Elementary Differential Geometry 0387142223, 1852331526

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Andrew Pressley

Preface

Department of Mathematics, King's College, The Strand, London WC2R 2LS, UK Cover iJluslrutic}T/ elements reproduced by killdpermission ofi Aplech S)'lIteffill, Inc" PubJisha"a of the GAUSS Mathemuical and ｓｴ｡ｾ ｩ｣Ｎ｡ｬ

" System, 23110"l S.E. Kent-Kanglq Road, Maple Valley, WA 980)8,

USA. Tel: (206) 432·7855 PU(2C6l 432·7832 email: infothptech,com URL: www.aptech.com Amtrican Statilitical Association: Chance Vol 8 No I, 1995 article by KS and KW Heiner ｾｲｔＧ Rings ofthe Northern Shawangunks' page 32 fig 2 Springer-Verlag: Mathematlca in Education and Researcli Vol 4 155ue 31995 article by Roman E Maeder, BeaU'ke Amrhein and Oliver Gloor 'lUustC81ed Mathematics: Yisualizetion ofMathc:matical Ｇｾ｢ｏ page 9 fig II, originlllly published u a CD ROM 'Illwtrated Mathematica' by TELOS: ISBN 0-387-14222-3, german edition byBirkhauser: ISBN 3-7643-S1()(}'4. MatbematiCll in Education and Research Vol4 l.uue31995 article by Richard J Gaylord and Kuume Nishidate 'Traffic Engineering with Cellular Automata' page 35 fig 2. Mathematica in EdueadOll and Rellearch Vo15 Issue 2 1996llrticle by ､ｾｨ｣ｩｍ TrOll 'The Implicitization of a Trefoil Knot' page 14. Mathematica in EdUClition and Research Vol 5 Issue 2 1996lrticle by Lee deColi 'Coins, Trees, Ban lItId Bells; Simulation of the Binomial Proces.s page 19 fig 3. Mathernatica in Education and Research Vo15 Issue 2 1996llrticle by Richard GAylord and Kazume Nishidate 'Contligiow Spreading' page 33 fig 1. Mathematka in Education and Reelrch Vol 5 Is.me 2 1996 article by Joe BuMer and Sian Wagon 'Secrets of the Maddung Constant' page 50 fig 1.

Springer Undergraduate Mathematics Series ISSN 1615-2085 ISBN 1-85233-152-6 Springer-Verlag London Berlin Heidelberg British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library Library of Congress Cataloging-in- Publication Data Pressley. Andrew Elementary differential geometry I Andrew Pressley. p. em. - (Springer undergraduate mathematics series. ISSN 1615·2085) Includes index_ ISBN 1-85233-152-6 (alk. pap«) I. Geometry, Differential. I. Title. II. Series.

QA641 .P68 2000 516.3'6-dc21 00-058345

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© Springer-Verlag London Limited 2001 Printed in Great Britain The use of registered names, trademarks etc. in this publication does not imply. even in the absence of a specific statement, that such names are exempt from the relevant laws and regulations and therefore free for general use. The publisher makes no representation, express or implied, with regard to the accuracy of the information contained in this book and cannot accept any legal responsibility or liability for any errors or omissions that may be made. Typesetting: Camera ready by author Printed and bound at the Athenaeum Press Ltd.. Gateshead. Tyne & Wear 12/3830-543210 Printed on acid-free paper SPIN 10711768

The Differential Geometry in the title of this book is the study of the geometry of curves and surfaces in three-dimensional space using calculus techniques. This topic contains some of the most beautiful results in Mathematics, and yet most of them can be understood without extensive background knowledge. Thus, for virtually all of this book, the only pre-requisites are a good working knowledge of Calculus (including partial differentiation), Vectors and Linear Algebra (including matrices and determinants). Many of the results about curves and surfaces that we shall discuss are prototypes of more general results that apply in higher-dimensional situations. For example, the Gauss-Bonnet theorem, treated in Chapter 11, is the prototype of a large number of results that relate 'local' and 'global' properties of geometric objects. The study of such relationships has formed one of the major themes of 20th century Mathematics. We want to emphasise, however, that the methods used in this book are not necessarily those which generalise to higher-dimensional situations. (For readers in the know, there is, for example, no mention of 'connections' in the remainder of this book.) Rather, we have tried at all times to use the simplest approach that will yield the desired results. Not only does this keep the prerequisites to an absolute minimum l it also enables us to avoid some of the conceptual difficulties often encountered in the study of Differential Geometry in higher dimensions. We hope that this approach will make this beautiful subject accessible to a wider audience. It is a cliche, but true nevertheless, that Mathematics can be learned only by doing it, and not just by reading about it. Accordingly, the book contains V

VI

Elementary Differential Geometry

p

over 200 exercises. Readers should attempt as many of these as their st.amina permits. Full solutions to all the exercises are given at the end of the book, but these should be consulted only after the reader has obtained his or her own solution, or in case of desperation. We have tried to minimise the number of instances of the latter by including hints to many of the less routine exercises.

Contents

Preface

, _. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

v

1.

Curves in the Plane and in Space 1.1 What is a Curve? Arc-Length. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Reparametrization. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 1.4 Level Curves vs. Parametrized Curves

1 1 7 10 16

2.

How 2.1 2.2 2.3

23 23 28 36

3.

Global Properties of Curves 3.1 Simple Closed Curves 3.2 The Isoperimetric Inequality 3.3 The Four Vertex Theorem

47 47 51 55

4.

Surfaces in Three Dimensions 4.1 What is a Surface? 4.2 Smooth Surfaces 4.3 Tangents, Normals and Orientability 4.4 Examples of Surfaces 4.5 Quadric Surfaces 4.6 Triply Orthogonal Systems 4.7 Applications of the Inverse Function Theorem

59 59 66 74 78 84 90 93

Much Does a Curve Curve? Curvature. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Plane Curves Space Curves

VII

Elementary btfferenbal Geometry

VIII

5.

6.

7.

8.

9.

_ . --. . . .

97 97 101 106 112

Curvature of Surfaces 6.1 The Second Fundamental Form 6.2 The Curvature of Curves on a Surface -6.3 The Normal and Principal Curvatures 6.4 Geometric Interpretation of Principal Curvatures

. . . .

123 123 127 130 141

Gaussian Curvature and the Gauss Map 7.1 The Gaussian and Mean Curvatures 7.2 The Pseudosphere 7.3 Flat Surfaces 7.4 Surfaces of Constant Mean Curvature 7.5 Gaussian Curvature of Compact Surfaces 7.6 The Gauss map

. . . . . .

147

Geodesics 8.1 Definition and Basic Properties 8.2 Geodesic Equations 8.3 Geodesics on Surfaces of Revolution _ 8.4 Geodesics as Shortest Paths 8.5 Geodesic Coordinates -

. . . . .

171 171 175 181 190 197

_ . . . . -

201 201 207 217 219

. . . . .

229 229 238 240 244

. . . .

247

The First Fundamental Form 5.1 Lengths of Curves on Surfaces 5.2 Isometries of Surfaces _ 5.3 Conformal Mappings of Surfaces 5.4 Surface Area 5.5 Equiareal Maps and a Theorem of Archimedes

-

Minimal Surfaces _ 9.1 Plateau's Problem -' 9.2 Examples of Minimal Surfaces 9.3 Gauss map of a Minimal Surface 9.4 Minimal Surfaces and Holomorphic Functions

10. Gauss's Theorema Egregium

10.1 10.2 10.3 lOA

Gauss's Remarkable Theorem .. Isometries of Surfaces The Codazzi-Mainardi Equations Compact Surfaces of Constant Gaussian Curvature

11. The Gauss-Bonnet Theorem

11.1 11.2 11.3

Gauss-Bonnet for Simple Closed Curves Gauss-Bonnet for Curvilinear Polygons .. _ Gauss-Bonnet for Compact Surfaces -

11.4 11.5

Solutions __ Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11

116

147 151 155 161 164 165

247 252 258

Singularities of Vector Fields Critical Points __

Index

f··

-__

_

--

--

_ _ _

-

- . .. 269 275

- . . . . .. - . . . . . . . . .. . . . . . . . . . . . . . . . .. - . . . . . . . . . .. -

- . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..

_

-

281 272 275 279 280 286 289 294 299 307 311 314

- 329

1 Curves in the Plane and in Space

In this chapter we discuss two mathematical formulations of the intuitive notion of a curve. The precise relation between them turns out to be quite subtle, so we shall begin by giving some examples of curves of each type and practical ways of passing between them.

1.1. What is a Curve? H asked to give an example of a curve, you might give a straight line, say 2 y _ 2x = 1 (even though this is not 'curved'!), or a circle, say x + y2 ::::: 1, or perhaps a parabola, say y - x 2 = O.

y - 2x = 1

An of these curves are described by means of their cartesian equation I(x, y) = c,

where

f is a function of x and y and c is a constant. From this point of view, 1

Ei@iiiEiiCSfY b'ii@#eiTtid' GEO:ri@lIY

2

1'1 (t)

c=

I f(x,y)

{(X,y) E R 2

(1)

= c}.

R , but we can also consider curves These examples are all curves in ･ｮ｡ｬｰＢｾｨｴ 3 3 in R - for example, the x-axis in R is the straight line given by 2

{(x,y,Z)ER and more generally a curve in R

3

h(x,y,z) =

3

'Y: (-00,00) ---t R 2,

I y=z=O}, h(x,y,z) =

C2'

Definition 1.1

).

Example 1.2 Now we try the circle x 2 + y2 = 1. It is tempting to take x = t as in the previous (we could have taken y = -Jt=t2). So we get example J so that y = ｾ the parametrisation

'Y(t)

1'1

(t)2

+ '"Y2(t)2 = 1

(3)

for all t E (oJ (3), and such that every point on the circle is equal to C1'1 (t) ,1'2 (t» for some t E (a, ,8). There is an obvious solution to Eq. (3): 1'1 (t) = COS t and 'Yz(t) sin t (since cos 2 t + sin 2 t 1 for all values of t). We can take (a,{3) = (-00,00), alt.hough this is overkill: any open interval (a f3) whose length is greater than 21T will suffice.

=

I 0',< t < ,8}.

= (t, Jl=t2).

But this is only a parametrisation of the upper half of the circle, because JI=t2 is always ｾ O. Similarly, if we had taken y = ＭｾＬ we would only have covered the lower half of the circle. H we want a parametrisationof the whole circle, we must try again. We need functions 1'1 (t) and 1'2(t) such that

(a,{3) ---t R n , for some a,{3 with

The symbol (a, (3) denotes the open interval

(a,{3) = {t E R

2

But this is not the only parametrisation of the parabola. Another choice is = (t 3 ,t6 ) (with (a,f3) = (-00,00». Yet another is (2t,4t 2 ), and of course there are (infinitely many) others. So the parametrisation of a given level curve is not unique.

Curves of this kind are called level curves, the idea being that the curve in Eq. (1), for example, is the set of points (x, y) in the plane at which the quantity I(x, y) reaches the 'level' c. But there is another way to think about curves which turns out to be more useful in many situations. For this, a curve is viewed as the path traced out by a moving point. Thus, if 'Y(t) is the position vector of the point at time t, the curve is described by a function 'Y of a scalar parameter t with vector values (in R Z for a plane curve, in R 3 for a curve in space). We use this idea to give our first formal definition of a curve in R n (we shall be interested only in the cases n = 2 or 3, but it is convenient to treat both cases simultaneously):

A parametrised curve in R n is a map "y -00 ｾ 0 < {3 ::; 00.

'Y(t) = (t,t

'Y(t)

might be defined by a pair of equations Cl,

= t, '"Y2 (t) = t 2. To get

every point on the parabola we must allow t to take every real number value (since the x-coordinate of 'Y(t) is just. t, and the x-coordinate of a point on the parabola can be any real number), so we must take (a,f3) to be (-00,00). Thus, the desired parametrisation is

a curve is a set of points, namely

=

J

A parametrised curve whose image is contained in a level curve C is called a parametrisation of (part of) C. The following examples illustrate how to pass from level curves to parametrised curves and back again in practice.

Example 1.1

The next example shows how to pass from parametrised curves t.o level curves.

Example 1.3

Let us find a parametrisation 'Y(t) of the parabola y = the components '"Y1 and '"Y2 of 'Y must satisfy

'"Y2(t) = '"Y1(t)Z

x 2. If "y( t)

= (1'1 (t), 1'2 (t»

J

(2)

for all values of t in the interval (a, (3) where 'Y is defined (yet to be decided), and ideally every point on the parabola should be equal to (')'1(t),1'2(t» for some value of t E (a, (3). Of course, there is an obvious solution to Eq. (2): take

Take the parametrised curve (called an astroiri)

'Y(t) = (cos3 t,sin 3 t). Since cos2 t + sin2 t == 1 for all t, the coordinates x point 'Y(t) satisfy

= cos3 t, Y = sin 3 t

of the

Elementary Differential Geometry

4

1. Curves in the Plane and in Space

The following result is intuitively clear:

This level curve coincides with the image of the map 'Y. In this book, we shall be studying curves (and later, surfaces) using methods of calculus. To differentiate a vector-valued function such as 'Y(t) (as in Definition 1.1), we differentiate componentwise: if

then

Proposition 1.1 If the tangent vector of a parametrised curve is constant, the image of the curve is (part of) a straight line. Proof 1.1

d'Y (d'Yl d'Y2 d'Yn) dt = (ft'(ft""'dt '

ｾＧｙ

dt2

ＲｙＧｾ rP'Yn) = (d?'Y1 dt2' dt2 , ... , dt2 '

If ..y(t) = a for all t J where a is a constant vector, we have, integrating componentwise, etc.

To save space, we often denote d'Y/dt by ..y(t), rP'Y/dt2 by ;yet), etc. We say that "'( is smooth if each of the components 1'1,1'2,···, 'Yn of 'Y is smooth, i.e. if all the derivatives d'Yi/dt, d?'Yddt2, d3'Yddt3, ... exist, for i. = 1,2, ... , n. From now on, all parametrised curves studied in this book wW":be assumed to be smooth.

'Y(t)

=

Jｾ

dt

=

J

adt = ta + b J

where b is another constant vector. If a ::j:. 0, this is the parametric equation of the straight line parallel to a and passing through the point with position vector b:

Definition 1.2 If 'Y(t) is a parametrised curve, its first derivative d'Y/dt is called the tangent vector of 'Y at the point 'Y( t) .

a

To see the reason for this terminology, note that the vector

'Y(t

+ 1St) - 'Y(t)

1St is parallel to the chord joining the two points 'Y(t) and 'Y(t + 1St) of the image C

If a

o = 0, the image of'Y is a single point (namely, the point with position vector

ｾＮ

0

ofT

EXERCISES

= (t 2 , t 4 ) a parametrisation of the parabola y

1.1

Is

1.2

Find parametrisations of the following level curves: (i) y2 _ x 2 = 1;

"'(t)

2

We expect that, as 1St tends to zero, the chord becomes parallel to the tangent to C at 'Y(t). Hence, the tangent should be parallel to lim 'Y(t + 1St) - 'Y(t) = d'Y. 6t-+O

1St

dt

f

= x2?

= 1. (ii) z4 + 1.3 Find the cartesian equations of the following parametrised curves: (i) 'Y(t) = (cos 2 t, sin 2 t)j (ii) 'Y(t) = (e t , t 2 ). 1.4 Calculate the tangent vectors of the curves in Exercise 1.3.

Elementary Differential Geometry

6 1.5

Sketch the astroid in Example 1.3. Calculate its tangent vector at each point. At which points is the tangent vector zero?

1.6 If P is any point on the circle C in the xy-plane of radius a > 0 and centre (0, a), let the straight line through the origin and P intersect the line y = 2a at Q, and let the line through P parallel to the x-axis intersect the line through Q parallel to the y-axis at R. As P moves around C, R traces out a curve called the witch of Agnesi. For this curve, find (i) a parametrisation; (ii) its cartesian equation. Q

:¥----------.R

1. Curves in the Plane and in Space

1.10 For the logarithmic spiral 'Y(t) = (e t cost, et sin t), show that the angle between 'Y(t) and the tangent vector at -y(t) is independent of t. (There is a picture of the logarithmic spiral in Example 1.4.)

1.2. Arc-Length If v = (Vi, ... , v n ) is a vector in R n , its length is

II v II = Jv? + ... + ｶｾＮ is another vector in R n, II u - v II is the length of the straight line segment

If u joining the points in R n with position vectors u and v. To find a formula for the length of any parametrised curve 'Y, note that, if 8t is very small, the part of the image C of'Y between 'Y(t) and 'Y(t + M) is nearly a straight line, so its length is approximately

o 1.7

A cycloid is the plane curve traced out by a point on the circumference of a circle as it rolls without slipping along a straight line. Show that, if the straight line is the x-axis and the circle has radius a > 0, the cycloid can be parametrised as

'Y(t) = a(t - sin t, 1 - cos t). 1.8

Generalise the previous exercise by finding parametrisations of an epicycloid (resp. hypocycloid), the curve traced out by a point on the circumference of a circle as it rolls without slipping around the outside (resp. inside) of a fixed circle.

1.9

Show that 'Y(t) = (cos 2 t sin t cos t, sin t) is a parametrisation of the curve of intersection of the circular cylinder of radius ｾ and axis the z-axis with the sphere of radius 1 and centre (-!' 0, 0). (This is called Viviani's Curve).

7

Again, since c5t is small, length is approximately

II 'Y(t + M) - 'Y(t) II· ("{(t + at) - -y(t))/M is II i'(t) II 8t.

nearly equal to i'(t), so the

(4)

If we want to calculate the length of (a not necessarily small) part of C, we can

divide it up into segments, each of which corresponds to a small increment rSt in t, calculate the length of each segment using (4), and add up the results. Letting c5t tend to zero should then give the exact length.

!'

J!!!if This motivates the following definition:

Definition 1.3 The arc-length of a curve 'Y starting at the point "Y(to) is the function s(t) given

9

Elementary Differential Geometry

8

Definition 1.4

by

set) =

If 'Y : (a, f1) ---t R n is a parametrised curve, its speed at the point 'Y( t) is II -y{ t) I\, and 'Y is said to be a unit-speed curve if-y(t) is a unit vector for all t E (a,,B).

t II ..y(u) 11 duo ltD

Thus, s(to) = 0 and set) is positive or negative according to whether t is larger or smaller than to. If we choose a different starting point 'Y(to) , the lI..y(u) II duo resulting arc-length s differs from s by the constant ｊｴｾｏ

We shall see many examples of formulas and results relating to curves that take on a much simpler form when the curve is unit-speed. The reason for this simplification is given in the next proposition. Although this admittedly looks uninteresting at first sight, it will be extremely useful for what follows.

Example 1.4

Proposition 1.2

For a logarithmic spiral

Let net) be a unit vector that is a smooth function of a parameter t. Then, the dot product n(t).n(t) = 0

for all t, i.e. net) is zero or perpendicular to net) for all t. In particular, if'Y is a unit-speed curve, then;Y is zero or perpendicular to

Proof 1.2 We use the 'product formula' for differentiating dot products of vector-valued functions a( t) and b (t ): d da dt (a.b) = dt· b

Using this to differentiate both sides of the equation n.n gives

we have

..y = (e t ( cos t .. II

t 11 2 =

sin t) , et (sin t + cos t)) ,

e 2t(cost - sint)2

+ e2t (sint + cost)2

Hence, the arc-length of'Y starting at 'Y(O) s

=

it

V2e 2u du

= (1,0)

= h(e

t

-

db

+ a. dt .

= 2e

2t

= 1 with respect to t

n.n + n.n = 0,

•

so 2n.n = O. The last part follows by taking n =

(for example) is

o

..y.

1).

EXERCISES If s is the arc-length of a curve 'Y starting at 'Y(to), we have

ｾＺ

=

:t ltDr II

1.11 Calculate the arc-length of the catenary 'Yet)

t(u)

II du = 11..y(t) 11·

(5)

Thinking of 'Y(t) as the position of a moving point at time t, ds/dt is the speed of the point (rate of change of distance along the curve). For this reason, we make

= (t, cosh t) starting at

the point (0, 1). 1.12 Show that the following curves are unit-speed:

(i)

'Y(t) =

(ii) 'Y(t)

0(1 + t)3/2, t(1 - t)3/2, ［Ｉｾ

= Ucos t, 1 -

sin t, Ｍｾ

cos t).

Elementary DifferentIa

10

eometry

1.13 Calculate the arc-length along the cycloid in Exercise 1.7 corresponding to one complete revolution of the circle.

1.3. Reparametrisation We saw in Examples 1.1 and 1.2 that a given level curve can have many parametrisations, and it is important to understand the relation between them.

1. Curves in the Plane and in

pace

11

Definition 1.6 A point "((t) of a parametrised curve "( is called a regular point if ..y(t) f:. 0; otherwise 'Y(t) is a singular point of 'Y. A curve is regular if all of its points are regular. Before we show the relation between regularity and unit-speed reparametrisation, we note two simple properties of regular curves. Although these results are not particularly appealing, they will be very important for what is to follow.

Proposition 1.3

Definition 1.5 A parametrised curve;y: (a,[J) -t- R n is a reparametrisation of a parametrised curve "( : (a, (3) ---t R n if there is a smooth bijective map ,p : (a, [J) ---t (a, (3) (the reparametrisation map) such that the inverse map ,p-l : (a,;3) -t- (a, fJ) is also smooth and ;y(i) = "((,p(t)) for all

i E (a,[J).

Any reparametrisation of a regular curve is regular. Proof 1.3

Suppose that "( and ;y are related as in Definition 1.5, let t = ,p(l), and let 'IjJ = $-1 so that i = 'IjJ(t). Differentiating both sides of the equation ,p(t{;(t)) = t with respect to t and using the chain rule gives

d,pd'IjJ -1

Note that, since ,p has a smooth inverse, "( is a reparametrisation of;y:

;y(,p-l(t)) = "((,p(¢-1(t))) = "((t) for all t E (a,;3). Two curves that are reparametrisations of each other have the same image, so they should have the same geometric properties.

di dt -

.

This shows that dl'/di is never zero. Since .y(i) = "((l'(i)), another application of the chain rule gives d.y d'Y d¢ di=dtdi'

Example 1.5

which shows that d;y/di is never zero if d'Y/dt is never zero.

In Example 1.2, we gave the parametrisation "((t) = (cos t, sin t) for the circle x 2 + y2 = 1. Another parametrisation is

Proposition 1.4

.y(t)

= (sint,cost)

(since t+ t = 1). To see that .y is a reparametrisation of "(, we have to find a reparametrisation map ,p such that sin 2

cos2

(cos ¢(t), sin l'(t)) One solution is ,p(t)

= 1r/2 -

= (sin t, cos t).

t.

As we remarked in the previous section, the analysis of a c;urve is simplified when it is known to be ｵｮｩｴｾｳｰ･ ､Ｎ It is therefore important to know exactly which curves have unit-speed reparametrisations.

o

If 'Y(t) is a regular curve, its arc-length s (see Definition 1.3), starting at any point of 'Y, is a smooth junction of t. Proof

L4

We have already seen that (whether or not 'Y is regular) s is a differentiable function of t and

ｾＺ

= II .y(t) II·

To simplify the notation, assume from now on that "( is a plane curve, say "((t) = (u(t),v(t)),

Elementary Differential Geometry

12

injective, that its image is an open interval (ii,13), and that the inverse map 8- 1 : (6,,8) --t (a,,8) is smooth. (Readers unfamiliar with the inverse function theorem should accept these statements for now; the theorem will be discussed informally in Section 1.4 and formally in Chapter 4.) We take 1> = S-l and let l' be the corresponding reparametrisation of 'Y, so that

2

where u and v are smooth functions of t. Define f : R --t R by

f(u,v) = Ju 2 + v 2 , so that

ds dt = f('-) u,v.

(6)

2

The crucial point is that f is smooth on R \{(0,O)}, which means that all the partial derivatives of f of all orders exist and are continuous functions except at the origin (0,0). For example,

of

au =

vu

of

u 2

+

av

v2 '

=

vu

t(s) = 'Y(t). Then,

d1'ds ds dt

v 2

II d1' II ds

+ v2 '

ds

= =

are well defined and continuous except where u v 0, and similarly for higher derivatives. Since 'Y is regular, u and v are never both zero, so the chain rule and Eq. (6) shows that ds/dt is smooth. For example,

af..

tPs

af ..

= au U + 8v v,

dt 2

and similarly for the higher derivatives of s.

0

II

dt

d'Y dt' =

［ｾ II =

II

d'Y dt

II =

ds dt

(by Eq. (5»,

o

1.

The proof of Proposition 1.5 shows that the arc-length is essentially the only unit-speed parameter on a regular curve:

Corollary 1.1 Let 'Y be a regular curve and let l' be a unit-speed reparametrisation of 'Y.'

The main result we want is

i'(u(t» = "(t)

Proposition 1.5 A parametrised curve has a unit-speed reparametrisation if and only if it is regular. Proof 1.5 n

Suppose first that a parametrised curve 'Y : (0:,,8) --t R has a unit-speed reparametrisation 1', with reparametrisation map 0 for all t and s -+ 0 as t -+ =f00 if ±k > 0, and express s as a function of t. Show that the signed curvature of'Y is 1/ ks. Conversely, describe every curve whose signed curvature, as a function of arc-length 8, is I/ks for some non-zero constant k. 2.6 A unit-speed plane curve 'Y has the property that its tangent vector t(8) makes a fixed angle () with 'Y(s) for all 8. Show that

2.5

where a >

°is a constant, is

0 < t < 21r,

E(t) ::::a(t+sint,-I+cost) (see Exercise 1.7) and that, after a suitable reparametrisation, f can be obtained from 'Y by a translation of the plane. 2.10 A string of length £ is attached to the point 8 :::: 0 of a unit-speed plane curve "1(8). Show that when the string is wound onto the curve while being kept taught, its endpoint traces out the curve

£(8) :::: 'Y(S) where 0

+ (£ - shes),

< s < l and a dot denotes d/ds. The curve £ is called the

involute of'Y (if "I is any regular plane curve, we define its involute

36

Elementary Differential Geometry

*

2. How Much

ｾ･ｯｄ

37

a Curve Curve?

of R 3 , and is right-handed, i.e.

to be that of a unit-speed reparametrisation of 'Y). Suppose that the signed curvature "'8 of'Y is never zero, say 1\;8(8) > 0 for all s. Show that the signed curvature of " is 1/(£ - 8).

b = t x n, n = b x t, t = n x b.

2.11 Let 'Y be a regular plane curve. Show that (i) the involute of the evolute of'Y is a parallel curve of 'Yi (ii) the evolute of the involute of'Y is 'Y. (These statements might be compared to the fact that the integral of the derivative of a smooth function J is equal to J plus a constant, while the derivative of the integral of J is J.)

b

n

2.12 Show that applying a reflection in a straight line to a plane curve changes the sign of its signed curvature. 2.13 Show that, if two plane curves 'Y(t) and i(t) have the same nonzero curvature for all values of t, then i can be obtained from 'Y by applying a rigid motion or a reflection in a straight line followed by a rigid motion.

Since b(s) is a unit vector for all 5, b is perpendicular to b. Now we use the 'product rule' for differentiating the vector product of vector-valued functions u and v of a parameter s:

d du dv -(u x v) = -d x v + u x -d .

\

ds

=t

Applying this to b

5

5

x n gives

b = i x n + t x 0 = t x 0,

2.3. Space Curves Our main interest in this book will be in curves (and surfaces) in R 3 , Le. space curves. While a plane curve is essentially determined by its curvature (see Theorem 2.1), this is no longer true for space curves. For example, a circle of radius one in the xy-plane and a circular helix with a b 1/2 (see Example 2.1) both have curvature one everywhere, but it is obviously impossible to change one curve into the other by any combination of rotations and translations. We shall define another type of curvature for space curves, called the torsion, and we shall prove that the curvature and torsion of a curve together determine the curve up to a rigid motion (Theorem 2.3). Let "y( s) be a unit-speed curve in R 3 , and let t = l' be its unit tangent vector. If the curoature 1\;(s) is non-zero, we define the principal normal of 'Y at the point 'Y( 8) to be the vector

= =

1 .

n(8) = "'(s) t(8). Since II i II = "', n is a unit vector. Further, by Proposition 1.2, and n are actually perpendicular unit vectors. It follows that

b=txn

since by the definition (7) of n,

i

x n

so t (8)

is a unit vector perpendicular to both t and n. The vector b(s) is called the binormal vector of 'Y at the point "y( s). Thus, {t, n, b} is an orthonormal basis

= II:n X n

= O.

Equation (9) shows that b is perpendicular to t. Being perpendicular to both t and b, b must be parallel to n, so

b = -Tn,

(10)

for some scalar 1', which is called the torsion of 'Y (the minus sign is purely a matter of convention). Note that the torsion is only defined if the curvature is non-zero. Of course, we define the torsion of an arbitrary regular curve 'Y to be the torsion of a unit-speed reparametrisation of "y. As in the case of the curvature, to see that this makes sense, we have to show that if we make a change in the unit-speed parameter of "y of the form

(7)

t.i = 0,

(9)

u = ±s + c, where c is a constant, then l' is unchanged. But this change of parameter has the following effect on the vectors introduced above:

t

I-t

±t,

i

It follows from Eq. (8) that

I-t

i,

n ｾ

'T I-t 1',

n, b I-t ±b,

as required.

b I-t b.

Then,

Just as we did for the curvature in Proposition 2.1 l it is possible to give a

formula for the torsion of a space curve "( in terms of"( alone, without requiring

. ds, "(=dt"('

a unit-speed reparametrisation:

Proposition

2.3

Let "((t) be a regular curve in R' with nowhere vanishing curvature. Then, denoting d/dt by a dot, its torsion is given by 7=

(1' x i)·'f Ih x il1 2 '

... _ (dS)'", "( dt "(

+

l' x i

3ds rl?s -" dt dt 2 '

+

d's, dt'''('

=

ＨｾＺＩＧ

"('

xi',

... (."( x "(") = (dS)6 '" (' "(. dt "(. "( x "(") , and so ';;.(1' x i) _ "("'.("(' xi') x "(" 11 2 •

o

II l' x i II' - II "('

We could 'derive' Eq. (11) by imitating the proof of Proposition 2.1. But it is easier and clearer to proceed as follows, even though this method has the disadvantage that one must know the formula for 7 in Eq. (11) in advance. We first treat the case in which "( is unit-speed. Using Eqs. (7) and (10),

= ｾｩ

+ rl?s_/ dt 2 " ,

(11)

Proof 2.3

Now, n

.. _(dS)2" "( dt "(

Hence,

Note that this formula shows that 7(t) is defined at all points ,,((t) of the curve at which its curvature I«t) is non-zero, since by Proposition 2.1 this is the condition for the denominator on the right-hand side to be non-zero.

7

39

2. How Much Does a Curve Curve?

Elementary Differential Geometry

38

Example

2.4

We compute the torsion of the circular helix "((8) = (acos8,asin8,bO)

= -n.b = -n.(t x nr = -n.(t x n + t x ill = -n.(t x ill. = ｾ［ｙＬ

studied in Example 2.1. We have

so 7

=

1.. (."( x -K"('

1.. (."( x = --"(. K,

(1..)) K"(

1'(8) = (-asin8,acos8,b),

(1... i< .. )) -"( - -"(

';;(8) = (a sin 8, -acos8,0).

d dt

i(8) = (-acos8,-asin8,0),

K.,..,2

Hence,

1 ... (. ..) = 2"(' "( x "( , I
O. Now, t=O

2

= II UU 11 = (it + vi').(-), + vi') = -)'.-)' + 2v.y.;Y + v 2 ;y.i' = 1 + v2 K,2, F = Uu.U v = (-)' + v;Y).-)':::: .y..y + v.y.i' = 1, G = II U 11 2 = .y.-)' = 1, E

t::::

0.6

V

since -)'.-)' = 1, -)'.;Y = 0, ;Y.1' :::::: developable is

",2.

So the first fundamental form of the tangent

t

= 0.2

t = 0.8

t:::: 0.4

t=l

(4)

We are going to show that (part of) the plane can be parametrised so that it has the same first fundamental form. This will prove the proposition. By Theorem 2.1, there is a plane unit-speed curve i' whose curvature is '" (we can even assume that its signed curvature is K). By the above calculations, the first fundamental form of the tangent developable of i' is also given by (4).

ementary Differential Geometry

5.8 Consider the surface patches u(ulv)

= (coshucosv,coshusinv,uL

u(u,v) = (u cos v, u sin v, v),

0

0'(E1 du 2 + 2Fl dudv + G l dv 2 )

E2uti + F2 (ut + tiv) + G2 vt cos () = .2 . . .2 2 (E2u + 2F2uv + G2v2 )l12 (E2u + 2F2uv + G 2v )l12 >..El uft + >..F1(itt + tlv) + >'G l vt 2 2 2 = (>..E l u + 2>..Fl uv + >"G l v 2 )1/ 2(>..E 1 tl + 2>..F1 tlt + >..G l t )l/2 E l uti + F1 (ut + tv) + G l vt 2 2 = (E1 u + 2F1 uv + G1 v 2 )1/2 (E l t/ + 2Fl fit + Gl t )l/2 '

+ Fl

2

+ 2F2 dudv + G 2 dv 2

for some smooth function >.(u, v), where (u, v) are coordinates on U. Note that >.. > 0 everywhere, since (for example) E 1 and E 2 are both> O. If 'Y(t) = O'I(U(t),v(t)) and .y(t) = al(ii.(t),v(t)) are curves in SI, then f takes "y and i to the curves a2(u(t), v(t)) and a2(ii.(t), vet)) in 52, respectively. Using Eq. (5), the angle 0 of intersection of the latter curves on 52 is given by

Ii =

Substituting in Eq. (6) gives

are proportional, say

E 2 du 2

0,

ｲ］ｾｅＲｃｏｓﾢＫｐＮ［ＬｳＢｩｮ

+ Gl

sin 2 ¢) (7)

VE (E 2

2

cos 2 ¢

+ 2F2 sin ¢ cos ¢ + G 2 sin 2 ¢) .

Squaring both sides of Eq. (7) and writing (El cos ¢ + F] sin ¢)2 = E 1 (E 1 cos 2 ¢ + 2Fl sin ¢ cos ¢

+ Gl

sin 2 ¢)

- (El G l

-

F't) sin 2 ¢,

we get

(EIGI-F12)E2(E2 cos 2 ¢ = (E2 G 2

-

+ 2F2 sin¢cos¢ + G 2 sin 2 ¢) F:j)E l (El cos 2 ¢ + 2F1 sin 41 cos ¢ + G 1 sin 2 41),

or, setting>.. = (E 2 G 2

-

Fi)EtI(E1 G 1

2

(E2 - >..Ed cos ¢

+ 2(F2

-

-

(16)

F{)E2 ,

>..Fd sin¢cos¢ + (G 2 - >"Gd sin z 41 = O.

Taking 41 = 0 and then ¢ = IT /2 gives E z = >..E1 , G 2 = >"G 1, and then 0 substituting in the last equation gives F 2 = >.F1 .

Example 5.7 We consider the unit sphere x 2 + y2 + Z2 = 1. If P = (u, v, 0) is any point in the xy-plane, draw the straight line through P and the north pole N = (0,0,1).

Elementary Differential Geometry

110

This line intersects the sphere at a point Q, say. Every point Q of the sphere arises as such a point of int.ersection, with the sole exception of the north pole itself.

111

As to

we get

0'1,

2(V (ud" = ( (u2 0'

N

_

(

( d.. -

2

-

u2

+ 1)

-4uv

4U)

+ v2 + 1)2 , (u 2 + v 2 + 1)2' (u2 + v2 + 1)2 -4uv 2(u 2 - v 2 + 1) 4v ) (u2 + v2 + 1)2' (u:.! + v 2 + 1)2' (u 2 + v 2 + 1)2

' .

This gives

E = (ud"·(O'd,, = 1

4(v 2

-

u2

+ 1)2 + 16u 2 v 2 (u 2 + v 2 + 1)4

+ 16u 2

=

which simplifies to E 1 4/(u 2 +v 2 + 1)2. Similarly, PI = 0, G 1 = E l . Thus, the first fundamental form of 0'2 is .\ times that of 0'1, where .\ = t(u 2 + v2 + 1)2.

EXERCISES

The vector NQ is parallel to the vector NP, so there is a scalar, say p, such that the position vector q of Q is related to those of Nand P by q - n = pep -

0),

5.9 Show that every isometry is a conformal map. Give an example of a conformal map that is not an isometry. 5.10 Show that the curve on the COne in Exercise 5.2 intersects all the rulings of the cone at the same angle. 5.11 Show that Mercator's parametrisation of the sphere

and hence

D'(u, v) = (sech u cos v, sech u sin v, tanh u) q= (O,O,I)+p«u,v,O) - (0,0,1)) = (pu,pv,l-p).

Since Q lies on the sphere,

is conformal. 5.12 Let

lex)

p2 U2 + p2 V2 + (1 _ p)2 = 1 which gives p = 2/(u 2 + v 2 + 1) (the other root p = 0 corresponds to the other intersection point N between the line and the sphere). Hence, q

=

2u ( u 2 + v2

2 2 2v U +V -1) + 1 ｾ u 2 + v 2 + l' u 2 + v 2 + 1 .

If we denote the right-hand side by O'l(U,V), then 0'1 is a parametrisation of the whole sphere minus the north pole. Parametrising the plane z = 0 by 0'2(U,V) = (u,v,O), the map that takes Q to P takes O'l(U,V) to 0'2(U,V). This map is called stereographic projection. We are going to prove that it is conformal. According to Theorem 5.5, we have to show that the first fundamental forms of 0'1 and 0'2 are proportional. The first fundamental form of 0'2 is du 2 + dv 2 .

be a smooth function and let O'(u,v) = (ucosv,usinv,f(u))

be the surface obtained by rotating the curve z = f(x) in the xzplane around the z-axis. Find all functions f for which 0' is conformal. 5.13 Let

0'

be the ruled surface

O'(U, v)

= "'(u) + v6(u),

where "'( is a unit-speed curve in R 3 and 6(u) is a unit vector for all u. Prove that 0' is conformal if and only if 6(u) is independent of u and 'Y lies in a plane perpendicular to 6. What kind of surface is 0' in this case? 5.14 Show that the surface patch O'(u,v) = (J(u,v),g(u,v),O),

Elementary Differential Geometry

112

where 1 and 9 are smooth functions on the uv-plane, is conformal if and only if either 111. = 911 and 111 = -911.,

113

5. The First Fundamental Form

Recalling that the area of a parallelogram in the plane with sides a and b is II a x b II, we see that the area of the parallelogram on the surface is approximately

II O'u Llu x O'v Llv [I = II au x O'v II

or

/11.

= -gv

and /v

= guo

The first pair of equations are called the Cauchy-Riemann equa· tions; they are the condition for the map from the complex plane to itself given by u + iv f.4 I(u, v) + ig(u, v) to be holomorphic. The second pair of equations says that this map is anti-holomorphic, i.e. that its complex-conjugate is holomorphic. We shall say more about holomorphic functions in relation to surfaces in Section 9.4.

5.4. Surface Area Suppose that q : U -+ R 3 is a surface patch on a surface S. The image of a is covered by the two families of parameter curves obtained by setting u = constant and v = constant, respectively. Fix (uo,vo) E U, and let.1u and Llv be very small. Since the change in 0'(u, v) corresponding to a small change Llu in u is approximately O'uLlu and that corresponding to a small change Llv in v is approximately O'v Llv, the part of the surface contained by the parameter curves in the surface corresponding to u = Uo, U = Uo + Llu, v = Vo and v = Vo + Llv is almost a parallelogram in the plane with sides given by the vectors O'u Llu and O'vLlv (the derivatives being evaluated at (uQ,vo)):

Llu.;lv.

This suggests the following definition.

Definition 5.3 The area k(R) of the part O'(R) of surface patch a : U --+ R 3 corresponding to a region R ｾ U is k(R) =

JLII

O'u x O'v

II dudv.

Of course, this integral may be infinite - think of the area of a whole plane, for example. However, the integral will be finite if, say, R is contained in a rectangle that is entirely contained, along with its boundary, in U. The quantity II O'u X a v II that appears in the definition of area is easily of the first fundamental form Edu 2 + 2Pdudv + Gdv 2 of 0': computed in ｴ･ｲｭｾ

Proposition 5.2

II00u x O'v II =

(EG - P 2)1/2.

Proof 5.2

We use a result from vector algebra: if a, b, c and d are vectors in R 3 , then

(a Applying this to II

0'11.

X

II au

X

0'11

b).(c x d) = (a.c)(b.d) - (a.d)(b.c). x O'v 2

11 =

11 2 = (O'u

X

(1v).(O'u

X

(1v), we get

((1U'O'U)(O'11'O'V) - (O'u.(1v)2 = EG ｾ p 2.

Note that, for a regular surface, EG - p2 surface 0'11. x O'v is never zero. Thus, our definition of area is k(R) =

JL

o

> 0 everywhere, since for a regular

(EG - p 2 )1/2dudv.

(8)

We sometimes denote (EG - P2)1/2dudv by dAq. But we have still to check that this definition is sensible, Le. that it is unchanged if 0' is reparametrised.

Elementary Differential Geometry

114

This is certainly not obvious, since E, F and G change under reparametrisation (see Exercise 5.4).

Proposition 5.3

5. The First Fundamental Form

5.16 A surface is obtained by rotating about the z-axis a unit-speed curve "( in the xz-plane that does not intersect the z-axis. Using the ｳｴ｡ｮｾ dard parametrisation of this surface, calculate its first fundamental form, and deduce that its area is 27f /

The area of a surface patch is unchanged by reparametrisation. Proof 5.3

Let u : U -t R 3 be a surface patch and let iT : U -t R 3 be a. reparametrisation of u, with reparametrisation map ｾ : U -t U. Thus, if 4i( i/., v) = (u, v), we have a(ii.,v) = u(u,v).

Let R ｾ

(j be a region, and let R = ifJ(R) ｾ

!hi'

Uu

x

UU

II dudv =

U. We have to prove that

!kII

iTf> x ail

where

ｊＨｾＩ

det(J(tll)

Uu

x ail

II

di/.dv =

21ta(sl Uu

= "(8) + a(n(s) cos 8 + b(s) sin 0).

Give a geometrical description of this surface. Prove that u is regular if the curvature J\, of"( is less than a-I everywhere. Assuming that this condition holds, prove that the area of the part of the surface given by'so < 8 < Sl, 0 < e < 27f, where 80 and 81 are constants, is

x uu,

!kｉ､･ｴＨｊｾﾻｬ

5.17 Let "(8) be a unit-speed curve in R 3 with principal normal n and binormal b. The tube of radius a > 0 around "( is the surface parametrised by

u(8,8)

is the jacobian matrix of ifJ. Hence,

!kII au

p( u) du,

where p(u) is the distance of "(u) from the z-axis. Hence find the area of (i) the unit sphere; (ii) the torus in Exercise 4.10.

II dii.dv.

We showed in the proof of Proposition 4.2 that

au x ail =

115

x

UU

80).

II di/.dv.

By the change of variables formula for double integrals, the right-hand side of this equation is exactly

/hi'

Uu

x Uv

11

dudv.

o

This proposition implies that we can calculate the area of any surface S by breaking S up into pieces, each of which are contained in a single surface patch, calculating the area of each piece using Eq. (8), and adding up the results (cf. Section 11.3, where an analogous procedure is carried out).

EXERCISES 5.15 Determine the area of the part of the paraboloid z = x 2 + y2 with 2 Z :S 1 and compare with the area of the hemisphere x + y2 + Z2 = 1, z < o.

The tube around a circular helix

Elementary Differential Geometry

116

5. The First Fundamental Form

117

z

5.5. Equiareal Maps and a Theorem of Archimedes We are going to use the formula (8) for the area of a surface to prove a theorem due to Archimedes which, legend has it, was inscribed onto his tombstone by the Roman general Marcellus who led the siege of Syracuse in which Archimedes perished. Naturally, since calculus was not available to him, Archimedes's proof of his theorem was quite different from ours. From his theorem, we shall deduce a beautiful formula for the area of any triangle on a sphere whose sides are arcs of great circles. In modern language, the Theorem of Archimedes asserts that a certain map between surfaces is equiareal, in the following sense:

Definition 5.4 Let 8 1 and 8 2 be two surfaces. A diffeomorphism f : 8 1 -t 8 z is said to be equiareal if it takes any region in 8 1 to a region of the same area in 8 2 , To find a {orrI\ula for f, let (x, y, z) be the cartesian coordinates of P, and (X, Y, Z) those of Q. Since the line PQ is parallel to the xy-plane, we have Z = z and (X, Y) = -X(x, y) for some scalar .\. Since (X, Y, Z) is on the cylinder, 1 = X 2 + y Z = .\2(X Z +y2),

We have the following analogue of Theorem 5.1.

Theorem 5.3 A diffeomorphism f : 8 1 -t 82 is equiareal if and only if, for any surface patch 0'( u, v) on 8 1 , the first fundamental forms z E 1 du 2 + 2F1 dudv + G 1 dv 2 and E 2 du z + 2F2 dudv + G 2 dv of the patches

(1

on 8 1 and f

-

Taking the

+ sign gives the point Q,

ｆｾ

=:

E 2GZ

-

F:j.

(9)

The proof is very similar to that of Theorem 5.1 and we leave it as Exercise 5.22. For Archimedes's theorem, we consider the unit sphere x Z + y2 + z2 = 1 and the cylinder x 2 + yZ = 1. The sphere is contained inside the cylinder, and the two surfaces touch along the circle x Z + y2 = 1 in the xy-plane. For each point P on the sphere other than the poles (0,0, ±1), there is a unique straight line parallel to the xy-plane and passing through the point P and the z-axis. This line intersects the cylinder in two points, one of which, say Q, is closest to P. Let f be the map from the sphere (minus the two poles) to the cylinder that takes P to Q.

so we get

f(x, y, z) = Cx 2 +Xy2 )l/2' (x 2 +Yy2 )1/2

on 8 2 satisfy

0 (1

E 1G 1

.\ = ±(x2 + yZ)-1/2.

I

z) .

We shall show in the proof of the next theorem that f is a diffeomorphism.

Theorem 5.4 (Archimedes's Theorem) The map f is equiareal. Proof 5·4

We take the atlas for the surface Sl consisting of the sphere minus the north and south poles with two patches, both given by the formula

°

a1(O,tp) = (cos cos tp, cos 9sintp,sinO), and defined on the open sets

{-7r!2
'1 - '"

I

o

0

A2 -

K

I = 0,

ｾＲＩ

(17)

where the last equality uses the fact that FIl is symmetric. Hence, Eq. (17) gives

because B is orthogonal, and CtFIlC

= 0,

are invertible. Obviously then, if T is any 2 x 1 column

since t 1 and t2 are perpendicular unit vectors. Let

OjI

say. Then

Ct(FIl - KFI)C = 0,

By multiplying out the matrices, it is easy to check that AtFIA = (T{FITI T 2FIT1

I'C,

= A2 = /\, and Eqs.

ＮＩＲｾ

1]1

135

-

I'L (

ｾ ｾＩ

==

o.

Example 6.3 It is intuitively clear that a sphere curves the same amount in every direction, and at every point of the sphere. Thus, we expect that the principal curvatures of a sphere are equal to each other at every point, and are constant over the sphere. To confirm this by calculation, we use the latitude longitude parametrisation as usual. We found in Example 5.2 that

E

= 1,

F

= 0,

G

= cos 2 8,

136

Elementary Differential Geometry

ay

137

6. Curvature of Surfaces

To find the principal vectors t 1 and t2, recall that ti

and in Example 6.2 that

L::; 1, M::; 0, N::; cos 2 e.

Ti

= ＨｾＺＩ

= £iO'u + r/iO'v, where

satisfies

So the principal curvatures are the roots of

[I ｾ

cos 2

K

e _0" cos2 eI::; 0,

i.e. K. ::; 1 (repeated root), as we expected. Any tangent vector is a principal vector.

i.e. For

1';,1

(=

1),

we get

Example 6.4

6

We consider the circular cylinder of radius one and axis the z-axis, parametrised in the usual way: O'(u, v) = (cos v, sin v, u). We found in Example 5.3 that

So T 1 is a multiple of

ＨｾＩＬ

and in Example 6.2 that L = 0, M = 0, N

= 1.

o.

and hence t 1 is a multiple of 00' u

(- sin v, cos v, 0). Similarly, for

"2

and hence t 2 is a multiple of Uu

E ::; 1, F = 0, G = 1,

=

+ 10' v

=

0' v

=

ｾ

) ,

(= 0), one finds that T 2 is a multiple of (

= (0,0,1).

As we mentioned above, one reason for introducing the principal curvatures and principal vectors is contained in the following result, which shows that, if we know the principal curvatures and principal vectors of a surface, it is easy to calculate the normal curvature of any curve on the surface:

So the principal curvatures are the roots of

° 1=0

°

[ O-K.

1-11:

11:(1';, K.

1)

Corollary 6.1 (Euler's Theorem) '

Let 'Y be a curve on a surface patch 0', and let 11:1 and K.2 be the principal curvatures of 0', with non-zero principal vectors t 1 and t2. Then, the normal curvature of'Y is

= 0,

= 0 or 1.

where () is the angle between -j and t 1 · Proof 6.1 ｾ

1

- 7 tt

We can assume that 'Y is unit-speed. Let t be the tangent vector of -y, and let t

= ｾｏＧｵ

+ 'f/U v , T

= ＨｾＩＮ

Suppose first that

1';,1

= "2 = 11:, say. By Proposition

6.3(ii), the normal curvature of'Y is 1';,1'1

= T t FIlT =

I';,

T t FIT =

II:

t.t =

11:.

This agrees with the formula in the statement of the corollary, since K.1

cos2 ()

+ 1';,2 sin2 8 = l';,(cos2 (} + sin2 9) = 1';,.

138

Elementary Differential Geometry

Assume now that K1 =f. K2, so that t 1 and t 2 are perpendicular by Proposition 6.3(iii). We might as well assume that t1 and t2 are unit vectors. Let

= {iD'u + rJiD'11'

ti for i

Ti

= Ｈ rJiｾｩ )

'

= 1, 2. Now "y = COS8t1

+ sin8t 2 ,

+ sin8(6D'u + fJ2D'...,) = ｾ｡ｵ + 11D'v, cos 8 + 6 sin 8 = e, 7]1 cos 8 + 7]2 sin8 = 11,

cos8(6D'u + 711D'11)

T

6. Curvature of Surfaces

139

Proof 6.2

If the principal curvatures 1t1 and K2 are different, we might as well suppose that K1 > K2. Let K n be the normal curvature of a curve 'Y on the surface. Then, since 2 2 2 K n = 1t1 cos 0 + K2 sin (} = 1t1 - (K1 - K2) sin 8, it is clear that Itn :5 1t1 with equality if and only if (} = 0 or 1r, i.e. if and only if the tangent vector i' of'Y is parallel to the principal vector t1- Similarly, one shows that Kn ｾ K2 with equality if and only if i' is parallel to t 2 . If K1 = K2, the normal curvature of every curve is equal to K1 by Euler's Theorem and every tangent vector to the surface is a principal vector by Propo0 sition 6.3(ii).

so ... 6

4

= cos8Tl + sin8T2 •

We conclude this section with the following computation which will be very useful on several occasions later in the book.

Proposition 6.4 Let N be the standard unit normal of a surface patch a( u, v). Then,

(20) where (:

Hence, by Eq. (10), the normal curvature of'Y is It n

= (cos8T; + sin OTi)Fl/(cos OT1 + sinOT2 )

+ cosO sin o(T;:Fl/T2 + TjFl/Td

+ sin 2 (}TiFl/T2 .

(18)

By Definition 6.2 and Eq. (9),

Tl :FIlTj

= KiTl:FITj =

{OKi

1

= -:Fi :Fl/.

The matrix :Fi 1:FII is called the Weingarten matrix of the surface patch and is denoted by W.

= T t FIlT = cos 2 0T;:FIIT1

ｾＩ

if i = j, otherwise. Substituting this into Eq. (18) gives the result.

(19)

o

Proof 6.4

Since N is a unit vector, we know that N u and N..., are perpendicular to N, hence are in the tangent plane to a, and hence are linear combinations of D' u and a...,. So scalars a, b, c and d satisfying Eq. (20) exist. To calculate them, note that N.au = 0 implies, on differentiating with respect to ti, that

Corollary 6.2 The principal curvatures at a point of a surface are the maximum and minimum values of the normal curvature of all curves on the surface that pass through the point. Moreover, the principal vectors are the tangent vectors of the curves giving these maximum and minimum values.

D',

Nu.D'u+N.D'uu = 0, Nu.D'u Similarly,

= -L.

Elementary Differential Geometry

140

Taking the dot product of each of the equations in (20) with gives -L = aE + bF, -M;:::aF+bG,

- M ;::: cE

tlu

and

tllJ

¥

6. Curvature of Surfaces

thus

6.19 Show that a curve on a surface is a line of curvature if and only if its geodesic torsion vanishes everywhere (see Exercise 6.11). 6.20 Two surfaces 8 1 and 8 2 intersect in a curve C that is a line of curvature of Sl. Show that C is a line of curvature of S2 if and only if the angle between the tangent planes of 8 1 and ｳＬｾ is constant along

+ dF,

-N=cF+dG.

These four scalar equations are equivalent to the single matrix equation

-(t

ｾＩ［Ｚ

(;

ｾＩ

-FII;::: fl ( :

(:

(:

C.

Ｉｾ

6.21 Let (f : W ｾ R 3 be a smooth function defined on an open subset W of R 3 such that, for each fixed value of u (resp. v, w), tI(u, v, w) is a (regular) surface patch. Assume also that

ｾＩ

(21)

ｾＩ

= -fIlfI/.

o

EXERCISES 6.15 Calculate the principal curvatures of the helicoid and the catenoid, defined in Exercises 4.14 and 4.18, respectively. 6.16 Let -y(t);::: CT(U(t),v(t)) be a regular, but not necessarily unit-speed, curve on a surface CT, and denote d/dt by a dot. Prove that the normal curvature of -y is n It

141

=

Lu 2 + 2Muv + NiP Eit 2 + 2Fitv + Gv 2 .

6.17 By using the results of Exercises 5.4 and 6.3, show that the principal curvatures of a surface either stay the same or both change sign when the surface is reparametrised, according to whether the unit normal stays the same or changes sign. Show also that the principal vectors are unchanged by reparametrisation.

6.18 A curve C on a surface 5 is called a line of curvature if the tangent vector of C is a principal vector of 5 at all points of C. If 'Y is a parametrisation of the part of a curve C lying in a surface patch tI of 5, and if N is the standard unit normal of tI, show that C is a line of curvature if and only if

N;::: -Ai', for some scalar A, and that in this case the corresponding principal curvature is A. (This is called Rodrigues' Formula.) Show that the meridians and parallels of a surface of revolution are lines of curvature.

I.

This means that the three families of surfaces formed by fixing the values of u, v or w is a triply orthogonal system (see Section 4.6). (i) Show that tlu.tl lJW ;::: tlv.tl uw ;::: (fw.(fuv = O. (Differentiate the Eqs. (21).) (ii) Show that, for each of the surfaces in the triply orthogonal system, the matrices fl and fII are diagonal. (Note that the standard unit normal of the surface obtained by fixing u is parallel to tIu, etc.) (iii) Deduce that the intersection of any surface from one family of the triply orthogonal system with any surface from another family is a line of curvature on both surfaces. (This is called Dupin's Theorem.) 6.22 The third fundamental form of a surface patch tI(u, v) is 2

II N u 11 du

2

+ 2Nu ·N v dudv+

2

2

II N v 11 dv ,

where N(u,v) is the standard unit normal at tI(u, v). Let :FIll be the symmetric 2 x 2 matrix associated to the third fundamental form in the same way as fI and fII are associated to the first and second fundamental forms (see Section 6.3). Show that FIll = fIIFil fH. (Use Proposition 6.4.)

6.4. Geometric Interpretation of Principal Curvatures The relative values of the principal curvatures at a point P of a surface patch tell us much about the shape of the surface near P. To see this, note first that, by applying a rigid motion of R 3 and a reparametrisation of (f (which does not change the shape of the surface), we can assume that

Elementary Differential Geometry

142

(i)

(1

at P is the xy-plane;

(iii) the vectors parallel to the x and y-axes are principal vectors at P, corre-

sponding to principal curvatures

and

K,1

K,2,

6. Curvature of Surfaces

143

if we neglect terms of order greater than two. We distinguish four cases:

P is the origin and u{O, 0) = Pj

(ii) the tangent plane to

¥

say.

The unit principal vectors can be expressed in terms of "11. and

Uv

by

(i) K,1 and K,2 are both> 0 or both < O. Then, (22) is the equation of an elliptic paraboloid (see Proposition 4.5) and one says that P is an elliptic point of the surface. (ii) 1\;1 and K2 are of opposite sign (both non-zero). Then, (22) is the equation of a hyperbolic paraboloid and one says that P is a hyperbolic point of the surface.

say. Then, if

(iii) One of K,1 and K,2 is zero, the other is non-zero. Then, (22) is the equation of a parabolic cylinder and one says that P is a parabolic point of the surface.

the point (x, y, 0) in the tangent plane at P is equal to

X(6CT u

+ 1}1(1v) + y(6uu + 1}2C1v)

+ Y6)uu + (X1}1 + Y1}2)U v

= (x6

= sU11.

+ tuv,

say. Thus, neglecting higher order terms,

u(s,t) =

(1(0,0)

1 2 + sUu + tu v + 2(s Uuu + 28tuuv + t 2Uvv )

1 2 a + 2sta + t 2 a ) , = (x, y, 0) + 2(8 vv uu uv

(iv) Both principal curvatures are zero at P. Then, (22) is the equation of a plane, and one says that P is a planar point of the surface. In this case, one cannot determine the shape of the surface near P without examining derivatives of order higher than the second (in the non-planar case, these terms are small compared to K,IX2 + K2y2 when x and yare small). For example, the surfaces below (the one on the right is called the monkey saddle) both have the origin as a planar point, but they have quite different shapes.

all derivatives being evaluated at the origin. Thus, neglecting higher order terms, the coordinates of u(s,t) are (x,y,z), where z = u{s, t).N

+ 2Mst + Nt 2 )

= !(Ls 2 2

］ｾＨｳ

ｴＩＨｾ

ｾＩＨＺＮ

Now,

x({I) + (6) = (8)t = (x6 ++ Y6) Y1}2 =

X1}1

'11

Y

1}2

xTI

+ yT2 ,

so z

1 t ) = 2(xT I + yTz ) FIl(xTI + yT2 2 = !(x T{FIlTI + xy(T{FIlT2 + ｔｾｆｉＩ 2

Note that this classification is independent of the surface patch CT, since reparametrising either leaves the principal curvatures unchanged or changes the sign of both of them.

+ y 2 T!;.FIlT2 )

2 + K,2Y 2) , = 21 (K,lX using Eq. (19). We conclude that, near a point P of a surface at which the principal curvatures are K,1 and K,2! the surface coincides with the quadric surface z =

ｾＨｋＬｬｘＲ

+ K, 2y 2)

(22)

Example 6.5

On the unit sphere, K,1 = K,2 = ±1 (the sign depending on the parametrisation) so all points are elliptic (and umbilics). On a circular cylinder, K,1 = ±l, K,2 = 0,

Elementary Differential Geometry

144

so every point is parabolic (and there are no umbilics). On a plane, so all points are planar (1) (and umbilics).

K:l

¥

145

5. Curvature of Surfaces

so u(U) is part of the sphere with centre /\,-l a and radius /\,-1. We have now proved the proposition when S is covered by a single surface patch. For an arbitrary surface S, the preceding argument shows that each patch in the atlas of S is contained in a plane or a sphere. But clearly two overlapping patches must then be part of the same plahe or the same sphere. It follows that the whole of S is contained in a plane or a sphere. 0

= K:2 = 0

We conclude this chapter with the analogue for surfaces of Example 2.2, which tells us that a plane curve with constant curvature is part of a circle.

Proposition 6.5 EXERCISES

Let S be a (connected) surface of which every point is an umbilic. Then, S is either part of a plane or part of a sphere.

6.23 Find the elliptic, hyperbolic and parabolic points on the torus described in Exercise 4.10.

Proof 6.5

6.24 Show that every point on the surface of revolution

Let U : U ---+ R 3 be a surface patch in the atlas of S with U a (connected) open subset of R 2 • Let K: be the commmon value of the principal curvatures of (I. By Proposition 6.3(ii),

u(u, v)

so the Weingarten matrix

W;

1

fi FII =

Ｈｾ

i '

ｾＩＮ

x2

(23) Hence,

so Since U is regular, (I,. and u" are linearly independent, so the last equation implies that K:,. = K:" ; O. Thus, K: is constant. There are now two cases to consider. If K = 0, Eq. (23) shows that N is constant. Then,

= N.u,. =

0, (N.u),,; N.u"

= 0,

so N.(I is a constant, say c. Then (I(U) is part of the plane r.N = c. If K: f:. 0, Eq. (23) shows that N; -K:u+a,

where a is a constant vector. Hence,

II u-

I I I 2 2 -a 11 ;11--N 11 = 2' K:

K:

is parabolic if and only if u is part of a circular cylinder or a circular cone. 6.25 Show that, if p, q and r are distinct positive numbers, there are exactly four umbilics on the ellipsoid y2

Z2

2+2"+2";l. p q r

By Proposition 6.4,

(N.u),.

= (f(u) cos v, feu) sin v, g(u))

K:

(Use Proposition 6.3(ii).)

7 Gaussian Curvature and the Gauss Map

We shall now introduce two new measures of the curvature of a surface, called its gaussian and mean curvatures. Although these together contain the same information as the two principal curvatures, they turn out to have greater geometrical significance. The gaussian curvature, in particular, has the remarkable property, established in Chapter 10, that it is unchanged when the surface is bent without stretching, a property that is not shared by the principal curvatures. In the present chapter, we discuss some more elementary properties of the gaussian and mean curvatures, and what a knowledge of them implies about the geometry of the surface.

7.1. The Gaussian and Mean Curvatures We start by defining two new measures of the curvature of a surface.

Definition 7.1 Let Kl and K2 be the principal curvatures of a surface patch. Then, the gaussian curvature of the surface patch is

and its mean curvature is 1 . H = '2(Kl 147

+ K2)'

148

Elementary DMerentlal Geometry

Note Some authors omit the 1/2 in the definition of H, even though this conflicts with the usual meaning of 'mean'. Note from Exercise 6.17 that the gaussian curvature stays the same when a surface patch is reparametrised, while the mean curvature either stays the same or changes sign. It follows that the gaussian curvature is well defined for any surface S. It is easy to get explicit formulas for H and K:

I. UaUSSi5i1 Cdl05tdi@alld til@ daMs tvlap

Example 7.2 In Example 6.2 we considered the surface of revolution /7(U, v) = (f(u) cosv,j(u) sin v,g(u)), where we can assume that f > 0 and j' d/du). We found that

L=jg-jg, M=O, N=fg. By Proposition 7.1(i), the gaussian curvature is

Let /7( u, v) be a surface patch with first and second fundamental forms

LN - M' (ig - Ig)fg K= EG-F' = f'

+ 2Fdudv + Gdv' and Ldu' + 2Mdudv + Ndv',

respectively. Then, M'. (,') K -- LN EO F7.1 'MFf,NE. (ii) H -- LG2(EG:2) , (iii) the principal curvatures are H ±

v' H'

L-I t2)

161

7. Gaussian Curvature and the Gauss Map

= i1(u, v)

for (u, ii) in some open subset Z of (; containing (uo, vo). This patch has the property that Ph :::::: el everywhere and Pt2 :::::: e2 when tl = O. The parametrisation we want is D'(u,v), where O'(U, v) is the intersection of the curve 82 t-+ .\(U,82) with the curve h t-+ P(tl,V). Thus, we consider the equations

EXERCISES 7.14 Let P be a hyperbolic point of a surface S (see Section 6.4). Show that there is a patch of S containing P whose parameter curves are asymptotic curves (see Exercise 6.12).

7.4. Surfaces of Constant Mean Curvature We now consider surfaces whose mean curvature H is constant. As we shall see in Chapter 9, such surfaces are encountered in real life as the shapes taken up by soap films. We shall discuss the surfaces for which H is everywhere zero in detail in Chapter 9. In this section, we are going to describe a construction which gives a correspondence between surfaces of constant non-zero mean curvature and surfaces of constant positive gaussian curvature.

Definition 7.2

iFrom Eq. (9),

au

-=a,

au

Let a be a surface patch with standard unit normal N, and let A be a constant scalar. The parallel surface 0'>" of 0' is

af! _ b

au - ,

(7)''

and similarly

au av = c, af; av

= d

=

0'+ AN.

.

Hence, the jacobian matrix

lJU) ｾＡ =

(a c) b

d

.

As usual, the fact that this matrix is invertible means that (u , v) can be expressed as smooth functions of (u, ii), for (u, ii) in some open subset V of W n Z containing (uo,iio), and we get a reparametrisation O'(u,v) of u(u,ii). Finally, the equation O'(u, v) = p(tl, v) implies that 0'"

and similarly

=

at l

au Ph

atl

=

au el,

tT

Roughly speaking, IT>' is obtained by translating the surface a a distance A perpendicular to itself (but this is not a genuine translation since N will in general vary over the surface).

162

Elementary Differential Geometry

Proposition 7.5 3,

Let 1\.1 and 1\.2 be the principal curvatures of a surface patch a : U --; R and suppose that there is a constant C such that 11\.11 and 11\.21 are both ｾ C everywhere. Let -\ be a constant with 1-\[ < l/C, and let all. be the corresponding parallel surface of a. Then, (i) a), is a (regular) surface patch; (ii) the standard unit normal of a>' at a),(u,v) is the same as that of a at a(u, v), for all (u, v) E U; (iii) the principal curvatures of (I), are I\.d(1 - AI\.t} and K2/(1 - AK2), and the corresponding principal vectors are the same as those of a for the principal curvatures 11:1 and K2, respectively; (iv) the gaussian and mean curvatures of a>' are

K 1- 2AH + A2K

The principal curvatures of a>' are the eigenvalues of the Weingarten matrix W>. of a A• By Proposition 6.4, this is the negative of the matrix expressing ｎｾ and N; in terms of ｡ｾ and a;. Equation (10) says that the matrix expressing Ｈｉｾ and Ｈｉｾ in terms of (I u and 0'11 is I - AW, and the fact that N l\ = N implies that - W is the matrix expressing N ｾ and N; in terms of (I u and iT 11' Combining these two observations we get

W A = (I - AW)-lW. If T is an eigenvector of W with eigenvalue 11:, then T is also an eigenvector of W A with eigenvalue 11:/(1 - AII:). The assertions in part (iii) follows from this. Part (iv) follows from part (iii) by straightforward algebra. 0

Corollary 7.1

H - >.K 1- 2AH + A2K'

and

163

7. Gaussian Curvature and the Gauss Map

If 0' is a surface patch with constant non-zero mean curvature H, then for A = 1/2H, 0'). has constant gaussian curvature 4H 2 . Conversely, if 0' has constant positive gaussian curvature K, then for A = ±l/VK, a>' has constant

respectively. Proof 7.5

mean curvature

=f! vK.

By Proposition 6.4, (I u

= au + AN = (1 + Aa) au + Ab a v,

>.

= a + ANv = AC(lu + (1 + Ad) (Iv,

l\

av

Proof 7.1

tJ.

(5)

11

This follows from part (iv) of the proposition by straightforward algebra.

0

where the Weingarten matrix

W=-(:

ＮＩｾ

EXERCISES

Hence, ｾｉＨ

= (1 + A(a + d) + A2(ad -

x Ｈｉｾ

bc)) au x (Iv.

Since 11:1 and 1\.2 are the eigenvalues of W (see Section 6.3), and since the sum and .product of the eigenvalues of a matrix are equal to the sum of the diagonal entnes and the determinant of the matrix, respectively, 1\.1

+ K2

= -(a

+ d),

7.. 15 The first fundamental form of a surface patch O'(u, v) is of the form E(du 2 + dv 2 ). Prove that (luu + O'vv is perpendicular to O'u and (111' Deduce that the mean curvature H = 0 everywhere if and only if the laplacian O'uu

+ a 1111 = O.

Show that the surface patch

1\.11\.2 = ad - be.

3

Hence,

(I(U, v)

ｾｉＨ

x a;

= (1 -

AKt}(l - A1I:2) au x (Iv.

(6)

Since IAI < 11C and 111:11 and 11\.21 are ｾ C, it follows that IAll:1I and 1-\1\.21 are < 1, so (1 - AI\.I)(1 - AK2) > 0, and Eq. (6) shows that (I), is regular and that its standard unit normal is NA_

>.

(lu

- II ｾｉＨ

X

),

(111

x (I;

_

11 - II

(lu X (111 (lu x lTv

= ( U - 3u3 + uv 2 ,v - 3v + U 2 V,u 2 -

= N.

2)

has H = 0 everywhere. (A picture of this surface can be found in Section 9.3.) 7.16 Compute the mean curvature of the surface with cartesian equation

z =f(x,y)

II

v

164

Elementary Differential Geometry

where f is a smooth function of x and y. Prove that H = 0 for the surface

z = In (cosY). cos x (A picture of this surface can also be found in Section 9.3.)

7.17 Let .,.(u,v) be a surface patch with first and second fundamental forms Edu 2 + Gdv 2 and Ldu 2 + N dv 2 , respectively (cf. Proposition 7.2). Define E(u, v, w) = .,.(u, v)

+ wN(u, v),

where N is the standard unit normal of .,.. Show that the three families of surfaces obtained by fixing the values of u, v or w in E form a triply orthogonal system (see Section 4.6 and Exercise 6.21). The surfaces w = constant are parallel surfaces of.,.. Show that the surfaces u = constant and v = constant are flat ruled surfaces.

165

7. Gaussian Curvature and the Gauss Map

and Q in X such that f(Q) $ f(R) $ f(P) for all points R in X, so that f attains its maximum value on X at P and its minimum at Q. Proof 7.6

Define f : R 3 ---> R by f(v) = II V 11 2 Then, f is continuous so the fact that S is compact implies that there is a point P in S where f attains its maximum value. Let P have position vector p; then S is contained inside the closed ball of radius II p II and centre the origin, and S intersects its boundary sphere at P. The idea is that S is at least as curved as the sphere at P, so its gaussian curvature should be at least that of the sphere at P, i.e. at least 1/ II p W· To make this argument precise, let 'Y(t) be any unit-speed curve in S passing through P when t = O. Then, f('Y(t)) has a local maximum at t = 0, so :/('Y(t)) = 0,

::2 f('Y(t))

$ 0

at t = 0, Le.

'Y(0).-y(0) = 0, 'Y(O)·i(O)

7.5. Gaussian Curvature of Compact Surfaces We have seen in Section 6.4 how the relative signs of the principal curvatures at a point P of a surface S determine the shape of S near P. In fact, since the gaussian curvature K of S is the product of its principal curvatures, the discussion there shows that (i) if K > 0 at P, then P is an elliptic point; (ii) if K < 0 at P, then P is a hyperbolic point; (iii) if K = 0 at P, then P is either a parabolic point or a planar point (and in the last case we cannot say much about the shape of the surface near P). In this section, we give a result which shows how the gaussian curvature influences the overall shape of a surface. We shall give another result of a similar nature in Section 10.4.

Proposition 7.6 If S is a compact surface, there is a point P of S at which its gaussian curvature K is > O.

We recall that a subset X of R 3 is called compact if it is closed (i.e. the set of points in R 3 that are not in X is open) and bounded (i.e. X is contained in some open ball). In the proof, we shall make use of the following fact about compact sets: if f : R 3 ---> R is a continuous function, then there are points P

+ 1 $ o.

(12)

The equation in (12) shows that p = "1(0) is perpendicular to every unit tangent vector to S at P, and hence is perpendicular to the tangent plane of Sat P. Choose a surface patch .,. in the atlas of S containing P, and let N be its standard unit normal. By the preceding remark, p

N =

±TIPli'

(13)

The inequality in (12) implies that the normal curvature "n = i(O).N of "1 at P (computed in the patch.,.) is $ -1/ II p II or ::': 1/ II p II, according to whether the sign in Eq. (13) is + or -, respectively. By Corollary 6.2, the principal curvatures of.,. at P are either both $ -1/ II p II or both 2': 1/ II p II· In each 0 case, K ::': 1/ II p II' > 0 at P.

7.6. The Gauss Map Proposition 2.2 shows that, if 'Y(s) is a unit-speed plane curve, its signed curvature ". = d 0 and ＨＭｾｦ + = 1 (see Examples 4.13 and 6.2 - note that in these examples we used a dot to denote d/du, but now a dot is reserved for djdt, where t is the parameter along a geodesic). We found in Example 6.2 that the first fundamental form of II is du2 + J(u)2dv 2. Referring to Eq. (2), we see that the geodesic equations are

(*)2

U

= J{u) :

i?,

:t

(f(U)2i.J)

= O.

(7)

We might as well consider unit-speed geodesics, so that

i.J.2

+ J(u)2i.J2

= 1.

(8)

l.From this, we make the following easy deductions:

Proposition 8.5

•

8. Geodesics

183

Proof 8.5

On a meridian, we have v = constant so the second equation in (7) is obviously satisfied. Equation (8) gives i.J. = ±1, so i.J. is constant and the first equation in (7) is also satisfied. For (ii), note that if U = Uo is constant, then by Eq. (8), i.J = ±l/f(uo) is non-zero, so the first equation in (7) holds only if df /du = O. Conversely, if df /du == 0 when u = uo, the first equation in (7) obviously holds, and the second holds because i.J = ±1/ f{uo) and feu) = f(uo) are constant. 0 Of course, this proposition only gives some of the geodesics on a surface of revolution. The following result is very helpful in understanding the remaining geodesics.

Proposition 8.6 (C1airaut's Theorem)

On the surface of revolution

lI{U, v)

= (f{u) cosv, J{u) sin v, g{u)),

(i) every meridian is a geodesic; (ii) a parallel u == Uo (say) is a geodesic if and only if df /du == 0 when U == Uo, i. e. Uo is a stationary point of f·

Let 'Y be a geodesic on a surface of revolution $, let p be the distance of a point of $ from the axis of rotation, and let 1/J be the angle between "t and the meridians of $. Then, p sin 'l/J is constant along 'Y. Conversely, if p sin 1/J is constant along some curve 'Y in the surface, and if no part of'Y is part of some parallel of $, then'Y is a geodesic.

geodesics

In the second paragraph of the proposition, by a 'part' of'Y we mean 'Y(J), where J is an open interval. The hypothesis there cannot be relaxed, for on a

p

184

Elementary DifFerential Geometry

8. Geodesics

185

parallel ¢ = 11'/2, so p sin 'IjJ is certainly constant. But parallels are not geodesics in general, as Proposition 8.5(ii) shows.

u = Uo

Proof 8.6

.. dp'2 u=p-v du ' showing that the first equation in (7) is indeed satisfied.

Parametrising 5 as in Proposition 8.5, we have p = f(u). Note that u ti / II Uti II = U u and CTvl II CT v II = p-1CT v are unit vectors tangent to the parallels and the meridians, respectively, and that they are perpendicular since F = O. Assuming that 1'(t) = u(u(t), v(t)) is unit-speed, we have

l' = cos 1/J Uu + p-l sin 'l/J U v (this equation actually serves to define the sign of 'I/J, which is left ambiguous in the statement of Clairaut's Theorem). Hence, Uu

Since "t = ita.. +

l' =

x

p-l sin 1/J au x

a v'

vU v , this gives VU11. x

CT v

Clairaut's Theorem has a simple mechanical interpretation. Recall that the geodesics on a surface 5 are the curves traced on 5 by a particle subject to no forces except a force normal to 5 that constrains it to move on 5. When 5 is a surface of revolution, the force at a point P of 5 lies in the plane containing the axis of revolution and P, and so has no moment about the axis. It follows that the angular momentum {} of the particle about the axis is constant. But, if the particle moves along a unit-speed geodesic, the component of its velocity along the parallel through P is sin tP, so its angular momentum about the axis is proportional to p sin 1/J.

Example 8.8

Hence,

We use Clairaut's Theorem to determine the geodesics on the pseudosphere (Section 7.2):

psin 1/J = p2 iJ. But the second equation in (7) shows that this is a constant, say {}, along the geodesic. For the converse, if p sin 'ljJ is a constant {} along a unit-speed curve 'Yin 5, the above argument shows that the second equation in (7) is satisfied, and we must show that the first equation in (7) is satisfied too. Since .

u(u, v) = (e U cos v, e11. sin v,

(10)

2 .

P

Differentiating both sides with respect to t gives . ..

2{}2.

2{}2

7

'2)

dp u. (..u-p-v du

U

)).

du 2 + e2u dv 2 . It is convenient to reparametrise by setting w = e- U • The reparametrised

u(v,w) = (.!.cOsv,.!.sinv,jI- 12 +COSh-1W)

{}2

uu= ----p3p=

+ cosh- 1 (e-

We found there that its first fundamental form is

w

2

2u

(9)

Eq. (8) gives

=1_

VI - e

surface is

{} = -sin'ljJ =-, P p2

i;.2

0

= p-l sin 1/J au x a v ,

piJ = sin t/J.

v

when It - tol < to, contrary to our assumption. Hence, the term in brackets must vanish everywhere on 1', i.e.

dp . du U '

=a.

If the term in brackets does not vanish at some point of the curve, say at 1'(to) = u(uo, vo), there will be a number f > a such that it does not vanish for It - tol < f. But then u = a for It - tol < f, so l' coincides with the parallel

w

w

and its first fundamental form is

dv 2

+ dw 2

w2 We must have w > 1 for u to be well defined and smooth. If 1'(t) = u(w(t),v(t)) is a unit-speed geodesic, the unit-speed condition gives (11)

and Clairaut's Theorem gives

1 . •1,

-

W

Sin 'f/

= W21.V = n

J£ 1

(12)

ｌＮＱｾｉＢｲｬｧｊ

ｾｵＬ

...... 1.,;...,.1.1'1;,.1.1' ｌｊＮｉｾ｜ＬＢ［ｲ

where rl is a constant, since p = l/w. Thus, iJ = [}w 2 • If [} = 0 we get a meridian v = constant. Assuming now that fl i- a and substituting in Eq. (ll) gives

Returning now to an arbitrary surface of revolution 5, we describe how Clairaut's Theorem allows us to describe the qualitative behaviour of the geodesics on S. Note first that, in general, there are two geodesics passing through any given point P of S with a given angular momentum fl, for v is determined by Eq. (9) and u up to sign by Eq. (10). In fact, one geodesic is obtained from the other by reflecting in the plane through P containing the axis of rotation (which changes n to -n) followed by changing the parameter t of the geodesic to -t (which changes the angular momentum back to n again). The discussion in the preceding paragraph shows that we may as well assume that n > 0, which we do from now on. Then, Eq. (10) shows that the geodesic is confined to the part of S which is at a distance ｾ n from the axis. If all of S is a distance > {} from the axis, the geodesic will cross every parallel of S. For otherwise, 'U would be bounded above or below on S, say the former. Let 'Ua be the least upper bound of'U on the geodesic, and let {} + 210, where f > 0, be the radius of the parallel 'U 'Ue. If 'U is sufficiently close to 'Uo, the radius of the corresponding parallel will be ｾ n + f, and on the part of the geodesic lying in this region we shall have

Hence, along the geodesic, dv = dw

i!.. = w

(v - va) = =F ｾ

(v -

±

)1-

Vo )

2

flw , [}2 W 2

..;'1 -

/

fl2 W 2,

+ w2

1 = [}2'

(13)

where Va is a constant. So the geodesics are the images under i1 of the parts of the circles in the vw-plane given by Eq. (13) and lying in the region w > 1. Note that these circles all have centre on the v-axis, and so intersect the v-axis perpendicularly. The meridians correspond to straight lines perpendicular to the v-axis.

=

w

lui

V Ｈｮｾｦｲ 1-

>0

by Eq. (10). But this clearly implies that the geodesic will cross u = 'Uo, contradicting our assumption. Thus, the interesting case is that in which part of S is within a distance {} of the axis. The discussion of this case will be clearer if we consider a concrete example whose geodesics nevertheless exhibit essentially all possible forms of behaviour.

1

v

Since w > 1, any geodesic other than a meridian has a maximum value of w, which it attains, and a maximum and minimum value of v, which it approaches arbitrarily closely but does not attain (see the diagram below). This shows that the pseudosphere is 'incomplete', i.e. a geodesic on the pseudosphere cannot be continued indefinitely (in one direction if it is a meridian, in both directions otherwise) .

ｾ

, I

Example 8.9

We consider the hyperboloid of one sheet obtained by rotating the hyperbola x2

-

z2

= 1,

x

> 0,

in the xz-plane around the z-axis. Since all of the surface is at a distance 2:: 1 from the z-axis, we have seen above that, if 0 :5 {} < I, a geodesic with angular momentum n crosses every parallel of the hyperboloid and so extends from Z = -00 to z = 00.

p

188

Elementary Differential Geometry

8. Geodesics

189

1 passing through a point P. If P is on the waist r of the hyperboloid (i.e. the unit circle in the xy-plane), which is a geodesic by Proposition 8.5(ii), then p = 1 at P so ¢ = 1f /2 and C is tangent to r at P. It must therefore coincide with r.

0< Suppose now that regions

n 1. Then the geodesic is confined to one of the two z

2:

J n2 -

1,

z

ｾ

-

J n2 -

{}>1 1,

which are bounded by circles r+ and r- , respectively, of radius n. Let P be a point on r-, and consider the geodesic C that passes through P and is tangent to r- there. Then, ¢ ::::: 1f /2 and p n at P, so C has angular momentum n. Now C cannot be contained in r-, since r- is not a geodesic (by Proposition 8.5(ii)), so C must head into the region below r- as it leaves P. Moreover, C must be symmetric about P, since reflection in the plane through P containing the z-axis takes C to another geodesic that also passes through P and is tangent to r- there, and so must coincide with C by the uniqueness part of Corollary 8.1. Since u :j; 0 in the region below r- by Eq. (10), the geodesic crosses every parallel below r- and z -+ -00 as t -+ ±oo. Suppose now that C is any geodesic with angular momentum n > 1 in the region below r-. Then a suitable rotation around the z-axis will cause C to intersect C, say at Q, and so to coincide with it (possibly after reflecting in the plane through Q containing the z-axis and changing t to -t). We have therefore described the behaviour of every geodesic with angular momentum > 1 that is confined to the region below r-. Of course, the geodesics with angular momentum n > 1 in the region above r+ are obtained by reflecting those below r- in the xy-plane. Suppose finally that n = 1. Let C be a geodesic with angular momentum

=

n

Assume now that P is in the region below r. Then 0 < 'IjJ < tr/2 at P, so as it leaves P in one direction, C approaches r. It must in fact get arbitrarily close to r. For if it were to stay always below a parallel i' of radius 1 + E, say (with t;: > 0), then we would have

lui ?-

J Cｾ J 1-

2

everywhere along C by Eq. (10), which clearly implies that C must cross every parallel, contradicting our assumption. So, if n = 1 the geodesic spirals ｡ｲｯｵｾ､ the hyperboloid approaching, and getting arbitrarily dose to, r but never qmte reaching it.

EXERCISES 8.13 There is another way to see that all the meridians, and the parallels corresponding to the stationary points of f, are geodesics on a surface of revolution considered in this section. What is it? 8.14 A surface of revolution has the property that every parallel is a geodesic. What kind of surface is it ?

p 190

the part of'Y between p' and q' by this shorter path, thus giving a shorter path from p to q in S. We may therefore consider a path 'Y entirely contained in a surface patch u. To test whether "'( has smaller length than any other path in (1 passing through two fixed points p and q in u, we embed 'Y in a smooth family of curves on u passing through p and q. By such a family, we mean a curve 'YT on (1, for each 7 in an open interval (-0,0), such that

8.15 Show that a geodesic on the pseudosphere with non-zero angular momentum {) intersects itself if and only if n < (1 + 11"2)-1/2. How many self-intersections are there in that case? (The condition for a self-intersection is that, for some value of w > 1, the two values of v satisfying Eq. (13) should differ by an integer multiple of 211".)

8.16 Let f: a(v,w) i-7 a(v,w) be an isometry ofthe pseudosphere, where the parametrisation a is that in Example 8.8. (i) Show that f takes meridians to meridians, and deduce that u does not depend on w. (Use the fact that meridians are the only geodesics on the pseudosphere that can be extended indefinitely in one direction.) (ii) Deduce that f takes parallels to parallels. (Parallels are the curves that are perpendicular to every meridian.) (iii) Deduce from (ii) and Exercise 7.11 that w = w. (iv) Show that f is a rotation about the axis of the pseudosphere or a reflection in a plane containing the axis of rotation.

191

8. Geodesics

Elementary Differential Geometry

/

/

(i) there is an E (-0,0);

E

>

0 such that 'YT(t) is defined for all t E (-f,f) and all

7

(ii) for some a, b with

'YT(a)

-f

=p

< a < b < f, we have and

'YT(b)

=q

for all

7

E (-0,8);

(iii) the map from the rectangle (-8,0) x (-E, E) into R 3 given by

(7, t) H "'(T(t) is smooth;

8.17 What do the geodesics on a pseudosphere correspond to in the disc model (Exercise 7.12(ii»? (Use the solution to Exercise 7.12 and the fact that the Mobius transformation z H ｾＺ is a conformal map from the (complex) plane to itself that takes lines and circles to lines and circles.)

(iv) 'Yo = 'Y.

8.18 Describe the geodesics on (i) a spheroid, obtained by rotating an ellipse around one of its axes; (ii) a torus (Exercise 4.10).

,'\

8.4. Geodesics as Shortest Paths

The length of the part of 'YT between p and q is

Everyone knows that the straight line segment joining two points P and Q in a plane is the shortest path between P and Q. It is almost as well known that great circles are the shortest paths on a sphere. And we have seen that the straight lines are the geodesics in a plane, and the great circles are the geodesics on a sphere. To see the connection between geodesics and shortest paths on an arbitrary surface S, we consider a unit-speed curve "'( on S passing through two fixed points p and q on the surface. If "'( is a shortest path on the surface from p to q, then the part of"'( contained in any surface patch u of S must be the shortest path between any two of its points. For if p' and q' are any two points of 'Y in (1, and if there were a shorter path in u from p' to q' than 'Y, we could replace

l II t b

£(7) =

T

II

dt,

where a dot denotes d/dt.

Theorem 8.2 With the above notation, the unit-speed curve 'Y is a geodesic d

d7£(T) = 0

Note Although we assumed that 'Y 'YT is unit-speed if T "I O.

if and only if

when T = O.

= 'Yo is unit-speed, we

do not assume that

p 192

Elementary DifFerential Geometry

Proof 8.2

We use the formula for 'differentiating under the integral sign': if f(T, t) is smooth,

d

= -d

dT

=

r

b

d

dT leT) = dT

1a II ",1' II

l

b

(l

-(geT, t)1/2) dt ar

= "21(1

a

g(r, t)-1/2 ｡ｾ

U(r, t) :::: ＲｴｩｵｅＨ ＯＱＭｧｾ VCr, t)

8E' 8F.. 8G' .8u = aT U 2 + 2 ar uv + aT v 2 + 2Eu ar

+ E v ｾＩ

. au 2

+ 2Eu arat

+ 2F

ｾ

2 it + 2 (Fu

(

au

+ Fv ｾＩ

itil

av )

2

aTat il

(au + 2F ar iI

ail) + it ar

+

(Gu ［ｾ

+ G v ｾＩ

lb

= ｾｧＭＱＯＲＨｅｴｊｩ

+ it aTat + 2Gil aTat

-lb

(:t {g-1/2(Eu

+ (Fu + Gil) ｽｾ + Fv)}

ｾ

+

:t

l Ｈｕｾ b

:t

=0

when r:::: O.

+ ｖＺｾＩ

U (0, t) > 0

ＺｾＩ

dt = 0

when r = 0

(18)

for all families of curves "(", then U :::: V = 0 when r = 0 (since this will . prove that 'Y satisfies the geodesic equations). Assume, then, that condition (18) holds, and suppose, for example, that U i=- 0 when r = O. We will show that this leads to a contradiction. Since U i=- 0 when T = 0, there is some to E (a, b) such that U(O, to) i=- 0, say U(O, to) > O. Since U is a continuous function, there exists 11 > 0 such that

Ｚｾ ｉ {g-1/2(Fu + Gil) }

{g-1/2(Fu + Gil)} .

-

(17)

For the converse, we have to show that, if

g-1/2 { (Eu + FiI) ::;t +(Fit + Gil) ::;t} dt

+ FiI) ｾ

+ 2Fvuil + Gvii)

d drl(r)

2

uU

= g-1/2 { (Eu

{g-1/2(Eu + FiJ) } ,

=

iJ2

av

2

+ 2Fu uv + G u v 2) - ｾ

Now 'Yo = 'Y is unit-speed, so since II ",1" 11 2 = g(r, t), we have g(r, t) = 1 for all t when r = O. Comparing Eqs. (17) with the geodesic equations in (2), we see that, if"( is a geodesic, then U v = 0 when r :::: 0, and hence by Eq. (16),

ail + 2GiI ar

2 2F"UV G .2)aV '2+2F" G u V.2)aU uUV + -aT + (E' vU + v + vV -aT 2 2 au av + 2(Eit + FiJ)a r a t +2(Fu+GiJ)--. aTat The contribution to the integral in Eq. (14) coming from the terms involving the second partial derivatives is

=(E

(16)

dt,

where

and a dot denotes dldt. Now,

ｾ

b

(14)

dt,

g(r, t) = Eu 2 + 2Fuil + Gil 2

= ( Eu

av) 1(1r (au U ar + v ar

d drL(r) =

where

ag ar

T

au :::: av :::: 0 when t = a or b. ar ar Hence, the first term on the right-hand side of Eq. (15) is zero. Inserting the remaining terms in Eq. (15) back into Eq. (14), we get

(Eu 2 + 2Fuil + Gil 2)1/2 dt

(l

(b

1

using integration by parts. Now, since "Y1"(a) and "Y1"(b) are independent of (being equal to p and q, respectively), we have fJ'yT -ar = 0 when t :::: a or b. Since

we see that

dt

ba

l

193

8. Geodesics

if t E (to - 11, to

+ 11)·

Let IjJ be a smooth function such that

dt, (15)

¢(t)

>0

if t E (to -11, to

+ 11)

and ¢(t) = 0 if t ｾ (to -11, to

+ 11)·

(19)

p 194

Elementary Differential Geometry

(The construction of such a function ¢ is outlined in Exercise 8.20.) Suppose that 'Y(t) = 0'( U (t), v (t)), and consider the family of curves 'YT (t) 0'(U(7, t), V(7, t)), where u( 7, t)

Then,

= u(t) + n;f>(t),

v(r, t)

1

/t

6

a

7

ar

T=O

Thirdly, in general, a shortest path joining two points on a surface may not exist. For example, consider the surface S consisting of the xy-plane with the origin removed. This is a perfectly good surface, but there is no shortest path on the surface from the point p (-1,0) to the point q (1,0). Of course, the shortest path should be the straight line segment joining the two points, but this does not lie entirely on the surface, since it passes through the origin which is not part of the surface. For a 'real life' analogy, imagine trying to walk from p to q but finding that there is a deep hole in the ground at the origin. The solution might be to walk in a straight line as long as possible, and then skirt around the hole at the last minute, say taking something like the following route:

=

= v(t).

au/ar = ¢ and av/ar = 0 for all T and t, so Eq. (18) gives o (au +'1 U(O,t)¢(t)dt. 0= Ua+V av) I dt=

195

8. Geodesics

(20)

to-1j

But U(O, t) and ¢(t) are both> 0 for all t E (to -1], to + TJ), so the integral on the right-hand side of Eq. (20) is > O. This contradiction proves that we must have U(O, t) = 0 for all t E (a, b). One proves similarly that V(O, t) = 0 for all t E (a, b). Together, these results prove that "y satisfies the geodesic equations.

o

Ｍ ｡Ｎｦ Ｚ ＾ｾ ｊＨ｜ｌ Ｑｬｯ

It is worth making several comments on Theorem 8.2 to be clear about what

it says, and also what it does not say. Firstly, if'Y is a shortest path on 0' from p to q, then £(T) must have an absolute minimum when T = O. This implies that ､ｾﾣＨｔＩ = 0 when T = 0, and hence by Theorem 8.2 that "y is a geodesic. Secondly, if "y is a geodesic on 0' passing through p and q, then £( T) has a stationary point (extremum) when T = 0, but this need not be an absolute minimum, or even a local minimum, so 'Y need not be a shortest path from p

•




q

I

This path consists of two straight line segments of length 1 a semicircle of radius 1:, so its total length is 2(1 -

---------- - - " '-----.... - -

=

1:)

+ 11"1: = 2 + (11" -

10,

together with

2)t:.

Of course, this is greater than the straight line distance 2, but it can be made as close as we like to 2 by taking t: sufficiently small. In the language of real analysis, the greatest lower bound of the lengths of curves on the surface joining p and q is 2, but there is no curve from p to q in the surface whose length is equal to this lower bound. Finally, it can be proved that if a surface 5 is a closed subset of R 3 (i.e. if the set of points of R 3 that are not in 5 is an open subset of R 3 ), and if there is some path in 5 joining any two points of 5, then there is always a shortest path joining any two points of S. For example, a plane is a closed subset of R 3 , so there is a shortest path joining any two points. This path must be a straight line, for by the first remark above it is a geodesic, and we know that the only geodesics on a plane are the straight lines. Similarly, a sphere is a closed subset of R 3 , and it follows that the short great circle arc joining two points on the sphere is the shortest path joining them. But the surface S considered above is not a closed subset of R 3 , for (0,0) is a point not in 5, but any open ball containing (0,0) must clearly contain points of 5, so the set of points not in S is not open.

¥ 196

197

8. Geodesics

Elementary Differential Geometry

Another property of surfaces that are closed subsets of R 3 (that we shall

(iv) Show that the function

also not prove) is that geodesics on such surfaces can be extended indefinitely, Le. they can be defined on the whole of R. This is clear for straight lines in the plane, for example, and for great circles on the sphere (although in the latter

(t) = 1jJ

surface.

-1)-

has the properties we want.

case the geodesics 'close up' after an increment in the unit-speed parameter

equal to the circumference of the sphere). But, for the straight line 'Y(t) = (t-l,O) on the surface S defined above, which passes through p when t = 0, the largest interval containing t = 0 on which it is defined as a curve in the surface is (-00,1). We encountered a less artificial example of this 'incompleteness' in Example 8.8: the pseudosphere considered there fails to be a closed subset of R 3 because the points of its boundary circle in the xy-plane are not in the

(t-to)

/

/

8.5. Geodesic Coordinates The existence of geodesics on a surface S allows us to construct a very useful

atlas for S.

a(u,v)

EXERCISES 8.19 The geodesics on a circular (half) cone were determined in Exercise 8.7. Interpreting 'line' as 'geodesic', which of the following (true) statements in plane euclidean geometry are true for the cone? (i) There is a line passing through any two points. (ii) There is a unique line pa..sing through any two points. (iii) Any two distinct lines intersect in at most one point. (iv) There are lines that do not intersect each other. (v) Any line can be continued indefinitely. (vi) A line defines the shortest distance hetween any two of its points. (vii) A line cannot intersect itself transversely (i.e. with two nonparallel tangent vectors at the point of intersection). 8.20 Construct a smooth function with the properties in (19) in the following steps: (i) Show that, for all integers n (positive and negative), tn.-,/t' tends to 0 as t tends to o. (Use L'Hopital's rule.) (ii) Deduce from (i) that the function

Ott) =

{.-I/t' o

Proposition 8.7 With the above notation, there is an open subset U of R 2 containing (0,0) such that a : U --> R 3 is an allowable surface patch for S. Moreover, the first fundamental form of a is du 2

+ G(u, v)dv 2 ,

where G is a smooth function on U such that G(O, v)

ift$O

= 1,

Gu(O, v)

= 0,

whenever (O,v) E U.

+ t)O(l -

is smooth everywhere, that 1jJ(t) 1jJ(t) = otherwise.

°

,,(V)

For this, let P be a point of S and let 'Y, with parameter v say, be a unit-speed geodesic on S with 'Y(O) = P. For any value of v, there is a unique unitspeed geodesic iV, with parameter u say, such that iV(O) = 'Y(v) and which is perpendicular to 'Y at 'Y(v). We define a(u, v) = i V (u).

if t ::': 0,

is smooth everywhere. (iii) Show that the function

1jJ(t) = 0(1

p

t)

> 0 if -1 < t < 1, and that

Proof 8.7

The proof that a is (for a suitable open set U) an allowable surface patch makes

Elementary Differential Geometry

198

use of the inverse function theorem, and is similar to the proof of Proposition 4.1 (see Section 4.6). Note first that, for any value of v, Uu(O,v) = dd i'V(U)!

Uv(O,v) = ddVi'V(O) =

'

ｾ

(

g1£

If P is a point on the equator of the unit sphere 8 2 , take -y to be the equator with parameter the longitude I.{J, and let 1'''' be the meridian parametrised by latitude 8 and passing through the point on the equator with longitude I.{J. The corresponding geodesic patch is the usual latitude longitude patch, for which the first fundamental form is d82

has rank 2 when u = v = O. Hence, at least one of its three 2 x 2 submatrices is invertible at (0,0), say

11£

A surface patch a constructed as above is called a geodesic patch, and u and v are called geodesic coordinates.

hv

u

199

Example 8.10

lugu gvIv)

(h

8. Geodesics

-y(v),

1£=0 and that these are perpendicular unit vectors by construction. If u(u, v) (J (u, v), g(u, v), h(u, v)), it follows that the jacobian matrix U

p

Iv). gv

+ cos2 8 dI.{J2 ,

in accordance with Proposition 8.7. We shall give an application of geodesic coordinates in the proof of Theorem

(21)

10.4.

By the inverse function theorem 4.2, there is an open subset U of R 2 such that

the map given by

EXERCISES F(u, v) = (J(u, v), g(u, v))

is a bijection from U to an open subset F(U) of R 2 , and such that its inverse map F(U) ｾ U is also smooth. The matrix (21) is then invertible for all (u, v) E U, so U1£ and U v are linearly independent for (u, v) E U. It follows that u : U -l' R 3 is an allowable surface patch. As to the first fundamental form of a, note first that

E =

2

II au 11 = II

ｾ､

.yV(u)

2

11 = 1

because .yv is a unit-speed curve. Next, we apply the second of the geodesic equations (2) to i'v. The unit-speed parameter is u and v is constant, so we get Fu = O. But when u = 0, we have already seen that au and a v are perpendicular, so F = O. It follows that F = 0 everywhere. Hence, the first fundamental form of u is

8.21 Let P be a point of a surface S and let v be a unit tangent vector to

S at P. Let ')'9 (r) be the unit-speed geodesic on S passing through P when r = 0 and with tangent vector v there. Let a(r,8) = ')'9 (r). It can be shown that a is smooth for - f < r < f and all values of 8, where f is some positive number, and that it is an allowable surface patch for S defined for 0 < r < f and for 8 in any open interval of length::::; 211". This is called a geodesic polar patch on S. Show that, if 0 < R < f,

l

R

II

II

uv(O,v)

2

11 = II

ｺｾ

2

11 =

11

2

dr = R.

By differentiating both sides with respect to 8, prove that

a r .U9

We have

G(O,v) =

ｾＺ

1

because'Y is unit-speed. Finally, from the first geodesic equation in (2) applied to the geodesic -y, for which u = 0 and v is the unit-speed parameter, we get G1£(O, v) = 0. 0 geodesics

= O.

p

200

Elementary Differential Geometry

This is called Gaus.'i Js Lemma - geometrically, it means that the

parameter curve r = R, called the geodesic circle with centre P and radius R, is perpendicular to each of its radii, i.e. the geodesics passing through P.

/

9

MInimal Surfaces

Deduce that the first fundamental form of a is dr 2

+ G(r, B)dB 2 ,

for some smooth function G(r, B).

\

In Section 8.4 we considered the problem of finding the shortest paths between two points on a surface. We now consider the analogous problem in one higher dimension, that of finding a surface of minimal area with a fixed curve as its boundary. This is called Plateau's Problem. The solutions to Plateau's problem turn out to be surfaces whose mean curvature vanishes everywhere. The study of these so-called minimal surfaces was initiated by Euler and Lagrange in the mid-18th century, but new examples of minimal surfaces have been discovered quite recently.

9.1. Plateau's Problem In Section 8.4, we found the condition for a curve on a surface to mininiise distance between its endpoints by embedding the given curve in a family of curves passing through the same two points, and studying how the length of the curve varies as the curve varies through the family. Accordingly, we shall now study a family of surface patches aT : U --+ R 3 , where U is an open subset 2 of R independent of 7', and 7' lies in some open interval (-6,6), for some 6 > O. Let a = aD. The family is required to be smooth, in the sense that the map (u, v, 7') --+ aT(u, v) from the open subset {(u, v, 7') 1 (u, v) E U, 7' E (-6, 6)} of R 3 to R 3 is smooth. The surface variation of the family is the function 'P: U --+ R 3 given by f{J

=

.TI r=O'

iT

201

202

Elementary Differential Geometry

9. Minimal Surfaces

203

where here and elsewhere in this section, a dot denotes d/dr.

and this energy is proportional to its area. A soap film spanning a wire in the

Let lI" be a simple closed curve that is contained, along with its interior int(lI"), in U (see Section 3.1). Then lI" corresponds to a simple closed curve ...,T = aT oll" in the surface patch aT, and we define the area function A(T) to be the area of the part of aT inside ...,T:

shape of a curve C should therefore adopt the shape of a surface of least area with boundary C. By Corollary 9.1, this will be a minimal surface. More generally, if the soap film separates two regions of different pressure, the film will adopt the shape of a surface of constant mean curvature. This is the case for a soap bubble, for example, for which the air pressure inside the bubble is greater than the pressure outside. To see this, we apply the principle of 'virtual work'. This tells us that, if the soap film is in equilibrium, and we imagine a ('virtual') change in the surface, the change in the energy of the film must be the same as the work done by the film against the air pressure. If p is the pressure difference, the force exerted by the air on a small piece of the surface of area LlA is pLlA, so the work done when it moves a small distance a perpendicular to itself is apLlA. On the other hand, the formula in Theorem 9.1 shows that the change in area of the surface is proportional to aHLlA (note that a is the component of the variation cp perpendicular to the surface). So p is proportional to H. Since the pressure difference must be the same across the whole surface, so must the mean curvature H. Surfaces of constant non-zero mean curvature were discussed in Section 7.4.

A(T) =

I!.

dAa·.

int(1f)

/

Note that, if we are considering a family of surfaces with a fixed boundary curve ..." then...,T for all T, and hence cpT(U,V) 0 when (u,v) is a point on the

=...,

=

curve 11'.

Theorem 9.1 With the above notation, assume that the surface variation cpT vanishes along the boundary curve lI". Then,

A(O) = -2

j' (

H(EG - p2)'/2 a dudv,

(1)

lint(1f)

where H is the mean curvature of a, E, P and G are the coefficients of its first fundamental form, a = cp.N, and N is the standard unit normal of a. We defer the proof of this theorem to the end of this section. If a has the smallest area among all surfaces with the given boundary curve ..." then A must have an absolute minimum at T = 0, so A(O) = 0 for all smooth families of surfaces as above. This means that the integral in (1) must vanish for all smooth functions a : U -t R. As in the proof of Theorem 8.2, this can happen only if the term that multiplies a in the integrand vanishes, in other words only if H = O. This suggests the following definition.

For the moment, we give only one eXll.mple of a minimal surface; others will be given in the next section. This example already shows, however, that the converse of Corollary 9.1 is false.

Example 9.1

"

A catenoid is the surface obtained by revolving the curve x = ｾ cosh az in the xz-plane around the z-axis, where a is a non-zero constant (a picture of the catenoid can be found in Example 9.2). We take a = 1 for simplicity. The catenoid can be parametrised by

a(u,v) = (coshucosv,coshusinv,u).

Definition 9.1 A minimal surface is a surface whose mean curvature is zero everywhere.

Then, (It/,

Theorem 9.1 and the preceding discussion then give

= (sinhucosv,sinhusinv, 1), (Iv.

x

(Iv

av

= (-coshusinv,coshucosv,O),

= (-coshucosv, - coshusinv,sinhucoshu),

N = (-sech u cosv, - sech usin v, tanhu),

Corollary 9.1 If a surface S has least area among all surfaces with the same boundary curve, then S is a minimal surface. Minimal surfaces have an interesting physical interpretation as the shapes taken up by soap films. A soap film has energy by virtue of surface tension,

D'uu (luu

= (coshucosv,coshusinv,O),

= (- sinh usin v, sinh ucosv, 0),

a vu = (- cosh u cos v, - cosh usin v, 0).

This gives the coefficients of the first and second fundamental forms of a as 2 E=G=cosh u, P=O, L=-I, M=O, N=l.

204

Elementary Differential Geometry

The first three of these equations show that the parametrisation u is conformal, and Proposition 7.1(ii) gives

H= LG-2MF+NE =0 2(EG - P2) , showing that the catenoid is a minimal surface.

9. Minimal Surfaces

205

The graphs of 1 + e- 2a and 2a as functions of a clearly intersect in exactly one point a = ao, say, and the inequality (2) holds if a > ao. If this condition is satisfied, the catenoid is not area minimising. It can be shown that if a < no the catenoid does have least area among all surfaces spanning the circles C+ and C-. It is time to prove Theorem 9.1.

Proof 9.1

Let cpr = iT"T, so that cpo = cp, and let NT be the standard unit normal of u T. There are smooth functions a"T ,j3"T and "'( of (u, v, T) such that tp"T

= aTN"T + j3"T ｾＧｏ + ,"TＬｾｵ

so that a = aO. To simplify the notation, we drop the superscript T for the rest of the proof; at the end of the proof we put T = O. We have Fix a > 0, and let b = cosh a. The surface S consisting of the part of the catenoid with Izl < a has boundary the two circles C± of radius b in the planes z = ±a with centres on the z-axis. Another surface spanning the same two circles is, of course, the surface So consisting of the two discs x 2 + y2 ｾ b2 in the planes z = ±a. The area of S is, by Proposition 5.2, {21r

r (EG _ p

i o La

2 )1/2dudv

=

(21r

r

cosh 2 ududv

= 21f(a + sinh a cosh a).

i o La The area of So is, of course, 21fb2 = 21T cosh 2 a. So the minimal surface S

will not minimise the area among all surfaces with boundary the two circles C± if cosh 2 a < a + sinh a cosh u, i.e. if 1 + e- 2a

< 2a.

(2)

A(T) =

/1

int(1l')

so

A=

110''1,1. x Uv II dudv =

/1 : int(1f)

/1

int(1l')

N·(O'u x O"v) dudv,

(N·(O'u x O'v)) dudv.

Now,

ｾ

(N.(uu x O"v)) = N.(O'u x Uv ) + N.(uu x O"v) +-N.(u" x uv).

Since N is a unit vector,

N.(O'u x O'v)

= N.N II 0"'1,1. x O'v II =o.

On the other hand, . N •(0"'1,1.

X Uv

2a

) _ ｾ

_ _ -

x O'v).(u" X O'v) 11 uuxUv II (uu·U'l,l.)(Uv.O'v) - (u"'O"v)(uv'O''') Iluuxuvll G(uu'u,,) - F(O'v.uu) (EG - F2)l/2 ' (0''1,1.

using Proposition 5.2. Similarly, N ( . 0''1,1. a

(3)

T

x·) _ E(O"v.uv) - F(u".uv) O"V (EG _ F2)l/2

(4)

Elementary Differential Geometry

206 Substituting these results into Eq. (4), we get

8«

8T N. /T u X /Tv

9. Minimal Surfaces

207

since such a parallel variation causes the surface to slide along itself and will

)) _ E(/Tv.lTv ) - F(iTu./T v + /Tu.iT v ) + G(/Tu.iTu) -

f

(EG _ F2)1/2

(5)

not change the shape, and in particular the area, of the surface. Thus, the main point is to prove Theorem 9.1 for normal variations, i.e. those for which (3 "I 0 everywhere on the\surface. Making this restriction simplifies the

= =

Now iru ::: CPu ::: nuN + f3u tl u

/Tu.iT u = EfJu Since O'u.Nu :::

above proof considerably.

+ lut1v + aN u + {3tT uu + 'Ytruv,

+ F"'{u + (o-u.Nu)a + (/Tu./Tuu)fJ + (/Tu·/Tuvh·

-(J'uu.N :::

-L,

/Tu.iT u = E(3u

(111..(111.11. :::

+ F"'{u

ｾｅｵ

- La +

and

Uu.t1uv :::

ｾｅｵＨＳ

ｾｅｶＬ

EXERCISES

we get

+ ｾｅｶＢＧｻＮ

9.1 Show that any rigid motion of R 3 takes a minimal surface to another minimal surface, as does any dilation (x, y, z) >4 a(x, y, z), where a is a non-zero constant.

Similarly,

/Tv.iTu = F(3u

+ G"'{u -

/Tu.iT v = E(3v

+ F"'{v

/Tv.iTv = F(3v

+ G"'{v -

Ma

- Ma Na

+ (Fu - ｾｅｶＩＨＳ

+ ｾｇｵＢＧｻＬ

+ ｾｅｶＨＳ

+ (Fv - ｾｇｵｨＬ

+ ｾｇｵＨＳ

+ ｾｇｶＢＧｻＮ

9.2 Show that z = l(x, y), where 1 is a smooth function of two variables, is a minimal surface if and only if

(1

Substituting these last four equations into the right-hand side of Eq. (5), simplifying, and using the formula for H in Proposition 7.1(ii), we find that

:T (N.(/T u x /Tv)) = ((3(EG - F 2)1/2)u + ("'{(EG - F 2)1/2) v - 2aH(EG - F 2)1/2.

(6)

Comparing with Eq. (3), and reinstating the superscripts, we see that we must prove that

If.

{((3°(EG - F 2)1/2)

IOt(1r)

+ ("'{O(EG -

F 2)1/2) } dudv = O.

11.

+ 1;)1.. - 21.1.1•• + (1 +

mI•• = o.

9.3 Show that every umbilic on a minimal surface is a planar point (see Proposition 6.3). 9.4 Show that the gaussian curvature of a minimal surface is S 0 everywhere, and that it is zero everywhere if and only if the surface is part of a plane. (Use Proposition 6.5.) We shall obtain a much more precise result in Corollary 9.2. 9.5 Show that there is no compact minimal surface. (Use Proposition 7.6 and Exercise 9.4.)

(7)

v

But by Green's Theorem (see Section 3.1), this integral is equal to

L

(EG - F 2)1/2(fJodv - "'{Odu) ,

and this obviously vanishes because (30 = "'{o = 0 along the boundary curve 1I'. This completes the proof of Theorem 9.1. 0

9.2. Examples of Minimal Surfaces The simplest minimal surface is, of course, the plane, for which both principal curvatures are zero everywhere. Apart from this, the first minimal surfaces to be discovered were those in the following two examples. Example 9.2

Note that we did not quite use the full force of the assumptions in Theorem 9.1, since they imply that aO (= a) vanishes along the boundary curve, and this was not used in the proof. So Eg. (1) holds provided the surface variation I{) is normal to the surface along the boundary curve. Note also that Theorem 9.1 is intuitively obvious for variations I{) that are parallel to the surface, i.e. those for which a = 0 everywhere on the surface,

A catenoid is obtained by rotating a curve x = ｾ cosh az in the xz-plane around the z-ij,Xis, where a > 0 is a constant. We showed in Example 9.1 that this is a minimal surface (we only dealt there with the case a = 1, but the general case follows from it by using Exercise 9.1).

\

208

Elementary Differential Geometry

y I

9. Minimal Surfaces

209

We suppose now that, for some value of u, say u = uo, we have g(uo) f:. O. We shall then have iJ(u) 1- 0 for u in some open interval containing uo. Let such interval. Supposing now that u E (a,{3), the unit(a,!3) be the ｬ｡ｲｾ･ｳｴ speed condition P + g2 == 1 gives-Cas in Example 7.2)

.

..

!

/9 - fg

= --;-, g

so we get

H Since

tl =

==! 2

(! _1) f

iJ

•

1 - j2 , S is minimal if and only if

f!=l-j2.

(8)

To solve the differential equation (8), put h

!

= dh

==

dt

== j,

and note that

dh df = h dh df dt dj"

Hence, Eq. (8) becomes The catenoid is a surface of revolution. In fact, apart from the plane it is the only minimal surface of revolution:

dh fh df Note that, since iJ 'f; 0, we have h 2 follows:

Proposition 9.1 Any minimal surlace 01 revolution is either part by applying a rigid motion to part 01 a catenoid.

01 a plane

or can be obtained

Prool9.1 By applying a rigid motion, we can assume that the axis of the surface S is the z-axis and the profile curve lies in the xz-plane. We parametrise S in the usual way (see Example 6.2):

(T(u, v) = (f(u) cos v, feu) sinv, y(u», where the profile curve u H (f(u) , 0, y(u)) is assumed to be unit-speed and I > O. From Example 6.2, the first and second fundamental forms are du 2 + l(u)2dv 2 and (jg - !iJ)du 2 + fiJdv 2, respectively, a dot denoting djdu. By Proposition 7.1(ii), the mean curvature is .. iJ) . H="21 ( /9-fiJ+7 .

2

=::

1- h .

f:. 1, so we can integrate this equation as

hdh =Jdf f' J 1- h 2 1 Ｎｾ］｡ｦＬ v1- h_ v"a2"--:f"'""2---1 af ' h where a is a non-zero constant. (We have omitted a ±, but the sign can be changed by replacing u by -u if necessary.) Writing h = df/ du and integrating again,

ｦｬ［］ ｾ ＧＺ ｡ｲｶＭｊ

-=1 =

f

J du,

1

= (tVl + a2 (u + bp,

where b is a constant. By a change of parameter b == O. So

U H

u + b, we can assume that

--------------------------aaya Elementilry Differentiill Geometry

210

'2 _

1_

/'2 _

.

du -

1_

-

dg _ ± ｾ

g=

..

211

line starts along the x-axis, at time v the centre of the line is at (0,0, a.v) and it has rotated by an angle wv. Hence, the point of the line initially at (u, 0, 0) is now at the point with position vector

To compute g, we have g -

9. Minimal Surfilces

h2 _ _ 1_ - a 2 j2'

u(u,v) = (u'coswv, usinwv,O'v).

1

VI + ｡ｾＺｌｵＲ

t

ﾱｾ

We leave it to Exercise 9.6 to check that this is a minimal surface.

sinh-1(au) a au = ± sinh(a(g -

+c

(where c is a constant),

en,

f

1 = - cosh(a(g - c)).

a

Thus, the profile curve of S is 1

x = - cosh(a(z - c)). a

By a translation along the z-axis, we can assume that c = 0, so we have a catenoid. We are not quite finished, however. So far, we have only shown that the part of S corresponding to u E (a, (3) is part of the catenoid, for in the proof we used in an essential way that 9 ¥= O. This is why the proof has so far excluded the possibility that S is a plane. To complete the proof, we argue as follows. Suppose that (3 < 00. Then, if the profile curve is defined for values of u ｾ (3, we must have 9«(3) = 0, for otherwise il would be non·zero on an open interval containing (3, which would contradict our assumption that (a, {3) is the largest open interval containing UQ on which il ¥= O. But the formulas above show that ·2

1

9 = 1 + a2 u 2

if

U

E

(a,I3),

so, since il is a continuous function of u, 9(13) = ±(l + a 2/32)-1/2 ¥= O. This contradiction shows that the profile curve is not defined for values of U ｾ (3. Of course, this also holds trivially if /3 = 00. A similar argument applies to a, and shows that (a., (3) is the entire domain of definition of the profile curve. Hence, the whole of S is part of a catenoid. The only remaining case to consider is that in which g(u) = 0 for all values of U for which the profile curve is defined. But then g(u) is a constant, say d, and S is part of the plane z = d. 0 Example 9.3

A helicoid is a ruled surface swept out by a straight line that rotates at constant speed about an axis perpendicular to the line while simultaneously moving at constant speed along the axis. We can take the axis to be the z-axis. Let w be the angular velocity of the rotating line and a its speed along the z-axis. If the

We have the following analogue of Proposition 9.1.

Proposition 9.2 Any ruled minimal surface is part of a plane or part of a helicoid. Proof 9.2

We take the usual parametrisation

t7(u, v) = "Y(u)

+ v6(u)

(see Example 4.12), where 'Y is a curve that meets each of the rulings and 6(u) is a vector parallel to the ruling through 'Y( u). We begin the proof by making some simplifications to the parametrisation. First, we can certainly assume that 116(u) II = 1 for all values of u. We assume also that 6 is never zero, where the dot denotes d/du. (We shall consider later what happens if 6(u) = 0 for some values of u.) We can then assume that 6 is a unit-speed curve (we do not assume that 'Y is unit-speed). These assumptions imply that 6.6 = 6.6 = O. Now we consider the curve

.y(u)

= 'Y(u) -

(;-.6)6(u).

;-.ts,

If ii = v + the surface can be reparametrised using" and the parameters u and v, namely

u(u, v) = ,,(u)

+ v6(u) ,

212

Elementary Differential Geometry

i'"

but now we also have

:.. (,\,.6)6 -

normal is -6. Hence, its binormal is 6 x (-6), and since .. d . -d (6 x 6) = 6 x 6 + 6 x 6 = -6 x 6 = 0 u ' it follows that the torsion of 6 is zero. Hence, 6 parametrises a circle of radius 1 (see Proposition 1.5). By applying a rigid motion, we can assume that 6 is the circle with radius 1 and centre the origin in the xy-plane, so that

(,\,.6)6).6 = 0,

since 6.6 = 0 and 6.6 = 1. This means that we could have assumed that '\'.6 at the beginning, and we make this assumption from now on.

=0

We have

'Y + v6, (Iv = 6, E = II'Y + v611 2 , F = ('\' + v6).6 = '\'.6, G = 1.

ｾ

213

Equation (12) shows that the curvature of the curve 6 is 1, and that its principal

+6 = ('\' -

au

9. Minimal Surfaces

=

6(u) = (cosu,sinu,O).

i,From Eq. (12), we get 6.(..y x 6)

Let A = (EG - F 2 )'(2. Then,

i.(6 x 6) =

N = A-'(,\, + v6) x 6. It follows that

Next, we have D'uu

=

L = M =

i' + v6,

6, A-'(i + v6).((,\, + v6) A-'6.((..y + v6) x 6) = tT uv

=

(1V11

= -6.(,\, x 6) = 0, so by Eq.

i

(10),

O.

is parallel to the xy-plane, and hence that -y(u) = (f(u), g(u), '!u

= 0,

+ b),

where f and 9 are smooth functions and a and b are constants. If a = 0, the surface is part of the plane z = b. Otherwise, Eq. (9) gives

x 6),

A-'6.(..y x 6),

9 cosu -

N=O.

i sin u = 2(j cosu + gsin u).

(13)

We finally make use of the condition ..y.6 = 0, which gives

Hence, the minimal surface condition

isinu =

H = LG - 2MF+NE =0 2A2

isinu + icosu = gcosu -; gsinu.

(i + v6).((,\, + v6) x 6) = 2(6...y)(6.(..y x 6». This equation must hold for all values of (u, v). Equating coefficients of powers of v gives (9)

i.(6 x 6) + 6.(,\, x 6) =

0,

(10)

6.(6 x 6) =

o.

(11)

Equation (11) shows that 6,6 and 6 are linearly dependent. Since 6 and 6 are perpendicular unit vectors, there are smooth functions a(u) and f3(u) such that

6 = a6 + 136. But, since 6 is unit-speed, 6.6 = O. Also, differentiating 6.6 -6.6 = -1. Hence, a = -1 and 13 = 0, so

6= -6.

(14)

Differentiating this gives

gives

i.('\' x 6) = 2(6.'\')(6.('\' x 6»,

gcos u.

= 0 gives 6.6 = (12)

(15)

Equations (13) and (15) together give fcosu

.,

+ gsinu =

O.

i

and using Eq. (14) we get = g = O. Thus, f and 9 are constants. By a translation of the surface, we can assume that the constants f, 9 and b are zero, so that -y(u) = (0,0, au) and a(u,v) = (vcosu,vsinu,au), which is a helicoid. We assumed at the beginning that 6 is never zero. If 6 is always zero, then 6 is a constant vector and· the surface is a generalised cylinder. But in fact a generalised cylinder is a minimal surface only if the cylinder is part of a plane (Exercise 9.8). The proof is now completed by an argument similar to that used

+ Elementary Differential Geometry

214

215

9. Minimal Surfaces

at the end of the proof of Proposition 9.1, which shows that the whole surface is either part of a plane or part of a helicoid. 0 After the catenoid and helicoid, the next minimal surfaces to be discovered were the following two. Example 9·4 Eimeper's minimal surface is

D'(u, v)

= (u - ｾｵＳ

+ uv 2tV -

ｾｖＳ

+ vu 2,u 2 - V2) .

It was shown in Exercise 7.15 that this is a minimal surface.

The white squares have centres of the form (mll', mr), where m and n are integers with m + n even. Since, for such m, n,

+ nrr) cosy cos(x + mrr) = cosx' cos(y

a surface patch in the sense used in this book as Strictly speaking, this is ｴｾｮ it is not injective. The self-intersections are clearly visible in the picture above. However, if we restrict (u t v) to lie in sufficiently small open sets, (1 will be injective by the inverse function theorem. Example 9.5 Scherk's minimal surface is the surface with cartesian equation _ In (COSY) z-- . cos x It was shown in Exercise 7.16 that this is a minimal surface. Note that the surface exists only when cos x and cos y are both > 0 or both < 0, in other words in the interiors of the white squares of the following chess board pattern, in which the squares have vertices at the points (rr /2 + mrr, rr /2 +mr), where m and n are integers, no two squares with a common edge have the same colour, and the square containing the origin is white:

it follows that the part of the surface over the square with centre (mn, nrr) is obtained from the part over the square with centre (0,0) by the translation (x, y, Z).H (x + mrr, y + nrr, z). So it suffices to ･ｘｽｾ｢ｩｴ the part of the surface over a smgle square: C

Elementary Differential Geometry

216

(ii) the parameter curve u = IT is a parabola; (iii) the parameter curve v = 0 is a cycloid (see Exercise 1.7). Show also that each of these curves, when suitably parametrised, is a geodesic on Catalan's surface.

EXERCISES 9.6

Show that the helicoid is a minimal surface.

9.7

Show that the surfaces u t in the isometric deformation of the helicoid into thr catenoid given in Exercise 5.8 are minimal surfaces.

9.8

Show that a generalised cylinder is a minimal surface only when the cylinder is part of a plane.

9.9

A translation surface is a surface of the form

z

where f and 9 are smooth functions. (It is obtained by moving the curve u I-t (u, 0, f(u)) parallel to itself along the curve v t-+ (O,v,g(v)).) Use Exercise 9.2 to show that this is a minimal surface if and only if d 2f/dx 2 ｾｧＯ､ｹＲ

=

9.3. Gauss Map of a Minimal Surface 3

= f(x) + g(y),

1 + (df /dx)2

217

9. Minimal Surfaces

Recall from Section 7.3 that the Gauss map of a surface patch u : U -+ R associates to a point u(u, v) of the surface the standard unit normal N (u, v) regarded as a point of the unit sphere 52. By Eq. (15) in Chapter 7,

where K is the gaussian curvature of u, so N will be regular provided K is nowhere zero, and we assume this for the remainder of this section.

1 + (dg/dy)2'

Deduce that any minimal translation surface is part of a plane or can be transformed into part of Scherk's surface in Example 9.5 by a translation and a dilation (x,y,z) I-t a(x,y,z) for some non-zero constant a. 9.10 Verify that Catalan's surface

u (u, v) = (u - sin u cosh v, 1 - cos u cosh v, -4 sin ｾ sinh

¥)

is a conformally parametrised minimal surface. (As in the case of Enneper's surface, Catalan's surface has self-intersections, so it is only a surface if we restrict (u, v) to sufficiently small open sets.)

Proposition 9.3 Let O'(u, v) be a minimal surface patch with nowhere vanishing gaussian curvature. Then, the Gauss map is a conformal map from u to part of the unit sphere.

We should really be a little more careful in the statement of this proposition, since for us conformal maps are always diffeomorphisms (see Section 5.3). However, even if N u >< N" is never zero, it does not follow that the map u(u, v) I-t N(u, v) is injective (see Exercise 9.12(ii)). Nevertheless, the inverse function theorem tells us that, if (uo, vo) E R 2 is a point where 0' (and hence N) is defined, there is an open set U containing (uo,vo) on which u is defined and on which N is injective. Then, N : U --+ 52 is an allowable surface patch on the unit sphere 52, and the Gauss map is a diffeomorphism from O'(U) to

N(U). Proof 9.3

By Theorem 5.2, we have to show that the first fundamental form (N u .Nu )du 2

+ 2(N u .Nv )dudv + (N".N,,)dv 2

of N is proportional to that of 0'. Form the symmetric 2 >< 2 matrix Show that (i) the parameter curve on the surface given by u = 0 is a straight line;

218

Elementary Differential Geometry

in the same way as we associated symmetric 2 x 2 matrices F j and Fj1 to the first and second fundamental forms of (f in Section 6.3. Then, we have to show that

219

9. Minimal Surfaces

. I·

(16) I

for some scalar A. By Proposition 6.4 l

We saw in Exercise 5.14 that a conformal parametrisation of the plane is necessarily holomorphic or anti-holomorphic, so this proposition strongly suggests a connection between minimal surfaces and holomorphic functions. This connection turns out to be very extensive, and we shall give an introduction to it in the next section.

I

N".Nu

where (:

EXERCISES

= a2 tlu.(f... + 2aba....tTv + b2 tlv.tlv = a2 E + 2abF + b2G ,

ｾＩ

= -Wand W = Fi 1 Fll is the Weingarten matrix. Computing

9.11 Show that the scalar A appearing in the proof of Proposition 9.3 is equal to -K, where K is the gaussian curvature of the surface.

N ....N II and NII.NII in the same way gives

:FIll = (

=

2

2

a E + 2abF + b G acE + (ad + be)F + bdG

Ｈｾ

ｾＩ

:) (;

Ｈｾ

9.12 Show that (i) the Gauss map of the catenoid is injective and its image is the whole of the unit Jlphere except for the north and south poles; (ii) the image of the Gauss map of the helicoid is the same as that of the catenoid, but that infinitely many points on the helicoid are sent by the Gauss map to any given point in its image.

acE + (ad + bc)F + bdG) c2 E + 2edF + d2G

ｾＩ

= (-W)tF1(-W) 1

= (-:Fi :FII)t:Fr(-:Fi 1:F11) 1

1

= FII:Fi F1:Fi :Fll =

FlI Fi 1 :F11.

9.4. Minimal Surfaces and Holomorphic Functions

Hence, Eq. (16) is equivalent to

:Fi 1 :F11:F;1 F11 = AI, i.e.

W2

In this section, we shall make use of certain elementary properties of holomorphic functions. Readers without the necessary background in complex analysis may safely omit this section, whose results are not used anywhere else in the book. We shall need to make use of special surface patches on a minimal surface. Recall from Section 5.3 that a surface patch tI : U -t R 3 is called conformal if its first fundamental form is equal to E(du 2 + dv 2 ) for some positive smooth function E on U.

= AI.

But, W2

=

(a

b

C)2 = ( a2 + bc

b(a + d)

d

c(a + d)) ｾ + be .

Now, recall from Section 6.3 that the principal curvatures K:l and I\, are the . I 2 elgenva ues of W. Since the sum of the eigenvalues of a matrix is equal to the sum of its diagonal entries, Kl

+ K:z =

-(a + d).

If tI is minimal, the mean curvature H = ｾＨｋＺｬ

+ 1\,2) vanishes, so a + d =

Proposition 9.4 0 and

Every surface has an atlas consisting of conformal surface patches.

hence

as we want.

o

"

We shall accept this result without proof (the proof is non-trivial). Let tI : tJ - t R 3 be a conformal surface patch. We introduce complex coordinates in the plane in which U lies by setting (= u+iv

for (u,v) E U,

220

Elementary Differential Geometry

and we define

+

221

9. Minimal Surfaces

We compute

lp«)

= tT u -

.:::ltT.tT u :::::: tTuu.tT u

(17)

itT v •

Thus, I{) = (If!l, 1f!2, 1f!3) has three components, each of which is a complex-valued function of (u, V),I i.e. of (. The basic result which establishes the connection between minimal surfaces and holomorphic functions is

'P3

+ (O'v.tTu)v 0'1I.tT1I)u

tTu.(fu :::::: O'v.tTv

and

-

(O'v.O'uv)

+ (0'1I. tTu)v. O'u.tTv ::::::

O. Hence,

.:::lO'.O'u ::::::

O. 0

3

U : R be a con!ormal surface patch. Then tT is minimal if and only zf the functzon I{J defined m Eq. (17) is holomorphic on U. tT :

"2(tT u .tT u )u

1 = "2(O'u.tTu -

But, since 0' is conformal, Similarly, .:1tT.tT 11 :::::: O.

Proposition 9.5 ｾ･ｴ

=

+ tTvv·tTu

1

Saying that lp is holomorphic means that each of its components /I' '" and . h 1 h' .,...1,.,...2 IS 0 omorp IC.

The holomorphic function

K2 at P 1 may assume that 1\.1 > "-2 everywhere.

10.11 Suppose that the first and second fundamental forms of a surface patch are Edu 2 + Gdv 2 and Ldu2 + N dv 2 , respectively (cf. Proposition 7.2). Show that the Codazzi-Mainardi equations reduce to

21Ev (LE + N) G

' N.. =

21 G.. (LE + N) G

= ＺｾＨＱｴＲ

-Itl),

(1t2) ..

= ｾｇＨｬｴ

-

Edu 2 + Gdv 2 respectively. By Exercise 10.8, 2E Ev = K1 -

.

Deduce that the principal curvatures Itl = LjE and satisfy the equations (ltl)v

by shrinking U if necessary, we

By Proposition 7.2, we can assume that the first and second fundamental forms of u are

+ dv 2 ,

respectively.

Lv =

SO

K2

and

Ldu2 + N dv 2 ,

G.. =

(ltl)v,

2G

(1t2) .. , "-2

K1 -

and by Corollary 1O.2(ii), the gaussian curvature 1t2

= NjG

1t 2).

K = -

We conclude this chapter with a beautiful theorem that is the analogue for surfaces of the characterisation given in Example 2.2 of circles as the plane curves with constant curvature.

(:u (vc;G)

Since P is a stationary point of 1 o. The proof of this theorem depends on the following lemma.

we have

2G "'1 -

ＨｶｾｇＩ

K,2

. (1t1)V

(1t2) ....

-

=

(1t2) ..

= 0, and

2E) Itl -

K2

(ltdvv

(again dropping terms involving E v , G. and the first derivatives of 1 O. Choose a patch t1 : U ---> R 3 of S containing P, and and 1t2 be its principal curvatures. Since 1t,lt2 > 0, by reparametrising let if necessary, we can assume that and 1t2 are both> 0 (see Exercise 6.17). Suppose that > 1t2 at Pi then by shrinking U if necessary, we can assume that > 1t2 everywhere on U. Since K is a constant> 0, the function (x _ ｾＩ 2 increases with x provided that x > K / x > O. Since > K / = 1t2 > 0, this function is increasing at x = It" so must have a local maximum at P, and must have a local minimum there. By Lemma 10.2, K :::; 0 at then 1t2 = K / P. This contradicts the assumption that K > O. 0

It,

It,

It,

It,

It,

It,

It,

It,

The Gauss-Bonnet theorem is the most beautiful and profound result in the

EXERCISES

theory of surfaces. Its most important version relates the average over a surface

of its gaussian curvature to a property of the surface called its 'Euler number' which is 'topological', i.e. it is unchanged by any continuous deformation of the surface. Such deformations will in general change the value of the gaussian curvature, but the theorem says that its average over the surface does not change. The real importance of the Gauss-Bonnet theorem is as a prototype of analogous results which apply in higher dimensional situations, and which relate geometrical properties to topological ones. The study of such relations is one of the most important themes of 20th century Mathematics.

10.12 Show that a compact surface with gaussian curvature> 0 everywhere and constant mean curvature is a sphere. (As in the proof of Theorem 10.4, if has a local maximum at a point P of the surface, then 1t2 = 2H has a local minimum there.)

It,

It,

J

11.1. Gauss-Bonnet for Simple Closed Curves The simplest version of the Gauss-Bonnet Theorem involves simple closed curves on a surface. In the special case when the surface is a plane, these curves have been discussed in Section 3.1. For a general surface, we make

Definition 11.1 A curve ,),(t) = a(u(t), vet)) on a surface patch t1 : U ---> R 3 is called a simple closed curve with period a if 1I'(t) = (u(t), vet)) is a simple closed curve in R 2 with period a such that the region int(1I') of R 2 enclosed by 11' is entirely 247

248

Elementary Differential Geometry

contained in U (see the diagrams below). The curve 'Y is said to be positivelyoriented if 1r is positively-oriented. Finally, the image of int(1r) under the map q is defined to be the interior inth) of "Y.

+ I

249

11. The Gauss-Bonnet Theorem

integral:

(Wf)

I=

J

e/,e"ds

=

1

･ＧＮＨｾｵ

o {Wy}

+ ･ｾｶＩ､ｳ

L 0

=

Ｈ･ＯＮｾＩ､ｵ

+ Ｈ･ＧＬｾＩ､ｶ

By Green's theorem (see Section 3.1), this can be rewritten as a double integral: i

I =

I·{

ｻＨ･ＯＮｾＩｵ

Ｈ･ＯＬｾＩＱｊｽ､ｵｶ

-

Jint(1r)

=

f· r

Jint('1r)

=!{

Jint('1r}

=

f· r Jirlt('1r)

allowed

not allowed

=

We can now state the first version of the Gauss-Bonnet Theorem.

Theorem 11.1 Let 'Y(8) be a unit-speed simple closed curve on a surface assume that 'Y is positively-oriented. Then,

1

£("Y) Kgds = 211" o

11

(1

of length l('Y), and

KdAq,

int('Y)

where Kg is the geodesic curvature of "Y, K is the gaussian curvature of tr and dAq = (EG - F2)1/2dudv is the area element on q (see Section 5.4).

We use s to denote the parameter of'Y to emphasize that 'Y is unit-speed. Proof 11.1

As in the proof of Theorem 10.1, choose a smooth orthonormal basis {e', e"} of the tangent plane of tr at each point such that {e', e", N} is a right-handed orthonormal basis of R 3 , where N is the unit normal to tr. Consider the following

I·(

ｻＨ･ｾＮＩ

-

Ｈ･ｾＮＩｽ､ｵｶ

LN - M2/2dudV (EG - F2)1

(by Lemma 10.1)

LN - M: (EG - F 2 )1/2dudv EG - F (1)

KdAq.

Jint('1r)

Now let O( s) be the angle between the unit tangent vector 1 of'Y at ,,( 8) and the unit vector e l at the same point. More precisely, (J is the angle, uniquely determined up to a multiple of 211", such that

1 = cos Be' + sin (Je" .

(2)

Then,

N x

'Y =

- sin (Je '

N

+ cos (Je'l.

(3)

Elementary Differential Geometry Now, by Eq. (2),

i' =

cos (}e

l

+ sin (}i/' + 8( -

sin Be'

+ cos Be"),

(4)

11. The Gauss-Bonnet Theorem

251

Note, however, that it is crucial that the interior of 11" is entirely contained in U, otherwise such a family will not exist, in general:

so by Eqs. (3) and (4) the geodesic curvature of"Y is /'\,9

(see ｮｯｩｴ｣ｾｓ

= (N x 1').'5' =

6.2)

8(- sin Be' + cos Be"). (- sin eel + cos (je")

+ (-

=iJ + cos

2

+ cos (Jell). (cos Oe' + sin (jell)

sin 8e'

8( e .e") - sin 2 B( e".e' ) l

+ sinO cosO(elJ.e" - e/.e /)

(by Eqs. (2) and (3)).

l

Since e and e" are perpendicular unit vectors,

e'.e' = e".e"

= 0,

e'.e" = -e'.e

ll

.

Hence,

I'll , = O· - e.e

/'\,g

and by the definition of I, rl('*Y)

I

=1

(8 - /'\,g)ds.

0

Thus, to complete the proof of Theorem 11.1, we must show that ri("Y)

10

Ods=2rr.

(5)

Equation (5) is called 'Hopf's Umlaufsatz' - literally 'rotation theorem' in German. We cannot give a fully satisfactory proof of it here because the proof would take us too far into the realm of topology. Instead, we shall justify Eq. (5) by means of the following heuristic argument. The main observation is that, if it is any other simple closed curve contained in the interior of 'Y, there is a smooth family of simple closed curves "Y T , defined for 0 :$ T :$ 1, say, with '}'O = 'Y and 'Y 1 = i (see Section 8.4 for the notion of a smooth family of curves). The existence of such a family is supposed to be 'intuitively ohvious'.

Observe next that the integral ｉＺＨＧｙｾＩ 0 ds should depend continuously on T. l T Further, since 'Y and e return to their original values as one goes once round '}'T, the integral is always an integer multiple of 211". These two facts imply that the integral must be independent of T - for by the Intermediate Value Theorem a continuous variable cannot change from one integer to a different integer without passing through some non-integer value. To compute Iol('Y) iJ ds, we can therefore replace "Y by any other simple closed curve i in the interior of '}', since this will not change the value of the integral. We take"" to be the image under t1 of a small circle in the interior of 11". It is 'intutively clear' that

l

l (1')

o

Ods = 2rr,

because

(i) e/ is essentially constant at all points of"" (because the circle is very small), and (ii) the tangent vector to l' rotates by 27f on going once round 'Y because the interior of l' can be considered to be essentially part of a plane, and it is 'intutively clear' that the tangent vector of a simple closed curve in the plane rotates by 2rr on going once round the curve. This completes the 'proof' of HopPs Umlaufsatz, and hence that of Theorem 11.1.

252

Elementary Differential Geometry

+

253

11. The Gauss-Bonnet Theorem

EXERCISES 11.1 A surface patch t1 has gaussian curvature :$ 0 everywhere. Prove that there are no ｳｩｭｰｬｾ｣ｯ･､ geodesics in iT. How do you reconcile this with the fact that the parallels of a circular cylinder are geodesics? 11.2 Let ')'(8) be a simple closed curve in R 2, parametrised by arc-length and of total length i(,),). Deduce from Hopf's Umlaufsatz that, if 11: 8 (8) is the signed curvature of ')', then

( ')') 1 o

"r-

It.(s)ds:::: 27f.

(Use Proposition 2.2.)

11.2. Gauss-Bonnet for Curvilinear Polygons For the next version of Gauss-Bonnet, we shall have to generalise our notion of a curve by allowing the possibility of 'corners'. More precisely, we make the following definition.

Now let iT : U -t R 3 be a surface patch and let 1r : R -t U be a curvilinear polygon in U, as in Definition 11.2. Then, 'Y = u 0 1r is called a curvilinear polygon on the ,.Surface patch tT, int('Y) is the image under iT of int(n"), the vertices of'Y are the points "Y(ti) for i = 1, ... , n, and the edges of iT are the segments of it corresponding to the open intervals (ti-1, ti)' Since u is allowable, the one-sided derivatives

Definition 11.2 A curvilinear polygon in R 2 is a continuous map 1r : R -t R 2 such that, for some real number a and some points 0 = to < t1 < ... < t n = a, (i) 1r(t):::: 1r(t/) if and only if t' - t is an integer multiple of a; (ii) 1r is smooth on each of the open intervals (to, t1), (h, t2), ... , (tn-I, tn); (iii) the one-sided derivatives ?i'-(td :::: lim 1r(t) -1r(t i ), ttt,

t - ti

?i'+(ti):::: lim 1r(t) -1r(ti) tJ,t, t - ti

(6)

exist for i = 1, ... ,n and are non-zero and not parallel. The points 'Y( ti) for i :::: 1, ... , n are called the vertices of the curvilinear polygon 'Jf, and the segments of it corresponding to the open intervals (ti-1 , t i ) are called its edges. It makes sense to say that a curvilinear polygon 'JI' is positively-oriented: for all t such that 'Jf(t) is not a vertex, the vector n. obtained by rotating ?i' anticlockwise by 7f/2 should point into int('Jf). (The region int('Jf) enclosed by 'Jf makes sense because the Jordan Curve Theorem applies to curvilinear polygons in the plane.)

exist and are not parallel. Let be the angles between -y±(ti) and e/, defined as in Eq. (2), let 8i = 8t - 0i be the external angle at the vertex ')'(ti), and let ai = 7f - 8i be the internal angle. Since the tangent vectors -y+ (ti) and -y- (ti) are not parallel, the angle r5i is not a multiple of 7f. Note that all of these angles are well defined only up to multiples of 27f. We assume from now on that 0 < ai < 27f for i = 1, ... ,n. A curvilinear polygon 'Y is said to be unit-speed if II -y II = 1 whenever -y is defined, i.e. for all t such that 'Y(t) is not a vertex of 'Y. We denote the parameter of 'Y by s if 'Y is unit-speed. The period of 'Y is then equal to its length l(-y), which is the sum of the lengths of the edges of 'Y.

Or

Theorem 11.2 Let 'Y be a positively-oriented unit-speed curvilinear polygon with n edges on a surface u, and let aI, a2, . .. , an be the interior angles at its vertices. Then, ("Y)

1 o

n

II: g

dS=L a i-(n-2)7fi=l

!r

Jint('Y)

Kdk-

Elementary DifFerential Geometry

254

..,..---------------11, The Gauss-Bonnet Theorem

255

Proof 11.2

Exactly the same argument as in the proof of Theorem 11.1 shows that

l

i ('Y)

o

Ｇｾｳ､ｧ

l

,"

i ('Y)

Ods -

J[

, !

KdAo-.

Jint('Y)

0

We shall prove that

(7) Assuming this, we get (('Y)

Ino

---_ ... -

n

Kgds = 2'11" = 2'11" -

LIS, - /'Jint('Y) r KdAa ,=1

t

,=1

n

=

L

Q , -

(-11' - ct,) - /'

ｾ

Le. 'Y and i agree except when s belongs to a small interval ＨｳｾＬ ＧＩ taining Si, so the contribution from the ith vertex is

K dAo-

Ｑ

Jint('Y)

(n - 2)'11" - /' ｾ

KdAo-.

Ods -

ｲｾｩ

Ods -

Jsi.

ＱｾＧ

Ods. 81.

The first integr81 is the angle between ｾＨｳｮ and ｾＨｳｩＩＬ which as si and si' tend to s, becomes the angle between -Y+(Si) and -y-(Si), Le. h On the other Si) and (8i, the last hand, since 'Y(s) is smooth on each of the intervals ＨｳｾＬ two integrals go to zero as si and si' tend to 8i. Thus, the contribution to the expression (9) from the ith vertex tends to c5i as si and sil tend to Si . Summing

Jint('Y)

,=1

Ｑｾ S,

say, con-

To establish Eq. (7), we imagine 'smoothing' each vertex of'Y as shown in the following diagram.

sn,

over all the vertices, we get

l

i ('Y) .

o

,

Ods -

l

i ("1)

{Jds =

0

L lSi. n

i=1

o

Equation (7) now follows from this and Eq. (8). '

Corollary 11.1 If'Y is a curvilinear polygon with n edges each of which is an arc of a geodesic, then the internal angles aI, a2, ... , an of the polygon satisfy the equation

If the 'smoothed' curve ;Y is smooth (l), then, in an obvious notation,

(('1) .

J

Ods = 211".

o

Since

'Y

t a i = (n - 2)'11" +

1o

Ods -

1£('Y) .

Ods

/1

KdAo-.

mt('Y)

Proof 11.1

and ;Y are the same except near the vertices of 'Y, the difference

£('1) ;.,

i=1

(8)

(9)

0

is a sum of n contributions, one from near each vertex. Near 'Y(s,), the picture is

This is immediate from Theorem 11.2, since ｾｧ

= 0 along a geodesic.

0

As a special case of Corollary 11.1, consider an n-gon in the plane with

256

Elementary Differential Geometry

straight edges. Since K = 0 for the plane, Corollary 11.1 gives n

L>l!i = (n -

2)11",

t

Finally, for a geodesic n-gon on the pseudosphere (see Section 7.2), for which K = -1, we see that E ai is less than (n - 2) 11" by the area of the polygon. In particular, for a geodesic triangle ABC on the pseudosphere,

ｬ］ｾ

a well known result of ｹｲｾｴｮ･ｭｬ

257

11. The Gauss-Bonnet Theorem

A{ABC) = 11' - LA - LB - LC.

geometry.

EXERCISES . For a curvilinear n-gon on the unit sphere whose sides are arcs of great cIrcles, we have K = 1 so E ai exceeds the plane value (n - 2)11' by the area II dAo- of the polygon. Taking n == 3, we get for a spherical triangle ABC whose edges are arcs of great circles, A(ABC) = LA + LB + LC -

11.3 Suppose that the gaussian curvature K of a surface patch (f satisfies K :::; -1 everywhere and that 'Y is a curvilinear n-gon on (f whose sides are geodesics. Show that n ;::: 3, and that, if n = 3, the area enclosed by 'Y cannot exceed 11'. 11.4 Consider the surface of revolution

1r. (f(U,

This is just Theorem 5.5, which is therefore a special case of Gauss-Bonnet.

v)

= (f(u) cos v, f(u) sin v, g(u)) ,

where 'Y(u) = (f(u),O,g(u)) is a unit-speed curve in the xz-plane. Let Ul < U2 be constants, let 'Yl and 'Y2 be the two parallels u = Ul and u = U2 on (f, and let R be the region of the uv-plane given by

Compute ("'(1)

o

1

Itgds,

1(('Y2) Itgds 0

and

J{J R

KdAo-,

and explain your result on the basis of the Gauss-Bonnet theorem.

258

11. The Gauss-Bonnet Theorem

Elementary Differential Geometry

The version of the Gauss-Bonnet theorem we are aiming for is obtained by covering a compact surface S with curvilinear polygons that fit together nicely, applying Theorem 11.2 to each one, and adding up the results. We begin to make this precise with

11.3. Gauss-Bonnet for Compact Surfaces The most important version of the Gauss-Bonnet Theorem applies to a compact surface S. It is a surprising' result that there are very few compact surfaces in R 3 up to diffeomorphism, and they can all be described explicitly. The simplest example is, of course, the sphere. The next simplest is the torus, which can be obtained by rotating around the z-axis a circle in the xz-plane which does not intersect the z-axis:

Definition 11.3 3

Let S be a surface, with atlas consisting of the patches l1i : Ui ---+ R • A triangulation of S is a collection of curvilinear polygons, (the interior of) each of which is contained in one of the (1i(Ui) , such that (i) every point of S is in at least one of the curvilinear polygons; (ii) two curvilinear polygons are either disjoint, or their intersection is a common edge or a common vertex; (iii) each edge is an edge of exactly two polygons. Thus, situations like

One can also join such tori together:

are not allowed. A triangulation of the unit sphere with eight polygons is obtained by intersecting the sphere with the three coordinate planes: z

,\-

This surface is denoted by T g , where 9 is the number of holes, called the genus of the surface (we take 9 = 0 for the sphere). We accept the following theorem without proof:

Theorem 11.3 For any integer 9 ｾ 0, T g can be given an atlas making it a smooth surface. Moreover, every compact surface is diffeomorphic to one of the T g •

We state without proof:

Elementary Differential Geometry

260

ＮＺｩｾ Ｍ ｦ

ｴｾ･ ｮ Ｘ ｂ Ｍｾｳ ｵ ｡ ｇ ･Ａｨｾｲ

Ｍ ｾ ｾ ］ Ｚ Ｚ ］ ＺＡ ｍ｀ｾｆ Ｖ｀ｾｈ ｩＡ

Theorem 11.4

where X is the Euler number of the triangulation.

Every compact surface has a triangulation with finitely many polygons.

We need to explain what is meant by the left-hand side of the equation in Theorem 11.5. Fix a triangulation of S with polygons Pi, say. Each Pi is 3 contained in the image of some patch (f i : Ui -t R in the atlas of S, say Pi = (fi(R i ), where R i ｾ Ui. Then, by definition,

We introduce the following number associated to any triangulation: \

Definition 11.4 The Euler number X of a triangulation of a compact surface S is

X=V-E+F, where

v=

the total number of vertices of the triangulation,

E = the total number of edges of the triangulation, F = the total number of polygons of the triangulation.

For the triangulation of the sphere given above, V = 6, E = 12 and F = 8, so X = 6 - 12 + 8 = 2. The importance of the Euler number is that, although different triangulations of a given surface will in general have different numbers of vertices, edges and polygons, X is actually independent of the triangulation and depends only on the surface. For example, we can get another triangulation of the sphere by 'inflating' a regular tetrahedron:

where K is the gaussian curvature of (fi. Unfortunately, we have to show that this is a good definition, Le. that it does not depend on our choice of patches (fi (since, even for a given triangulation, Pi may be contained in more than one patch), nor does it depend on the triangulation itself. To see this, we note first that, if Ui : Ui -t R 3 is a reparametrisation of (fi and if Pi = O'i(R i ), where Ri ｾ Ui, then

because both the area element dAtr and the gaussian curvature K are unchanged by reparametrisation (see Proposition 5.3 and Exercise 6.17). Next, if {Pi} and {PH are two triangulations of S, it is intuitively clear that we can find a third triangulation {Pk} of S such that each Pi is the union of some of the Pk', as is each Pi- For example, if some ｾ and some Pj overlap as follows,

ｾｰ

This time, V = 4,' E = 6 and F = 4, so X = 4 - 6 + 4 = 2, the same as before. This property of X is a consequence of the following theorem, which is the desired extension of Theorem 11.2 to compact surfaces:

J

Theorem 11.5 Let S be a compact surface. Then, for any triangulation of S,

f1

KdA = 2rrx,

then by inserting additional vertices we can create the appropriate polygons

P"· k •

262

Elementary Differential Geometry

¥ I

11. The Gauss-Bonnet Theorem

263

I I

where ni is the number of vertices of Pi, 'Yi is the curvilinear polygon that forms the boundary of Pi, l('Yi) is its length, and L i is the sum of its interior angles. We must therefore sum the contributions of each of the three terms on the right-hand side of Eq. (10) over all the polygons Pi in the triangulation. First, Ei L i is the sum of all the internal angles of all the polygons. At each vertex, several polygons meet, but the sum of the angles at the vertex is obviously 21T, so (11) where V is the total number of vertices. It is then clear that (in an obvious notation)

since both sides are equal to

L/ｌｉｋ､ａ｡ｾＮ k

Jo

This is just because the integral of K over the union of finitely many polygons, all contained in a single surface patch and any two of which are either disjoint or intersect only in a common edge or vertex, is simply the sum of the integrals of K over each polygon. Together with the fact that lIs KdA is independent of the triangulation, Theorem 11.5 thus implies

Next,

ｾＨｮｩ

- 2)11'::::

ＩｩｮｾＨ

11' - 211'F:::: 211'E - 211'F,

(12)

where F is the total number of polygons and E the total number of edges, since in the sum Li ni each edge is counted twice (as each edge is an edge of exactly two polygons).

Corollary 11.2 The Euler number X of a triangulation of a compact surface S depends only on S and not on the choice of triangulation.

We now give the proof of Theorem 11.5. Proof 11.5

As above, we fix a triangulation of S with polygons Pi, say, each of which is contained in the image of some patch t1 i : Ui -t R 3 in the atlas of S, say Pi :::: t1i(Ri), where R i ｾ Ui . By Theorem 11.2,

JrRi KdAa.

J

:::: L i

-

(ni - 2)1T +

l

0

l

Finally, we claim that

('Y;) "'9

ds,

(10)

Ll i

0

l ('Yi)

K.gds

= O.

(13)

264

Elementary Differential Geometry

Indeed, note that in the sum in Eq. (13), we integrate twice along each edge, once in each direction. By Eq. (5) in Chapter 6, "'9 changes sign when we traverse a given curve in the opposite direction, so these two integrals cancel out. The various contributiolls to the sum in Eq. (13) therefore cancel out in pairs, thus proving Eq. (13). Putting Eqs. (10), (11), (12) and (13) together gives

JIs

KdA

. JL.

= l;. =

ｾｌｩ

KdAq,

••

-2)rr+ ｾ

= 2rrV - (211B - 211P) = 2rrx,

+0

[[("til

•

10

To see why Theorem 11.5 is so remarkable, let us apply it to the unit sphere

KdA = 41T.

265

lIs KdA = 41T. (More generally, this discussion shows that the Euler number of any compact surface is unchanged when the surface is deformed without tearing.) We complete the picture by determining the Euler numbers of all .the compact surfaces.

Proof 11.6

52. Then, X = 2 so we get

Jｉｳｾ

11. The Gauss-Bonnet Theorem

The Euler number of the compact surface T g of genus g is 2 - 2g.

"'gds

o

proving Theorem 11.5.

I

Theorem 11.6

'

- ｾＨｮｩ

f

The formula is correct when 9 = 0, since we know that X ::;:; 2 for a sphere. We now prove it for the torus T 1 • To find a triangulation of the torus, we use the fact that it can be obtained from a square in the plane by gluing opposite edges:

(14)

Of course, this result is not remarkable at all because K = 1 so the left-hand side of Eq. (14) is just the area of the sphere. But now suppose that we deform the sphere, i.e. we think of the sphere as being a rubber sheet and we pull and stretch it in any way we like, but without tearing:

1[\

/1\

ｾ

We subdivide the square into triangles as shown:

For such a deformed sphere S, K will not be constant and the direct computation of the integral K dA will be difficult. But if we start with a triangulation of the undeformed sphere, then after deformation we shall have a triangulation of the deformed sphere with the same number of vertices, edges and polygons as the original triangulation. It follows that the Euler number of the deformed sphere is the same as that of the undeformed sphere, i.e. 2, so by Theorem 11.5,

IIs

t.lementary UlTlerentlal l.Jeometry

Lt>t>

This leads to a triangulation of T 1 with V = 9, E = 27 and F::;:; 18. One must count carefully: for example, the four circled vertices of the square correspond to a single vertex on the torus. Note also that not just any subdivision of the square into triangles is acceptable. For ｾｸ｡ｭｰｬ･Ｌ the subdivision

1

267

11, The Gauss-Bonnet Theorem

Suppose we carry out the gluing by removing a curvilinear n-gon from T g and T1 and gluing corresponding edges (having fixed suitable triangulations of Tg and T 1 ). If V', E ' and pi are the numbers of vertices, edges and polygons in the triangulation of T g , and V", E" and pI! those for T 1 , the numbers V, E and F for T g + 1 are given by

V ::;:; Vi - n + VII - n + n = Vi + V" - n, E = E' - n + EI! - n + n = E' + E" - n, F ::;:; pi - 1 + F 'I

is not acceptable, since after gluing, the two shaded triangles intersect in two vertices, which is not allowed:

-

+ pI! -

1 ::;:; F '

2.

Indeed, V is the number V' of vertices in T g plus the number V" in T 1 , except that the n vertices of the polygon along which T1 and T g are glued have been counted twice, so V = V' + V" - n; a similar argument applies to the edges; and F is as stated because the polygon along which T 1 and T g are glued is not part of the triangulation of Tg+l' Hence,

X(Tg+l}

=V =.

(Vi

= Vi -

E

+F

+ V" E'

n) - (E'

+ pi + VII

+ E"

- Ell

- n)

+F

II

-

+ (F' + F" -

2)

2

= X(Tg) + X(Tt} - 2 = 2 - 2g + 0 - 2 (by the induction hypothesis) ::;:; 2 - 2(g + I}, But the finer subdivision above does work, and gives

x ::;:; 9 -

o

proving the result for genus 9 + 1.

27 + 18 "" 0 ::;:; 2 - 2 xl,

proving the theorem when 9 ::;:; 1. We now complete the proof by induction on g, using the fact that Tg +l can be obtained from T g by gluing on a copy of Tl:

Corollary 11.3 We have

JlTr

KdA::;:; 41T(1- g).

g

Proof 11.3

Just combine Theorems

n.5 and

o

11.6.

EXERCISES 11.5 T.

Show that, if a triangulation of a compact surface with Euler number X by curvilinear triangles has V vertices, E edges and F triangles, then

3F:=2E, E=3(V-X),

ｖｾ ＨＷＫｊＴＹＭＲ ｘｽＬ

Elementary Differential Geometry

268 11.6

A triangulation of the sphere has n curvilinear triangles, and r triangles meet at each vertex. Explain why there are 3n/r vertices altogether in the triangulation, and write down the total number of edges. Show that

6 4 ---=1. r n Deduce that r ::; 5, and sketch triangulations of the sphere corresponding to r 3,4 and 5. 11.7 Show that, given 5 points on a sphere, it is impossible to connect each pair by curves on the sphere that intersect only at the given points. (Such a collection of curves would give a triangulation of the sphere for which 2E ;::: 3F, since each face would have at least 3 edges.) Deduce that the same result holds if 'sphere' is replaced by 'plane'. (Use stereographic projection.)

=

11.8 Let Pl,P2,P3 and Ql,Q2,Q3 be points on a sphere. Show that it is impossible to join each Pi to each Qj by 9 curves on the sphere that intersect only at the given points. (This is sometimes called the 'Utilities Problem', thinking of PI, P2 and P3 as the gas, water and electricity supplies to three homes Ql, Q2 and Q3') 11.9 Show that, if a compact surface S is diffeomorphic to the torus T 1 , then

JIs

KdA=O.

Can such a surface have K = 0 everywhere? 11.10 Show that, if S is the ellipsoid x2

+ y2 a2

Z2

+ b2

= 1,

where a and b are positive constants,

JIs

K dA

j

ab2 cos 9

--."--::--,,",,,:,,"",:::-:-;:: d()

-,,/2 (a 2 sin 2 9 + b2 cos2 9)3/2

(ii) x 2 + y2 + Z4 = l. Which is which, and why? Sketch the compact surface and write down its Euler number.

11.4. Singularities of Vector Fields Suppose that S is a surface and that V is a smooth tangent vector field on S. This means that, if (l : U ｾ R 3 is a patch of Sand (u,v) are coordinates on U, then V

= a(u, v)O'u + f3(u, v)O'v,

where a and f3 are smooth functions on U. It is easy to see that this smoothness condition is independent of the choice of patch 0' (see Exercise 11.14).

Definition 11.5 If V is a smooth tangent vector field on a surface S, a point P of S at which V = 0 is called a stationary point of V.

The reason for this terminology is as follows. We saw in the proof of Proposition 7.4 that, if P is any point of 5, there is a unique curve 'Y(t) on S such that "I = V and '1(0) = P; '1 is called an integral curve of V. We can think of '1 as the path followed by a particle of some fluid that is flowing over the surface. If V = 0 at P, the velocity "I of the flow is zero at P, so the fluid is stationary there. We are going to prove a theorem which says that the number of stationary points of any smooth tangent vector field on a compact surface S, counted with the appropriate multiplicity, is equal to the Euler number of S. To define this multiplicity, let P be a stationary point of V contained in a surface patth 3 0' : U ---t R of S, say, with 0'( Uo, vo) = P. Let be a nowhere vanishing smooth tangent vector field on O'(U) (for example, we may choose = O'u or O'v), and let 1/J be the angle between V and

e

= 47r.

By computing the above integral directly, deduce that "/2

269

11. The Gauss-Bonnet Theorem

e.

e

= 2.

11.11 Suppose that S is a compact surface whose gaussian curvature K is> 0 everywhere. Show that S is diffeomorphic to a sphere. Is the converse of this statement true? 11.12 One of the following surfaces is compact and one is not: (i) x 2 - y2 + Z4 = 1;

Definition 11.6 With the above notation, the multiplicity of the stationary point P of the tangent vector field V is 1

JJ(P) = 211"

rl('Y) d'IjJ

Jo

ds ds,

Elementary Differential Geometry

270

where '1(s) is any positively-oriented unit-speed simple closed curve of length l('Y) in O'(U) with P in its interior.

¥

n. The Gauss Bonnet Theorem

271

(iv) V(x, y) :::: (x, -y); IJ.:::: -1

It is clear that IJ.(P) is an integer, ｡ｾ､

an argument similar to our heuristic proof of Hopf's Umlaufsatz in Section 1i.l shows that I-'(P) does not depend on the choice of simple closed curve 'Y. It is also easy to see that it is independent of the choice of 'reference' vector field (see Exercise 11.15).

e

Example 11.1

The following smooth tangent vector fields in the plane have stationary points of the indicated multiplicity at the origin (we have shown the integral curves of the vector fields for the sake of clarity); (i) V(x,y):::: (x,Y)j J.I.:::: +1.

The stationary point in examples (i), (ii), (iii) and (iv) is called a source, sink, vortex and bifurcation, respectively. Let us verify the multiplicity in case (iv), for example. Take the 'reference' tangent vector field to be the constant vector field (1,0). Then, the angle 'l/J is given by

e:: :

(cos 1/', sin 'l/J) ::::

ｾ IIVII == ( .jx2 +y2 X

I -

Y

.jX2 +y2

)

.

Taking 'Y(s) :::: (cos s, sin s) to be the unit circle, at 'Y( s) the angle 1/' satisfies (cos 'l/J, sin 1/')

so 1/' :::: 21T - s. Hence, 1

2

(l(0,0) = 21T

(ii) V(x,y):::: (-x,-y); J.I.:::: +1

1

:::: (cos s, - sin s),

0

11"

d

ds (271" - s) ds :::: -1.

Theorem 11.7 Let V be a smooth tangent vector field on a compact surface S which has only finitely many stationary points, say P1 ,P2 , ••• ,Pn . Then, n

L{l(Pi ):::: X, i=l

(iii) V(x, y) :::: (y, -x)j J.I. :::: +1

the Euler number of S. Proof 11.7

Let 'Yi be a positively-oriented unit-speed simple closed curve contained in a patch O'i of S with Pi in the interior of 'Yi' Assume that the 'Yi are chosen so small that their interiors are disjoint. Choose a triangulation of the part Sf of S outside 'Yu 'Y2,'" ,'Yn by curvilinear polygons j . Note that the edges of some of these curvilinear polygons will be segments of the curves 'Yi:

r

Elementary Differential Geometry

272

+

273

11. The Gauss-Bonnet Theorem Combining Eqs. (15), (17) and (18), we get

Ln i;;:l

l

i (";.)

' (f'.f" -

e ' ,e") ds = 27rX·

(19)

0

But, from the proof of Theorem 11.2,

,./1 e.e .',"

Note also that, when these polygons are positively-oriented, the induced orientation of the 'Yi is opposite to their positive orientation (see the diagram above, in which the arrows indicate the sense of positive orientation). We can regard the curvilinear polygons in $I, together with the simple closed curves 'Yi and their interiors, as a triangulation of 8, so by Theorem 11,3,

[ K dA +

15'

t [

K dA = 27rX,

(15)

i;;:l llnt('YJ

where X is the Euler number of 8. On 8', we choose an orthonormal basis {e', e"} of the tangent plane of 8 at each point so that e' is parallel to the tangent vector field V, Arguing as in the proof of Theorem ILl, we see that

1

l

K dA = L

5'

l(rj)

e'.e" ds,

(16)

0

j

where s is arc-length on r j and f(r}) is its length. Any common two of the curvilinear polygons j is traversed once in each direction contributions to the sum in Eq. (16) cancel out. What remains is the along the segments of the curves 'Yi that are part of the polygons j ' of the remark about orientations above, we get

r

r

1

KdA=-L

l

i=l

0

5'

n

l ('Y)

. e'.e"ds,

edge of so their integral In view

(17)

where s is arc-length along 'Yi and f('Yi) is its length. Now choose an orthonormal basis {fl, f"} of the tangent plane of 8 on each patch C1i. By the proof of Theorem 11.1,

1

int{'Y,)

K dA =

l

0

l('Y;)

•

f',f" ds.

(18)

n

=11-/\'g,

f l' f '=tp-K,g, '''

where ,,"g is the geodesic curvature of 'Yi and () and tp are the angles between 'Yi and e ' and f', respectively, Then, 'I./J = tp - () is the angle between e ' and fl , Le. between V and the 'reference' tangent vector field fl on (Ii. So the left-hand side of Eq, (19) is

as we want.

o

We now give some simple examples of vector fields on surfaces (we show their integral curves for clarity). Example 11.2

A vector field on the sphere with 1 source and 1 sink: X = 2

Elementary Differential Geometry

274

T I

11. The Gauss-Bonnet Theorem

275

of S, the three components of V at the point u(u, v) are smooth functions of (u, v).

(f

Example 11.3

A vector field on the torus with no

I

ｳｴ｡ｾｩｯｮ ｲｹ

11.15 Show that the Definition 11.6 of the multiplicity of a stationary point of a tangent vector field V is independent of the 'reference' vector field (. (If is another reference tangent vector field, and () is the angle between ( and then d(}jds -(1- p2)-1/2 p, where p = cos(). Now use Green's theorem to show that ｊｾＨＢ Ｉ ､Ｈ ｪ ｳＩ､ = 0.)

points: X = 0

e

........... , .....

...........

ｾ

.

e,

=

)

11.5. Critical Points If I (ti, v) is a smooth function defined on an open subset U of R 2 , we say that a point (uo, vo) is a critical point of I if al jau and allav both vanish at (uo, vo). Equivalently, the gradient of I,

Example 11.4 A vector field on the double torus T2 with 2 bifurcations: X = -2

VI =

(a l al) au' av '

should vanish at (uo, va). If now F : S --+ R is a smooth function on a surface S (see Exercise 4.11), and if q : U --+ R 3 is a surface patch of 5 then I = F 0 q is a smooth function on the open subset U of R 2 . This suggests I

Definition 11.7 Let 5 be a surface and F : S --+ R a smooth function on S. A point P of Sis a critical point of F if there is a surface patch q of S, with P = q(uo, vo), say, such that I = F 0 (T has a critical point at (uo, vo).

EXERCISES 11.13 Let k be a non-zero integer and let V (x, y) field on the plane given by a

'f3_{(X+i y )k

+t

-

(x

. )-k - ty

= (a, f3)

be the vector

ifk>O, 'f k O. Since t = 'Y/ 11"1 11= ｴｳｯ｣ｫＨＱＫｾ Ｇｶ - sint,ksint + cost), we have n s = ｶＧｫｾＫＱ (-ksin t - cost, kcost - sin t). So dt/ds = (dt/dt)/(ds/dt) = ke

e( -kt k'+1 - k sint-cost, k cost-sint ) = Jk2+1nS so"'s=1 / ks.ByTheorem 2.1, any other curve with the same signed curvature is obtained from the logarithmic spiral by applying a rigid motion. 2.6 (i) Differentiating 'Y = rt gives t = ft + ｾｂｲｮｓＮ Since t and n s are perpendicular unit vectors, it follows that ｾｳ = 0 and 'Y is part of a straight line. (ii) Differentiating 'Y = rn s gives t = rD s + rn s = fn. - ",.rt (Exercise 2.3). Hence, f = 0, so r is constant, and "'s = -l/r, hence "'s is constant. So 'Y is part of a circle. (iii) Write'Y = ret cos O+ns sin 0). Differentiating and equating coefficients sin 0 = 1, f sin (J + ｾｳｲ cos 0 = 0, from which of t and D s gives f cos 0 - ｾｳｲ = - sinO. From the first equation, r = scosO (we can f = cosO and ｾｳｲ assume the arbitrary constant is zero by adding a suitable constant to s) so ｳｾ = -l/scotO. By Exercise 2.5, 'Y is obtained by applying a rigid motion to the logarithmic spiral defined there with k = - cot (J. 2.7 We can assume that 'Y is unit-speed. Then "I>' = (1 - AK:s)t, and this is non-zero since 1 - AK: s > O. The unit tangent vector of 'Y>' is t, and the

Elementary Differential Geometry

286

arc-length s of 'Y'\ satisfies ds/dt = 1 - AK,. Hence, the curvature of 'Y'\ is II dt/ds II = II i II /(1 - AK,) = K,/(l - AK,). 2.8 The circle passes through 'Y(s) because II < - 'Y II = II ••'n • II = l/IK,I, which is the radius of the circle. It is tangent to 'Y at this point because < - 'Y = -L n , is perpendicular to the tangent t of 'Y. The curvature of the '. circle is the reciprocal of its radius, Le. IK,I, which is also the curvature of 'Y. - ｾｮＬK. = ＭｾｮＬ K. so its arc-length 2.9 T he tangent vector of < is t + -L(-K,t) n. is u = J II E II ds = J ｾ､ｳ = Uo where Uo is a constant. Hence, the unit tangent vector of < is -n, and its signed unit normal is t. Since • ' the signed curvature 0 f < IS . K,3/'K,. -dn,/du = K,t/(du/ds) = r,-t, Denoting d/dt by a dash, 'Y' = a(l - cost,sint) so the arc-length s of 'Y is given by ds/dt = 2asin(t/2) and its unit tangent vector is t = 'Y = (sin(t/2), cos(t/2)). So n, = (- cos(t/2), sin(t/2)) and (cos(t/2), - sin(t/2)) = -1/4asin(t/2)n" i = (dt/dt)/(ds/dt) = Ｔ｡ＧｩｾｻｴＩＲ so the signed curvature of'Y is -1/4asin(t/2). Its evolute is therefore < = a(t-sin t, I-cos t) -4a sin(t/2)(- cos(t/2), sin(t/2)) = a(t+sin t, -1+ cost). Reparametrising by t = 71'H, we get a(t-sint, 1-cost)+a( -71', -2), which is obtained from a reparametrisation of'Y by translating by the vector a(-1r, -2). 2.10 The free part of the string is tangent to 'Y at 'Y(s) and has length t-s, hence the stated formula for t(s). The tangent vector of t is 'Y - 'Y + (t - s)'Y = K,(t - s)n, (a dot denotes d/ds). The arc-length v of t is given by dv/ds = K,(t - s) so its unit tangent vector is n, and its signed unit normal is -to Now dn,/dv = •• d_,jti., = f-:';t, so the signed curvature of tis l/(t - s). 2.11 (i) With the notation in Exercise 2.9, the involute of f is

,L,

dE

1

du

"'s

4

Solutions

2.14 (i) t = ＨｾＱ

= tLo -

Jz) is a unit vector so 'Y is unit-speed;

W '/

2

= -K.t + Tn is easily checked. (ii) t = sin t, - cos t, ｾ sin t) is a unit vector so 'Y is unit-speed; i = cos t, sin t, ｾ cos t), so K ｾ II i II = 1; n = ｾｩ = (- cos t, sin t, ｾ cos t); b = t X n= ＨＭｾＬｏ ＭｴＩＬ so b=O and 7 =0. Show that 7 = 0 or observe that x = y = t + 1, z = It' satisfy x - y - z = O. By Proposition 2.5, 'Y is a circle of radius I/K = 1 with centre 'Y + ｾｮ = (0,1,0) in the plane passing through (0,1,0) perpendicular to b = ＨＭｾＬ 0, -t), i.e. the plane 3x + 4z = O. Let a = K/(K 2 + 7 2 ), b = 7/(K 2 + 7 2 ). By Examples 2.1 and 2.4, the circular helix with parameters a and b has curvature a/Ca' + b') = K and torsion b/(a 2 + b2 ) = 7. By Theorem 2.3, every curve with curvature K and torsion 7 is obtained by applying a rigid motion to this helix. This follows from Proposition 2.3 since the numerator and denominator of the expression in (11) are smooth functions of t. Let a dot denote d/dt. Then, 6.=;y = Kn, so the unit tangent vector of 6 is T = n and its arc-length s satisfies ds/dt = K. Now dT Ids = ti./(ds/dt) = ｋＭｬＨ｟ｾｴ + 7b) = -t + ｾ｢Ｎ Hence, the curvature of 6 is II -t + ｾ｢ II = (1 + ;;'-)1/2 = 1', say. The principal normal of 6 is N = 1'-1 (-t + ｾ｢Ｉ and its binormal is B = T x N =.= IJ-l(b + ｾｴＩＮ The torsion T of 6 is given by dB/ds = -TN, Le. K- 1B = -TN. Computing the derivatives and equating coefficients of b gives T = (KT - 71 0 in R 3, and its inverse is the projection (x, Y, z) >-+ (Y, z). Similarly for ｾＷｴ and t7±. A point of the sphere not in the image of any of the six patches would have to have x, y and z all zero, which is impossible. 4.4 Multiplying the two equations gives (X 2_Z 2) sin 0 cos 0 = (1_ y2) sin 0 cos 0, so x 2 +y2 - z2 = 1 unless cosO = 0 or sinO = 0; if cosO = 0, then x = -z and y = 1 and if sin 0 = 0 then x = z and y = -1, and both of these lines 4.1

are also contained in the surface.

290

Elementary Differential Geometry

The given line L. p T = 0. (ii) Let N the a unit 'normal of S. Then, K = ¢=> N 1. (t x N 1) = O. Since N1 is perpendicular to Nl and N 1 is perpendicular to t, this condition holds ¢=> Nt is parallel to t, i.e. ¢=> N1 = -Ai' for some scalar A_ Now use Exercise 6.18. 2 7.5 Using the parametrisation iT in Exercise 4.10, we find that E = b , F = O,G:::: (a+ bCOSU)2 and L = b,M = O,N = (a + bcosu) cosu. This gives K ::::Gosulb(a + bcosu), dAD- = (EG - F 2 )1/2dudv = b(a + bcosu)dudv. Hence, 7.4

=

=

°

°

!!

11 21r

K dAD-::::

21r

cos u dudv

= 0.

304

Elementary Differential Geometry

The first part follows from Exercises 5.3 and 6.4. The dilation multiplies E,F,G by a 2 and L,M,N by a, hence H by a-I and K bya- 2 (using Proposition 7.1). 7.7 Since tT is smooth and (fu X (fu is never zero, N = (fu X uu/ II UU x UU II is smooth. Hence, E,F,G,L,M and N are smooth. Since EG - F2 > (by the remark following Proposition 5.2), the formulas in Proposition 7.1(i) and (ii) show that H and K are smooth. By Proposition 7.1(iii), the principal curvatures are smooth provided H 2 > K, i.e. provided there are no umbilics. 7.8 At a point P of an asymptotic curve, the normal curvature is zero. By Corollary 6.2, one principal curvature "'1 ｾ 0 and the other "'2 ｾ O. Hence K = "'1"'2 ｾ O. On a ruled surface, there is an asymptotic curve, namely a straight line, passing through every point (see Exercise 6.12). 7.9 By Exercise 6.22, FIll = FIIFi 1 FIl. Multiplying on the left by :Fi 1 , the given equation is equivalent to

¥

Solutions

7.6

(v, w) is a reparametrisation map. The first fundamental form in terms of v, w is given by

Ｈｾ Ｉ］ｊｴ ［

°

A2 +2HA+KI=0, where A = -Fi 1 FIl =

Ｈｾ

ｾＩＮ

By the remarks following Definition

6.1, the principal curvatures are the eigenvalues of -A. Hence, 2H = sum of eigenvalues of -A = -(a + d), K = product of eigenvalues of -A = ad - be. Now use the fact stated in the question. 7.10 By Eq. (9) in Chapter 6, -y/y = TtFlT; by Eq. (10) in Chapter 6, N.-y = -N..:y (since N.-Y = 0) = -"'n = -TtFIlT. Now, N.N = (uN u + vNu).(uNu+vN v ) (Nu.Nu)U2+2(Nu.Nv)uir+(Nv.Nu)V2 TtFIIlT. Hence, multiplying the equation in Exercise 7.9 on the left by Tt and on the right by T gives N.N+2HN.-Y+K-Y.-Y = 0; J(.")' is an asymptotic curve, = 0 so N.-y = O. So, assuming that 'Y is unit-speed, we get N.N = -K. But Exercise 6.12 gives N = ±b, so N = =f1'n and N.N = 1'2. 7.11 The parametrisation is tT(u, v) (f(u) cosv, f(u) sin v, g(u», f(u) e", g(u) = v'1 - e 2u - cosh-\e- U ), -00 < u < O. (i) A parallel u = constant is a circle of radius f(u) = eU , so has length

=

305

ｾＩｊ = (_Oi

Ｉｾ ｾＨ

ＩＲ ｾＨｦ

so the first fundamental form is (dv 2 (ii) We find that the matrix

j=

Ｈｧｾ

ｧｾ

Ｆｾ

ｾＩ

=(

ｾＨ

ｾ

+ dw 2 )/w 2 .

v(w+l) - (w + 1)2)

｟ｾＨｶＲ

-o!) = (* ?), Hv 2 -(W+l)2)) v(w

+ 1)

,

so the first fundamental form in terms of U and V is given by

jt Ｈ ｾ

o ｾ

0) j = (v

2

+ (w + 1)2)2 1= 4w 2

1 I (1 - U2 _ V2)2 '

after some tedious algebra. In (i), u < 0 and -tr < v < rr corresponds to -rr < v < rr and w > 1, a semi-infinite rectangle in the upper half of the vw-plane.

=

"'n

=

=

2rre" .

(ii) From Example 7.2, E = 1, F = 0, G = f(U)2, so dAo- = f(u)dudv and the area is ］ｾｉ e"dudv = 2rr. (iii) From Example 7.2, the principal curvatures are = jg - jt; -jig = _(e- 2u _1)-1/2, ｾＲ = fiJIP = g/f = (e- 2u _1)1/2. (iv) Ｑｾ < 0, "'2 > O. 7.12 (i) Setting ii. = v,v = w = e- u , we have u = -lnv,v = it so, in the

I:'"

notation of Exercise 5.4, J =

Ｈｾ

"'I

-}i). Since J is invertible, (u, v)

I-t

To find the corresponding region in (ii), it is convenient to introduce the complex numbers z = v + iw, Z = U + iV. Then, the equations in (ii) are equivalent to Z = ｾＫＬ z = ｩＨｾＡｬＩＧ The line v = rr in the vw-plane ｣ｯｲ ･ｳｰ ｮ､ｾ to z + Z = 2rr (the bar denoting complex conjugate), i.e. Ｉｬ＿ＡｾＨｩ - i(1!.11 ) = 2rr, which simplifies to IZ - (1- ｾＩＱＲ = ｾ［ so v = 7T corresponds to the circle in the UV -plane with centre 1 - i1f and radius .!. 1f • Similarly, v = -rr corresponds to the circle with centre 1 + i1f and radius

ｾＮ

Finally, w = 1 corresponds to z - z = 2i, i.e. ＩＱ＿ＡｾＨｩ + ｩＨｾＡＮＱＩ = 2i. 2 This simplifies to IZ - tl so w 1 corresponds to the circle with centre 1/2 and radil.;ls 1/2 in the UV-plane. The required region in the UV -plane is that bounded by these three circles:

= !;

=

Elementary DifFerential Geometry

306

•

307

Solutions

v Proposition 7.4 gives the result. 7,15 See the proof of Proposition 9.5 for the first part. The first fundamental form ofthe given surface patch is (1+u 2 +v 2)2 (du 2+ dv 2 ), so it is conformal, and U uu + t1 1111 = (-2u, 2v, 2) + (2u, -2v, -2) = O. 7.16 Parametrize the surface by t1(u, v) = (u, v, feu, v)). By Exercise 7.3, E = 1 + ｦｾＬｆ == fufll,G = 1 + f;,L == (1 + ｦｾ + f;)-1/2 fuu,M == (1 + ｦｾ + !J)-1/2 fuv, N = (1 + ｦｾ + ｦｾＩＭＱＯＲ 11111' Hence,

+NE 2(EG - F2)

H = LG - 2M F

Taking f(u,v) = In ＩＺｾＨ

H

7.13 Let "Y(u) == (f(u), O,g(u)) and denote d/du by a dot; by Eq. (2), I+K 1 == O. If K < 0, the general solution is f = ae-.;=1?u + be.;=Ku where a, b are constants; the condition f(7t /2) = f{ -7t /2) = forces a := b == 0, so "Y coincides with the z-axis, contradicting the assumptions. If K = 0, f == a + bu and again a = b == 0 is forced. So we must have K > 0 and f = acosVKu+bsinVKu. This time, f(7t/2) == f(-7t/2) == 0 and a,b not both zero implies that the determinant

°

cosVK7t/2 \ cos../K7t/2

sin../K7t/2 \=0. - sin ../K7t/2

This gives sin.JR7t = 0, so K = n for some integer n :j=. O. If n = 2k is even, f = b sin 2ku, but then f{O) = 0, contradicting the assumptions. If n = 2k + 1 is odd, f = acos(2k + l)u and f(7t/2(2k + 1)) = 0, which contradicts the assumptions unless k = 0 or -1, Le. unless K = (2k+l)2 == 2

1. Thus,

1 == a cos u, 9 ==

J1 - j2

=

vii - a2sin

2

u. Now, "y =

= sec

= fuu(l + ＯｾＩ

2(1

2fullfufll + fllll{1 + ｦｾＩ + ｉｾ + J;)3/2

gives

+ tan 2 v) - sec2 v{1 + tan 2 u) = 0 2(1 + tan 2 u + tan 2 v)3/2 . E v = t1 + wN ll > E w == N. Eu·E w =

2

u{l

/1 11

N u x N ll II dudv

= / 11K!11 t1 u

x U ll 11 dudv

= /l1K1dAo-.

7.18 From the formula for K in the solution of Exercise 7.5, it follows that S+ and S- are the annular regions on the torus given by -7t /2 ::; u ::; 7t/2 and 7t /2 ::::; u ::::; 37t/2, respectively.

(j, 0, g) is

perpendicular to the z-axis {=}!J :::: 0. So the assumptions give ｾ == 0, i.e. a:::: ±1. Then, "Y(u) == (± cosu, 0, ± sinu) (up to a translation along the z-axis) and S is the unit sphere. 7.14 Let 0'(1£, v) be a patch of S containing P = O'(uo, vo). The gaussian curvature K of S is < 0 at P; since K is a smooth function of (u, v) (Exercise 7.7), K(1£, v) < 0 for (u, v) in some open set U containing (Ua, vo); then every point of O'(U) is hyperbolic. Let 1\;1, 1'02 be the principal curvatures of 0', let 0 < () < 7t /2 be such that tan e = -1\;1/ 1"2, and let e1 and e2 be the unit tangent vectors of 0' making angles () and -0, respectively, with the principal vector corresponding to K1 (see Corollary 6.1). Applying

°

since 7.17 E u = t1 u + wN u , 11 t1 u .N = Nu.N = 0, and similarly E ll .Ew = O. Finally, 11".Ell = t1 U 'U ll + 2 w(O'".N ll + D'l1. N U) +w 2 N".N ll = F - 2wM + w2 N u .N v = w N u .N ll ; by Proposition 6.4, N u = -ft1 u , N ll ::::: ＧＱｴｾＭ so N u .N ll = ｾｆ = O. Every surface u = Uo (a constant) is ruled as it is the union of the straight lines given by v = constant; by Exercise 7.4(ii), this surface is flat provided the curve "Y(v) = t1(uo, v) is a line of curvature of S, Le. if 0'11 is a principal vector; but this is true since the matrices :F1 and :FII are diagonal. Similarly for the surfaces v = constant. 7.18 By Eq. (15), the area of t1(R) is

v'

S+

S-

Elementary Differential Geometry

308

It is clear that as a point P moves over S+ (resp. S-), the unit normal at P covers the whole of the unit sphere. Hence, JJs + IKldA = JJs - IKldA = 411' by Exercise 7.18; since IKI = ±K on S±, this gives the result.

Chapter 8 By Exercise 4.4, there are two straight lines on the hyperboloid passing through (1,0,0); by Proposition 8.3, they are geodesics. The circle given by z = 0, x 2 + y2 = 1 and the hyperbola given by y = 0, x 2 - Z2 = 1 are both normal sections, hence geodesics by Proposition 8.4 (see also Proposition 8.5). 8.2 Let II, be the plane through 'Y(s) perpendicular to t(s); the parameter curve s = constant is the intersection of the surface with IIs . From the solution to Exercise 5.17, the standard unit normal of 0' is N = - (cos 8 n + sin 0 b). Since this is perpendicular to t, the circles in question are normal sections. 2 2 2 . 8.3 Take the ellipsoid to be + ｾ + ;.- = 1; the vector (-to, ;"0-) is normal to the ellipsoid by Exercise 4.16. If 'Y(t) = (f(t), g(t), h(t)) is a curve On the ellipsoid, R = (f; + ｾ + ｾＩＭＱＯＲＬ S = ＨｾＫ ｾ + ｾＩＭＱＯＲＮ Now, 'Y is a

8.1

f.,

?'

geodesic ;Y is parallel to the normal (j,y,h) = some scalar A(t). From

f;+ ｾ Ｋ

ｾ

+ｾ +ｾ

= 1 we get ｾ

+ ,. + ｾ +A ＫｾＨ ｾ ＩｾＫ

Ｋｾ = 0, i.e. ｦ［Ｋｾ 2 2 which gives A = _5 / R • The curvature of'Y is f;?+"

II ;Y II = (p

+ y2 + h 2)'/2

= IAI ＺｾＨ

Finally, 1 d (

"2 dt

1

R2 5

) = 2

A(f,., #t, f,-)

+ :: + ｾＺ

t

2

ｉｾ

for

= 0, hence = 0,

= ｾＲＧ

(P

2 fj+ gg. hi.) g2 i. ) + -+-+p' ( p4 q4 r 4 q2 r2

2 +(1' +g2 +h ) (i/+gy+i.h) p4

q4

r4

p2

q2

r2

4

309

solutions

1 O. Then, a is the maximum distance of a point of the spheroid from the z-axis, ｾｯ the angular momentum n of a geodesic must be :::; a (we can assume that n ｾ 0). If n :::: 0, the geodesic is a meridian. If 0 < n < a, the geodesic is confined to the annular region

q;.,

and the on the spheroid contained between the circles z :::: ±bJl discussion in Example 8.9 shows that the geodesic 'bounces' between these two circles:

Elementary Differential Geometry

312

n=

3d

Soiubons

a, Eq. (10) shows that the geodesic must be the parallel z = O. (ii) Let the torus be as in Exercise 4.10. If [l = 0, the geodesic is a meridian spirals around the torus: (a circle). If 0 < n < a - b, the ｧｾｯ､･ｳｩ｣

8.19 From Exercise 5.5, the cone is isometric to the 'sector'S ofthe plane with vertex at the origin and angle 1l'V2:

If n = a - b, the geodesic is either the parallel of radius a - b or spirals around the torus approaching this parallel asymptotically (but never crossing it):

Geodesics on the cone correspond to possibly broken line segments in S: if a line segment meets the boundary of S at a point A, say, it may continue from the point B on the other boundary line at the same distance as A from the origin and with the indicated angles being equal:

If

o

B

A

If a - b

< n < a + b, the geodesic is confined to the annular region

consisting of the part of the torus a distance :2: {l from the axis, and bounces between the two parallels which bound this region:

(i) TRUE: if two points P and Q can be joined by a line segment in S there is no problem; otherwise, P and Q can be joined by a broken line segment satisfying the conditions above:

q,

o

p

If {l = a + b, the geodesic must be the parallel of radius a + b.

s

Elementary Differential Geometry

314

315

Solubons

To see that this is always possible, let PI, P2, q, and q, be the indicated distances, and let R and S be the points on the boundary of the sector at a distance (p,q, + P1q2)/(P, + q,) from the origin. Then, the broken line segment PR followed by SQ isthe desired geodesic. (ii) FALSE: Q

a

°

°

since e- 1 / t ' --+ as t --+ 0. We prove that 8.20 (i) This is obvious if n ｾ n 1 t t- e- / ' --+ as t --+ by induction on n > 0. We know the result if n = 0, and if n > we can apply L'Hopital's rule: . t- n nt- n - 1 n t-(n-2) . . hm -1 = lun = hm Ｍｾ Ｍ t2 t-.o e / t--+O 2 e1/t2 t--+o 2 e 1/ t2 ,

°° °

p

".

(iii) FALSE: many meet in two points, such as the two geodesics joining P and Q in the diagram in (ii).

(iv) TRUE: the meridians do not intersect (remember that the vertex of the cone has been removed), and parallel straight lines that are entirely contained in S do not intersect. (v) TRUE: since (broken) line segments in S can clearly be continued indefinitely in both directions. (vi) TRUE: a situation of the form

which vanishes by the induction hypothesis. (ii) We prove by induction on n that 0 is n-times differentiable with dnO = { ｾＬｴＩ e- 1/ t ' if t ¥- 0, dtn if t = 0, where Pn is a polynomial in t. For n = 0, the assertion holds with Po Assuming the result for some n ｾ 0,

°

ｏＡＫｾ

_ (-3nPn dtn+1 -

if t

¥- 0, so we take Pn+ 1 dn+!O _. dtn+! ＭＱｾ

a

t3n+1

ｐｾ

= l.

2Pn) -l/t'

+ t 3n + t 3n + 3 e

= (2 - 3nt2 )Pn + t 3ｐｾＮ Pn(t) -l/t' _ . t3n+! e - Pn(o) Ｑｾ

If t = 0, e- 1/ t' _ t 3n +! -

°

by part (i). Parts (iii) and (iv) are obvious. 8.21 Since "Ie is unit-speed, Ur.U r = 1, so foR Ur.U r dr = R. Differentiating with respect to 0 gives foR Ur.Ure dr = 0, and then integrating by parts gives r=R

O'e.D'r[r=O -

in which the indicated angles are equal is clearly impossible. But the answer to this part of the question depends on the angle of the cone: if the angle is a, instead of rr/4, lines can self-intersect if a < rr/6, for then the corresponding sector in the plane has angle < rr:

R

lo

0

U()·f1rr

dr = O.

Now U(O, 0) = P for all 0, so Ue = 0 when r = 0. So we must show that the integral in the last equation vanishes. But, U rr = t e , the dot denoting the derivative with respect to the parameter r of the geodesic "re, so tT rr is parallel to the unit normal N of U; since ue·N = 0, it follows that US.a rr = O. The first fundamental form is as indicated since ur.u r = 1 and ur.ue = 0.

316

Elementary Differential Geometry

Solutions

Chapter 9 9.1 9.2 9.3 9.4

This follows from Exercise 7.6. This follows from Exercise 7.3. 1t1 + 1t2 = 0 and 1t1 = 1t2 ==> 1t1 = 1t2 = O. 1t1 + 1t2 = 0 ==> 1t2 = ==> K = 1t11t2 = :s O. K = 0 ¢==} = o ¢==} 1t1 = 1t2 = 0 ¢==} the surface is part of a plane (by Proposition 6.5). 9.5 By Proposition 7.6, a compact minimal surface would have K > 0 at some point, contradicting Exercise 9.4. 9.6 By the solution of Exercise 7.2, the helicoid /7(u,v) = (vcosu,vsinu,Au) has E = A2 + v 2,F = O,G = I,L = O,M = A/(A 2 +V 2)1/2,N = 0, so

-It,

-It;

317

which integrates to give f == ｾＭ In cos a(x + b), where b is a constant; we can assume that b = 0 by translating the surface parallel to the xaxis. Similarly, 9 == ｾ In cos ay, after translating the surface parallel to the

y-axis. So, up to a translation, we have y _ 11 (cosa z--n - - ), a cos ax which is obtained from Scherk's surface by the dilation (x, y, z) >-+ a(x, y, z). If a = 0, then / = g" = 0 so f = b + cx, 9 = d + ey, for some constants b, c, d, e, and we have the plane z = b + d + ex + ey. 9.10 The first fundamental form is (cosh v + l)(cosh v - cos u) (du 2 + dv 2 ), so /7

It;

is conformal. By Exercise 7.15, to show that • I

that

U uu

- LG-2MF+NG_ 2(EG _ F2) - 0. H -

9.7

tT uu

A straightforward calculation shows that the first and second fundamental forms of /7' are cosh 2 u(du 2 +dv 2) and - cos t du 2 - 2 sin tdudv + cos t dv 2, respectively, so 2

2

H = - cos t cosh u + costcosh u = O. 2 cosh4 u 9.8 From Example 4.10, the cylinder can be parametrised by /7(u, v) = ')'(u) + va, where')' is unit-speed, II a 11= 1 and ')' is contained in a plane II perpendicular to a. We have /7 u = t = t (a dot denoting d/du) , /7 v = a, so E == 1, F == 0, G == 1; N == t x a, O'uu == t == Kn, tT uv == t1 vv == 0, so L = ItD.(t x a), M = N = O. Now t x a is a unit vector parallel to II and perpendicular to t, hence parallel to Dj so L = ±K and H = ±K/2. SO H = 0 ¢==} K = 0 ¢==} ')' is part of a straight line ¢==} the cylinder is part of a plane. 9.9 Using Exercise 9.2, the surface is minimal ¢==} (1

+ gI 2)/ + (1 + p)g" =

0,

where a dot denotes d/ dx and a dash denotes d/ dy; hence the stated equation. Since the left-hand side of this equation depends only on x and the right-hand side only on y, we must have /

1 + j2 = a,

g"

1 + g12 = -a,

for some constant a. Suppose that a i O. Let r = j; then / = rdr / df and the first equation is rdr/df = a(1 + r 2), which can be integrated to give af = ｾ In(1 + r 2 ), up to adding an arbitary constant (which corresponds to translating the surface parallel to the z-axis). So df / dx = ±v' e2af - 1,

iT

is minimal we must show

+ tT vv == 0; but this is so, since = (sin u cosh v, cos u cosh v, sin

ｾ sinh ｾＩＬ

/7vv = (- sin u cosh v, - cosu cosh v, - sin ｾ sinh ｾＩＮ

(i) /7(0, v) = (0,1 - cosh v, 0), which is the y-axis. Any straight line is a geodesic. (ii) /7(1T, v) = (1T, 1 + cosh v, -4 sinh ¥), which is a curve in the plane x = 1T such that Z2

= 16sinh2 ｾ = 8(coshv -1) = 8(y -

2),

i.e. a parabola. The geodesic equations are

ｾ

(Eu)

= ｾｅｵＨｕＲ

+ v 2 ),

ｾ

(Ev)

= ｾｅｶＨｵＲ

+ v 2),

where a dot denotes the derivative with respect to the parameter t of the geodesic and E = (cosh v + 1) (cosh v - cos u). When u = 1T, the unit-speed condition is Ev 2 = 1, so v = 1/(cosh v + 1). Hence, the first geodesic which holds because E u = sin u(cosh v + 1) = 0 equation is 0 = ｾｅｵｖＲＬ when u == 1I"j and the second geodesic equation is d

dt (cosh v

+ 1) =

(cosh v

+ 1) sinh v v2 =

sinh v v,

which obviously holds. (iii) /7(u, 0) = (u-sin u, I-casu, 0), which is the cycloid of Exercise 1. 7 (in the xy-plane, with a = 1 and with t replaced by u). The second geodesic equation is satisfied because E v = sinhv(2coshv + 1- cosu) = 0 when v = O. The unit-speed condition is 2(I-cosu)u 2 = 1, so u = 1/2 sin ¥. The first geodesic equation is 1t(4sin 2 ｾｵＩ = sinuu 2 , Le.1t(2sin ¥) = cos ¥u, which obviously holds. 9.11 A = a2 + be = -(ad - be) (since d = -a) = - det W = -K.

Elementary Differential Geometry

318

9.12 (i) From Example 9.1, N = (-sechucosv,-sechusinv,tanhu). Hence, if N(u,v) = N(u',v'), then u = u l since u 14 tanhu is injective, so cas v = cas Vi and sinv = sinv', hence v = Vi; thus, N is injective. If N = (x, y, z), then x' + y2 = sech 2 u i' 0, so the image of N does not contain the poles. Given a point (x, y, z) on the unit sphere with x' + y' i' 0, let u = ±sech -1 v' x' + y', the sign being that of z, and let v be such that cosv = -xl .jx2 + y2, sin v = -y/ v'x 2 + y2; then, N(u, v) = (x, y, z). (ii) By the solution of Exercise 7.2, N = (A 2 +V')-'/'( -A sin u, Acos u, -v). Since N(u,v) = N(u+2krr,v) for all integers k, the infinitely many points u(u+2k7l",v) = u(u,v)+(0,0,2k7l") of the helicoid all have the same image under the Gauss map. If N = (x, y, z), then x' + y' = A' /(A 2 + v') i' 0, so the image of N does not contain the poles. If (x, y, z) is on the unit sphere and x' + y' i' 0, let v = - AZ/ .jx' + y' and let u be such that sin u = -xl v'x' + y', cosu = -y/ .jx' + y'; then N(u, v) = (x, y, z). 9.13 The plane can be parametrised by u(u, v) = ub + ve, where {a, b, e} is a right-handed orthonormal basis of R 3. Then, '(J = uu - iuv = b - ie. The conjugate surface corresponds to if{) = c + ib; since {8, C, - b} is also a right-handed orthonormal basis of R 3, the plane is self-conjugate (up to a translation). 9.14 '{J = ＨｾｦｬＭ g'), ｾｦＨＱ + g2),fg) ==} i'{J = ＨｾｩｦｬＭ g'), fif(1 + g2),ifg), which corresponds to the pair if and g. 9.15 By Example 9.6, '(J(() = (sinh(, -icosh(, 1). From the proof of Proposition 9.7, f = 'P1 - i'P2 = sinh ( - cosh ( = -c', 9 = 'P3/f = -e'. 9.16 '{J = u. - iu v = (1 - u 2 + v 2 - 2iuv,2uv - i(1 - v 2 + u'), 2u + 2iv) = (1 - (2, -i(1 + ('), 2(). So the conjugate surface is o7(u,v) = '.Ro /(i(1- (2), 1 + (2, 2i() d( = '.Ro (i (( = (-v

ｾＩ

+ u'v _

, (+ v;, u

ｾ

+ ｾＳ

(up to a translation)

o7(U,v) =

Ｈｾ

(U - V

+ UV

- U'V

+ｾ

ｾｕＳＩ

3

V -

.2-. (U+ V+UV' +U2V _ ｖＳｾ V2

9.17 '{J = ＨｾＱＭ

(-4)(1_ (2), ｾＨＱＭ

ＳＨ｟ ｾ ｯｒＮＧ］ｵ

2

= '.Ro

Ｈ ｾＭ

(-4)(1 + (2),((1 _ (-4)), so

-

C ' )3, ＫＨ ｾ

C ' )3, ＫＨ ｾ

up to a translation. Put (= e 0; since K < 0, we have (n - 2)rr > 0 and hence n ｾ 3; and ifn = 3, then JJ.nt('Y) (-K)dAu < rr so

11.

int('Y)

L) Ev ( ) E = 2E 1 -1 ｾ x,y,z ｾ 1. The surface in (ii) is obtained by rotating the curve x 2 + Z4 = 1 in the xz-plane around the z-axis:

Solutions

327

rl{'y)

Jo

ｾ ､yl_p2 ｳ

= 0. Using Green's theorem, this integral is equal to

{ Pudu + Pvdv

J'Jr ｾ

{ a( = Jint('Jr) 8u ｾ

Pv

a(

)

- 8v

ｾ

Pu

)

,

where'Jr is the curve in U such that -yes) = O'('II'(s))j and this line integral vanishes because

8 (

8u

Pv

ｾ

)

8 ( =8v ｾ

Pu

)

11.16 Let F : S --+ R be a smooth function on a surface S, let P be a point of S, let (1 and f1 be patches of S containing P, say O'(uo, vo) = u(Uo"vo) = P, lIu &vand let I F 0 0' and 1 = F 0 0'. Then, fu ｾ Ｎｩｾｶｉ + 11l7ffJ.' I" = lu ｾ + ｉｬｾＧ so if lu = Iv = 0 at (uo, vo), then Iii. = Iv = 0 at (uo, vo). Since 11.1. = Iv = 0 at P, we have

=

fuii. It is clearly diffeomorphic to the sphere, so X = 2. 11.13 Take the reference tangent vector field to be ( = (1,0), and take the simple closed curve 'Y(s) = (coss,sins), At 'Y(s), we have V = (a,I1), where OR _

a

+ 1/-'

-

{(COsS+isinS)k ( . ) Ie cos S - 1.sm s-

if k > 0, 0 1'f k a - O!&u + /-' &u' /-' - a 8v (3 &v' Hence, it and i3 are smooth if a and /3 are smooth. Since the components of the vectors 0' u and 0' v are smooth, if V is smooth so are its components. If the components of V = aO'v + /3O'v are smooth, then V.O'u and V.O'v are smooth functions, hence

G(V.O'u) - F(V.O'v)

a=

EG-F2

(3 = B(V'O'Il) - F(V.O'u) EG-F2

are smooth functions, so V is smooth. 11.15 If;j, is the angle between V and (, we have -if; - t/J = 8 (up to multiples of 271" ) j so we must show that Jot (-y) iJ ds = 0 (a dot denotes d/ ds) ° This is not obvious since 8 is not a well defined smooth function of s (although d9/ ds is well defined). However, p = cos 9 is well defined and smooth, II ( 1111 ( II· Now, p = -iJsin9, so we must prove that since p =

e.V

8u 8v = luu (8U)Z 8u + 2/uv 8ft 8ft + Ivv

(8v)2 8ft

'

fuv and ivv. This gives, in an obvious

with similar expressions for

ｾｩＨ

tation, il = J t 1lJ, where J ::::

7ffJ.

i!)

no-

is the jacobian matrix ·of

｟ ｾ

the reparametrisation map (ft,ii) I-t (u,v). Since J is invertible, 11. is invertible if 11. is invertible. Since the matrix 11. is real and symmetric, it has eigenveGtors,vl,v2, with eigenvalues AI, AZ, say, such that vfvj = 1 if i = j and 0 ifi¥- j. Then, if v = al VI + a2V2 is any vector, where 01, a2 are scalars, v t 1iv = AlaI + ＾Ｇｺ｡ｾ［ hence, v t 1lv > 0 (resp. < 0) for all v ;l '0 A!1and A2 are both> 0 (resp. both < 0) {::::::} P is a local minimum ＧＨｲｾＮ ilaca:J. maximum); and hence P is a saddle point {::::::} v t 1ivcan be br:Jth > Il!) and < 0, depending on the choice of v. Since J is invertible,a vecter v ;l 0 {::::::} v = Jv ;l OJ and vtili) = iJt J t 1lJv = v t1lv. ｾｨ･｡ｳｲｴｩ＼ｭ in the last sentence of the exercise follow from this. 11.17 (i) Ix = 2x ｾ 2y, 111 = -2x +8y, so Ix = 111 = Oat the 0r.igin. ixx = 'bl'eso 11'e . h :rn:Igm ... 2 , 1X1/ = -2 , 11 1=8 so 1l = -2 8 . '1J" n IS mvertL· 11'

,( 2 -2)

is non-degenerate; and the eigenvalues 5 ± is a local minimum.

(ii)

Ix

=

11/

= 0 and 1l =

Ｈｾ

ｾＩ

VI3 of 1l are ibdtih > (@, 'so;it

at the origin; det1i = -16< i(iJ, Ｐｓｾ

the eigenvalues of 1l are of opposite sign and the origin is a saddle \point. (iii) Ix = 11/ = 0 and 11. : : : at the origin, which is therefol'e;a·degenerate critical point.

°

328

Elementary Differential Geometry

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11.18 Using the parametrisation iT in Exercise 4.10 (with a = 2, b = 1) gives 1(0,1{J) = F(iT(O,I{J)) = (2 + cosO)COSI{J + 3. Then, 19 = -sinOcosl{J, f", = -(2 + cosO)sin