46243922 geometry from a differentiable viewpoint

John McCleary

Geometry from a Differentiable Viewpoint

Geometryfrom a Differentiable Viewpoint

JOHN McCLEARYVassar College

CAMBRIDGEUNIVERSITY PRESS

PUBLISHED BY THE PRESS SYNDICATE OF THE UNIVERSITY OF CAMBRIDGEThe Pitt Budding, Trumpington Street, Cambridge CB2 IRP, United Kingdom

CAMBRIDGE UNIVERSITY PRESSThe Edinburgh Building, Cambridge CB2 2RU, United Kingdom

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® Cambridge University Press 1994

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no reproduction of any part may take place withoutthe written permission of Cambridge University Press.

First published 1994Reprinted 1996, 1997

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ISBN 0-521-41430-X hardbackISBN 0-521-42480-1 paperback

To my sistersMarv Ann, Denise, and Rose

Contents

Introduction page ix

PART A

Prelude and themes: Synthetic methods and resultsI. Spherical geometry 3

2. Euclid 10

Euclid's theory of parallels 16

Appendix. The Elements: Book 1 21

3. The theory of parallels 24Uniqueness of parallels 24Equidistance and boundedness of parallels 26On the angle sum of a triangle 28Similarity of triangles 31

4. Non-Euclidean geometry 1 34The work of Saccheri 34The work of Gauss, Bolyai, and Lobachevskii 39

5. Non-Euclidean geometry 11 45

The circumference of a circle 56

PART B

Development: Differential geometry

6. Curves

Early work on plane curves (Huygens, Leibniz. Newton, Euler)The tractrixDirected curvatureDigression: Involutes and evolutes

7. Curves in spaceAppendix: On Euclidean rigid motions

8. SurfacesThe tangent planeThe first fundamental formArea

8°i. Map projectionsStereographic projection

Vii

viii Contents

Central projection 123

Mercator projection 124

Lambert's cylindrical projection 126

Azimuthal projection 126

Sample map projections 127

9. Curvature for surfaces 131

Euler's work on surfaces 131

The Gauss map 134

10. Metric equivalence of surfaces 145

Special coordinates 151

11. Geodesics 157

Euclid revisited I: The Hopf-Rinow theorem 165

12. The Gauss-Bonnet theorem 171

Euclid revisited I1: Uniqueness of lines 175

Compact surfaces 176

A digression on curves 180

13. Constant-curvature surfaces 186

Euclid revisited III: Congruences 191

The work of Minding 192

PART C

Recapitulation and coda

14. Abstract surfaces 201Hilbert's theorem 203Abstract surfaces 206

15. Modeling the non-Euclidean plane 217The Beltrami disk 220The Poincar6 disk 224The Poincar6 half-plane 227

16. Epilog: Where from here? 242Manifolds (differential topology) 243Vector and tensor fields 247Metrical relations (Riemannian manifolds) 249Curvature 252Covariant differentiation 261

Riemann's Habilitationsvortrag: On the hypotheses which lie at the foundationsof geometry 269

Appendix: Notes on selected exercises 279Bibliography 297

Symbol index 303Name index 304Subject index 305

IntroductionArEaMETPHTOE 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 agreat story - it spans more than two millennia, stars an impressive cast of characters, andcontains some of the most beautiful results in all of mathematics. My immodest goal forthis book is to tell this story.

Another role of history is to focus our attention and so to provide a thread of unity througha parade of events, people, and ideas. My goal grows small and quite modest before all ofGeometry, especially its recent history. A more modest goal then is to provide a focus inwhich 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 thesubject.

In recent years, it has become a luxury to offer a course in differential geometry in anundergraduate curriculum. When such a course exists, its students often arrive with a modernintroduction to analysis, but without having seen geometry since high school. In the UnitedStates geometry taught in high schools is generally elementary Euclidean geometry basedon Hilbert's axiom scheme. The beautiful world of non-Euclidean geometry is relegated toa footnote, enrichment material, or a "cultural" essay. This is also the case in most currentintroductions to differential geometry. The modern subject turns on problems that haveemerged 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 toa meaning of geometry in differential geometry, or to identify their activity as somethingdifferent, unconnected.

This book is an attempt to carry the reader from the familiar Euclid to the state ofdevelopment of differential geometry at the beginning of the twentieth century. One narrowthread that runs through this vast historical period is the search for a proof of Euclid'sPostulate V. the Parallel Postulate, and the eventual emergence of a new and non-Euclideangeometry. In the course of spinning out this tale, another theme enters - the importance ofproperties of a surface that are intrinsic, that is, independent of the manner in which thesurface is embedded in space. This idea, emphasized by Gauss, provides an analytic keyconcerning 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 doseof spherical geometry, in which some of the important ideas of non-Euclidean geometryare 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 inChapters 4 and 5, which are an exposition of synthetic non-Euclidean geometry as introducedby Saccheri, Gauss, Bolyai, and Lobachevskii. I have tried to follow the historical path inPart 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 areal-valued measure of rigor, then it would be seen to be monotonically increasing throughPart A. In Chapter I. I take a deliberately informal tone to invite the reader to glimpse a newworld without the burden of exacting detail. What fails to be completely proved here willbe established in Part B with the methods of differential geometry. I have taken the sametone as Euclid, his critics, and the founders of non-Euclidean geometry in the rest of PartA. 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 fromthe hindsight of more formal studies. The excitement of their achievements goes furtherwithout the burden of complete rigor.

What remains unresolved at the end of Part A is the existence of a concrete representationof non-Euclidean geometry, that is, a rigorous model. The development section of thissonata is dedicated to making this possible. One of the key features leading to the foundingof non-Euclidean geometry is the introduction of analysis into the development of thefoundations. Trigonometric formulas such as the Lobachevskii-Bolyai theorem (5.8) andtheir proofs reveal the basic role that analytic notions play in geometry. Trust in analysisis what convinced Lobachevskii and Bolyai of the correctness of their ideas, and it is toanalysis 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 coor-dinates and measures such as curvature and torsion. There is a brief interlude in Chapter 6where the story of involutes, evolutes. and Hugyens's clock is told. Though it is out of theflow of the main story, Hugyens's work is paradigmatic fordifferential geometry -questionsof an applied nature press the geometer to find new ways to think about basic notions.

Chapter 8 presents the basic theory of surfaces in R3. Another interlude follows on map

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 asurface. This curvature, the Gaussian curvature, is shown to be independent of the mannerin which the surface lies in R3, that is, it is an intrinsic feature of the surface. A newtheme emerges from these considerations - in order to know what is truly geometric about asurface, 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 12leading to the Gauss-Bonnet theorem and its global consequences. In Chapter 13 we finallyobtain an analytic recipe for our desired non-Euclidean plane - it is a complete, simplyconnected surface of constant negative Gaussian curvature.

The final part of the sonata is a recapitulation of themes from Pan A. The developmentspills 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 usto consider something more general. abstract surfaces, an idea that is possible if the in-trinsic 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 acoda on the theme of the intrinsic; Riemann's visionary lecture of 1854 is discussed alongwith 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 Lorentzmetrics, vector fields and tensor fields, the Riemann curvature tensor, covariant differen-tiation, and Levi-Civita parallelism. Chapter 16 is followed by a translation of Riemann'sHabilitationsvortrag, "On the Hypotheses which Lie at the Foundations of Geometry." Mytranslation is based on Michael Spivak's found in Spivak (1970, vol. 2). My thanks go tohim for permission to use it. I have cleaned up some of the language and restored Riemann'srhetorical structure.

Exercises follow each chapter. Those marked with an asterisk have a solution in the finalappendix prepared by Jason Cantarella.

The ideal of presenting a strict chronology of ideas throughout the book would havelimited 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 allorders, though this restriction is not required to prove most of the theorems. The interestedreader should try to find the degree of differentiability needed for each construction ortheorem to obtain the most general results. This concession is to uniformity and simplicityin 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 allof the book in a semester. In order to use this book in one term, I recommend the followingchoices:

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 differentialgeometry. This serves the student who is learning the subject in order to study generalrelativity.

The short course requires the reader to be acquainted with multivariable calculus, withelementary linear algebra including determinants and inner products, and with advanced

xii Introduction

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

Acknowledgments

Many folks have offered their encouragement, time, and advice during the preparation ofthis book. Most thanks are due to my grammatical and mathematical conscience, JasonCantarella, who combed the text for errors of expression and exposition. A small part ofhis effort was supported by a grant from the Ford Foundation and by research positionsat Vassar College. Thanks to MaryJo Santagate for taking my handwritten class notes andproviding 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, andJohn Stillwell for hard-to-get materials that filled in gaps in my understanding. Thanks toJohn Jones for introducing me to the publisher, to Lauren Cowles of Cambridge UniversityPress for her patience and understanding, and to Greg Schreiber for extending the work ofa copy editor far beyond this author's expectations. Special thanks to Jeremy Gray whosebooks, 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 themesSynthetic methods and results

111

Spherical geometryGeometry is the art of good reasoning from poorly drawn figures.

Anonymous

The word geometry is of Greek origin, yewuEtpia, to "measure the earth" In antiquitygeometric techniques were used by the Egyptians to assess taxes fairly according to area. Thegeometry chronicled and developed by Euclid (ca. 300 B.C.) in his great work The Elementsbegins with the geometry of the plane - the abstract field of a farmer. Before discussingEuclid's work and its later generalizations, let us make a short detour into spherical geometryto "measure the Earth;' idealized as a sphere. The earliest science of astronomy and theneed 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 planepassing through the center of the sphere (for example, the equator and the lines of constantlongitude).

Early geometers understood that great circles share many formal properties with linesin the plane making them a natural choice for lines. For example, given any pair of non-antipodal points on the sphere, there is a unique great circle joining that pair of points; toconstruct it. take the intersection of the sphere and the plane containing the pair of pointsand the center of the sphere. Following Archimedes (287-212 B.c.) we could define a lineto be a curve that is the shortest path joining any two points that lie on it. In fact, greatcircles satisfy this criterion (as we prove in Chapter I I ). To get an intuitive sense of whythis is true pin down a piece of a rubber band on an orange and snap it tight. The rubberband 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 intersectinggreat circles as the angle formed by the intersection of the defining planes of the greatcircles with the plane tangent to the sphere at the point of intersection. With this definitionof 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 seethis 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 propertieswe need to define area for polygonal regions on the sphere are few, however, and we takethem 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 greatcircles and the antipodal points where they cross. Antipodal points or figures are those thatcorrespond to opposite ends of a diameter of the sphere. A lune is given by an angle B atthe center of the sphere between the two defining planes, which is also the angle betweenthe great circles. Assumption (4) implies

area of the lune =B

4n R2 = 20 R2.2n

Three lunes.

We assume our triangle lies in one hemisphere - if not, subdivide it into smaller trianglesand argue on each piece. A triangle is the intersection of three lunes. Extending the threelunes giving AABC to the rest of the sphere gives rise to antipodal limes and an antipodalcopy 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 angleexcess of the spherical triangle.

Since every triangle on the sphere has area, every triangle on the sphere has interior anglesum greater than n. If the radius of the sphere is very large and the triangle very small inarea - as a triangle of human dimensions on this planet would be - there is negligible angleexcess.

Spherical coordinates.

Astronomers track the positions of heavenly objects andtimekeepers calculate the positions of the sun and moon. Formost methods of calculation an analytic expression for pointson the sphere is desirable. Denote the sphere of radius R,centered at the origin in R3. by ER. In terms of rectangu-lar coordinates, ER is the set of points (x, y, z) satisfyingx2 + y2 + z2 = R2. The familiar spherical coordinates forER 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 isthe measure in radians of the angle made by the radii at each point multiplied by the radiusof the sphere. Notice that this length is unchanged when we rotate the sphere around someaxis 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 ona sphere of radius R with right angle at vertex C and sides of length a, b, and c (as shownin the diagram)

c a bcos R = cos R cos R

A right triangle.

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

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


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

A B R2cosacos$cosy = = = cos a cos /i.

IIAII IIBII R2

If we express angles a, Q, and y in radians, we get a = a/ R, i = b/ R. and y = c/ R. andso the theorem is proved.

We call this the Pythagorean theorem because it relates the hypotenuse of a right spher-ical triangle to its sides. To see a connection with the classical Pythagorean theorem (seeChapter 2) recall the Taylor series for the cosine:

x2 x4 x6cosx=l- 2 +-- -6!+....

The spherical Pythagorean theorem gives the equation

2R2+..._rl_22+...1(I_ R2+...1

a2 b`- a'-b'`=I_2R2-z22+4R4 +....

Subtract I and multiply by -2R2 to obtain

2 stuff 2 other stuffc +RZ

=a +b,`+ R2

Let R go to infinity and we deduce the classical Pythagorean theorem. Since the Earth is asphere of such immense radius compared to everyday phenomena, ordinary triangles wouldseem to obey the classical Pythagorean theorem.

In the previous proof we used the dot product on R3 and the embedding of the spherein IR3. We go further with this idea and prove another of the classical formulas of sphericaltrigonometry 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 asphere 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 lyingentirely within a quarter of a hemisphere. Suitable modifications of the proof can be madefor larger triangles.

1. Spherical geometry 7

Let DA BC be a right triangle with right angle at C. Extend the radius OA to OA' whereBA' 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' andOBC are perpendicular, that is, lines in each plane that intersect on the line of intersectionof the planes and are perpendicularto that line are perpendicular to each other. Since L ACBis a right angle, planes OBC and OAC are also perpendicular. We leave it to the reader toshow 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 anglesat the center of the sphere determined by L BOC, L AOC, and LAO B, respectively. By thedefinition of the trigonometric functions we find

A'C' A'C' BA'sin # = = = sin L B sin

OA' BA' OA' y

sin(b/R)and so = sin y. By reversing the roles of A and B we obtain in a similar fashion

sin LBsin(a/R) sin(a/R)

=sin(b/R)

= sin y, and sosin LA sin LA sin LB

For an arbitrary triangle we can construct the spher-ical analog of an altitude to reduce the relation for twosides to the case of a right triangle. For example, in theadjoining figure we insert the altitudes from A and B.By the right angle case we see

Altitudessin(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 canbe proved: There are six related pieces of data determining a spherical triangle, namely, thethree sides and three angles. Given any three pieces of these data, one can determine theother three. It follows that similar triangles on a sphere are congruent. The solution of thisproblem 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 geometryof our school days set the stage for the rest of the book. The sphere has some strikinggeometric 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 sharedby the sphere and the plane in an analytic fashion by viewing the plane as a sphere of infiniteradius. Other analytic tools, such as trigonometry, coordinates, and the embedding of thesphere in R3, open up the geometry of this surface for us. All of these ideas will return toguide 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 satisfyingx2 + 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 alonggreat 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 andf13. 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 notnecessarily determine a great circle. Give necessary and sufficient conditions to obtain agreat circle.

1.5 A pole of a great circle is one of the endpoints of a diameter of the sphere perpendic-ular 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 ofthe great circle, there is a unique great circle through the given point perpendicular to thegiven 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 byL = 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 thecongruence criterion for spherical triangles, Angle-Angle-Angle.

2Euclid

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 couldbe applied to determine the positions of the sun and stars or to measure land areas. Merchantsbrought these ideas from Egypt and Babylonia to ancient Greece where the prevailing ideasof 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 andinfluenced all subsequent generations.

The Elements proceeds by the axiomatic method - definitions and axioms are presentedfirst, then propositions are shown to follow from these assumptions and from each otherthrough logical deduction. When later mathematicians sought a model for the rigorousdevelopment 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, thedefinition of point holds little meaning - Heath, the editor of Euclid (1956). explains it as adevice to avoid circularity. The modern view is to take such terms as point, line, and planeas undefined and give axioms to determine the relations among them. One can begin witha set S (space) and two classes of subsets, C, the lines, and P. the planes. A point P. thatis, 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 thefive postulates and five common notions. We discuss these next. A list of the propositionsof Book 1 (1.1-1.48) follows in the appendix to this chapter.

10

2. Euclid 11

Postulate I. 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 strongenough 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 clausein axiom (1); Euclid tacitly assumes this, and he applies uniqueness in the proofs of certainpropositions (for example, Proposition 1.4).

The axiomatic approach proceeds without interpretation of the basic terms. This al-lows for different realizations of the axioms. For example, certain finite sets with subsetsand relations are incidence geometries and their properties are used to study problems incombinatorics and algebra.

In Postulate I Euclid described a tool for drawing, the ideal straight edge. In Postulate IIEuclid 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 furtherprimitive notions: Given three distinct points P. Q, R lying on a line 1. we want to saywhen Q lies between P and R. This relation is denoted B(P, Q, R) and can be definedset-theoretically as a ternary relation, B C S x S x S, satisfying certain axioms (to be givenlater) which make abstract the notion of betweenness. A line segment PQ C I is definedas 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 ofa 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 IIbecomes

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


The notions of betweenness and congruence for Postulate 11 are realized if the set S isequipped with a distance function, d: S x S -+ R. making S a metric space. Recall thata 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 ternaryrelation

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 geometrybased on metric postulates).

Alternatively, if we have a notion of betweenness and congruence for an incidencegeometry (S, C, P), then we can ask for a measure of length. This is a real-valued function0 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 measuresof length are proportional and, given a particular line segment AB, there is a measureof 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 functionDot, : I -' R that determines a ruler or coordinate system on 1. The point 0 is the originof the coordinate system and the ray 00' determines the positive part of the line. In laterchapters we will assume the existence of rulers on lines in order to introduce functionsdefined on lengths, that is, on congruence classes of line segments. For a proof of theexistence 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 Greeksallowed 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, onecannot 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 withoutcollapsing 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 meetone 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 otheris 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 notionof congruence of angles. Euclid imagined congruence of figures to be a motion in space ofone figure that superimposes it on another figure such that corresponding points and linesegments coincide. Bertrand Russell (1872-1970), in his 1902 article for the EncyclopediaBritannica Supplement, points out that "(a)ctual superposition ... is not required; all thatis required is the transference of our attention from one figure to another." We take acongruence, 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 anddeduces Postulate IV as a theorem.

From the modem viewpoint Euclid's Postulate IV is a statement about the existence ofenough congruences to compare every pair of right angles. For example, to superimposeadjacent right angles, we can reflect one right angle through space onto its neighbor. Euclidassumes the existence of a congruence that we call reflection across a line in his definitionof right angle. This implies the "homogeneity of space." The motion of a figure determinedby two line segments at right angles onto another that preserves angles and distances impliesthat 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 thesense of superposition. Consider the case of a torus (an inner tube): The meridians (verticalslices) and the innermost and outermost horizontal circles on the torus may be taken forlines. There is no "motion" of the torus to itself, however, such that the right angle at thepoint X of intersection of a given meridian with the innermost circle and the one at thepoint 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 inChapter 10.


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

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

PROOF. Let AB be the given line segment. By Postulate IIIthere is a circle with center A and radius A B as well as a circlewith center B and radius BA. Since AB is contained in bothcircles they meet at a point C. By Postulate I we can formthe line segments AC and BC and so a triangle. Now BC isa radius and so it is congruent to BA = AB: AC is a radiusand so it is congruent to AB. Thus an equilateral triangle isconstructed.

REMARKS. (I) Notice how the axiomatic method is applied here. only postulates anddefinitions are used.(2) Euclid assumes that the constructed circles must meet at the point C. Criticism of thisassumption goes back before the fifth century A.D. (see Proclus (1970)). The problem isovercome in Hilbert's scheme (Hilbert 1970) by requiring the axiom of continuity. This ax-iom and a related statement, Pasch's axiom, guarantee the existence of points of intersectionsfor many figures in Euclid.

Axiom of continuity (Archimedes). Given line segments AB and CD there are pointsA1, A2, ... , An lying on Ah such that CD = AAt AI A2 = = An_i An and B liesbetween A and An.

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

To illustrate why these axioms are necessary, suppose the plane is simply the set ofordered pairs Q x Q. If we take A = (0,0) and B = (1.0) in Proposition 1.1, then thepoint C = (i, ) in the proof is missing. The axiom of continuity provides the basis forthe completeness of the real number line, the property of R that all Cauchy sequences ofreal numbers converge to a real number. With this property the planes of Euclid are like R2and 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 thefact that we can bisect a line segment (Proposition 1.10) and Postulate 11, we can apply theaxiom of continuity to coordinatize a line by making points correspond to dyadic expansionsof real numbers. The notion of Dedekind cuts may also be used to coordinatize lines (seeBorsuk and Szmielew (1960)).

It is also possible to introduce a measure of angles, that is, protractors. Such a measuresatisfies properties analogous to those satisfied by a measure of lengths. Once again theaxiom 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 sidesis extended, the exterior angle formed is greater than either of the opposite interior angles.

PROOF. Consider the extraordinary figure.Given AABC, extend BC along to D. Wewant to show LBAC is less than LACD. LetM be the midpoint of AC (which can be con-structed as in 1.10) and join B to M. ExtendBM to a point E so that BM = ME (Pos-tulate II). Now CM = MA, BM - ME,and LAMB = LCME (these are vertical an-

gles 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 andD is on the ray LECM is inside LDCM and so LBAM is less than LMCD. ThusLBAC is less than LACD. To argue that LACD exceeds LA BC, construct the analogousdiagram extending AC.

Though we are stepping out of Euclid for a moment, we next derive an important theoremfor the theory of parallels. The assumptions are in place for the proof of this result, and themethod is analogous to the proof of Proposition 1.16.

Theorem 2.2 (the Saccheri-Legendre theorem). The sum of the angles interior toa 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 anglesis two right angles plus an angle congruent to LBAZ. Suppose LBAC is the least of thethree interior angles. As in the proof of 1.16, let M be the midpoint of BC and construct the


-4 ----4ray AM. Mark DI on AM so that AM MD1, and so ACMA - ABMD1. Notice thatL B A M + LCAM = L B AC so one of these angles is less than or equal to i L BAC, wherethe meaning of i L BAC is clear. Thus either L BADI or L B Di A is less than 1 L BAC. Nowobserve

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 thesmallest angle of AABC. If we iterate this procedure n times, we obtain a triangle whosesmallest 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 thisstage still has angle sum two right angles plus LBAZ. Thus the two remaining angles inthe triangle must sum to more than two right angles which contradicts 1.17 - the sum of anytwo 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 satisfiedby spherical geometry, since triangles on the sphere have angle sum greater than two rightangles. One of the points of departure for the sphere is Postulate 11; it is not possible to definea natural notion of "between" on a great circle and so the modern version of the postulatecannot be realized. The uniqueness assertion of Postulate I also fails on the sphere; whentwo 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 thatEuclid begins to construct a theory of parallels and area in order to prove the Pythagoreantheorem (1.47). Euclid defines parallel lines as follows:

(23) Parallel straight lines are straight lines which, being in the same plane and beingproduced 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 fallingon two straight lines makes the alternate interior angles congruent to one another, thestraight lines will be parallel to one another.

A E B PROOF. Let the line' cross lines Ah and t

so that L BEF is congruent to LCFE. SupposeY X AmeetsonthesideofBandDatX.Con-

sider A EFX. Now LCFE is an exterior angleC F D and hence it is greater than LFXE and LXEF

(1.16). But LXEF = LBEF = LCFE. Thisis a contradiction. We can argue similarly that A and t do not meet on the side of Aand C.

2. Euclid 17

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

Postulate V. That, if a straight line falling on two straight lines makes the interior angleson 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 simplymore difficult to state than the others and it uses loose words such as "indefinitely." PostulateV became the focus of criticism of later generations of mathematicians who accused Euclidof introducing an "unproved theorem" to simplify proofs. This criticism is the source ofcenturies 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 canbe compared. He introduces a different use of the word "equals" (in 1.35) to mean "ofequal area." Along the way he shows that the interior angle sum of a triangle is two rightangles, that corresponding angles and segments of a parallelogram are congruent, and thatparallelograms may be constructed with area that of a given triangle. After showing thatsquares 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 squareon the side subtending the right angle is equal in area to the squares on the sides containingthe right angle.

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

1g Prelude and themes: Synthetic methods and results

Ah is the line . Now LDCB + LBCA = LKCA + LBCA and sides DC = CBand AC = CK. By Side-Angle-Side (1.4) we have ACKB = OCDA. If AM is analtitude from A (that is, AM is perpendicular to BC) consider the rectangles CDNMand 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 oABFGhas the same area as 0MNEB and the result follows by putting CDNM and 0MNEBtogether.

Reading the propositions of Book I that follow 1.27, one is struck by the role that parallelsplay in the determination of relations among the areas of elementary figures. This use ofareas is in part due to the lack of symbolic algebra in Greek mathematics. The square of anumber was interpreted as the area of a square with side the given number. There are manyother proofs of the Pythagorean theorem, some of which are based on simple algebraicformulas.

Before leaving Euclid let us record the rest of Hilbert's system for describing planargeometry. This provides the modern means to make Euclid's work precise as well as anexample 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 andE C S x S x S x S whose properties are intended to encode our ideas of betweennessand congruence. If a triple of points (P, Q, R) lies in B. we say that B(P. Q, R) holds, orthat 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) andB(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 apoint 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 19

H -)determined by the line O'P'. Then there is a unique ray 0'Q' in the chosen half-plane 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-Angle-Side).

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 areindependent and so all are required. The axiom for parallel lines will be discussed in thenext chapter.

Exercises

2.1 From the axioms for an incidence geometry show that two lines I and I', 1 # I'. eithershare 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 regionscalled half-planes.

2.3 From the congruence axioms prove 1.4, that is, if given AA BC and tA'B'C' such thatAB = A'B', AC = A'C', and LBAC = LB'A'C'. then BC = B'C'. Prove the othercongruence criteria - Angle-Angle-Side. Angle-Side-Angle (1.26). and Side-Side-Side(1.8) - from the axioms.

2.4 (a) Given congruent angles OQ = LP'O'Q' and a ray 0' 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' andLXOQ = 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 andQ' 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 incidencegeometry. (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 equalto 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 theorigin as a set of lines. we can define G as the set of planes through the origin in 1R3 withE meaning I C P. Show that this set S, the set of lines L. and E so interpreted give anincidence 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 appendixto 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.


2.9' Prove that the segment joining the midpoints of two sides of a triangle has length lessthan 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 sidecongruent. 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 proofdid you need Postulate V? (Hint: Take two bisectors that meet in a point. Join the thirdvertex to this point and consider the segments perpendicular to the sides through thispoint. 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. Showthat 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 con-structions for the sum and product of the lengths of two line segments, the reciprocal ofthe 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.)

AppendixThe 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 thegiven line segment as base.

Proposition 1.2. Given a point and a line segment, one can construct a line segment withthe 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 linesegment congruent to the lesser.

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

Proposition 1.5 (pons asinorum). In an isosceles triangle the angles at the base are con-gruent. If one of the congruent sides is extended to form a new triangle and the other side isextended 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 thecongruent angles are congruent.

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

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 pointon the line.

Proposition 1.12. To a given line and a point not on the line one can construct a linethrough the point perpendicular to the given line.

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

21


Proposition 1.14. If two lines through a point on a given line form adjacent angles withthat 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 ex-tended, 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 aregreater than the remaining one.

Proposition 1.21. If a triangle is constructed inside a given triangle on its base, then thetwo sides of the new triangle are less than the corresponding sides of the given triangle andthey form a greater angle.

Proposition 1.22. Given three line segments so that any two taken together are greaterthan 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 agiven angle.

Proposition 1.24. If two triangles have two sides congruent, then the greater includedangle is opposite the greater base.

Proposition 1.25. If two triangles have two sides congruent, then the greater base isopposite the greater included angle.

Proposition 1.26 (Angle-Side-Angle, Angle-Angle-Side). If two triangles have two an-gles and one side congruent, then the triangles are congruent.

Proposition 1.27 (the alternate interior angles theorem). If a line falling on two straightlines makes the alternate interior angles congruent to one another, the straight lines are parallelto one another.

Proposition 1.28. If a line falling on two straight lines makes the exterior angle congruentto 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 congru-ent, the exterior angle congruent to the interior and opposite angle, and the interior angles onone 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 parallelto the given line.

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

Appendix. The Elements: Book/ 23

Proposition 1.33. The line segments joining corresponding endpoints of congruent andparallel line segments are themselves congruent and parallel.

Proposition 1.34. In a parallelogram the opposite sides and angles are congruent and thediameter bisects the area.

Proposition 1.35. Parallelograms on the same base and in the same parallels have equalareas.

Proposition 1.36. Parallelograms on congruent bases and in the same parallels have equalareas.

Proposition 1.37. Triangles which are on the same base and in the same parallels haveequal areas.

Proposition 1.38. Triangles which are on congruent bases and in the same parallels haveequal 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 ofequal areas are also in the same parallels.

Proposition 1.41. If a parallelogram has the same base with a triangle and they are in thesame 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 thearea of a given triangle.

Z7 Proposition 1.43. In any parallelogram the complements of the paral-lelogram about the diameter are equal in area.

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

Proposition 1.45. One can construct in a given angle a parallelogram of area equal to agiven 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 hy-potenuse 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 sumof the squares on the remaining two sides of the triangle, then the angle contained by the twosides is a right angle.

3

The theory of parallelsThis 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 isclear then from this that we should seek a proof of the present theorem, and thatit is alien to the special character of postulates.

Proclus (410-85)

The most reliable information about Euclid and early Greek geometry is based on thecommentaries of Proclus whose objections to Postulate V are stated in the quote. To itsauthor 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 andstraight edge. Furthermore, Euclid studiously avoided using Postulate V for the first 28propositions of Book 1. The first application was to prove 1.29, the converse of 1.27 and1.28. Several of the previous propositions are related to their neighbors as converses withproofs 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 toprove it?

To eliminate this "blemish" on Euclid's great work, subsequent generations heededProclus's call and either sought a proof of Postulate V from the other assumptions, or theytried to replace it with a more self-evident assumption. Some of the titles of their effortsindicate 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. latergenerations, seeking to avoid this fate, corrected the work of their predecessors only tocommit 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 theassociated 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 byProclus (1970).

24

3. The theory of parallels 25

Proposition 1.29. A line cutting two parallel lines makes alternate angles congruent, theexterior angle congruent to the interior and opposite angle, and the interior angles on thesame side congruent to two right angles.

PROOF (Ptolemy). A straight line which cuts parallels must make the sum of the interior

A F/ B angles on the same side equal to, greater than, or less than tworight angles. "Greater than" is not allowed because, following the

figure, if L BFG and L DG F sum to more than two right angles,so also must LA FG and LCG F because A and CMG are noDmore parallel than FB and G. "Less than" is argued similarly.

Proclus identified the source of Ptolemy's petitio principii which has since become knownas 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 throughthe point parallel to the given line.

A FI

C IG D

In the figure we see how two parallels through the point F. ifthey existed, would ruin Ptolemy's argument. In fact, Playfair'saxiom is equivalent to Postulate V. and so Ptolemy had actuallyassumed 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 thatthe interior angles on one side sum to less than two right angles. Construct line ti throughF 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 andfurthermore, D is inside L EFM. Therefore tb meets A on the side of B and M.

CF M

L

nA E B

In the other direction, if we are given a point F not on a line A construct (1.12) the linerMso that EF is perpendicular to rM and EF is perpendicular to Ah. By Proposition1.27, rM is parallel to A4-h. Suppose tb is any line through F. If t 54 t, makesan angle on one side less than a right angle. By Postulate V. C meets Ah. Hence rM isthe unique parallel.


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 distancebetween 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 Fand 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 greaterthan 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 lessthan two right angles, construct the line n such that L BFP and L DPF sum to two rightangles. Proposition 1.28 implies that Ah is parallel to and so T-6 cuts a parallel at thepoint 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 theperpendicular 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 fifthpostulate due to Aghanis who asserts that equidistant straight lines exist. This assumptionis 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 medievalIslamic geometers. The philosopher is Aristotle (384-322 n.c.) who wrote on parallels asa source of logical errors. The Jesuit priest Christopher Clavius (1537-1612) asserted thatit needed proof and he offered that it is false of other species of curves (such as a hyperbolaand 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 themidpoint of that segment, then if BD is the perpendicular to AC meeting at D. B D is notgreater than half of CM.

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


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 iterateddoubling and the axiom of continuity, we can constructa perpendicular from M to AC that exceeds any givenlength.

The assumptions that parallel lines were a fixed distanceC P apart or that they were no more than a bounded distance

apart find their way into many of the attempts to prove Pos-

tulate V. Equidistance may have figured in Archimedes's definition of parallels (Rosenfeld1988, 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 wellas 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 equivalentto Postulate V.

PROOF. From Postulates I through V. one can prove Euclid's proposition 1.34, which assertsthat, in a parallelogram, the opposite sides and angles are congruent to each other. ConstructP perpendicular ton and ) perpendicular to k. Also construct (1.12) V throughS 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 arecongruent by 1.34).

By the congruence criterion Side-Angle-Side,ADBAiscongruenttotHBMandsoDA = HMandLB H M is a right angle. Hence LDHK is greater than aright angle. The reader can now prove (or look in Chap-ter4) that LCKH is congruent to LDHK. This givesus a quadrilateral with angle sum greater than four rightangles, which is not possible by the Saccheri-Legendretheorem (Theorem 2.2). Hence twice BD is less thanor equal to CM.

S P in

P R

n

I

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


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 aline, 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 translationand preservation of texts by the people of medieval Islam. The distinguished geometers ofthis culture also tried to improve Euclid by proving Postulate V or replacing it with a morereasonable 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 talways make with Ah unequal angles always acute on the side toward A and obtuse on theside toward B, then the lines approach continually nearer in the direction of the acute anglesand diverge in the direction of the obtuse angles, and the perpendiculars will diminish andincrease in the respective directions. Furthermore, the converse holds.

This assumption is found in one form or another in the work of Thabit ibn Qurra (836-90 1), 'Umar Khayyiim, and Nasir al-Din al-Tusi. The most important consequence of thisassumption is the following:

Lemma 3.5. Under the previous assumption, if AC and B D are at right angles to A Band 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-

n

tuse. In either case, by the assumption, AC could not becongruent to BD because the sides AC and BD are dimin-ishing or increasing .

A BNotice that this can be applied to show that there are

triangles for which the three interior angles sum to tworight angles. Take the diagonal of the quadrilateral discussed in the lemma and apply theSaccheri-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.

n

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


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. ThusLCAB + LCBA + LACB = LECA + LACB + LBCD = two right angles.

In the other direction, we are given lines 1, cut by a line ACS with A perpendicular

to n and LCAB acute. Let GI be on AB and construct GI H, perpendicular to AC' atHt. Now Hi falls on the side of V toward C. If AH, is congruent to or exceeds AC, thenwe are done by Pasch's axiom.

n

Suppose H, falls between A and C. Construct points H2, H3, .... H on A with

AHi = HI H2 = = H and AH greater than AC. This is possible by the axiomof continuity. Construct points G2, G3, ... , G along A with AGI GIG2

Let G2 K be the perpendicular to A_ from GZ at K. We show that K = H2. LetAL = G, H, be perpendicular to AC and join LG,. Since the angle sum in a triangle is tworight angles we see that L ALG, = L H, G, L = a right angle, and AAG, L /G, HI A.Thus AH, = LG,. Let M on G2K be such that KM = G, H,. By the same argumentL K MG I = L HI G I L = a right angle and G I Al = H, K. Since L, G, . and M are now seento be collinear, L LG, A = LG2G, M and so by Angle-Angle-Side, AG I MG2 - LAG, L AandH,K - GIM - LGI - AH, =H,H2.Therefore K=H2.

By iterating the argument we prove that Gk HA. and in particular. G H,,, is perpendicular

to Thus t is inside DAG N and n is parallel tot . Pasch's axiom implies

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 rightangles. 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 interiorangle .sum equal to two right angles, then every triangle has angle sum two right angles.

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


LBTC + LTCB + LCBT + LBTA + LTAB + LABT = four right angles. By theSaccheri-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 righttriangle. We now show that every right triangle has angle sum two right angles.

M

SB

A T N

Construct S, as in the figure shown, so that S is not on the same side of the line A as Tand AS - BT, BS - AT. Then AA BT - 6BAS and AT BS is a rectangle. If AA MNis 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.

T' Now let A PQR be any right triangle and construct AA'QT'congruent to A A N M so that A' Q > P Q and T' Q > R Q. Thetriangle 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 ofAPR 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 arbitrarytriangle 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 trianglehas angle sum strictly less than two right angles, then all do. Also, if one wants to provePostulate V from Postulates I through IV, then it suffices to produce one triangle with anglesum two right angles. Saccheri tried in vain to prove this (Saccheri 1733). His efforts willbe discussed in the next chapter.


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 atheorem 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 mag-nitude. That is, given a triangle & ABC and any line segment DE there is a point Fso that AABC is similar to ADEF, that is, LABC = LDEF, LBCA = LEFD, andLCAB = LFDE.

Wallis knew that he had assumed Postulate V in his postulate, which he considered morenatural. Euclid's Postulate III provides similar circles of arbitrary radius. Wallis's postulateis 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?) tothe point C as follows - choose points Bt on A and Ai on A as in the figure. To anypoint 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. Thusthe 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 on t.5Now construct the triangle

with base AC similar to AC1 CE. This new triangle has sides that are segments along Aand t and so these lines meet.


In the other direction we present a proof due toSaccheri (1733, Scholion III). Suppose we are givenL1 A BC and segment DE. By Euclid's 1.23 constructDk andESsothat LEDR-LBACandLDES-LABC. By I.17 these base angles sum to less thantwo right angles and so by Postulate V, 5 meetsES at say F. It suffices to show L DFE - LACB.

B D E Let N be on DF so that NF - AC. By Playfair'saxiom 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 andPostulate V holds, LNMF = LDEF. By Angle-Angle-Side ANMF = ABC and soLACB - LNFM LDFE and AABC is similarto ADEF.

This theorem is startling in its consequences. Suppose Postulate V were false. Thenno similar triangles could exist and Angle-Angle-Angle is a congruence criterion. AsJ. 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 suchan angle, which could be constructed anywhere, could be a standard of length, and geo-metric construction would replace the need for a Bureau of Standards. To contrast withmore familiar notions, we use an absolute unit, "1 /360 of a circle;' to measure angles (thedegree). Similarly, given a line segment, we could construct an equilateral triangle on itby Proposition 1.1. The amount that this differed from n in angle sum could be fed into afunction which gave out the length of a side, since the angle defect of an equilateral trianglecompletely characterizes it by AAA.

The complete lack of similarity has a major drawback - no one could carry a reliablestreet map for it would have to be the size of the city. Architectural plans would need to bethe same size as the building!

Many other equivalents to Postulate V can be given. In the following theorem we mentionsome 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.


(10) There exists an acute angle such that every point in the interior of the angle is on aline 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 remoteexterior angle.

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

3.3 Suppose all triangles have an angle defect, that is, rr- the interior angle sum of anytriangle is a positive number. Let S(AABC) = a - LABC - LCAB - LBCA denotethe 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 triangleis the sum of the angle defects of the constituent parts.

3.4' Criticize the following "proof" of Postulate V (due to Legendre, 1833) and determinewhich of the equivalents of Postulate V he assumed:

F "Given DABC, construct on BC the triangle LBCDcongruent to AABC with LDBC = LBCA, LDCBLCBA. Then draw through D any line which cuts A andAh. If the angle sum of DABC is jr - S. then the anglesum of AAEF is less than or equal to rr - 28. Repeatingthis process n times gives a triangle with angle sum lessthan or equal to n - 2"S. If S > 0, then n can be chosenlarge 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 thatPostulates I through V cannot hold. Show how the various equivalents of Postulate V failon the sphere and interpret their failure as a fact about the geometry of the sphere.

4Non-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 Geometryappeared, most of the critics of the Elements were concerned with the "purification" ofEuclid's work from its perceived imperfections. So strong was the conviction that Postu-late V depended on Postulates I through IV that some of them did not see in their work thebasis 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 ofGauss, Bolyai, and Lobachevskii, the founders of a "new geometry."

What distinguishes this chapter from the next is the role of analysis in the discussion, aswell as the role of space (three-dimensional). In this chapter we restrict ourselves, for themost 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 Euclidesab omni naevo vindicates ("Euclid vindicated of every flaw," Saccheri (1733)) marks atriumph of logic in the pursuit of a proof of Postulate V. It also contains the beginning ofthe 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 accomplishedthrough a reductio ad absurdum argument; assume Postulates I, II, Ill, and IV and thenegation of Postulate V and then reason to a contradiction. Saccheri's great insight is hisidea 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 hadbeen introduced much earlier in the work of 'Umar Khayyam. It also figured in a work byGiordano Vitale (1633-1711) who gave another petitio principii proof of Postulate V in his1686 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 formthe 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.

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.

C PROOF. Consider the line segments AC and B D. Since A D- BC

A B

We need the next observation in a later result.

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

ND

I/N ABCN. By Side-Side-Side we find AAMN - ABMN and so1A MN = LBMN and MN is perpendicular to AB. By joiningD to M and C to M the same argument applies to show that MNis perpendicular to CD.

C PROOF. Join AN and BN. By Side-Angle-Side, LADN -

A M B

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 takeHOA 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 baseAB. Under the assumption HRA. HOA, or HAA we have AB = CD, AB > CD, orA B < CD, respectively, and the angle sum of a triangle is =, >, or < two right angles,respectively.

NU

A1n

M

PROOF. Let M and N denote the midpoints of AB and CD, re-spectively. We consider the assumption HAA since HRA is knownand the argument for HOA is similar. Suppose CD < AB, thenCN < BM and so extend CD to T with NT - MB. ThenL 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

C T

exterior angle L BCN. By HAA, L BCN is acute. Since L BT N = L T BM > a right anglewe get a contradiction to the exterior angles theorem (Euclid 1.16). Thus AB < CD.

36

D

Prelude and themes: Synthetic methods and results

C Now suppose we have AABC. a right triangle with right an-gle at B. Construct AD perpendicular to AB with AD = BC.Assuming HAA, CD > AB. Thus LDAC > LACB since thegreater side subtends the greater angle (Euclid 1.25). But LCA B +

A B LDAC =a right angle. Therefore, the angle sum of AA BC is lessthan two right angles.

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

C X Y D Consider two parallel lines A and 5, and construct

lines AU and YV perpendicular to --h from X. Y onConsider L U XY and L V Y X. If both angles are acute, then,

k by continuity. A and have a common perpendicularh UX d VY If f /" VVb

H IV V I DLV YX is right and the other is acute (by HAA), then CDand 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 YXmay change to a right angle (giving a common perpendicular) or remain obtuse.

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

PROOF. If. for some V'Y' perpendicular to Al to the right ofW 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 assumptionLV YW is acute. However, LV'Y'Y is obtuse and the exterior

V V' 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 thiswe use the fact that, under HAA, all triangles, and hence all quadrilaterals, have an angledefect. We also assume that we can measures angles. By Theorem 4.3 the sum of the interiorangles in a triangle is less than two right angles and so the sum of the interior angles ofa quadrilateral is less than four right angles. Denote the angle defect of a quadrilateralKLMN by

tween t e segments an one o ore

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 Vto 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 toAl 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 mustbe some Vk Vk+1 Yk+l Yk with S(Vk Vk+1 Yk+l Yk) < 7r/n. Since VkYk > ST we can alwaysconstruct a quadrilateral VkS'T'Vk+l with VkS' = T'Vk+l = VkVk+I = ST andeverysuchquadrilateral is congruent to any other. Let S (Vk S' T' Vk+l) = p > 0. We now choose n largeenough 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 impliesHRA and it cannot be greater because LVnYnC + LVn_IYn_1D would be greater thantwo right angles, which would violate HAA. Saccheri shows that asymptotic lines as inTheorem 4.4 exist under HAA.

Theorem 4.5 (Proposition XXXII). Given a point P not on a line A1 there are threeclasses 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 thatLAPD' < LAPD" < a right angle.

38

By the exterior angles theorem L PD" D' is acuteand LA PD" is acute; by continuity of the changeof 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'saxiom, 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,

PP'{ meets A1, that is. PAL is the upper limit of lines through P meeting A41. Let PZ' bethe line so that if Pte' is any line with L A PV acute and L A PV > L A PZ', then PV has acommon 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'.

Claim: Pd = PZ'.

Suppose PL 0 PZ' Since LZPZ' is acute, the distancebetween Pl and PZ' can be made as large as desired(Corollary 3.3). So choose S' on PZ and Son Pal so that

SS' > AP. Construct the perpendicular from S' to A at

T. Since PZ' is asymptotic to AP, AP > S'T. Now thedistance between a line and a point is given by the perpen-

dicular and since S' is on the opposite side of Pal from T.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 itis 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 flawlesslogic, he attributes properties at infinity that are only true in finite ranges. The flaw in hiswork was observed by J. H. Lambert (1786).

From Saccheri's work we have a wealth of theorems true for the system of geometrydetermined 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 anew world he had effectively explored. This new world waited for almost another 100 yearsfor giants to lay claim to it.

Prelude and themes: Synthetic methods and results

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, Lam-bent, Taurinus, Schweikart, F. Bolyai, and others (see Bonola (1955) or Gray (1979)) to getto 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 excerptfrom 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 natureof Euclid's geometry. Gauss's unwillingness to publish his work left the public introductionof this "curious geometry" to Janos Bolyai (1802-60) and Nicolai 1. Lobachevskii (1792-1850) 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 perpendicularfrom P to Al A line "is parallel to A1 through P if for any line Iq with S lyingin 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 parallelis the "first" line in the set of lines through P, sweeping

up from not meeting A4-h. Notice that without PostulateV we do not know if this first line is unique (Playfair's axiom),and so it depends on which direction we sweep the lines. Toinclude 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 theseparallels.

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

PROOF (Gauss, 1870, VIII, pp. 202-5 (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 definedHfor lines in a direction. Consider any line through P', P'C, such that C is inside BA P'Q.

Join P to C and by assumption Pt meets A *-h. But now P'C has entered a triangle and, byPasch'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

so that AR is perpendicular to . 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 itdoes at point D. Let E be such that R E - RD, E # D. Then LEAD = L SA B. Choose D kwith L A D F = L A ED. The line F is parallel to Ah and so DDD' meets A-h at say G. Let Hbe on " so that EH - DG; then by Side-Angle-Side, AA EH is congruent to AA DG.Thus LEAH =LDAG. Now LEAN = LEAD+LDAH = LSAB+LDAH = LDAGwhich 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 isparallel 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 QA QM r

P S T

U

A

B

Case 2. Ah is between " and UV. If Al is not parallel to UV, there is a line Uwith Ah parallel to U{'. Now W is not inside UV BA since "is parallel to UV, whichimplies 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 ", andso UV meets P, a contradiction.

From this notion of parallel we define a function of a perpendicular line segment to agiven line:

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

A

p Under the assumption HRA, I1(AP) equals a right angle forany AP; HAA implies that fl (AP) is strictly less than a right

R(AP) angle. We record some elementary properties of P(AP) here. Inthe next chapter we will determine an analytic expression involvingfl(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).

Then there is a line R1 so that LC R T- L A PQ and

a point, say U. Now let E be on Awith A E = CU and consider the triangle AA PE. SinceAP = CR and LPAE = LRCU, by Side-Angle-Side, RAPE = ACRU. Then LAPE = LCRU,which equals the angle of parallelism, and the linesand P coincide, which is impossible. The case forfl(AP) > fl(CR) is symmetric.

Suppose we have chosen a unit length and there is a measure of length on line segments.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 themeasure.(2) If HAA holds and A P > A R. then Il (A P) < fl(AR), that is, the angle of parallelismis 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 itto MS'. Since Il(AP) = fl(AR), LMPQ' = LMRS'. We also have AMR - MP andL PMQ' = L RMS', so APQ'M = L RS'M. It follows that MS' is perpendicular to .

This implies that n and share a common perpendicular. Since is parallel to.under HAA this cannot happen.

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

P

M

S.

A

Q'

Q

S

(3) In fact, any acute angle is the angle of parallelism for some length. To see this, fix theangle L BA X. Suppose R, S, is perpendicular to A at R, and St lies on. The triangleAA R, SI has some angle defect, say S > 0. Mark R2 along Ah with R, R2 = A Ri. If theperpendicular to Ah from R2 meets at S2, then R2 S2 > R, S1 and we can construct thepoint S, along R2S2 such that R2S, = R1 S. It follows that AAR2S2 contains two copiesof LIAR, S, and so the angle defect of AAR2S2 is at least 28. If every perpendicular to A1within L BAX meets Al, then we can iterate this procedure to construct a triangle withangle 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.


We now introduce an object which has the property that it is absolute, that is, it does notdepend on HAA or HRA. With it we can organize many properties of parallels. We beginwith a pencil of lines, which is a set of lines satisfying one of two properties - the pencilof 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 setof lines passing through the same point "at infinity" to gainan analogy to the pencil of lines through a point. Notice thatevery 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 withrespect to the pencil, if the segment XY makes congruent angles with the unique linesin the pencil through the points. If one has a pencil of parallel lines, the set of all pointscorresponding to a given point is called a horocycle.

If one has a pencil of lines through a given point, then itis a simple consequence of the converse of pons asinorum(Euclid 1.7) that the set of points corresponding to a givenpoint 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 thepencil of parallel lines that determine the horocycle. The corresponding angles at A and Bhave measure fl (AM) = fl (BM), where M is the midpoint of AB. This will be useful inChapter 5.

We say that two horocycles are concentric if they are determined by the same pencil ofparallel lines. The statements in the following lemma are elementary to prove and are left tothe reader. Henceforth we will assume that we have chosen an interval as unit length and thatall 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 thelonger there corresponds a longer arc.(2) Segments of lines in the pencil intercepted between two concentric horocycles arecongruent.(3) The arcs of concentric horocycles intercepted between two lines in the pencil decreasein 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 interceptedbetween two lines in the pencil is expressible in terms of an exponential function of thedistance 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 then

congruent segments along the parallels AA', BB', and CC' induce equal subdivisions of

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

A'B', and son

AB n BC BCA'B' m B'C'

='B'C'

n

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

B' B"

sl .s2 s3

A'Y

A..

Now suppose sl = AB, S2 = A'B', and S3 = A"B". We can write s,/s2 = f(x) ands2/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'(x) = lim f(x + Ax) - f(x)=

f(Ax) - Iex-O Ax - f(x) ex-O Ax


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

limf (AX) - 1 = I

ax-.0 Ax k

Then f(x) =k

f(x) and so f(x) = e /' or AB = A'B'eAA'l

The value I/k is chosen with malice of forethought, we will see in Chapter 13 how itfigures in other situations. Before introducing trigonometry and the further properties of theangle of parallelism, we move on to some fundamental constructions due to Lobachevskiiand 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, showthat 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, underHAA, I and m diverge.

4.3 Complete the proof of part (b) in Theorem 4.7 where you consider a ray A that doesnot meet the line ".

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

4.5' Given a pencil of lines parallel to a given line, if P. Q, and R are points, and P and Qcorrespond and Q and R correspond with respect to this pencil, then show that P and Rcorrespond 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 agiven line. Show that the set of points that correspond with respect to this pencil is thecurve of points equidistant to the given line. Observe that HRA holds if and only if thiscurve is a line.

4.9 Fix a line Al and a perpendicular segment OA. Suppose I is a line through O. andM is a point on I in the direction of X. Suppose that the perpendicular to A' through Mmeets A ' at P. Show that the length of PM is a continuous function of the lengthof the segment OM. (Hint: Use Saccheri quadrilaterals.)

511

Non-Euclidean geometry III have created a new universe from nothing. All that I have hitherto sent youcompares 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 predecessorsin that they were forging a new geometry, not repairing an old one. They also developedmethods in a three-dimensional space where HAA holds in each plane leading them toformulas 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 spacewe need to explore certain relations between pairs of lines, lines and planes, and pairs ofplanes. To begin, two lines that do not meet in such a space may be skew or they may becoplanar, that is, they lie in the same plane. If the latter occurs, then we can speak of thelines 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 twoplanes 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 12-

MQ

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

S 12. Let To denote the plane on which 11 and 12 lie. If12 m meets 11 at a point P, then 12 and P lie on both the

planes To and T2, which contradicts the assumptionIt

B S that T1 96 T2. Similarly m does not meet 12. HA A' Now let Q and Q' lie on m, let A lie on 11 = AA',

and let B lie on 12 = BB', so that AQ is perpendic-ular 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 isparallel to 11. Similarly m is parallel to 12.

45


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

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

DeSnitlon 5.3. A line ! Is perpendicular to a plane T if ! meets Tat a point P and everyline 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 andonly if there are two lines m i and m2 lying on T, passing through P, and perpendicularto 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 exitsthrough a point Son R1 R2. Choose points Qi and Q2 on ! with P between Q1 and Q2 andPQr = Q2 P. Because m r and m2 are both perpendicular to 1. the right triangles formedsatisfy

OQ1 PRi = AQ2PRi, and OQ1 PR2 - LQ2PR2.

This implies RI Q1 - RI Q2 and R2Qi = R2Q2. By the congruence criterion Side-Side-Side, ARI Q1 R2 = ARI Q2R2. This implies that LQr Ri R2 = LQ2 Ri R2 and LQr R2RrLQ2R2Ri. By Side-Angle-Side ARuSQu = AR1SQ2 and so SQi = SQ2. ThusA Q1 SQ2 is an isosceles triangle with the segment PS a bisector of the base Q i Q2. HencePS is perpendicular to !.

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

Corollary 5.5. Given a line I and a point P, then there is a unique plane T containing Pand perpendicular to 1.

PROOF. We prove the case when the point P lies on the line l. The case P ' I is similar andis left to the reader. Because we have an incidence geometry there is some point Qj not on


the given line !. Let Tt be the plane generated by the line ! and the point Qt. On Ti thereis a line 1i perpendicular to ! through P. Now the assumption of an incidence geometryfurther implies the existence of another point Q2 not lying in the plane T1. Let T2 be theplane generated by / and Q2. On T2 let 12 be the line perpendicular to ! through P. Theplane 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 theplane T" generated by I and Q meets the plane T in a linen and the plane T' in a line n'. However, because Q lies onn', n 54 n'. This gives two lines on the plane T" passingthrough 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 agiven plane passing through a given point is solved with thehelp 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 anyline on T and rA the line on T perpendicular to m with A on m, then if B # P is anyother point on 1, then the line segment A B is perpendicular to m.

PROOF. 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 soPQ = PQ'. Again by Side-Angle-Side the triangles APBQ and APBQ' are congruentfrom which it follows that B Q = B Q'. Finally triangles ABA Q and ABAQ' are congruentby Side-Side-Side and so LBAQ = LBAQ', that is, AB meets m at right angles.

Corollary 5.7. Given a plane T and a point P there is a unique line I perpendicular toT and passing through P.

PROOF. If P E T, then let !t be any line on T passing through P. By Corollary 5.5 thereis a plane Ti containing P and perpendicular to !i. On Ti let 1 be the line through Pthat is perpendicular to the line m given by the intersection of T and T1. This line is alsoperpendicular to !i and so I is perpendicular to T.

If !' is another line perpendicular to T passing through P, then let T" denote the planedetermined by ! and 1'. This plane meets T in a line m and m has the property that 1 and


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

If P ¢ T, then suppose m is any line on T. On the plane determined by P and m formthe line segment PQ with Q on m and PQ perpendicular to m. On T construct the linen perpendicular to m through the point Q. In the plane determined by n and P let A P bethe line segment perpendicular to n with A on n. By construction the plane of P, A. andQ is perpendicular to the line m. Now apply Lemma 5.6 with P and Q reversed to seethat the line I = AP is perpendicular to the plane T. The uniqueness of I follows from theuniqueness 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 planesbeing perpendicular.

Definition 5.8. A plane T is perpendicular to a plane T' if there is a line I on T' that isperpendicular to T.

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

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

Some further properties of perpendicular planes and lines in space are stated in thefollowing 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 isperpendicular 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 andnonintersecting.


Proposition 5.11. Given a line ! and a plane T with I not lying on T there is a uniqueplane T' containing I and perpendicular to T. Given two distinct planes T1 and T2, thereis a plane T" perpendicular to both T1 and T2.

PROOF. Let A be any point on I that does not lie in T. By Corollary 5.7 there is a linem passing through A and perpendicular to T. The plane T' determined by ! and m isperpendicular to T. By the uniqueness of the line m and Proposition 5.10 the plane T' isthe 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 anyplane containing I is perpendicular to both T1 and T2. If 154 m, then the plane determinedby 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 givenplane T, then the line m = T f1 T' is called the perpendicular projection of l onto T. Aline ! is parallel to a plane T if ! is parallel to its perpendicular projection onto T. Twoplanes are parallel if the intersections with some common perpendicular plane are parallellines.

Notice that two planes that intersect cannot be parallel; a common perpendicular isperpendicular to the line of intersection. Thus parallel planes do not intersect. We nextprove 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 containsI and is parallel to T.

PROOF. Let fl be the unique plane containing ! and perpendicular to T. Let in = n n T so1 is parallel tom. Let T' be the plane containing ! perpendicular to fl. By definition T' isparallel 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

direction of parallelism consider the plane perpendicular to fl containing A B. This planeseparates space into two half-spaces. The half-space containing A' and B' is the directionwhere intersections will occur between planes.

Suppose U is another plane containing I. Let Z denote the point on U with BZ per-pendicular to U at Z. We first observe that under HAA Z lies in the half-space of A' andB' between T and T'. Consider the plane determined by Z and m. The intersection of thisplane with U is a line that is parallel to ! and m by Lemma 5.1. Furthermore the angleLB'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 showthat U meets T we consider the plane Il' determined by A B and Z; fl' is perpendicular toU. Let BB be the intersection of fl' and T; also Al = Il' fl U. We show that L BAZ <

-81LBAY = fl(AB). It then follows that AAlr meets from the definition of the angleof parallelism. To compare the angles join B to Y and consider the triangles ABAY andABAZ. Since BZ is perpendicular to U, ABZY is a right triangle and so, by Euclid'sProposition 1.18, BZ < BY. By Lemma 5.6 LBYA is a right angle. In the plane of ABYAconstruct a triangle ABZ'A = ABZA. Since BZ < BY, LAYB = LAZ'B, and bothtriangles are on the base AB, a judicious use of Euclid's 1.19 implies that L BAZ < L BAYand the theorem is proved.

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

BFA rn

5. Non-Euclidean geometry !! 51

Definition 5.14. The set of points corresponding to a given point in space with respect toa given pencil of parallel lines is called a horosphere.

A reasonable property to be expected of the horosphere is that it be "homogeneous" inthe sense that it can be defined by any of its points. This is made precise by the followingresult.

Proposition 5.15. Two points in space corresponding to a third correspond to each other.

PROOF. In Lobachevskii's Theory of Parallels this is Proposition 34 and Proposition 10 inBolyai's The Science Absolute of Space (see Bonola (1955)). Suppose A corresponds to BHand A corresponds to C. Let D be the midpoint of A B and E the midpoint of AC. Let AA',H H H HBB', CC', DD', and EE' be lines through these points in the pencil of parallels. By thedefinition 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 toHA B. and lying in the plane of A BC. The line FF' perpendicular to the plane of A BC is partHof the pencil of lines since it is parallel to DD' by the choice of the length of FD. It follows4- 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 isperpendicular 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 andconsider 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 henceto 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 pencilof parallels as the axis of the horosphere and any point as the initial point to define it (allother points on it correspond to the initial point).


If we intersect a plane containing one of the parallels in the defining pencil with thehorosphere, the intersection is a horocycle. Furthermore, any two distinct points on thehorosphere 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 horocyclesand points on the horosphere is Euclidean. This result was established before the workscited 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 measureof length on line segments.

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

tan(n(x)/2) = -xlk.

Except for the determination of the constant k, this formula gives an analytic expressionto 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 thehyperbolic trigonometric functions sinhx, cosh x, and tanhx. These functions arose asanalogs of the ordinary trigonometric functions with which one can express the area un-der a semicircle as an integral f a - x 2 dx. Area under a hyperbola is given by theintegral f x - a dx whose solution shares many properties with ordinary trigonomet-ric functions. In 1768, J. H. Lambert produced a comprehensive study of the hyperbolictrigonometric functions containing the work of Euler and the Bemoullis.

Formally introduce a complex variable into the Maclaurin series for ex to get

i) _ Y2 iY3 v4e = l +iy -2!

-3! 4!

/ 2 a 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 functionsex -x ex _ e-x

cosh x =

2e

, sinh x2

, and tanh x =cosh x

It is interesting to note that

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 anygood 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 thelengths of each segment, then evl k = cosh(u/k).

PROOF. Consider the diagram. Here TT is parallel to both VX and -0 which are takento be perpendicular. Choose some length u > 0. Along 01 let A denote the point withOA 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' theparallel 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 theircommon 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.HLet B denote the point along AA' where OC crosses. Let V denote the point correspondingHto B along UU'. Write BN = u + v. and lets denote the length of OB; we then have

a = (a - s)et°+Olk oror - s = e_(u+v)/k.

a

Since JIM = u - v and we have the concentric arcs V C = a + s and ME = a, thenwe 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 + e"''''.

a or

Multiplying through by evl k gives

evl k =eulk

2e-ul k = cosh(u/k).

This proves the lemma.


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)/kor a or

_(eu/k - e-u/k)e-v/k.

/k - -u/kS eu/k = S cosh(u/k) = -e"a a 2

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

Therefore,

= sinh(u/k)

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

pendicular to the plane of DABC. Lines CC'

andBHB'

are the lines through C and B in theHpencil of lines in space parallel to AA' in the di-

rection AA'. These lines meet the horospherethrough A associated to this pencil at pointsCl and B1, giving a triangle DABC ( on thehorosphere. 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 s( =a tanh(a/k)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 - 1 Substituting wex

findcoshZ

Z coshZ(a/k) - I cosh2(b/k) - 1tank (c/k) =

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 -cosh clk)

and so we have proved

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 havea tanh(b/k) = a tanh(c/k) cos(LCAB) or

tanh(b/k)cos(LCAB) =

tanh(c/k)

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

tanh(a/k)cos(LABC) =

tanh(c/k)

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

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

cosh(b/k)

ortanh(a/k)

sin(LCAB) =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

71

x xx = 1,limo tanhx = -colym e +e-

C a B we see thentanh(b/k)

cos 11(b) = lim cos(LCAB) = lim = tanh(b/k),aoo a-.oo tanh(c/k)

and

sin 11(b) = limtanh(a/k)

=1

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


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) - I - tanh(b/k)

= cosh(b/k) - sinh(b/k)sin fl(b) (1/cosh(b/k))

- eb/k + e-b/k _ eb/k - e-b/k = e_blk2 2

The circumference of a circle

As an application of the trigonometry associated to the geometry of Lobachevskii andBolyai, we compute the circumference of a circle of radius R. The resulting formula can becompared 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 ofLobachevskii, and in a particularly interesting form as Proposition 25 of J. Bolyai's pa-per (in Bonola (1955)).

Theorem 5.20. In a triangle DABC with sides of length c for AB. b for AC, and a forBC 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 triangleand leave the general case to the reader. Suppose that the right angle is at C. In the discussionafter the proof of the hyperbolic Pythagorean theorem we proved that

tanh(a/k)sin LA =

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

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

Rearranging the terms gives one of the equations. Because there is nothing special aboutchoosing 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 regularn-gon inscribed in a circle of radius R. Let S denote half of the length of a side of theregular inscribed n-gon. A right triangle is formed by the center of the circle, the midpointof a side, and the adjacent vertex. The interior angle at the center is n/n for this triangle,and Theorem 5.20 implies

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

It follows that 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

sinh xlim = I.x-.0 X

This follows easily from the Maclaurin series sinh x = x + 7T + + If limn 2nSnexists, then it follows from the elementary theory of limits that

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

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) limsin(n/n)

n-- (n/n)= 2rrksinh(R/k).

Theorem 5.21 (Gauss 1831). The circumference of a circle of radius R is given by2nk sinh(R/k).

If we denote the circumference of a circle of radius R by circum(R), then notice that theformula leads to a series expression:

circum(R) = 2nksinh(R/k)/ 3 R5=2nk k+6k3+120k5+

it R3=2nR+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 parallellines in space we have taken some elementary notions and derived formulas in a mannerconsistent with analysis. These formulas contain an unknown, k, and they coincide withEuclidean formulas when k goes to oo. If k # oo, we have developed a non-Euclideangeometry and, because our formulas do not lead to contradictions, we are led to abandonthe idea that Postulate V follows from Postulates I through IV. Thus, through the power ofthe axiomatic method allied with analysis, we have reached a turning point in the 2000-year-long history of Postulate V. What remains to finish the story is a concrete realization ofnon-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 goalwe turn for support and foundations to deeper analysis and its relation to geometry.


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 aresmall 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 thefigure the plane can be covered without overlapping. Under HAA show that any convexpolygon with an even number of sides with angle sum equal to a submultiple of 27r tilesthe 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

Sun a Earth

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

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

sin Il (b) sin f1(c)cos(LBAC)cosfl(b)cosfl(c)+ = I.

sin fl (a)

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 ): LetAE and BD be perpendicular to A B. Let A E have length x. Let DE be perpendicular toAE. 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 y acos sin 2 cosh -.

2

5.9 If AA BC is given, show that the perpendicular bisectors of three sides belong to a pencilof 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 finitearea, say A. We prove a theorem of Gauss in this exercise that there is a constant A suchthat the area of an arbitrary triangle is ,l()r - a.- fi - y), where a, fi. y are the measuresof 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


and the parallel to both rays (figure A). Then f (¢) + f (x - 0) = A as in figure B. FigureC 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. Sincef(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 isadditive, that is, if a triangle is decomposed into subtriangles, the angle defect of the largetriangle is the sum of the angle defects of the parts. Show that this implies that there is anupper bound to the area of any triangle. In particular, a figure made up of three mutuallyparallel lines encloses a finite area.

PART B

DevelopmentDifferential geometry

6

CurvesI have been occupied with a new discovery ... in order to make my clock evenmore exact .... What, however, I never had expected I would discover, I have nowhit upon, the undoubtedly true shape of curves ... I determined it by geometricreasoning.

C. Huygens (6 December 1659)

The history of differential geometry begins with the subject of curves. Notions such astangents to curves can be found in Euclid, Archimedes, and Appolonius. The calculus isfounded on geometric ideas and so it is natural to find investigations of curves among thetopics 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 animportant 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. Moregenerally we can speak of curves of class k, that is, the mapping a : (a, b) --+ R" has con-tinuous derivatives up to order k. One then requires the appropriate order of differentiabilityin each definition and theorem. To focus more on the geometry than the analysis we haveignored 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 providingthe 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 producton 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". satisfiesthe Leibniz rule:

d+

rfa(r) ,B(r)] = a'(r) ,B(t) a(r) 14'(r).

63

64 Development: Differential geometry

Definition 6.2. Given a parametrized, differentiable curve, a : (a. b) - R". the tangentvector 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. Thisgives 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 isthe 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 tracebut 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 theright 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 andg: (c, d) - (a, b) a real-valued function. If g is smooth and g has a smooth inverse, thenwe 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 curveand 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.0o o

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

u

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

dv Ila'(r)Ildr = s(t),ca du as

since I dgI 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 theopposite 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 astandard 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 =ds/dt'

and we see that g is smooth. Let (3 = a o g. It follows

that

0

s g(s)

f IIf'(w)lldw =fs

Ila'(r)III aw

I dw = f Ila'(r)lldr = s(g(s)) = s.0

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 Ila'(t)II = Ila'(t)II = 1ds Ids/dtl Ila'(t)II

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


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 = a + b and

(s)acosl a +b I'asin( a +b )ba +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 = 4-22)(b2 - a2) cos2 t.

The resulting length function,

s(t) =J

/ a2 + (b2 - a2) cos2 rdr.0

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

Early work on plane curves (Huygens, Leibniz, Newton, Euler)

Now that we have introduced curves, their parametrizations, and their tangent vectors, weconsider 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 muchthe curve is "bending" at each point. The classical (Greek) curves of the straight lineand the circle are understood to "bend uniformly," that is, the straight line "bends" notat 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 wantK±(t) - 0. For a circle we want Kt(t) = 11r, where r is the radius of the circle, thatis, small circles bend more than large circles. The problem of generalizing this functionto an arbitrary curve is implicit in work of Appolonius (third century n.c.) and in workof Johannes Kepler (1571-1630). It is taken up by Gottfried Leibniz (1646-1716) in apaper 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 (seeChapter 9) is due to Leonhard Euler (1707-83).

Let a: (a, b) - 1R2 be a regular plane curve, parametrized by arc length. In the Euclideanplane, three noncollinear points lie on a unique circle. For s in (a, b), take si, s2, and s3 nears 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 threepoints. We use the dot product to form a function approximating the square of the radius ofthis 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 thereare points tl E (St, s2) and 12 E (s2, s3) with p'(tt) = p'(12) = 0. Applying Rolle's theoremagain 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 goto s, so p'(s) = 0 and a'(s) (a(s) - Ca (s)) = 0. Furthermore, a"(s) (a (s) -Ca (s)) = -1.

a(s) a(s)

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 isa limit of the circle based on three nearby points on thecurve.

The point Ca(s) is called the center of curvature ofa at s, and the curve given by C. (s) is called the curve of centers of curvature. We definethe (unsigned) plane curvature of a at s by the reciprocal of the radius of the osculatingcircle,

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

PROOF. Since a'(s) a'(s) = 1, by differentiating we have that a'(s) a"(s) = 0, thatis, a"(s) is perpendicular to a'(s). Since a(s) - Ca(s) is perpendicular to a'(s) as well, itfollows 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


We observe immediately that the definition and theorem generalize our conventions forthe straight line and the circle. Given the arc-length parametrization of the circle of radiusr, 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 anyregular curve a(t).

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

xy, yxK+(1) _ ((x')2 + (y')2)'/2

PROOF. 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

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) atso, for so = s(to). Now

z 2

fi'(s) = a'U(s))ds andfl (s) = a"(t(s)) (ds) +a'(t (s))dsz

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

dt I d2tx,x"

+ V'y"

ds x'(t)2+y'(1)2and ds2 =-('(t)2+y'(1)2)2'

By Euler's formula, we find

K±(10) = K±(fl; so) = Ix'(t)2 + Y'(t)2

o (I" Y11)

(x (1)2 + '(1)2)2 (x . Y )II

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

(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 wehave

Kf(t)ab

(a2

I

=sin2 t + b2 cos21)3/2

ahI

(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 minimumwhen 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 ofdescribing the curve followed by a weight being dragged on the end of a fixed straightlength, the other end of which moves along a fixed straight line. Leibniz and Huygensunderstood that such a curve had the property that at every point the tangent line met thefixed line a fixed distance away. Huygens named such curves Traktorien (in a 1693 letterto Leibniz). The name that has stuck is the tractrix (Loria 1902).

We suppose the v-axis to be the fixed line, the fixed lengthto be a, and the curve to begin at (a, 0) on the x-axis. By theanalytic characterization, the tractrix (x (t), y(t)) satisfiesthe equation

dv a - .rdx X' X

The tractrix.

Squaring both sides of the equation and manipulating the terms wefind

(.r')2 + (v1)2 =(a- )2

(x')2.X

Taking the derivative of both sides of the first equation yields

n " 21 2 . 3xy - yx _ -axthat is,x"v'-v"x'=

a (x)(x')2 .r2 a - X x'- a - X

This leads to the following expression for the plane curvature as a function of x:

K± (X, v) =a2(x3 X3 x

x2 a- x a3(x')3 I= la a- xAnother parametrization of the tractrix results from the integration

Y(x) = 1a - x

d x.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

Another consequence of the previous equations is an arc-length parametrization of thel2

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

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

= x. Forx a

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

dy I-x dx-I-e2$

ds x dsand 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

K±(s) = IIO (s)II =e -S

1

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

Directed curvature

At present, K±(t) is simply the reciprocal ofa length. We now introduce a sign which makesplane curvature more sensitive. We want to be able to distinguish a curve a: (a, b) - R2from its reverse &(t) = a(b - t), and to distinguish convexity on either side of a pointwhere the curvature vanishes (for example, a point of inflection). To make these distinctionswe introduce an orientation:

Definition 6.8. A ordered pair of nonzero vectors [a, v], u, v E R2, is said to be instandard 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 by114

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 thefact 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 notationi 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) tangentvector a'(s) by T(s). Let N(s) denote the unique unit vector perpendicular to T(s) withstandard 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 theidentity

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

It follows from Theorem 6.6 that IK(s)I = K±(s) and so the same formula for thedirected 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 functiona(t) = (1, f(M illustrates the choice. Here Ila'(t)II = I + (f'(t ))2, and we can compute

I f(t)I1

(f(t))21

(f(t))2

-f (t) 1

I +(f(t))2 I +(f(t))2N(t) -- (

T'(t)- r' (t) -f(t) 1

I + (f'(t))2 ( I + (f'(t))2 1 _+(f'(t))2The sign of the curvature is determined by the second derivative f"(0. which is positiveif f(t) is concave up, negative if f(t) is concave down. Since any curve in R2 is locallythe graph of a function, we see that the directed curvature at a point is positive if the curveturns 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 isgiven in the following theorem.

Theorem 6.10 (fundamental theorem for plane curves). Given any continuousfunction K: (a, b) - R, there is a curve a: (a, b) -. R2, which is parametrized by arclength, such that K (s) is the directed curvature of a at s for all s E (a, b). Furthermore, anyother curve a: (a. b) -, R2 satisfying these conditions differs from a by a rotation followedby a translation.

PROOF. To prove the theorem we turn to the theory of differential equations. The conditionsof 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.


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-matrixgiven 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 integrateT(s) to obtain a(s). Notice that we can choose a(ao) anywhere in R2, and that we canchoose u' = T(ao) as any unit vector. Changing i at a(ao) involves a rotation and thatrotation passes through the differential equation so that another solution would appear asT(s) = peT(s), where pe is the matrix

rcos0 -sin0pe 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 wewrite 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),

we can take x(s) = r2 cos(r2s). Then y(s) =,2

sin(r2s) by the condition x'2 + y'2 = 1.

This is a parametrization of the circle of radius12

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 planethrough 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 moreaccurate clocks. Practical designs were based on the motion of a pendulum, requiring care-ful 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 1673treatise Horologium oscillatorium sive de mote pendulorum ad horologia aptato demon-strationes geometricae. Some of the ideas introduced in Huygens's classic work, such as theinvolute and evolute of a curve, are part of our current geometric language. We next sketchthe principal ideas of Huygens's pendulum clock and collect the main geometric ideas inthe context of extending the theory of plane curves. For a thorough historical study of thispart 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 regardlessof the initial point on the curve. If a circle were a tautochrone, then a rigid pendulum wouldhave the same period independently of the choice of initial angle. This is not true, as isshown in any elementary physics text, though Galileo observed that it is approximately truefor small angles. If a pendulum were made to swing along the path of a tautochrone, thenit would deliver a reliable period without worry about the initial angle. This is the basis ofan 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 itrolls along a straight line.

Of particular interest is the cycloid for a circular wheel rolling "under" the line, and thepoint chosen on the circle. Referring to the figure, take -n < 9 < it and let O' be the pointon the circle that began at the point 0 after rolling through 9 radians. We denote the radiusof 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' Kalso has length r9, LO'C'K has measure 9. By projecting onto the diameter LK from thepoint 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.


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 farafield (see Dugacs (1955) or Gindikin (1988)). Let us assume that a tautochrone is beingfollowed 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 thependulum from an initial point. Resolving vectors into horizontal and vertical componentswe find the acceleration due to gravity determines the differential equation

d2ad!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 harmonicmotion:

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

gcos0dLo=Pdo.

Consider the following diagram. Since the tangent to the curve followed by the bob ofthe pendulum is perpendicular to the pendulum, 0 is also the angle that the tangent makeswith the horizontal. Therefore, we have

dx da dy dodt

=cos0dt.dt

=sin0dt

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),

cos 20.y(4') - yo = - 02

For 0 = 0, xo = 0, and yo = -02 the origin is midway between the two cusps of a

cycloid generated by a circle of radius 02

Having determined the curve that we would like the "ideal" pendulum to follow, we nextconsider 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 stringalong 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 iscalled an evolute of Q.

A pendulum with a flexible string constrained to unwind along a fixed curve a followsa path given by an involute of a. The practical problem of building the "ideal" pendulumclock requires that we construct the evolute of a cycloid. We next develop some of theproperties 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 onlyif, 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 eachpoint. 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).


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 associatedto a(s). Recall from earlier in the chapter that a unit-speed curve a(s) has an associatedcurve - the curve of centers of curvature, Ca(s).

Proposition 6.16. The curve of centers of curvature Ca(s) associated to a unit-speedcurve a(s) is an evolute of a.

PROOF. Recall some of the properties of Ca(s):

I(1) K(s) =

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-speedcurve. However, the property of being an involute is characterized by the facts that the pointof the curve lies on the tangent of the evolute and, at associated points, that the tangents areorthogonal. To prove the proposition then, it suffices to show that Cc(s) 1 a'(s) and thata(s) = Ca(s) + A(s)C., (s).

Since a(s) is a unit-speed curve we associate to each point of the curve the orthonormalpair (T(s), N(s)}, where T(s) = a '(s) and [T(s). N(s)] is in standard orientation. ByEuler's formula (Theorem 6.6) we know that T'(s) = K(s)N(s). We take this relation onederivative 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

C s) K'(s) N(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 determinethe curvature and hence the curve of centers of curvature. We fix the radius of the generatingcircle to be I - the general case will be clear. Apply the formula for the plane curvature ofan 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) = 2+2cos9(-sin0. 1 +cos9)f ( singI +cos9

2 \ l +cos9 )_ f -2 sin(0/2) cos(0/2) f cos(0/2)l

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

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 precedingdiscussion, using a pair of plates curved to follow a cycloid between which the pendulumwould swing. Thus the bob followed the path of a tautochrone. His proof of Theorem6.18 used synthetic and infinitesimal ideas; he did not have the calculus at his disposal. Thisexample of an applied problem leading to developments in differential geometry is repeatedthroughout 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) (0. 1) given by g(t) = t2 + 1

a reparametrization?

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

fi(t) = 2 (t + t2 + I, 2In(t + t2 + 1)t + t +l ///

6. Curves 79

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

6.5 Let k: (a, b) -+ R be a function with k(t) > 0 for all t. Let 9(s) = f k(s)ds be anyantiderivative 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. Shown dp

. Alsothat the arc length along the curve is given by p2 + (02d9, where p' =+a

show that the directed curvature is given by the formula

K(0) = 2(P )2 - PP" + p2

((P')2 + p21312.

ae

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 thecycloid 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 ouranalytic 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 areauxiliary 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 satisfyingdydt

(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 spaceSur 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 wassuccessful in characterizing plane curves and so we seek to generalize the idea to curves inspace. In two dimensions the normal to the tangent is easy to describe. In three dimensionsthere is a continuum of choices of a normal vector. Furthermore, we must contend with anew 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 crossproduct 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 (ax 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 differentiablefunctions.

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, letN(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 perpen-dicular to T(s). This definition fixes the choice of unit normal by choosing the derivativeof 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, andfurther

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 takesT(s) H el = (1. 0, 0). N(s) ez = (0, 1, 0), and B(s) r e3 = (0, 0, I) has positivedeterminant. When this condition holds we say that the ordered basis (T(s), N(s). B(s)) isright-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 bysaying 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, thelimiting process used to define the plane curvature determines an osculating circle and anosculating plane for space curves. The osculating plane is spanned by T (s) and N(s). Sincethe cross product is orthogonal to the tangent and normal vectors, changes in the binormalmeasure how "nonplanar" a curve is.

Proposition 7.2. Suppose a(s) is a regular unit-speed curve. Then B(s) is a constantvector 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) arecontained 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) B(s) + (a(s) - a(0)) B'(s) = T(s) B(s) = 0.Ys-

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 planeperpendicular to q'.


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 followingequation:

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 vectoris 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), wetake 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) =

Iacosl a +b I'asin( a +b )b a +b /

Let w denote l Then.1a2 +b

T (s) = (-aw sin ws, aw cos ws, wb),

and T'(s) = (-awe cosws, -awe sin ws, 0), so K(s) = aw2.

It follows that N(s) = (- cosws, - sin ws, 0) and the binormal is given by

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 planeat a constant rate.


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 calleda moving frame or moving trihedron along the curve.

The moving trihedron with its curvature and torsion were introduced in the Toulousethesis of F. Frenet (1816-68) in 1847. Independently, J. A. Serret (1819-92) publishedsimilar results in the Journal de Liouville (1851) before Frenet's results had received widerecognition. 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 withnonzero 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, thenan arbitrary vector u" E R3 can be written

'u = ( v " wi)wi + (v w2)iu2 + (i . 1 7 V 3 ) @ 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).

Differentiation Ogivesusthat -K(s).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) _-(-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:

0 K(s) 0 T(s) T1 S)

-K(s) 0 r(s) N(s) = N'(s)0 -r(s) 0 B(s) B'(s)

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


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

PROOF. Consider the matrix-valued function

0 k(s) 0

g(s) _ -k(s) 0 f(s) _ (a;j(s)).0 -f(s) 0

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

3 Vj VI

v; _ Fa;jvj or v2' = g(s) v2 (*)j=1 v3 V3

and, by Lemma 6.11, there is a solution V1 (S)- v2(s), u3(s), dependent upon the initialconditions. We may take (VI(so), V2 (SO), v3(so)I, to be a choice of an orthonormal basisfor 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 +ajkflikk=1

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

1, i=j,S;j =

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

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 differentialequation (**) and so, by uniqueness of solutions to differential equations, we have a movingframe.

7. Curves in space

To define the curve a(s) we integrate

a(s) =J

s vt(t)dt.SO

We leave it to the reader to verify that a(s) has the correct curvature and torsion.

85

The assumption that K (s) i4 0 is necessary in order to obtain uniqueness in the sense ofthe theorem. If K(s) = 0 along a subinterval (at, bt) of (a, b), then a line segment in 1R3realizes this part of the curve. To see how uniqueness fails, suppose M is the midpoint ofthe 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 Mperpendicular to the line segment for at - e < s < at and on the other side of this planefor bt < s < bt + e. Rotate the solution around the axis given by the line segment in thehalf-space containing a(bt) through some nonzero angle while leaving the other half-spacefixed. This will preserve the curvature and torsion but there is no way to superimpose thisnew 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 itscurvature (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) isa 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 acircle of radius I /K.

The power of the Frenet-Serret apparatus lies in the manner in which it organizes theprincipal analytic properties of a curve. Applications of this viewpoint are found in theexercises. 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 thetangent T (s) and u maintain a constant angle with respect to each other. The vector u iscalled the axis of the helix. We can rake 0 to be unit length.


For example, u = (0, 0, I) for the helix a(s) = (toss, sins, s). Observe that if a(s) is ageneral 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 thebinormal B(s) playing the role of the constant vector u. The Frenet-Serret apparatus allowsus to characterize general helices.

Proposition 7.8. A unit-speed, regular curve a(s) is a general helix if and only if thereis a constant c such that r(s) = cK(s).

PROOF. If i is the axis of a general helix a(s), then T(s) v = cos9, a constant, and thisimplies T'(s) v = 0, orK(s)N(s) v = 0, that is, u 1 N(s). Apply the facts that we havean 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

dd-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. Fur-thermore 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 whenthe curve is regular but not unit speed. The relevant formulas are given in the followingproposition.


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

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

and r(t) =(a'(1) x a"(t)) a,,,(t)

IIa'(t)II3 IIa'(t) x a"(1)112

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

s(t) is the arc-length function andt(s)isitsinverse. Here d` = x'(t)2+y'(t)2+z'(1)2 =

Ila'(t)II and so Ws =Ila'(r)p.

Furthermore,

d2t 12x'x" + 2y'y" + 2z'z" dt a'(t) a"(t)ds2 - 2 (x12 + yi2 + 2'2)3/2 ds - 11a'(t)114

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

K(to)2 = IIfl"(so)112 = a (to) (ds)2 +a(to)d-Z

2

II

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

Ila'(to)112 - \ IIa'(t)II4 ) a (1)II

_ IIa,,(to)112 (a'(to) a"(to))2 (a'(to) . a"(to))2

Ila'(to)114- 2

IIa'(to)116 + 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)

xB"(so) and N'(so) =

fl (so) - fl"(so) dKK(SO) K(SO) K(SO)2 dS

From these expressions we find

r111 11 _(6'(so) x fl"(so))

fl,,,(so)

K (SO) 2

Since fl'(so) = T(so) =Ila'(to)II

and 0"(so) =Ila'(ta)IIZ

+a'(to)ds2 we obtain

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

Ila'(to)113


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

C2 and so

r(t )-o -K (s0)

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

(10) Ila'(to)1l6

Ila'(to)116 Ila'(to) xa"(to)112,

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

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

2

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) isplanar. After a tedious computation, we also find

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

Ila'(t)113 (2a2 sin2 t + b2 cos2 t)3/20.

This curve has the same curvature as an ellipse in the standard plane. By the fundamentaltheorem for curves in R3, a(t) is an ellipse lying in some plane determined by the constantsc and d.

Much more can done in developing special properties of curves from the Frenet-Serretapparatus. We include some of these ideas in the exercises. The successful classification ofthe smooth curves in R3 via curvature (K # 0) and torsion establishes a paradigm that onecould hope to generalize to other geometric objects, for example, two-dimensional objects.We turn to this in the chapters that follow.

AppendixOn Euclidean rigid motions

A missing ingredient in our discussion of curves in R" is a notion of equivalence - given twocurves a(t), #(t), when should we consider them geometrically equivalent? Reparametriza-tion is one reduction; certainly a curve and any oriented reparametrization (one for whichthe derivative is positive) ought to be considered equivalent. This focuses our attention onthe trace of the curve with a direction along it. Consider the directed trace of a curve as arigid object in R" and move it about in space by a motion of R" preserving rigid bodies. Forexample, translation of R3 by the vector (I, 0, 0) corresponds to "moving one unit to theright;' and we can expect such a motion to preserve a rigid body. In this appendix we deter-mine those mappings of R" to itself that are the rigid motions. Those properties of curvesthat 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 translatesrigidity 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

by

The most obvious examples are translations, t,;,(v) = u + w. In R2 the rotations given

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 isometrythat 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, iv-in 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


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 onR". In particular, we find that an orthonormal basis (i1, ... , 6,1 is carried by F to anotherorthonormal 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 F(v1)+...+a.F(Vn)

Write F(a1 i1 + +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-onemapping. An isometry fixing the origin, a linear isometry, has a like inverse. We denote byO(n) the orthogonal group of all linear isometries of R". Choosing a basis for R". we canwrite 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 lineartransformation 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) andsimilarly 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 areinvariants in the sense of equivalence under rigid motion. One should compare this fact tothe 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 Frenet-Serret apparatus. What is the nature of the curve?


7.4 Let a(s) be a unit-speed curve with domain (-e, e), and with K > 0 and r > 0. Let

fl(s) = J B(u)du.0

(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,

f = K, T = B, N = -N, and T, and (K, r, T. N, B) is the Frenet-Serretapparatus 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 deter-mines 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 thecurve at a point a(so) can be given by

2 \a(so + As) = a(so) + T(s)As (1 - 6 (As)2 1 + N(s)(As)2 (2 + 6 As )

+ B(s)(As)3 6\+

R4,

where the remainder R4 satisfies limR4

= 0. Projecting into the various planeso.-.o (As)-'

{T, N). (T. B), { N, B). determine the algebraic curves on which the curvature and torsionof a curve must lie.

7.8 The tangent vector T(s) to a unit-speed curve a(s) determines a curve on the unit spherein 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 analogousprocedure 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 betweenthe functions j and f2 may also characterize curves. Prove that a nonplanar curve liesentirely on a sphere if and only if the following equation holds between the curvature andtorsion:

Appendix. On Euclidean rigid motions 93

7.10 Given a unit-speed curve a(s), determine the equation satisfied by the osculating planeto the curve as a function of s. Show that a curve is planar if and only if the osculatingplane 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)), notnecessarily unit speed. Prove the following formula for the torsion of a:

I Ix" y z

det I x y" z"r KZ(S')6 x"i yn, Z",

7.12 Take the theory of the previous chapter of involutes and evolutes and generalize it tocurves in R3. In particular, show that an evolute /3(s) of a curve a(s) with Frenet-Serretapparatus (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 fixedreal numbers a, b, then show that there is another curve (the Bertrand mate) fl(s) suchthat 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 unitspeed.

7.15 Suppose L: R" - R" is a linear transformation. Define the adjoint L' of L to be thetransformation of R" determined by L(v) w = v L' (w). Show that for a particularorthonormal basis, V is represented by a matrix that is the transpose of the matrixrepresenting L. Define the Rayleigh quotient of a nonzero vector "u as

R(v) _L(v)-vv-v

.

Show that R attains a maximum and minimum on R" - (0). Show that R is differentiablewith 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 subspaceK C R" such that (a) dim K = I or 2. and (b) L(u) E K for all u" E K, that is. F hasan 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. LetK' be the subspace of R" satisfying K' = (v E R" 10 w for all w E K). Show that ifK is an invariant subspace of R" with respect to a linear isometry F, then Kl is also aninvariant subspace. Show also that F 1K is an isometry. If K has dimension 2 and F 1K


is a rotation, then K is called a rotation plane. Prove that for n > 3 a linear isometry hasa rotation plane. It is also possible that F I K is a reflection across a line contained in K oracross the origin. Prove that any linear isometry of R" can be expressed as a compositionof n or fewer reflections in hyperplanes. (See the article by Lee Rudolph, The structureof 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-speedcurve in R3. Compute the curvature and torsion of A o a in terms of the curvature andtorsion of a(s).

SurfacesEt quia per naturam superficiemm quaelibet coordinate debet esse functio binanimvariabilium.

Euler. Opera posthuma, vol. 1. p. 494

To build a foundation for geometry we first define the structures where geometry can takeplace. Euclid tells us in Definition 5 of the Elements that "a surface is that which has lengthand breadth only." A plane is a special kind of surface and taking this point of view a littlefurther, we can expect that the non-Euclidean "plane" will not be like a Euclidean planeand 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 essenceof a surface is its two-dimensionality, that it is describable by two variables, length andbreadth.

Our goal in developing the classical topics of differential geometry is to discover thesurfaces where non-Euclidean geometry holds. On the way to this goal we examine themajor 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 foundedby 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 firstintroduce the relevant objects, structures, and examples.

U

.l ul)-ol

95


Definition 8.1. If S is a subset of R3, then S is a (regular) surface if for each point pinS. there is an open set V of R3 and a function x: (U C R2) -> V fl S, where U is an openset 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)), thefunc-tions 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 inversex - i : V fl S -+ U that is continuous. This means that x - I is the restriction to V fl Sof a continuous function W -+ R2, for some open set W C R3 that contains V fl S.

(3) The Jacobian matrix

aft

s(a' v)

aft(u, v)

a

J(x)(u, v) = f2(u, v) 22 (u, v)

au av

au (u, v)/

(u, v)at,

has rank 2. That is, for each value of (u, v) E U, J(x)(u, v) : R2 -. R3, as a linearmapping, is one-to-one, or equivalently, ker J(x)(u, v) _ {0). This condition on xis 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 neighborhoodof a given point. This makes precise the idea that a surface is locally two-dimensional. Wecan think of a chart x: (U C R2) - S as the correspondence that a cartographic mapmakes 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 vectorq" 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, s(ui - r)ql + (u2 - s)q2 +

U3q3

The mapping x satisfies the conditions x(ui, 142) = i and (x(r, s) - i) q' = 0. Thus xmaps R2 to 1714.4. Since x is an affine mapping, it is differentiable. The inverse of x is easilyseen to be given by x'1 (ui, 142- u3) = (U 1, 112), which is continuous. It follows that x isone-to-one and onto lg,y. Finally, the Jacobian is given by

afiar

afias 1 0

aft ah 0 1

ar as qI q2

afi afi q3 q3ar as

8. Surfaces 97

which has rank 2. This shows that the planes in R3 are regular surfaces. (For planes that areperpendicular to a vector q' 96 0 with 93 = 0, a change of names of the axes allows us toapply 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 themapping x : D -+ S2 by

x(u, v) = (u, v, l - u2 - v2).

The image of this patch covers only the upper hemisphere of S2. The mapping is differen-tiable and has inverse x- I (u i , u2, u3) = (u 1, 142). The Jacobian is of rank 2 and so the patchis regular. In order to cover all of S2 with similar hemispheric charts, we need five morepatches 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-studiedsurface. 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: Supposef: (U C R2) -+ R is a smooth function, where U is an open set in R2. Consider the graphof f, the set ((u, v, f(u, v)) I (u, v) E U}, as a surface in R3. Notice that derivatives ofall orders exist for the function x(u, v) = (u, v, f(u, v)) which is injective and onto thegraph. Finally, the Jacobian

I 0

J(x) = o0

ofhas rank 2.

au av

Thus smooth, real-valued functions of two variables provide us with many examples ofsurfaces. 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 chartfor 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

1 0

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. Be-

tween these three expressions, for each (u, v), at least one is non-zero so J(x)(u, v) hasrank 2. Since g(u) > 0 for all u E (a, b), (g(u) cos v, g(u) sin v) is injective as a function

of v. Since we have rotated the graph of a function, x is injective. To construct the inversefunction to x, suppose (a, b, c) is in the image of the chart x and solve for v:

arccos(b/(b2 + c2)), c > 0,V - { 27r - arccos(b/(b2 + c2)), c < 0.

We check that this expression is continuous at the only problematic point, that is, c = 0. Inthis 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 thatdoes not intersect the x-axis and does not intersect itself, then we obtain a surface by rotatingthe trace of a(t) around the x-axis. In fact, any line in the plane may be chosen as the axisof 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 generateanother class of surfaces, we recall two important theorems from the calculus of severalvariables (see Spivak (1965) for details).

The inverse function theorem. Let F: (U c R") R" be a differentiable mapping

and suppose, at u E U, the Jacobian matrix J(F)(i) =(aFj 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 anopen set around (i , v) and X. v) = 0. Let M be them x m matrix given by

of (u)M= 1 <i, j <<m.ax"+;

If det M 54 0, there is an open set U C R" containing u and an open set V containing v sothat, for each i E U. there is a unique s' E V such that f (i. s) = 0. If we define g: U -+ Vby g(F) = s, then g is differentiable.

These theorems are equivalent and they provide us with another method for generatingsurfaces.

Definition 8.2. Given a differentiable function f: R' -+ R' with n > m and its JacobianJ(f) = (a j/ax;), a point u E R" is a critical point off if J(f)(i). as a linear mappingR" - R'. is not onto. The point f (u) E R' is called a critical value of f. If v E R' isnot 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 regularvalue, 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

Jacobian of F has the form

I 0 0

J(F) = 0 1 0aj aj aj

ax ay az

99

and so det J(F)(p) 54 0. Apply the inverse function theorem to get neighborhoods of p andF(P), say V and W, respectively, such that F: V - W is invertible and F-1: W --- V isdifferentiable. 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 valueand so determine a regular surface in R3. namely, f-1 ({0)):

x2 y2 z2ellipsoids, j(x, y, z) = a2 +

b2+

c2- I ;

x2 y2 z2hyperboloids of l or 2 sheets, f (x, y, z) = a2 t

62- 2 - I ;

2

andfor 0 < r < a, the torus, f(x.y,z)=z2+( x2+y2-a) - 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 andsubstitute 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. Awayfrom the spine the pages are locally two-dimensional, but if three pages come together in aline 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 meansfor a real-valued function, f : S --+ R, or a function between surfaces, m : St -* SS, to be


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 isdifferentiable. However, this presents a problem - there may be infinitely many differentcoordinate patches containing a given point to check for differentiability at the point. Thisis rendered unimportant by the following proposition.

Proposition 8.5 (change of coordinates). Suppose S is a regular surface and x : (U CR2) -+ R3 and y: (V C R2) -+ R3 are coordinate charts with W = x(U) fl y(V) C Snonempty. 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 ofhomeomorphisms. Since h-I and h are symmetric in x and y, we only need to show thath is differentiable. Let (ro, so) E x-t (W) and write p = x(ro. so). Consider the followingtrick: y is a coordinate chart which we write as

Y(u, v) = (fi (u, v), f2(u, v), f3(u, v)),

and so af, af,au av

J(y) _ aft aft has rank 2.au av

af3 af3au av

aflaf2_af,af2W e assume thatau av au au

54 0 at (uo, vo) = h(r0, so) = y- I (p). Extend y to the

functionF: 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, af,au av

J(F) = aft aftau dv

af3 af3au 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 differen-tiable. 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 thirdcoordinate. 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 atp if for any coordinate chart x: (U C R2) -+ S with p E x(U), the mapping fox: (U CR2) -+ R is differentiable at x-gy(p). f is differentiable, if it is differentiable at everypoint of S.

Proposition 8.5 implies that if f: S --+ R is differentiable at a point in one coordinatechart, then it is differentiable at that point for any other chart containing the point. Theproposition also allows us to define easily what it means for a function between surfaces0: S, --+ S2 to be differentiable.

Definition 8.7. A function q5: S, --+ S2 between surfaces S, and S2 is differentiableat p in S, if there are coordinate charts x: (U C R2) - Si and y: (V C R2) -+ S2with p E x(U), O(p) E y(V). such that the composite function y 1 o 0 o x: U -. V isdifferentiable at x-1(p). 0: S, -- S2 is differentiable if it is differentiable at each pointin Si. A function 0: Si - S2 is a diffeomorphism if it is differentiable, one-to-one, andonto, and has a differentiable inverse function. Two surfaces S, and S2 are said to bediffeomorphic 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 diffeo-

morphic via the diffeomorphism ¢: S2 -+ the ellipsoid,111

O(x, y, z) = (ax. by, cz).

The notion of diffeomorphism yields an equivalence relation on the set of all regularsurfaces in R3. 71vo diffeomorphic surfaces are equivalent in the sense that they shareequivalent sets of differentiable functions and any construction involving the calculus onone carries over to the other. This equivalence relation is the basis for the differentialtopology of surfaces. However, it is too weak an equivalence to be useful in distinguishinggeometric objects.

The tangent plane

The derivative of a function is the best linear approximation of the function at a point. Ageometric realization of linear approximation is the tangent plane - a linear subspace of R3that is the best approximation to a surface at a point. To define the tangent plane we restrictour 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),


called the coordinate curves on S through p. Define

xu =a

(X (u, vo))I =ax

(uo. vo)au (uo vo) au

and x =a

(X (uo, u)) I =ax

(uo, uo);av

(U0. L-0)av

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 andonly 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, wehave proved the proposition.

It follows that for each coordinate patch on a regular surface in R3 we can define anormal direction, that is, x x xv. We define the unit normal vector to the surface Sat apoint 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 thechoice 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 andy: (V C R2) -r S are two coordinate charts containing p. then at p we have

XU xxv YuXYoIIXuXXuII IIYuxY6II

8. Surfaces 103

PROOF. Let W = x(U) fl y(V ). The subsets x-' (W) C U and y-' (W) C V are relatedby 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 thechart y by using x(u, v) = y(u(u, u), v(u, v)). The tangent vectors become

au au au avx°=Yu-+y +yeav.

We can compute x x x directly to get

au av au avXU X xa = Yaau +Yaau X Yaav +yuav

au as a;, au

au av - au avY a x Yo = det J(y ' o x)ya x

It follows that

XU xx _ detJ(y'ox) y;,xyD =f yaxyaIlxu x x,,ll IdetJ(y- ' o x)111 Y5 X YtIl IIYy X Yoll

On a given patch, the normal is simplyxu x x

We can "translate" the normal overIlxu X

the surface by moving patch to patch, where we choose the variables u and v so that theunit 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 thisconstruction gives a surface when we ignore the boundary (see the exercises). If we movethe normal at a point around the surface, when we arrive back at the first chart from whichthe 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 liesin the overlap of two charts, p E x# (Up) (1 xy(Uy), then the Jacobian J(xy' o xf) at phas positive determinant.

Since the sign of the determinant of the Jacobian determines whether the normals at apoint in the overlap of coordinate charts agree, the definition of an orientable surface letsus 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.

Definition 8.11. Let p be a point in a regular surface S. and x: (U C R2) -+ S anycoordinate chart containing p. The vector subspace of R3 of vectors normal to N(p) _

XU X xv

II Xv II

is called the tangent plane to Sat p and is denoted by Tp(S).x xNotice 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 thisfashion. However, it is more important to think of Tp(S) as a set of directions, admittingthe 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 nextsee that Tp(S) contains the directions of the tangent vectors to any regular curve on thesurface S through p.

Proposition 8.12. A vector v E R3 is an element of Tp(S) if and only if there is adifferentiable 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. Choosinge > 0 small enough we can take the image of a to lie in the image of a patch x: (U CR2) -+ 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,

ax du ax dv du dva'(t)= 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 curveon the surface.

There is another interpretation of the tangent plane that frees us of the dependence oncoordinate charts (though the proofs of the relevant properties go through charts). To discussthis we introduce an important object associated to a regular surface S. Define the set ofsmooth 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 Leibnizcondition:

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 satisfiesthe 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 ofan ambient space like R3 in which to define tangent vectors.

A differentiable mapping between surfaces induces a mapping of tangent planes at corre-sponding points. To define the mapping suppose F: S1 - S2 is a differentiable function andp 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

i-I--- (V C: R2).(UCIIt2)-$I

F F o x(u, v) = .(P1 (u, v), P2(u, v)). If acurve 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))).


Suppose 'v = a'(0) E Tp(S); with respect to the basis {x,,, x,,) we can write v = xdu

+dt

x dv . With all this in place we compute

d (aidu aFidu dE2du aFZdvdFp(u) = dt(Foa)l,_o = au dt + av dr' au dt + av di

aF, apt duau av r dt 11

a IP2 a P2r=0

1=0

This identifies dFp with the linear mapping given by multiplication by the matrix deter-mined by the Jacobian of r- 1 o F o x.

Notice that the previous argument applies to the identity mapping id: S -+ S and twodifferent 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 certainconstructions 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 turnto 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 innerproduct (u', v)p = u v is inherited from R3. This appears undesirable on first considerationas it seems not to be a feature of the surface but an artifact of the embedding. Later wewill show that important geometric properties of surfaces can be developed from the firstfundamental 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 associatedto curves a: (-e, c) - S and P: (-q,,) - S. If we write a(t) = x(u(t), v(t)) and

1 du

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))

_ du dv du dvIp

x°dr+xvdr

- x°drxvdr

dude dude dude dvdv= I p(xu, xu)

dr dr+ Ip(xu, xv) ( dr dr + dt dr

+ Ip(xt xt,)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 com-

ponent 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 chartx : R2 -+ 1R3, x(u, v) = (u, v, 0). For this chart x = (1, 0, 0) and xv = (0, 1, 0). Thecomponents 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 polarcoordinates (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 metricare 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><rsin0,cos0).

For this chart,

x, =(-sin >/isin 0,cos>Gsin 0.0) and xe=(cos*cos0.sin Vi cos0.-sin 0)

108 Development: Differential geometrn

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 coordinatechart 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 firstfundamental 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. Theseproperties, 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 tocome later (Chapter 14).

With the first fundamental form we can calculate the length of curves on a surface. Ifa: (a, b) -+ S is a curve on S. then the length of the curve between a(p) and a(t) along ais given by

s(t) =J

Ila'(r)lldr = f a'(r)) dr.1U s

In local variables, s(t) can be expressed as the integral

/ /E(u(r). v(r))

du2

( dr l + 2F(u(r). v(r))ddu

ddtr'

+G(u(r). Or))du

I dr I dr.r1i

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 = Iand F = 0. then ds2 = du' + dv2 expresses an "infinitesimal" form of the PythagoreanTheorem. Of course, this is shorthand for

ds _E

(du+ 2 F

drrdv+ G

(dvdt dt dt dt 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) E123}, 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 coordinatechart

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 -. R3givenbyx(O, v) = (cos9,sin8, v).Herexe = (-sinO,cos9,0)andx = (0,0. 1). HenceE(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, considerthe cylindrical rollers used to print a newspaper - no distortion of the print takes place asthey transfer print onto a planar page. The cylinder is a piece of the plane rolled up withoutaltering relations between points.

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

(4) Let f: (U C R2) -+ R be differentiable and consider the surface given by the graph off 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,u

) and1 -u -vx = (0, I, u

). When we compute the dot products and simplify we get theI -u -uline element

ds2 = 1 -u2

du2 +2uv

dudu + I - u2 dv2-u2-u2 T--;T--V2 I -u2-u2


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 invarianceof this local expression. Suppose a: (-e. e) - S is a curve on S lying in the intersectionof 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 curvea(t) in these different coordinates we get

'du dv _ du dD

a (t) = .rdt + x`. dt = y" dt + 1, dt

.

Write the component functions for the metric associated to x as E, F, and G and thoseassociated to y as E, P. and G. The change of coordinates y- 1 o x implies

an aD au anX. = y" au

+yp as .

x`, ,5 - + V-.au au aD aD+2Fauau+G(all )2

Fan an

FauaD auaD\ i)DdF.I+G.

au av au av av Ju/ au- -)V

Ju 2 ail aDG=.rf x,.=E(av) +2Favav+(an)2

av

au aD au anE F) _ all all E F i)u av

(F G aif aD (F ) au au

av av all at,E

=J(v cx' F)J(''ox).F GG

It follows that the change of component functions of the metric is made by multiplyingthe 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

duldtv/dt) = J(yI ox) (ddFildt

(d v/dt

8. Surfaces 111

This leads to the matrix equation

ds2 = (du dv)E F

CE F)du

dv

= CJ(y I o.r)Cdu))f CF G) CJO' ox)Cdv

\ E' F du=(du du) CF G)Cdo

Thus the line clement is preserved under change of coordinates. The transformational prop-erties of the matrix of component functions make it significant as a local object defined onthe surface. We will consider similar expressions throughout the text. We can think of thematrix of component functions as "invariant under change of coordinates" in an unusualway - the change does not preserve the form exactly, but the correction term, here theJacobian, preserves the relevant form. This kind of invariance is made precise through theuse of the tensor calculus, which is developed in Chapter 16.

EXAMPLE. In practice, the transformation of the line element is carriedout like a change of variable in integration. Consider the surface S givenby the graph of u2 + v2. One coordinate chart is given by

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 is = r cos 0, v = r sin 8. which implies

du = cosOdr - rsinOd8. dv = sinBdr +rcos8d8.

Substitute these expressions for it. v, du. and dv into the given line element to obtain thefollowing:

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 con-structions and define certain quantities. For example, suppose two curves on a surface S


intersect at a point p. The first fundamental form may be used to determine the angle be-tween the curves - we simply apply the definition of the angle between vectors in R3 fromlinear algebra. If a: (-E, e) -+ S and f : (-e, e) -- S are differentiable curves that satisfya(0) = f(0) = p, then the angle 0 between a and O at p is given by the angle between thetangent vectors to a and 6 at p:

cost =Ip(a'(0), )3'(0))

1Io(a'(0), a'(0))lp(f'(0). $'(0))

For example, consider the tangent vectors to the coordinate curves, x,,, xi,: the angle betweenthe coordinate directions is given by

0 = arccosIP(x,,, xv)

arccos ( F l(JP(T-'X-)IP(X,1'

x and x, are orthogonal when F = 0. We call such a coordinate chart anorthogonal parametrization.

EXAMPLE. In the earlier computation for the paraboloid ((u, v, u2 + v22)) the rectangularcoordinates give

F _ 4uv

EG (1 + 4u2)(1 + 4u2)

These coordinate curves are seen to be the intersection with the paraboloid of vertical planesthat are parallel to the u- or v-axis. The resulting curves on the surface meet in an anglegiven by arccos(F/ EG). The polar coordinates, on the other hand, give an orthogonalparametrization.

Area

Through the first fundamental form we have defined the are length of a curve and the anglebetween 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 thatis 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 coordinatechart. If R C S is a bounded region and R C x(U), then the area of the region R is givenby

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. Inthis case, we consider a transformation U -. S -+ Tp(S) for p a point in a small regionlying in U. The tangent plane Tp(S) is a linear approximation to S near the point p. Toprove the theorem one shows that the infinitesimal contribution to the distortion by thistransformation is given by Ilxu x xull. For details, see your favorite multivariable calculusbook (Spivak (1965)).

Recall that Ilx x xv 112 = Ilx. II21Ixv 112 - (x xv)2, so a local expression for area is givenby the integral

area(R) = r f E(u, v)G(u, v) - (F(u, v))2dudv.J x '(R)

The area of a bounded region is a measure that should be independent of the choice ofcoordinate chart. In order to see how our local expression for an area transforms withrespect 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.

OOF. If the variables u, v are transformed to variables u, D. then the expression dudePR

transforms to I det J(h)ldudv. Thus

f f IIyr, x yolldudu = f f Ilxv x xv11 I det J(h)ldudvA- (R) y (R)

Ilxv x xvlldudv.Ax '(R)

For a change of coordinates y ) o x, the component functions of the metric transformin such a way that

E F E Fdet F G = det p G (det J(y ) o x))2,

and so 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).


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

j

2a t

r

1Jf

EG - F2drd9 = r drd9R JO 1 r

= 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)) sinceE = G = I and F = 0. The expression EG - F2 is a measure of how much the surfacediffers from its parametrizing plane.

Exercises

8.1 Determine if the following functions qualify as coordinate charts for a surface in R3. Takex:(-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).(3) x(u, v) = (u, u + v, v3/2).

8.2 Let x: R2 --> R3 be given by x (u, v) = (u - v, u + v, u2 - v2). Show that the image ofx is a regular surface S, and describe S.

83' 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 atangent developable surface. Determine the component functions of the metric for sucha 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 thechange 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 thatx: V -+ R3 is a coordinate chart for a surface in R3. Determine the effect of the coordinatetransformation on the metric coefficients associated to x and to y = x o f.

8. Surfaces 115

8.6 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 generatedby the graph of a function f : (a, b) --. R, that is, the set ((u, v) I v = f(u)r.

8b

Map projectionsYou are quite right that the essential condition in every map projection is theinfinitesimal similarity; a condition which should be neglected only in veryspecialcases 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. Coordinatecharts for surfaces are named after cartographic maps. In this chapter we consider themotivating examples forcoordinate charts, the classical representations of the Earth, thoughtof 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 sphereS2) on a flat surface, that is, to give a mapping Y: (R C S2) - R2 that is injectiveand differentiable, and has a differentiable inverse. Such a mapping Y is called a mapprojection. A coordinate chart for the sphere, x: (U C R2) -> S2, with a differentiableinverse, determines a map projection by taking Y = x-t, and conversely, a map projectiondetermines a coordinate chart.

The purposes of cartography, such as navigation or government, determine the propertiesof interest of a map projection. An Ideal map projection is one for which all relevantgeometric features of the sphere are preserved in the image. Lengths, angles, and areason S2 are carried by an ideal map projection to identical lengths, angles, and areas foran appropriate choice of unit. More generally, such a map between surfaces is called anisometry. Many geographical features of interest could be found on an ideal map - shortestdistances between points would be represented by line segments. and so navigation wouldbe as simple as using a ruler and a protractor. The relative sizes of land areas would beapparent.

One of the first results in the study of mathematical cartography is due to Euler in hispaper 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 "infinitesimalsimilarity." Here a synthetic argument suffices. Let AABC be a triangle on S2 formed ofsegments of great circles. From Chapter I we know (informally) that great circle segmentsminimize lengths, and so the ideal map projection takes A B, BC, and AC to line segmentsin 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 tosuit to the purpose of the map and we must settle for the failure of other properties. T\voproperties serve most purposes:

Definition 81.2. A diffeomorphism f : S1 -+ S2 of surfaces (in particular, a map pro-jection) is conformal if it preserves angles, that is, given two curves, a: (-E, e) -+ S1and f : (-n, q) -+ S,, with a(O) = p = P (O), then the angle between a'(0) and f'(0) inTT(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 equiarealones. Before discussing the classical examples. we develop general criteria for a diffeomor-phism to be conformal and to be equiareal.

Proposition 81.3. Suppose f : Si -+ S2 is a diffeomorphism. Then f is conformal if andonly 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 theangle 01 between the corresponding curves on S2:

cos0rp(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))= cos0,.

VIf(p)(dfp(v), dfp(v)) i f(p)(dfp(W), 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. Thuswe can apply the same u, v coordinates on each surface. Furthermore, the local expressionfor 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 dipdfp (XU dt +x`'dt) = ti"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) andG(u, v).


Now suppose that f is conformal, and that ti and w are orthogonal unit vectors inTp(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 innerproduct we find

1 _ Ip(u,u+w) _ Ip(V,V+W)lp(u, u) Ip(v + w, u + 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

E = a2V V + 2abV W +b2W W

= P2(p)(a2 +b2) = P2(p)E.

Similarly, P = p2(p)F and G = p2(p)G. Varying the orthonormal basis smoothly in asmall neighborhood around p, we obtain this relationship between component functionsfor all points near p.

We now fix a representation of S22, a coordinate chart, against which we can compareother projections. The representation of choice dates back to the second century 13.c. andis standard in geography. A point on S2 is determined by its longitude A, measured fromthe Greenwich meridian, and its latitude 0, measured from the equatorial plane. Thesecoordinates are like spherical coordinates for S2 in R3 (see Chapter 1). but differ in thechoice of reference plane for the latitude. In the language of the Chapter 8. we have thecoordinate 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, re-spectively. This chart covers most of S2, leaving out the poles and the "International DateLine:" By rotating the chart, we could obtain other charts to cover S2 completely. For ourpurposes, 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 lineelement 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 Yis a map projection, then the composite

+ SZ >RZ(-7r, ;r) x (-2, 2 ___X

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 ofS is given by

area(S)=JJ EG-F2dAdO= rrcos0dAd4,.T r

Let E, F, and G be the component functions for the metric associated to the chart Y o x onR2. Expressed in this composite chart the area of Y(S) is given by

area(Y(S)) _ (J EG - F2 dAdm..1 r

This leads to the criterion for being equiareal:


Proposition 8"`.5. A map projection is equiareal if the induced coordinate chart onR2 C R3 satisfies to - P2 = cost 0.

More generally, a diffcomorphism of surfaces is equiareal if the induced metric satisfiesEG - F2 El; - P2. Notice how these criteria are fashioned out of the basic stuff of thedifferential geometry of surfaces, the first fundamental form and the component functionsof the metric associated to a coordinate chart. We now turn to the examples.

Stereographic projection

This map projection, applied to star charts, was known to Hipparchus of Nicae (190-125 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) sothat the plane tangent to the south pole is the plane z = 0. A point P on this sphere satisfiesP = (x, y, I + z) with x2 + y2 + z2 = 1. Let T = (0, 0, 2) denote the north pole.The rayT A meets the plane tangent to the south pole in a point Q = Y(P) = (u, v, 0). As vectorsin R3. P - T and Q - T are linearly dependent and so (P - T) x (Q - T) = 0, where xdenotes 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).

Thus u =2x

, v =2Y

. Introducing the longitude and latitude coordinates we haveI-z I-z(x, y, z) = (cos x cos 0, sin ), cos 0. sin 0), and so the formula for stereographic projectionis

o _ 2cosx cos0 2sinl cos0Y x(J1,0) ( 0'

I -sin0 ' I - sinm J

Proposition 81.6. Stereographic projection is conformal.

8bit. Map projections 121

PROOF. We view R2 C R3 as a surface with a coordinate chart given by Y ox. The tangentvectors to the coordinate curves are

2sinA cosh 2cosA cosm(Yo.r)a =

I -sinm ' I -sin4,,0

2 cos A 2 sin L'0(Y o x)m =

I - sin 0' 1 - sin 0

The component functions of the metric are

z

E(A,0)= (I - sin 0)2 (1 -sin 0)2

The function p(A, 0) = I - sinis nonzero for 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 aredeveloping. Notice that the previous computations also show that stereographic projectionis not equiareal. The metric coefficients associated to Y o x satisfy

EG p2 = 16 cost tb# cost 0.

(1 -sin O)°

More directly, the image of stereographic projection is all of R2; such a mapping simplycannot be equiareal.

Stereographic projection comes up often in other parts of mathematics, especially com-plex analysis. Euler. Lagrange, and others introduced complex variables into the study ofmap projections where the new point of view proved useful. We next record a remarkablegeometric 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 understereographic projection is a circle in the plane. If a circle on S2 passes through the northpole, 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 considerthree cases: (1) a circle passing through the north pole. (2) a circle that does not pass throughthe north pole and is not a great circle, and (3) a great circle that does not pass through thenorth pole.

In the first case, the image of the circle under stereographic projection is the intersectionof 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 apoint S that is the vertex of a cone tangent to the sphere and containing C. The ray TS isnot perpendicular to the z-axis and so it meets the plane z = 0 in a point M.


Suppose P is a point on C and Q = Y(P) is the image of P under stereographicprojection. In the plane determined by T, Q, and M consider LT QM and LQPS. ExtendST' 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. NowT 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 toATQM. 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 SP' SPTM MQ MQ-

This can be rewritten as MQ = TS . Now SP has the same length for all points P

on the circle C, and so the length MQ is the same for all points on C, that is, the image ofthe circle under stereographic projection is a circle with center M determined by the conepoint S.

In the case of a great circle not passing through the north pole the argument just givenfails - there is no tangent cone and hence no cone point S. To remedy this consider theunique 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 thesphere. Since TM is perpendicular to the plane in which the great circle lies, TM is parallelto the tangent plane at P and so parallel to the line I lying in the plane f> (T, Q, M) tangentto the sphere at P. The line I meets the line through T tangent to the sphere at the northpole and lying in fl (T, Q, M) at a point R. The pairs of alternate interior angles L P RT andLTMQ 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 andso 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 itwould be more convenient to represent great-circle routes as straight lines on a map. Thisis the attractive and unique feature of central projection.

To construct the projection, fix the tangent plane at a point R on the sphere and joina point P in the adjacent hemisphere to the center of the sphere. Extend this segment tothe tangent plane and this is the image of the point P. Central projection is defined on theopen hemisphere with R as center. The analytic expression for this map projection is easilyderived 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 animportant 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 thecenter of the sphere. The image of the great circle under central projection is the intersectionof 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 canbe 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 thecylinder can be unrolled to give a planar map - such projections are called cylindrical. Anappropriate piece of the plane can also be fashioned into a cone that may be placed on thesphere as the target for projection - such projections are called conical (see the exercisesfor examples). We next consider some examples of cylindrical projections.


Mercator projection

This map projection is the best known. Its appearance in 1569 marks the beginning of anew 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 aworld of commerce that was growing to include the entire globe. To understand the problemsolved 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 constantangle with respect to the family of meridians.

Since the meridians determine the North-South directions, the course set by a navigatoron a constant compass heading follows a loxodrome. In general a loxodrome is not theshortest path between two points on the sphere, but it is simplest to navigate. Extendingmost loxodromes determines a spiral path to one of the poles.

The problem solved by Mercator is to find a map pro-jection such that: (I) Meridians and parallels project to or-thogonal 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

A loxodrome.

satisfies Y o x(A, 0) = (A, v(A, 0), 0). The conditionv(A, m) = v(4,) guarantees the rest of the first desiredproperty. For the second, we characterize loxodromes.

Proposition 8".10. A unit-speed curve a: (-e, e) - S2 is a loxodrome if and only ifa(s) = x(A(s), 0(s)), where A'(s) =

cosCl

and 4,'(s) = c2 for some constants c1, c2.0 (s)

PROOF. Suppose a : (-r, c) - S2 is a unit-speed curve and a(s) = x(A(s), 4,(s)) in thelongitude-latitude coordinate chart. Then we can write a'(s) = A'(s)xA + O'(s)xm. Thecondition that the angle made by a with all meridians be constant is equivalent to

a'(s) xx a'(s) x0=

Ila'(s)II Ilxxll =ci, a constant,

Ila'(s)II II-Toll -c2, a constant.

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 meridiansand parallels are perpendicular, the constants are the cosines of complementary angles soci+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. Wecan write AX(s) + Bv(O(s)) + C = 0 for some A, B, and C. This equation leads to thecondition

v'(O(s))O'(s)x'(s)

= a, a constant.

By the proof of the previous proposition.

c2d v/d4 d v=

b

ci/cos0(s)=a or

do cos0

Integrate this differential equation by separating variables. We are free to choose the un-known 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, and0 = (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 theinitial 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 cylindricalprojections of the sphere. Consider the sphere S2 inside aninfinite cylinder tangent to the equator. Central projection ofthe sphere to this cylinder is not conformal, but adjusting it to"make the cylinder infinitesimally close" leads to the Merca-tor projection. Arguing with infinitesimal triangles, as Mer-cator 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 theinfinitesimal corrections needed to discover his projection were at hand.


Lambert's cylindrical projection

All of the examples presented so far have not been equiareal.Furthermore, they have all been constructed via projectionfrom a point. In the next classical example, due to J. H. Lambert,we consider an orthographic projection, that is, a mapping forwhich all lines of projection are orthogonal to the plane of pro-jection. Instead of a plane we substitute a cylinder of heighttwo, exactly fitting around the sphere. The lines of projectionare required to be perpendicular to the cylinder at each point.

From this description, the Lambert projection is expressedby Y o x(x, 0) = (>`, sin 0, 0). Rolling out the cylinder on theplane we get the image of the Lambert projection, the openrectangle (-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 functionsof 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 verticallines and parallels to horizontal lines. In fact, it is unique as an equiareal projection withY o x(x, 0) = (x, u(0), 0).

The possibility of such a cylindrical equiareal projection may have been known toArchimedes, who had computed the area of a sphere and the area of a cylinder of thesame height. Lambert's work appeared in his Beytrage zum Gebrauche der Mathematikand deren Anwendung (1772) in which Chapter VI is dedicated to finding the general formof a conformal map projection. This is one of the first mathematical treatises on the subjectof cartography. Lambert gives several new projections which bear his name and are in usetoday.

Azimuthal projection

Map projections are a particular use of change of coordinates. From our experience of planegeometry we can try to set up polar coordinates on the sphere. This leads to the so-calledazimuthal projection: Fix a point P on the sphere and a point Q in the plane. Send points ona circle of radius p centered at P on the sphere to the points on a circle of radius p centeredat Q in the plane.

To obtain an analytic expression, we choose a convenient point as P. the north pole, andone 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 arbitrarypoint (A, ¢) is measured along a meridian and is given by (n/2) -0. We now simply changethe coordinates on the sphere from (A, 0) to (A, p), where p = n/2 - ¢. The change ofcoordinates 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 andso distances are preserved. If we rotate the sphere to center the projection at an importantpoint, then the distances from the center of the coordinates to other points on the globeare compared easily by following the new set of meridians. Notice that in these rotatedcoordinates, the line element remains ds2 = sin2 p dA2 + dp2, the polar line element ons2.

Many other map projections exist, suited to the specific needs of map makers. Themathematical problem of characterizing all conformal and equiareal map projections wassolved 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 developmentsof mathematical cartography which include the work of N. A. Tissot (1824-1904), whogave a uniform theory of distortion for map projections (see Robinson (1960)).

In later chapters we will consider the more general problem of developing surfaces, thatis, 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 restrictour attention to Australia and its neighbor New Zealand. Since these countries are in theSouthern Hemisphere, projections such as stereographic or central projection do not producecompletely unrecognizable maps. The grid lines are 10° apart. The maps were producedusing Mathematica and John Novak's package called WorldPlot.

Stereographk projection.


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 ofS2. Do the same with the coordinates determined by the inverse of stereographicprojection.

8'.2' Suppose f: S, -+ S2 is a diffeomorphism between surfaces that is both conformaland equiareal. Show that f is an isometry, that is, there are coordinate patches aroundcorresponding points on the surfaces with ds, = ds2.

S'63 Apply the criterion forconformality to Lambert's cylindrical projection anddeterminethe 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 isorthographic. 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 mapprojection is equiareal or conformal.

g"'.5 Project a loxodrome on the sphere orthographically to the plane tangent to one of thepoles. 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 withthe meridian and ends up by winding around the pole. Contrast this with the lengthof the great-circle route.

8'.7' The plane and the cylinder were used in map projections because there is an idealmap 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 coordinatecharts for it. Show that the coordinate charts you have chosen provide an ideal mapprojection 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. Thusone could fashion a map projection by mapping the sphere to a cone and then to theplane. Choose a parallel 4$o in the northern hemisphere and construct the cone tangentto the sphere at this latitude. Project to the cone from the center of the sphere. Give ananalytic expression for the resulting map projection and determine if it is conformalor equiareal as a function of the latitude.

8/e.9 Consider the following mapping to the cone tangent to the parallel at 00: Send apoint 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 . Using the results

of the previous exercise, show that this projection is equiareal.

9

Curvature for surfacesInvestigations, in which the directions of various straight lines in space are tobe 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, andsuppose the different points of the sphere to represent the directions of straightlines 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 thefirst fundamental form, was defined. We now define analogs of curvature and torsion forsurfaces in R3. The goal is to describe how a surface "curves" at a point. The first measurewe introduce is naive - it will depend on curves in the surface. Later in the chapter, weassociate a more appropriate, two-dimensional measure to the same task. Based on theseideas, the generalization for surfaces of the fundamental theorems for curves is realized inChapter 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 unittangent 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 withS. This gives a curve on S near p.

Suppose co(s) is a unit-speed parametrization of the curve so that We definethe normal curvature of S in the u direction at p, k (v), to be the directed curvature K (s)

131


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 lacourbure des surfaces (Omnia Opera, volume 28, pp. 1-22) Euler published the followinguseful theorem.

Theorem 9.1 (Euler 1760). If the normal curvature k,, (u) is not a constant function ofv, then there are precisely two unit tangent vectors X) and X2 such that k (XI ) = k) ismaxima! 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 iscompact, the directions X1 and X2 exist. We prove that if X( is the direction in whichthe maximal normal curvature occurs and v makes an angle 0 with X), thenk(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 bethe graph of a function, ((x, y, z) I z = f (x, y)}. A particular choice of a coordinate chartfor 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/

= 0, andaf I

= 0.ax

I(0.0)

aY (o.o)(*)

Another trick is needed - we rotate the xy-plane until a`IfI = 0. Rotations are given

byaxay (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. Thenafe = - sin 0

of+

av ax

cos B of , anday

a2le -sin0cos0a-jj+(cosz0-sin20) +sinBcosBa?jaxay 8x2 ax ay ay2

a2 f (sin 29 (a2 j a2 fI=cos2B-ay +2/ / \ ay ax2

.

ax \

To obtain axay = 0 when axe

a

, solve for 0 in the equation

2azjtan 20 =

axayja2f

axe - are (0.0)

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 setf to be fo as constructed.

When the surface S intersects the xz-plane, the resulting curve is parametrized (notnecessarily 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 directed2

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 Scan be parametrized by c;,(t) = (t. f (t cos 0, t sin 4,)). The directed curvature is given by

k (v) = dt f (t cos 0, t sin 0) = cost 4 axf (0, 0) + sine 4 a-f (0, 0)2

11=0 2 Y

= (cost m)ki + (sin2 O)k2.

As a function of 0, 0 < m < 2tr, achieves its minimal value at k2 and its maximalvalue 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 ofthe surface near a point. From the previous proof we can write the equation of the surfaceS near p as a graph z = f (x, y) with

z = kix2 + k2y2 + R(x, y), where limR(x,

y) = 0.x2 + y2

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 quadraticsurface approximating S at p is a paraboloid. The level sets given by a constant z valueare 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 lieson 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 surfaceapproximating Sat p is a hyperbolic paraboloid. The level sets given by a constant z valueare hyperbolas. In this case, near p, there are always points of S on either side of the tangentplane.

Case 3. ki > 0, k2 = 0 or ki = 0, k2 < 0. Such a point is called parabolic and thequadratic surface approximating Sat p is a parabolic cylinder, z = kix2 or z = k2y2. Nearp, 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 orf (X, y) = ±xa show that the behavior of S with respect to the tangent plane near p mayvary.

When k1 = k2 i4 0, then every direction is a principal direction. Such a point is called anumbilic point and we may take any pair of orthogonal directions at such a point as principaldirections. For example, all points on a sphere are umbilic.

When we have developed further tools, we will prove this result in a manner that doesnot use a special choice of coordinates. We now jump forward 68 years to the pioneeringwork of Gauss on surfaces.

The Gauss map

For curves, a measure of curvature is given by changes in the normal vector. This notion wasgeneralized to surfaces by Carl Friedrich Gauss (1777-1855) in his seminal 1827 paper,Disquisitiones generales circa superficies curvas (Werke, volume 4, pp. 217-58). Gausstook his inspiration from astronomy and the idea of the "celestial sphere" Like many othercontributions of Gauss, Disquisitiones had a profound effect on subsequent work on thissubject. 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 mappingN: S -. S2, the unit sphere in R3 centered at the origin, given by associating to eachpoint in S the unit normal at the point.

Locally we have the formula

N(p)_ xuXxt,Ilxu X XV11


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 payattention that results depending on the Gauss map in fact do not depend on the choice ofcoordinates.

For a regular surface, we leave it to the reader to show that N is differentiable and so ithas a differential map dNp: Tp(S) - T,u(p)(S2). What makes the Gauss map useful is thefact that the tangent plane to S at p is the plane normal to N(p) and it is a property of S2that Tq(S2) is the plane normal to q. Thus Tp(S) = TN(p)(S2), and dNp can be taken to bea 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 asthe tangent to a curve, that is, if v is in Tp(S), then there is a curve a: (-e, e) - S witha'(0) = ti. It follows that dNp(u') = (N o a)'(0). This is the rate of change of the normalvector 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 constant vector, and so d N = 0.a +b +c

(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).


It follows that(-2uo, 2vo. 1)

N(p) = N(x(uo, vo)) =4u0+4vO2+1

To obtain an expression for d Np at p = (0, 0, 0), let a: (-e, e) -* S be a unit-speed curvewith 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

dN(o.o,o) = C-2 2)

.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 thefirst fundamental form.

PROOF. Because Ip is an inner product, it is linear in each variable and so it suffices tocheck the proposition for the basis vectors, that is, lp(xu, dNp(x,,)). Todetermine d Np we compute

(N o x(u, vo))'(uo) =IN J = N, (N o x(uo. v))'(vo) =

INJau (Uo.un) av (U0.vo)

= N,,,

thus we need to check that l p(N,,, x,,) = Ip(xu, N,.). We know that l p(N, 0 =lp(x,,, N). Taking partial derivatives, we obtain

Ip(N0, Ip(N, x,,,,) = 0 = l p(N,,. Ip(N. x,,,,).


By the differentiability of the coordinate patch, x,,,,. Thus,

Ip(x.,dNp(xo))

What about self-adjoint linear mappings? We record certain useful properties of thesemappings. 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 Ahas two real, not necessarily distinct, eigenvalues.

PROOF. Apply the defining equation (A(v), w) = (u, A(w)) to the elements of any basisfor V and the resulting matrix representation for A is given by

(a bA- _b

dl.

To compute the eigenvalues of A we solve for the real roots of the characteristic polynomialdet(t I - A) = t2 - (a +d )t + (ad - b2). When we substitute these values into the quadraticformula, 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 eigenvectorsel, 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 andassociated eigenvectors et and e2. Take ei and e2 to be of unit length. Suppose v E V isalso 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 orit 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 sin2(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

the appropriate setting for the Gauss map. We now introduce the function that plays forsurfaces the role that curvature and torsion play for curves. Recall that a quadratic formon a real vector space V is a function B: V - R that satisfies 8(r v) = r2 B(V) for allV 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 isthe 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 withS is a curve c;,(s) on S near p = taken to be unit speed with u. Sinceca(s) is also a plane curve, we can talk about its directed curvature K(s), which is given byc"(s) = K(s)N(s). By the definition of the normal curvature in the direction v, we havec;;(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' (s)) + lc;.ts)(N(c-v(s)), c "(s)) = 0, or6 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, theproperties of such operators imply properties of Iip, In particular, Property 3 together withTheorem 9.6 give another proof of Theorem 9.1.

There are many other planar curves through p on S with vas tangent vector. The followinggeneralization 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. Meusnierpublished only one paper in mathematics and went on to publish important papers in otherscientific 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 tov E Tp(S) and w makes an angle f with N(p), 0 < rq < ,r/2, then the unit-speed curvec,;, (s) through p with v given by intersecting the plane determined by [v, iu1 withS has directed curvature

IIp(i )Kp, _

cos 0

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 matrixin 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 andplanar points. The nature of the approximating quadratic surface is determined by the signof the Gaussian curvature. Another important observation is that the Gaussian curvaturedoes not depend on the choice of coordinates around p; any change of coordinates leadsto a change of basis for the tangent plane at p, but K(p) remains the same because it is adeterminant.


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 xVII

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))) = u'(0)xu + v'(0)xv,r=0

Since N and N lie in Ta(S)

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 wewrite 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:

u'(0)Nu + v'(0)N,..dN.(iu) = dt N ox(u(t). v(t))11=0

e (N., xu) (N, xuu).

f (Nv, xu) (N. xuv) = (N, xvu) (N., x,,),

g (Nv, xv) (N. xvv)

Nu=allxu+a2lxv,Nv = al2xu + a22xv,

We can now write Ilp(w) = e(u')2 + 2 fu'v' + g(v')2. In terms of the functions (ark ) wehave

11P60 = - all021

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.

9. Curvature for surfaces

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

a ,re f1 a21 E F=( :2a22f g a F G

from which it follows that

all a21 _ -1 e

(f G F

atza22

EG - F2 f g -F E

Theorem 9.9. The following equations hold:fF-eG gF- fG fF-gEeF- fE_

all EG-F2 0212

_EG-F2' a21

_ _EG- 'azz EG-F2.

Corollary 9.10. K(p) = det d Np = det(aj j) =",g

f 2'EG - Fz

As an example, we calculate the Gaussian cur-vature for a surface of revolution (Example (4) inChapter 8). Let

c(t) = (x(t). µ(t)): (-r, r) -+ R2

be a regular curve that does not cross itself or the t-axis. A coordinate chart for the surface of revolutionobtained by revolving the curve c(t) around the t-axis is given by

x(u, v) = (X(u) cos v, a(u) sin v, µ(u)),

for-r<u<rand0<v<2r.We begin the calculation with the basis for the tangent vectors:

xu = (),'(u) cos v, X'(u) sin V, A, (u))

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.l (A')2 + (A17). The otherthe coordinate chart has length Ilxu x xvll = EG - =

auxiliary functions are given byxu x xv

e = (N. x,,) xuu)Ilxu x xvll'

')2 + (µ')z(xp x xv, xuu)

(A')2 +(l')2

.


By similar calculations we obtain f = 0 and g = p In the basis {x,,,(p')2

dNx(,,,,,) takes the form

X"pr - A'p"

d Nx(u.,,) _ ((A')2 + (l2')2)3I2

0

0

- px (A')2 + (p')2

In this form it is evident that x and x are the principal directions with principal curvatures"

z p - z p and-p

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 Gaussiancurvature is given by the determinant of dNp:

,L"(p')2K(p)

- 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. Theformula 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( ) -PA'p'p" - x"(p')2x((A')2 + (p')2)2

- COS ucost u cos u(sin2 u + 1) + sin u

cos4 U

sin u sing a sing U = -1- -(sin u t cost

COS42

u +/sing u

u

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, thereis another geometric interpretation, in fact, the one that guided Gauss.


Proposition 9.11. Let p be a point in a surface S with K(p) # 0. Then

K(p) = limarea N(A)

A- p area A

where A is a connected region around p and the limit is taken through a sequence of suchregions 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 linearityof 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 Aaround p is given by the formula

f rarea(A) =

JJIlxu x

X (A)

Mapping the region to S2 by the Gauss map we get

area(N(A)) = jf. IINk x Nvlldudvx '(A)

Kllxu x xvlldudu.x '(A)

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

tp f f - i (A) Ilxu x xvllduduxA

ffx-'(A) Kllxu x xuIldudv

(I/area(A)) ffx-'(A) Kllxu x xvlldudu

areaA-.0 (I/area(A))ffx_'(A) Ilxu x xvlldudu

limKllxu x xv11 I = K( ).

area A-.O llxu X X, 11 some point in Ap

lim

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 seethat the bigger the sphere, the smaller its curvature and the better it approximates the plane.

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 asmooth 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 k2satisfy Iki l < 1 and Jk21 < 1 at every point in S. If a(t) is a regular curve in S, does itfollow 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 conjugatehyperbolas, 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 ofthe surface. Interpret the property of convexity in terms of the principal curvatures andGaussian 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 chartx(u, v) = (f,(u. v), f2(u, v), f3(u. v)). that ds2 = Odu2 + 4dv2, for some functionm. Show that the image of this coordinate chart is minimal if and only if the functions jare harmonic, that is,

a2f a2jau2 + av2 = 0.

9.9 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 generatedby rotating the catenary curve y = acosh(z/a) around the z-axis. Prove that M is aminimal 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 ofthe principal directions of curvature for each t. Prove Rodrigues's formula: a(t) is a lineof 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 thiscase is -x(t).

10

Metric equivalence of surfacesIt is evident that any finite part whatever of the curved surface will retain the sameintegral 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 ofthe existence of an ideal map projection (Chapter 8"'), that is. a mapping Y: (R C S2) - R2preserving lengths, angles, and area. The archaic terminology for such a map projection is adevelopment 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 isa development of a given surface onto another.

A diffeomorphism between two surfaces, 0: S( - S2, identifies the analytic structureof the surfaces, such as coordinate charts, at corresponding points. However, in order toidentify the geometric structures of two surfaces, concepts such as arc length and area mustalso 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 andiui, 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 theappendix to Chapter 7 we showed that any rigid motion F: R3 -+ R3 is given by aninvertible 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 bean isometry of surfaces.

If we further restrict our attention to isometrics of R3 that fix the origin, the linearisometrics, 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 andreflection across a plane containing the center of the sphere.

Not every isometry between surfaces is the restriction of a rigid motion of R3. Forexample, 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 arisefrom rigid motions of R3 are the most interesting.

145


Furthermore, diffeomorphic surfaces need not be isometric. For example, two spheresof different radii are diffeomorphic, but not isometric - the arc length of a complete greatcircle depends on the radius of the sphere. The notion of isometry determines an equivalencerelation on surfaces.

The notion of isometry makes precise the ideas of rigid motions and congruence. Wedefine 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 madeup of segments of curves on a surface may be thought of as rods in a configuration and theterm rigid motion is synonymous with congruence.

It follows from the definition that the inverse of an isometry is also an isometry and so theset of congruences of a surface forms a group. The importance of this observation cannot beoverestimated. It is the basis of another approach to geometry via so-called transformationgroups, initiated by Felix Klein (1849-1925) and Sophus Lie (1847-99). (An introductionto this approach is found in Ryan (1986).)

Properties that are preserved under isometrics are the most important to the geometry ofsurfaces. We call such properties intrinsic. Properties that depend on the particular descrip-tion of a surface are called extrinsic. For example, the fact that the z-axis is asymptoticallyclose to the surface of revolution of the tractrix is an extrinsic property of the surface. Wewill 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 functionsassociated to a coordinate patch.

Proposition 10.2. If 0: Si -+ S2 is an isometry, and p E S1, then there are coordinatecharts x: (U C R2) - Si around p and x: (U C R2) - S2 around 0(p) such thatthe 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 tobe the coordinate chart given by x = Oox. This satisfies the properties of a coordinate patchby virtue of the properties of a diffeomorphism. Now i, = and zv =Since 0 is an isometry, E(u, v) = E(u, v), F(u, v) = F(u, v), and G(u, v) = ((u, v) bydirect calculation.

The proposition gives an important pointwise property of an isometry. An immedi-ate corollary is that length, angle, and area are preserved by an isometry. For propertiessufficiently local, it is enough to have a local version of isometry to compare surfacesgeometrically.

Definition 10.3. A mapping 0: (V C Si) -> S2 of a neighborhood V of a point p inSi is a local isometry at p if there are neighborhoods W C V of p and 1v of 0(p) with01 w : iV - W an isometry. Two surfaces are locally isometric if there is a local isometryat every point for each of the surfaces.

A local isometry may fail to be an isometry. Forexample, a cylinder is locally isometric tothe plane (roll it out). However, because the plane and the cylinder have different topologicalproperties, no single isometry identifies the cylinder with part of the plane.


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 thatE = E, F = F, and G = c on U, then m = . o x-I : (x(U) CSI) -s S2 is a localisometry.

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 polar-ization 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 correspondingpoints, then we expect them to share all other intrinsic properties as well. We now cometo the central role of curvature in geometry. At any point of a regular surface S C 1R3there is a (not necessarily orthogonal) basis for the R3 directions from the point that canbe taken conveniently to be {xu, xu, N) with the choice of a coordinate chart. Consider thederivatives 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 ofS in the parametrization x. The indices i, j, and k vary over I and 2 where I correspondsto the variable u and 2 to the variable v. The lower indices ij refer to the second partials ofthe 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 afunction f (u, v), we also have

Eu = 2(xuu.xu). Ev = 2(xuv,xu).

F. = (x,u,xv) + (xu.xuv), Fu = (xuv,xv) + (X,,Xuv),

Gu = 2(xuv, xv), Gv = 2(xuv, xv).


Expanding these inner products in terms of the Christoffel symbols we obtain the followingrelations, here written as equations and in matrix form:

f,', E+rIIF =2'E,,.

I'i,F+r11G=Fu-7' E,,,

E'2E+f12F=2E1it r12F+ri2G='Gu,

r'22 E + r22 F = FL, - I' Gu.

{ r'22F+ r2 G = 2G22

E\F

G/(rl

r?i / - \F.EEJ

\F GJ \riz) "GU)

2 = ).\F G)(r22)

(F2'G,

Thus the functions T are expressible in terms of the metric coefficients E, F. and G. andtheir derivatives. Proposition 10.2 implies that at corresponding points in isometric surfacesthe Christoffel symbols agree. Any expression given in terms of Christoffel symbols that isindependent of the choice of coordinates is then intrinsic.

We now record the most important theorem of Gauss's Disquisitiones (§ 12). The resultmakes curvature and the intrinsic properties of surfaces the principal objects of study fordifferential geometry.

Theorema egregium. Gaussian curvature is intrinsic.

PROOF. We give two proofs. In each we show that the Gaussian curvature K can be expressedentirely in terms of the component functions of the metric and their derivatives. We beginwith 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

xuu x xo + xu x xu a I

N,, = + (XU X Xv).EG - F au EG - F

xuv X Xt. + Xu X Xi.u aNV = + av (xu x xu).,I- av EG-F

Substituting these expressions in the equation for curvature and simplifying, we obtain thefollowing expression, where the column vectors shown are in fact 3 x 3 matrices and so wehave 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 Xu l Xuu

= del Xv ) det ( 1, - det Xuv ) del xvXU v xv I XU

Xuu (XvV)11 Xuu (XUV)']= det x x - del x xXv Xvxv xv

xuu XVV xuu . XU Xuu Xv= det XU xvv E F

Xv xuu F G

Xuo Xuv Xuv Xu xuv Xv- det XU xuv E F

xv Xuo F G

149

We can express the second partials, xuu, xvv, and xvv, in terms of the component functionsof the metric and the Christoffel symbols. Furthermore, we can rewrite xuu xuu - xuv xuvas follows: Since x xvv = F - 21 Gu we find x,,,, xvv + xu xvvu = Fvu - 2Guu. AlsoXU xuv = i Ev implies that xuv xuv + x xuuu = 11 E. Since the coordinate patch isassumed 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 - I Guu- I Evv2 2

Putting this all together, we get the formula for Gaussian curvature

- 'Evv + Fuv - 2Guu IEu F. - 2' E,K = I

(EG - F2)2I ( 2'

det Fu - I' Gu E F

Gv F G

7 0 Ev 2' Gu

- det i Ev E F

Gu F G

Corollary 10.5. Isometric surfaces have the same Gaussian curvature at correspondingpoints.

This proves that Gaussian curvature is truly a geometric feature of a surface. The def-inition of the Gaussian curvature relies on many features of the embedding of the surfacein 1t3 such as the Gauss map. It is remarkable (egregium) that this function is independentof these trappings. Euler's theorem on the nonexistence of an ideal map projection followsdirectly from theorema egregium since the sphere and the plane have different curvatures.

There are further relations between the Christoffel symbols that follow from the differ-entiability of the parametrization. They also lead to our second proof of theorema egregiumwhich is based on a calculation similar to the one at the heart of the Frenet-Serret theorem.


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)

)ux,, +r2 (r,2xa+r2 xo+fN)+ fuN+ f(aiixu+a2lxu)+(r212 12 1 12

_ ((r,2)- + rizri, +ri2ri2 + fai,)xu+ 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.

fF-gEG - F2

and a21 = EG -F2 and so from theFrom Theorem 9.9 we know that a22 =E

second line we obtain the relation

(ril) , - (ri2)u+rl l r12+r2

l r22 - r12r11 - ri2ri2 = fa21 - ea22_ eF- fE fF-gE-fEG-F2eEG-F2

=Eeg-f2 =EK.EG - F2

Once again we have proved that K is expressible in terms of Christoffel symbols and thecomponent functions of the metric.

How well does the apparatus of functions (E, F, G, e. f, g) and their associated func-tions, the Christoffel symbols, characterize a surface? This natural question extends therelation of the Frenet-Serret apparatus to a curve in R3, that is, the fundamental theorem for


curves (Theorem 7.6). Let the Gauss equations refer to the relations among the Christoffelsymbols obtained from (x,,,,) - (xuv)u = 0 and (xuv)v - (xvv)u = 0:

(a) (rii)v - (r1 2)u + rii rig+p2

i rz2 - ri2ri i - ri2ri2 = EK.(b) (ri2)u - (rii)v + ri2ri2 - ri i rz2 = FK.

(c) (r22)u - (1-124 + ri2rii + rz2ri2 - ri2ri2 - ri2rz2 = GK,(d) (ri2)u - (ri2)u + & 12' 12 - r22ri i = FK.

Let the Mainardi-Codazzi equations refer to the relations coming from the differentiabilityof N, that is, Nu, - Nvu = 0. They are

ev - fu = er;2 + f(r12 - l' ) - gri m11

fv - 91, = er22 + f(ri2 - rig) - gri2

G. Mainardi (1800-79) and D. Codazzi (1824-75) were colleagues at the Universityof 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, andg in 1856. Ten years later. Codazzi published his account of these relations in terms ofdirectional cosines. In this equivalent form, the equations became widely known as partof his second-prize winning memoir submitted to the Paris Academy in 1860. In fact, theequations were known in 1853 by K. Peterson (1828-81), a Russian geometer who studiedin Dorpat under Senff and Minding (Reich 1973).

The Gauss equations and the Mainardi-Codazzi equations together are called the com-patibility equations for surfaces. Ossian Bonnet (1819-92) proved the following versionof the fundamental theorems we have proven for curves.

Theorem 10.6 (fundamental theorem for regular surfaces). Let E, F, G, e, f, g bedifferentiable functions defined on an open set V C R2 with E > 0 and G > 0. Suppose thatthese functions satisfy the compatibility equations, and EG - F2 > 0. Then for each q E Vthere is a neighborhood U of q, U C V, and a diffeomorphism x: U - x(U) C R3 suchthat the regular surface x(U) has E, F, G, e, f, and gas component functions of the firstand second fundamental forms. Furthermore, if U is connected and z: U -+ . (U) C R3 isanother such diffeomorphism, then there is a vector v E R3 and a matrix A in 0(3) suchthat g=Aox+v.

The proof, like that of the previous fundamental theorems, is a side trip into the theory ofdifferential equations and would take us too far afield (see Spivak 1975, volume 3, pp. 79-85). It follows from the theorem that the six functions E, F, G, e, f, and g determine thelocal theory of a surface.

Special coordinates

The simplifying feature of curves is the arc-length parametrization. There is no such canon-ical choice of coordinates for surfaces. This fact is a source of both difficulty and freedom.

We use it to our advantage and try to construct coordinate charts whose properties are closerto the geometric features of the surfaces. This will simplify the computation of importantmeasures such as the Gaussian curvature. Artful choices of coordinates play a crucial rolein certain arguments in later chapters. Here we introduce two useful constructions to givethe flavor of these ideas.

A familiar feature of Cartesian coordinates in elementary geometry is that coordinatelines meet at right angles. On a surface, this condition implies that the second componentfunction of the metric, F, is identically zero. There are many ways to construct such acoordinate patch for a surface. We now consider a particular choice that arises from thesurface'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 di-rections are orthogonal. A patch for which the coordinate curves are lines of curvature isorthogonal, 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,for all t E (a, b), 0.

Asymptotic lines do not always exist at each point in a surface. To see this write thedifferential 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 ateach point in a neighborhood of a point in a surface, then there are two asymptotic directionsat each point. Thus if K(q) < 0 for all q in a neighborhood of a point in a surface, then wecan try then to construct a coordinate chart with asymptotic lines for coordinate curves.

There are appropriate differential equations for lines of curvature and asymptotic lineswhose solutions lead to the coordinate charts we seek. However, we present a more generalmethod for constructing special coordinate patches. If W is an open subset of a surface S andV (q) is a smooth assignment, q r-> V (q) E Tq (S), for each q E W. then we call V a vectorfield on W. In a coordinate patch x: (U C R2) -+ S this can be expressed by a pair of smoothfunctions a(u, v), b(u, v) such that V (x(u, v)) = a(u, v)x,, +b(u, v)x,,. An Integral curvefor 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 integralcurves for the fields of principal directions - these vector fields (X1, X2) are orthogonaland of unit length and so F = 0 and E = G = 1. Thus the line element takes the formds2 = 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 thecoordinate curves for our charts. More generally, given a vector field V (q) on an open set

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 curvatureare lines in the direction of a principal vector and asymptotic lines are lines in asymptoticdirections.

Theorem 10.9. Suppose p is a point in a regularsurface S, and in a neighborhood W C Sof p two linearly independent vector fields X (q) and Y(q) are defined. Then there is a patcharound 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 Uand 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 isu'/u'; this gives the ordinary differential equation

du a(u, v)dv = .fx(u, v) = b(u, v)

By Theorem 6.19 on the existence of solutions of such differential equations, there is asolution 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 equationsassociated 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 curvesux(v; 0, vo), and u y (v; uo, 0). Consider the function sending (r, s) to (uo, vo). By the linearindependence of the vector fields, and the uniqueness of solutions of differential equations,this change of coordinates is well defined and smooth, and determines a diffeomorphismof a neighborhood of the origin in U to some open set in R2. Let 0 be an inverse to thisfunction. The composite y = x o m is a coordinate chart for a neighborhood of p in S. Thecoordinate curves of this patch run along curves that satisfy the differential equations givenearlier. 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) andv' = 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 < 0in a neighborhood around p, then there is a patch around p whose coordinate curves areasymptotic lines.

PROOF. We leave it to the reader to show that the set of umbilic points in a regular sur-face is closed. Around a nonumbilic point p, then, there is a neighborhood of nonumbilic


points. The principal directions determine a pair of linearly independent vector fields onthat neighborhood to which Theorem 10.9 applies.

If the curvature in a neighborhood around p is negative, then there are two linearlyindependent 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 mayapply 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 tothe surface are simplified. As we have already said, the metric component function F = 0by Euler's theorem. We can also deduce that the component function f of the secondfundamental 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. Thisyields

( Oa rau) (./ g/ 0 E/ EG

and since E i4 0 G on a regular surface, f = 0. This matrix equation also implies theformula K = eg/EG.

If the coordinate curves are asymptotic lines, then Il ll p(x,.) = 0. Expressing Hpin terms of the basis (xv, xv,) this implies that e = 0 = g. This gives the simpler expression

_fzfor Gaussian curvature K = EG - F2

The simplification of the components of the first and second fundamental forms leads toa simplification of the associated Mainardi-Codazzi equations. We record the results herefor use later in Chapter 14. If the coordinate curves are lines of curvature, the discussionabove 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), fo = f(r2 - r;z).22

(EGf F2)(!GEU - FF + z FEv + Z FED, - 2

(EGfSimilarly, the form of fv is given by

.fv (EGf FZ) (' aThese expressions play an important role in later proofs.


Exercises

10.1 Show that a surface S is isometric to a surface of revolution if it has a coordinate chartaround 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 conformalmapping of the surface to the plane.

10.2 Determine the Christoffel symbols r, f for the polar-coordinate parametrization of theplane.

103' Suppose that F = 0 for some coordinate chart. Prove that, on this chart.

aE aGI a TV a auK_2EG au EG au EG

10.4 Looking forward to the generalization of the differential geometry of surfaces to theirhigher-dimensional analogs (Riemannian manifolds), we present another version oftheorema egregium, notation-wise. All indices shown vary over the set (1, 2) corre-sponding to coordinates (u, v). Let u = u1 and v = u2. Consider the matrices

(gil) =E F(F G (g'1) = (go -I.

Show that the Christoffel symbols are defined by

ri gi` + - agk2

(at1au; auk aul

Define the Rlemann curvature tensor to be the function

r'; + rkr' - r, rk,Riskau rik aau k

k

along with the transformed version

0oi.2

R'jki = gi.Rikrm=i.2

Show that theorema egregium now becomes

_ R1212K

det((g,j))

10.5* A surface S in R3 is ruled if through each point p E S there is a line in R3 entirelycontained in S. Show that the line through p lies along an asymptotic direction. Provethat if a surface is ruled, then K < 0 at each point. Give an example of a ruled surfacethat 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

x = ui cos vi. v = ur sin vr, z = av,. the right helicoid.

Show that the mappings v = yr and u = F_2 + a2 0r ill = u - a determine anisometry between these two surfaces.

10.8 Suppose that the functions e, f, and g are all zero for a surface S. Show that the surfaceis 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, andthat the mean curvature H(p) 96 0 for all p in W. Show that the unique asymptoticcurve through each point p E W is a straight line in 1R3, and that the tangent plane isconstant for all points along any such line.

11

GeodesicsBut if the particle is not forced to move upon a determinate curve, the curvewhich it describes possesses a singular property, which had been discovered bymetaphysical considerations; but which is in fact nothing more than a remarkableresult of the preceding differential equations. It consists in this, that the integralf v ds, comprised between the two extreme points of the described curve, is lessthan on every other curve.

P. S. Laplace, Mfcanique CEleste (1799)

In the last few chapters we developed some of the analytic and geometric properties ofsurfaces in ]R3 that are consequences of the properties of the first and second fundamentalforms. The important geometric properties are intrinsic, that is, preserved by an isometryof the surface. To develop further elementary geometric notions we next introduce a notionof "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 sur-face should be intrinsic. We first try to adapt the condition of zero plane curvature. Leta: (-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 thepoint a(s), T'(s) has a component pointing in the normal direction and a component lyingin the tangent plane. The intrinsic part of this vector will be seen to be the component lyingin 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


Theorem 9.6 tells us that the magnitude of the component in the direction of N. (T'(s), N), isthe plane curvature of the curve given by the intersection of the plane spanned by (T(s), NIwith the surface S. This is the normal curvature of the surface at a(s) in the directionT(s), given by lla(s)(T(s)) = (T'(s), N). The other component, lying in thetangent 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 Gaussand later professor of Mathematics at Dorpat (now Tartu in Estonia). This university wasfar removed from the research centers of mathematics, but it was an outpost for work indifferential 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 hislifetime 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 thefunction

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 toS 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 featureof a curve, that is. it is dependent only on the component functions of the metric, and so itis 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 inthe na(s) direction and so it is perpendicular to N. Ignoring the component in the normaldirection 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'rz + (v')2rz )x,,, N).iz z2

ToTo 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)

u' u"+(u')zrli+2u'v'rl2+(v)2r22EG-F2det(,)z z z z )V/ v + (u r + 2u v r12 + (VI)2r22

If 0: S -> S' is an isometry, then, at O(x(uo, vo)), the expression for kg is the samefor the curve $ (x(u(s). v(s))) since it depends only on the components of the metric, theChristoffel symbols, and (u(s), v(s)).

It is useful to have an expression for geodesic curvature when a curve fl(t) is not unitspeed. This is made easy by the properties of the cross product in llt3. Write the arc-lengthparametrized curve associated to ,8(t) as a(s) = 6(t (s)), where t(s) is the inverse functionof the arc length. Then

z z

T (s) = a'(s) = 0'(t (s)) d and T'(s) = a" (s) = P"(t (s)) (T) + (t (s)) dsz


It follows that T(s) x T'(s) = [#'(1 (s)) x 6"(t (s))]()3.

ds Expressing everything in sight

in terms of to = t(so) we find

(f'(to) x fl"(to), N) = (#'(Io) x #"(to). N)kglBuor = (T(so) x T'(so). 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 u' u" + (u')2I'i r + 2u'v'f 12 + (v')21'22kglBt1or (#'(to), fi'(to))312

detV, 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 partof T'(t) vanishes. An inhabitant of the surface would see T'(t) = 0, since such an observercould only perceive acceleration in the tangent plane. This leads to the following class ofcurves on a surface.

Definition 11.4. A geodesic on a surface S is a curve on S whose geodesic curvature isidentically zero.

For example, on the unit sphere S2 c 1R3, consider a curve of constant latitude. Since alatitude is the intersection of a plane fl with the sphere, it is a planar curve and T'(s) lies in11. If the normal at a point at this latitude does not lie in fl, then T'(s) has a component inthe normal direction and kg 54 0. It follows, by rotating the sphere if necessary, that planarsections 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 graphis 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 wheneverthe following differential equations hold

u"+(u')2r11 +2u'v'r12+(v')2ri2 =0, 11f

= 0.v" + (u')2r + 2u''ri + (v')2ri22i

I

(*)

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 equationsgiven 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) =INand 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 definedin the proposition is also a geodesic.

We next consider a deeper application of the local formulas and show that geodesics arethe curves that locally minimize distance. Thus they satisfy the condition of Archimedesfor being lines.

Theorem 11.6. If a: (-e. F) - S is a unit-speed curve and for any a, bin (-e. e) witha(a) = p, a(b) = q, a is the curve of shortest distance joining p and q in S, then a is ageodesic.

PROOF. The proof is a page out of the calculus of variations, a subject that finds manyapplications 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 coordinatepatch 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).

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 ofcurves 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

daP L(r) = f (ds ' d

das)ds.

By assumption L(r) has a minimum at r = 0. Consider L'(r).

n a a dar darL'(r) = d J(dalr dar = ds

dr f ds 'ds)ds

la ar (ds ' ds

=

L n

((asa2C

ar' as) fas' as )) ds.

cFor r = 0,

ac= I. and so applying integration by pans we obtain

as 'aas

r_o

ac0 = L'(0) =

fh(a2casar' as)

r_o ds

-f b[ds(aC' as)Ir=v(aC' 8sC)r-olds

b a2C

1 a ' as )Ir-ola -j.b (8C

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

cSince na(s) is perpendicular to

s)= a'(s), we know T ,

aas= 0. Hence

=o

= =ac a2c k

0 L'(0) -fb

(ar . 2 )I d s = - (.k(s)na(s),

k

=_f

,l(s)kg(s)ds < 0.

11. Geodesics 163

This contradiction follows from the assumption that kg(so) 0. Thus we have shown thata is a geodesic.

The term geodesic comes from the science of geodesy, which is concerned with mea-surements of the Earth's surface. Several mathematicians have been involved with geodesy,including Gauss. The Konigsberg mathematician F. W. Bessel (1784-1846) published ananalysis of several geodetic surveys to determine more precisely the shape of the Earth asan ellipsoid of rotation. C. G. Jacobi (1804-51) made a study of the "shortest curves" on anellipsoid of rotation which he referred to as "geodesic curves:' motivated by such curves onthe 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 choiceof direction along a great circle joining two points on a sphere. In 1844 the term "geodesiccurve" replaced "shortest curve" in the influential work of J. Liouville (1809-82) who tookthe term from the work of Jacobi.

The equations (*) associated to a geodesic curve are grist for the mill in the theoryof differential equations. The following general theorem leads to many useful geometricresults. 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 differentialequation

d z /

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 > 0such 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 mappingf: 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 sowe can apply it to determine the existence of geodesics. The restriction to a neighborhoodin the theorem implies a restriction on the length of tangent vectors to a surface. This ismade precise by defining the associated norm on the tangent space Tp(S) determined bythe 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 andan e > 0 such that if q E U and i E TqS with 116 11 < e, then there is a unique geodesicy;,: (-1, 1) -> S with y5(0) = q and 4(0) = U. Moreover the mapping

C:((q,15)IgEU.vETq(S), and 11i)11 <e) x(-1,I)- S

given by C(q, u, t) = yo (I) is smooth.

By applying the homogeneity property of the equations (*) we can reparametrize theresults of the corollary as follows:

If p is a point in a surface S, then there is a neighborhood U of p and an e > 0 such thatif q E U and i E Tq S with 11011 = 1, then there is a unique geodesic y;,: (-e. e) -+ S withy;,(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 greatcircle segments are all of the geodesics on the sphere. To wit, if y (s) is a unit-speed geodesicthrough a point p = y (O) with v = y'(0), then by an isometry of S2 we can take this point tobe on the equator and v pointing due north. The known geodesic satisfying these conditionsis the meridian through this point on the equator, a great circle. Thus the transformed y liesalong a great circle. The isometry may be given by a rotation or two so its inverse takesgreat circles to great circles. Hence, y lies along a great circle.

We now obtain some useful consequences of the existence of geodesics. In the normedlinear 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 pdetermined 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 completeproof 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 neighbor-hood 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 lessthan 2ev.

The exponential map gets its name from its form in a special higher-dimensional contextwhere 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 notionsfrom the classical theory of surfaces. In Chapter 13 we will consider useful coordinatecharts constructed from the exponential map. These coordinates appear in the work ofGauss (1828) and Riemann (1854) and so the exponential map may be traced back toGSttingen. We now turn to some synthetic notions that can be expressed analytically viathe 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 anyunit-speed geodesic emanating from p. Euclid's Postulate 11 requires that any straight linesegment be extendable in a straight line any given length. In the language of surfaces weinterpret this postulate as the following property of surfaces.

Definition 11.11. A surface S is said to be geodesically complete if even. geodesicy: [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 mappingexpp. Geodesic completeness implies that, in any direction from p, expp may be applied toa 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 constructexamples where geodesic completeness fails: For instance, let S = S` - Isouth pole); thenfrom 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. Forany pair of points p, q r= S, there exists a unique length-minimizing geodesic joining pand q. The choice of the term length-minimizing is important: contrast the two geodesicsjoining New York City and Chicago on the great circle (kg = 0) through these cities. Onthe sphere, if the two points are antipodes, there are infinitely many length-minimizinggeodesics joining them.

Counterexamples to the existence part of Postulate I are also simple to construct. ConsiderS = 1R2 - ((0, 0)), the Euclidean plane with the origin removed. Then there is no geodesicjoining points (-1, 0) and (1, 0).

In what follows, we examine the question of existence of geodesics joining points (theuniqueness question is considered in the next chapter). Our discussion leads to a conditionthat 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, fromthe definition of a surface, connectedness implies that for any pair of points in S, there issome piecewise regular curve joining the points. To a regular curve a : [to. tj J --> S, denotethe 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 isgiven 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 andthe definition of distance as an infimum. Certainly d (p, q) = d (q. p), so it suffices to showthat 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 thanep from p lie in expp(BEP(0p)), that is, for r E exp(BE0(0p)) there is a u E BP(Op) suchthat 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 thecontradiction. 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 metricspace (M, d) is a Cauchy sequence if, for any E > 0, there is an index N = N(E) withd(ak, at) < E whenever k,1 > N. A metric space M is complete if every Cauchy sequenceof points in M converges to a point in M. To see that R2 - ((0, 0)) is not complete, considerthe Cauchy sequence ((I/n, 0)).

In fact, metric completeness of a surface as a metric space (S, d) is the key to PostulatesI and II.

Theorem 11.15 (the Hopf-Rinow theorem). Let S be a connected surface. Then thefollowing 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 thatS is geodesically complete. Let p = d(p.q). Consider ep > 0 small enough so thatexpp: B.,(Op) -. S is a local diffeomorphism and between any pair of points inexpp(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 thecompactness of E. By the choice of ep, po = expp(8uo) for some unit vector vo. We claimthat 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 anddistance 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] I 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 neigh-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(s'.q)foralls'E V.

We write again

d (Y(ro), q) = min (d(y(ro), s') + d(s', q)) = 8' + d (p , q),s'E E'

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 geodesicfrom y(ro) to po' has length ro +8'. This curve achieves the minimal length. We show in alemma and corollary to follow that a piecewise-geodesic curve of minimal length must be, infact, 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 mustbe 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 settopology, any Cauchy sequence of points in S converges to a point in S. Hence (S. d) is acomplete 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 assumptionimplies that the sequence is a Cauchy sequence in S and so it has a limit, whichwe write as y(b). By Corollary 11.8 (existence and uniqueness of geodesics), since y(b) isdefined and y'(b) can be computed as a left limit, we can extend y beyond b. Continuingthe argument along in both directions we can extend the domain of y to all of R and S isgeodesically 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 asegment 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

aA du aAa'(t) = au dt + at

Since IIX(t)II = I for all t, IIaA/auII = I by the construction of the exponential map.

Furthermore, (aA 1A) = 0 by an argument similar to the proof of Theorem 11.6 (see11 au at

also the proof of Lemma 13.2). It follows that

(a'(t). «'(t)) _Z +

(dA' )

12

dt Iat atdt

ll. Geodesics 169

and soa ja df (a'(t),a'(t))dt di Iu(b) -

a

To obtain equality we need 8A/et = 0, that is, X'(t) = 0 or X(t) is constant. We alsoneed 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

rf (y'(t), y(1)) d, < J (a'(t).a'(t))dt0 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 Pos-tulates I and 11 from the point set topology of the surface S. For example, if S is compactand connected, then it is a complete metric space and so it is geodesically complete. Globalproperties such as metric completeness or compactness cannot be determined merely fromlocal information. However, they are important ingredients in the construction of modelsfor 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 ageodesic if Il is a plane of symmetry of S. that is, the two sides of S are mirror imagesacross 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 ageodesic.

11.4' Let a: (0, 11 - S2 be the curve a(t) = (cos e' , sine, 0). Show that a is a geodesic onS2, but in the latitude-longitude parametrization of S2, a does not satisfy the differentialequations (*).

11.5 Determine the analog of equations to be satisfied by a geodesic, not necessarilyunit speed.

11.6 Let 11(0, w) = 3'(11(0 + ti) - 11(0) - 11(w)) denote the inner product determined bythe second fundamental form. Prove the analog of the Frenet-Serret theorem for curvesa 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.

11.7 Suppose that S is geodesically complete. Show that for any point p E S, the mappingexp, has domain all of Ta(S).

11.8 For the reader with some knowledge of topology: Show that the topologies on a surfaceS 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 thereis 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 satis-fies the property that every pair of points is joined by a unique geodesic. If ¢: S, - S2is a local isometry. show that it is a global isometry.

12The Gauss-Bonnet Theorem

This theorem, if we mistake not, ought to be counted among the most elegant inthe 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. InDisquisitiones generales circa superficies cun'as (Gauss 1828). Gauss deduced some of thebasic properties of geodesic triangles. In particular he proved a general relation between areaand 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 forgeodesic curvature kg and applied Green's theorem to prove a far-reaching generalizationof Gauss's theorem for polygons whose sides are not necessarily geodesic segments. Forcompact, orientable surfaces the theorem of Gauss and Bonnet has remarkable topologicalimplications.

In order to state the main theorem of the chapter we need to set out a handful of assump-tions. Fix an orientable surface S C R2 and suppose R C S is a region in S satisfying thefollowing properties:

(1) The region R is simply connected in S. That is, any closed curve in R can becontracted to a point without leaving the region. Thus R is like a convex subset ofS. Furthermore, the interior of R in S is connected. This condition excludes caseslike the region enclosed by a figure 8.

171

(2) The boundary of R is given by a curve a : [0. s ] -+ S. Furthermore, a is piecewisedifferentiable 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 generalsort 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) anda;, exist and the angle between them is well defined. We call this angle theexterior 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 previousequation determines e, up to sign. A cusp is a vertex where a! (s,) _ -a, +i (s,) andso 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 theordered basis [a, (si). a'

tt (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 nearbypoint on ai, say a, (s, - >)) for n > 0 andclose to 0. choose a nearby point on a,+1,say a,+i (s, + r)), and consider the orderedbasis[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 welldefined near the vertex and its handednessremains the same as q goes to zero. Thisdetermines 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 satisfyF = 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 piecewisedifferentiable curves on a surface like the boundary of the region R. Though out of placehistorically, this theorem due to H. Hopf (1894-1971) has a natural place in our discussionand it was quietly assumed by Gauss and Bonnet.

To state the result we introduce functions ¢, : [s, _ s, ] - R along the curves a, . Thesefunctions measure how the tangent a' (s) turns around the region R. Let 0, (s) measure theangle from x,, at ai (s) to a,(s). Such a function is determined up to multiples of 27r and itsatisfies the formula

cosOi(s) = Ia,tsi (a;(s).

To determine the values i, (s), begin by choosing an angle to represent Ot (0). Thenextend ¢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 determinedby the equation

02(sl) = EI + 01(sl)

Now extend h2(s) along a2(s) by continuity, and so on. In this fashion we can constructall of the functions 0i exactly. We now state the key topological result for such curves on asurface (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 belongin a course on topology. In order to avoid such a long detour we postpone a sketch of theproof to the end of the chapter. For now we conclude a useful corollary that follows fromthe 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 takesthe form (Exercise 10.3)

K2 EG (a,v \ EG/ + 8u ( EG)/

We next obtain a local expression for the geodesic curvature along each curve ai using thefunctions Oi and the fact that F = 0. This convenient expression is related to the Gaussiancurvature via Green's theorem.

Lemma 12.3. Along the curve ai (s),

kg(s) =I Gudv - E,Vdu + dOi

2 EG ds ds ds

PROOF. Let e _ and e = 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 obtainthe 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)d

ieu +cos0i(s)e;, +cosOi(s) d' e +sinOis)e;,

=dOi

na;(s)+cosOi(s)e;,+sinOi(s)e'v.


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,,e° ds ds fE

_ I ( du dv E du E. dv(x°Ods +xuvds -xi' (i:5Th;E+ 2E3/2 TS

Since F = 0, F. = 0 and so xv) _ E. It follows that

I du I dv(e°, ev) =

EG(xuu- xv) ds + 17M

(xuv, xv) ds

I dv duGdv-Evds2 EG

T-s ds


We now compute J kg(s)ds, the total geodesic curvature along a, the boundaryJ=1 9

of the region R. By Lemma 12.3 we havenEJ

kg(s)ds= J f I IGTs

dp1.j_) ai=) aj i=) , 2 EG L 7,

Recall Green's theorem for a region W C R2 with boundary \\a simple closed curve aW:

J Pdu + Qdv= jj (- )w w au av /

Substituting the integral of interest, we get

I Gdv - Ev du

2 Jax-'(R) EG EG

2 JJ'1'(R > (a,V ( EG)+ u ( EG)/ dude

- (R> 2 EG (av ( EG) + au (:))jJ.r ,.Tdude

= -jj KdA.R

This observation with Lemma 12.3 proves the main theorem.


Theorem 12.4 (the Gauss-Bonnet theorem). If R is a simply connected region in aregular 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 Disqui-sitiones (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. LetR = 6ABC be such a triangle with sides given by geodesic segments. The interior anglesare 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 fromDisquisitiones (§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) Ona surface of constant positive curvature, the area of a geodesic triangle is proportional tothe angle excess. (c) On a surface of constant negative curvature, the area of a triangle isproportional to the angle defect.

PROOF. 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 Gaussiancurvature

ARKdA=KARdA=Karea(R).

This corollary reproves an equivalent of Postulate V for the plane (Theorem 3.6), andit recovers the formula for the area of spherical triangles (Proposition 1.2). It also leadsto a condition on a surface that models non-Euclidean geometry: In order to obtain theformula for the area of a non-Euclidean triangle (Exercise 5. 10), the surface must haveconstant negative curvature. In fact, more can be said. In Chapter II we discussed Euclid'sPostulate I and the existence of geodesics joining two distinct points in a surface. Thisleft open the question of the uniqueness of geodesic segments joining two points. Thereare infinitely many geodesics joining the north and south poles of a sphere. However, thiscannot happen on surfaces of nonpositive curvature.

Corollary 12.7. On a surface of nonpositive curvature, two geodesic segments yi, y2meeting at a point p cannot meet at another point q so that the curves form the boundaryof 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, soc 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 ageodesically complete surface of nonpositive curvature satisfies Euclid's Postulate I. thatis, 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. Compactnessimplies 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 caninterpret the formula of Gauss and Bonnet as an expression for the total geodesic curvaturealong the boundary of such a region.

We suppose further that the surface S has a triangulation,

P=(A,CSIi=1,...,m),


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 thatS = 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 forgranted in the nineteenth century. Gauss, as Director of theGottingen Astronomical Observatory, had considerable expe-rience doing geodetic surveys in which large geographical re-gions are broken up into geodesic triangles (see volume 9 ofhis collected works). For small enough regions of a surface, thegeodetic approach of laying out geodesic triangles can lead toa triangulation. To prove that all compact surfaces have a trian-

gulation requires the use of topological results such as the Jordan curve theorem. The firstcomplete proof was due to Tibor Radb in 1925. The interested reader may find a proof inMoise (1977) and Doyle and Moran (1968).

Using the orientation we can take each triangle to be oriented as follows: Each boundary2A; = ai is parametrizable as a unit-speed, piecewise differentiable curve. Choose thedirection of the parameter so that (a' (s), nQ, (s), N(a; (s)) is a right-handed frame at eachregular 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 Ebut each edge is shared by two triangles. Thus 3F = 2E.


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 tri-angles. It follows that the integral Er f,,,, kg(s) ds van-ishes. This implies that

AS

F

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 is27rV. Also, since 3F = 2E, we have 7r(3F - 2E) = 0, and so

AS= 2rr(V - E + F) = 27rXp(S).

Gauss referred to the integral f fs K d A as the total curvature of the surface S. Sincethe total curvature does not depend on the triangulation, we have proved the independenceof the Euler-Poincare characteristic.

Corollary 12.9. Given two triangulations P and P' of a compact, oriented surface Swithout boundary, Xp(S) = Xp-(S).

We can write X (S) for the value 2L JJ K d A or for V - E + F associated to anyAS

triangulation of S. Euler introduced his formula, V - E + F = 2, in 1758 for polyhedra thatare homeomorphic to a sphere. This led to a combinatorial proof that there are only fivePlatonic solids. In 1639, however, Rend Descartes (1596-1650) had proved a result thatanticipated Euler's formula and the Gauss-Bonnet theorem. To state his result, considera polyhedron that is homeomorphic to a sphere, such as the icosahedron or the cube. Ateach vertex the sum of the interior angles of the faces that meet at that vertex exceeds 27rby 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 whichDescartes proved this result appeared in print in 1860 and so it had no direct influence onthe developments just given or on Euler's work. The correct context for Euler's formula wasfound by Jules Henri Poincare (1854-1912) in 1893. Using his newly defined homologytheory. Poincare showed that the alternating sum V - E + F is a topological invariant ofa surface. This implies its independence of the choice of triangulation. In this context theGauss-Bonnet theorem implies the topological invariance of the total curvature. This result

KdA =2rrV-nF=2trV-nF+JrOF -2E)


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 theoremimplies that there is no change in the Euler-Poincar6 characteristic. The invariance may beinterpreted as a kind of rigidity. For example, if you poke your finger into a balloon (idealizedas S2), it swells on the side to decrease the Gaussian curvature away from your finger wherethe curvature has increased. However, if you require that the deformation maintain thecondition of constant curvature, the class of deformations is severely restricted. In the caseof S2, it contains only the translations, rotations, and reflections.

The Euler-Poincarr characteristic also restricts the ways a surface may lie in R3. Forexample, the torus T can be shown to have Euler-Poincare characteristic X (T) = 0. SinceT is compact, it has a point that is furthest away from the origin. At that point, the Gaussiancurvature is positive. This implies that there are points on the torus where the Gaussiancurvature is negative since f fTZ K d A = 0. Thus there is no way for the torus to lie in R3with constant curvature.

We now turn to a surprising application of the Gauss-Bonnet theorem to the theory ofcurves 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 theFrenet-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.

PROOF. Let R denote one of the regions into which N(s) divides S2. Suppose a is orientedso that N(s) is the boundary of R with R lying to the left of N(s). By the Gauss-Bonnettheorem we have

fgJJ=R R

Here kg(s) is the geodesic curvature of the curve N(s) on S2. From this formula the theoremfollows 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 Ifor 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 (Theo-rem 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

To compute the integral f o kg (s) d s we effect the following change of variables. The functionl

arctan (-I has as derivativeKr

d Tl KT' - K'Tdarctan (- l =

s K KZ +r 2

Denote the arc length along N(s) by a. Thends =

I .Thusdo K +r

d T KT' - K'Tarctan (-> =

da K (K2 + T2)3/2

Since a(s) is a closed curve,

This proves the theorem.

Jr r)

J kg(s)ds =L(o)

od arctan (T) da = 0.

0 da K

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 digressionwe record the key properties that are needed to prove the Gauss-Bonnet theorem. Theseproperties rely upon certain topological features of curves.

Suppose a : 10, r] R2 is a closed, differentiable, unit-speed curve in the plane. Sup-pose further that a(s) is one-to-one and differentiable, and that a'(0) = a'(r). The Jordancurve 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) pointsinward.

The principal example of a closed curve is the unit circle S' C R2. The circle comesequipped 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 anopen set 9 c_ U C St. called an elementary neighborhood of 9, such that p t (U) C R is aunion 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, theopen set around the point (I.0) given by the angles between -n/6 and n/6 has inverseimage

oc

p-t ((cos0, sin 9) 1 -7r/6 < 0 < n/6(= U (2nk - (tr/6), 27rk + (,r/6)).k=-oo

A special property of this mapping is that it allowsus to work on the circle or on the line when we workin the small. This leads to a factorization of a curve onS' through the mapping p. If we are given a mappingf : [a, bJ St. we can construct a lift of f, that is, amapping f : [a, b] -* ][$ such that p o f = J'. To con-

p-1(U) struct the lift, choose a point a in in with p(a) = (a).

(=Take an elementary neighborhood U around .f (a) andconsider its inverse image under f. This is an open setaround a in [a. b]. By assumption there is a diffeomor-phism that identifies U with the piece of p-I (U) con-taining a. Consider the composite of f with the diffeo-morphism. The composite is defined on f-t (U) andgives a mapping J: f-t (U) -> ]R with f = p o f. Wenow move along [a. bJ toward the edge of f-t (U) andextend f a bit further by the same argument. Because[a, b] is compact, the iterated extension eventually givesus 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 satisfiesTO = T (r). Construct a lift of T. say T : [0. r] -> R. Then p o T(0) = p o T (r) and soT(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 thata(s) is one-to-one and differentiable, and a'(0) = a'(r), and if a(s) is oriented so that theenclosed region is on the left, then ?(r) - T(0) = 27r.

The mapping H is continuous on Sr. and H(s, s) = T (s). By taking the family of piecewiselinear curves pictured and composing with H, we obtain a continuous deformation of themapping 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:

(r,r)A SKETCH OF A PROOF. To prove the theorem we introduce anauxiliary function H : Sr R2. where Sr = ((a, b) I 0 < a <b < r}. Define H by

ifx < y and (x, y) 54 (0, r).

ifx =y=s,-T(0). ifx=Dandy=r.

of (Y) - a(x)

[la(y) - a(x)II 'H(x. y) = T(s)

Fact 1. The difference of the values at the endpoints of a lift of a closed curve is unchangedunder deformation.

To see this define another auxiliary function

H(0, 2s), if t = 0 and 0 < s < r/2,

H(2s, r/),

\ \if t = 0 and r/2 < s < r,

A,(s)= Hls,l? -Ilsl, if0<t<land0<s<tr

Hls2

ts+2r(

2

l-t)1 if0<t<land2 <s<r.\\ -t -t J -

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 begeneralized to lift the entire family x,(s) to a mapping ;,(s). The difference of the valuesat the endpoints of the lifts, that is, A, (r) - A, (0), varies continuously overt. However, thisdifference is also a multiple of 2rr. Since the set of multiples of 2n is discrete, the valueA,,(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 byrestricting 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 value0 < so < r for which the trace of a lies entirely on one side of the tangent line to a througha(so). Extend the domain of a to all of IR by a(s+r) = a(s). Now work with the one-to-onecurve a: (so, so + r] - R2 and reparametrize it to have domain [0, r]. By a rotation andtranslation we can arrange the particular conditions desired. The parameter for the particularcase may now be rotated back to the original curve a by a family of one-to-one curves allwith the same trace as a, and the argument given in Fact I can be applied to this family toshow 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. Asimilar argument for HIlO,rixlri gives Hllo.rlxlr)(r) - HII0.rlxirl(0) = n. Putting thecontributions 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 closedcurves. There we would speak in terms of winding numbers, covering spaces, and homotopyclasses of curves. Since we will not need these ideas later, we refer the interested reader toMassey (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 thelift at the endpoints can be written as the integral

T(r) - T(0) =J

r T'(s)ds = I K(s)ds.0 0

This value is called the total curvature of the curve a.

Our next task is to generalize Theorem 12.11 to a case more suitable to generalizedpolygons. 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, unit-speed, differentiable curve. Then a; (s;) = a;+, (s;) and a a i (0). Assume that a isone-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 upof curved sides. At each vertex, we define an external angle as we did for surfaces at thebeginning 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 acounter-clockwise direction to a'+l (s;): if the answer lies between 0 and n, the sign ispositive. 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, thatis, a'(s;) and a,+, (s;) point in opposite directions. To settle this case as n or -n, choosepoints near the vertex along a; and a;+i . respectively. The tangent vectors at these points, asvectors 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 thatsign for the cusp.

We can associate a lift of the tangent directions along this piecewise differentiable curveas before: Choose an angle that represents Ti (0) = a, (0), say a, and let T, (0) = a. Bythe same argument as given before, we can extend this lift of T, (s) to T, (s, ). To crossover to a2, the angle of the tangent vector turns through the exterior angle el. DefineT2(sl) = T, (s,) + El. This determines the representative for T2(s,) and so we can continueto lift along a2. Continuing in this manner, we can lift each T (s) ending with T

Proposition 12.12. + E = T, (0) + 27r.

PROOF. 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)and the point of departure along a;+, by Q; = a;+, (u;). As theradius 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;). LiftingTp, (s) so that Tp, (t;) = T, (ti), we get that the total change along fi; of the tangent vectorfrom t; to u, is close to e,.

By construction, the curve made up of the portions of the a; outside the circles and f; inthe circles is sufficiently differentiable to apply Theorem 12.11 and get a difference in thelift of the endpoints to be 27r. Since the contributions of the f; is close to the sum of theexterior angles, the lift we constructed for the piecewise differentiable curve has a differenceat the endpoints close to 27r minus the contribution of fi,,. This proves the theorem.


Finally, we complete the proof of the Hopf Umlaufsatz (Theorem 12.1). One of theassumptions set out in the beginning of the chapter is that the region R, a subset of aregular orientable surface S, the boundary of which is a generalized polygon on S, isentirely 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 theassumptions that R does.

The coordinate chart x induces an inner product on the tangent space at each point of Uthat is different from the Euclidean inner product as follows: If rri (t) = (ui (t), vi (t)) and172(t) = (u2(t), V2 (t)) are curves in U that meet at Y11 (0) = 112(0), then the inner productbetween the tangents to the curves may be taken to be

(i (0), (0),xuu2(0)+xv v2(0))

For example, the cosine of the angle between the line parallel to u-axis and the line parallelto the v-axis through a point in U is given by F(uo, vo)/ E(uo, vo)G(uo, uo). Denote theusual Euclidean inner product (the dot product) on R2 by (, )E. We can interpolate betweenthe 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. Supposethat the boundary of W is a piecewise differentiable and closed curve 6: (0, R2. We

can form the functions Oj'(s) that lift the tangents along the curve with respect to the innerproduct ( , ),. Notice that the angles between xu and a '(s) may differ as t varies. Whent = 0 the inner product is Euclidean and so our argument above gives the Hopf Umlaufsatzfor the surface R2. As the inner product changes continuously, the functions 0,(s) changecontinuously, and so the difference of the values of the lift at the beginning and endpoint ofthe curve plus the last angle gives a continuous mapping to multiples of 2,r. Since this setis 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 simplyconnected region in S turns through 2,r.

This last part of the proof anticipates a more general notion of surface. For some valuesoft, the inner product ( , ), may not correspond to the inner product at a point in somesurface lying in IR3. The pair (U, (, ),) for a fixed value oft is a coordinate chart for anabstract 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 joiningtwo points (x. y, z), (x', y', z') with z 54 z'. We know that K - 0 on the cylinder. Isthis a contradiction to Corollary 12.7?

12. The Gauss-Bonnet Theorem 18S

123' Let R' C S be a subset of a surface with boundary 8R' a piecewise differentiable curveon S. Suppose further that R' can be triangulated so that each triangle is simply connectedand lies in a coordinate chart. Show that the Gauss-Bonnet theorem for R' takes the form

> ej+ J kg(s)ds+ KdA=2trX(R').i aR

If! R

12.4 On a surface S with Gaussian curvature everywhere nonpositive, show that a closedgeodesic, 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 thesphere 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 onegets a surface that is homeomorphic to the torus. Show that this gluing procedure gives anorientable 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 eachother. Is it possible to construct an analogous tiling of the sphere using congruent geodesicquadrangles so that four such quadrangles meet at each vertex?

13

Constant-curvature surfacesI remark only that I have developed Astral Geometry so far that I can completelysolve 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 conditionsa surface must satisfy in order to provide a model of the geometries of interest. We take ourcue from the metric-space properties of surfaces found in Chapter I I. In particular, it nowmakes 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 isan empty set for certain radii. When such a circle determines a curve on S, then we maycompute 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 ittakes the form

1

27r R sin (R) = 27rr - R= + , on a sphere of radius R (for r < n R), and

2 rk sinh (k) = 2irr + -Sky + . in hyperbolic geometry.YTr1

The Taylor series for the non-Euclidean cases illustrates the deviation from the familiarEuclidean formula. In synthetic and Cartesian geometry circles enjoy many useful proper-ties. We examine some of these properties and interpret them on a general surface in whatfollows. 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 ofa surface. We now introduce another pair of systems that generalize rectangular and polarcoordinates on the Euclidean plane. In the plane the pencil of lines through the origin leadsto polar coordinates. The pencil of lines perpendicular to a given line leads to rectangularcoordinates. On an arbitrary surface we consider the pencil of geodesics through a givenpoint. 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). Leta: (-tr, n) Tn(S) be a smooth mapping that satisfies the property that 11a(0) 11 = Iand Ip(m, a(8)) = cos8 for all 0 E (-rr, n). The image of (-rr, rr) under the mappinga(0) is the unit circle in Tp(S) (minus one point) and a(0) parametrizes the circle accordingto 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 aneighborhood of p E S (Theorem 11.10). This leads to the following coordinate chart.

186

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 rfrom p. Alternatively, it is the image of the set (u E Ta(S) I Ili II = r} under the exponentialmap. For a fixed angle 9o, the geodesic given by y: (0, r] - S, y(s) = expp(sa(Bo)) is ageodesic 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. Therestriction to radii of length less than ep allows usto do some analysis through the exp map. We firstrecord an important geometric result generalizing aproperty 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 geodesicradii.

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 acurve in the tangent plane satisfying Ila(t)II = r < E for all t E 1, an interval. The curvefl(u) = expp(a(t)) is a portion of a geodesic circle of radius r. Define the family of curves

a,(s) = exp,(sa(t)). 0 < s < I.

Then a,(I) = [3(t), and, for a fixed to E 1, a,o(s) is a geodesic. Our goal is to prove thatfor each to E I

d da,0) a,a(s) .( dt r=rp ds L) 0

The idea of the proof is to consider the family of curves a, (s) as a variation of a fixedgeodesic 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 anenergy function defined by

E(t) =dcrldj(L.

d dThis function is constant because it is related to the length of the geodesic radius from p to


)S(t). Taking the derivative with respect tot we obtain

id da, dor, )0 = E'(ro) =d1 fo (ds ' _Ws )ds

=10

1 a do, do, I a2a, dor

= fo at `ds ds)dsl f_1p = 2 `alas' ds )ds11.10

d dor L_, a2o,2J (do:dt' ds)I,=rO-cat' as=)

Jds

r=10

s=1

- 2f' '(aa, a20,,

)1'=10 ds.o at ' aszs=2 l aa, Loll)

1 at ' as

aa, aa,

at as 11=10

=10

ao, (1) aaro (801(0) aa,0(to) I

S=0as Hat as

Since a,(0) is just the point p, the second term is 0. The first term is the inner productwe wish to compute. It suffices then to show that the integral that is the last term in thepenultimate 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.

Let us now consider the line element for geodesic polar coordinates. The coordinatecurves are given by geodesic circles when r is constant, and geodesic rays emanating fromp when 0 is constant. By the definition of the exponential map, the curves with 0 heldconstant are unit-speed geodesics and so E = I. By the Gauss lemma, the coordinatecurves 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 easilycompute the associated apparatus of intrinsically interesting functions. It is also a generalresult, 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 atthe conditions on G, recall a strange property of polar coordinates for R2; the origin is asingular point for this coordinatization of the plane. It is, however, a kind of false singularityas 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 fromrectangular 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 singularpoint for rectangular coordinates. We next study what happens to geodesic polar coordinatesat p if we pass to the limit as r goes to 0 and use rectangular coordinates to smooth out thesingularity.

Since the change of coordinates given by X is an isometry, we can recompute ds2 in x

and y: Recall that dr=xdx + ydy

d8-xdy -ydx

and sor r2

ds2 = Edx2 +2Pdxdy+Gdy2

= dr2 + G(r, 8)d82

(XdX+YdY)2+G(rO)(XdY_YdX)2r r2

= 1 2 + G 42 dx2 + 2 ( 1 - Z ) Z dxdy + ( 2 + G 42 ) d y2.

It follows from this expression and x2 + y 2 = r2 that

E- I (*), and d - I = x2r (G(r, 8)

)I =r2

2(G r28)

1)

and so x2(E - 1) = y2(G - I). Since this equation holds for any x, yin the region ofdefinition, it follows that t - I = my2 + and C; - I = mx2 + . Substituting thisinto the equations (*) we find that

G(r,8) 2 m 3r2 = I+mr + and so G(r,8)=r+ 2r +

This implies the following conditions on the function G:

Proposition 13.3. For geodesic polar coordinates with ds2 = dr2 + G(r. 8)d82, and forany fixed angle 8o, -n < Bo < rr, we have

alim G(r, So) = 0, and lr-Oim

ar 1

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:

K= - I aZ V'U(-r-,O)

(r, 9) art

Substituting our results for the series associated to G we get the equation

m; aZ.- K +...)=-KI-G =

arz=3mr+...,

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 ofthe unit vector iii which determines 9 = 0 gives another form of the polar coordinates forwhich the variable r remains the same. Thus these formulas do not depend on a choice of0 along which r goes to 0.

We can now compute the circumference, denoted by circump(r), of a geodesic circle ofradius r centered at a point p. In geodesic polar coordinates, for fixed r < e p, the geodesiccircle of radius r is the coordinate curve x(r, 9), -n < 9 < n, and its circumference isgiven by

circump(r) = G(r.9)d9IF

r3(r-K(p)6 +... d9

rK(p) 3

In 1848, J. Bertrand (1822-1900) and V. Puiseux (1820-83) gave the following very geo-metric interpretation of the Gaussian curvature.

Proposition 13.4. if circump(r) is the circumference of a geodesic circle of radius rcentered at the point p in a surface S, then

K(p) = lim6trr - 3circump(r)

r-.O 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 afamily of curves

Ye(s) = export, (sr(N x ap(t))) = expa(,)((sr)na(t))

for r < e. the bound associated with a(O) by exp in this neighborhood. The curves i Hy, (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 theproof of Lemma 13.2 with little change, they can be shown to be perpendicular to thegeodesic parallels. This leads to a version of rectangular coordinates in a neighborhood ofp = a(0) with the curve att) acting as the x-axis and the normal geodesics acting as thelines x = xo. The associated chart gives coordinates called geodesic parallel coordinatesand 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 ofsurfaces. On a surface S, suppose p e S is taken to be the center and q E S is joined to pby a geodesic segment taken to be the radius of a circle in S. When the exponential map isdefined on C Tp(S), where r > d(p. q). we can construct the circle centered at pwith radius pq as a geodesic circle. In general. however, this construction is restricted tosmall 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 isone-to-one, the image of the set of points a fixed distance from Op in Tp(S) maps via expto 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, ona general surface the circumference of a circle of radius r may depend on the Gaussiancurvature at its center. This leads us to a further assumption about the surfaces we willconsider as models of non-Euclidean geometry - such a surface must have constant Gaussiancurvature.

From the formula for the circumference of a circle in non-Euclidean geometry (Theo-rem 5.21) and the theorem of Bertrand and Puiseux (Proposition 13.4). we can give finally aninterpretation for the mysterious constant k of non-Euclidean geometry found in the work ofGauss, Lobachevskii, and J. Bolyai. To wit, comparing Taylor series for the circumferenceof a circle of radius r we get

2rrksinh(r)=2rrr+3 ,+...=2nr-n 6(p)r;+....

k A-

Thus 2/k2 -K(p) _ -K, a constant. This sets another condition on a surface modelingnon-Euclidean geometry - it must have constant negative curvature. This is also apparentfrom the formulas for area given by Corollary 12.5 and the relationship between area andangle defect in non-Euclidean geometry.

Using geodesic polar coordinates and the assumption of constant Gaussian curvature wecan also make precise Euclid's Postulate IV. Given two right angles at different points in asurface, then near the vertices of the angles we can construct a local isometry by composing

geodesic polar coordinates and their inverses carrying one point to the other and the rightangles onto one another by making the right angles correspond to the geodesic curves 0 = 0and 9 = n/2. To say that the angles are congruent, however, we require that this localisometry extend to an isometry of the whole surface to itself. Thus Euclid's Postulate IV isa statement about the existence of many isometrics of the surface to itself. If for each pair ofpoints p, q E S, and pair of unit tangent vectors i E Tp(S), w E Tq (S), there is an isometrytaking p to q and ii to iuu, then we call the surface S point transitive. The development ofthese ideas is part of the theory of transformation groups; we will not develop these issuesfurther, except to observe that certain surfaces important for non-Euclidean geometry havethis property (see Chapter 15). Notice that a point-transitive surface is a surface of constantcurvature.

The work of Minding

A general question discussed in Chapter 10 is whether two surfaces can be comparedgeometrically; 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 Siand $z, when does dsl = dsz, where dsi is the line element on S,? A necessary conditionfor a solution is equality of the Gaussian curvature at corresponding points. The solution tothis problem in the special case of constant-curvature surfaces is due to Minding, who alsointroduced geodesic curvature.

Theorem 13.5 (Minding 1839). 7ivo surfaces S1 and Sz with constant Gaussian cur-vatures 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 thesame Gaussian curvature at corresponding points. To prove the converse, choose p E Si andq 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 I a2,/rC2art

_ -K' = -K2 - ,2 art

a2This is a special case of the differential equation are + K f = 0 with the initial conditions

lim,,o f = 0 and limr-.oa

ar= 1. where K = Ki = K2, a constant. The uniqueness

of solutions to differential equations with initial conditions implies GI = G2. The desiredlocal isometry is the composition of expp I with the coordinates around q.

To extend our understanding of constant-curvature surfaces, let us consider particularsurfaces in R3 where K = 0, K > 0, and K < 0. Minding's theorem tells us that we can


compare locally any two surfaces sharing the same constant curvature. A given constant-curvature surface then is a local model for all surfaces of that curvature. One route toconstructing model surfaces is the differential equation to be satisfied by ,V,

art+Kf=0 such that of =0 and limo I.

Surfaces with K = 0a2

In this case8r

G= 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/ R2a2

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 sucha 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, givenby longitude and distance from the north pole, that is, the azimuthal projection. We alreadyknow that such a sphere is a surface of constant curvature I/ R2. By Minding's theorem wetake 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 thatgeodesics are closed curves. If we look for those surfaces in R3 of constant Gaussiancurvature 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 withconstant 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

on a compact space and so it achieves a maximum at some point p E S. At such a pointof maximal height the tangent plane Tp(S) bounds the surface, that is, S lies entirely onone side of this tangent plane. Thus K(p) must be nonnegative by the characterization ofthe 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 wouldviolate 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 therewould be points p where the principal curvatures k1 > k2 were locally a maximum and aminimum, respectively. The following lemma, due to David Hilbert. shows that this cannothappen.

Lemma 13.7 (Hllbert 1901). Suppose p is a nonumbilic point in a regular surface Ssuch that k 1 > k2 are the principal curvatures on S. Suppose k 1 achieves a local maximumat p and k2 achieves a local minimum at p. Then K(p) < 0.

PROOF. Since p is a nonumbilic point of S. there is a coordinate patch around p withcoordinate lines given by lines of curvature (Corollary 10.10). For such coordinates wehave

k1 =Ee

and k2 =gG .

Furthermore, the Mainardi-Codazzi equations take the form

Eve+g)= Ev(k1+k2). Gv(egGu

e,, - 2 E G 2 gu 2 F_+ G

2(k1 + k2)

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

8v 2E02 -kt)

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 computedfrom the formula

K 2 EG \ ( EG ) + a ( '))Expanding the derivatives and simplifying we get the expression

Gu

K_

2EG(Evv+Guu)+4EG (E + EG + G + EE )

Thus, at p, the Gaussian curvature takes the form K = 2EI (Evv + Guu).

Since k, is a local maximum and k2 a local minimum at p, the second partials of ki andk, at p satisfy

a'k-< 0 and

a2k20.>

av2 au

We relate E,,,. and G,,,, at p to the principal curvatures by

E,,,, _a2ki 2E _ a'-k2 2G

>0, G",

>0.av2 (k2-ki) - 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 umbilicpoint, then S is a sphere. Suppose x : (U C R2) -> S is a coordinate patch around a pointp. Since every point is an umbilic point, every curve is a line of curvature, and from thedefinition of lines of curvature we can apply Rodrigues's formula (Exercise 9.10), whichgives

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 expressionsN,,,, = +Ax,,,. and N,,,, _ Ax,,,,. Now x,,,, = x. and N,,,. = N,,,,, and so weobtain

--uxr + A,,x = 0.

Since x,, and x, are linearly independent, A = A, = 0 and so x is a constant. IntegratingRodrigucs'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 ofS 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 islocally isometric to any other, so we need only one model to study the geometry of such asurface near a point. For convenience fix R = I. In Chapter 9 we showed that the surfaceof revolution of the tractrix 0(t) = (sin(s). Intan(r/2) +cos(l)) is a surface of Gaussiancurvature K = -1.


The differential equation that determines geodesic polar coordinates on a surface of

constant negative curvature is 3rfR2

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, withoutconstructing geodesic polar coordinates on the surface of revolution of the tractrix, we haveobtained the line element that would result.

The surface of revolution of the tractrix is called the pseudosphere in the literature ofthe late nineteenth century. As we will see in later chapters, the local geometry of thissurface 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 itssurface area is 4n.

The pseudosphere.

Is this the surface that we have been seeking as amodel of non-Euclidean geometry? In fact, the pseudo-sphere falls short of ourgoal - it is not complete. Abovethe x)-plane. we could continue the pseudosphere byreflection, but on the circle of points where z = 0, thesurface is manifestly singular. Geodesics starting nearthis "equator' are unable to continue past it smoothlyand so the surface is incomplete.

We begin the next chapter by trying to correct thisfailure of the pseudosphere. That is, we consider com-plete surfaces of constant negative curvature in 1R3. Thegeneral problem of finding such a surface has a surpriseending due to David Hilbert.

Exercises

13.1 Suppose that a coordinate chart can be chosen for a neighborhood of a point in a surfacesuch that the coordinate lines u = constant or v = constant are geodesics. Prove thatthe line element can be brought into the form ds2 = due + duZ. that is, the surface isdevelopable on the plane near the point.

13.2 Prove the theorem of Diguet (1848) that

K(p) = lim12(,rr2 - areap(r))

r-O 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 Prove that the tangent developable surfaces (x(s, 1) = a(s)+ta'(s))defined in Chapter 8are locally isometric to the plane.

13.4' Prove one of the important properties of the exp mapping as follows. Suppose that p andq 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 theparticular form of the metric in geodesic polar coordinates to show that a variation of thegeodesic 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 surfacesthat arise if r I (Wilsttp).

13.6` Prove the analog of the Gauss lemma for geodesic parallel coordinates, and obtain theassociated line element ds2 = dx2 + G(x. ))d v2. What form does G(x, y) take on thesphere of radius R?

13.7' Show that the principal curvatures on a surface are differentiable as follows: Let H denotethe mean curvature and K the Gaussian curvature of a regular surface. Prove the formulafor 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 surfacearea of the surface of revolution of a tractrix determined by dragging a weight on the endof a fixed length, say a.

PART C

Recapitulation and coda

1411

Abstract surfacesThe concept "two-dimensional manifold" or "surface" will not be associated withpoints in three-dimensional space; rather it will be a much more general abstractidea.

Herman Weyl(1913)

In the previous chapters we showed that a surface serving as a model of non-Euclideangeometry must be geodesically complete and have constant negative curvature. We nextsearch for surfaces in R3 satisfying these conditions. One candidate, the surface of rotationof the tractrix, has constant negative curvature, but it fails to be complete. In fact, we will findthat these conditions cannot be met by any surface in R3! This leads us to consider a moreabstract notion of a surface, that is, an object that need not be a subset of some Euclideanspace. 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 (toldin detail in Scholz (1980)); we give the modem version here. Theorem 14.4 frames ourstory, 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 pwith asymptotic lines as coordinate curves. We first show that these coordinates have someeven nicer properties.

Proposition 14.1. if p is a point in a surface S C R3 of constant Gaussian curvatureK- -C2, then there is a neighborhood of p with all coordinate curves unit speed andasymptotic.

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 - F2) + FE -

(EG f F2) (EG - F2 ) + FG - G E 1 .

Furthermore, e = 0 = g and so the value of the Gaussian curvature is given by

-f2K EG-F2.

201

202 Recapitulation and coda

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 otherMainardi-Codazzi equation. Multiplying these equations by the appropriate functions weget

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--+J

`G(r)1j2dr.0 0

These are just the arc-length functions of the coordinate curves and the inverse of this coor-dinate 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 withline element given by

ds2 = due + 2Fdudv +dv2.

Any such system of coordinates is called a Tchebychev net and it solves the problemof finding a parametrization x: (a, b) x (c, d) -+ S of a neighborhood of a point inS with arc length as parameter along the coordinate curves. This problem was studied byP. L. Tchebychev (tr`ebylev 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 patchof S without stretching the fibers.

In a coordinate patch given by a Tchebychev net the Gaussian curvature takes on a simpleform:

Corollary 14.2. Let S be a surface in JR3 and x : (a, b) x (c, d) -+ S a Tchebychev neton S. Let w(u, v) denote the angle between the coordinate curves x,, and x,, at (u, v). Thenthe 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:

I

a2coSw0

3Cos(0

auav auK

TEG--F2)2detl acosw

avI Cos w

0 cosw 1 /1 L

cos2 w)a2cosw acoswacoswl(I - + cosw jsin4 w auav au av

1 3 a2w 2 aw aw 2 aw awsin w--sin WCOS(--+sin wCosw--sin° w L auav au av au av

_ -I a2w

sinw auav

We next use the hypothesis that S is geodesically complete to extend a single specialcoordinate 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. Thenthere 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 1there is an asymptotic Tchebychev net x : (-r, r) x (-r, r) - S with x(0. 0) = po}. Sincethe Tchebychev net has coordinate curves parametrized by arc length, the images of x areopen squares on S. Consider the union of all of these patches, which gives a coordinatepatch x : (-1. l) x (-!,1) -+ S. Let R denote the image of x. Consider the boundary ofthe subset R, denoted bdy R. It is closed, bounded, and hence compact, and by the metriccompleteness of S, bdy R C S. Around each point q E bdy R there is an asymptoticTchebychev 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 tocover the boundary. Taking E to be the minimum value of the Eq,, we can extend x to asquare of side I + e, which contradicts the least upper bound property of 1. Since we knowat least one asymptotic Tchebychev net exists around each po. we find that the largest suchcoordinate chart has domain all of R2.

Hilbert's theorem

We now come to the remarkable theorem due to Hilbert. Hilbert's 1901 proof utilizes themachinery of covering spaces - a worthy but lengthy detour to our story. We present a proofdue 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 ofconstant negative curvature K = -C2 is equipped with a function w: R2 - R such that0<w(u,v)< rand

a2w= C2 sin w. (O)

FU -8V

Since the sine is positive over the values of w, the differential equation satisfied by w impliesthat aw/au is an increasing function of v, and so for all u we have

aw(u, v) >

aw(u. 0) for v > 0.

all au

Integration over the interval [a, b] with respect to u gives the inequality

Jn 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 havew(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'12C2 sinw(u, v)dudv =

fT fu2 8wdudv

, , auav

= w(u2, T) - w(u,, T) - w(U2, 0) + w(u,, 0).

Rearranging terms we get

T fu_n > w(u2, T) - w(ul, T) = w(u2, 0) - w(u,, 0) + J 2 C2 sinw(u, v)dudv

v u,

> w(u2, 0) - w(u,, 0) + C2T(u2 - ut) sine.

However, this cannot be true for large T, so there is no function w defined on all of R2satisfying the differential equation (O).

This result puts the search for a model of non-Euclidean geometry into perspective. Ournormal 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 conse-quence of the existence of an asymptotic Tchebychev net.

Proposition 14.5 (J. N. Hazzidakis 1880). if S is a surface of constant negative cur-vature and x : (a, b) x (c, d) -a S is an asymptotic Tchebychev net, then any quadrilateralformed by coordinate curves in the image of x has area satisfying

area = - > a; - 2rr <27r

K i=1 -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 thesides 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 u2 V2 u2

C2area=J J C2dA=JU

C2sinwdudvv, u, v, u,

V2 U

wuvd ud v= L, L,ui u,

= w(u2, V2) -w(u2, u,) -w(ui, v2)+w(ui, vi)4

O N

i=1

Since the interior angles ai are all less than 2r, it follows that C2area < 2n.


The formula of Hazzidakis may be applied to obtain another proof of Hilbert's Theoremby 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 asymptoticsides, but this contradicts Proposition 14.3.

In order to find a model of non-Euclidean geometry we next turn to a generalization ofsurfaces in R3.

Abstract surfaces

The concept of a surface in 1R3 described in Chapter 8 is concrete; the definition refers toa subset of R3. The important geometric tools, such as the first and second fundamentalforms, are derived from the Euclidean geometry of R3. The real workhorse for the geometryof surfaces in R3, however, is the coordinate chart. To generalize the concept of surface wefocus on these constructs.

Definition 14.6. A smooth abstract surface (or smooth two-dimensional manifold) isa set S equipped with a countable collection of one-to-one functions called coordinatecharts 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 collectionA generates a maximal such set called an atlas of charts on S. That is, if x : U -, Sis another chart such that xa h o x and x - o xa are smooth for all Of E A, then x is inthe collection generated by A. The atlas generated by A is called a differentiablestructure on S.There is a further more technical assumption about the set S with atlas generatedby A:

(4) The collection of subsets determined by the x (U) C S for U in the atlas is a subbasisfor a topology on the set S. which is required to satisfy the Hausdorfcondition. Thatis, 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 secondcountable, that is, a countable collection of charts from the atlas determine the opensets in S such that any other open set is a union of open sets from the countablecollection.


The reader who is unacquainted with point set topology can find a discussion of theHausdorff condition, second countability, and subbases for topologies in any good intro-ductory topology book (e.g., Munkres (1975)). An atlas for a surface may be likened toa cartographic atlas - each coordinate chart is a page in the atlas, and on intersections ofcharts, there is a differentiable correspondence between pages. The abstract surface, how-ever, need not float in space like the Earth. It is not required to be a subspace of someparticular 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-ing the equivalence classes of 3-tuples, (x, y, z) (rx, ry, rz)whenever r 96 0. In each equivalence class there are two repre-sentatives satisfying x2 + y2 + z2 = I. If we take a coordinatechart for S2 such that xa (Ua) n -xa (Ua) = 0, then this definesa coordinate chart on RIP2 by identifying lines with their repre-sentatives on S2. The surface RIP2 with the atlas generated bythese charts is called the real projective plane.

Having defined the objects of interest, we next identify the important mappings betweenthem.

Definition 14.7. Given two surfaces S and S', a function 0: S - S' is differentiableat a point p E S if, for any coordinate charts xa : (Ua C R2) -* S around p andyy : (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 function0: 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 differentiableinverse 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. Themeaning of differentiability here is determined by composing each function with a chart orits inverse and applying the notion of differentiability between portions of R and R2. Wedenote 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 pin 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


Euclidean space, the idea of a tangent vector to a curve must be based on some notion otherthan the "direction" of the derivative of the curve. As we saw in Chapter 8, the directionalderivative 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, definethe 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 byTp(S). the tangent space of S at p.

This generalizes the ideas from Chapter 8. However, it is difficult to make geometricsense of such a tangent space. In order to make some connection with our intuition, weprove the following result.

Proposition 14.9. Tp(S) is a two-dimensional vector space. If X is a tangent vector, thenX(fg) = f(p)X(g) + g(p)X(f). that is, X satisfies the Leibniz rule for the product offunctions in CO°(p).

PROOF. 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 CR2) -* S is a coordinate chart around p = x(0, 0). We can write a curve x: (-e, E) -+ Swith 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 ona smooth function:

d afdu dvx'(0)(f) = d f(x(u(t), v(r)))I

au dt +of

o av dio r- =0

= u (0) au + v'(0)av .

We now introduce the analog of the coordinate tangent vectors xu and.rv to a surface inR3. The coordinate curves x(u. vo) and x(uo, v) are particular examples of smooth curvesin S and their tangent vectors satisfy

x(u, v(,)'(f) = a-u and.x(uo, 0'(J) = av

a aWe name these particular tangent vectors as

auand

avIn 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)


for a curve X by simply taking the line through (0, 0) E U of the required slope composedwith x. The sum and scalar products of tangent vectors are clear from this representationand Tp(S) is a two-dimensional vector space.

The Leibniz rule follows directly from this local description as it is a property of partialderivatives.

Finally, if y: (V C R2) -+ S is some other coordinate patch around p, then the transitionfunction y-1 o x : U fl x- I (y(V )) -+ V fly 1 (x(U)) determines a change of basis forTp(S). To see this, write x(u, v) and y(r, s) as coordinates. Then the coordinate curves forx become smooth curves in the patch determined by y and so

a = a a +ba and a =ca +a.au ar as av 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 ox(u, 0). Given a smooth function f on S, we have

a(fox) a(fo y) ar + a(fo y) as a(fox) _ a(fo y) ar + a(fo y) asau ar au as au' av ar av as av'

We denote this transformation in the notation of the tangent space by

ar aa as

Here J(y 1 o x) denotes the Jacobian. Since x-1 o y is also differentiable, we obtain theinverse 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 functionbetween abstract surfaces.

Definition 14.10. Given a differentiable mapping 0: S -+ S' and a point p in S, thedifferential of ¢ at p,

d4p: Tp(S) --j- TO(p)(S),

dis 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 localform: If x: (U C R2) -+ S is a coordinate patch around p, a(t) = x(u(t), v(t)), and


y: (V C R2) - S' is a coordinate patch around q(p) with coordinates y(r. s), then, in the

basis aa

for T#tpt(S') we havear' as 111

ara asa ara asad¢p(1'(0)) =u'(0) (au

ar + au as + v'(U) (av ar + av as)ar ar

= au av u(o)) _ J(v o 0 o x)(a'(Op.as as v'(O)

aal av

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 mappingid: S - S. In different coordinate charts around a point, the previous computation showsthat d(id),, = Ay- t o x) in local coordinates.

What remains lacking in this course of generalization is the line element. For surfacesin R3. ds was induced by the usual dot product for R3. Since tangent vectors were, infact, vectors in R3. it was easy to see that the first fundamental form was well defined andindependent of the choice of coordinate chart. On an abstract surface we have considerablefreedom in choosing a line element, as long as the choice is made coherently with respect tothe coordinate charts. We next define the analog of the first fundamental form for abstractsurfaces, a Riemannian metric, and then discuss what properties such a thing must have tobe well defined.

Definition 14.11. A Riemannian metric on an abstract surface S is a choice of positive-definite inner product (. ) p on each tangent plane, Tp(S) for P E S, such that the choicevaries 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 acoordinate chart x : (U C R2) - S, we have a basis for the tangent space at each point,

a aThe inner product is determined by linearity in terms of this basis and by the

au' avfunctions Ea a t

. F=(a a) .G= a a )P.

au' au fp auavp avavA Riemannian metric requires that these functions E, F, and G be smooth functions ofu and v. Another requirement is that they "match" on the overlap of two patches. This is


made precise by showing how these component functions of the metric transform undertransition functions.

Lemma 14.12. If x : (U C R2) - S and y: (V C R2) - S are two coordinate chartsfor 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 mappingy tox:x-t(x(U)fly(V))->V.

The proof is left to the reader. The transformational properties of the local description ofa metric make it an instance of a "tensor field" on S. In Chapter 16 we will consider morefully the analysis and algebra of tensors, which form the basis of an approach to the localaspects of differential geometry for higher-dimensional manifolds.

Having defined a Riemannian metric, it is not immediate that such a construct existsin general. The Riemannian metric on a surface in R3 obtains the correct properties byinheriting them from the ambient space. In order to establish the existence of Riemannianmetrics for abstract surfaces, one must utilize all the topological assumptions we made inthe definition. We refer the interested reader to Warner (1971) for details.

T T

EXAMPLES. (I) The flat torus: Let I = (0, 11 denote the closedunit interval in R and consider the equivalence relation on I xgiven by (0, v) (1, v) and (u. 0) (u. 1). The quotient spaceof equivalence classes is a space homeomorphic to the usual torus

St x St. We now endow T2 = I x // with a Riemannianmetric 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 toruswill 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 andnegative curvature.

(2) Unrolling a surface of revolution: Suppose we take the graph of a single-variable functiony = f(x) > 0 over an open interval (a. b). Rotating this curve around the x-axis in R3gives 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 givenby the identity. The Riemannian metric is given by taking any curve in S' and composingit 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.


Thus we have "unrolled" the geometry of the surface of revolution onto the stripe (a. b) x Rin R2. This construction has the property that it preserves the local geometry of the surfaceof revolution, but it changes the topology; the abstract surface gotten from unrolling thesurface 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-latitudecoordinates on the sphere. In fact, we can view the subset (0, 2n) x (-n/2, n/2) C R2 asan 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 candefine the are length of a curve a : (a, b) -- S by

s(t) =J

1 (a'(r), a'(r))a(T) dr.

and the angle 0 between curves a1 (t) and a2(t) is determined by

{aI(tP), a' (t p)) p

(aj UP),.011

(t11))p(a2(tp). c4(tp)) p

where ai (t,) = a2(tp) = p.The notion of area generalizes to abstract surfaces as well. Suppose R C S is a region

in S lying entirely in a coordinate chart x : (U C R2) -+ S. Then the area of R is given bythe integral

ff dA = f f -1(R) EG - F2dudu.

The geometric features of a surface in R3 discussed in Chapters 8 through 13 were de-duced from the associated first and second fundamental forms. In the case of an abstractsurface, however, there is no normal direction defined on the surface and so the second fun-damental form does not exist. The work of Gauss and, in particular, his theorenw egregiumfree 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 systemsof equations

(EF

F ) ( I'l 21 E

"I ) ( Fu - I E,

* F r,2 ) ' E

\ F G) r'2 ) -(FoGiGI-

2

The Gauss-Riemann curvature is defined on the image of x as the function

-2' Ec,,,+ F,,,, - 2'G,,. 2' E. F. - 2' E

K=(EG

I- F2)2

den F 'G E F

G F G

0 21 E" 11 G

- det 2 E1, E F

G,, F G

From our second proof of theorema egregium we could also have defined K as theexpression

K = E21)v - 22)u + rl 22

+21

r22 - r12r1t - r12r12)

Example (1). the flat torus, has constant zero curvature by the previous formulas. This isvery 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 forthe 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 answeris local and so we can argue in a single coordinate chart. The argument carries over fromthe 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 acoordinate chart around y (t ), the following differential equation is satisfied:

I

Uu" + (U,)21-', + 2u'v'r112 + (t")2r2, = 0,

v"+(u')2F11 +2u'dr12+(v')2r22 =0.

EXAMPLES. Let us consider the flat torus. In this case, the Christof-

fel symbols vanish and so the geodesic equations become u" =0 = v". In the chart for the torus, this is simply a straight Euclideanline along the torus. When the line has a rational slope, notice thatthe geodesic eventually rejoins itself and is closed. When the slopeis irrational, the line has an image dense in the torus, that is, everypoint 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 symbolsbecome

f(tt)f"(u) 2 f"(u)1 -f'(tr)f(u)

+, I 2122=

2r12=

+I U (u)) I(rn ' I (f (u))The geodesic equations become

I

u", + (u')2 f (u)f'(u) - (v')z 1'(u)f(r1) - 0,I + (f (u)) 2 1 + (f'(u))2

(u)v"+2rr''f =0.j(u)


Vertical lines, t i-+ (uo, t), are geodesics whenever uo is a critical point of f(u). Byparametrizing the horizontal lines

t H (r(t), vo) such that r" = -(r')2J'(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 obtainall the familiar local results about them. For example, there is a unique geodesic in eachdirection from a point. The exponential mapping can be defined as in Chapter 11, andgeodesic polar coordinates exist. The arc length provides the structure of a metric space onan abstract surface and the topology induced by this metric space structure coincides withthe 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 wemean by an isometry.

Definition 14.15. A one-to-one, onto, differentiable function m : S -+ S' between abstractsurfaces, 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 generalizedversion of the inverse function theorem for abstract manifolds tells us that an isometry is adiffeomorphism. It follows that we can use the charts from S to give charts on S' and so at apoint, there are charts on each surface such that the isometry sends the associated functionsE, F. and G on S to identical functions on S'. It follows that the Gauss-Riemann curvatureis preserved by isometries.

For example, consider T2. the flat torus, and ask if there is a surface in R3 that isisometric to it. Since the flat torus is a quotient of a compact space, it is compact, andsince the curvature is constant on it, a surface in R3 that is isometric to the flat torus wouldbe 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 tothe 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. Thecoordinate charts on S1 x S' are images of open intervals (a, b) x (c, d) under the samemapping >' 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 componentfunctions of the metric:

E=(a'

a)au au (uo.ib)

=at

(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.


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 Riemannianmetric from the dot product on R". If this occurs, we say that the surface S with the inheritedRiemannian metric is embedded in R". When an abstract surface S' is diffeomorphic toa surface S embedded in R", then we say that S' has an embedding in R". Finally, if thediffeomorphism 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 embeddedin R". Having an isometric embedding identifies the geometry of the surface S' with thegeometry of a surface in R", in particular, the Riemannian structure on S' is identified viathe isometry with the canonical Riemannian structure on subsets of R".

By definition all of the classical surfaces in R3, if taken abstractly, have isometric em-beddings in R3. The remarks above show that there is no isometric embedding of the flattorus 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-curvaturesurface 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 inR6. 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 geometryby 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 therelation on differentiable structures by A A' if the union AtJ.A' defines a differentiablestructure. Show that is an equivalence relation. Show that an atlas generated by A isthe 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 r1 r3 11 x,OI(x) _ - - . ' 2(X) = - ..I . 3(X) _ -.- .

x1' X1 X2 r2 \r3 XI,defined for 4, : V1 - R2. The mappings 4'; satisfy the property 0, (.x. v. z) =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 RIP2.

14.3 Define what it means for an abstract surface to be orientable.

14.4 Prove Lemma 14.12.


14.5 Given a differentiable mapping 0: S -+ S'. prove that do, is linear. Also, prove thechain 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 alsoa Riemannian metric on S.

15

Modeling the non-Euclidean planeIn recent times the mathematical public has begun to occupy itself with somenew concepts which seem to be destined, in the case they prevail, to profoundlychange the entire order of classical geometry.

E. Beltrami (1868)

The notion of an abstract surface frees us to seek models of non-Euclidean geometry withoutthe restriction of finding a subset of a Euclidean space. A set, not necessarily a subset ofsome 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 geometryof Lobachevskii, Bolyai, and Gauss. In this chapter we present the well-known models ofnon-Euclidean geometry due to E. Beltrami (1835-1906) and J. Henri Poincard.

This chapter contains many computational details, like a lot of nineteenth-century math-ematics. The foundations for these calculations lie in the previous chapters. It will be thesmall details that open up new vistas.

We begin by considering a problem Beltrami posed and partially solved in a paper of1865. He asked for local conditions on a pair of surfaces, Sr and S2, that guarantee that thereis a local diffeomorphism of Si - S2 such that geodesics on Si are taken to geodesics onS2. Such a mapping is called a geodesic mapping. Beltrami solved the problem when oneof the surfaces is the Euclidean plane. He gave conditions for the existence of a mappingtaking 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 theEuclidean plane, then the Gaussian curvature of the surface S is constant.

PROOF. Suppose f : (W C S) -+ lR2 is a geodesic mapping, a local diffeomorphismdefined from an open set W C S to the plane. Let U C R2 be an open set lying in the imageoff and take x: (U C R2) - S to be the chart given by x = f-r. Since f is a geodesicmapping, straight lines in U go to geodesics in S. Write ds2 = Edu2 + 2Fdudv + Gd v2for 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 0)u"

and so u'v" - u"v' = 0.

217


The general condition for a curve to be a geodesic is given by the differential equation(Proposition 11.5)

U, u"" + (u')2r11 + 2uv'r;2 + )-I"12)0 = det

v' 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 andGaussian curvature:

(a) (r?) - (ri2)u + ri l riz + ril rz`z - rizri l -22 = EK.(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.

The relations associated to a geodesic mapping to the plane imply some simplifications

(a)

(c)

EK =ri2ri2 - (r2l2)u

GK = rizriz - (r12)1

(b)

(d)

FK = (ri2)u - (2r12)v + ri2r12

FK = (1 2)v - (2r,12),, + r12ri2.

Subtracting (d) from (b) we get (riz). = (ri2)L,. Thus we can rewrite (b) and (d) as

(b) FK = rizr12 - (ri2)v (d) FK = ri2ri2 - (r:2)u

We now use the equations (r12)u0 = (r12), and (rig),,v = (ri2)vu. Taking the partialderivative of line (a) with respect to v, and of line (b) with respect to u, we get

a(EK)= E K + EKv = 2r122 (r122)u - (r122

av °)uu,

a(FK)au = FK r12(r12)u + r12(r12)u - (r12)uu.

and so EKv - FK + K(E, - Fu) = 2r12(r12)v - r12(r12)u - r12(r12),,. Substitutingfrom (a), (b), and (d), we find

EK - FK K(EL, - 2r12(r12r12 - FK)

- ri2(r12r12 - EK) - r12(r12r12 - FK)

K(E - K(r12E - r12F).

1 5. Modeling the non-Euclidean plane 219

Since I'11 = 0 and I'i i = 2r12, we can write (see Chapter 10)

r 1'2E - I'2F = I'i2E +r22F-2r 2 2

1 1 12F - ri i G

= r 12E+f 1G)

=

Thus EKv - FKu = 0. By computing a(au)a(GK)

we carry out a similar derivation

and obtain FK, - G K = 0. In matrix form this is

EG)(

KK

u)-(0)(F

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 ofsurfaces 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 takesgreat 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 sphereof radius R centered at the origin to the plane tangent to the south pole (0, 0, - R) has theform x : IIt2 - S2,

x(u, v) =R

(u, v, -R)./R2 +u +v

PROOF. Let Q = (u, v, -R) denote a point in the plane tangent to the south pole andconsider the line segment in R3 joining Q to the origin. This passes through a point P onthe sphere. Write the coordinates of the pointx(u, v) = P = (r, s, t). The linear dependenceof OP and OQ implies

(0,0,0) = OP x OQ = (-Rs - tv, Rr + tu, rv - su).


From this equation it follows that r = tu

and s = tv

. The condition r2 + s2 + t2 = R2

- R2implies that t = and the proposition follows.

R +u +vFrom 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 thesphere is a surface in R3 we compute directly:

Rxy =

(R2 + u2 + v2)3/2(R2 + V2, -uv, Ru),

(R2 + u2 + uz)3/z(-uv, R2 + u2, Rv).

These coordinate vectors determine the following line element on the sphere and hence onthe 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 constantGauss-Riemann curvature I / R2. Notice that as R goes to infinity, this metric converges todu2 + dv2.

The Beltrami disk

Computing the curvature associated to the metric induced by central projection generates amorass of algebra that arrives at K(p) = I/ R2 for all p. The key feature of the calculationis that it depends only on R2 and not on R. From this observation, Beltrami made the leapthat brings us to the first model of non-Euclidean geometry. He replaced R with R toobtain the line element

ds2 = -R2(v2 - R2)du2 - 2uvdudv + (u2 - R2)dv2

(-R2 + u2 + u2)2

= R2(R2 - v2)du2 +2uvdudv + (R2 - u2)du2

(R2 - u2 - v2)2

This formula determines a Riemannian metric on the abstract surface given by the interiorof 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 equationsr;, = 0 = r22, 2r12 = r;,, and 2r12 = rig continue to hold. Thus the geodesics on thisabstract surface are Euclidean line segments.

Recall that the desired surface to model non-Euclidean geometry must have constantnegative curvature and be geodesically complete. In fact, the surface we have just constructedis geodesically complete. However, we prove this property by a roundabout route. What

R

15. Modeling the non-Euclidean plane 221

follows up to the proof does not depend on geodesic completeness, so the reader mayproceed 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 concreterealization of their non-Euclidean plane. Fix the value of R as 1. Our abstract surface is theinterior 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):

(I - v2)due+2uvdudv+(I -u2)dv2ds8 -(1 - U2 - v2)2

Let IIDe = (11), dsB) denote the Beltrami model, that is, the abstract surface 1D with theBeltrami metric.

It is immediate that Euclid's Postulate V doesnot hold on DB; consider the accompanying di-agram, where we have pictured a point not on agiven line and an infinite family of lines (that is.geodesics) through the point, each of which doesnot intersect the given line.

To derive results of a more analytic but localnature we choose a convenient point in lB. Laterargument.,, will show that these particular compu-tations hold everywhere in 1DB. When u = 0 orv = 0 the middle term of dsB vanishes and sothe geodesics u H (u, vp) and v H (uv, v) areperpendicular to the v-axis or u-axis, respectively.

At the center of the disk the axes themselves are perpendicular geodesics. Our convenientpoint is the origin (0, 0). To discuss distance from (0, 0) and circles centered at (0, 0) polarcoordinates 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 substitutethe 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,


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) =J , I

dt12 = 2 In( I ±r)

Notice that as r approaches I the distance goes to infinity. Thus lines through the originare infinite in length. Later we will construct all of the isometrics of another model thatis isometric to IDB and the previous computation will prove the geodesic completeness ofboth models.

By fixing r and varying 0 we get a circle of radius p = 2 In (._±L) in D. Itscircumference is given by the integral

IT rd0 tarcircum(p) = -

0 1-r 1-r

We next solve for r in terms of p:

p = 2 In I r implies + r = e2p, so I + r = e2° - re2o

e2p - 1 eP - e-Pandr= _ =tanhp.e2p + I eP + e-P

The circumference becomes

2a tanh pcircum(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 paral-lelism. Choose (0. 0) as the point of intersection of the perpendicular geodesics given by

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 thev-axis is the first nonintersecting line. This line passes through the point on the limit circleS2 = (1.0). The parallel line is determined by the line segment joining A to 12, which canbe 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 isparametrized by C2(t) = (u2(t). v2(t)) = (0, r - t) with tt2 = 0 and v2 = -1. Weput these data into the Beltrami metric to compute the cosine of the angle between thecurves C( and C2 at A, that is, the angle of parallelism:

E(O,r)uju',+F(0.r)(uivZ+u,vi)+G(0.r)v'u;fl (p) _cos

4O.rI(Cj(0). Cj(0)) 1(O.r)(C,(0), C,(0))

r/(I - r22)22

V(11(1 - r2)) + (r2/() - r2)2) 1/(1 - r2)2r

I -,.2+,.2= r = tanh p.

This implies the Lobachevskii-Bolyai theorem:

11(p) I - cos l"I(p) I - tanh(p)tan

2 sin 11(p)1 - tanh-(p)

2e-P

= e-P(eP + e-P)2 - (eP - e-p)2

We close our discussion of the Beltrami model byestablishing the theorem of Saccheri (Theorem 4.4)that non-Euclidean parallels are asymptotic. Con-sider the point A = (0. r) and the parallel to theu-axis through A given by the line segment A S2 withS2 = (1.0). Choose a point P = (a. 0) on the u-axis and consider the line segment joining P to theline AS2, meeting at the point Q = (a. r - ra). Theparametrization of this segment given by C(1) =(a, r(1 - a)!), 0 < t < 1, allows us to compute thelength of PQ. Since u' = 0 and v' = r(1 - a) forthe curve C. the arc length of PQ is given by theintegral

J G(C(!))(u')2dt =J

(r(I - 1 - a2di

U 0 -a - - r2(1 - a)2!2)2

I=r - a- (dt = I

InI+a+r 1 -a

' 0 (I +a)-r2(l -a)!22 2 1 +a - r1 -a'


As a goes to I, the limit of this expression is zero, that is, the length of PQ goes to zeroand parallels are asymptotic.

We postpone a discussion of horocycles and distance in the Beltrami disk until we havedeveloped 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 Euclideanline segments, but the angles can be far from their Euclidean appearance in measure. Fixa Euclidean angle between two line segments, say n/2, and place this Euclidean figureat different points in IDa. The angle measure depends on the position of the vertex. Forexample, 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 non-Euclidean geometry that has the interior of a Euclidean disk as underlying abstract sur-face. In this model, however, angle measurement agrees with its Euclidean representation.Of course, geodesics will fail to remain Euclidean straight lines. However, they are trans-formed into reasonable classical curves.

We begin with another visit to the sphere. Recall from Chapter 8" that orthographicprojection is a mapping from 11D C R2 to the lower hemisphere of S2 for which all lines ofprojection are perpendicular to the plane containing D.

Proposition 15.3. Orthographic projection of the Beltrami model IDS to the lower hemi-sphere 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),

(u, V) i-+ (u, v, I - I-u2-v2).

In Chapter 8 we computed the line element for the lower hemisphere of S2 C R3 with thecoordinate 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 - v2)ds2B.

By Proposition 8"'.3, the mapping is conformal.

Orthographic projection takes the geodesics in De, Euclidean line segments, to semicir-cles on the sphere that meet the equator at right angles. Follow this mapping by stereographicprojection from the north pole to the plane that is tangent to the south pole. The lower hemi-sphere 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 diffeomor-phism given by the composite of orthographic and stereographic projection. Parametrizethe 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 stereographicprojection to (x. y. 0) given by

2u 2v

I + 1-rr -u- I+ I - u-- v-

If we write w = - I - u22 -_P', then it follows that

dw= udu+vdv anddu12 _ (udu+vdu)21-u--v- w'- (I -rr2-v2)2

This implies that we can write

ds2 - (I - rr2 - u2)(du2 +dv`) + (udu + vdv)2e (1 - u2 - v2)2

(du- + d v- + du12 ).

u, wStereographic projection is given analytically by x = 12u and Y = 1

2v.This gives

dx = 2(1 - w)du + 2udw dv - 2(I - w)dv + 2vdw(I - w)2 (l - w)2


It follows that

dx2+dy2 =4

(I - I(I -w)2(du2+dv2)+2(1-w)(udu+vdv)dw+(u22+v2)dw2].

Since w2 = I - u2 - v2 we can substitute I - w2 for u2 + v2 and -wdw for udu + vdv.

dx2 +dy24

= (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

Since

ds2 = (1- w)2(dx2

+ d 2).R - 4w2 Y

x2 + y2 _ u2 + v2 _ 2w(l - w) 2w

4 -1 (I-w)2-1 (1-w)2 1-w,the metric induced by the mapping on 1)2 takes the form

ds2dx2 + dy2

R =

\ I

- x2 + y2 )2.

4

The subscript R refers to Riemann; a metric of this form appears in the middle of hisfamous Habilitation lecture as part of a very general discussion of higher-dimensionalmanifolds with constant curvature. We will discuss these ideas in the next chapter. We writeDp to denote the abstract surface with Riemannian metric (D2. dsR). which is called thePoincare 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 theplane (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 conformalrepresentation of angles. The metric is a multiple ofthe Euclidean metric and so angles between curveshave the same measure as their Euclidean representa-tion. This allows us to visualize properties of anglesmore directly. For example, we prove That there is aregular octagon in Dp with all right angles. From thecenter of the disk IiD2 send out eight rays each n/4apart. At infinity, the angle between the line parallelto an adjacent pair of rays and each ray is zero. Con-sider the family of (Euclidean) circles concentric tothe limit circle. For each such circle construct the

regular octagon with the intersection of the circle with the eight rays as vertices. As thecircles decrease in radius, we obtain a regular octagon with increasing angles. The metricconverges to the Euclidean metric at the origin, and so near the origin, the octagons are


nearly Euclidean, that is, the angles are close to 37r/4. By continuity there is a regularoctagon for some concentric circle with all right angles.

Poincare arrived at this model of non-Euclidean geometry from an entirely differentdirection than Beltrami. His investigations of complex differential equations, based on thework of L. I. Fuchs (1833-1902), led him to consider figures in a disk made up of arcs ofcircles meeting the boundary at right angles. The trick of sending such circles to straight linesegments in the plane by composing stereographic projection with orthographic projectionled him to a result he needed. He did not make the connection with the Beltrami disk untillater 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 expeditionorganized by the Ecole des Mines. The circumstances of the journey made me forget mymathematical work; arrived at Coutances we boarded an omnibus for I don't know whatjourney. At the moment when I put my foot on the step the idea came to me, without anythingin my previous thoughts having prepared me for it; that the transformations I had made useof to define the Fuchsian functions were identical with those of non-Euclidean geometry. Idid not verify this, I did not have time for it, since scarcely had I sat down in the bus than Iresumed the conversation already begun, but I was entirely certain at once. On returning toCaen I verified the result at leisure to salve my conscience."

This account is often quoted in discussions of mathematical discovery and the psychologyof mathematical invention. For a detailed discussion of the related mathematics see Gray(1986). In fact, Beltrami had presented this model in his 1869 paper Teoriafundamentaledegli spazii di curvatura constante (Beltrami 1869). That paper was concerned with ex-tending the results of the Saggio (Beltrami 1868) to higher dimensions in the context ofRiemann's theory of manifolds. Poincare established another context, complex functiontheory, where the non-Euclidean plane appeared naturally and fruitfully. The popular nameof the Poincare disk model is due to Poincare's efforts to bring the new geometry into themainstream of nineteenth-century mathematics.

Poincare's inspiration, around the year 1880, continued after the disk model. Furtherstudies 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 (then28 years old) applied the methods of stcreographic and orthographic projection to provideanother conformal model of the non-Euclidean plane.

To construct the model, project the Beltrami disk IDB C R2 x 10) C R3 orthographicallyto the lower hemisphere of the sphere of radius one centered at (0, 0. I) as we did to constructI)p. Now rotate the sphere around the axis through the center parallel to the x-axis througha right angle to put the lower hemisphere in the half-space y > 0. This is accomplished bythe mapping

(u, v, I - 1 - u2 - v2) H (u, 11 - u2 - v2, v + 1).


Finally stereographically project from the north pole. This takes the "right hemisphere" ofthe sphere to the upper half-plane in R2 given by H = ((u. v) I v > 0). The compositemapping IIDa -> H is seen to be

2u 2 1 - 'u-- u-(u, U) H -v, I - u

What is the metric on H induced by this mapping? Let w = I - u - u so that-wdw = udu + vdv, and the Beltrami metric can be expressed as

dse = 12 (dug + d v2 + d w2 ).w

Write x = 2u and y = 2w to getI-u 1-v

dx = 2(1 - v)du+2udvanddY =

2(1 - v)dw+2wdv(I - v)2 (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)dv2J4

_ (I - u)4J(1 - v)2(du2 +dw2) - 2v(l - v)dv2 + (1 - v2)dv2J

2

_ (1 4v)2(du2+dv2+dw2)= (14wv)2ds2=y2ds2.

2Thus we obtain the induced metric dsP

dx2 + dy= Y2 . Let H now denote the abstract

surface with Riemannian metric (H, dsP), which is called the Poincare half-plane. This


surface is isometric to DB and so models the geometry of Lobachevskii and Bolyai. Thegeodesics are the images of geodesics in De, that is, either vertical lines or semicircleswhich meet the x-axis at right angles. Like the Poincart disk, the Poincard half-plane is aconformal 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, well-known methods of complex analysis can be brought to bear. We next take advantage of thisserendipity. 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 thefollowing 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 theirproofs 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)`I = TA_i.

(2) The following matrices and their associated linear fractional transformations generateall linear fractional transformations. That is, every T,4 can be written as a composite of thefollowing:

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 exceptz = -d/c. By moving our discussion to the Riemann sphere, we can extend the domain toall complex numbers. The Riemann sphere is the result of mapping the complex numbersto the sphere by stereographic projection and adding the north pole as infinity. We extenda linear fractional transformation to a mapping of the Riemann sphere to itself by lettingTA(-d/c) = no and TA(oo) = a/c. In this formulation a line in the complex plane is acircle through no on the Riemann sphere and so we can talk of both circles and lines ascircles. This brings us to a useful proposition:


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 caseA=0.

To prove the proposition we merely apply Proposition 15.5 and check the generatorsof the set of linear fractional transformations. Clearly translation takes circles to circles. IfTA(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

+B- +B- +A=0.II

z z

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 projectivegeometry and gives the foundation for another approach to non-Euclidean geometry thatwas developed by Felix Klein (1849-1925). To state the property we need the followingnotion.

Definition 15.7. Given distinct complex numbers zi, z2, z3, and z4, their cross ratio isgiven by

(ZI, z2: z3, Z4) =ZI - Z2 Z4 -3

.

ZI - Z3 Z4 -Z2

The cross ratio depends on the ordering of the complex numbers zj. It is left to the readerto determine the effect on the value of the cross ratio of a permutation of the values z;. Weextend the cross ratio to the Riemann sphere by allowing one of the zi to equal oo:

zI - z2(zI. z2; z3, oo) =

ZI - Z3

The relation between the cross ratio and linear fractional transformations is given in thefollowing result.


Proposition 15.8. Given a linear fractional transformation TA and four distinct pointson 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 _ (ad - bc)(zr - z2)TA(zr) - TA(z2) =

czr + d cz2 + d (czl + d)(cz2 + d)

The cross ratio determines a linear fractional transformation as follows: Given threedistinct points on the Riemann sphere z2, z3, and 24, define

7x2.23.24(2) = (z, z2; 23, 24) =Z - Z2 Z4 - Z3

Z-Z3 24-Z2In this form we can see that the linear fractional transformation Tz2,Z,,Z4 takes Z2 to 0, Z3 tooo. and z4 to 1. From this fact we can take any circle (in the extended sense) to any other circleby 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 ofthe points to be 22, z3, and Z4, then the transformation T22.2,.%4 takes these three points to0, oo, and 1, respectively. These are points on the extended circle that is the real line in thecomplex plane. The inverse of this transformation takes the real line to the circle determinedby 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 isreal.

We now concentrate on the Poincare half-plane H. The main object of study in Poincare'spaper 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 = I a b 1 , a, b, c, d E R, and ad - be = I.d

Notice that any real 2 x 2 matrix with determinant positive determines an element inSL2(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 thesame linear fractional

transformation), and inversion. The important property of these transformations is given bythe following result.

Theorem 15.9. A real linear fractional transformation in SL2(R) is an isometry of thePoincarf half-plane K Furthermore, any isometry of H is given by such a real linearfractional transformation.

PROOF. We first check that these real linear fractional transformations take H to itself. IfTA is in SL2(R), then

TA (u + iv) _au+aiv+b = (au+b)+aiv (cu+d) -civcu+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'


Since ad - be = 1, the transformation takes H to itself. It also takes the u-axis (the realline), the boundary of H. to itself. Since the singular point of TA, z = -d/c, is a realnumber, 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 atcorresponding 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-idvanditfollows that dx2 + dy2 = dzd2. Solving for y we get y = z

2iz and so

2 dx2+dy2 dzd2dsP = y2 = 4 (Z - 2)2.

At w = TA (z), the change of coordinates takes the form

dw =a(cz+d)dz-c(az+b)dz - (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 = d` We can now(c2 + d)2

compute

dwdtu dzd2

(w - tp)2_

((az+b)(c2+d) - (a22+b)(cz+d)12dzd2

[(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 Hand any pair of unit-speed geodesics, yi passing through P, n through Q, can be mappedby a real linear fractional transformation taking P to Q and yi onto Y2. The existence anduniqueness of geodesics through a given point and the fact that isometrics take geodesicsto geodesics imply that we have constructed all of the congruences. Geodesics in H maybe vertical lines or semicircles with centers on the real axis. We treat the case of a pair ofsemicircles 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 Qlies on the semicircle of radius r2 and center c2. Let al and bi denote the intersection of the

first semicircle with the real line and orient (al. P, bi) in the direction of the geodesic yithrough P. Likewise, orient the geodesic y2 through Q as (a2. Q, b2). We may take bothgeodesics oriented counterclockwise.

Apply real translations to bring the centers of both semicircles to the origin. Now applyreal similarities to make the radius of each semicircle unity. The points P and Q becomee'm and e'" on the unit semicircle. It remains to show that there is a real linear fractionaltransformation taking the unit semicircle to itself and e'9 to e'". To construct such a trans-formation, form an intermediate transformation taking - I to 0, e'0 to i, and I to oo. Usingthe cross ratio, this can be done by

z + I er0-1I e 'a+l

Write era = cos 9 + i sin 0 and compute

ei9-I (cos9-t)+isin9re'B+1 (cos9+1)+isin9

_ -sin9(cos0+ 1)+sin9(cos9 - I)+i(cos29 - I + sin' 0)(cos 9 + 02 + sine 9

_ -singcos9+I ER

This shows that the intermediate transformation has determinant which is posi-c

2os9sin+9

I

five forO < 9 < n, and so it determines an element of SL2 (R). Similarly, the transformationtaking - I to 0, e"? to i, and I to oo is in SL2(R). The composite of the first with the inverseof 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 Beltramidisk is infinite. Geodesic completeness of the three models follows from the theorem. Theimage of a line through the origin in ID5 is a geodesic in H. Since the transformation isan isometry, there is at least one geodesic in H that is the diffeomorphic image of a one-dimensional subspace of T(o.o)(®a) under the exp map. Applying isometrics we get thatevery geodesic in H is infinite, and finally, reversing the isometry to DB and followingthrough 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 geodesicallycomplete abstract surface of constant negative curvature implies that Postulate V cannot beproved from Postulates I through IV. Euclid is vindicated in his choice to call Postulate Van assumption, and a search stretching over two millennia is ended, not with a proof, butwith a new geometric universe to study.

(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 geodesicP = (a, y j) and Q = (a, Y2). The path C(t) _ (a, y i + t (.v2 - YO), 0 < t < 1. providesa parametrization of the geodesic with C'(t) _ (0, v2 - yi). The distance between thepoints P and Q in H is denoted dH(P. Q) and is given by

dH(P, Q) _dx2+dy2

cut y2

- I (y2 - vi)dtyr+t(y2-YO

= 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 Euclideanprocedure; consider the perpendicular bisector of the Euclidean line segment PQ. Theintersection 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 pointsc-r and c+r. Consider the following transformation that takes c-rto0, P = zi = xi+iyito 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) (xi - c)+ivi - r i(xi -c)2+yj -r2+i(2rvi)zt -(c-r) (xi -c)+ivl +r (xi -c+r)2+ vi

Since zt lies on the semicircle of radius r and center c. (xj - c)2 + v = r2. and wezi - (c+r) -2rvi

see that i = 2. The determinant of the transformation iszi - (c - r) (xl -c+r)_+yi

4r2yta positive real number.(xi -c+r)2+yi '

If we apply the transformation to Q we get

T(Q)=.z2-(c-r) zl -(c+r)

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 pointof the form iy with y = (z2, c - r; c + r, zl ). By the earlier formula for vertical lines, weget

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)/


This form of the distance function relates directly to the cross ratio. By manipulating theexpressions a little, one can simplify this expression to prove, when P = (xi,yi) andQ = (x2, Y2), that

dH(P,Q)= In ri -c+r y2(XI-C+1. yl

Another simplification comes by the fact that the cross ratio (z2. C - r: c + r. zi) is areal number (since the points lie on a circle). This number times its conjugate is simply thecross ratio squared, but the complex expression becomes

I:2-(c-r)1'- Izi -(c+r)1? _ (dE(:2.(c-r)) dE(z1,(c+r))2Iz2 - (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

d)HI(P,Q)= IndE(:2.(c-r)) dE(z1.(c+r))\11

dE(:1.(c-r))/ .

The other models of the non-Euclidean plane are isometric to H and so we can relate thedistance function on H to the distance functions on D8 and ®p. We treat the case of ID8here and leave Dp to the reader.

Suppose P and Q are points in ®8. Let the distance function on 11D8 be denoted byd8(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 circleat points A and S2. The points A and n go to the intersection ofthe geodesic in H through q5 (P) and 0(Q) and the real line. Theformula 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

2u 2 1-u --v- u' 1-v

Write P = (U I. vi), Q = (u2, 112), A = (uo, vo), and S2 = (uOO, voo). To computede(P, Q) we need to simplify the following horrendous expression involving the squaresof the distances:

z z

2112 2uo 2 + 2 I - u; - 112 2u2 2 1 tr i - vv2

I-vu} 1-v2(2141I-vi I - uOO) I- vi

2 ?'

2u2 2u, 2 + 2 1 - uZ - v? - 2uo2 + (211 - u1 - vi

(1-v2 1 -vOC, 1-v2(2uI

1-vi 1-uo, l-vi


Now we can use u2 2+ vo = I = u + v;o, and the fact that P. Q, A, and n lie on a line, sayv = mu + b, to significantly simplify this expression (a vertical line in D8 needs anothersimple argument). It becomes, in fact,

1 /u2-up Ui - Uoo\dB(P,Q)=- Inl2 U2 - Uoo ui - up

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) _ U2 rtp ui - uoodE(Q, S2) dE(P, A) U2 Uoc III - Up

Substituting this into the formula for distances in DB we get

dE(P, S2)ldB(PQ) In

(dE(Q.A)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 inEuclidean 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 suggestconnections to projective geometry. These connections were made by Felix Klein in hispaper of 187 I, Uber die sogenannte Nicht-Euclidische Geometrie. He based his work on anidea 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 modelis most often referred to as the Klein model, or the Klein-Beltrami model, or even theCayley-Klein-Beltrami model. The modern term for non-Euclidean geometry, hyperbolicgeometry, was coined by Klein in this paper. Our emphasis on the differential geometricapproach, and the dates of publication of the relevant papers of Beltrami and Klein led to ourchoice 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 ableto determine the circumference of a circle of radius p and center (0. 0). Such a circle inDB 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 specialproperties of stereographic projection is that it takes circles on the sphere to circles (in theextended sense) in the plane (Proposition 8'.7). Thus these circles of any radius centeredat (0, 0) in D8 map isometrically to non-Euclidean circles in H which have the form of


Euclidean circles but with different centers. By applying an isometry of H we can movethe center of a given circle to any point of H. Since linear fractional transformations takeEuclidean circles to Euclidean circles, we have shown that all non-Euclidean circles in Hare 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. TheEuclidean center, (a, b), is not the non-Euclidean center. To determine the center in H wecan choose a diameter and halve it. The most convenient diameter is the vertical one, theline ((a, y) I y > 0).

Let B = (a, b - r) and T = (a, b + r) denote the bottomand top, respectively, of the circle. The line segment BT is adiameter and so the midpoint, P = (a, p), is the center of thecircle 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

-r b+rp - In

bb

- r =1n( b-rThese 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 ofradius psinh(p) and center (a. pcosh(p)).

Similar results hold for the circles in Dp. The Beltrami model is more difficult - circleswith center away from (0, 0) are not Euclidean circles. In fact, non-Euclidean circles inHB are represented by Euclidean ellipses. This is made apparent by carefully studyingorthographic projection of circles on the sphere to the disk. The appearance of conics inthis 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 verticallines that are all parallel to each other in the "up" direction. Ifwe choose a point P in H and take any semicircular geodesicI through P, it forms an angle a with the vertical line throughP. To find the other point on 1 that corresponds to P (seeChapter 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 solvethis 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 throughthe corresponding point. This shows that the points that correspond to P all share the samey coordinate, and so a horocycle with respect to the pencil of vertical lines is a horizontalEuclidean line.


Every other horocycle can be constructed by applying the isometries of H. A real linearfractional 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 withone point on the limit circle. The case of D8 is similar to Dp but involves Euclidean ellipsesinstead 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-Euclideantriangle is given by the angle defect. The easiest model in which to manipulate triangles isthe Beltrami model. Since we know all the isometries of H l and, in principle, of HD8, we onlyneed to consider a right triangle with one vertex at the convenient point (0, 0) E D8. Set outtwo 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-Euclideanangles coincides with the Euclidean measure since dse = due +dv2 there. Let a denotethe angle measure of angles LPOQ - LP'Q'O. and let fi denote the angle measure ofLPQO = 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 withwhich to compute the non-Euclidean distance along the legs:

-u-va=de(P,Q)=ln

I +( 1-u -v

b=d8(O,P)= I In+u2

(j),I

vandc=de(O,Q)= I

2lnI +

u u2

-1- u +v


This implies that u = tanh b, u +v = tanh c, and V= tanh a. Since sech2 b =

1-u1 - tanh2 b, we can write cosh b = l and so tank a = v cosh b.

uThe point P _ (0, v') is determined by

Ia=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 computesines and cosines from the Euclidean right triangles:

sin a =v

cos 16 =v' = v cosh b

= cosh b sin a.u +v u +u u +u

Away from the origin we may use Euclidean polar coordinates. The line element ds2transforms into

2 dr2 r2d02ds = (I - r2)2 + I - r2

Following Chapter 14, to compute the area of A0PQ, we integrate

QEG - F2drdOarea(AOPQ) = ff

OP

a unhbsec0 r

0 02)3/2drdO(l- r

a

O('ll_tanh2bsec2g_l)d0;ktt5

1sins dl

IS" (1 - tank b) - r2

_ -a + arcsinGsin a

1 - tanh2 b

= aresin(cosh b sin a) - a = aresin(cos 0) - a

=2- -a=n-(a+Q+2and this is the angle defect of AOPQ.

Since 1)B is isometric to Dp and H, the same result applies for triangles with a vertexat the image of (0. 0). In H we can apply the elements of SL2(R) to prove the result for allright triangles in H. and hence in DB and Dp. By dividing an arbitrary triangle into righttriangles we obtain the result for any triangle.

We can prove many other results of non-Euclidean geometry through the analytic meth-ods developed here. The methods of differential geometry provide the foundation on whichEuclidean and non-Euclidean geometry are special cases of a vastly richer set of geometricobjects, each with its own story to tell.

Though we have seen an end of the long-standing question of the Parallel Postulate, wehave by no means seen the end of the development of differential geometry. In the nextchapter we explore a few directions that we can follow from here.

Exercises

15.1 Suppose there is a geodesic mapping of a surface S to the plane. Taking the inverse ofthe mapping as coordinates on S, show that the induced metric on the plane satisfiesthe differential equations:

0.

15.2 By a change of scale on the coordinates u and v, show that a solution to the differentialequations given in Exercise 15.1 is the following set of metric functions where R anda are constants:

_ 2 a'- + u2 2 uu 2 (a2 + u2)

E -R

(a2 + u2 +v2)2. F = - R (a2

+ u2 +v2)2. G = R (a2

+ u2 +v2)2.

15.3 From the rectangular parametrization of the lower hemisphere of the sphere of radius Rcentered at the origin in R3 by the inverse of central projection, compute the componentfunctions 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 <

ds2 - (1 -v2)du2+2uvdudv+(I -u2)dv2a - (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 linethrough 0 not containing A. Suppose I is a line through A that does not intersect theu-axis but is not one of the two parallels through A to the u-axis. Show that ! and theu-axis share a perpendicular.

15.6 Prove the assertion for abstract surfaces made by Riemann in his celebrated Habilita-tionsvortrag that the metric

ds2 =due + dv2

1 I + . (u2 +u2)lz

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 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 thegeodesics that are not vertical lines conveniently. Recall that cosh2 t - sinh2 t = 1. Di-viding by cosh2 t and rearranging terms we get tanh2 t + sech2 t = 1. Since tanh t > 0for all real t, the following is a parametrization of the semicircle with center c on thereal 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 distancebetween 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) denotetheir cross ratio. Determine, in terms of w, all of the values taken on by the cross ratiowhen 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 theisometry 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 theEuclidean 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 centerand 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 innon-Euclidean trigonometry - Bolyai's theorem (Theorem 5.20):

sin A _ sin B _ sin C

sinh a sinh b 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 haveplayed 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 thePhilosophical Faculty of the University in Gottingen to fulfill the requirements of promotionto 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 forpromotion), and the third with the foundations of geometry. Against usual practice, Gausschose the third topic and Riemann offered the lecture Ueber die Hypothesen, welche derGeomerrie 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 ofRiemann's words follows this chapter.

The text of the lecture did not appear in Riemann's lifetime. Its posthumous publicationin 1868 (in the 1866/67 Abhandlungen der Koniglichen Gesellschaft der Wissenschaften zuGottingen) brought a nearly immediate response from the community of mathematiciansworking on differential geometry. The period of the subject driven by fundamental ele-mentary questions, especially non-Euclidean geometry, ends coincidentally in 1868 withthe publication of Beltrami's Saggio di interpretazione della geometria non-euclidea (Bel-trami 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 ofLobachevskii and Bolyai, and, in fact, he thought that such a space was impossible. Shortlyafter Riemann's lecture appeared, Beltrami published an analysis of spaces of constantcurvature in many dimensions (Beltrami 1869) based on Riemann's ideas and including adetailed version of non-Euclidean three-dimensional space.

The reader will find that the mathematical content of Riemann's lecture is sparse andcondensed. However, there was enough detail present to give direction to subsequent gen-erations of geometers. One of the key points is the separation of the point set propertiesof a higher-dimensional space from its possible metric properties as we have seen for anabstract surface with or without a Riemannian metric. Many metrics may be defined on asingle 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 philosoph-ical confusion over the primacy of Euclidean or non-Euclidean geometry. Riemann hadsteeped 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, theappropriate notion of a point set for higher dimensions.

Manifolds (differential topology)

What is a many-dimensional space? Riemann proposed the concept of "multiply extendedquantities;' which was refined for the next fifty years and is given today by the notion of amanifold.

Definition 16.1. An n-dimensional manifold is a set M (often denoted M") equippedwith 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 collec-tion A generates a maximal such set of charts with respect to this property calledan 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 fora topology on the set M. which is required to satisfy the Hausdorff condition. Thatis, if p and q are in M, then there are coordinate charts in the atlas y: V Mand =: 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 two-dimensional manifold.

If P E M" is a point in the image of a coordinate chart x: (U C R") -- Al, thenwe denote x-1 (p) = (X1, ... , .r" ). This superscript convention is part of the algebraicformalism of tensor analysis on which the modern local theory of manifolds is based. Wewill describe tensor analysis in more detail later.

EXAMPLES. (1) The simplest n-dimensional manifold is R" with the atlas generated by theidentity chart. More generally. the atlas containing the identity chart makes any open subsetof R" an n-dimensional manifold.

(2) All of the abstract surfaces of Chapter 14. and hence all of the classical surfaces in R3of Chapter 8, are two-dimensional manifolds.

(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 thesphere an (n - l)-dimensional manifold.

(4) Let RP" denote the set of lines through the origin in R"+1. One can define algebraiccoordinates for RP" through the equivalence relation (xo..... x") (rxo, , rx")whenever r # 0. In each equivalence class there are exactly two representatives such thatEi(xi)2 = I. Coordinate charts for S" that satisfy x(U) fl (-x(U)) = 0 determinecoordinate charts on RP". The n-dimensional manifold RP" with the atlas generated bythese charts is called the real projective n-space.

The notion of differentiable mappings between manifolds is a direct generalization ofthe definition for mappings between abstract surfaces, with the added richness of possiblydifferent 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 chartsxa : (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 differentiableif it is differentiable at every point p E M. A function 0: M -> M' is a diffeomorphismif 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 ofpin 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, definethe tangent vector to A at t = 0 as the linear mapping

A'(0): C00(p) -+ R, A'(0)(f) =dt (f o

11=0


The collection of all such linear mappings for all smooth curves through p is denoted byTp(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 Leibnizrule 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 chartwhose tangent vectors can be expressed as the partial derivative operators

a-, i = 1,...,nau'l

Definition 16.5. Given a differentiable mapping 0: M -+ M' and a point p in M, thedifferential 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 CR") -. M is a coordinate patch around p, x(t) = x(x1(t),....x"(t)), and y: (V CR") --, M' is a coordinate patch around O(p) with coordinates y(y', .... Y"). then

d0p(A'(0)) _ E(x')'(0)a(yo8x0ox)i

8ai./ .

It follows from the linearity and invertibility of the differential associated to a diffeomor-phism m: M -- M' that dim M = dim M'.

EXAMPLES. (5) A source of manifolds with beautiful properties is the family of groupsof matrices. In particular, we may consider the collection of all n x n matrices with realentries as the space R"2. The determinant mapping det : Rn2 - R is smooth and the inverseimage of the regular value 0 is a closed subset of R"2. The complement of det- 1 ((0)) is ann2-dimensional manifold, GL" (R). The symmetric n x n matrices, that is, those that satisfyA = A'. form a subset of

R'2of 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. Thisdetermines a subspace of the manifold GL, (R) of invertible symmetric matrices. Define themapping F: GL"(R) -> GL"(R) by F(A) = AA'. This mapping is smooth, and it takesa matrix and makes it symmetric since (A')' = A and (AB)' = B'A'. In the discussionat the end of Chapter 7, we proved that the group O(n) of rigid motions of R' is made upof invertible n x n matrices satisfying A-1 = A' (Proposition 7.12). This condition allowsus to express O(n) as F-1(1) where I is the identity matrix. To determine the dimensionof 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. SinceF(GL"(R)) lies in the symmetric matrices, the image of the tangent space has dimensionat most n(n + 1)/2. By applying a version of the implicit function theorem (Chapter 8) one


can prove that the dimension of O(n) is as large as it can be, that is, n(n - 1)/2, which isn2-n(n+ 1)/2, the dimension of the kernel ofd FA. Thus O(n) is an n(n- 1)/2-dimensionalmanifold.

(6) Given a manifold M" we can associate to it a 2n-dimensional manifold by "gluingtogether" all of the tangent spaces of M: Let TM = UPEM TT(M). A point in TM has theform (p, X) where p e M and X E Tp(M). If x : (U C R") -+ M is a coordinate chartcontaining p, then we can define a chart for TM containing (p, X) by letting i : (U x R" CR2') - TM be given by

a(x(xI n), Za .

ax'

To prove that we have a manifold, we check the transition functions between charts. Ify: (V C R") -- M is another chart containing p, then we can write p = x(x 1 , ... , x") _

n jy(y1, ... , y") and x' = xi (y1.... , y"), so

a = J'+ L- aThe effect on the coordi-

axi J=1 ax' ayj

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

J(.r-o},)=fJ(.r-toy) 0H 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 sothe transition functions are smooth and nondegenerate. Thus we have shown that T U is a2n-dimensional manifold.

The manifold TM is equipped with a smooth mapping A : TM -+ M given byrr (p. X) = p. We call this mapping n : TM - M the tangent bundle of M. The manifoldTM 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 generatedby a single chart, TR" may be expressed as R" x R", that is, a product. This simple formdoes not hold for most n-dimensional manifolds. The failure is often a consequence ofthe topology of the manifold. The study of the topological properties of manifolds is thethrust of differential topology, which has developed into an important area of mathematicsin the century after Riemann's lecture appeared (see Dubrovin et al. (1984,1985,1990) orGuillemin and Pollack (1974)).

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 mappingX : 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 mani-folds 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) _ a'ax,

for functions ai thati-

depend on p. Equivalently, if we choose a set of functions (at: U -+ 1t I I < i < n)on each coordinate chart x: U -+ M and require on the overlap of charts that the sets offunctions transform into one another coherently, then we have determined a vector field.The transformation rule for coherence is derived as follows: For two coordinate patchesx: U -+ M and y: V -+ M we may write

X(P) _ a1(p)ax, = av(P)aaY;i j

where a''U : U -. IR and aiV : V -+ )R are differentiable functions. By the chain rule wej

havea

= aY aand so, for all j,

i axi aa j'x yaY'

a`' = > 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, thatdetermines the transformation rule, not the geometry.

The coordinate charts provide enough analytic structure on a manifold to speak locallyof derivatives, vector fields, and integrals. In a given coordinate chart, such expressions maybe convenient to manipulate while being meaningless globally due to a lack of coherenceon overlapping charts. The problem we consider next is how to decide when a collection ofanalytic expressions is not dependent on the choice of coordinates. We first consider someexamples:

(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' = L9vaxY

iThis follows from the rule dx' _ Ek ax, dyk, familiar from the calculus. A col-

lection of sums >i BUdx', one for each coordinate chart, is called a (?)-tensor field

(or a 1-form) on M, if the coefficient functions on overlapping charts are related bythe transformation rule.

This example is important to the calculus on a manifold. If f : M -+ R is adifferentiable 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'ayj ax ayj'

i

where f (YI ..... Y") = BY I(x 1, ... , x"), ... Y"(x 1.... x')). To recover ffrom its derivatives, we apply Stokes's Theorem and integrate the I-form

df-`ax'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 componentfunctions 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 lineelement on overlapping charts as

ds2 =>g,sdx'dxs =

Lemma 14.12 implies that

ax' axsgij = E gis ayj

- - .r.s ayj

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 - , - R I 1 < i1..... is. jl.... , jr :.-S n)

constitutes an (")-tensor field on M if on overlapping coordinate charts, we have thetransformation rule

k,...k, i,...i, ayki ayk, axi' axi,TI,...t, = T;,...,, axi, axi, ayt, ayi, .

i, ,....1,.ji.... ,j,

where T denotes the aggregate of functions associated to the coordinate chart y: (V CR") - M.


In order to build the natural algebraic home for tensors we would need a sizeable detourthrough the linear algebra of tensor products and the dual and tensor product bundlesassociated to the tangent bundle. Riemann and his immediate successors did not have thesenotions, 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 impliesthat the tensor field does not depend on a choice of coordinates. The founder of this algebraicstructure 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 decalcul difft rentiel absolu et leur applications that appeared in 1901 in the MathematischeAnnalen at the invitation of Felix Klein. The absolute differential calculus of the title refersto the independence of choice of coordinates. The absolute differential calculus was renamedtensor analysis by Einstein in 1916. We will discuss the calculus of tensor fields in whatfollows.

Metrical relations (Riemannian manifolds)

In the second part of his lecture, Riemann turned to the problem of making elementarygeometric constructions possible on a manifold. In making such constructions, one requireslines 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 easilyfor any dimension.

Definition 16.8. A Riemannian metric on a manifold M is a choice of positive-definiteinner product ( , ) p on each tangent space Tp(M), for P E M, such that the choice variessmoothy from point to point. A manifold with a choice of Riemannian metric is called aRiemannian manifold.

For a particular coordinate chart x : (U C R") -+ M, the inner product may be repre-sented 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 functionsg,, form a (2)-tensor.

Lemma 16.9. If x : (U C R") - M and y: (V C R") -i M are two coordinate chartsfor 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 thematrix associated to v, and J(y -' o x) is the Jacobian matrix associated to the mappingv-' 0 x: x - ' (x (U) fl y(V )) -s y ' (x (U) n y(V )).


The proof is left to the reader. Component-wise, this matrix equation gives us

F,ayr ays

gii = L.rs ax' ax7r, s

that is, {gii) is a (0)-tensor field on M. The fact that the metric is positive definite impliesthat the matrix (gii) is invertible; we denote its inverse by (gki). The Jacobian is also aninvertible matrix:

J(y lox)laxil and soJ(y-lox)-J(x-1oy)=(aYj).

r k

These relations imply the identities Ekay,

axY' = SS and Ek gikgki = S where Sm is

the Kronecker delta function, S, = 1, and. if n m, Sm = 0. We may apply these rules toform new tensor fields.

Proposition 16.10. Suppose (T"""') is an field on M, an n-dimensional Rie-mannian manifold. Then the following expressions

nil...jp...it Til...i'_,nlip+,...1,Uxi1-J'gxm il...i,

m=1

Mvxil..i, LmTi,...

Jl... iq...i, JA...iq-ImJq+1... JM=1

constitute an ("+-I,)- and an (r-+:) -tensor field, respectively. The operation of contractionof indices on a tensor field is defined by

wil...op ...(, - Ti, ...Ip-.I mlp+I...j,il... iq...i, Jl...iq Imiq+'...i,

m=1

The contraction of an (')-tensor field is an -tensor field.

Here-

means to omit the index; for example, ii i2 ... in = i2 in.

PROOF. It suffices to demonstrate the transformation rules. We prove one case and leavethe others to the reader.

n Tkl...kp-lmkp+l...k,

m=1

axu (1xu il...i, ayklg,, a},x aym Til ...i, axi'

u.v.m.i1.....i,.A...-J,

k,Til...i, aygulp 11...i, axi'

8Vkp-1 aVm aVkp11 ayk. a.YJ1

axi, aytlaxi'ayt'axip-1 axip axip+'

aykp-I dVkp+1... axip-I

axip'ayk, axn axis axi'ax', aVx avtl

...avt.

Uil...jp...i, aVk1uil..i, aXll

a)'k,, ' aykp41. ayk, axu axJ1axip axip+I ... axiayx avtl

ax/'ayt'

u.i,.....i,


n ax, av'n uHere we used = = 3' and the definition of U.avm ax'o 0nr-1

These procedures of "raising;' "lowering," and contracting indices allow us to changethe types of tensors, sometimes arriving at geometrically significant tensors from other suchones (see Proposition 16.14).

In the presence of a Riemannian metric we define the arc length along a curve a(t) ina manifold M by

fs(1)= `p

and the angle 0 between curves a( (t) and A2(t) is determined by

cosh = (); (1p), A'2(tp))p

(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

The transformation rule (based on the chain rule) dyka y

= -dx 1, givesax ./

ds2 g,,dx' dx' y'd y'.r,l 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 Riemannianmanifolds

(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 thegeometry of Bolyai and Lobachevskii. Riemann attributed the confusion over Postulate Vto the assumption that the plane, as a two-dimensional manifold, has a unique geometry. Byseparating the metric relations from the space itself, he sought to eliminate the confusion. Forthree-dimensional space R3, Riemann showed that there are many metrics and so geometryas 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. HermannMinkowski (1864-1909) lectured to the 80th Assembly of the German Scientific Union inKiiln 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 ofexperimental physics, and therein lies their strength. They are radical. Henceforth spaceby itself and time by itself are doomed to fade away into mere shadows, and only a kindof union of the two will preserve an independent reality.

The union of space and time that lies at the basis of special relativity maybe described by aslight generalization of the Riemannian metric. We drop the condition that the inner productbe 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 (Lang1971) implies that, at a particular point p E M, there is a change of coordinates in whichthe line clement takes the form

ds2 = Eg;jdr'dx1 = >dvkdyk - dymdym.i k=1 m=/+I

In this case we say that the index of (g;1) is n - 1. A pseudo-Riemannian manifoldis a manifold with a nondegenerate metric of constant index. A Lorentz manifold is apseudo-Riemannian manifold with a metric of index 1.

The simplest Lorentz manifold was introduced in Minkowski's lecture. Minkowski spaceis 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 threeclasses

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 lightis constant in every frame of reference. A curve or world line of a physically meaningfulobject must have its tangent vector everywhere timelike or null in order to preserve thisassumption.

The study of differential geometry and, in particular. pseudo-Riemannian manifolds wasspurred on by the development of relativistic physics. The interested reader may consultO'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 ques-tion for Riemannian geometry called the local equivalence problem. Several constructionsmay 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 solutionto the local equivalence problem. Certain analytic questions from the study of the heatequation can also be reduced to this problem. On this question Riemann offered a countingargument that was a key step in later research.

A metric is determined by n(n + 1)/2 functions on each coordinate chart, namely, thecomponent 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)/2choices. Riemann argued that there must be some set of n(n - 1)/2 functions that determinethe metric. When n = 2, there are three functions, E. F. and G. that make up the com-ponents of the metric. TWo may be fixed by a change of coordinates, leaving one functionthat fixes the metric completely. That one function is the Gauss-Riemann curvature. Thework of Minding (Chapter 13) bears this out. The general case is answered by the Riemanncurvature 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 Frenchgeometer E. Cartan (1869-1951). The special cases of n = 2 and n = 3 were solved byG. 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 ho-mogener 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 theend of the paper, where he thanked Dedekind for making unpublished papers of Ricmannavailable. The Christoffel symbols (Chapter 10) were introduced in this paper. Christoffel'scomputations form the basis for the invariant approach to Riemannian geometry whichcharacterized 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,

aye a,gij = grs Vi axe

r.s

This expression may be differentiated to give the equation

ag.- - r ages aye ays ay` a'ye avs ayr a-v'axle . < < ay' ax' axe axt + ` s gas ( axi axle ax% + axi ax%axA

These derivatives may be permuted and summed with signs, to obtain the Christoffelsymbols of the first kind associated to (g, j ):

k ! 1

ag) ag,k _ agile[i 1=2 axk + axi ax'


These functions are sometimes denoted f jk,i . Now let [St. r] denote the Christoffel symbolsof the first kind associated to the metric Substituting the analogous expressions forthe [st, r] into the definition of the [jk, i] and simplifying we get

I agi; ag k agjk[Jk, i 1 =

2 axk + ax) - ax'-lay, ays ay' ayr a2xs= [s, r]

ax' axj aXk+ E grs

az' axjaxk .rs.l r.s

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

ayr ays ay' r- ayr a2ysr,k

r.s,r,l g[Sr r]

aX1 axe aXk +rr ; 9' grs aXt aXjaXk

ax' ax' ayr ays ay' ax' axr ayr a2 ys-'IS1, r]--- + E k""krs---ay ay" a.' aXi aXk ay" ay, axt axjaxk

", 1ays y' ax' ax' a2 ys_ [Sr. rJ-- + grs-

r.s.t.u aXj aXk ay" r.s.u ay" aXjaXk

r" ays ay' ax' + ax' a2 ys

_ S i u aXj aXk ay" S ays axiaxk'

where we have applied the relationsay°

ay,

= Sr and k"rkr, = S u. Fix an index

s = a; we can use the previous equation to isolate a second partial derivative.

a2ya a a2ys aya axr a2w'

axiaxk = Ea$ axiaxk - ax' (ays axiaxk )aya

(lT'r" ay., ay' axr

ax'ik _ s i u st xjs axle ay")

I' aya ra ays ayli /k ax' 3.1 s' axe aXk

We have proved the following result.

Lemma 16.11 (Christoffel 1869). Let r;k denote the Christoffel symbols of the secondkind associated to the metric (g j) and coordinates (x 1, .... x") and f sf those associatedwith (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,

82yaaxjaxk Jk ax' s.f s` axj aXk

.

This equation shows that the second mixed partials of the change of coordinates areexpressible in terms of first partials and Christoffel symbols. It follows that this is true ofall higher iterated partial derivatives. By his previous work on invariant theory. Christoffelreduced the local equivalence problem to an algebraic problem that he had already solvedin 1864.

To discover the Riemann curvature we take another derivative.

a3 ya (arjk aya + rm a2 ya

axjaxkaxl = 1: ( axl ax"' kax"'ax

ataf ay" ays ayf a2ys ayf ay., a2yf

- ayu axl axj aXk - St axjaxl aXk + aXj aXkaXls.f.u

Expanding the second partial derivatives according to Lemma 16.11, we obtain the (some-what horrifying) expression

a3yk I _ arm k 8ymk

`3ya

aXjax aX = aX ax +r rMI

a ayf ay' ara ay" a c ay'rjkmrs' _aX"f x, s ayu Dxt Dxi aXk

ra rr ays ay' + ra , ay" ay' av'rs.r s` )l aXr aXk s,f,u,u

sf UV -,XI aXl aXk

-ra rr

ayf ays +Tar'

ayuDv" art

r.s.fsf kf ax" Dxl su S I uu aXk axl aXj

The point of all this computation is to compare what we get when we equate the mixed3 a 3 a

partialsaxk

(recall that the Gauss equations leading to theorema

egregium arose this way). Set on the left everything that is made up of expressions relatedto the coordinate chart x, and on the right everything related to y; we obtain

!rj[( aXlk aXkl/

+ (rjkl'rl - r;ll'rk)j ax"'r

r are; aru a a -", ays avf av= s I (ayu - (Iy,) + (Fuf,rsf - HIV Su I) axj a.rk a.rl

m

If we multiply both sides byax

aand sum over all a, we obtain the equation

m

aXIk - aXk +> (r%krrt - r;,r,0r

_ arsr _ arsu ' i a o a al'S aye ay° axe"ua (ay° ay +v' (rstruu - I'Surr.) ax%axk axe aya

Definition 16.12. The Rlemann curvature tensor is the (j)-tensor field given by (RRtkI < m, j, k, l < n), where

mrRrk - [(axrk axkr) + `r', rrl - r; rrk

r

The previous derivation gives the transformation rule:

R,rkmaay`ay`ayuax'"s.r.u.a

RSul a% axk ax, ay,a

A necessary condition for a solution to the local equivalence problem emerges fromthe transformation rule - in order for two Riemannian metrics to be transformable, oneto 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 thecomponent functions of the metric and their derivatives, we can test a pair of metrics forequivalence directly.

A special case of the local equivalence problem was treated in one of Riemann's unpub-lished works, his Paris Academy paper of 1861. The problem set by the Academy in 1858concerned heat conduction and systems of isothermal curves. In his paper Riemann recastthe problem in geometric terms which required that a given metric be transformed into thestandard Euclidean metric. His essay lacked certain details and the Academy decided notto award the prize until the essay was revised. Riemann's failing health prevented him fromcompleting the essay and the prize was withdrawn in 1868.

If a metric takes the form of the standard Euclidean metric in a coordinate system(y" y"), then the Christoffel symbols P' = 0 for all a, s, and t. Thus a necessarycondition for a metric (gig) to be transformable to the Euclidean metric is the vanishing ofthe 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 followingRiemann's formulation. If (x 1.... , x") give the coordinates for the given metric (gjj ), thenlet ()''..... y") be the desired coordinates for the standard Euclidean metric (kid = Sid).Riemann derived the particular case of Lemma 16.11, namely,

av=

a ayJ3fir'- axtr=I tk aXr aXJaxk = 3 5


We can write this in a convenient matrix form as the equation

rik rik ... rik aYg lax' aY /ax'

r2k r2k r2k avR/ax2 a

axk

aye/ax2

rnk rnk... rnk aya/ax" aye/ax"

or Fk(x)(°y0) = 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 bythe equation of mixed partials which leads to the Riemann curvature tensor:

aat axk -azk aat (°)",

which implies Rjmlk = 0 for all m, j, k, and 1.

To prove sufficiency one generalizes a familiar integrability result from two dimensions:

FACT. If f, g: R2 R satisfy of =ax

in an open set containing (0, 0), then, for

zo E R, there is a function F: (U (: R2) - R with U an open set containing (0, 0) suchthat 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. Werefer the reader to Spivak (1970, vol. 2) for a choice of seven proofs.

Theorem 16.13 (Riemann 1861). An n-dimensional Riemannian manifold for whichall the functions making up the Riemann curvature tensor vanish, that is, RRkt = O for allchoices of i, j. k, and 1, is locally isometric to R" with the standard Riemannian metricds2 = E. dx'dx'.

A Riemannian manifold that satisfies 0 for all i. j. k. and / is called flat. TheRiemann curvature tensor measures the deviation from flatness of a manifold.

Associated to the Riemann curvature tensor is a (4)-tensor field obtained by lowering anindex

R(jkt = grr

r-1

Notice that if Rjkt = 0 for all choices of indices, then R;jkt = 0 as well. We will tease outa 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


manifold. Suppose X and Y are linearly independent tangent vectors at a point p in an n-dimensional Riemannian manifold M. We may choose the vectors X and Y as the coordinate

vectors asli

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 thatpart of the x'x2-plane in R" lying in U. By applying linear isomorphisms to R" we canalways arrange for any given coordinate chart around p to have such properties. The rest ofthe charts for M transform smoothly when restricted to the appropriate subspaces of theirdomains and so we obtain an abstract surface SXy containing p. Since the tangent plane ofSXy is a linear subspace of the tangent space of M at p, we may restrict the metric on M toa 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 pin an Riemannian manifold M, the Gauss-Riemann curvature at p of thesurface SXy, denoted KXy (p), is given by

81212XyP -EG-F2'

where R1212 is a particular component of the lowered Riemann curvature tensor restricted

=to SXy and E = g11 = (axi axl) ' F = 912 = axl 'aXz) and G = 922P P

a a )P.

1TX2' TX-2

PROOF. The proposition is a restatement of the formulas from Theorenra Egregium as theyapply to abstract surfaces. To wit we have

1 1_ arzz _ ar21 1 1 1 2 1 2 1

R2,2 ax1 aXz + (r22r1, - r12r12 + r22r12 - r12rz2) = GK

ar2 are8212 = 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 inthe directions X and Y.

Not all of the component functions Rjkt or Rijkt of the forms of the Riemann curvaturetensor are independent. In fact, there are many symmetries indicating that a deeper algebraicstructure must be enjoyed by this tensor. We record these symmetries in the followingproposition. The proof is left to the reader.

Proposition 16.15. The following equations hold:

(1) Rjkt = Rijkt = -Rjikt(2) Rjkt + Rktj + RI. = 0.Ijk(3) Rijkt = Rktij


By exploiting the symmetries of the curvature, one proves that then(n2

1) = (n)

functions R,,,, determine R,jtt and hence R" k1. These functions are the missing data thatdetermine a metric conjectured to exist by Riemann.

A Riemannian manifold is said to have constant curvature if the sectional curvaturesKXy = c, a constant, for all linearly independent X and Y. Constant Gauss-Riemanncurvature 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 theirconfiguration, so that every line can be measured by every other." The local equivalenceproblem and a version of the exponential mapping imply that such a manifold M has constantcurvature. Riemann also proposed examples of manifolds of constant curvature.

Proposition 16.16. The subset ((x 1.... , xn) I 1 + 4 E, (X,)2 > 0) of Rn with themetric

ds -E; dx'dx'

=z

t l + 4 r (x` )2

is an n-dimensional Riemannian manifold of constant curvature.

PROOF The component functions of this metric may be written g,,a,,

= F,, , where F =

I + 4 (x')2. If we write f = In F, then we have

ag,, - 2 a F -28,, of3Xk F3 axk = F2 ark .

This leads to the following expression for the Christoffel symbols of the first kind:

1 of of of 1[jk. i ] F,2 S,

axt+ S,t

axJ S, ax' )

The Christoffel symbols of the second kind are given by

Clt = Eg+m[jk nt] _ F`S'm[jk.ml = F2[jk, i].,n=1 nr=1

Similarly we can relate the lowered form of the Riemann curvature tensor with its usualform by

n n

L'n+ n I

r rR,1,, = gmRJ,t = F,2 RJ,J =F2

R),1.M=1 n,=1

To compute the sectional curvature we use the fact that Kxy(p) is linear in X and Y and


so it suffices to compute K a where i j.

ari.. ar!.Ri =.. Ji- -r'" r'iij axi axi M

ii M, J, 1-i

a2f + a2faxiaxi axiaxi

n

+ E F°(Ijj, m]Imi, i) - [ii, m)Imj, i))M=1

a2f= axiati + axiaxi + (axi

)2+ (af\21af2sIaY )

In our case f(x1, ... , x") = In l I + a (Xs)2) from which it follows that` 4 s=I

of _ axsand

a2f= , (2aF - (ax"')2).

ax' 2F' a.r'axs 4F-

Substituting this into the preceding expression we obtain Riiij = a/F°, and since gii =gjj = I/ F2,

K = Ril1ia a - =a.

To '. giigjj

When n = 3 and a = -I we obtain a model for Bolyai's and Lobachevskii's non-Euclidean space. In particular, the model is represented by the interior of the ball of radius 2in R3 and so, topologically, it is simply connected. Furthermore, the boundary of the modellies an infinite distance from any point in the space and so it is complete. The argumentsof Bolyai and Lobachevskii are realized in this manifold by restricting to planar sectionsof the ball; these are non-Euclidean planes that are isometric to the Poincart disk. Theseremarks were established by Beltrami in his paper on spaces of constant curvature (Beltrami1869). He also presented new examples of constant-curvature manifolds by applying thehigher-dimensional analogs of stereographic and orthographic projection. In particular, heidentified 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 theplanes given by ((x1 , x2, c) I c a positive constant) are horospheres and it is immediatefrom the form of the line element that the induced geometry is Euclidean. Thus Beltramishowed analytically that the Euclidean plane lies inside a three-dimensional non-Euclideanspace (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 toa manifold. To extend our understanding of these analytic expressions, we turn next toderivatives of the functions making up a tensor field. First consider a vector field givenlocally on coordinate charts x : (U C I[.') ---> M and r: (V C r) -+ M by X =

ai; aa, = > ai- Suppose we take the partial derivative of a'. with respect toa I,]

a coordinate direction This operation implies a transformation rule as follows: Since_ r r a1'ar

atax,.

aa'`. aa',, ax' a i , 02 i,r a.x'

ai.9 axj ayy a.Y' +a ax'a.YI ati'y

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 a would transform as a (:)-tensor. In the preliminary stages of his devel-aVq

opment of general relativity, Einstein struggled with this problem, and for a brief time hepostulated that the only physically interesting changes of coordinates were linear.

The restriction to linear changes of coordinates is unreasonable for geometry and physicsand so another approach is needed here. Let us expand the previous expression with thehelp of Lemma 16.11.

aai aa'L a.Y, al?, 321.r ax'a1 (! ax, avy ax' +RUa.Y'a.Y! al'g

aat; ax aI"*

kv , a , a v' ax,axl aY'

+aL, r",a.Y! rs'ax' axj a>y

as'ax, aV ai.ivy

where we have used r,, al a'-. Rearranging the terms we findax,

J°1' , r aaU 1 r; , 1 av" a.Ytdry

+a f,y = E axe +aL,i fax' a}'9

aar. /that is, the expression

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.


The case of a (1)-tensor (0j) is similar. The analogous calculation leads to the expression

ae; F Br fr9i:j=axj-r

Oil

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 offunctions that behaves as a well-defined field (of tensors) on a manifold, and the covariantderivative once again determines a well-defined field on the manifold.

Finally, we give the definition of the covariant derivative for a general tensor field. Itis made up of parts depending on the upper and lower indices and behaves like a partialderivative following the Leibniz rule with respect to multiplication by a function on themanifold.

::. ,Definition 16.18. The covariant derivative with respect to xk ofan (;)-tensor I

is given by

ar''..., r n

axk+ T'I...j.

1=1 m=1s It

A=1 µaI

++

rkm

The functions (T.'' '' constitute an (S )-tensor.

So far we have presented a formalism to generalize the partial derivative and stay withinthe space of tensors. To put some geometry back into the discussion let us consider aspecial case of interest, geodesics on an n-dimensional manifold M. We begin by recallingthe 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 vectormay be written

'(t) _ dx' a dyj aa

, dt axi

=,1 dt ayj'


dV'- dC' ayrand the ordinary rules of differentiation implydt = 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 d2 r' al .r d.-(' dxJ (121-r

dt-' dt'- ax' + dt lit ax'a.Y'r r.J

d'x' a'r dx' dxt BV' ay'' aV'E - (F-dt' ax' + dt dt " a.Ck - r" ax' axt

From this expression we obtain

d2 i" r dy'' dy' d'xt t dx' dx1 aVr

dt'- r°' d d = dt' + 'J dt dt axi .

This is the defining property for covariant differentiation with respect tot; it takes a particularvector field, the field of tangents to the curve, and gives back a vector field of secondderivatives. We denote this operation by

Da' d'-xt dx' dxt adt - ddt dt) ait

We generalize the covariant derivative along a curve a(t) to give the derivative of a vectorfield 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 = a' ar definedonaneighborhood contain-sing the trace of a (t), the covariant derivative of X along a (t) is given by the expression

DX da' d.Y't a

dt -' 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 thisformula we prove a somewhat technical lemma. This result lies at the heart of the derivationof 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)

PROOF. We simply compute

1

8j(dY ` r

k ! dj

kdd dr dti, jdl + L+

k.f

d2x' dx' i dxj dxk dx'&j t2 di + &jrkl dt dt dt

d2x' dxj dxi dxk dx'= gii dt2 dt + F(kl, j'1

dr dtT

'.1 j.k.t

d2x' dxi 18gjk dxi dxk dx'= gii dr2 dt +E 2 axk dt dt dr

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.

This property ties curves withDa'

= 0 to arc length. Analysis similar to the proof of

Theorem 11.6 shows that such curves in fact provide the shortest distance between pointsalong 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 aresatisfied for i = 1..... n:

d2x' dxi dxk

dt + F_ rjk dr dt= 0.

j.kAll 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 andthe special coordinate systems based on exp.

The geodesic equations bear a formal identity with the equations arrived at through thegeodesic curvature definition for geodesics on surfaces in R3. Is there a sense in which theseequations and covariant differentiation are "geometric"? The last stage of the developmentof differential geometry that we will discuss in this book begins with an answer to thisquestion. 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 pointis simply a copy of fl and we may identify every Tp(Il) as Il. This makes taking thederivative of a vector field simple. If X : fl - T Il is a vector field and x : (U C R2) -+ Ila coordinate chart, then X(x(a, b)) and X(x(a + t. b)) can be thought of as being in thesame plane. Thus

X(x(a + t, b)) - X(x(a, b))=

ax1-o t

ax'(x(a.b))

makes sense.

When we consider a more general surface S C R3, the vectors X(x(a + t. b)) andX(x(a, h)) are in different tangent planes which are usually not parallel. Hence the dif-ference quotient with limit definition of the derivative does not make sense. Every tangentplane to the surface S. however, is abstractly isomorphic to every other. The problem is todecide 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 bygeneralizing the parallelism enjoyed by Euclidean space. A similar approach was arrivedat 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 Rie-maannian 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 thecovariant derivative in a Euclidean space coincides with an ordinary derivative, we see thata vector field is parallel along a curve if it is constant along it. To see how this leads to aninterpretation 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 ina 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 (seeSpivak (1970, vol. 2)). We note that the equations are linear and so if X = rxi + X2. thenX = 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. Thusour definition makes sense. The mapping r,,. is called parallel transport along a(t) and itis a linear isomorphism. The inverse of r,, is defined by reversing the curve and linearityfollows 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-dimensionalRiemannian 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 art

defined in a neighborhood

around p, we have

Et d rte (X((-t(h)))-X(p)a s

d.rt 0r- h


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)+ Ebi(h. 0L C' If'f(ck(tD = 0.

,./

Recall that the derivative with respect to t along ck (t) is the partial derivative with respectto xk so this differential equation becomes

abt (h, 1)+ 'V'bj(h t)I'k (ck(t)) = 0, 1 .

axk i

By the mean value theorem we may write

bt (h, h) = bt (h, 0) + hdb'(h, t) I - , for some ur E (0, h).

t-u,

With this in mind, we compute

limrh'(X(ck(h))) - X(ck(0)) = lim (bt(h,0) -a(ck(0))) a

h-0 h h-.o It ax1

= limoh

V" (br(h. h) - hdbt(t

t) Ia

- at(ck(0))) aartU

at(ck(h)) - at(ck(0)) dbt(h.1) a- h-0

h dt I,_ axr

- (aal

axk+h-0 bt(h,ur)l1j(ck(ur)) Xr

,r 5-a

= ` axk + ua/r j XIr i

r a=Ea:kax

The parallelism introduced by Levi-Civita has some further properties that are exploredin the exercises. Not only does it give an interpretation of the covariant derivative, but it

uncovers a new foundation for differential geometry. At the heart of our flurry of indices isChristoffel's lemma (16.11) which gives a kind of transformation rule for the rilk:

a2 a- L. rik aya ayIax ay,%ax ax aXj ax

Suppose (G'. I I < i. j, k < n) is a collection of n3 functions that satisfy the sametransformation rule as the r, jk. Using the theory of differential equations as a tool, one candevelop a notion of covariant differentiation and parallel transport by following the formaldevelopment 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. Inparticular, this choice of structure is prior to the choice of Riemannian metric and belongs tothe analytic machinery associated to the manifold. This level of generality was introducedby H. Weyl (1885-1955) in 1918, who was motivated by his attempt to unify relativity andelectromagnetism.

Among the possible connections associated to a Riemannian manifold, the one based ona choice of Riemannian metric and its associated Christoffel symbols is unique. This is thecontent of the following result:

Theorem 16.25 (fundamental theorem of Riemannian geometry). Given aRiemannian manifold M there is one and only one connection on M such that paralleltransport 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 thecanonical connection on (M, ( , )).

The coordinate viewpoint that we have presented gave way in the twentieth century tocoordinate-free constructions that do the work of the tensor calculus globally. The basic ideasfor these developments lie in Levi-Civita's work and parallelism. We close with Gauss'smotto from his Copenhagen Preisschrift, 1822, a paper in which he solves "the problem ofmapping the parts of a given surface onto another such that the image and the mapped partare similar in the smallest parts."

Ab his via sternitur ad maioraFrom 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 currentformulation of differential geometry with a sense of where the subject has been and a deeperintuition 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 ofdimension m + n.

16.3 Show directly that the upper half-space,

(H" = ((x' .... , x") E R" I x" > 0), ds2 = (dx1)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 tensorof 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 <i,j,k<n.16.7 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 thefollowing identity due to Ricci:

- ' iai:I:k - 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) xr(TM) - 1'(T M) is defined by

a k a

V(52;7 .

ax) Gi;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, forthe choice of the canonical connection (f j'k) on a Riemannian manifold, the covariantderivative 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 basicprinciples for constructions in space. It gives only nominal definitions of these things,while their essential specifications appear in the form of axioms. The relationship betweenthese presuppositions is left in the dark; one does not sec whether, or to what extent, anyconnection between them is necessary, or a priori whether any connection between them ispossible.

From Euclid to Legendre, to name the most famous of the modern reformers of geome-try. this darkness has been dispelled neither by the mathematicians nor by the philosopherswho have concerned themselves with it. This is undoubtedly because the general conceptof multiply extended quantities, which includes spatial quantities, remains completely un-explored. I have therefore first set myself the task of constructing the concept of a multiplyextended quantity from general notions of quantity. It will be shown that a multiply extendedquantity is susceptible of various metric relations, so that Space constitutes only a specialcase of a triply extended quantity. From this however it is a necessary consequence that thetheorems of geometry cannot be deduced from general notions of quantity, but that thoseproperties that distinguish Space from other conceivable triply extended quantities can onlybe inferred from experience. Thus arises the problem of seeking out the simplest data fromwhich the metric relations of Space can be determined, a problem which by its very natureis not completely determined, for there may be several systems of simple data that sufficeto determine the metric relations of Space; for the present purposes, the most importantsystem is that laid down as a foundation of geometry by Euclid. These data are - like alldata - not necessary. but only of empirical certainty, they are hypotheses; one can thereforeinvestigate 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 immeasurablysmall.

1. Concept of an n-fold extended quantity

In attempting to solve the first of these problems. the development of the concept of multiplyextended quantity, I feel particularly entitled to request an indulgent criticism, as I am littletrained in these tasks of a philosophical nature where the difficulties lie more in the conceptsthan in the construction, and because I could not make use of any previous studies, exceptfor some very brief hints that Privy Councillor Gauss has given in his second memoiron biquadratic residues, in the Gottingen Gelehrte Anzeige in his Jubilee-book, and somephilosophical 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 whichallows different realizations. Depending on whether or not a continuous transition of in-stances can be found between any two of them, these realizations form either a continuousor a discrete manifold; individual instances in the first case are called points and in the lattercase elements of the manifold. Concepts whose instances form a discrete manifold are sonumerous that some concept can always be found, at least in the more highly developed lan-guages, under which any given collection of things can be comprehended (and in the studyof discrete quantities, mathematicians could unhesitatingly proceed from the principle thatgiven objects are to be regarded as all of one kind). On the other hand, reasons for creatingconcepts whose instances form a continuous manifold occur so seldom in everyday lifethat the position of sensible objects and colors are perhaps the only simple concepts whoseinstances form a multiply extended manifold. More numerous reasons for the generationand development of these concepts first occur in higher mathematics.

Distinct portions of a manifold, distinguished by a mark or by a boundary, are calledquanta. Their quantitative comparison is effected in the case of discrete quantities by count-ing, in the case of continuous quantities by measurement. Measuring involves the super-position of the quantities to be compared; it therefore requires a means of transporting onequantity to be used as a standard for the others. Otherwise, one can compare two quan-tities 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 ageneral division of the science of quantity, independent of measurement, where quantitiesare regarded, not as existing independent of position and not as expressible in terms of aunit, but as regions in a manifold. Such investigations have become a necessity for severalparts of mathematics, for the treatment of many-valued analytic functions, and the dearthof such studies is one of the principal reasons why the celebrated Theorem of Abel and thecontributions of Lagrange, Pfaff and Jacobi to the general theory of differential equationshave remained unfruitful for so long. From this general part of the science of extendedquantity 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 makeclear the generation of the concept of a multiply extended manifold, the second reducingposition fixing in a given manifold to numerical determinations and will make clear theessential character of an n-fold extension.

2

In a concept whose instances form a continuous manifold, if one passes from one instanceto another in a well-determined way, the instances through which one has passed forma simply extended manifold, whose essential characteristic is that from a point in it acontinuous movement is possible in only two directions, forwards and backwards. If onenow imagines that this manifold passes to another, completely different one, and once againin a well-determined way, that is, so that every point passes to a well-determined point ofthe 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 cancontinue 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 asa synthesis of a variability of n + I dimensions from a variability of n dimensions and avariability 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 oneconsiders an arbitrary piece of a manifold of one dimension - with a fixed origin, so thatpoints of it may be compared with one another - varying so that for every point of the givenmanifold it has a definite value, continuously changing with this point, or in other words, onetakes within the given manifold a continuous function of position, and, moreover, one suchfunction that is not constant along any part of the manifold. Every system of points where thefunction has a constant value then forms a continuous manifold of fewer dimensions than thegiven 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 speakingthis will occur in such a way that every point of the first passes to a definite point of anyother: 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 numericaldetermination and to the determination of position in a manifold of fewer dimensions. Itis now easy to show that this manifold has n - I dimensions, if the given manifold is ann-fold extension. By an n-fold repetition of this process, the determination of position inan n-fold extended manifold is reduced to n numerical determinations, and therefore thedetermination of position in a given manifold is reduced, whenever this is possible, to afinite number of numerical determinations. There are, however, also manifolds in which thefixing of position requires not a finite number, but either an infinite sequence ora continuousmanifold of numerical measurements. Such manifolds form, for example. the possibilitiesfor 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, onthe 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 essentialcharacteristic has been found in the fact that position fixing in the manifold can be reducedton 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 theconditions that suffice to determine them. These metric relations can be investigated only inabstract terms, and represented in context only through formulas: under certain assumptions,however, one can resolve them into relations which are individually capable of geometricrepresentation. and in this way it becomes possible to express the results of calculationgeometrically. Thus, to put this work on solid ground, although an abstract investigationwith formulas certainly cannot be avoided, the results can be presented in geometric garb.


The foundations of both parts of the question are contained in the celebrated treatise ofPrivy Councillor Gauss on curved surfaces.

1

Measurement requires an independence of quantity from position, which can occur in morethan one way; the hypothesis that next presents itself, and which I shall develop here, is justthis, that the length of lines be independent of their configuration, so that every line can bemeasured by every other. If fixing of position is reduced to numerical determinations, sothat the position of a point in the given n-fold extended manifold is expressed by n varyingquantities xi, x2, x3, and so forth up to x,,, then specifying a line amounts to giving thequantities x as functions of one variable. The problem then is to set up a mathematicalexpression for the length of a line, for which purpose the quantities x must be thought of asexpressible in units. I will treat this problem only under certain restrictions, and I first limitmyself to lines in which the ratios of the quantities dx - the increments in the quantitiesx - vary continuously; one can then regard the lines as broken up into elements withinwhich the ratios of the quantities dx may be considered to be constant and the problem thenreduces to setting up at every point a general expression for the line element ds, which willtherefore contain the quantities x and the quantities dx. I assume, secondly, that the lengthof the line element remains unchanged, up to first order, when all the points of this lineelement suffer the same infinitesimal displacement, whereby I simply mean that if all thequantities dx increase in the same ratio, the line element changes by the same ratio. Underthese assumptions, the line element can be an arbitrary homogeneous function of the firstdegree 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 simplestcases, I next seek an expression for the (n -1)-fold extended manifolds that are everywhereequidistant from the origin of the line element, that is, I seek a continuous function ofposition that distinguishes them from one another. This must either decrease or increasein all directions from the origin; I want to assume that it increases in all directions andtherefore has a minimum at the origin. Then if its first and second differential quotientsare finite, the first-order differential must vanish and the second-order differential cannotbe negative; I assume that it is always positive. This differential expression of the secondorder remains constant if ds remains constant and increases quadratically if the quantitiesdx, and thus also ds, all increase in the same ratio; it is therefore equal to a constant timesds2, and consequently ds equals the square root of an everywhere positive homogeneousfunction of the second degree in the quantities dx, in which the coefficients are continuousfunctions 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. Thenext simplest case would perhaps include the manifolds in which the line element may beexpressed as the fourth root of a differential expression of the fourth degree. Investigationof this more general class would actually require no essentially different principles, but itwould be rather time consuming and throw proportionally little new light on the study ofSpace, especially if the results cannot be expressed geometrically; I consequently restrictmyself to those manifolds where the line element can be expressed by the square root of

Riemann's Habilitationsvortrag 273

a differential expression of the second degree. One can transform such an expression intoanother similar one by substituting for the n independent variables, functions of n newindependent variables. However, one cannot transform any expression into any other inthis way; for the expression contains n(n + 1)/2 coefficients which are arbitrary functionsof the independent variables; by the introduction of new variables one can satisfy only nconditions, and can therefore make only n of the coefficients equal to given quantities. Thereremain n (n - 1)/2 others, already completely determined by the nature of the manifold to berepresented, and consequently n (n - 1)/2 functions of position are required to determine itsmetric 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 beinvestigated 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 ofindependent differentials, I propose to call flat. In order to survey the essential differencesof all the manifolds representable in the assumed form, it is necessary to eliminate thefeatures depending on the mode of presentation, which is accomplished by choosing thevariable quantities according to a definite principle.

2

For this purpose, suppose one constructs the system of shortest lines emanating from agiven point; the position of an arbitrary point can then be determined by the direction ofthe 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 quantitiesdx at the origin of this shortest line, and by the length s of this line. In place of thedx° one now introduces linear expressions At formed from them in such a way that theinitial value of the square of the line element will be equal to the sum of the squares ofthese expressions, so that the independent variables arc: the quantity s and the ratio of thequantities At; finally, in place of the da choose quantities xi, x2, ... , .r proportionalto 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 thenext order term in its expansion equals a homogeneous expression of the second degreein the n (n - 1)/2 quantities (x i dx2 - x2dxi ). (x i dx3 - x3dx, ), .... and is consequentlyan infinitely small quantity of the fourth order, so that one obtains a finite quantity if onedivides it by the square of the infinitely small triangle at whose vertices variables have thevalues (0, 0. 0, ... ). (x, , x2, X3, ... ), (dx,, dx2, dx3.... ). This quantity remains the sameas long as the quantities x and dx are contained in the same binary linear forms, or as longas the two shortest lines from the initial point to x and from the initial point to dx remainin the same surface element, and therefore depends only on the position and direction ofthat element. It obviously equals zero if the manifold in question is flat, that is, if the squareof the line element is reducible to E(dx)2, and can therefore be regarded as the measureof deviation from flatness in this surface direction at this point. When multiplied by -3/4it becomes equal to the quantity that Privy Councillor Gauss has called the curvature ofa surface. Previously, n(n - 1)/2 functions of position were found necessary in order todetermine 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, thenthe metric relations of the manifold may be determined, provided only that no identicalrelations can be found between these values, and indeed in general this does not occur. Themetric relations of these manifolds, in which the line element can be represented as thesquare root of a differential expression of the second degree, can thus be expressed in away completely independent of the choice of the varying quantities. A similar path to thesame goal could also be taken in those manifolds in which the line element is expressed ina less simple way, for example, by the fourth root of a differential expression of the fourthdegree. The line element in this more general case would not be reducible to the square rootof a quadratic sum of differential expressions, and therefore in the expression for the squareof the line element the deviation from flatness would be an infinitely small quantity of thesecond dimension, whereas for the other manifolds it was an infinitely small quantity ofthe fourth dimension. This peculiarity of the latter manifolds therefore might well be calledflatness in the smallest parts. For present purposes, however, the most important peculiarityof these manifolds, on whose account alone they have been examined here, is this, that themetric relations of the doubly extended ones can be represented geometrically by surfacesand those of the multiply extended ones can be reduced to those of the surfaces containedwithin them, which still requires a brief discussion.

3

In the conception of surfaces the inner metric relations, which involve only the lengths ofpaths within them, are always bound up with the way the surfaces are situated with respectto points outside them. One can, however, abstract from external relations by consideringdeformations that leave the lengths of lines within the surfaces unaltered, that is, by con-sidering arbitrary bendings - without stretching - of such surfaces, and by regarding allsurfaces obtained from one another in this way as equivalent. For example, it follows thatarbitrary cylindrical or conical surfaces are equivalent to a plane, since they can be formedfrom 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 theother hand, they count as essentially different from the sphere, which cannot be transformedinto the plane without stretching. According to the previous investigations, the inner metricrelation at every point of a doubly extended quantity, if its line element can be expressedas the square root of a differential expression of the second degree, which is the case withsurfaces, is characterized by the curvature. This quantity can be given a visual interpre-tation for surfaces as the product of the two curvatures of the surface at this point, or bythe fact that its product with an infinitely small triangle formed from shortest lines is, inproportion to the radius, half the excess of the sum of its angles over two right angles. Thefirst definition would presuppose the theorem that the product of the two radii of curvaturesis unaltered by mere bendings of a surface, the second, that at each point the excess overtwo right angles of the sum of the angles of any infinitely small triangle is proportional toits area. To give a tangible meaning to the curvature of an n-fold extended manifold at agiven point and in a given surface direction through it, one must proceed from the fact that ashortest line emanating from a point is completely determined if its initial direction is given.


Consequently one obtains a certain surface if one prolongs all the initial directions from thegiven point that lie in the given surface element into shortest lines, and this surface has adefinite curvature at the given point, which is equal to the curvature of the n-fold extendedmanifold at the given point in the given surface direction.

4

Before applying these results to Space, it is still necessary to make some general consider-ations about flat manifolds, that is, about manifolds in which the square of the line elementcan 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 itsuffices to know that at each point the curvature is zero in n(n - 1)/2 independent surfacedirections. The manifolds whose curvature is everywhere zero can be considered as a specialcase of those manifolds whose curvature is everywhere constant. The common characterof those manifolds whose curvature is constant can be expressed as follows: Figures canbe moved in them without stretching. For obviously figures could not be freely shifted androtated in them if the curvature were not the same in all directions, at all points. On the otherhand, the metric properties of the manifold are completely determined by the curvature; theyare therefore exactly the same in all the directions around any one point as in the directionsaround any other, and thus the same constructions can be effected starting from either, andconsequently, in the manifolds with constant curvature figures can be given any arbitraryposition. 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 valueby a, then the expression for the line element can be put in the form

I E(dx)2I+Or 24

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 asphere whose radius is the reciprocal of the curvature; but in order to survey the multiplicityof these surfaces, let one of them be given the shape of a sphere, and the others the shapeof surfaces of rotation that touch it along the equator. The surfaces with greater curvaturethan the sphere will then touch the sphere from inside and take a form like the portionof the surface of a ring, which is situated away from the axis; they could be developedupon zones of spheres with smaller radii, but would go round more than once. Surfaceswith smaller positive curvature are obtained from spheres of larger radii by cutting out aportion bounded by two great semicircles, and bringing together the cut-lines. The surfaceof curvature zero will be a cylinder standing on the equator; the surfaces with negativecurvature will touch this cylinder from outside and be formed like the part of the surface ofa ring which is situated near the axis. If one regards these surfaces as possible positions forpieces of surface moving in them, as Space is for bodies, then pieces of surface can be moved


in all these surfaces without stretching. The surfaces with positive curvature can always beso formed that pieces of surface can even be moved arbitrarily without bending, namely onspherical surfaces, but those with negative curvature cannot. Aside from this independenceof position for surface pieces, in surfaces with zero curvature an independence of positionfor 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-foldextended quantity, the conditions may be given which are sufficient and necessary for thedetermination of the metric relations of Space, if we assume the independence of lines fromconfiguration and the representability of the line element as the square root of a second-orderdifferential expression, that is. flatness in the smallest parts.

First, these conditions may be expressed by saying that the curvature at every pointequals zero in three surface directions, and thus the metric relations of Space are impliedif the sum of the angles of a triangle always equals two right angles. But secondly, if oneassumes with Euclid not only the existence of lines independently of configuration, but alsoof bodies, then it follows that the curvature is everywhere constant, and the angle sum inall triangles determined if it is known in one.

In the third place, finally, instead of assuming the length of lines to be independentof place and direction, one might assume that their length and direction is independentof place. According to this conception, changes or differences in position are complexquantities expressible in three independent units.

2

In the course of the previous considerations, the relations of extension or regionality werefirst of all distinguished from the metric relations, and it was found that different metricrelations were conceivable along with the same relations of extension: then systems ofsimple metric specifications were sought by means of which the metric relations of Space arecompletely 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 theseassumptions are borne out by experience. In this connection there is an essential differencebetween mere relations of extension and metric relations, in that among the first, wherethe possible cases form a discrete manifold, the declarations of experience are surely nevercompletely certain, but they are not inexact, while for the second, where the possible casesform a continuous manifold, every determination from experience always remains inexact -be the probability ever so great that it is nearly exact. This circumstance becomes importantwhen these empirical determinations are extended beyond the limits of observation into theimmeasurably large and the immeasurably small; for the latter may obviously become evermore inexact beyond the boundary of observation, but not so the former.

When constructions in Space are extended into the immeasurably large. unboundednessand infinitude are to be distinguished; one belongs to relations of extension, the other to


metric relations. That Space is an unbounded triply extended manifold is an assumptionwhich is employed for every apprehension of the external world, by which at every momentthe domain of actual perception is supplemented, and the possible locations of a sought-after object are constructed; and in these applications it is continually confirmed. Theunboundedness of space consequently has a greater empirical certainty than any experienceof the external world. But its infinitude does not in any way follow from this; quite to thecontrary, Space would necessarily be finite if one assumed independence of bodies fromposition, and thus ascribed to it a constant curvature, as long as this curvature had ever sosmall a positive value. If one prolonged the initial directions lying in a surface direction intoshortest lines, one would obtain an unbounded surface with constant positive curvature, andthus a surface which in a flat triply extended manifold would take the form of a sphere, andconsequently 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 theexactness with which we pursue phenomenon into the infinitely small, does our knowledgeof their causal connections essentially depend. The progress of recent centuries in under-standing the mechanisms of Nature depends almost entirely on the exactness of constructionthat has become possible through the invention of the analysis of the infinite and through thesimple principles discovered by Archimedes, Galileo, and Newton, which modem physicsmakes use of. By contrast, in the natural sciences where the simple principles for such con-structions are still lacking, to discover causal connections one pursues phenomenon intothe spatially small, as far as the microscope permits. Questions about the metric relationsof Space in the immeasurably small are thus not idle ones.

If one assumes that bodies exist independently of position, then the curvature is ev-erywhere constant, and it then follows from astronomical measurements that it cannot bedifferent from zero; or at any rate its reciprocal must be an area in comparison with whichthe range of our telescopes can be neglected. But if such an independence of bodies fromposition does not exist, then one cannot draw conclusions about metric relations in the in-finitely small from those in the large; at every point the curvature can have arbitrary valuesin three directions, provided only that the total curvature of every measurable portion ofSpace is not perceptibly different from zero; still more complicated relations can occur ifthe line element cannot be represented, as was presupposed, by the square root of a differ-ential expression of the second degree. Now it seems that the empirical notions on whichthe metric determinations of Space are based, the concept of a solid body and that of alight ray, lose their validity in the infinitely small; it is therefore quite definitely conceivablethat the metric relations of Space in the infinitely small do not conform to the hypothesesof geometry, and in fact one ought to assume this as soon as it permits a simpler way ofexplaining phenomena.

The question of the validity of the hypotheses of geometry in the infinitely small isconnected with the question of the basis for the metric relations of space. In connectionwith this question, which may indeed still be reckoned as part of the study of Space, the


above remark is applicable, that in a discrete manifold the principle of metric relations isalready contained in the concept of the manifold, but in a continuous one it must comefrom something else. Therefore either the reality underlying Space must form a discretemanifold, or the basis for the metric relations must be sought outside it, in binding forcesacting upon it.

An answer to these questions can be found only by proceeding from the conceptionof phenomena which has hitherto been proven by experience, for which Newton laid thefoundation, 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 onlyto insure that this work is not hindered by the restriction of concepts, and that progress incomprehending 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 whichthe 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 asensual passion; it will deprive you of health, leisure, and peace - it will destroyall joy in your life. These gloomy shadows can swallow up a thousand Newtoniantowers and never will there be light on earth: never will the unhappy human racereach 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 thebook. In most cases, instructions for constructing solutions stand in for fully worked-outanswers. In several cases, the reader is provided only with the necessary identities and ageneral strategy for completing the problem. I hope that the reader will turn to these pagesnot in despair but with the hope that the discussion presented here will lead to a deeperunderstanding of the subject. Several of the more complicated graphics in the text wereproduced using Mathentatica.

Chapter 1

1.2. Consider points on the sphere as vectors in R3. The standard formula for the anglebetween 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 distancebetween 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 vectorsare 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 line11i n fl3 and m is parallel to the line f12 n 1713. From the properties of the cross product (seethe 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 intellectualconversation during the preparation of this manuscript, and Annie Campisi for inspiring him to complete theprocess This work is dedicated to the memory of his father. Robert Georgc Cantarella.

279

280 Appendix: Notes on selected exercises

1.7. On the diagram, let c be the center of the spherical circle, a be a point on the sphericalcircle, and 0 be the center of the sphere. Note that LaOc is the anglesubtended by the radius p at the center of the sphere. Use this fact to finda formula for the length of ab (the planar radius of the circle) in terms ofp and R. This leads to the expression given in the exercise. If we rewrite

2n R sin A as 27rsin(p/R)

then L'Hopital's rule implies that the limitR (1/R)

as R -+ oo is 27rp.

Chapter 2

2.9. Before attempting this problem, we recommend that the reader examine the first few

C pages of Chapter 4 to introduce the Saccheri quadrilateral. We refer to

E Gthe diagram on the left. Mark off D and E, the midpoints of sides AC andF D11J\ t7tBC. Extend up, in both directions and drop perpendiculars to fromA, B, and C meeting in F. G. and H. Now show tADF = ACDHand ABEG = L\CEH. This proves that PG = 2 DE and that-FA-

HC = 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 theD left. Construct AC and BD, each unit length and each perpendicular

to AB. Then connect CD forming a rectangle. Connect the diagonalAD. Now mark off point E on AB such that AE AC.Construct

A E the perpendicular to AB at E. meeting Al) at G, and CD throughF. 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 demonstratethat the area of HI BA is equal to the area of unit square ACFE. Proposition 1.43 may behelpful.

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

h

F

d e

does not hold. We refer to the diagram on the left. Fix a line I and apoint a not on 1. Construct the perpendicular to / through a, meeting

PP' in a point b on I and draw the first parallel P to 1 through a. By1 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 perpendicularto I at e, meeting Pat f. Prove, by continuity, that some line through a, say P', is perpen-dicular to of at some g E ef. Now use the Saccheri-Legendre theorem (Theorem 2.2)

Appendix: Notes on selected exercises 281

to conclude that P' is parallel to 1. By construction, P was the first such parallel. Thiscontradiction proves the result.

3.4. The error here is subtle. The key step in the proof is the construction through D of a linewhich cuts AC and A B. The assumption that such a line always exists is the assumption thatLBAC is an angle such that every point in the interior of LBAC is on a line intersectingboth rays of the angle away from the vertex. By Theorem 3.9 (10), this is equivalent toPostulate V.

Chapter 4

4.2. Suppose 1, m have common perpendicular ab with a on 1. Follow the diagram on theleft. Choose any con ! and drop the oernendicular tom from c. meeting

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

f to in from e, meeting in at f. We must show of > cd. Extend cd anddrop 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 suchD 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

A BBE, so AABE is isosceles. Thus LEAB = LEBA, so LDABLCBA.

p S Now P, Q, and R are all on the horocycle of points correspond-ing to Q. Fix Son the parallel through P, and T corresponding toSon the parallel through R. The horocycle of points corresponding

RT

to S is concentric with the horocycle of points corresponding to Qso 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 provedthat P and R correspond.

4.6. Let 1, m, n be parallel lines. Locate corresponding points a, b, c on !, in, and n

or respectively, and locate d, e, and f on 1, m, n away from a, b, and c.n Suppose that HRA holds. Prove that ab, be are in a line by proving thata c

Labe and Lcbe are right angles. To do this, use the assumption that a, b,

d e f 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 thisimplies that the horocycle connecting a and c and the line connecting a and c coincide. Toprove 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.


Chapter 5

5.4. Construct the altitude BD of AABC. Consider the case shown in the figure, where

B BD is inside ABC. By the formulas leading up to the proof oftanh((b - x)/k)

Theorem 5.19, we have that cos LCAB =tanh(c/k) .

Further,

cos fl(b) = tanh(b/k) and cos fl (c) = tanh(c/k). ThusC

cosLCABcosfl(b)cosfi(c)=tanhl b k x Itanhl k I. (I)

Further. forx E 1R, sin fl(x) = I/cosh(x/k). Thus \sin 11(b) sin 11(c) _ cosh(a/k)

sin fl (a) cosh(b/k)cosh(c/k)

By the hyperbolic Pythagorean theorem (5.19). cosh(a/k) = cosh(p/k)cosh(x/k) andcosh(c/k) = cosh(p/k) cosh((b - x)/k). Thus

cosh(a/k) _ cosh(x/k)cosh (b/k) cosh(c/k) cosh((b - x)/ k) cosh(b/k) (2)

We observe that cosh(x/k) = cosh (__ib k x - k I since cosh(- ) is an even function. Using

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 Pythagoreantheorem (5.19) to get the following identities:

cosh(DA)

= cosh (k) cosh (k) ,

cosh(k)=cosh(2)coshBB(v).These identities imply

1 _ cosh(OB/k)cosh(x/k) cosh(DB/k)

Suppose we have a right hyperbolic triangle DABC with sides opposite vertices given bya. b. and c and right angle at C. The formulas for cos(L ABC) and sin(LCAB) in Chapter 5

imply cosh(b/k) =cos(LABC)

sin(LCAB)Apply this to triangles LAOB and AADB to see that

cosh(OB/k) _ cos(LOAB)sin(LADB)cosh(DB/k) sin(LAOB)cos(LDAB)'


Now apply Bolyai's theorem (5.20) to triangles AADB. AAOB, and L DA E to see

sin(LADB) _ sinh(AB/k)/sinh(DA/k) =1.sin(LAOB)cos(LDAB) (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 involutedetermined by the general formula

p(t)=a(t)+`1.-

IIa()

)II)a'(t).

In this case, the result is

r I + 4x2dxI0(t)=(t,12)+

!1c-fOj /(1,2t).\\ 1 + 4t'-

Involute of a circle. Evolute of an Ellipse

6.10. Referring to the discussion following Theorem 6.18 for the general formula for thecurve 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.

3ab(a2 - b2) sin 2tK'(t)=-2(a2 sine t + b22 cost /)5/2

This is zero at t = 0, rr, 37r/2.

6.11. For any e > 0, define hE: (a + e, b - eJ x B"(0, R) --i. R", where B"(0. R) isthe 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 valuetheorem, 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) -- Rdefined 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 existssome M2 such that H(t, vr, v2) < M2 for all (t, yi. Y2) in D x B"(0, R). Of course, thisM2 satisfies the conditions of our claim.

We now apply Theorem 6.19. Choose A > 0 so that R > AMi. Clearly, A and Rare 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 functionthrough 0, denoted vo(t): b(to, A) --# R", such that vo(t) = hE Q. "vo(t)). Here b(to, A) isthe open ball around to of radius A in R. Now let ti = to + A/2 and apply the theoremfor initial conditions (ti,. o(tr)). By uniqueness, the resulting function vr(t) extends thedomain of definition of vo to (a + e, a + e + 3A/2). Repeating this process finitely manytimes 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 completesthe 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 writ-ten in terms of B. However, T = N x B. so T' = N' x B + N x B'. Since N is a unitvector, the Frenet-Serret theorem implies that K = T' N.

7.7. The expansion is the Taylor series for a(s), that is,

a"(so)(As)2 of (s0)(ts)3a(so+Os)=a(so)+a'(so)(As)+ 2 +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):

K 2 2 2r2 3 KT 3n=2r. b2 =9K

n, b= 6t-.

7.9. Since the curve is spherical, there must exist some point c E R3 and a real number rsuch that (a(s) - c) (a(s) - c) = r2 for all s. Differentiate both sides three times to obtain


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

K.4ar(s) =I/IIA(a"(s))II

= 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

r (s) = -det(A)(a'(s) x a"(s)) a"'(s))

.4.a 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

avu

(a1 + tal) (vaj ())aI + vaj ai(u)

r(aI + ua,)a

v(va;(W)) = a; + vat a; (U)

a a aj + va3 aj(u)

au (a3 + va')at

(vai(n))

Since K(u) 96 0, we know that a"(u) 96 0. Thus, since v # 0. this matrix has rank two. Asfor 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.

8,4. We make two claims. First, we show that for 7r/2 < u < 2n, u = u - n/2, and u = v

The Mobius Strip

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). Wemust restrict the domain of these assertions to prevent attemptingto evaluate x(u, v) or i (u, v) outside their domains of definition. Inthe first case, r/4 + u/2 = n/4 + (u - rr/2)/2 = u/2. Also, wehave

IT

rrsin u =sin (u - 2) = sin u cos

2- cos u sin

2= - cos u,

n It ITcos 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

m(u, v) =

nx(u, v). for

4< u < n,

i(u - 2, v). fore <u-<2a +4.

By our earlier work, this function is well defined on r x (- 1, 1). Interestingly, m (r/4, v) =m(27r + n/4, -v), as i(37r/2 + 7r/4, -v) = x(n/4, v). This is exactly the condition thatdefines 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)= -- ai) au

au av 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 usef to induce a coordinate chart on S2 from a coordinate chart x: U -+ S1. Let E. F. and Gbe the component functions of the metric on S1 with respect to x and E, F. and G be thecomponents of the metric on S2 under fox. Now, by Proposition 8.3. there is a nonzerofunction 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 functionsA (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,


assume that A(p) is greater. Since these functions are continuous, there is some boundedneighborhood N of p where A(q) > B(q). Suppose x - ' (N) = V and compute the area ofN and the area of f(N). By our assumption that f is equiareal, area(N) = area(f(N)).But

(x(u, v))duduarea(N) = j'J A

(x(u, v))dudv = area(f(N)).> Jj B

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 ofrevolution of the function f(u) = u tan s. This provides a coordinate chartfor 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 and AV1 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 thatthey are equal to the metric coefficients for the plane under polar coordinates. This provesthat the map projection is ideal, since the metric coefficients determine angles, lengths, and

areas on a surface.

Chapter 9

9.2. The function z = /'(u. v) leads to the coordinate chart x: (U C R2) R3 given byx(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

. z

au2 8u2 (auav)K(P) =


9.5. Recall that IIp(v) = dNp(v) u. Now write v in polar coordinates in the basis ofprincipal directions X1. X2. Thus v = 11i 11(Xi sin0 + X2 cos0) for some angle 0. Ifu E Dp, then

12IIp(v) = ( ki 0 sing sinO

ii i0 -k2) (cos 0) 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, k2have 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 directionassume that d Na(,) (a' (t )) = AU)a'(t ). Expand this out in terms of a coordinate chart wherea(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')2E+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 rwhen

(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).

Chapter 10

10.3. Recall from theoretna egregium that the Gaussian curvature K is given by the fol-lowing formula when F = 0

(EG)2

- 72 2 2EOP + - 1 G«<, !El, -

7

Ev

3G E 0

G,, 0 G

Expand the determinants and simplify to obtain the result.

0 ; E,, g G«

E,, E 0

G 0 G I.

10.5. There are three results to prove here. To prove that the line in R3 through p containedin S is in an asymptotic direction, suppose that this line is in the direction of the unittangent 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 < 0at each point p E S. we observe that for some c' at p. k (p) = 0. This means that k, > 0 andk2 < 0 so k, k, < 0. To construct a nontrivial family of ruled surfaces. choose a continuousfunction f: IR --. R such that 1(x) = - J'(tr + x). Using cylindrical coordinates. define Sby S = {(r. H, :) j z = r( J'((9))}. For example. take /'(.r) = (1/2)sin(3x). The resultingsurface is ruled.

10.6. Suppose S is minimal. At a point p E S let X1. X2 denote the principal directionsand 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 solutionsto this equation. n/4, 3;r/4. 57r/4, and 72r/4. Of course. this applies to the other asymptoticdirection, as well. Apply the intermediate value theorem to prove that if k, 0 k2, thena 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.


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 flby considering the reflection R: R3 -s R3 across f1. Observe that R restricted to S is amapping from S to itself. In fact, for a given patch x: (U C R2) -> S around a(s), we canform another patch, R o x: (U C R2) -+ S, by composing with the reflection. Prove thatthe normal in this patch is the reflection of the normal in the original patch. Proposition 8.9implies that R(N(a(s)) = bN(a(s)) for some b E R. Thus N(a(s)) E R. By the definitionof 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 inter-section 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 bya multiple of arc length. But a'(t) = (-e' sin e', e' cos e', 0), so

wt(a'(1), a'(1)) = dt

(ez,)=

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) - fQ kg(s)ds.

Thus we have

F Fkg(s)ds.f f KdA=F(iti+i,2+i,3-a)-Y

f8A,Rr=I ,=1

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)-nF- kg(s)ds.1_1 aR'

Now each triangle contributes 3 edges to the triangulation, while each edge that is not onthe boundary of R' is shared by 2 triangles. Thus, if C is the number of edges along theboundary of R', we have 3F = 2E - C. Also, at each vertex not on the boundary, theinterior angles sum to 27r. At each vertex on the boundary, they sum to jr or to rr - Ej for

some j. Thus, defining e = 0 for J < i < C, and observing that there are precisely Cvertices on the boundary, we have

F C

L(tt I + it2 + it3) = 2rr(V - C) + L(rr - Ej ).t=1 j=I

Putting this all together, we get

C

ffKdA=27r(V-C)+(7r-Ej)-1lF-r

J k(s)ds' j=1 aR'

=27rV -rrC-trF+n(3F-2E+C)-EEj- J kg(s)dsj=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 characteris-

Prism

tic 0. The extension to g handles follows by induction. First, take any triangulationof an empty prism. Observe that the prism has Euler characteristic 2. This followsfrom the fact that the prism is homeomorphic to the sphere. Take any triangulationof the sphere, and join the top face and the bottom face of the prism to two facesof the triangulation of the sphere. The Euler characteristic of the resulting figuremust 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). Thusthe 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 curvehas length sa(S) > Ilwll, and further, that if sa(S) = Ilwll, then a(t) is the geodesic radiusconnecting p and q. Of course, if a([0, S]) 1Z exp(B((OP)) then sa(S) > Ilwll. We dealwith the case where a([0, S]) C exp(BF). Suppose a(t) = (r(t), 9(1)) in geodesic polarcoordinates with respect to a fixed vector in the direction of w. The arc length is given by

sa(S) = Ja (drs)2d1 = Ja (d 12 +G(r(t), 9(i)) (de 12dt.

But G(r(t), 0(1))dle

is nonnegative for all t, so

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 somereparametrization of the geodesic radius connecting p and q.

13.6. Suppose that exp is a diffeomorphism on the open ball B0(0) E Ta(S). Choosesome a: R -+ S with a(0) = p and establish geodesic parallel coordinates around p witha as the base curve. Further, suppose that fi: (1 C R) -- Tn(S) is a curve in Tn(S) withIIfi(t)II < e and P(t) = (sona(r)) for some so E R. Clearly y(t) = expa(,)(#(t)) is aportion 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 foreach to E 1,

dtar(SO)11_10, dsa,0(s)s-so I = 0.

Defineda,' da,

E(t) _ ds.o ds ds

Observe that this function gives the distance from a(t) to y(t) along geodesic perpendic-ulars. This is obviously a constant, so we can apply the methods given in the proof of theGauss lemma (13.2) to arrive at the identity

a,

' 'or,0 -

(aat

as 1,=,0

We can evaluate the first term to see

aa, aa,

(at' as ,_,0

s.s0aa,

-a2o,

s=o Jo 1 at ' a-s2 I,=to ds.

(dot(s0) da,,, (do(0) da,°dt (to). ds (so)I dt (to). ds (o)

!dar(so) dat° 'dt (to). ds (so) - (a (to). Ha(lo))

dot(so) da,°

dt(to). ds (so)

s=s0

s=0

Of course, this is what we are looking for. It remains to show that

(30 aa, a2a, 1

J1-.-1I.Ods=0.

at as2

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. Tocompute the line element for this coordinate chart on the unit sphere, sup-pose a(t) is a geodesic. Then, following the diagram to the left, we see thatthe geodesic parallel a, (y) is the geodesic circle on the sphere with centerP 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( - y),so G(x. y) = sin('w - y).


13.7. To prove the first identity, recall that H = -7tr(dNp). But dNp can be written as

dNp = I pZi 122 Theorem 9.9, we have

1 I (Eg-2Ff+eG).H=-2(111+122)=2 EG-F2

To show that the principal curvatures k, are given by H ± H- - K recall that the Gaussiancurvature is given by K = k1 k2 = det(dNp). The determinant of dNp - t1, where I isthe identity transformation and t a variable, is given by t2 + 2Ht + K. Expressing dNp interms 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 mustshow that they satisfy the following properties.

(1) They are defined on open sets U, in R2, such that the Oi 1 are injective and satisfy

U1o,1(Ui)=RP2.(2) The transition functions O, o 0 1 are smooth.

(3) The images of open sets in R2 under the 0 form a subbasis for a topology on RIP2

that is Hausdorff and second countable.

The first property is easy to prove. Obviously them' 1 are each defined on R2 and they areinjective on this domain. To see that U1 0,.- 1 = RP2, realize that for each i in question.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 setS 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-countableHausdorff space because (R2, usual) is a second-countable Hausdorff space. We now definethe topology Ton RP2 by S C RP2 is open if and only if. for i E (1.2, 3), S n V, is openin Ti. This is a topology. To prove that it inherits second-countability and the Hausdorffcondition from the Ti, use the following lemma:

For i # j, Sc V1 n Vj is open in (Vi, T) if and only if S is open in (Vj, T j). SupposeS 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(0j o = Oi o 0,7' is smooth (and thus obviously continuous) by property (2). But

4j o,0T 1(0j(S)) = rbj(S). so Oj(S) is open in R2 and hence S is open in (Vj,T ). Theconverse is similar.

Chapter 15

15.5. Suppose that ! has slope m. Reflect over the v-axis, if necessary, to arrive at the case

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 numbery(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 theunit 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 Rsuch that p = (r tanh to + c, r sech to) and q = (r tanh t1 + c, r sech t1). The distancebetween p and q is given by

(p, q)I dx + (LY)'] di4 dt 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 identitiesimply 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

(xo-c+r ylld}p(P.9) = In x1 -c+r yo/

where m < 0. Since! does not intersect the u-axis and I is not thefirst 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)


Substitute the expressions for the x, and' v, given earlier into this formula and simplify it toverify 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 triangleL 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!-u +r

and b= lln(l+tt).

2 I-u--v 2 l - tt

Also, we can prove that v' = tanh a = v cosh b. We need a formula for it' in terms of it andv. Now

b = dB((O. v'). (tt'. v'))

=

2

Ind, (W. v'), (- I v')) dE((O. v'). (V/1 - (v')'. a'))

dE((u'. v'), (1/-1 - (v')2. v')) dE((O, v'), (- l - (v')2, v'))

l it'+ l-(v')2 I I+u'coshaIn - =-In{ I_tr'cosha)=-

it, - I - 2 \

Thus u' = u sech a. With this result in hand, we are prepared to compute. Since the non-Euclidean length of lines through the origin depends only on their Euclidean length, wehave that (u')- + (v')2 = u- +v-. Further, since angle measure is Euclidean at theorigin, we can measure sin A and sin B using the Euclidean triangles we have drawn. Thus

sin A = t and sin B =it sech a

. Using this, and the fact that it = tanh b. weu--+v'- u--+v2

see that

sin B U I

sinh b cosh a sinh b tt- '+v- cosh a cosh b u'- + v2v

sinh a u- + v-sin A

sinh a '


It remains only to show thatI = sin B

We know that tanh c = u + v2. Further.sinh c sinhb

by a standard hyperbolic trigonometric identity, csch2 c = coth2 c - 1. Thus I =sinh c

1-u--v. Choosing a convenient term from the previous chain of equalities, we claim

U +vthat

I _ 1-u -u- 1

sinh c ' u'- + u2 cosh a cosh b u2 + u

2-v-V

and sech2 a = 1 - tanh- a, so sech a =1 - u

. Further,Now tanh a =71 - u2' 1 - u

we know that sech b = Substituting these expressions into the preceding equa-tion proves our claim. Thus we have established Bolyai's theorem for right triangles. Theextension to the more general case is not difficult.

To prove the hyperbolic Pythagorean theorem, we use the foregoing work to concludethat

cosh a=1- u

cosh b=I

and cosh c= II-u2-v2 1-u 112 - v

The result follows directly.

BibliographyThis text owes an enormous debt to the vast literature on geometry. I would like to call attention to thesources most relevant to each chapter for the reader who wishes to learn more about a particular topicfrom 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 ofBusemann (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 followsthe outline of the excellent treatments in Spivak (1970.1975), do Carmo (1976), Struik (1961), andHsiung (1981). 1 was lucky to have Yoder (1989) and Schroder (1988) to review when I was writingthis book. These books inspired the last part of Chapter 6 and all of Chapter 8"u. Chapter 15 is inspiredby reading Beltrami's Saggio and Poincare's Theorie des groupes Fuchsiens with some help fromCoolidge (1909). Chapter 16 is my attempt to outline what Riemann said in the lecture based on hintsfound in Pinl (1981) - it may be seen as an elementary companion to the chapter "What did Riemannsay?" 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 usedin preparing the text. The bibliography at the end of Spivak (volume 5) will aid the reader in goingon in the modern subject of differential geometry.

Books

Ahlfors, L. V., Complex analysis: An introduction to the theory of analytic functions of one complexvariable, 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 andthe 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, Wash-

ington, DC, 1967. Includes essays: What is analysis in the large? by M. Morse: Curves and surfacesin 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.

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Symbol indexa' f.261

BS (6,), SS (6,). 164

C°O(S).C"(p). 104, 204

Ca (s). 67

cj (s), 131

d(p. q). 166d Fp. 105

dNp Tp(S) - Tp(S). 135ds2. 108

Dp. 141

E. F. G. 107e. f. g. 140H(p). 141J(x)(u. v). 96K(p). 139

K. 93k1.k2, 132kg(s). 158kn(3). 131

na(s). 157O(n). 90R. 256S2.97T(s). N(s). 70T(s). N(s). B(s). 80T4(:). 211

In. i 1.70circum(R), 57cosh x. 52

8(ABC). 338`t, 250

8,,, 94

D/dt. 263

rk 147, 254

n(1). 69area(R). 113

expp(u-). 164

11p(i ). 138

]p(ii, CO. 106

K(s). 71, 80

K±(t), 66(v. u')p. 106

11(AP).4011(x ). 41

1*, 89

smh x, 52

r(s). 82rgit), 1679,.,. 262Ftt1P,70axv.80fl u-. 63

Ta(S). 104 Z(9), 73

x,,, x,.. 102 IIi I. 163

Irk. i 1.253

303

Name indexAghanis. 26 La place. P.S. (1740-1827), 157Appolonius (third century B.C.). 66 Legendre. A.M. (1752-1833). 33. 39Archimedes (287-212 B.C.). 3. 27, 57, 126 Leibniz. Gottfried (1646-17161, 66Aristotle (384-322 B.C.). 26 Levi-Civna, T. (1873-1941), 249, 265

Lie, Sophus (1847-99), 146. 242Bartels, Martin (1769-1836), 158 Liouville. J. (1803-82), 163Beltrami, E. (1835-1906). 217, 242, 260 Lobachevskii, Nikolai I. (1792-1850), 34, 39Bernoulli. Jacob (1654-1705). 66Bernoulli, Johannes (1627-1748), 66 Mainardi, G. (1800-79), 151Bertrand. J. (1822-1900). 190 Mercator (Kramer. Gerardus) (1512-94). 124Bessel. F.W (1784-1846). 163 Meusnier, Jean-Baptiste (1754-93), 138Bianchi, L. (1856-1928). 253 Minding. Ferdinand (1806-85). 158, 192Bianu?ia, D.. 215 Minkowski. Hermann (1864-1909). 251Bolyai, W.F (1775-1856). 39. 279 Monge, Gaspard (1746-1818). 95. 138Bolyai, Janos (1802-60). 39. 45Bonnet. Ossian (1819-92) 151, 171.

Cartan, E. (1869-1951). 253

Nasir al-Din al-TUsi (1201-1274). 24. 28-9

Cayley, Sir Arthur (1821-91). 236 Perrault. Claudius (1613-88). 69

Christoffel, E.B. 11829-1900). 253 Peterson, K. (1828-81). 151

Clavius, Christopher (1537-1612), 26 PotncarC J. Henri (1854-1912), 178. 217. 227

Codazzi. D. (1824-75). 151 Posidonius (ca 135-50 B.C.). 27Proclus (410-85). 24

D'Aigutllon, F. (1566-1617), 120 Ptolemy, Claudius (100-78), 25

Darboux. G. (1842-1917), 253 Puiseux. V. (1820-83), 190

Descartes. Rend (1596-1650), 178Diguet, 196 Qurra, Thiibit Ibn (836-901). 28

Einstein, Albert (1879-1955), 249. 251 Rado, Tibor, 177Euclid (ca. 300 B.C.). 3. 10 Ricci-Curbastro, G. (1853-1925). 249Euler. Leonard (1707-83). 4, 66-7. 95, 116, 132 Rtemann, G.F. Bernhard (1826-60). 226, 242

Rodrigues. O (1794-1851), 134Frenet. F (1816-68), 83 Russell, Bertrand (1872-1970), 13Fuchs, L.I. (1833-1902), 227

Sacchen, Girolamo (1667-1733). 26, 29. 34Gauss. Carl-Fricdrich (1777-1855), 39. 57. 116. Schouten, J.A. (1883-1971), 265

131. 134. 163. 171, 186. 242, 267 Schweikart. F.K. (1780-1859). 39Graves, Carlie R., xii Serret. J.A. (1819-92), 83

Simplicius, 26Hilbert, David (1862-1943). 11. 193. 203Hipparchus of Nicae 1190-125 B.C.). 120Holmgren E 203

Sind, Ibn (980-1037). 27

, ,

Huygens, Christian (1629-95), 73Taurinus, F.A. (1794-1874). 39, 56Tchebychev. P.L (1821-94). 202Tissot. N.A. (1824-1904), 127

Jacobi, C.G.J. (1804-51), 163. 179

Kepler, Johann (1571-1630), 66Vitale, Giordano (1633-1711). 34

Khayyam,'Umar (1027-I123), 27, 28, 34Klein, Felix (1849-1925). 146, 230, 236. 249 Wachter. FL. (1792-1817), 52. 260

Wallis, John (1679-1703), 31

Lagrange. J.L. (1736-1813), 127 Weyl, Hermann (1885-1955). 201, 267

Lambert, J.H. (1728-77). 32. 38-9. 52, 126

304

11

Subject index1-form. 248

absolute property. 42adjoint of a linear mapping. 93alternate interior angles theorem. 16Angle-Angle-Angle. 32angle between curves on a surface. 112angle defect. 33. 36, 59, 175, 239angle excess, 5. 175angle of parallelism. 40arc length

along a curve. 65along a great circle. 5on an abstract surface, 212

on a manifold, 251on a surface. 108

arc-length parametrization, 65area

in the Beltramt disk. 231on the sphere, 4on a surface. 113

asymptotic line. 152atlas. 206. 243Australia. 127-9axiom of continuity, 14axiomatic method. 10axioms of congruence. 18axioms of order. 18

Beltrami model. 221betweenness. I I

Bianchi identity, 268binormal vector. 80bounded region. 112

cartographic coordinates, 118catenary curve, 144Cauchy sequence. 166Cayley-Klem-Beltramt model. 236center of curvature. 67change of coordinates, 100Chrtstoffel symbols, 147. 212

of the first kind, 253of the second kind. 254

circle in a surface. 186circumference of a circle. 57. 190compatibility equations. 151component functions of the metric. 107cone, 98. 130conformal mapping, 117. 224congruence. 11. 13, 146connection, 267

Koszul, 268Levi-Civita, 267

constant curvature manifold, 259contraction of a tensor, 250convex surface. 144coordinate chart. 96. 206coordinate curves, 102corresponding points. 42. 50covariant derivative. 261-3critical point, 98critical value, 98cross product, 80cross ratio. 230curvature

directed curvature, 71Gauss-Riemann curvature, 213Gaussian. 139geodesic, 158mean, 144normal, 131of a space curve. 80plane, 67sectional. 258total, 178

curve (parametrized. differentiable, regular). 63of centers of curvature, 67spacelike. null, timelike. 252

cycloid. 73

Darboux vector, 92development of surfaces, 127. 145diffeomorphism, 101. 207. 244differentiable function

between abstract surfaces. 207between manifolds, 244between surfaces, 101

differentiable structure, 206differential of a mapping. 209, 245distance function, 12distance

on the Poincarc half-plane, 235on a surface, 166

dot product, 63Dupin mdicatrix, 144

elementary neighborhood. 180ellipsoid, 98elliptic point. 133embedding, 215Euler-Pomcare characteristic, 177evolute. 75exponential map. 164exterior angles theorem, 15extrinsic property, 146

Fano plane. 19

305

306

first fundamental form, 106flat manifold, 257frame, 81

moving, 83Frenet-Seret apparatus. 83Frenet-Serret theorem. 83Fundamental Theorem

for curves in R3, 84for plane curves, 71for regular surfaces. 151of Riemannian Geometry, 267

Gauss equations, 151Gauss lemma, 187Gauss map, 134Gauss-Bonnet theorem, 175generalized polygon, 183geodesic circle. 187geodesic curvature, 158geodesic completeness. 165geodesic mapping. 217geodesic parallel coordinates. 191geodesic polar coordinates, 187geodesic radius. 187geodesic, 160. 213, 264graph, 64. 97great circle, 3

HAA, HOA, HRA, 35handle. 185Hausdortf condition, 206Hazzidakis's formula. 205helix, 64

general, 85Hopf-Rinow theorem. 166Hopf's Umlaufsatz. 173horocyclc, 42, 237

concentric. 42horosphere, 51

hyperbolic geometry, 236hyperbolic point. 134hyperbolic sine theorem, 56hyperbolic trigonometric functions, 52hyperboloid, 98

implicit function theorem. 98incidence geometry, I Iinner product, 63, 106

index of. 252nondegenerate, 252nonsingular. 108positive-definite, 108

integral curve, 152intrinsic equations. 92

Subject index

intrinsic normal, 157intrinsic property. 146inverse function theorem, 98involute, 75isometric embedding, 215isometry, 89, 116. 145, 214

Jacobian, 96joke. 306

Klein model. Klein-Beltrami model, 236Kronecker delta function, 84

Lagrange's formula, 91Lambert quadrilateral. 44latitude. 118Lie group, 245Liebmann's theorem, 193lift of a mapping. 181line, 10

of curvature. 144, 152in the direction of a vector field. 153parallel to a plane. 49perpendicular to a plane. 46

line element. 108, 251line segment, I Ilinear fractional transformations. 229

real, 231

Lobachevskii-Bolyai theorem. 52. 222local equivalence problem. 252local isometry, 146longitude. 118Lorentz manifold, 252loxodrome, 124June. 4

Mainardi-Codazzi equations. 151manifold. 243

two-dimensional. 206map projection, 116

azimuthal. 126central. 123. 219conformal. 117conical. 123. 130cylindrical, 123equiareal, 117ideal. 116Lambert cylindrical. 126Mercator, 124orthographic, 126. 224stereographic, 120-2, 225

Matheinamca, 127, 279mean curvature. 144measure of length on lines. 12, 271

meridian. 1 19on surfaces of revolution, 142

metric geometry. 12metric space. 12, 166

complete, 166minimal surface, 144Mtnkowski space. 252Mi bms band, 103. 114

New Zealand, 127-9non-Euclidean circles, 236nonintersecting lines, 39normal curvature. 131normal neighborhood, 165normal vector

to a curve. 70. 80to a surface. 102

norm. 164

orthogonal group, 90orthogonal parametrization, 112, 152osculating circle. 67

parabolic point. 134paraboloid, I I I

parallel lines, 16. 39in space, 45

parallel planes. 49parallel transport. 265Pasch's amom, 14patch. 96pencil of lines. 42perpendicular planes, 48Philosopher's Principle. 26planar point, 134plane curvature, 67Playfatr's axiom, 25Poincare disk, 226Poincare half-plane, 228point transitive mapping. 192polar line element on the sphere, 127polarization identity, 90pole, 8Postulate I. 11. 165. 176Postulate 11, 11, 166Postulate III. 12, 191-2Postulate IV. 13. 191Postulate V, 17, 221principal curvatures, 133principal directions, 133protractor. 15pseudo-Rtemannian manifold, 252pseudosphere. 196

Subject index

Pythagorean Theoremhyperbolic, 55Proposition 147. 17spherical, 5

quadratic form. 138

Rayleigh quotient, 93real projective n-space, 244real projective plane. 207regular value, 98regularity, 96reparanietrtzation, 64reverse of a curve. 70Ricct's lemma. 268Riemann curvature tensor. 155. 256Riemann sphere. 229Riemannian manifold. 249Riemannian metric. 210. 249right-hand orientation, 81rigid motion, 146

of Euclidean space, 89Rodrigues's formula, 144rotation plane, 94ruler. 12

Sacchen-Legendre theorem. 15Sacchen quadrilateral, 34second countability. 206second fundamental form, 138self-adjoint linear mapping, 136simple connectivity, 171speed. 64

spherical coordinates. 5spherical sine theorem, 6standard orientation, 70surface. 96

abstract, 206orientable, 103ruled, 156topological. 98

Sylvester's law of inertia, 252

tangent bundle, 246tangent developable surface, 114

tangent plane, 104tangent space. 208, 245tangent vector. 64. 208, 244tautochrone, 73Tchebychev net. 202tensor field. 248theorema egregium, 148Three Musketeers theorem, 29

309

Subject index

torsion, 82geodesic torsion, 169

torus, 98flat, 211

trace, 63tractrix, 69. 142. 196transition function, 243triangulation. 176

unrolling a surface. 211upper half-space model. 260

vector field. 152, 247

Wallis's Postulate, 31Weierstraw coordinates, 241world-line, 252

umbilic point, 134unit-speed curve. 65

/ ifferential geometry has developed in many directions

since its beginnings with 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 sur-

faces 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 non-

Euclidean geometry and the modern ideas of abstract surfaces and manififlds.

CAMBRIDGE ""mtanibr '°`'°

UNIVERSITY PRESS

ISBN 0-521-42480-1

46243922 geometry from a differentiable viewpoint

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