous results to conclude that p2(p) = I

for all p E St. 86b.7. Suppose the cone is in angle *. As the illustration shows, the cone is a surface of revolution of the function f(u) = u tan s. This provides a coordinate chart for the cone of the form

x(u,01) = (f(u)cos01, f(u) sin 01. u). To prove that this chart yields an ideal map projection from the cone to the plane. make the substitutions u = r cos

V1

and A

1

Coordinatize the plane in polar coordinates.

This new chart maps the set (0. oo) x (0, 27r sin t/r) onto the cone by the chart

x(r.02)= (rsinrcos(Snz*l.rsinVisinl Snz*).rcos When you determine the component functions of the metric of this surface you find that they are equal to the metric coefficients for the plane under polar coordinates. This proves that the map projection is ideal, since the metric coefficients determine angles, lengths, and

areas on a surface.

Chapter 9 R3 given by 9.2. The function z = /'(u. v) leads to the coordinate chart x: (U C R2) x(u, v) = (u, v, f(u, v)). Computing directly in terms of this chart the functions E, F, G. and e, f, g. and using Corollary 9.10, one arrives at .

K(P) =

au2 8u2

(auav)

z


288

9.5. Recall that IIp(v) = dNp(v) u. Now write v in polar coordinates in the basis of principal directions X1. X2. Thus v = 11i 11(Xi sin0 + X2 cos0) for some angle 0. If u E Dp, then 12 ii

IIp(v) =

i

(

0

ki

0

sing

-k2) (cos 0)

sinO cos 0

)

= -ki sin2 0 - k2 cos2 0.

In polar coordinates, this implies that the parametric equation for Dp is

Dp(9) _ 0 - ki sin2 9 - k2 cos2 0 -' , B). The graph of this function is: an ellipse if ki, k2 have the same sign, a hyperbola if ki, k2 have different signs, a pair of parallel lines if one of ki , k2 is zero, and empty if ki = k2 = 0.

9.10. Assuming that a(t) is a line of curvature in S, write a'(t) in the basis of principal directions of curvature. Now notice that dt N(a(t)) = dNa(,)(a'(t)). Finally, write dNa(,) in the basis of directions of principal curvature, one of which is a'(1). In the other direction assume that d Na(,) (a' (t )) = AU)a'(t ). Expand this out in terms of a coordinate chart where

a(t) = x(u(t), v(1)) to get N,, u' + Nov' =

x,, v').

Take the dot product of xu and x with the previous equation to obtain the equations:

(e+AE)u'+(f +AAF)v' = 0.

(f +kF)u'+(g +.kG)v' = 0. It follows from Theorem 9.6 that we can write

e + 2 f(v'/u') + g(v'/u')2

E+2F(v'/u')+G(v'/u')2' when u'

0. Letting r = v'/u', then k. (u', v') =

has a critical point in terms of r

when

(e -

(f -

0,

(f - k,,(ro)F)u' + (g - kn(ro)G) v' = 0. that is, the equation -x is a solution to the critical-point problem, and so -A(t) is a principal curvature along a(t).


289

Chapter 10 10.3. Recall from theoretna egregium that the Gaussian curvature K is given by the following formula when F = 0 - 72 EOP + - 12G«<,

(EG)2

!El, 2

-

Ev

0

; E,,

g G«

7

3G

E

0

E,,

E

0

G,,

0

G

G

0

G

I.

Expand the determinants and simplify to obtain the result. 10.5. There are three results to prove here. To prove that the line in R3 through p contained

in S is in an asymptotic direction, suppose that this line is in the direction of the unit tangent vector u in Tp(S). Observe that 0 since the line has plane curvature 0. Now use Theorem 9.6 (llp(i>) = to conclude that llp(v) = 0. Rodrigues's formula (Exercise 9.10) implies that the line is asymptotic. To prove that a ruled surface S has K < 0 at each point p E S. we observe that for some c' at p. k (p) = 0. This means that k, > 0 and k2 < 0 so k, k, < 0. To construct a nontrivial family of ruled surfaces. choose a continuous

function f: IR --. R such that 1(x) = - J'(tr + x). Using cylindrical coordinates. define S by S = {(r. H, :) j z = r( J'((9))}. For example. take /'(.r) = (1/2)sin(3x). The resulting surface is ruled.

10.6. Suppose S is minimal. At a point p E S let X1. X2 denote the principal directions and a,. 52 asymptotic directions. To prove that the asymptotic directions are perpendicular. observe that if a, is an asymptotic direction, and a, makes the angle p with X,, then

0 = k, (at) = k1(cos2 0) + k,(sin26).

But k, = -k2. so this implies sin 2 0 = cost 0 or sin 0 = ± cos 6. There are four solutions to this equation. n/4, 3;r/4. 57r/4, and 72r/4. Of course. this applies to the other asymptotic direction, as well. Apply the intermediate value theorem to prove that if k, 0 k2, then

a 1 0 +62,soa, 1a,. Now suppose a, I a2. If a, makes angle 01 with X1, and a'2 makes angle 0, with X,, then we have sin 01 = cos 02 and cos 01 = sin B,. Consider the following derivation:

0=

kit (6,,)

=k,(cos29,)+k2(sin'0,)+k,(cos2fl,)+k2(sin 20,) = k, + k2. It follows that S is minimal.

290


Chapter 11 11.2. Without loss of generality, take a to be unit speed. Since a(s) is a planar curve, a'(s) and a"(s) are in the plane Ii. The normal N(a(s)) to the surface S is also in the plane fl by considering the reflection R: R3 -s R3 across f1. Observe that R restricted to S is a mapping from S to itself. In fact, for a given patch x: (U C R2) -> S around a(s), we can form another patch, R o x: (U C R2) -+ S, by composing with the reflection. Prove that the normal in this patch is the reflection of the normal in the original patch. Proposition 8.9 implies that R(N(a(s)) = bN(a(s)) for some b E R. Thus N(a(s)) E R. By the definition of geodesic curvature conclude that a(s) is a geodesic.

11.4. To show that a(t) is a geodesic, observe that a(t) is the curve formed by the intersection of S2 and the plane (x. y, 0). Obviously, this plane is a plane of symmetry for S2. To show that a does not satisfy the equations (*), proceed by contradiction. If a(t) satisfied (*), then by Proposition 11.5 for some e > 0, a: (-e, e) -- S would be parametrized by a multiple of arc length. But a'(t) = (-e' sin e', e' cos e', 0), so wt

(a'(1), a'(1)) =

(ez,)

dt

2e2r.

=

Thus dt (a'(t), a'(t)) i4 0 for all t. This contradicts our assumption that a(t) is param-

etrized by a multiple of arc length in some neighborhood.

Chapter 12 12.3. Triangulate R' using F triangles A 1,, A ... jr. Immediately, we have f fR' Kd A = EF i f fo KdA. For each triangle Aj, let the interior angles of Aj be given by iji, ij2, ij3. By the Gauss-Bonnet theorem (12.4), for each j, we have

j'f KdA = (iii + 1)2 + ij3 - n) -

f

Q

kg(s)ds.

Thus we have F

Rr=I

F

f f KdA=F(iti+i,2+i,3-a)-Y

kg(s)ds.

,=1 f8A,

Since the triangles are oriented, the contributions of each side to the last integral cancel out,

except on the boundary of R'. Thus EF I fao kg(s)ds = faR, kg(s)ds, and we have that

JfKdA=(irl+r2+1,3)-nF1_1

kg(s)ds. aR'

Now each triangle contributes 3 edges to the triangulation, while each edge that is not on the boundary of R' is shared by 2 triangles. Thus, if C is the number of edges along the

boundary of R', we have 3F = 2E - C. Also, at each vertex not on the boundary, the interior angles sum to 27r. At each vertex on the boundary, they sum to jr or to rr - Ej for


291

some j. Thus, defining e = 0 for J < i < C, and observing that there are precisely C vertices on the boundary, we have F

C

L(tt I + it2 + it3) = 2rr(V - C) + L(rr - Ej ). j=I

t=1

Putting this all together, we get

r

C

ffKdA=27r(V-C)+(7r-Ej)-1lF- J j=1

'

k(s)ds

aR'

=27rV -rrC-trF+n(3F-2E+C)-EEj- J kg(s)ds j=1

aR

= 2Jrx (R') - Efj - f ks(s)ds. j=1

R'

12.5. We prove that adding one handle to a sphere results in a figure with Euler characteristic 0. The extension to g handles follows by induction. First, take any triangulation of an empty prism. Observe that the prism has Euler characteristic 2. This follows from the fact that the prism is homeomorphic to the sphere. Take any triangulation of the sphere, and join the top face and the bottom face of the prism to two faces of the triangulation of the sphere. The Euler characteristic of the resulting figure Prism must be the Euler characteristic of the sphere plus the Euler characteristic of the prism minus the four faces lost in the joining (two on the sphere, two on the prism). Thus the resulting figure has Euler characteristic 0.

Chapter 13

13.4. Suppose some curve a(t): [0, S] -> S connects p and q. We prove that this curve has length sa(S) > Ilwll, and further, that if sa(S) = Ilwll, then a(t) is the geodesic radius connecting p and q. Of course, if a([0, S]) 1Z exp(B((OP)) then sa(S) > Ilwll. We deal with the case where a([0, S]) C exp(BF). Suppose a(t) = (r(t), 9(1)) in geodesic polar coordinates with respect to a fixed vector in the direction of w. The arc length is given by

sa(S) = Ja But G(r(t), 0(1))

(drs)2d1 = Ja (d 12 +G(r(t), 9(i)) (de 12dt. is nonnegative for all t, so

dle

sa(3)

Jb

(d Y dt = r(S) - r(0).

Since a(S) = q, r(S) = Ilwll, and a(0) = p,, so r(0) = 0. Thus sa(S) > I1t"u11. So suppose

some a(t) has sa(S) = Ilwll It follows that, for this a(t), dB = 0. But this means that 9(t) .is a constant. Since a(S) = (11@11, 0), this means that 9(t) = 0, and a(t) is some reparametrization of the geodesic radius connecting p and q.

292


13.6. Suppose that exp is a diffeomorphism on the open ball B0(0) E Ta(S). Choose some a: R -+ S with a(0) = p and establish geodesic parallel coordinates around p with a as the base curve. Further, suppose that fi: (1 C R) -- Tn(S) is a curve in Tn(S) with

IIfi(t)II < e and P(t) = (sona(r)) for some so E R. Clearly y(t) = expa(,)(#(t)) is a portion of a geodesic parallel to a(t). Define a family of curves at (s) = exp.,,) (s n. (t)).

Then a,(0) = a(t), a,(so) = y(t), and the a,0(s) are the geodesic perpendiculars to a(t). As in the proof of the Gauss lemma for geodesic polar coordinates, we must prove that for each to E 1, dtar(SO)11_10, dsa,0(s)s-so I = 0. Define

da,' da, ds. ds ds

E(t) _ o

Observe that this function gives the distance from a(t) to y(t) along geodesic perpendiculars. This is obviously a constant, so we can apply the methods given in the proof of the Gauss lemma (13.2) to arrive at the identity 0

-

s.s0

(aat a,

' 'or, as 1,=,0 s=o

aa, a2o,

-

Jo

1 at '

a-s2

I,=to ds.

We can evaluate the first term to see s=s0

aa, aa,

(at'

as

,_,0

(dot(s0) da,,, dt (to). ds (so)I

(do(0) dt

da,°

(to). ds (o)

s=0

'

dat° !dar(so) dt (to). ds (so) - (a (to). Ha(lo))

dot(so)

dt

da,°

(to). ds (so)

Of course, this is what we are looking for. It remains to show that (30

J

aa, a2a, 1 at as2

1-.-1I.Ods=0.

This can be accomplished easily by following the proof of Lemma 13.2. The derivation of the general form of the line element is also simple. To compute the line element for this coordinate chart on the unit sphere, suppose a(t) is a geodesic. Then, following the diagram to the left, we see that the geodesic parallel a, (y) is the geodesic circle on the sphere with center P and radius n/2 -y. Using the formula for the circumference of a geodesic circle on the sphere derived in Exercise 1.7, we see that ff" G(x, y) dx = 27r sin(

so G(x. y) = sin('w - y).

- y),

293


13.7. To prove the first identity, recall that H = -7tr(dNp). But dNp can be written as

dNp = I pZi

Theorem 9.9, we have

122

I

1

H=-2(111+122)=2

(Eg-2Ff+eG). EG-F2

To show that the principal curvatures k, are given by H ± H- - K recall that the Gaussian

curvature is given by K = k1 k2 = det(dNp). The determinant of dNp - t1, where I is the identity transformation and t a variable, is given by t2 + 2Ht + K. Expressing dNp in terms of the basis of principal directions puts the principal curvatures, the eigenvalues of -dNp, along the diagonal. The formula fork; follows from the quadratic formula.

Chapter 14 14.2. To show that these functions determine a differentiable structure on RP2, we must show that they satisfy the following properties. (1) They are defined on open sets U, in R2, such that the Oi are injective and satisfy 1

U1o,1(Ui)=RP2. (2) The transition functions O, o 0 are smooth. 1

(3) The images of open sets in R2 under the 0 that is Hausdorff and second countable.

form a subbasis for a topology on RIP2

The first property is easy to prove. Obviously them' are each defined on R2 and they are = RP2, realize that for each i in question. injective on this domain. To see that U1 0,.- 1 we have V1 1(R2) = ((x1, x2, x3) E RP2 I xi # 0). It follows that u1 V1 = RP2. We prove the second property for 02 o 0, 1. The extension to the other cases is trivial. Observe that 02 o 0, 1(u, v) = ¢2(I, u, v) = (I /u, v/u). Obviously, this is smooth. The third property is more interesting. On each V1, define a topology Ti as follows. A set S C V1 is open in (V,, Ti) iff 01(S) is open in (R2, usual). Notice that under this definition, 0i is a homeomorphism of (R2, usual) and (Vi, 7). Thus (V1, T) is a second-countable Hausdorff space because (R2, usual) is a second-countable Hausdorff space. We now define the topology Ton RP2 by S C RP2 is open if and only if. for i E (1.2, 3), S n V, is open in Ti. This is a topology. To prove that it inherits second-countability and the Hausdorff condition from the Ti, use the following lemma: 1

For i # j, Sc V1 n Vj is open in (Vi, T) if and only if S is open in (Vj, T j). Suppose S is open in (V1, T j). Then 01(S) is open in R2, so Oj o Or 1(¢1(S)) is open in R2, since = Oi o 0,7' is smooth (and thus obviously continuous) by property (2). But (0j o

4j o,0T 1(0j(S)) = rbj(S). so Oj(S) is open in R2 and hence S is open in (Vj,T ). The converse is similar.


294

Chapter 15 15.5. Suppose that ! has slope m. Reflect over the v-axis, if necessary, to arrive at the case where m < 0. Since! does not intersect the u-axis and I is not the first parallel to the it-axis we know that m > -r. Parametrize I s,r+sm) as 1(0 = Q. r+tm). Recall that the line given by Cs(t) = (s, t) is perpendicular to the u-axis at (s, 0).

The lines ! and Cs intersect at (s, r + sm ). We must show that there exists some so such that ! and C, are perpendicular at

(so, r+som). We must also verify that the point (so, r+som) is in the unit disk. Now if !(t) = (ut(t), vi Q)) and C. Q) = (us(t), vs(t)), then we have U; (1) = I,

U; (1) = 0,

VI(I) = M.

v;(t) = I.

Thus, if 9(s) is the angle between these two lines at (s, r + sm), then

Cos0(s)=

F(s, r + sm) + mG(s, r + sm) 4s.r+sm)U'(s),!'(s)) 1(s..+sm)(Cs(r + sm), C(r + sm))

After simplifying this expression one discovers that there exists a positive real number y(s) such that y(s)cos9(s) = ms2 + 2rs + m. We immediately observe that the right-

hand side of this equation equals m at s = 0. Thus, if m = 0. the proof is complete. If not, the right-hand side is negative at s = 0. If (so, r + sum) lies on the unit circle, then so =

-mr

+m21+i

-r

. Substitute this into the foregoing polynomial to see that

mso + 2rso +m > 0. By continuity, there must be some s1 such that (s1, r +s1m) is in the unit disk and y(sl) cos 0(s1) = 0. This completes the proof.

15.9. Suppose p and q are points on a semicircular geodesic with center c and radius r. Using the parametrization provided in the exercise, we see that there exist some to, tl E R such that p = (r tanh to + c, r sech to) and q = (r tanh t1 + c, r sech t1). The distance between p and q is given by (p, q)

I

4

dx dt

+ (LY)'] di dt

Here, we have dt

dt =

(r tanh t) = r sech2 l

and

d` = dt (r sech t) = -r sech i tank t.

Substitute these values into the above expression and the hyperbolic trigonometric identities imply the surprising fact that 8(p, q) = t1 -to. The distance formula forpoints p = (xo, yo), q = (x 1. y 1) on a semicircular geodesic with center c and radius r is given by

d}p(P.9) = In

(xo-c+r yll x1 -c+r yo/

295


Substitute the expressions for the x, and' v, given earlier into this formula and simplify it to

verify that dH(p, q) = ti - to. You will need to use the fact that sinh t + cosh t = e`. 15.14. We first prove Bolyai's theorem for a right triangle. Set out copies of a right triangle L A BC on the Beltrami disk as shown in the figure.

Following the discussion of distance in the Beltrami disk, we can compute

a=

I In

2

!-u +r

I-u--v

and b=

ln(l+tt).

l 2

l - tt

Also, we can prove that v' = tanh a = v cosh b. We need a formula for it' in terms of it and v. Now b = dB((O. v'). (tt'. v'))

=

In

d, (W. v'), (- I dE((O. v'). (V/1 - (v')'. a')) v')) dE((u'. v'), (1/-1 - (v')2. v')) dE((O, v'), (- l - (v')2, v'))

2

=-l In - it'+ it, -

l-(v')2 I

I+u'cosha

I

=-In{ 2 \I_tr'cosha)

-

Thus u' = u sech a. With this result in hand, we are prepared to compute. Since the nonEuclidean length of lines through the origin depends only on their Euclidean length, we have that (u')- + (v')2 = u- +v-. Further, since angle measure is Euclidean at the origin, we can measure sin A and sin B using the Euclidean triangles we have drawn. Thus

sin A = see that

t

u--+v'-

and sin B =

sin B sinh b

it sech a

u--+v2

. Using this, and the fact that it = tanh b. we

U

cosh a sinh b

I

tt- '+v-

cosh a cosh b u'- + v2 v

sinh a sin A

sinh a '

u- + v-

296


= sin B

I

We know that tanh c = u + v2. Further. sinhb by a standard hyperbolic trigonometric identity, csch2 c = coth2 c - 1. Thus I =

It remains only to show that

1-u--v that

U +v

sinh c

sinh c

. Choosing a convenient term from the previous chain of equalities, we claim I

_

sinh c ' Now tanh a =

V

71 - u2'

1-u -uu'- + u2

1

cosh a cosh b u2 + u

and sech2 a = 1 - tanh- a, so sech a =

1

2-v-

-1u- u

. Further,

we know that sech b = Substituting these expressions into the preceding equation proves our claim. Thus we have established Bolyai's theorem for right triangles. The extension to the more general case is not difficult. To prove the hyperbolic Pythagorean theorem, we use the foregoing work to conclude that

cosh a=

1- u

I-u2-v2

The result follows directly.

cosh b=

I

1-u

and cosh c=

I 112

-v

Bibliography This text owes an enormous debt to the vast literature on geometry. I would like to call attention to the sources most relevant to each chapter for the reader who wishes to learn more about a particular topic from a possibly different viewpoint. The whole project owes much to Gray (1979), Struik (1933-

4; 1961), and Spivak (1975, especially vols. 1, 2. and 3). Chapter I was inspired by the article of Busemann (1950). Chapters 2 through 5 owe much to Euclid (1956), Gray (1979). Bonola (1955). Greenberg (1993), and Rosenfeld (1988). Reading Saccheri (1920), Bolyai (1832), and Lobachevskti (1829-38) (found in Bonola (1955)) is great fun and highly recommended. The middle section of the book in which classical differential geometry is developed follows the outline of the excellent treatments in Spivak (1970.1975), do Carmo (1976), Struik (1961), and Hsiung (1981). 1 was lucky to have Yoder (1989) and Schroder (1988) to review when I was writing this book. These books inspired the last part of Chapter 6 and all of Chapter 8"u. Chapter 15 is inspired

by reading Beltrami's Saggio and Poincare's Theorie des groupes Fuchsiens with some help from Coolidge (1909). Chapter 16 is my attempt to outline what Riemann said in the lecture based on hints found in Pinl (1981) - it may be seen as an elementary companion to the chapter "What did Riemann say?" in Spivak (1970, vol. 2). 1 have tried to include all the historical references made in the text as well as all the books I used in preparing the text. The bibliography at the end of Spivak (volume 5) will aid the reader in going on in the modern subject of differential geometry.

Books Ahlfors, L. V., Complex analysis: An introduction to the theory of analytic functions of one complex variable, 2d ed.. McGraw-Hill, New York. 1966. Bianchi, L., Vorlesungen fiber Dii ferentialgeometrie (trans. from Italian by M. Lukat), Teubner, Leipzig, 1899.

Bishop, R. L., and Goldberg, S. I., Tensor analysis on manifolds. Macmillan, New York, 1968. Reissued by Dover, New York, 1980. Bonola, R., Non-Euclidean geometry (trans. H. S. Carslaw), Dover, New York, 1955. Borsuk, K., and Szmielew, W., Foundations of geometry: Euclidean and Bolvai-Lobachevskian geometry. Projective geometry (trans. from Polish by E. Marquit), North-Holland, New York, 1960.

Butzer, P. L., and Feher, F., eds., E. B. Christoffel: The influence of his work on mathematics and the physical sciences, Birkhauser, Boston, 1981. Carmo, M. P. do, Differential geometry of curves and surfaces, Prentice-Hall. Englewood Cliffs, NJ. 1976. Carmo, M. P. do, Riemannian geometry. Birkhauser, Boston, 1992. Cartan, E., Lefons sur la gdometrie des espaces de Riemann. Gauthier-Villar, Pans, 1928. Chern, S: S., Studies in global geometry mid analysis, Mathematical Association of America, Washington, DC, 1967. Includes essays: What is analysis in the large? by M. Morse: Curves and surfaces in Euclidean space, by S. S. Chern; Differential forms, by H. Flanders; On conjugate and cut loci, by S. Kobayashi; Surface area, by L. Cesari; Integral geometry, by L. A. Santalo. Coolidge. J. L., The elements of non-Euclidean geometry, Clarendon Press. Oxford. 1909. Coolidge, J. L., A history of geometrical methods. Oxford University Press. Oxford, 1940. Reissued by Dover, New York, 1963. Coxeter, H. S. M., Non-Euclidean geometry, 5th ed., University of Toronto Press, Toronto, 1968. Coxeter, H. S. M., Introduction to geometry. Wiley, New York, 1961. Darboux, G., Lej ons sur la thr orie generale des surfaces et le.s applications gdometriques du calcul infinitesimal, Gauthier-Villars, Paris. 1887-1896. In four volumes, 3d ed.. Chelsea. New York. 1972. 297

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Prenowitz, W., and Jordan, M., Basic concepts of geometry, Ardsley House. New York, 1965. Proclus, A Commentary on the first book of Euclid's Elements [trans. G. R. Morrow). Princeton University Press. Princeton, NJ. 1970. Richards, J. L., Mathematical visions: The pursuit of geometry in Victorian England, Academic Press. New York. 1988. Riemann, B., Gesammelte mathen,atische Werke and ivissenschaftlicherNachlass, ed. R. Dedekind and H. Weber, Teubner. Leipzig, 1892. Reissued by Dover. New York, 1953. Robinson, A., Elements of cartography, Wiley, New York. 1960. Rosenfeld, B., A history of non-Euclidean geometry. (trans. A. Shenitzer). 1st English ed.. SpringerVerlag. New York. 1988. Ryan, P. J., Euclidean and not-Euclidean geometry: An analytic approach. Cambridge University Press, New York. 1986. Saccheri, G., Euclides ab omni Naevo Vindicates. 1st ed.. Milan, 1733; 1st English ed. (trans. G. B. Halstead). Chicago. 1920: 2d English ed. (trans. F. Steinhardt), Chelsea, New York, 1986. Scholz, E., Geschichte des Marnigfaltigkeitsbegriffs von Riemann bis Poincare, Birkhauser, Boston. 1980.

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Riemann, B., Ober die Hypothesen. welche der Geometric zu Grunde liegen. Abhandlungen der Kiiniglichen Gesellschaft der Wissenschaften zu Gottmgen 13(1868). Rodrigues, 0., Recherches sur la theorie analytique des lignes et des rayons de courbure des surfaces, et sur la transformation d' une classe d' int6grales doubles, qui ont un rapport direct avec les formules de cette theorie, Ecole Polytechnique Corresp. 30814-16), 162-82. Serret, J. A., Sur quelques formules relatives a double courbure, J. Math. p. appl. 16(1851), 193-207. Struik, D. J., Outline of a history of differential geometry, 1, Isis 19(1933), 92-120; Ii. Isis 20(1934), 161-91. Taurinus, F. A., Theorie der Paraliellinien, published in 1825. In Engel and Stackel (1895). Tchebychev, P. L., Sur la coupe des vetements. Euvres. volume 2, 1878.708.

Symbol index a' f.261 BS (6,), SS (6,). 164

C°O(S).C"(p). 104, 204 Ca (s). 67

cj (s), 131

d(p. q). 166 d Fp. 105

In. i 1.70 circum(R), 57 cosh x. 52 8(ABC). 33 8`t, 250

8,,, 94

dNp Tp(S) - Tp(S). 135

D/dt. 263 rk 147, 254

ds2. 108

n(1). 69

Dp. 141

area(R). 113 expp(u-). 164

E. F. G. 107 e. f. g. 140 H(p). 141 J(x)(u. v). 96 K(p). 139

K. 93 k1.k2, 132 kg(s). 158 kn(3). 131

na(s). 157 O(n). 90

R. 256 S2.97 T(s). N(s). 70 T(s). N(s). B(s). 80

11p(i ). 138

]p(ii, CO. 106 K(s). 71, 80

K±(t), 66 (v. u')p. 106 11(AP).40 11(x ). 41

1*, 89 smh x, 52

r(s). 82 rgit), 167 9,.,. 262 Ftt1P,70

axv.80 fl

u-. 63

T4(:). 211 Ta(S). 104

Z(9), 73

x,,, x,.. 102

IIi I. 163

Irk. i 1.253

303

Name index Aghanis. 26 Appolonius (third century B.C.). 66 Archimedes (287-212 B.C.). 3. 27, 57, 126 Aristotle (384-322 B.C.). 26

Bartels, Martin (1769-1836), 158 Beltrami, E. (1835-1906). 217, 242, 260

La place. P.S. (1740-1827), 157

Legendre. A.M. (1752-1833). 33. 39 Leibniz. Gottfried (1646-17161, 66 Levi-Civna, T. (1873-1941), 249, 265 Lie, Sophus (1847-99), 146. 242 Liouville. J. (1803-82), 163 Lobachevskii, Nikolai I. (1792-1850), 34, 39

Bernoulli. Jacob (1654-1705). 66 Bernoulli, Johannes (1627-1748), 66 Bertrand. J. (1822-1900). 190

Mainardi, G. (1800-79), 151

Bessel. F.W (1784-1846). 163 Bianchi, L. (1856-1928). 253 Bianu?ia, D.. 215 Bolyai, W.F (1775-1856). 39. 279 Bolyai, Janos (1802-60). 39. 45 Bonnet. Ossian (1819-92) . 151, 171

Minkowski. Hermann (1864-1909). 251

Cartan, E. (1869-1951). 253 Cayley, Sir Arthur (1821-91). 236 Christoffel, E.B. 11829-1900). 253 Clavius, Christopher (1537-1612), 26 Codazzi. D. (1824-75). 151

D'Aigutllon, F. (1566-1617), 120

Darboux. G. (1842-1917), 253

Mercator (Kramer. Gerardus) (1512-94). 124 Meusnier, Jean-Baptiste (1754-93), 138 Minding. Ferdinand (1806-85). 158, 192 Monge, Gaspard (1746-1818). 95. 138

Nasir al-Din al-TUsi (1201-1274). 24. 28-9 Perrault. Claudius (1613-88). 69 Peterson, K. (1828-81). 151 PotncarC J. Henri (1854-1912), 178. 217. 227 Posidonius (ca 135-50 B.C.). 27 Proclus (410-85). 24 Ptolemy, Claudius (100-78), 25 Puiseux. V. (1820-83), 190

Descartes. Rend (1596-1650), 178

Diguet, 196

Qurra, Thiibit Ibn (836-901). 28

Einstein, Albert (1879-1955), 249. 251

Rado, Tibor, 177 Ricci-Curbastro, G. (1853-1925). 249 Rtemann, G.F. Bernhard (1826-60). 226, 242 Rodrigues. O (1794-1851), 134 Russell, Bertrand (1872-1970), 13

Euclid (ca. 300 B.C.). 3. 10 Euler. Leonard (1707-83). 4, 66-7. 95, 116, 132

Frenet. F (1816-68), 83 Fuchs, L.I. (1833-1902), 227 Gauss. Carl-Fricdrich (1777-1855), 39. 57. 116. 131. 134. 163. 171, 186. 242, 267

Graves, Carlie R., xii

Sacchen, Girolamo (1667-1733). 26, 29. 34 Schouten, J.A. (1883-1971), 265 Schweikart. F.K. (1780-1859). 39 Serret. J.A. (1819-92), 83 Simplicius, 26

Hilbert, David (1862-1943). 11. 193. 203 Hipparchus of Nicae 1190-125 B.C.). 120

Sind, Ibn (980-1037). 27

Holmgren , E , 203 Huygens, Christian (1629-95), 73

Taurinus, F.A. (1794-1874). 39, 56 Tchebychev. P.L (1821-94). 202 Tissot. N.A. (1824-1904), 127

Jacobi, C.G.J. (1804-51), 163. 179 Kepler, Johann (1571-1630), 66

Khayyam,'Umar (1027-I123), 27, 28, 34 Klein, Felix (1849-1925). 146, 230, 236. 249 Lagrange. J.L. (1736-1813), 127 Lambert, J.H. (1728-77). 32. 38-9. 52, 126

304

Vitale, Giordano (1633-1711). 34 Wachter. FL. (1792-1817), 52. 260 Wallis, John (1679-1703), 31 Weyl, Hermann (1885-1955). 201, 267

11

Subject index 1-form. 248 absolute property. 42 adjoint of a linear mapping. 93 alternate interior angles theorem. 16

Angle-Angle-Angle. 32 angle between curves on a surface. 112 angle defect. 33. 36, 59, 175, 239 angle excess, 5. 175

angle of parallelism. 40 arc length along a curve. 65 along a great circle. 5 on an abstract surface, 212

on a manifold, 251 on a surface. 108 arc-length parametrization, 65 area in the Beltramt disk. 231 on the sphere, 4 on a surface. 113 asymptotic line. 152 atlas. 206. 243 Australia. 127-9 axiom of continuity, 14 axiomatic method. 10 axioms of congruence. 18 axioms of order. 18 Beltrami model. 221 betweenness. I I Bianchi identity, 268 binormal vector. 80 bounded region. 112 cartographic coordinates, 118 catenary curve, 144

Cauchy sequence. 166 Cayley-Klem-Beltramt model. 236 center of curvature. 67 change of coordinates, 100 Chrtstoffel symbols, 147. 212

of the first kind, 253 of the second kind. 254 circle in a surface. 186 circumference of a circle. 57. 190 compatibility equations. 151 component functions of the metric. 107 cone, 98. 130 conformal mapping, 117. 224 congruence. 11. 13, 146 connection, 267 Koszul, 268 Levi-Civita, 267

constant curvature manifold, 259 contraction of a tensor, 250 convex surface. 144 coordinate chart. 96. 206 coordinate curves, 102 corresponding points. 42. 50 covariant derivative. 261-3 critical point, 98 critical value, 98 cross product, 80 cross ratio. 230 curvature directed curvature, 71 Gauss-Riemann curvature, 213 Gaussian. 139 geodesic, 158 mean, 144 normal, 131

of a space curve. 80 plane, 67 sectional. 258 total, 178 curve (parametrized. differentiable, regular). 63 of centers of curvature, 67 spacelike. null, timelike. 252 cycloid. 73

Darboux vector, 92 development of surfaces, 127. 145 diffeomorphism, 101. 207. 244 differentiable function between abstract surfaces. 207 between manifolds, 244 between surfaces, 101

differentiable structure, 206 differential of a mapping. 209, 245 distance function, 12

distance on the Poincarc half-plane, 235 on a surface, 166 dot product, 63 Dupin mdicatrix, 144 elementary neighborhood. 180 ellipsoid, 98

elliptic point. 133 embedding, 215 Euler-Pomcare characteristic, 177 evolute. 75

exponential map. 164 exterior angles theorem, 15 extrinsic property, 146 Fano plane. 19

305

306

first fundamental form, 106 flat manifold, 257 frame, 81

moving, 83

Frenet-Seret apparatus. 83 Frenet-Serret theorem. 83

Subject index

intrinsic normal, 157 intrinsic property. 146 inverse function theorem, 98 involute, 75 isometric embedding, 215 isometry, 89, 116. 145, 214

Fundamental Theorem for curves in R3, 84 for plane curves, 71 for regular surfaces. 151 of Riemannian Geometry, 267

Gauss equations, 151 Gauss lemma, 187 Gauss map, 134 Gauss-Bonnet theorem, 175 generalized polygon, 183 geodesic circle. 187 geodesic curvature, 158 geodesic completeness. 165 geodesic mapping. 217 geodesic parallel coordinates. 191 geodesic polar coordinates, 187 geodesic radius. 187

Jacobian, 96

joke. 306

Klein model. Klein-Beltrami model, 236 Kronecker delta function, 84

Lagrange's formula, 91 Lambert quadrilateral. 44 latitude. 118 Lie group, 245 Liebmann's theorem, 193 lift of a mapping. 181 line, 10 of curvature. 144, 152

in the direction of a vector field. 153 parallel to a plane. 49 perpendicular to a plane. 46

great circle, 3

line element. 108, 251 line segment, I I linear fractional transformations. 229

HAA, HOA, HRA, 35 handle. 185

Lobachevskii-Bolyai theorem. 52. 222 local equivalence problem. 252

Hausdortf condition, 206 Hazzidakis's formula. 205

local isometry, 146 longitude. 118 Lorentz manifold, 252 loxodrome, 124 June. 4

geodesic, 160. 213, 264 graph, 64. 97

real, 231

helix, 64 general, 85 Hopf-Rinow theorem. 166 Hopf's Umlaufsatz. 173 horocyclc, 42, 237 concentric. 42 horosphere, 51 hyperbolic geometry, 236 hyperbolic point. 134

hyperbolic sine theorem, 56 hyperbolic trigonometric functions, 52 hyperboloid, 98

implicit function theorem. 98 incidence geometry, I I inner product, 63, 106 index of. 252 nondegenerate, 252 nonsingular. 108

positive-definite, 108

integral curve, 152 intrinsic equations. 92

Mainardi-Codazzi equations. 151 manifold. 243 two-dimensional. 206 map projection, 116 azimuthal. 126 central. 123. 219 conformal. 117 conical. 123. 130 cylindrical, 123

equiareal, 117 ideal. 116 Lambert cylindrical. 126

Mercator, 124 orthographic, 126. 224 stereographic, 120-2, 225

Matheinamca, 127, 279 mean curvature. 144 measure of length on lines. 12, 271

Subject index

meridian. 1 19 on surfaces of revolution, 142 metric geometry. 12 metric space. 12, 166 complete, 166 minimal surface, 144 Mtnkowski space. 252 Mi bms band, 103. 114

New Zealand, 127-9 non-Euclidean circles, 236 nonintersecting lines, 39 normal curvature. 131 normal neighborhood, 165 normal vector to a curve. 70. 80 to a surface. 102 norm. 164 orthogonal group, 90 orthogonal parametrization, 112, 152 osculating circle. 67 parabolic point. 134 paraboloid, I I I parallel lines, 16. 39 in space, 45 parallel planes. 49 parallel transport. 265 Pasch's amom, 14 patch. 96 pencil of lines. 42 perpendicular planes, 48 Philosopher's Principle. 26 planar point, 134 plane curvature, 67 Playfatr's axiom, 25 Poincare disk, 226 Poincare half-plane, 228 point transitive mapping. 192 polar line element on the sphere, 127 polarization identity, 90 pole, 8 Postulate I. 11. 165. 176

Pythagorean Theorem hyperbolic, 55 Proposition 147. 17 spherical, 5 quadratic form. 138 Rayleigh quotient, 93 real projective n-space, 244 real projective plane. 207 regular value, 98

regularity, 96 reparanietrtzation, 64 reverse of a curve. 70 Ricct's lemma. 268 Riemann curvature tensor. 155. 256 Riemann sphere. 229 Riemannian manifold. 249 Riemannian metric. 210. 249 right-hand orientation, 81

rigid motion, 146 of Euclidean space, 89 Rodrigues's formula, 144 rotation plane, 94

ruler. 12 Sacchen-Legendre theorem. 15 Sacchen quadrilateral, 34 second countability. 206 second fundamental form, 138 self-adjoint linear mapping, 136 simple connectivity, 171 speed. 64

spherical coordinates. 5 spherical sine theorem, 6 standard orientation, 70 surface. 96 abstract, 206 orientable, 103 ruled, 156 topological. 98 Sylvester's law of inertia, 252

tangent bundle, 246

Postulate 11, 11, 166

tangent developable surface, 114

Postulate III. 12, 191-2

tangent plane, 104

Postulate IV. 13. 191 Postulate V, 17, 221 principal curvatures, 133 principal directions, 133 protractor. 15 pseudo-Rtemannian manifold, 252

tangent space. 208, 245 tangent vector. 64. 208, 244 tautochrone, 73 Tchebychev net. 202 tensor field. 248 theorema egregium, 148 Three Musketeers theorem, 29

pseudosphere. 196

309

Subject index

torsion, 82 geodesic torsion, 169 torus, 98 flat, 211

trace, 63 tractrix, 69. 142. 196 transition function, 243 triangulation. 176 umbilic point, 134 unit-speed curve. 65

unrolling a surface. 211 upper half-space model. 260

vector field. 152, 247 Wallis's Postulate, 31 Weierstraw coordinates, 241

world-line, 252

/ ifferential

geometry

since its beginnings with

has

developed

in

many

directions

F.uler and Gauss. This often poses

a

problem for undergraduates - which direction should one follow? what do these ideas have to do with geometry? This book offers a more focused treatment of the topic, one which is designed to make differential geometry an approachable subject for advanced undergraduates.

Professor McCleary considers the historical development of non-Luclidean geometry, placing differential geometry in the context of the syn-

thetic geometry students will recognize from high school. The text serves as both an introduction to the classical differential geometry of curves and surfaces and as a history of a particular surtace, the non-Euclidean or hyperbol-

ic plane. The main theorems of synthetic Euclidean and non-Euclidean geometry are presented along with their historical development. The author

then introduces the methods of differential geometry and develops them toward the goal of constructing models of the hyperbolic plane. Interesting diversions are offered, such as Iluvgens's pendulum clock and mathematical

cartography; however, the ti0cus of the hook is on the models of nonEuclidean geometry and the modern ideas of abstract surfaces and manififlds.

ISBN 0-521-42480-1

CAMBRIDGE UNIVERSITY PRESS

""mtanibr

'°`'°

46243922 Geometry From a Differentiable Viewpoint

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