a course in differential geometry thierry aubin

A Course inDifferentialGeometry

Thierry Aubin

Graduate Studiesin Mathematics

Volume 27

American Mathematical Society

Selected Titles in This Series

27 Thierry Aubin, A course in differential geometry, 200126 Rolf Berndt, An introduction to symplectie geometry, 200125 Thomas } iedrich, Dirac operators in Riemannian geometry, 200024 Helmut Koch, Number theory: Algebraic numbers and functions, 200023 Alberta Candel and Lawrence Conlon, Foliation I. 200022 Gfinter R. Krouse and Thomas H. Lenagan, Growth of algebras and Gelfand-Klrillov

dimension, 200021 John B. Conway, A course In operator theory, 200020 Robert E. Gompf and Andrda I. Stlpsics, 4-manifolds and Kirby calculus, 199919 Lawrence C. Evans, Partial differential equations, 199818 Winfried Just and Martin Weese, Discovering modern set theory. IT: Set-theoretic

tools for every mathematician, 199717 Henryk Iwanlec, Tbpies in classical automorphic forms, 199716 Richard V. Kadison and John R. Ringrose, Fundamentals of the theory of operator

algebras. Volume 11: Advanced theory, 199715 Richard V. Kadison and John R. Ringrose, Fundamentals of the theory of operator

algebras. Volume I: I oentary theory, 199714 Elliott H. Lieb and Michael Loss, Analysis, 199713 Paul C. Shields, The ergodic theory of discrete sample paths, 199612 N. V. Krylov, Lectures on elliptic and parabolic equations in Holder spaces, 199611 Jacques Dhunier, Enveloping algebras, 1996 Printing10 Barry Simon, Representations of finite and compact groups, 19969 Dino Lorenalni, An invitation to arithmetic geometry, 19968 Winliried Just and Martin Weese, Discovering modern set theory. 1: The basics, 19967 Gerald J. Janusz, Algebraic number fields, second edition, 19966 Jens Carsten Jantzen, Lectures on quantum groups, 19965 Ride Miranda, Algebraic carves and Rlernann surfaces, 19954 Russell A. Gordon, The Integrals of Lebesgue, Denjoy, Perron, and Henstock, 19943 WIlliam W. Adams and Philippe Loustaunau, An introduction to Grdbner bases,

1994

2 Jack Graver, Brigitte Servatius, and Herman Servatius, Combinatorial rigidity,1993

1 Ethan Akin, The general topology of dynamical systems, 1993

A Course inDifferentialGeometry

Thierry Aubin

Graduate Studiesin Mathematics

Volume 27

Editorial BoardJames Humphreys (Chair)

David SaitmanDavid Sattinger

Ronald Stern

2000 Mathematics Subject Clamfiication. Primary 53B05, 53C05, 53C22, 53C40, 58A17,580.05, 58C25, 58C35, 58305.

Assmc-r. This book provides an introduction to differential geometry, with prinicpal emphasison Riemannian geometry . It covers the essentials, concluding with a chapter on the Yamahaproblem, which shows what research in the Said looks like. It is a textbook, at a level which isaccessible to graduate students. Its aim is to facilitate the study and the teaching of differentialgeometry. It is teachable on a chapter-by-chapter basis. Mary problems and a number of solutionsare included; most of them extend the course itself, which is confined to the main topics, suchas: differential manifolds, submanifolds, differential mappings, tangent vectors, differential forms,orientation, manifolds with boundary, Lie derivative, integration of p-direction field, connection,torsion, curvature, geodesics, covariant derivative, Riemannian manifolds, exponential mapping,and spectrum.

Library of Congress Cataloging-In-Publication DataAubin, Thierry.

A course in differential geometry / Thierry Aubin.p. cm. - (Graduate studies in mathematics, ISSN 1065-7339; v. 27)

Includes bibliographical references and index.ISBN 0-8218-2709-X (alk. paper)1. Geometry, Differential I. retie. II. Series.

QA641.A795 2000516.3'6--dc21 00.068275

Copying and reprinting. Individual readers of this publication, and nonprofit librariesact lag for them, are permitted to make fair use of the material, such as to copy a chapter for usein teaching or research. Permission is granted to quote brief passages from this publication inreviews, provided the customary acknowledgment of the source is given.

Republication, systematic copying, or multiple reproduction of any material in this publicationis permitted only under license from the American Mathematical Society. Requests for suchpermission should be addressed to the Assistant to the Publisher, American Mathematical Society,P.O. Box 6248, Providence, Rhode Island 02910.6248. Requests can also be made by e-mail toreprint-porsissionSaas.org.

© 2001 by the American Mathematical Society. AU rights reserved.The American Mathematical Society retains all rights

except those granted to the United States Government.Printed in the United States of America.

® The paper used in this book is add-free and falls within the guidelinesestablished to ensure permanence and durability.

Visit the AMS home page at URL: http;//vsv.ara.org/10987654321 060504030201

A mon ProfesseurAndre' Lichnerowicz

(in memoriam)

Contents

Preface ix

Chapter 0. Background Material 1

'Ibpology 1

Tensors 3

Differential CalculusExercises and Problems

Chapter 1. Differentiable Manifolds 19

Basic Definitions 19

Partition of Unity 25

Differentiable Mappings 27

Submanifnlds 29

The Whitney Theorem 30

The Sard Theorem -34Exercises and Problems as

Solutions to Exercises 40

Chapter 2. Tangent Space 43

Tangent Vector 44

Linear T agent Mapping 46

Vector Bundles 48The Bracket [X, Y] 49

Exterior Differential 52

Orientable Manifolds 55

Manifolds with Boundary 58

Exercises and Problems fi4

Solutions to Exercise and Problems fib

Chapter 3. Integration of Vector Fields and Differential Forms 77

vii

viii -1. Contents

Integration of Vector Fields 77

Lie Derivative 79The Frobenius Theorem 81

Integrability Criteria. 85Exercises and Problems 87Solutions to Exercises and Problems 93

Chapter 4. Linear Connections 99First Definitions 99Christoffel Symbols 100Torsion and Curvature 101Parallel Transport. Gecxlesics 103Covariant Derivative 105Exercises and Problems 107Solutions to Exercises 108

Chapter 5. Riemannian Manifolds 111Some Definitions 11.1

Riemanmian Connection 114Exponential Mapping 117Some Operators on Differential Fbrms 121

Spectrum of a Manifold 125Fxereisp-q and Problems 129

Solutions to Exercises and Problems 145

Chapter 6. The Yamabe Problem: An Introduction to Resear ch 169

Bibliography 177

Subject Index 179

Notation 183

Preface

This book provides an introduction to differential geometry, with principalemphasis on Riemannian geometry. It can be used as a course for second-year graduate students. The main theorems are presented in complete detail,but the student is expected to provide the details of certain arguments. Weassume that the reader has a good working knowledge of multidimensionalcalculus and point-set topology

Many readers have been exposed to the elementary theory of curves andsurfaces in three-space, including tangent lines and tangent planes. Butthese techniques are not necessary prerequisites for this book.

In this book we work abstractly, so that the notion of tangent space doesnot necessarily have a concrete realization. Nevertheless we will eventuallyprove Whitney's theorem asserting that any abstract n-dimensional manifoldmay be imbedded in the Euclidean space RP if p is sufficiently large.

In order to develop the abstract theory, one must work hard at the be-ginning, to develop the notion of local charts, change of charts, and atlases.Once these notions are understood, the subsequent proofs are much easier,allowing one to obtain great generality with maximum efficiency. For exam-ple, the proof of Stokes' theorem-which is difficult in a concrete context-becomes transparent in the abstract context, reducing to the computationof the integral of a derivative of a function on a closed interval of the realline.

In Chapter I we find the first definitions and two important theorems,those of Whitney and Sard.

Chapter II deals with vector fields and differential forms.

ix

x Preface

Chapter III concerns integration of vector fields, then extends to p-planefields. We cite in particular the interesting proof of the Frobenius theorem,which proceeds by mathematical induction on the dimension.

Chapter IV deals with connections, the most difficult notion in differen-tial geometry. In Euclidean space the notion of parallel transport is intuitive,but on a manifold it needs to be developed, since tangent vectors at distinctpoints are not obviously related. Loosely speaking, a connection defines aninfinitesimal direction of motion in the tangent bundle, or, equivalently, aconnection defines a sort of directional derivative of a vector field with re-spect to another vector. This concrete notion of connection is rarely taughtin books on connections. In our work we devote ten pages to developing theseideas, together with the related notions of torsion, curvature and a workingknowledge of the covariant derivative. All of these notions are essential tothe study of real or complex manifolds.

In Chapter V we specialize to Riemannian manifolds. The viewpointhere is to deduce global properties of the manifold from local properties ofcurvature, the final goal being to determine the manifold completely.

In Chapter VI we explore some problems in partial differential equationswhich are suggested by the geometry of manifolds.

The last three chapters are devoted to global notation, specifically to us-ing the covariant derivative instead of computing in local coordinates withpartial derivatives. In some cases we are able to reduce a page of computa-tion in local coordinates to just a few lines of global computation. We hopeto further encourage the use of global notation among differential geometers.

The aim of this book is to facilitate the teaching of differential geometry.This material is useful in other fields of mathematics, such as partial differ-ential equations, to name one. We feel that workers in PDE would be morecomfortable with the covariant derivative if they had studied it in a coursesuch as the present one. Given that this material is rarely taught, one mayask why? We feel that it requires a substantial amount of effort, and thereis a shortage of good references. Of course there are reference books such asKobayashi and Nomizu (5J, which can be consulted for specific information,but that book is not written as a text for students of the subject.

The present book is made to be teachable on a chapter-by-chapter basis,including the solution of the exercises. The exercises are of varying difficulty,some being straightforward or solved in existing literature; others are morechallenging and more directly related to our approach.

This book is an outgrowth of a course which I presented at the UniversitkParis VI. I have included many problems and a number of solutions. Someof these originated from examinations in the course. I am very grateful tomy friend Mark Pinsky, who agreed to read the manuscript from beginning

to end. His comments allowed me to make many improvements, especially inthe English. I would like to thank also one of my students, Sophie Bismuth,who helped me to prepare the final draft of this book.

Chapter 0

Background Material

In this chapter we recall some fundamental knowledge which will be used inthe book: topology, algebra, integration, and differential calculus.

Zbpology

0.1. Definition. A topology on a set E is defined by a family O of subsetsof E, called open sets, such that

a) The set E and the null set 0 are open sets.

b) Any union of open sets is an open set.

c) Any finite intersection of open sets is an open set.

(E, 0) is a topological space.

0.2. Examples. If 0 = {E, 0}, the corresponding topology is called trivial.If 0 consists of all subsets of E, the topology is called the discrete topology.

On Rn the usual topology may be defined as follows: Let x be a pointof R" and p > 0 a real number. We consider the open ball of center x andradius p, BS(p) = {y E R" I Hix - yul < p}. An open set in R" will be aunion of open balls or the empty set 0.

0.3. Induced topology. Let F be a subset of E endowed with a topology0. The induced topology on F is defined by the following set 0 of subsetsofF: AEOifandonlyifA=AflFwithAEO.

0.4. Example. Let F be a finite set of points in R. The topology on Finduced by the usual topology on R" is the discrete topology. We will findother examples in 1.16.

1

2 0. Background Material

0.5. Definitions. A neighbourhood of a point x in a topological space E isa subset of E containing an open set which contains the point x.

We can verify that a subset A C E is open if and only if it is a neigh-bourhood of each of its points.

B C E is closed if A = E\B is open.A topological space E is said to be connected if the only subsets which

are both open and closed are the empty set 0 and the space E itself.The closure of a subset B C E is the smallest closed set containing

B. The closure B always exists-indeed, the intersection of all closed setswhich contain B (E is one of them) is a closed set according to b) in 0.1.

0.6. Proposition. Any neighbourhood A of X E B has a nonempty inter-section with B.

Proof. Let Q C A be an open neighbourhood of x. If o n B = 0, then E\12is a closed set containing B; hence B C E\SZ and x f B, a contradiction.

00.7. Definitions. The interior B of B C E is the largest open set containedin B (B is the union of all open sets included in B.)

A topological space is separable if it has a countable basis of open sets{Ai}iEN. That means any neighbourhood of x contains at least one Ai withx E Ai.

A topological space is Hausdorff if any two distinct points have disjointneighbourhoods.

A family {Sl;}ieI of subsets of E is a covering of B C E if B C (Rt.A subcovering of this covering is a subset of the family, {S?L}1ej (with J c I),which itself is a covering. If J is finite the subcovering is said to be finite.

0.8. Definition. A subset A C E is a compact set if it is Hausdorff and ifany covering of A by open sets has a finite subcovering.

This definition implies the following necessary and sufficient condition:A c E, a Hausdorff topological space, is compact if and only if any familyof closed sets whose intersection is empty has a finite subfamily of emptyintersection.

0.9. Theorem. Let E be a Hausdorf topological space. If K C E is acompact set, K is closed. This condition is sufficient when E is compact.

Proof. We argue by contradiction. If K is not closed (K 96 7), thereexists x E K such that x 0 K. Now in a Hausdorff topological space, theintersection of the closed neighbourhoods of a point x is just the subset {x},which is closed. Indeed, for any y E E there exist disjoint open sets 9 and

3

12, neighbourhoods respectively of x and y. E\i2 is a closed set, which is aneighbourhood of x since A C E\fl, and y E\i2. Thus the traces on K ofthe closed neighbourhoods { V; },E I of x would have an empty intersection.So {E\Vf}jeI would be a covering by open sets of K.

Since K is compact, there would exist a finite set J C I such that{E\ Vi };EJ is a covering of K. Thus n,E J Vi =V would be a neighbourhoodof x and V fl K 0. Since x E I?, Proposition 0.6 gives a contradiction.

Suppose E compact and A C E closed. Then the closed seta for A areclosed sets for E, and compactness for A follows from the necessary andsufficient coo&tion for A to be compact (see above).

0.10. Definition. Let E and F be two topological spaces. A map f of Einto F is continuous if the preimage f -' (f) of any open set 1 C F is anopen subset of E.

0.11. Theorem. he image by a continuous map of a compact set is com-pact.

Proof. Let K C E be a compact set. Consider any covering of f(K) byopen sets 5 (i E I). (f -1(S2i)}iE j is a covering of K by open sets; thusthe exists a finite set J C I such that K C UiSJ f -'(W- So {ftifiEJ is acovering of f (K).

0.12. Definitions. A continuous map is said to be proper if the preimageof every compact set is a compact set.

If E and F are Hausdorff topological spaces, a continuous map f of Einto F is proper if E is compact. Indeed, let K C F be a compact set. SinceK is a closed set. f -1(K) is a closed set. So f -1(K) is compact, since aclosed set in a compact set is a compact set (Theorem 0.9).

Let E and F be two topological spaces. A map f of E onto F is ahomeomorphism if it is one to one and if f and f -1 are continuous maps.

Tensors

0.13. Definition. Let E and F be two vector spaces of dimension respec-tively n and p. The tensor product of E and F is a vector apace of dimensionnp, and is denoted by E ®F. A vector of E OF is called a tensor. Tb X E Eand y E F we associate x ®y E E ®F. This product has the followingpropertim-

a) (xI+xa)®y=x1®y+x2®yandx®(yi+ya)=x®yl+x@y2,where x, x1, x9 belong to E and y, yl, y2 belong to F.

b) If a E R (in this course the vector spaces are on R), then

(ax)®y=xs(ay)=a(x(9 y).


c) If { ei } 1 <i <n is a basis of E and { f j } i 5, , a basis of F, then ei ®f1is a basis of E (9 F.

d) If G is a third vector space, with E ®(F (9 G) _ (E ®F) ®G, then

(x®y)®z=x®(y(9 z),for z belonging to G.

The tensor product is associative; it is not commutative. Let x E E andyEF, x=xiei andy=yifj. Here we use

The Einstein Convention: When the same index (such as i or j) isabove and below, summation over this index is assumed (here from 1 to nfor i, and from 1 top for j). i (or j) is called a dummy index.

According to a), x ® y = xiy'ei ® fj. # = x'yj are the components ofx ®y in the basis lei ® fj } of E ® F.

What happens when we change basis?Let {e«}i<a<n and fj,3j.jS#,Sp be other bases respectively of E and F.

Then e« = a%ei and f S = c J j . Or, if we want, ei = b ff,, and f j = dfi fhere ((b;)) is the inverse matrix of and ((d3)) is the inverse matrixof

A tensor T = t«13e® ® fp = tdjei ® f2. Thus

t'j = t°pa ,d and t«'s = 04i 'do.

0.14. Definition. A (p, q)-tensor, attached to a vector space E of dimen-sion n, is a tensor of El ®E2 ® where E, = E for q values of iand Ej = E*, the dual space of E, for p values of j. We say that the tensor

is p times covariant and q times contravar cant. We denote by ® E*®

Ethe set of the (p, q)-tensors.

Let ( O } ' < < be the dual basis of {ei}i<i<n A (p,q)-tensor is a linearcombination of fl ®f2 ® ®f q, where fj = 0) for p values of j, andfi = e; for q values of i. If we consider another basis of E, say {e«}1<«<ne« = aaej, the new dual basis is given by B = bJ 8j, where ((by))is the inverse matrix of ((a')): bjas = o, the Kronecker tensor (b, = 0 ifi0jandb =1ifi=j).0.15. Examples. A (1,0)-tensor w is a 1-form: w = wj9j = w;99 ; thuswj = b9CDO. The components of w are wj in the basis {0'} and wy in thebasis {Bp}. The indices j and 0 are called covariant indices.

A (0,1)-tensor X is a vector: X = X'ei = X«e"«. Thus X' = aQXa.The components of X are X' in the basis {ei} and X« in the basis {ea}.The indices i and a are called contruvariant indices.

The contravariant indices are up, and the covariant indices are down.When we change the basis of E, { ei } -r {ea }, we express the components

of a (p, q)-tensor T (in the basis lei }) in terms of its new components (inthe basis {e}) by means of the matrix ((&)) for the covariant indices andthe matrix ((a )) for the contravariant indices.

0.16. Examples. In this book we will consider a Riemannian metric g. Ata point, g is a (2,0)-tensor on the tangent space which is isomorphic to R'.gij are the components of g in the basis {ei}. If ga,q are the components ina new basis {ea}, we get g,1 = b°bga,6, since g = 9ifel ® 8i = 9ayaea ®Bb

We will consider also the curvature tensor R, which is a (3,1)-tensor.We get

R ' = b° a,3bkbi

R in the first basis ands ra the componentsin the new basis.

0.17. Theorem. A system of nP+9 real numbers attachedto a basis 8`1®O'2 ®. ®8'p ®eiy+1®. are the components of a (p, q)-tensor on E if and only if in a new basis gal® ®9°p ®eal ®the system of real numbers ta,a2...ap°p}1"'Qp+ satisfies

iP+I...ip+9 = ba1bb ... b°p ro+1 a y+9 aP,...ap+9tili2...ip it i2 Lp aap+1 ... ap+9 OIOt2...ap

So we have

41 i76i 1 ®g`2 ®... ga O"P ®eip+1 ®, ...

talae...0p°P+l...ap+901 9012 ®... ® gap®eap.}1 ... ®eap+

Often it would require long computations to verify the above equalities.Fortunately there are tensoriality criteria.

0.18. 11msoriality Criteria Let {wi}1<,<n be a system w of n real num-bers attached to the basis lei) of E, and {wa}1<a<n the system w in thebasis Suppose that for any vector X E E, wiX' = °X°. Thenw is a 1-form.

Since X' = aa.tc (Example 0.15), we find that (wia, 0 forany X. Thus Wa = wia'a. Multiplying the equality by b7 , we get, aftersummation, wi = b,*(;).. So w is a (1,0)-tensor, that is a 1-form, accordingto Theorem 0.17. This result can be generalized.

0.19. Theorem. Suppose that a system T of nv+9 real numbersin the basis {ei}, in the basis {eep} sat-

isfies the following condition: for any vectors X1, , X p and any 1 -forms


1 v

ip+,...;p+aX+t X'2 ... X`pwl w44i i2...b1 s P w ... p+4

= ta, l .. . Xp p -1 9ap+l .. wo+v.

Then T is a (p, q) -tenor.

Proof. For 1 < j < p we have XX = aQX7 , and for 1 < k < q we haveb°wa. Putting these expressions in the condition of Theorem 0.19, we

get4Ii2

.ipy,l...aal ... aa;b

v+ t2 ... btr+v -Thus T is a (p, q)-tensor.

There are others tensoriality criteria.

0.20. Example. Suppose that a system T of n3 real numbers tip in thebasis {ei}, 11.0 in the basis {ea} satisfies the following condition: for anyvectors X and Y, t kX'X' = Zk are the components of a vector Z; that is,we also have Z"r Then T is a (2,1)-tensor.

Proof. We have X' = a,X°, Y' = a' ,Y'3 and Zk = t42-f. Thus tk?a,a', _aryP... Multiplying both sides by bk, we get, after summation, tija' a'b _

since bleary = 67. So T is a (2,1)-tensor.

0.21. Definition. The exterior product of two vectors x and y of E is theskew-symmetric tensor

xAy=x®y-yOx.If {x'} and {y'} are the components of x and y in the basis {e;}, then x'yf --xxy' = z'j are the components of x A y in the basis {ei A eJ }, 1:5 i < j < n,which has Cn elements.

In this book we will consider especially the skew-symmetric p-forms onR" (or on a vector space E of dimension n), that are the skew-symmetric(p, 0)-tensors. If {x'}1<i<n are the coordinates on R" (or E) correspondingto a basis {ei}1<i<n, we denote by dx' the 1-form such that dx'(ei) = 6ji,the Kronecker tensor. A 1-form is defined when we know what it giveson the vectors of a basis, so dx' is well defined. We denote by APE thespace of the skew-symmetric p-forms on E. More generally, the elements(dx''' Adxa2 A ... A dxJ'p)1<_A,<A2<...<A,<" form a basis of ARE.

These forms are well defined when we know what they give on a systemof p vectors of a basis. By definition,

dxal Adx\2 A..-Adxap(ea>,e,,... ,e,,,) =avid ...;

here 1 < Al < Az < < Ap < n and 1 < pi < µ2 < <.up < n. Indeed,we arrange the p vectors in order. When we interchange two of them, theresult changes sign.

0.22. Contraction of Indices. Let T be a (p, q)-tensor. The contractionconsists of choosing two indices, one covariant and the other contravariant.We set them equal and make summations, and thus get a (p-1, q-1)-tensorwhich is a tenser obtained from T by contraction.

Let be the components of T (see 0.17 for the nota-tions). Choose for instance it and ip+i. We obtain, after summation overj=it=ia+i,

Tip.. t7.i2...,ip9dP+z r+o,

which are the components of a (p -1, q -1)-tensor. Indeed, since a,9 Y3 } 1 =89.11w e have

r+a...QPTv

` 7 iP-l...iailsi2 ... acv b' b«p+2 ... b°'*l .i3."' .ry 0 as P ip+1 iP+2 ir"= ....... aa2 ... aDb°Fy ... b° q

Differential Calculus

0.23. Definition. Let f be a continuous map of an open set fl C R" intoRP. We say that f is diferentiable at xo E R" if there exists a linear mappingA E C(R', R") such that for any y E R" with xo + y E it we have

f (xo + v) - f(xo) = A(y) + IIvI{w(xo, y),

where w(xo, y) -+ 0 as y - 0.

We call A the differential of f at ;ro. If f is differentiable at every pointx E 12, we say that f is differentiable on S1. The differential off at x dependson x; we denote it by f'(x). If fl 3 x - f '(X) E C(Rn,RR) is continuous, wesay that f is continuously differentiable on f2, or f is Cr.

There are functions which are differentiable on fl and which are not Cr.Nevertheless, in this book when we talk about a differentiable function f , weassume that f is at least C1. We will never consider differentiable functionswhich are not C'.

0.24. Remark. We can define differentiable maps of a Banach space BIinto a Banach space B2. The definition is the same as above except that thelinear mapping A must be continuous (for finite dimensional Banach spaces,A linear implies A continuous). Recall that a Banach space is a completenormed vector space. Complete means that any Cauchy sequence converges.

0. Background Material

0.25. Let f be a differentiable map of 11 C R" into RP. f is defined by meansof a system f', f 2, JP P of p differentiable functions on Sl (real-valued):the coordinates of f (x) in RP are f' (x), f 2(x), , fP(x). The differentialf'(x) is represented by the Jacobian matrix ((8f°/8x'))x with n columnsand p rows.

Here {x`}1<i<" are coordinates on R' ({e;} is a basis of R"), and{y° } i <°<P are coordinates on RP. Indeed, set A = ((a)). From Defni-tion 0.23 we get

f°(x+te;) - f°(x) = to?+ltlw°(x,tei),which is defined for t in a neighbourhood of 0, and w°(x, tei) -' 0 whent - 0. Thus if ° is differentiable and a, = 8 f °/8x`.

f is C' on Sl if and only if the partial derivatives 8 f °/ex' (1 < o < p,1 < i < n) exist and are continuous on 12.

The rank of f at x is the rank of ((8f°/8xi))x; it is the dimension ofthe vector subspace of RP, the image of R" under f'(x).

0.26. Definition. We say that f is C2 on Il (f E C2(fl)) if the map

S2 3 x f(x) E G(R", RP)

is C' on Q.

By induction we define C'-functions (r E N). C'°-functions are funo-tions which are C'-functions for all r E N; in this case each f° has partialderivatives of any order. C" functions are analytic functions.

Let g be a C'-function of 0 C RP into R'n. Suppose f (11) C 0; then go fis a Cl-function of 11 into R' and (gof )' = go f'. We will write also df forf'; thus d(g o f) = dg o df . If f is a homeomorphism of fl onto f (S2) C R"(n = p), then f is a diffeomorphism if f and f -1 are differentiable. In thiscase (f')-1 = (f -' )', since d(f -1 o f) is the identity map.

0.27. Mean Value Theorem. Let a, b be two points in n C R" such thatthe line segment la, b) C S2, and let f be a Cl function of Cl into RP. Then

Ilf(b) - f(a)ll 5 MIIb- all,

where M = SUPXEf01 llf'(x) II

The set [a, b] = {a + t(b - a)/ t E [0,1)} is compact and x -' ll f'(x)ll iscontinuous; thus M is finite. Here the norm II II is the norm as operator: ifu E G(R", RP), then (lull = supllzU=1 llu(x) II.

Proof. Set $(t) = f (a + ty) with y = b - a. Then $ is a C' map of I0,1)into RP. According to the chain rule mentioned in 0.26, 4'(t) = f'(a+ty)oy.

Differential Calculus 9

Thus

and

1 1f (b) - f (a) = t(1) - t(0) = fo 4'(t)dt = fo f'(a + ty) o y dt

1

11f (b) - f(a)ll <- 1 IIf'(a +ty)IIIlyIIdt < MIlyll.

0.28. Proposition. The image of R' by a C1-mapping f into RD is of(Lebesgue) measure zero if n < p. The same result holds for an open setilcR".

We give this result because we will use it in Chapter 1.

Recall that a set A C RP is of measure zero (µ(A) = 0) if for an ar-bitrary e > 0 there exist balls Ba (a E N) such that A C UaEN Ba andE,..N FA(B,) < e.

Thus a countable union of sets of measure zero has measure zero.

Proof of Proposition 0.28. Let Kr = {x E R" I IX 'I < rfor I < i < n}({x'} are the coordinates of x). We slice each side of the cube Kr intok segments of the same length. We obtain k" little cubes, with sides oflength 2r/k. If we consider Kr C R" C RR, Kr has measure zero in R".Indeed, K, c 1 Rp1(p) with p = r//k, where Bp, (p) is the ball inP' of radius p = ry' /k and center Pj, the center of a little cube. Soµ(K,.) < k"µ(Bo(p)) = k"-nµ(Bo(r/i)), where u denotes the measure inP'. When k oo, the right hand side goes to zero; then µ(Kr) = 0.

Let f be a C1-mapping of R" into RP. On Kr we have Ilf'I1 S Mr,and by the mean value theorem, for x, y in Kr, 11f (x) - f (y) II <_ Mr II x - y1lThus the little cube Kj of center Pj is such that f (Kj) C B f(p,)(pM,). Thisimplies

jz[f(Kr)] <- k"µ[Bo(pMr)] = k"-,A[B0(rMrv1n-)],

which goes to zero when k -+ oo. So p[f (Kr)] = 0. As R" = Ur' 1 Kr,we see that f (R") = Vr=1 f (Kr). According to the theorem cited above,µ[f (R")] = 0, since µ[f (Kr)] = 0 for all r E N. For general fl, there existsa sequence of compact sets {K} such that K; C Ki+1 and S2 = U .6N K$We prove that tC[f (K1)] = 0, using the result above. For this we only haveto consider a C1-mapping f of R" into R" such that f /Ki = f /Ki. Thenµ[f (Ki)] = 0 for all i implies µ[f (0)] = 0.

0.29. Inverse Function Theorem. Let f be a C' -mapping of an open set0 C R" into R". If f'(xo) is invertible (rank f = n at xo E S2), there existsan open neighbourhood 0 of xo such that $ = f IB is a diffeomorphism of 9onto f (0) . If f E C'`, then 4' and 4-1 are C'` on 0.


Proof. Let us consider the mapping 9(x) = f'-1(xj) o f (x). We haveg'(xo) = Identity. Set h(x) = g(x) - x. h is C1 on f2, and since h'(xo) = 0,there exists r > 0 such that llx - xoll < r implies llh'(x) 11 < 1. We choose rsmall enough so that rank f = non the ball b C fl of radius r and centerxo. According to the mean value theorem, for x and x' in B

11h(x) - h(x')II < 'Ilx - x'II.

Let B be a ball of center yo = g(xe) and radius smaller than r/2. Considerthe equation x + h(x) = y for y E B given. We define by induction thesequence xk+l = y - h(xk) (k > 0). We have

llxk+1-xkll = llh(xk)-h(xk-1)ll <'Ilxk-xk-111 <' Ilxl-xoll =' lly-yollsince y E B implies xk E B. Indeed, x1 = y - [g(xo) - xoj and

k k

Ilxk - xoll <5Ilxl-x1_111 <Ily - yollE <211y-yoll7=1 =1

The sequence {xk} is a Cauchy sequence. So xk -. x E B. Since h iscontinuous, x = y - h(x). Soy g(x). Moreover, the solution is unique inB, since llh'(z)I1 < for z E B. Indeed, Ilx-xll = Ilh(x)-h(x)ll S 1Ilx-illimplies x = x.

Set 0 = B n g- (B); then = f Is is bijective. Let us show that -1 isdifferentiable on f (0). For all c > 0, there is q > 0 such that 11X - xol1 < r)implies

II g(x) - 9(xo) - x + xoll < Ellx - xoll,

since h'(xo) = 0. ThuslIf'-1(xo) o (y - yo) - $-1(y) + -1(yo) II < 2elly - yo Il.

This inequality proves that -' is differentiable at xo. Its differential isfi-1(xo), and rankt-1 = n at yo. As we can give the same proof atany point zo E 0, it follows that $-1 is differentiable on f (0). Since(V1)'(y) = f'-' [V'(y)l, it follows that V1 is C' on f (0). 4-1 is Ckif f is Ck

0.30. Remark. The inverse function theorem holds in Banach spaces. Theproof is the same.

And now a global result.

0.31. Theorem. Let f be a C' -mapping of ft C B into b, B and B beingtwo Banach spaces. Suppose that

a) f is injective, andb) f'(x) is an isomorphism from B onto b for every x E ci.

Then f is a difeomorphism from Il onto f (S2) C B.

Since f is injective, 4 = f It, : 0 -i f (Cl) is invertible; i.e., 4-1 exists.According to 0.29, $'1 is continuous and C1-differentiable (these are localproperties).

0.32. Let B be a Banach space, and let (t, x) -- f (t, x) E B be a con-tinuous function defined in a neighbourhood of (to, xo) in R x B. Since fis continuous, there exists a neighbourhood V of (to, xo) where f (t, x) isbounded.

Ilf(t,z)II < M for all (t,x) E V.

We consider a closed ball f Z of radius r and center xo in B, and I =[to - a,to + a] C R (a > 0) such that I x f2 C V, a and r satisfyingMa < r. Recall that a map h of an open set 0 of a Banach space B1into a Banach space B2 is Lipschitz in 0 if there exists k E R such thatiIh(a) - h(b)l < klla - bit for (a, b) E 0 x 0.

0.33. The Caudh Theorem. The differential equation

(*) a = f (t, x), z(to) = zo,

has a unique continuous solution x(t) if f (t, x) is Lipschitz in x on I x A.The solution is defined on a neighbourhood J C I of to, and its values arein n.

Proof. First of all, a continuous function x(t) satisfying (*) satisfies theintegral equation

x(t) =xo+J f(u,x(u))du,

and conversely. Let C(I, B) be the Banach space of all continuous functionsg on I into B endowed with the norm sups 11g11. Consider the mapping 0of C(I, 0) into itself defined by

t : C(l, fl) i) x(t) y(t) = xo + J f (u, x(u)) du.

Since f is Lipschitz in x, there exists k such that, for t E I and a, b in 11,

Ilf(t,a) - f(t,b)II 5 klla - b11-

Thus 0 is locally a contracting mapping. Indeed,

II$(x)--O(y)o <_ It-tolllf(t,x)-f(t,y)II <-kit-tolllx(t)-y(t)IIPick 0 (0 < 6 < a) such that ik < 1, and set J = [to - p, to + S] C 1.Then 4' is a contracting mapping of the Banach space C(J, B) into itself.The fixed point theorem then implies the existence of a unique z E C(J, B)such that $(z) = Z.


0.34. Remark. If f (t, x) is not Lipschitz, we cannot say anything in gen-eral. But if the Banach space is R", then there is a solution of (*), butit may not be unique. In that case there exist a solution greater than theothers, and a solution smaller than the others.

0.35. Example. n = 1, x' = 2 jxJ, x(0) = 0.

0

General solution ........ Greatest solution - - - - Smallest solution

0.36. Dependence on Initial Conditions. The solution of (*) (Theorem0.33) depends on the initial conditions (to, xo). Thus we can write it in thefoam x(t, to, xo).

If we choose (i,i) in a neighbourhood 0 of (to,x0), we can choose r anda small enough so that the closed ball f2j of radius r and center i, andI = [t - a, t + o), satisfy I x fit C V for any (t, i) E 0.

0.37. Theorem. Let f be a continuous function, Lipschitz in x, as in The-orem 0.33. Then there exists a neighbourhood 0 of (to, xo) in R x B such thatthe solution x(t, t, i) of the differential equation

x' = f (t, x), x(t) = z,

exists on [t' -,3, t + 01, 3 > 0 being independent of (t, i). Moreover, (t, t, s)- x(t, t, x) is continuous. If f is C-, this map is C°°.

Exercism and Problems 13

Exercises and Problems

0.38. Problem. Let t x(t) E R be a function defined on an interval ofR. Consider the family of first order differential equations E,\ (A E R):

X, = x2(1 + t2A2) - 1, x(0) = 2.

a) Show that E,\ has a unique maximal CO° solution xa defined on aninterval la = (aa, b.,).

b) Integrate the equation Eo explicitly.c) Let f (t, x) and g(t, x) be two continuous functions on R2 with value in

R, and suppose that f and g are uniformly Lipschitz in x. Let (to, xo)be initial conditions, y(t) the maximal solution of the equation

x' = f (t, x), x(to) = xo

(it exists on I), and z(t) the maximal solution of the equation

x = g(t, x) , x(to) = xo

(it exists on J). If f (t, x) < g(t, x) on R2, prove that z(t) > y(t) forto <t EInJandthat z(t) <y(t) for to>tEInJ.

d) If f (t, x) < g(t, x) on R2, prove that z(t) > y(t) for to < t E I f1 Jand that z(t) < y(t) for to > t E I fl J. Hint. Consider the family offunctions 9,, (t, x) = g(t, x) + .1 with n E N.

e) Show that bA is finite. What is the upper bound of ba?f) Prow that xa > 0 on I. Hint. Argue by contradiction, considering

the equation x' = x2(1 + t2A2).g) Establish that xa(t) < 2 for t < 0. Deduce from the previous question

that as = -oo.

0.39. Problem. Let f (t, x) be a continuous map of I x 1 C R x R" into R",where I=[to,to+a] and S1= B,,, (r)CR"(a>0,r>0,xoER",B,,,, (r)the ball with center xo and radius r). Set M = sup I If (t, x) I I for (t, x) E I x n,and choose a .

a) Consider for t E [to - a, to] the function

yo(t) = xo + f(to,xo)(t - to)and the functions yk ( < k E N) defined by yk(t) = yo(t) for t E[to - a, to] and yk(t) = xo + f t f (s, yk(s - ))ds for t > to. Showthat yk(t) is defined and continuous on [to - a, to + k], and then on[to-a,to+a].

b) Prove that the family yk(t) is equicontinuous on I-that is to say, forany e > 0, there exists q > 0 such that It-$I < q Ilyk(f)-yk(s)II < Ef o r all k > t ands belonging to I. Then apply Ascoli's theorem:


there exists a subsequence {yki} C {yk} which converges uniformlyon I to a function z. Show that z satisfies on I the equation E:

z' = f (t, z), z(to) = xo.

c) From now on n = 1 (y E R and f (t, y) E R). Consider the equationEp (p E N):

Y, = f (t, y) + p y(to) = xo-

Prove that Ep has at least one Cl-solution defined on I. Let p < q betwo integers, zp a solution of Ep and zq a solution of Eq. Prove thatzp(t) > zq(t) for t E I. Deduce that z(t) = limp,, , zp(t) is a solutionof E larger than any other solution of E. z is called the maximalsolution of E.

0.40. Problem. Let t - x(t) E R be a function defined on an open set ofR. Consider the differential equation

(E) X, = A - f (t)ex,

where f is a nonvanishing continuous periodic function on R with period 1,and A is a real parameter.

a) For to E R, show that there exists, in a neighbourhood of to, a uniquedifferentiable solution of (E) such that x(to) = xp, xo being a givenreal number.

b) Verify that if x is a solution of (E), then y = e-z is a solution of

= f(*) y' + AY

c) When A 0, prove that

YAM =ea

1 1 ( ea" f (t + u) du0

is the unique differentiable solution of (*) which is periodic of period1.

d) What can we say about the existence of a periodic solution of (*) inthe case A = 0?

e) When f > 0, deduce from c) that (E) has a periodic solution (ofperiod 1) if and only if A > 0.

f) Show that Aya (t) = f (t). Deduce that (E) has a periodicsolution (of period 1) for all A < 0 if and only if f (t) < 0 for all t.

g) If fo' f (t)dt < 0, establish the existence of e > 0 such that (E) has aperiodic solution (of period 1) for any A E (-e, 0).

0.41. Problem. Let

E={f ECl([O,11)1 f(0)=0and f(1)=1}.

What is the greatest real number m such that

M < I f'(x) - f(x)I dx

kw any f E E? Hint. Consider the function fe-=.

0.42. Problem. Let t -' x(t) E R be a function defined on an open set ofR. Consider the differential equation

(Ea) x'=A+l+t2, x(0)=0, AER.

a) Prove that there exists Ao such that (Ex) has a solution on [0, oo) ifand only if A < Ao. What is the value of AO?

b) When A > Ao, the maximal solution exists on [0, aa). Show that

sinh- < as < sinhar

FA-I.

0.48. Exercise. Let f be a real valued differentiable function at each pointof [a, b] C R, and suppose that f(a) < f'(b). Let It E (f'(a), f'(b)). Provethat there exists zo E (a, b) such that f'(xo) = yo.

0.44. Problem. Let A (n E N) be the positive solutions of tan x = x.

a) Show that°O 1 1?=10'n=1

b) Consider the equation

(E) -y" + b2y = f(x), y(O) = 0, 01) = y(1),with b > 0 and f a continuous function. Is there a solution? Is itunique?

c) Find a function G(x, t) on [0, 1] x [0, 1] such that the solution of (E)is

1) = G(x, t.) f (t)dt.fo

y(x

d) Is G(x, t) continuous? Does it satisfy G(x, t) = G(t, x)?

e) If the equation

-1!"+by=1iy, y(0)=0, y'(l) =y(1),has a non-trivial solution, what can we say about µ?


0.45. Exercise. Consider the differential equation

x' =X 2 + t, x(0) = 0,

where t -s x(t) E R is a function on a neighbourhood of 0 E it

a) Show that a solution exists on (-b, a) with a < 3.b) What can we say about b?

0.46. Exercise. Consider the differential equation y' = x - Then x -'Y(x) E R is a function defined on an interval of R. Prove that it has a uniquesolution on (0, oo) which is positive and which tends to zero when x -+ oo.

0.47. Problem. In this problem the given functions and the solutions aredefined on R, with values in it They are even and periodic of period 2ir. Eowill be the space of continuous functions on R which are even and periodicof p e r i o d 2sr. For k E N, we have Ek = { f E Eo I f E &I. Let C,06 bethe space of the bounded continuous functions on R endowed with the normIIf IIco = sup if I, and P,.(x) the set of the functions in El whose Fourier serieshave vanishing coefficients ak when k > n (PR(x) is a linear combination ofthe functions cos kx with 0 < k < n).

Part I

a) Let h E Eo. Show that for the equation y" + y' cotan x = h(x) tohave a solution in E2 on the open set Cl C R, where tanx 0, it isnecessary and sufficient that h(x) sin x dx. What can we say aboutuniqueness?

b) For p a real number, verify that

F:y -y"+i'cotanx-pyis a map inPP(x).

Let f, E PR(x), and prove that if p > 0 (which is assumed hence-forth), then the equation

(1) y"+y'cotanx-py= fR(x)has a unique solution in E2 (that means that the function in E2satisfies (1) on fl).

c) Let f E El, and denote by fa the partial sum up to order n of theFourier series of f. For each n E N, we consider the solution yn inE 2 of equation (1). Given k p o i n t s x1, .Z, . , xk of 10, ir], show thatthere exists a subsequence {yp} C {y.} which converges at these kpoints.

d) Prove that there exists a subsequence of {y,} which is a Cauchysequence in COD. Deduce that the equation

(2) y"+y'cotanx-py= f(x)

Exercises and Problems 17

has a unique solution in E2.

e) Since f E El, what is the regularity of the solution?Part H

We next study the equation

(3) z" + z' cotan x + h(x) = f (x)e"=,

where h, f belong to Ep and v E R, f 0 0 and v 96 0.

a) Reduce the study of (3) to the study of the equation

(4) y"+y'cotanx+a= f(x)ey,

where a is constant and f E Eu. When f has a constant sign, establishthat for (4) to have a solution in R2, it is necessary that a have thesign of f .

When a = 0, verify that for (4) to have a solution in E2, it isnecessary that f changes sign and that fo f (x) sin x dx > 0.

b) For the rest of Part II we suppose a>0and f(x)>0forall xER.Exhibit two real numbers m and M such that f (x)em < a < f (x)eMfor all x E R. Then consider the sequence of functions defined byinduction as follows: cpo = m and, for k > 0, pk is the solution in E2of the equation

(5) Wk + Vk cotan x - f (x)elk-' -- a - Wk-1,

where p > 0 is a real number. Prove that Wr > 00.c) If p is chosen large enough, establish that the sequence {Wpk} is in-

creasing and bounded by M.

d) Prove that {cpk} is a Cauchy sequence in C.Deduce that equation (4) has a unique solution in E2. What is

its regularity?

Part IIIIn this part we suppose a < 0.

a) Verify that for (4) to have a solution in E2, it is necessary that fbe negative at least somewhere. When f (x) - -2 and a = -2, theequation

(6) y" + y' cotan x + 2ey = 2

has an obvious solution yo. But in fact there exists a one-parameterfamily of solutions yt of (6) in E2, yt being Cr in a neighbourhood oft. Find the equation (7) satisfied by w = (dyt/dt)t--o.

b) Find the solutions of (7) in E2. Let tJi be one of them, i1' 0.


c) Find solutions of (6) in E2 of the form

y = k log [p(1 + E V,)],

where k, la and a are real numbers to be chosen. Find all the solutionsof (6) in F-1-

d) If equation (4) has one or more solutions in E2, let y be one of them.Prove the following identity:

ffIsir2xevix=_(1+a/2)jfev sin 2x dx.

When a = -2, show that equation (4) does not always have a solution,even if the necessary condition found in III a) is satisfied.

Specialists will recognize the Kazdan-Warner condition for theso-called Nirenberg problem (see Aubin [2]).

Chapter 1

DifferentiableManifolds

One begins a new field in mathematics with some definitions, and this courseis no exception. There are many definitions, especially at the beginning. Thesubject of our study is differentiable manifolds. It is necessary to understandwell what a differentiable manifold is.

We give the proof of the theorem on partition of unity, very useful indifferential geometry. This proof needs point-set topology. The reader isassumed to know the definition of a topology and that of a compact set (seeChapter 0). But what is useful throughout the book is differential calculus.One must know what a differentiable mapping is, and the Cauchy Theoremon ordinary differential equations.

This chapter continues with the definition of a submanifold. To provethat a subset of a manifold is a submanifold. using the definition, seems tobe difficult; fortunately we have at our disposal Theorem 1.19, which willbe very useful for applications.

We end the chapter with two basic theorems, Whitney's and Sard's.We give the difficult proof of Whitney's theorem, because it is a beautifulapplication of the knowledge already acquired. The reader may skip theproofs of Whitney's theorem and the theorem on partition of unity.

Basic Definitions

1.1. Definition. A manifold Mn of dimension n is a Hausdorff topologicalspace such that each point P of Mn has a neighbourhood Cl homeomorphicto Rn (or equivalently to an open set of Rn).

19

20 1. Differentiable Manifolds

More generally, we define a Banach manifold: each point has a neigh-bourhood homeomorphic to an open set of a Banach space. Here we willstudy only manifolds of finite dimension.

The notion of dimension makes sense because there is no homeomor-phism of Rn into RP if n 96 p. We do not prove this main result, becausewe will study differentiable manifolds, for which the proof of the notionof dimension is obvious. We will only consider connected manifolds. If amanifold has more than one component, we study one component at a time.

1.2. Proposition. A manifold is locally compact and locally path connec-ted.

By definition, locally compact (resp. locally path connected) means thatevery point has a basis of compact (reap. path connected) neighbourhoods(a family of neighbourhoods is a basis of neighbourhoods at P, if any neigh-bourhood of P contains a neighbourhood of the family). A set E is pathconnected if, given any pair of points P, Q in E, there is an are in E from Pto Q.

An arc of M is the image, by a continuous map, of [0,1] C R into Mn.If I' is an arc in R" and cp the homeomorphism of Definition 1.1, W-1(I') isan arc in Mn. If there exist an are from P to Q and another from Q to T,their union is an are from P to T.

Let P E Mn and let it be a neighbourhood of P homeomorphic to anopen set of Rn, fl i°- R". In this chapter, B,. will be the ball of R" withcenter 0 and radius r. As often, we suppose without loss of generalitythat cp(P) = 0. r = 1, 2, ... , p.... (p E N). form a basis of

Basic Definitions 21

neighbourhoods of P which are compact and path connected.

'p- I

1.3. Proposition. A connected manifold is path connected.

Proof. Let P E M,,, and let W be the set of points Q of Mn for whichthere is an are from P to Q. W is closed. Indeed, let T E W, and U aneighbourhood of T homeomorphic to R". We have U f1 W 0 0; thus thereexist Q E U fl W and an are from P to Q; there is also an arc from Q to T.Hence TEW.

W is open. Indeed, let Q E W; Q has a neighbourhood f2 homeomorphic toR" and 0 C W. Since M" is connected and W 0 0 (P E W), W = M".

1.4. Definition. A local chart on M" is a pair (ft, cp), where fl is an openset of M" and 9 a homeomorphism of i2 onto an open set of R". A collection((T.j, Wi)ie1 of local charts such that UfE1 "4 = M" is called an atlas. Thecoordinates of P E fl related to the local chart (12, co) are the coordinates ofthe point rp(P) in R".

1.5. Definition. An atlas of class Ck (respectively CO', C)? on M, is anatlas for which all changes of charts are Ck (respectively C, C). That isto say, if (fL,, spa), and (f2q, p,3) are two local charts with QC n Qq 0 0, then

22 I. Differentiable Manifolds

the map FAQ o cp,1, called change of charts, of VS(f Halo) onto cp,(i2Q nits)is a diffeomorphism of class CA (respectively C,Cw).

Rn D Vo(0Q n its) % 12« n fts f?. W.(n. n its) c Rn

We consider the following relation of equivalence between atlases of classCk on Mn: two atlases (U;, P+)iEI and (We, *Q)QEA of Class Ck are said to beequivalent if their union is an atlas of class Ck. That is to say that gyp, o * 1isCkon*,,(UtnWQ)when U,nWQ#0.

1.6. Definition. A differentiable manifold of class Ck (respectively. C° orCI) is a manifold together with an equivalence class of Ck atlases (respec-tively, C° or CO).

On a manifold there need not always exist a differentiable atlas (of classCk), but if there exists an atlas of class C', then time are atlases of class C°°(which are Cl-equivalent to it) if the manifold is paracompact. It is possiblenow to talk about differentiable functions Ck on a Cr-differentiable manifoldwhen k < p. A function f on Mn (unless we say otherwise, a function takesits values in R) is Ck-differentiable at P E M. if for a local chart (U, ip)with P E U the function f o V-1 is Ck-differentiable at W(P). We easilyverify that this definition makes sense--the notion of differentiability doesnot depend on the local chart. Indeed, let (i2, 0) be another local chart atP; then f otji-1 = f orp-1ocpoY,-1 is Ck-differentiable at 10(P) since Vo,1,-1

is Ck-differentiable because k < p.

1.7. Remark. We can define complex manifolds M. Consider an atlas oflocal charts (f)j, cpj)iE1, where Bpi is a homeomorphism of li onto an openset in C. If any change of charts Vj o cpi 1 is holomorphic on pi(S2j fl S2j),M is a complex manifold of complex dimension m (n = 2m).

1.8. Example. An open set Il of a differentiable manifold M,, is a differ-entiable manifold. It is endowed with the atlas (Ut, ' ){E, obtained fromthe atlas (UU, j)jE, of m. by setting U1 = U; fl 12 and letting cpj be therestriction of cp, to U1.

The sphere S, is a compact analytic manifold.

Proof. Let us consider the unit sphere S, C Rn+1 centered at 0 E R"+1,with P and T the north and south poles of coordinates zi+1 = ±1,zj = 0 for 1 < i < n in Rn+1 We define the charts (it, c) and (8, vp) asfollows: n = S, \{P}, 8 = &\{T}; for Q E 0, cp(Q) = Q1, the intersectionof the straight line PQ with the hyperplane II of equation zn+1 = 0 ({p isthe stereographic projection of pole P); and for Q E 8, r(Q) = Q2, theintersection of the straight line TQ with H.


Obviously (0, So) and (8, t1') form an atlas A for S,,: 0 U 9 = S and IIis identified with R". What is the class of A?

Let (r, a) be polar coordinates for Ql and (p,w) polar coordinates forQ2. Thus r = OQI, P = OQ2i a = w and rp = OP.OT = 1. On S2 f16, thefinction apoljr-1, which is defined by a = w and p r = 1, is analytic sincep and r are not zero on SZ (10. We can see that, in Cartesian coordinates{x'} for Q1i {y'} for Q2 we have y'/p = x=/r; thus

x'and x' - y`

y1(x)2 'j 1(yj)2

1.9. The real projective space P,,(R) is a compact analytic manifold.

Proof. Recall that Pn(R) = (Rn+1 - {O})/R, where R. is the followingrelation of equivalence: x - 1 in Rn+1- {O} if there exists p E R such thatx=pi.

Let U ; (i = 1, 2, , n + 1) be the set of points of Rn+1 whose Ohcoordinates are not zero.

The open sets Sgt = U,/1 cover P,,(R): L)7+1 SZt = ,,(R). We considerthe atlas A = {f?,, Sot}1<j<n+1, where v is defined as follows.

To the equivalence class t E Q., (x = (x1,2,... , xn+1)), we associateSoi(x) = yt E R" whose coordinates are y* = x?/x' for j < i and y = xj+1/x=f o r j > i, (y:,y2,... ,y -1,1,y ,... ,yi) E ±. The cpi (i = 1,2,... ,n + 1)are bijections; we endow P"(R) with the topology such that the gyp; arehomeomorphisms. What is the class of A?

Partition of Unity 25

To simplify the writing let us consider p o +t If a E thenx = {x'} with x" and, xx"+1 nonzero. For j < n we have yin+1 = z^I /x"+tand p = xi/xn. But yn+I = xn/xn+1 = 1/y;

+ 1ltn+1 = y /y for j < n.This change of coordinates is analytic since yn 36 0.

To prove that Pn(R) is compact, we consider in each Sti a compact Kisuet that P"(R) _ I 1"+'K- . For instance, lets E Pn(R), x = {x'} and x'Obe s u c h that Ix'I Ix - I f o r all 1 < i < n + 1. In fib, we consider Kc,, theset of s such that Ix'I < Ix'-) for all i. In R" the set K defined by Iy'I < 1for 1 < j < n is compact; thus K,. = W-1 (K) is compact.

Partition of Unity

1.10. To study a manifold, we will have to glue together the charts of anatlas, and use partitions of unity. This is why we will suppose the manifolds

compact (the weakest topological hypothesis which implies the existenceof partitions of unity).

A topological space E is paracompact if, to any covering of E by openset f2; (i E I), we can associate a covering, locally finite and thinner, byopen sets 0j. Locally finite means that any point has a neighbourhood Wsuch that Z n W # 0 except for a finite set of indices i. Thinner meansthat 9 C fl; (with this notation, some 0, may be the empty set). Eis countable at infinity if there is a family of compact sets Kq (q E N)

0 0such that KI CK2C ... C Kq CKq+tC ..., the union of the K. being E(E==1Kq)1.11. Theorem. A paracompact manifold is the union of a family of con-nected manifolds which are countable at infinity.

Proof. The manifold is the union of its connected components; let Vn beone of them. Let { flp } pE V be a family of open sets houteomorphic to R"with lip compact. Since V" is paracompact, there is a locally finite cover{0{} thinner than the cover {flp}. Pick 91 96 0 and set K1 = BSI; thenby induction we define Kq = U3EJa e', J. being the set of indices j forwhich 9 f n Kq_I 0. The set Kq is compact, as a finite union of compactsets; indeed we verify below that the set of open sets 0i which have anonempty intersection with a given compact set K is finite. Let Up be anopen neighbourhood of P such that 9; n Up = 0 except for a finite set ofindices i. The family { Up } covers K; thus there exist Up, , Up2,... , Up,which cover K. Now if 9, n K # 0, then 9, n up O 0 for at. least one Up,,and the set of these 0, is finite.


Moreover K._1 C UiEJq 9j =Kq, so W = t 1 Kq is open. But W isclosed too. Indeed, for P E W, let 9j, be one of the open sets of the coveringwhich is a neighbourhood of P. We have Wfl9;" 36 0, thus 0j,, has nonemptyintersection with a compact set Kq, and P E 0;° C K9+1 C W = V, sinceV" is connected and W * 0. Thus it is proved that V" is countable atinfinity.

1.12. Theorem (Partition of unity). On a paracompact differentiablemanifold of class Ck (respectively C°0), there exists a Ck (respectively C°O)partition of unity subordinated to a given covering.

From now on, manifolds are always paracompact; we will mention it nomore. Let {9;bE1 be a covering of a manifold by open sets. A partition ofunity subordinate to the covering {9f} is a family of functions {(x;}jE j withthe following properties:

i) suppa; C 9;.ii) Any point P has a neighbourhood U such that Unsuppoj = 0 except

for a finite set of ai.

iii) 0<a;<1;a-ejaa=1.

Proof of Theorem 1.12. Let us prove the result for a connected compo-nent Af of the manifold. According to Theorem 1.11, M,, is countable atinfinity: M = U,-1 Kp, Kp CK,+I, Kp being compact. Let {0,} be thegiven covering, and for Q E Kp+ 1 \Kp, let us consider a local chart (VQ, rpQ )such that VQl(B3) is included in an open set 0,; and in Kp}2\Kp_1 (here wesuppose that pQ(Q) = 0 E R" and pQ(VQ) = R"), Br C R" is the ball cen-tered at 0 of radius r. Now {,pQ1(B1)}QEM is a covering of K1; thus a finite

subset pQ,(B1), co (B1) of {VQ1(Bi)}QEM covers K1. By induction weQjjexhibit a sequence of points Qk such that the open sets VQ' (Bi) (Q E M)

0

cover Kp+l\ Kp; thus gyp-' (Bl ) , ... , .pqiI

(B1) cover Kp+t \ Kp. We ver-

ify that the open sets Sgt. = (B3) form a locally finite covering thinnerthan 0;. Let us consider the function

7(r) = [j2P(t)(tt}' j(t)d

(t) = e -1T for 1 < t < 2 and p(t) = 0 for t 0 ] 1, 2[. The functionwith p-1(r) is equal to 1 for r < 1 and vanishes for r ? 2. Set 7j (Q) = 7 (kPQ1(Q) I )when VQ1(Q) exists, zero otherwise. The functions a j = ryj ( 1 7i forma partition of unity subordinate to the covering {Oj}. Since 'y(r) is C. ifthe manifold is C' (respectively Ck), so is the partition of unity; see 1.6.

0

Differentiable Mappings 27

Differentiable Mappings

1.18. Definition. A Ck mapping f of a differentiable manifold Wp intoanother Mn is called C'-differentiable (r < k) at P E 9 C Wp if 0 o f o v-1is Cr-differentiable at ip(P), and we define the rank of f at P to be the rankof 1# o f o <p-1 at Sp(P).

Here (9, w) is a local chart of Wp at P, and (11, l') is a local chart ofM,a with f (P) E ft. We easily verify that this definition makes sense, sinceit does not depend on the local charts.

1.14. Remark. In the definition of a differentiable manifold, when we con-sider an atlas Vj)jEr, each function pt is a homeomorphism of 9; ontocpi(9i) C Rn. But now cpj is a diffeomorphism of ej onto V;(9i). Indeed,according to Definition 1.13, to see if joi is a differentiable mapping, we haveto verify that 0 o Bpi o V-1 is differentiable for a local chart (9, gyp) and a localchart (I t ). Choose (9, gyp) = (9;, gyp,) and = Identity; then obviously dpiis differentiable: Id o Vi o 4p,-1 = Id, which is differentiable.

1.15. Definition. A Cr differentiable mapping f of W. into M is an im-mersion if the rank of f is equal top = dim Wp at every point P of Wp. Itis an imbedding if f is an injective immersion such that f is a homeomor-phism of Wp onto f (Wp) endowed with the topology induced from M. ACr differentiable mapping f of Wp onto Mn is a submersion if the rank off is equal tan = dim Mn.

Since rank f < max(p, n), it follows that n > p if f is an immersion, andp >_ n if f is a submersion.

1.16. Examples. W. = R, Mn = R2.

CO t (ii) t (iii) f

P

(v)


In example (iii) the equation of the mapping for t < 0 is x(t) = t,y(t) = 0 and for t > 1 it is x(t) = 0, y(t) = t. In example (iv) theequation of the mapping is, in polar coordinates, 8 = -t, r = i for t > 1.(i) is not an immersion, rank f = 0 at 0. (ii) is an immersion; we haverank f = I everywhere; but it is not an injective immersion. (iii) is aninjective immersion, but it is not an imbedding. Indeed, f ((-e, +e)) for anye > 0 is not an open set in the induced topology. In the induced topology, aneighbourhood of 0 contains open sets of the form f ((-e, +e)) U f ((E , oo)).(iv) is an imbedding. Here there is no difficulty; in the induced topology,a neighbourhood of 0 contains open sets of the form f ((1, oo) ). (v) is animbedding.

1.17. Definition. Let M, M be two Ck differentiable manifolds. M iscalled a covering manifold of M if there exists a differentiable mapping II(called a projection) of M onto M. such that for every P E M:

i) II-1(P) is a discrete space F, and

ii) there exists a neighbourhood 11 of P such that 11' (Q) is diffeomor-phic to flxF.

Each point P E 11-1(P) has a neighbourhood f2' C M such that therestriction II' of II to SY is a diffeomorphism of IT onto Q.

The map II is a 2-sheeted covering if F consists of two points.

Af

!1

Submanifakls 29

Submanifolds

1.18. Definition. A submanifold of dimension p of a differentiable mani-fold Mn is a subset W of Mn such that for any point of W there exists alocal chart (fl, gyp) of Mn, where o(1l) is an open set of the form 8 x V with9 c RP and V c R"-P, such that V(flnW) = 8 x {0}. W is endowed withthe topology induced from Mn.

Thus there exists a system of local coordinates (x 1, , xn) on Sl suchthat Wp is locally defined by the equations xP+1 = xp+2 = ... = xn = 0. WPis endowed with a structure of differentiable manifold induced from Mn. Weconsider for Mn an atlas (fl,, V,);E, of local charts as above. Then (fli, p,),E jis an atlas for W of the same class with ft; = fli n w and Bpi the componenton 8i of the restriction of gyp; to f2,.

Considering this definition, it seems difficult to prove that a subset Wof Mn is a submanifold. Fortunately there is the following theorem, whichis very convenient.

1.19. Theorem. A subset W of M. defined by a set of n - p equationsf, (P) = 0, ... , fn-_p(P) = 0, where fl, , fn-p are C1-functions on Mn, isa differentiable aubmanifold Wp of Mn if the map of Mn into R"-P definedby P (f1(P), ... , fn-p(P)) is of rank n - pat any point P E W.

Proof. Let xo be a point of W C Mn, and (fl, gyp) a local chart of Mnwith xo E fl, (x1, x2, ..., x") the corresponding coordinates. One of thedeterminants of the (n - p) x (n - p) submatrices of the matrix ((8f/9z-'))is nonzero at xe. Without loss of generality, let us suppose that it is the onewhere j =p+ 1,p+2, n.

According to the inverse function theorem, there is a neighbourhood8 C fl of x0 on which as coordinates of P E ewe can take y1 = x1, ... , yP =xP, f+l = fl,... , y" = fn-p. Let aP be the homeomorphism defined on 0by P --- {y1} E R", 0(8 n W) CRP x {O}. Here W is a differentiablesubmanifold of Mn.

1.20. Example. An open set 9 of a manifold M. is a submanifold of di-mension n. Let us consider an atlas (Sli,,p,),EJ for Mn; then (fl,, p,),EJ withfli = fls n e and 0, _ w, /f4 is an atlas for e. gyp; (fl,) is an open set of Rn.

The set Sn(1) of the points x = {x'} E Rn+1 satisfying fl(x)=,'(x')' - 1 = 0 is a submanifold of Rn+l. The rank of the matrix((8f1/8x1)) is 1 on Indeed, the derivatives are 8fi/8xa = 20, andthe matrix is never 0 on Sn (1) since (xi )2 = 1.

The set of matrices T (n, p) with p rows and n columns is a normedvector space. If aj are the components of the matrix M E T(n, p), we set


JIM 11 = sup la-,')- So the bijection of T(n,p) onto ilt"P defined by ((a=)) -{xk} with xi+"(i-1) = a; is a homeomorphism. Here one chart covers themanifold T(n, p).

1.21. The set T(n, p, k) C T(n, p) of matrices of rank k is a submanifold ofT(n, p)

Proof. In order for the rank of M E T(n,p) to be greater than or equal tok. we have to have that one of the determinants DQ of the submatrices k x kof M not zero. The set Ea of the points satisfying Da # 0 is an open setof T(n,p), and their union is an open set of T(n,p); it is a submanifold ofT(n, p) which is of dimension np (an open set of a manifold is a submanifold).

Thus if k = inf (n, p), we have proved that T(n, p, k) is a submanifold ofT(n,p) of dimension np.

If k < inf (n, p), then at a point M E T (n, p, k) at least one of thedeterminants D(, is not zero. Let us suppose Dl 96 0; we have to show(and it is a necessary and sufficient condition) that all the (k + 1) x (k + 1)determinants of the type

1 k k+aa1 ... a1 a1

Da3(M) = al ak ak;k k

k+8 ... ak+Q ak+9

vanish with 1 < a < n - k and 1 < 0 < p - k.The map r : T(n,p) - W"-k)(n-k) defined by

M -i {D11(.M), 1 ) 1 2 ( M ) , . . . , D"-k,v---k(M)}

is of rank (p - k)(n - k) on El. Indeed, the partial derivatives with respecttoxx with A = k+v+n(k+,u -1), 1 : 5 1 <µ<p-k, are

°°3s= Dl if v = a and it =0; 0 otherwise.a

IN

Thus the map 1 is of rank (n - k)(p - k). According to Theorem 1.19,E1 nT(n, p, k) is a submanifold of dimension np-(n-k)(p-k) = k(n+p-k).But as T(n, p, k) C U EQ, according to Definition 1.18 (which is purely local)T (n, p, k) is a submanifold of dimension k(n + p - k) of T(n, p).

The Whitney Theorem

1.22. Theorem. Every parncompact differentiable and connected manifoldM may be imbedded in 1R2"+1


We will prove a weaker theorem: every connected C2 -differentiable man-ifold has an immersion in R2' and an imbedding in R2"+1. The steps of the

proof are Propositions 1.25 and 1.27.In the first lemma we will prove that we can perturb the map f a little

bit, locally in 9, so that the new map is an immersion on 11. With thesecond lemma we prove that if f is already an immersion on a compact K,we can do this perturbation so that the new map is still an immersion onK.

1.23. Lemma. Let f (x) be a C2-differentiable map of an open set Sl C R"into RP (with p > 2n). Then for any c > 0 there exists an n x p matrixA = ((a; )) with Iai I < e for all i and j so that x f (x) + A.x is animmersion.

Proof. If J(x) is the Jacobian matrix of f at x, we want J(x) + A to beof rank n for all x E S1. That is to say, A must be, for any x, different fromthe matrices of the form B - J(x) with B a matrix of rank < n.

The map F of 11 x T(n, p, k) into T(n, p) defined by (x, B) -p B - J(x)is C'. When k < n the dimension of fl x T(n, p, k) is n+k(n+p- k) < n+(n - 1) (p + 1) < np - 1 (by setting k = n - 1, then taking p > 2n).

Therefore when k < n the image of St x T (n, p, k) by F is of zero measurein T(n, p), identified with R"" according to a well known theorem of measuretheory:

The image by a C1-mapping of an open set of R" into R' (n < p) is ofzero measure in R".

Recall that a set A C RP is said to be of zero measure if for anyc > 0, there exists a sequence of balls A. such that A C U°° 1 Bi andi= , vol Bx < E. In particular, if the measure of A is zero, no open set,except the null set, is included in A. Thus the interior of E =F(f1 x Lrk--0 T(n, p, k)) is empty, and we can choose A E T(n, p) not inE as close as we want to the zero matrix.

1.24. Lemma. Let f be a Cl -map of M, into R" (p > n). If the rank of fis equal to n on a compact set K C 91, (11, gyp) a local chart, then there existsrl > 0 such that for any C'-map g satisfying IIJ(g)II < rl on K we have f +gof rank nonK.

Proof. We write J(g) for the Jacobian matrix of g. Let 6(x) be themaximum of the absolute values of the determinants of the n x n submatricesof J(f) at x. 6(x) is positive and continuous on K; thus there exists 6 > 0such that 6(x) > 6 for all x E K. A determinant is a continuous functionof its components; therefore there is an it > 0 such that, if IIAII < +l, thematrix J(f) + A is of rank n on K.


1.25. Proposition. Let f be a Ck-map (2 < k < oo) of the connected andCk-differeentiable manifold Mn into RP (p > 2n) and let be a continuousfunction everywhere positive on Mn. There exists a Ck-immersion g of M,into RP such that II f (P) - g(P)II <'P(P) for all P E M.

Proof. Let (Sti, Oj), j E N, be a sequence of local charts such that the) .

homeomorphic to B3, form a locally finite cover of Mn, like Vj = cps 1(131).For the notations, we refer to the proof of Theorem 1.12. Let us applyLemma 1.23 to f o VI 1 on B3 with a small enough. We set f1(P) = f (P) +y1(P)A1 o cp1(P). Since y1 < 1, y1 = 1 on Y1 and y1 = 0 outside wpi'(B2),it follows that f1 is of rank non V1 and 11h (P) - f(P)II < %'(P)/2, becausewe have chosen a small enough for that.

By induction we define a sequence of Ck-differentiable maps fr of rankn on K, U'=, VI. We set

f,+1(P) = fr(P) +yr+1(P)Ar+1 0 pr+l(P),

the A,.+1 being chosen according to Lemma 1.23 with Sl = 52,.+1 and a smallenough so that Ilfr+1(P) - fr(P)II < 2-r-'j,(P) and so that fr+1 is still ofrank n on Kr nfirt1 (Lemma 1.24).

The covering being locally finite, at a given point P, from some jo on wehave fj (P) = f?+1(P) for j > jo. Thus the limit of the sequence { fj } is a Ck-differentiable map g which is an immersion satisfying II f (P) -g(P)II < ri(P)for allPEMn.

1.26. Theorem. Every connected, paracompact, Ck-differentiable mani-fold has a Ck-immersion in R2n. The image of the manifold may be ina ball Bo C R2n of radius p > 0 as small as one wants.

Proof. We consider the map f of Mn into R2n defined by f (P) = 0 for allP E Mn, and we choose r'(P) = p. According to Proposition 1.25, thereexists an immersion g of Mn into R2n such that g(Mn) C B,,.

1.27. Proposition. Let f be a Ck-differentiable map (2 < k < oo) of theconnected Ck-differentiable manifold Mn into R' (p > 2n + 1) and let rji be acontinuous everywhere positive function on Mn. Then there is an injectiveCk-immersion h of Mn into RP satisfying Il f (P) - h(P) II < i(P) for allPEMn. -

Proof. According to Proposition 1.25, there is an immersion g of Mn intoR2n such that II f (P) - g(P)II < tP(P)/2 for all P E Mn. According to The-orem 1.19, g is locally injective: every point P E Mn has a neighbourhoodfl where g is injective.


Let {S4}, i E N, be a sequence of such compact neighbourhoods whichform a locally finite covering of M (as in Proposition 1.25), f i being ho-meomorphic to B3. Let {bi} be a sequence of points of R2"+1 which we willchoose later. Let us say already that we will choose bi satisfying

Ilbill <- 2-(i+21 inf i(P)PEtl;

and small enough so that the maps hr = g + r 1i=1 biyi are immersionsaccording to Lemma 1.24.

Let Dr C M x M be the set of pairs (P1. P2) for which yr(Pi) 96 ryr(P2).We consider the map Gr of the open set Dr into R2n+1 defined by

Gr : (Pi, P2) ---.hr(P1) - hr(P2)yr(P2) --Mj 1)

Gr is C2-differentiable and the dimension of Dr is 2n; therefore G, (D,) isof zero measure in R2n+i. So we can successively choose b,. not in Gr(Dr)and close enough to 0 E R2n+i, as we said above. The covering {S1 } beinglocally finite, at every point P, from some io on, we have hi(P) = h;+1(P)for i>io.

The limit of the sequence hi is a Ck-differentiable immersion h which isinjective. Indeed, by contradiction, let us suppose that there exist f1 andP2 such that h(P1) = h(P2).

Let jo be an integer such that for j > jo we have 'y, (P1) = y, (P2 )and hj(P1) = h(P1) = h(P2) = hl(P2). Now hr+1(Pi) = hr+i(P2) giveshr(Pi)+bryr(P1) = hr(P2) +brryr(P2). Since br ¢ Gr(Dr), this implies thatyr(Pi) = 7r(Pz) and hr(P1) = hr(P2). By induction, for any i, we havehi(Pi) = h,(2) and' (P1) ='yi(P2)

At least one y. (A) is not zero, say yip (Pi) 0 0. Thus yip (P2) is notzero. So Fl and P2 belong to S, where g = hl is injective, which is incontradiction with h1(Pi) 96 hi (Pi) if Pi # P2.

1.28. Theorem. Every Ck-diferentiable manifold (2 < k < oo) has a Ck-imbedding into R2"+1, and its image is a closed subset of R2"+1.

Proof. According to Proposition 1.27, there is an injective immersion hof M., into R2n+'. In order that h be an imbedding, it is necessary andsufficient that for any compact set K C R2"+n, h-1(K) is compact. Indeed,since h is injective, h will be then a homeomorphism of M" onto h(M").

We will choose f and ay for that. Pick 0 1 and all the components off (P) zero except the first; we choose f1(P) _ F 1 jy j (P). We have thath(P) E K implies IIh(P)lI 5 p = sup }IxII for x E K and IIf(P)II <_ p + 1.Thus P E U1<,< .17), which is compact (y, being equal to 1 on Vl, P f iwhen j>p+1).


The Sard Theorem

1.29. Definition. Let M and W p be two C°° differentiable manifolds ofdimension n and p respectively, and let f be a map of class Ck of M intoWp. The points of M" where rank f < p are called critical points of f .All other points of M are called regular points. A point Q E W. such thatf -1(Q) contains at least one critical point is called a critical value. All otherpoints of WP are called regular values.

When f is a smooth real valued function (WV = R), P is a critical pointoff if all the first derivatives off at P vanish, since (rank f)p = 0. Thegradient at P vanishes. A critical point P is called non-degenemte if andonly if the matrix ((8a f /8x' 8xi)) p is non-singular.

The index of f at P is the number of negative eigenvalues of((a'f/8x'8xi))p. M. Morse proved that any bounded smooth funetlonf : M R can be uniformly approximated by smooth functions whichhave no degenerate critical point (see Milnor 18]).

If n < p, all points of M,, are critical since rank f < inf(n,p) < p, andthe critical values form a set of measure zero. This is obvious according toProposition 0.28.

But the Sard Theorem asserts that this result holds in general if the mapf is Cl: the set of critical values is of measure zero. We are at the beginningof the course (in Chapter 1) and we have only seen some definitions and a fewtheorems, but nevertheless we were able to prove a very important theorem,the Whitney Theorem, which shows that a differentiable manifold, whosedefinition is abstract, is nothing else than a surface of dimension n in RPfor p large enough. And now it is possible to give a second very importanttheorem:

1.30. Theorem (The Sard theorem). Let M" and W. be two connectedC°G differentiable manifolds of dimension n and p respectively, n > p, andlet f be a map of class Ck (k > 1) of M into Wp. If k > it - p + 1, thecritical values form a set of measure zero.

In case the manifolds are only C'', we have Sard's theorem for any p ifand only if r > n.

Proof. Since the manifolds are separable (they are connected), we have onlyto prove the result for a local chart (fl, So) of M and a local chart (0,,0) ofW such that f (Sl) C 9.

Thus the theorem will be proved if the result holds when M is an openset of R" and W = R. Indeed, we will only have to consider the map

0 0 -of into tp(A) C RP.

The Sari Theorem 35

Thus henceforth f is a Ck map with k > max(n - p + 1, 1) of an openset fl C R" into RP, and A is the set of the critical points of f . When n < pwe proved the result in 0.28. Therefore we assume n > p.

First step. The Sard theorem is true for functions (p = 1).As in the proof of Proposition 0.28, we only have to prove the result

when Mn is a unit closed cube C of R". The proof is by induction on thedimension n. Suppose n = 1 and f E C'. Then x E S2 C R" is a criticalpoint off if and only if f(x) = 0. Moreover, as f is C' on the compactset C, for any c > 0 there exists m E N such that Iy - zI < 1/m impliesIf (v) - f'(z)I < E.

Thus Iy - xI < 1/m implies I f'(y)I < e, and according to the mean valuetheorem If (x) - f (y) I < e/m.

Now we proceed as in the proof of Proposition 0.28. We divide C intom intervals of length 1/m. Let J be one of them which has a critical point.We have mess f (J) < s/m. The set A of critical points of f is covered bysome intervals like J. The number of these intervals is at most m, of course.So mess f (A) < me/m = E. Since E is as small as one wants, meas f (A) = 0.

The result is true for n = 1. Now we suppose, by induction, that if 9is an open set in R"-1, and g is a Ci-1 map of 9 into R, then g(B) hasmeasure zero, B being the set of critical points of g. The result is also trueif 8 is a C°O separable manifold, according to the begining of the proof.

Let Ak (1 < k < n) be the set of critical points x of f such that allderivatives of f of order less than or equal to k vanish at x, with x Ak+1

when k < n. If x E An, for any c > 0 there exists m E N such thatIlv - xII S nl/2/m implies that the norm of the differential D"f (y) of ordern at y is less than E. Thus Ily - xII < n1/2/m implies

11f (y) - f(x)II << on-"n"/2.

Then we split the unit cube C into Mn cubes of side 1/m. If K, one of them,has a point z E A,1, we have

mess f (K) < 2Em-"nn/2.

At most m" little cubes have a non-empty intersection with An; thusmess f (An) < 2En"/2. Since c is as small as one wants, mess f (An) = 0.

When x E Ak (k < n), at least one derivative off of order k+ 1 does notvanish at x. Suppose without loss of generality that 01h(x) 0 0 with h somederivative off of order k, h(x) = 0. The function h, which is at least C', hasrank I at x, so in a neighbourhood e of x the set D = {y E 9 I h(y) = 0}is a submanifold of dimension n - 1. Obviously Ak n 0 C D. Let g = f ID;by the induction hypothesis messg(Ak n 0) = 0. The result follows sinceR" is separable.


Second step. Let

Aj = {x E f2 I rank D f (x) = j }, A= U Aj.0<j<p

Since R" is separable, we have only to show that for any j any xo E Ajhas a neighbourhood 0 C R" such that mess f (8 n Aj) = 0. Observe thatU0<1<j Ai is closed for each j. Hence Ao is measurable; thus Al is measurableand by induction any Aj is measurable. By the following lemma, f (0 n Ao)has measure zero.

1.31. Lemma. Let f : 0 - RP be a Ck map (k > n) and let Ao ={x E D I D f (x) = 0}. Then f (A,,) has measure zero in R.

Proof. Let f, (x) be the first coordinate of f (x). We have Aa C B ={x E f ( dfl(x) = 0}; hence f(A,) c f, (13) x RP- 1. By the first step,f1 (B) has measure zero in R", so that fl (B x R'1) has measure zero in RPaccording to Fubini's thearem.

1.32. Remark. When the regularity off is only k = max(n - p + 1,1), wecan prove the result of Lemma 1.31 by using Lemmas 1.33 and 1.34, but theproof is not so easy.

Third part of the proof of Sard's theorem. Let 0 < j < p and xo E Albe such that rank D f (xo) = j. We can suppose without loss of generalitythat ((8ff/8x`))2, (1 < a < j,1 < i < j) does not vanish. Accordingto the inverse function Theorem 0.29, in a neighbourhood of x, {yi} withyi = fi for 1 < i < j and y' = x' for j < i < n is a coordinate system.Set F(y) = f (x), where y = cp(x), V being the diffeomorphism of changeof coordinate systems. We have FF(y) = fi(x) = yi for 1 < i < j. F(y)is a Ck map of a neighbourhood V of yo = ip(xo) E R' into W. Since therank of F at yo is j, DF(yo) is represented by a matrix ((ad)) such thatail = (8FF/8y`)(yo) = b° for 1 < a < j. Since the rank of this matrix is j,we have all = 0 for i > j and l > j. Hence if we consider the map

G : RP-j D V. 3 z (Fj+1(a, z), Fj+2(a, z),..., Fp(a, z)),

where a E Rj and V. = {z E R"-f I (a, z) E V}, then DG(zo) = 0 if(a, zo) = yo. Let Ej be the set of critical points of F, where the rank of F isj. Up to the diffeomorphism of change of coordinate systems, Ej is nothingelse than Aj n V-1(V). By Lemma 1.31, F(VO n Ej) has measure zero inRP-j. Hence mess F(Ej) = 0 according to Fubini's theorem. This implies

meas f (Aj n rp-1(V)) = 0,

and then mess f (Aj) = 0. The proof of Sard's theorem is complete whenk > n. If not, we have two more lemmas to prove.

The Sard Theorem 37

1.33. Lemma. Let f be a function of class Ck on a closed unit cube in Rn,and A the set of critical points of f. Then A = A,, U A with Ao countableand A such that, for any pair (x, y) in A,

1f(x) - f(y)I <- a(Ilx - yID)IIx - yII c,

where a(E)-0 as e--0.

Proof. If n = 1, A is the set of points x where f'(x) = 0. Let A, c Abe the set of isolated points (A' is countable). If x E A - A!, there isa sequence of points xi in A such that xi -+ x. Also, f'(xi) = 0 impliesf"(x) = 0. Let Ao be the set of isolated points x E A - Ao where f"(x) = 0(Ao is countable) and so on until we consider fk(x). Then A. = U1u<k Aois countable, At being the set of isolated points x in A - U1<_i<! A, wherefl(x) =0. We setA=A - Aa. At xEAwe have fi(x)=0(1<j<k).Using the mean value theorem, since y --+ fk(y) is continuous, we get theinequality of the lemma. We can extend the proof in the case n > 1 ( seeSternberg [141).

1.34. Lemma. Let f be a Ck map of a closed unit cube in Rn into RP andlet A be the set of points when f has rank zero. Then f (A) has measurezero if k > n/p.

Proof. We saw that A = Ao U A (Lemma 1.33). f (Ao) has measurezero since A. is countable. For Awe split the unit cube into m" cubes Kof side 1/m (m E N). If x, y in A belong to the same little cube C, thenIly - x1 l < n1/2/m = e. Hence 11f (x) - f (y) II < a(E)Ek, and

mess f (C fl A) < wp_lap(e)Ekp.

Since there are at most Mn cubes K with K n A 9& 0,

measf(A) < wp-1ap(E)Ekp-nn'n/2.

When kp > n, the right hand side is as small as one wants. Hence mess f (A)=0.

1.35. Remarks. When k > max(n - p + 1, 1), we have k _> n/p, sincen - p + 1 > n/p if n > p. The assumption on kin Sard's theorem issharp; Whitney (see Sternberg [141) gave a counterexample in the case whenk < max(n - p + 1, 1).

1.36. Corollary. The set of regular values of a C°° map f of Mn ontoWp is everywhere dense in Wp. If Q is a regular value, then f (Q) is asubmanifold of Mn of dimension n - p.

V = f-1(Q) is defined locally by a set of p functions (f1,f2, ... fp) = f,and we know that rank f = p; so we have to consider a local chart at Q.


Thus, according to Theorem 1.19, V is a submanifold of M,a of dimensionn - p.


1.37. Exercise. C = {z E CJIzI = 1} being the unit circle centered at 0in C, we consider the map p of R into the torus T = C x C defined byR 3 u -+ V(u) = (e2"', e2'"u) E T. What can we say about <p (immersion,injective immersion, imbedding) and about the subset So(R) of V.

1.38. Exercise. On R2, endowed with the coordinate system (x, y), weconsider the function f defined by

f(x,y)=x3+xy+y3+1.a) For which points P E R2 is f -' [f(P)] a submanifold imbedded in R2?

b) Draw the complementary set of this set of points P.

1.39. Exercise. Let M, W be two differentiable manifolds. We suppose Wis compact and M is connected. We consider a differentiable map II of Winto M such that II is locally a diffeomorphism at any point Q E W.

a) Let P E M, and show that II-'(P) is a finite subset of W.b) P r o v e that the cardinality of 11 ' (P) does not depend on P.c) Verify that W is a covering manifold of M.d) Construct a counterexample which will establish that if W is not

compact, W may be not a covering manifold of M. One will chooseM = R.

1.40. Exercise. We consider the map w of R"+', endowed with the coor-dinate system {x'} (i = 1, 2,... , n + I), into R2n+1, whose coordinates are{y° } (a = 1, 2,... , 2n + 1), defined by

rp : {y°} with y' _ xkxlk+l=I+a

for l<a<2n+l, andy2"+1 =711(x')2

a) On which open set of R"+' is So an immersion?

b) From gyp, construct an imbedding of P"(R) into R.

1.41. Exercise. On C'"+' \ {0} consider the equivalence relation R definedby zl - z2 if there exists p E C, p 3A 0, such that zI = pz2. The quotient setof C'"+1 \ {0} by R is called the complex projective space Pm(C). As we didfor the real projective space, we define an atlas (fl , rA) (i = 1, 2, .. , m+ 1)

as follows. Let Ut be the set of points of Cni+1 \ {0} whose ate' complexcoordinate is not zero, and let fl = U,/R. Let z be the equivalence class of


a pointaEUi,aadlet -1, 1,e+1,..., -+')Ez. Vi (i) will be,the point f o f C- of c o m p l e x Coordinates 1 , t + 1 ,

a) We define a subset M of P3(C) by

M = {i E P3(C) I t2t3 = t1t4, (t2)2 = t1S3, (S )2 =ee}

where z = ((1, g2, 3 4) E i. Show that M is a compact submanifoldof dimension 2 of P3(C).

b) Consider the map 4 : P'(C) --+ P3(C) defined by

P1(C)32 -41(2)=iEP1(C)with Z = (rl1,f12) E C2 \ {0} and

z = [(t I)3, (g1)2g2,111(112)2, (172)3] E C° \ {0}.

Prove that -6 is a diffeomorphism of P1 (C) onto M.

1.42. Exercise. Let M,, be a COO compact manifold of dimension n.

a) Exhibit a finite cover of Mn by a family of open sets 9, (i = 1, 2,..., p)homeomorpllic to a ball of R" such that 9, C f2i with each fli homeo-morphic to a ball of R. Prove the existence of C°° functions f, onM" satin ying 0 < f= < 1, supp f; C 1'2; and f; (x) = 1 when x E 9s.

b) Deduce the existence of an imbedding ip of M" into R9 with q =(n + 1)p.

1.48. Eiunrcise. Show that a proper injective immersion is an imbedding.

1.44. Exercise. We identify R4with the set of 2 x 2 matrices.

a) Show that the set M2 of 2 x 2 matrices whose determinant is equalto 1 is a submanifold of R4. What is its dimension?

b) Prove that the tangent space to M2 at 12 = (o °) may be identifiedwith the set of matrices of zero trace.

C) Show that the set Mn of n x n matrices whose determinant is equal to1 may be identified with a submanifold of R"2 . What is its dimension?

d) Characterize the tangent space to Mn at the unit n x n matrix In.

1.45. Exercise. Let E be the set of straight lines in R3.

a) Establish a bijection between E and the quotient set of P2 x R4 byan equivalence relation (P2 the real projective space of dimension 2).

b) We endow E with the structure of a topological space (the finestpossible) such that a is continuous. Is ir open? (A map is open if theimage of any open set is an open set.)

c) Show that E is Hausdorg.


d) On E define a structure of an analytic manifold.Hint. Consider the open sets 0, = ir(fj x Rs), where the open

sets f2; (i = 1, 2, 3) cover P2 as in the course.

e) Using the proof of the previous question, show that E is a vector fiberbundle. Characterize its elements. What is its dimension?

f) Let S be the unit sphere in R3. Prove that the set of tangent straightlines to S is a compact differential submanifold of E. What is itsdimension?

Hint. Apply in $; a theorem of the course. For the compactnesslook at the proof of compactness of P2.

1.46. Exercise. Let Mp and W. be two C°° differentiable manifolds ofdimension p and n respectively, and let f be a CO0 map from M. into W.We say that f is a subirnnersion at x E Mp if there exist a neighbourhoodn C Mp of x, a neighbourhood 0 C W of f (x), a C°°° manifold V, andtwo C°° maps g and h such that g is a submersion of fE into V and h is animmersion of V into Wn with f lo = h o g.

a) Prove that f is a subimmersion if and only if the rank of f is constantin a neighbourhood of x.

b) Shaw that the function Mp ? x --+ r(x) =rank off at x is lowersemicontinuous.

c) Prove that the subset of M. where f is a subimmersion is dense inM.

Solutions to Exercises

Solution to Exercise 1.37.Sp is everywhere of rank 1; thus p is an immersion. If cp(u) = W(v), we

would have u-v E Z in order that e2`" = 0`", but also u-v E xZ in orderthat e2i" = e2". Since 7r is irrational, this is impossible except if u = v.

Therefore cp is an injective immersion. Let P = (e2i", eu"") be a pointof V(R). If cp is an imbedding, there would exist an open set Cl of T whoseintersection with o(R) would be a given connected arc ry of w(R) throughP. We can show that, as close as one wants to P, there are points of V(R)which do not belong to 7. It is well known that for any a > 0, there areintegers n and p such that tpsr - nJ < a. Set v = u + n, e2'"a = e2i'"' andJe2'" - e j = 1e24" -1J = Je*n ) -1J < Jean -11, which is as small asone wants.

Finally, let us prove the well known fact mentioned above. For P E Nwe set xp = inf(prr - n) for n E N, n < par. So xp = par - np, and the set{xp}pEN C 10, 1 (.


Let {xp; } be a subsequence which converges in (0, 11. So we have(P.+1 - Pi) r - (rb,+, - rip,) - 0.

Solution to Exercise 1.38.If (x, y) -- i f (x, y) is everywhere of rank 1 on Wp = f V (P)],(P)], then

Wp is a submanifold imbedded in R2 according to Theorem 1.19. Now8f /8x = 3x2 + y vanishes when y = -3x2, and Of/ft = 3y2 + x is zerowhen x = -3ya. These two equations have for solution (x = 0, y = 0) and(x = - j, y = -g ). Set f1= (0,0) and Q = (-', -' ).Then f(0,0) = landf(-,-3) =1+1/27.

Let us study Wn. WO \ fl is a submanifold. We have to see what happensat Q. The equation of Wn is x3 + xy + y3 = 0. If x is small with respectto y, the equation is xy + y3 = 0. y = 0 is impossible if x:1- 0, but we havean arc where x - -y2. By symmetry we have an arc where y - -x2. f2 isa double point. At f2 there are two arcs, one with tangent y = 0, the otherwith tangent x = 0. There is nothing more in the neighbourhood of Cl, sinceif we suppose x - ay with a 96 0, we find aye = 0, which is impossible. Letus study WQ. Wq \ Q is a submanifold. In a similar way as above, we provethat Q is an isolated point. So WQ is not a submanifold.

Therefore the set of points P for which Wp is a submanifold is A =f -'((-oo,1) U (1,1 + 1/27) U (1 + 1/27, oo)).

Let D be the line of equation x + y = 1/3. f (D) = 1 + 1/27. Thecomplementary set of A is Wn U WQ. That is, {Q} U D and a curve throughCl asymptotic to D. This set is symmetric with respect to the line y = x.


Solution to Exercise 1.39.II-1(P) is compact (a closed subset of a compact set), and II-1(P) is

a set of isolated points since n is locally a diffeomorphism. Thus II-1(P)is a finite subset of W. We will show below that cardII-1(P) is locallyconstant. This will imply that c a r d II-1(P) = Constant, since V is con-nected. L e t P i (i = 1, 2, , m) be the points of II-1(P). There are disjointneighbourhoods f2 of Pi which are homeomorphic to a neighbourhood 9of P. Let {Qk} be a sequence in 9 which converges to P. Obviouslylimsupk, card II-'(Qt) > card II-1(P). Let us prove the equality by con-tradiction. If we do not have equality, there is a sequence {xk} in W suchthat f(xk)=Qkand xkVS2ifor all hand 1.

Since W is compact, a subsequence, noted always {xk}, converges to apoint x of W. x is not a point Pi, and, by the continuity of II, II(x) = P, acontradiction. II-1(9) is diffeomorphic to F x 9, F a set of m points. Fora counterexample take W = ] - oo, 0[ U ] 1, oo[ and II the identity map.

Solution to Exercise 1.40.We have 8y2r+1/&zi = 2x` and ey° f 8x; = 2x1+'-' when a - n < i < a.

We verify that V is of rank n + 1 on R"+1 ` {0}. If x1 0, the first(n + 1) x (n + 1) determinant in Drp is equal to (2x1)"+1 # 0. If x1 = 0 andx2 3& 0, the first row is zero but the (n + 1) x (n + 1) determinant in Dcpwith the n + 1 following rows is equal to (2x2)"+1 76 0, and so on.

Let S be the unit sphere in Ri+1 and let do be the restriction of (pto S,,. We consider the map f of Sa into R20 defined by f°(P) = Co' for1 <a<2n.

On Sn. y2n+1 = 1. Thus rank f = rank iv = n. Let II be the naturalprojection Sn P,, (R). We verify easilrv that if f (P) = f (Q), then Q = Por Q is the point opposite to P on S . Thus we can write f = f o 11, with fan injective immersion of P(R) into R2". Since f is obviously proper (PR(R)is compact), f is a homeomorphism of P. (1t) on its image.

Chapter 2

Tangent Space

In this chapter we introduce many basic notions. First we will study tangentvectors, then differential forms. We will give two different definitions of atangent vector at a point P E M, (they are dual to each other). Then,of course, we will prove that the definitions are equivalent. M,, is a C'differentiable manifold (r > 1) and (Q, W) a local chart at P; {s{} are thecorresponding coordinates.

We dune the tangent space Tp(M), P E M. It is the set of the tangentvectors X at P, which has a natural vector space structure of dimension n.The union of all tangent spaces is the tangent bundle T(M). We will showthat if r > 1, T(M) carries a structure of differentiable manifold of classC", which is a vector fiber bundle (T(M), ir, M) of fiber R" and basis M(T(M) B X -i ,r(X) = P E M if X E Tp(M)). Likewise we define thecotangent bundle T*(M). A vector field on M is a differentiable map f ofM into T (M) such that 7r o C is the identity. Thus a vector field X on M isa mapping that assigns to each point P E M a vector X(P) of Tp(M), anassignment which satisfies some regularity condition.

Likewise we define differential p-forms, exterior differential p-forms, ... .

The notions of linear tangent mapping and linear cotangent map-ping (4)*)p associated to a differentiable map 4 of one differentiable manifoldinto another are very important.

The linear cotangent mapping V allows us to transport differentiablep-forms in the direction opposite to that of the map

This chapter continues with the definition of the bracket [X, Y] of twovector fields X and Y.

43

2. Tangent Space

We will define the exterior product, the inner product and the exteriordifferential on the direct sum of exterior differential forms; our definition isan extension to exterior differential forms of the usual differential of differ-entiable functions. We proceed with the study of orientable manifolds andof manifolds with boundary, and conclude with Stokes' formula.

Tangent Vector

2.1. Definition. Consider differentiable maps 7i of a neighbourhood of0 E R into Mn such that -t(O) = P. Let (it, jo) be a local chart at P.

We say that ryl - % if V o -n and V o rya have the same differential atzero. We verify that this definition makes sense (it does not depend on thelocal chart). It is an equivalence relation R. A tangent vector X at P toM,, is an equivalence class for R.

2.2. Definition. Let us consider a differentiable real-valued function f de-fined on a neighbourhood 0 of P E Q. We say that f is flat at P if d(f oW'1)is zero at W(P).

This definition makes sense; it does not depend on the local chart. If(S2, 3) is another local chart at P, then, on n fl fl,

d(f o'P-1) = d(f o ca ) o d(V o cp 1 }.

A tangent vector at P E M, is a map X : f --- X (f) E ]R definedon the set of the differentiable functions in a neighbourhood of P, where Xsatisfies the following conditions:

a) IfA,pER, then X(Af+ug)=AX(f)+AX(g).b) X(f)=0iffisflat atP.

It follows from a) and b) that

Tangent Vector 45

c) X(f9) = f(P)X(9)+9(P)X(f)Indeed,

X(f9) = X{(f -f(P)+f(P)](9-9(P)+9(P)]},X(f9) = X V - f (P))(9 - 9(P))] + f (P)X (9) + g(P)X(f)

since X(1) = 0 (the constant function 1 is flat).Now d((figi) o w-'],(p) = d((fi o'-')(g1 o V-1)],(p) = 0 if fi and 91

are zero at P. Thus (f - f (P))(9 - 9(P)) is flat at P, and c) follows.

2.3. Definition. The tangent space Tp(M) at P E M is the set of tangentvectors at P.

Using Definition 2.2, let us show that the tangent space of Definition 2.3has a natural vector space structure of dimension n. We set

(X + Y)(f) = X(f) + Y(f) and (AX)(f) = AX(f).

With this sum and this product, Tp(M) is a vector space. And now let usexhibit a basis. {x'} being the coordinate system corresponding to (f2, gyp),we define the vector (8/8x')p by

(0x;)p(f)_I

-9-T L(p)

The vectors (a/ax')p (1 < i < n) are independent since (a/ax')p(x3) _J,', and they form a basis. Indeed, as f - E 1(a f /ax') px' is flat at p,

X(f)

X(xt) a') P] (f).ax

The X' = X(x) are the components of X in the basis (a/ax')p. Observethat the expression of X (f) contains only the first derivatives of f.

2.4. Proposition. The two definitions of a tangent vector are equivalent.

46 2. Tangent Space

Let y(t) be a map in the equivalence class 7 (7(0) = P), and f a real-valued function in a neighbourhood of P.

Considering the map X : f -- (8(f o ry)/8t]t__o, we define a map 'Pof the set of tangent vectors (Definition 2.1) to the set of tangent vectors(Definition 2.2), ' : X. Indeed, since

d(f o7) = d(f ov-1 ocpo7) = d(fif -y1 - -t2 we have

ra(fat71)lc=o = (a(J)lc=o'

because by definition [d(v o yl)]t=o = [d(cp o 72)]t-o. Moreover, X is atangent vector (Definition 2.2): (a) is obvious, and if f is flat at P, than(8(f o -y)/&)t--o = 0 since (d(f o cp-1) pi = 0.

Let us show now that ' : ry --- X is one-to-one and onto. Let Xbe a tangent vector (Definition 2.2), X = E 1 X'(8/8ci)p. Consider themap ry: (-e, e) ? t -p y(t) E Mn, the point whose coordinates are {tX{}(we suppose that p(P) = 0 E ft"). Then

(8(f 0'Y)1 _ c 8(f o V-') 8(tXi) _ X(f}.at J t_o r 8x 8t

So 'P is onto. Moreover, if yl is not equivalent to y2i then [d(+p o y1)] t=6 #[d(+p o y2)]t=o, and it is possible to exhibit a function f such that

[d(f o -tl)]t= * [d(f o -2)]t=0.

2.5. Definition. The tangent bundle T(M) is UpEM Tp(M). If 7 (M) de-notes the dual space of Tp(M), the cotangent bundle T*(M) isUpEM Tp(M). If r > 1, we will show that T(M) carries a structure ofdifferentiable manifold of class C". Likewise for Tp* (M).

Linear Tangent Mapping

2.6. Definition. Let 4) be a differentiable map of Mn into Wp (two differ-entiable manifolds). Let P E M,,, and set Q = 4(P). The map d induces alinear map (4*)p of the tangent bundle Tp(M) into TQ(W) defined by

[(4)*)pX1(f) = X Y o fl,here X E Tp(M), ((P+)pX E TQ(W) and f is a differentiable function in aneighbourhood 9 of Q. We call the linear tangent mapping of $ atP.

To define a vector of TQ(W) (Definition 2.2), we must specify how itacts on differentiable functions defined in a neighbourhood of Q. Obviously,f - pX ] (f) is linear. Moreover, if f is flat at Q. then f o t, which is

Linear Tangent Mapping 47

differentiable in a neighbourhood of P, is flat at P, and we have X (f o4) = 0.So ((*)p is a linear map of Tp(M) into TQ(R').

(O*)p is nothing else than (d4i)p. Indeed, consider a local chart at Pwith coordinates {x*} and a local chart at Q with coordinates {y°}. 4P is de-fined in a neighbourhood of P by p real-valued functions 4'°(x1, x2, ... , x"),a = 1, 2, , p. Using intrinsic notations to simplify, we get

X(fo0)=d(fo-O)poX=(df)o(d4')poX=(dj)o($*)pX.Indeed, {X1} being the components of X in the basis {(8/8x`)p}, the com-ponents of Y = (41*) pX are

n °Y° I '''' X'

8x*

in the basis {(8/8y°)Q}. When we use intrinsic notation, we do not specifythe local charts. In the coordinate systems {x`} and {y°}, the equalityabove shows that (df)p = ((09*°/8x'))p = (4 )p. When we do not specifythe point P, we write L. instead of (4)p.

2.7. Definition. Linear cotangent mapping ($*)p. Let P E Mn andQ = 4'(P). By duality, we define the linear cotangent mapping (V) p ofTQ(W) into 7p(M) as follows:

7Q(W) 9 w , (4)p(w) E TA(M),(($*)p(w), X) _ (w, (4*)p(X)) for allX E Tp(M).

In case w = df we saw (Definition 2.6) that d(f o P) p o X = (df) o ($) pX.Thus

(4*)p(df) = d(f o ')p2.8. Proposition. 'l o ib* = (if o 4P)*.

Let V be a third differentiable manifold and V a differentiable mappingof W into V. If f is a differentiable function in a neighbourhood of T(Q)

48 2. Thngent Space

and X E Tp(M), then

['I'.(4 X)](f) = ['T (X)](f o'I') = X(f oI o$) = 01 o'P)*(X)](f)If $ is a diffeomorphism, we infer that 4,* is bijective and (-6-1) _('F.)-1.

2.9. Example. The tangent vector -2 to a differentiable curve -y(t) of M,,(y is a differentiable map of (a, b) C R into Mn). Let to E (a, b). By defini-tion ( )i is the tangent vector at 7(to) defined by (_2)t = [N(d)lt., (1)being the unit vector on R. For a differentiable function f in a neighbour-hood of ry(to), we have

dry (f) - d(f ° 7) = tim f [7(to + h)] - f [7(to)]h--.o hdt) w dt ) to

Vector Bundles

2.10. Proposition. The tangent bundle T(M) has a structure of differen-tiable manifold of class C''-1, if Mn is a differentiable manifold of class C''with r> 1.

Let (U, p) be a local chart on Mn and P E U. If {zt} are the coordinatesof Q E U and {e } the coordinates of p(Q) E Rn, then z' _ i' for 1 < i < n.This is the equality of two real numbers. If we consider the equality of twofunctions, we must write a' = ` o V. We have

V W.- ( 0 )PIndeed, (f being a differentiable function in ((a neighbourhood of Sp(P) in R',

I la) P1 1 P(f ° gyp) = \Lf ).{P)

=(')

(P)(f).

Thus ,p* is a bijection of T(U) onto V(U) x R" C R.

Let (U0, be an atlas for Mn. The set of ]T(UB), isan atlas for T(V). Let us show that this atlas is of class Cr-1. Sup-pose U. n Us # 0. On Sp.(U. n U8) set 0 = vo o cp;'; then we have8=(Q, X) = [e(Q), (dO)Q(X)], where Q E p0(Ua n Us) and X E TQ(RR).Thus (dO)Q is of class C''' 1.

2.11. Definition. A differentiable manifold E is a vector fiber bundle offiber the vector space F if there exist a differentiable manifold M (calledthe basis) and a differentiable map II of E an M such that, for all P E M,II'1(P) = Ep is isomorphic to F and there exist a neighbourhood U of Pin M and a diffeomorphism p of U x F onto R-I(U) whose restriction toeach Ep is linear, p satisfying 11 o p(P, z) = P for all z E F.

The Bracket [X, 11 49

2.12. Proposition. The tangent bundle T(M) is a vector bundle of fiberIIn.

M is the basis. If X E T(M) and X E Tp(M) for a unique point P E M,then the map n is x -+ II(X) = P. Thus II-1(P) = Tp(M), which is avector space of dimension n: F = R". If (U, rp) is a local chart at P, wesaw that ip is a diffeommorphism of T(U) = II-1(U) onto rp(U) x Itn. So we

can choose p = V. 1 o (V, Id), and we know that (rp- 1) p is linear. Moreover,p(P, z) E Tp(M), and thus ]a o p(P, z) = P, for all z E Rn.

2.13. Definition. Likewise we can consider the fiber bundles T*(M),A" r(M),77(M):

T*(M) = U 17(M),PEM

P

A(M),A7M= UPEM

where A'77(M) is the space of skew-symmetric p-forms on Tp(M), and

U 07 p(M) Tp(M),PEM

where 0 7P(M) 4 Tp(M) is the space of tensors of type (r, s), r timescovariant, s times contravarlant, on Tp(M).

2.14. Definition. A section of a vector fiber bundle (E, II, M) is a differ-entiable map { of M into E such that II o e = identity.

A vector field is a section of T(M).An (r, s)-tensor field is a section of 7, (T(M)).An exterior differential p-form is a section of AP 7' (M). In a local chart

an exterior differential p-form

17 = L aj,...jdx" A dxh A ... A dTJP,1j,<j2<...<jp<n

where a1 ...jp are real valued functions./FP(M) will denote the vector space of exterior differential p-forms.

r(M) will denote the vector space of vector fields.7 (M) will denote the vector space of (r, s)-tensor fields.

The Bradwt (X, Y]

2.15. Definition. The bracket [X, Y] of two vector fields X and Y is thevector field defined by

[X,Y](f) = X[Y(f)] - Y[X(f)],

50 2. Tangent Space

where f is a C2 function on M. Let us show that, in fact, the definitionis valid for CI functions and that we have defined above a new vector field[X, Y]. Using intrinsic notations,

[X,Y](f) = d[Y(f )].X - d[X(f )].Y = d[df.Y].X - d[df.X].Y= (d2f ) (X, Y) + df.dY.X - (d2f)(Y,X) - df.dX.Y= df[dY.X - dX.YJ = [dY.X - dX.Y](f ),

since d2 f is a symmetric bilinear form. From this expression, it is clear that[X, Y] satisfies conditions (a) and (b) of Definition 2.2. If the vector fieldsX and Y are C', then [X, Y] is a Cr-1 vector field. In a coordinate system{x' } corresponding to a local chart (S2, gyp),

2.16. Definition. Lie algebra. The map (X. Y) -+ [X, Y] is bilinear andantisymmetric.

a) [X, Y] = -[Y,X], and [X, Y] satisfies the Jacobi identity.

13) [X, [Y, Z]] + [Z, [X, Y]] + [Y, [Z, X]] = o.

A real vector space L, endowed with a bilinear map L x L into L satisfyinga) and 3), is called a Lie algebra. So the set of CO° vector fields is a Liealgebra. A straightforward computation proves,3), and a) is obvious.

2.17. Definition. Projectable vector field. Let M and W be two differen-tiable manifolds and W a differentiable map of M into W. We have definedthe linear tangent mapping, but in most cases it does not allow us to asso-ciate to a vector field X on M a vector field on W. There are two reasonsfor this. First, 4r (M) may be not all W; second, if %P is not injective, we mayhave at some points of W several tangent vectors images by +Yt of vectorsof the vector field X.

This is why we say that a vector field X is projectable by AY, if for allQ E W(M) and each P E M such that P(P) = Q, we haveindependent of i. P(M) is assumed to be W.

If 41 is a diffeomorphism, any vector field X on M is projectable by 41.

Let Y be a vector field on W. The vector fields X and Y are said to becompatible by 41 if for all P E M we have Y(T(P)).

2.18. Proposition. Let X1, Y1 and X2, Y2 be two pairs of vectors fieldscompatible by 40. Then (XI, X21 and [Y1,Y2] are compatible by 4 (see 2.17for the notations). Consider P E M and Q = 41(P), and let g be a C2

The Bracket [X, Y] 51

function in a neighbourhood of Q. Then

[Y1,Y2](g) = Y1{Y2(9)] _ Y2[Yl(9)] = X1[X2(9o'I')] _ X2[Xi(9o'I')]= [X1,X2](9o W).

Thus *.([X1,X2]) = ['1',(X1), `l',(X2)]-

In the proof above there are some subtle points. When we write for thedefinition of the linear tangent mapping Y2(g) = X2(9 o 41), we understandthe equality of two real numbers. But here they are functions. So we mustwrite

[1`2(9)] 0'1' = X2 (9 o ').

2.19. Definition. A differentiable manifold M of dimension n is paral-leliaable if on V there are n vector fields X1, X2i - , X,,, such that, at anypoint P E M. {Xl(P),X2(P),. , X"(P)} is a basis for Tp(M).

2.20. Definition. The product manifold M x W of two differentiable man-ifolds M. and Wp is a differentiable manifold of dimension n + p, defined bythe atlas (U, X 0j, (apt, O?)) ((i, j) E I x J) on the topological product spaceM X W, (Uj, pt)jEt being an atlas on M" and (0j, IPf)jEJ an atlas on Wp.

2.21. Proposition. A manifold M" is parallelisable if and only if its ton-gent bundle T(M) is trivial-that is to say, difeomorphic to M x R". thedfeomorphisrn p being linear on each fiber and satisfying R o p` 1(P, z) = P(see 2.12).

Let { e1, - - - , be a basis of R" and -b the diffeomorphism of T(M)onto M x R" if T(M) is trivial.

Set X; (P) = $-1(P, ei) for all P E M. and i = 1, 2,- - , n. Thenthe set{X{(P)}, i = 1, 2, - , n, is a basis of Tp(M), as otherwise a linearcombination e 0 of et would be such that -6 -1(P, e) = 0. That isimpossible, since 0-1 is a diffeomorphism and b-'(P,0) = 0. So M" isparallelisable.

Conversely, if there exist n v e c t o r fields Xi (i = 1 , 2, , n) such thatfor all P E M. the set {XX(P) } is a basis of Tp(M). let its consider the map9 of M x R" into T(M) defined by

9`P,"ate{/

.'X;(P),

where the At are n real numbers.9 is a differentiable bijection, linear for fixed P. If we prove that the

rank of 0 is 2n, then 0 is a di$eomorphism, fi = 0-1 and T(M) is trivial. Let

52 2. Tangent Space

{z'} be a coordinate system in a neighbourhood of P. {x'} and {aV} forma coordinate system in a neighbourhood of (P, R") in M x W". Moreover,

49

Xi(x) _ Xi (x)k-1

and

(gy)p = \(A) (XX(P)))Here (Id) is the n x n identity matrix, (0) is the n x n zero matrix, (A) isan n x n matrix, and the n x n matrix (X (P)), with components Xjk(P),is invertible. Thus the rank of (DO)p is 2n.

Exterior Differential

2.22. Definition. The algebra of exterior differential forms A(M).A(M) = (DP"-fl AP(M), with an exterior product defined below, is an

algebra. For simplicity we will say only "diferential form? instead of "ex-terior differential form" when no confusion is possible. We suppose M is aCOO manifold. The algebras A°(M) = C""- (M) and AP(M) for p > 0 are de-fined in 2.14. Given q E Aq(M) and E AP(M), we define qAf E AP'-'(M),the exterior product of in and t;, by

(n A C)(X1,... , Xp+q)

I E e(a)n(Xv(1), ... , X0(q)X(Xc(1+q), ... , XQ(P+q)),p'q' VE'P

where X1, , X,+q are p + q vector fields and the sum is over the set Pof permutations a, e(a) being the signature of a. The exterior product isassociative and anticommutative: l; A q = (-1)Pgq A l;.

We also define the inner product i(X)77 of a differential form q E Aq(M)(1 < q< n) by a vector field X. i(X)v7 is a differential (q - 1)-form definedas follows: If X; (i = 1, 2, ... , q -1) are q -1 vector fields , then

[i(X)17](X1,X2,... ,Xq-1) = FAX, X1,X2,... ,Xq-1)

We verify that if t E AP(M), then

i(X)(q At) _ [i(X)q] A1: + (-1)qq A [i(X).J and i(X)[i(X)q] = 0.

2.23. Proposition. Let 0 be a differentiable map of M,a into WP. To anydifferential q-form q on W, we can associate a differential q-form 0*q EAq(M), the inverse image of q by 4b, defined by

(4*q)p(X1, X2, ... , Xq) = fi(n) ('4X 1, ... ,44Xq),

Exterior Differential 53

X1, - , Xq being q vectors of Tp(M). For a function f on W (f E A°(W ),we sets*f =fo4.

Here there is no difficulty such as in Definition 2.17. Indeed if P E M,then Q = $(P) is unique and (4*ri)p = (,DP)r7i. We verify that

2.24. Definition. Exterior differential.To q E Aq(M), we associate the exterior differential form dri E Aq+l(M)

defined byq+1

rA

dr1(X1, ... , Xq+1) = E(-1)i-'Xi[r1(X1, .. , Xi, ... , Xq+l)]i=1

+,(_1)i+j]7([Xi,Xj],X1,... ,X;,... ,Xj,... ,X4+1)i<j

where Xi, . , Xq+i are vector fields. A caret over a term means that thisterm is omitted. According to this definition,

f EA°(M)df(X)=X(f),gEA'(M) di7(X,Y)=X[-n(Y))-Y[0(X)]-n([X,Y])

Let us show that dq is indeed a differential form. It is obvious thatdr7(X, Y) = -dq(Y, X) and that dri(Xi + X2, Y) = dq(Xj, Y) + dn(X2, Y).

But we have to prove the C°(M)-linearity: for any f E C°°(M),dri(fX,Y) = fd>7(X,Y)

According to the expression of dry,

di (fX,Y) = fX[r!(Y)] -Y[r!(fX)] - rl([fX,Y])= fX[t(Y)] -Y[fri(X)] - rl(f[X,Yj -Y(f)X),

since V X, X, YJ = dY.(fX) - d(f X ).Y = f [X, Y] - (cf.Y)X.As Yffr1(X)] = fY[rf(X)] +rl(X)Y(f), we have proved the announced

result for a differential 1-form.

For q > 1 the proof is similar, but we have a simpler expression of theexterior differential, from which it is obvious that dq is an exterior differential(q + 1)-form.

2.25. Local expression of the exterior differential. Let (i2, gyp) be a lo-cal chart, x1, ... , x" the corresponding coordinates, {8/8x`} (i = 1, - , n)the n vector fields of the natural basis, and (d-O} the dual basis.

The differential q-form n is written

37 _ E aj,...jq(x)d2J1A...Adxj-,jl <ja<yq

54 2. Tangent space

where aj,...jq are differentiable real-valued functions.

Since [8/ax=, 88x7] = 0, by definition

a a 1an (

q+1 ^

ll

k=1

5D; [77

X-J"-, ...

q+1%k-l aaj,...jk...jg},

k=1arjk

Thus, we easily verify that

dr7 = da j, .. yq A dx't A ... A dx'q.j j <j2 <... <jq

From this expression we see that

d(ii A ) = drl A t + (-1)1r) A

and then that d2 = 0. On differential forms, we will speak of the differentialinstead of the exterior differential.

2.26. Proposition. Let M and W be differentiable manifolds and 0 a dif-ferentiable Snap of M into W. For any exterior differential form rl E A(W),

d(4*r)) = 0*(dr!)

We saw this result on the functions f E A°(W):

d(f o -6) = -*(df) (see Definition 2.7).

For a differential 1-form ri = d f E Al (W) the proposition is trivially truesince d2 = 0. Moreover (Proposition 2.23), P(r) AC) = (*(rl) A 4b*(t;).

Choose local charts on M and W, {x'} being the coordinate system inthe considered chart on W. We have (see 2.25)

drt= dajl...jq Adx''

Thus

*(drl) = E (4'*daj,...jq) A 4 dx't A ... A -t*dxjqi t <i2 <jq

= E d(aj,...jq o 4) A d(xi' o 6) A A d(x'v o ).j l <j2 <... <jq


On the other hand,

'Vii = E (aj,...j4 o 4')d(xl' o ID) A ... A d(xj4 0 4'),

and according to 2.25

d(Vt1) _ E d(aj,...3, o A d(x" o $) A ... A d(ziQ o 4)).J1 «s<.. <jv

Orlentable Manifolds

2.27. Definition. A differentiable manifold is said to be orientable if thereexists an atlas (in the equivalence class) all of whose changes of charts havepositive Jacobian.

Given two charts of the atlas, (52, gyp) and (0, 0), with fin8 0, denote by{s'} the coordinates corresponding to (0, ip) and by {y°} those correspond-ing to (9, o). In n n 9, let A; = ay°/Ox' and Bay = 8xj/W. According tothe definition, the Jacobian matrix A = ((Ai')) E GL(R')+, the subgroupof GL(R) consisting of those matrices A for which det A = JAI > 0.

2.28. Theorem. A differentiable manifold M is orientable if and only ifthere exists a differential n form that is everywhere nonvanishing.

Suppose M. is orientable. Let be a locally finite atlas, all ofwhose changes of charts have positive Jacobian, and {a,} a partition of unitysubordinate to the covering {R}. Let x,, xf, , xi be the coordinates oni2{, and consider the differential n-form w, = a,(x)dz' A dx, A .. A dx?.

Let us verify that w = L_,Cl wi is nowhere zero. A given point P belongsonly to a finite number of f2,; let f21, 522, ... , ft,,, be these f4-.

Write all wi(P) in the same coordinate system {xi}:m 1w(P)=1a1(P)+

(I iI a,(P)Jdx;Adx A A dx;.

l f=2 l P

w(P) does not vanish, since each term in the bracket is nonnegative andsome of them are strictly positive. Recall that a, (P) = 1, a, (P) > 0and 84/8I>0.

Conversely, let w be a nonvauishing differential n-form, and A =(t2;, Vi)jer an atlas such that all the i2, are connected. From A we willconstruct an atlas all of whose changes of charts are positive. On ft, thereexists a nonvanishing function fi such that w(x) = f; (x)dxi A dxi A .. A dx;.Since w does not vanish and since 52, is connected, f, has a fixed sign. If f{is positive, we keep the chart (52,, gyp;). In that case we set c3; = Bpi. Other-wise, whenever fj is negative, we consider yaj, the composition of fpj with

56 2. Tangent Space

the transformation (x1, x2, x") -i (-xl, x2, 'X") of R". So from Awe construct an atlas A = (f-4, c )iEI

Now if f, is equivalent to f, but in the chart (f4-, Bpi), then f, = fi ifVi = Vii, fj = -fJ otherwise. So fj and f, are positive. At X E fki nf2j, denoting by Al Cthe determinant of the Jacobian of pj o Bpi 1, we havefjJAI = fi. Thus JAI > 0 and all changes of charts of A have positiveJacobian.

2.29. Definition. Let M be a connected orientable manifold. On the setof nonvanishing differential n-forms, consider the following equivalence re-lation: wl - w2 if there exists f > 0 such that wl = fw2. There are twoequivalence classes. Choosing one of them defines an orientation of M; thenM is called oriented. There are two possible orientations of an orientableconnected manifold.

Some examples of orientable manifolds are the sphere, the cylinder, thetorus, real projective spaces of odd dimension, the tangent bundle of anymanifold, and complex manifolds.

Some examples of nonorientable manifolds are the Mobius band, theKlein bottle, and real projective spaces of even dimension (see 1.9).

We can see the Mobius band in R3.(1,

a rectangle ABCD in R2:[-1, 1) X ] - E, E[, c > 0, A = (1, E), B = (1, -E), C = (-1, -E), D = (-1, e),and identify the segment AB with CD. If we do this by identifying A withD and B with C, we have a cylinder. If instead we identify A with C andB with D, we get a Mobius band.

Let us consider the atlas A with two charts, (fZl,col), P2, IM, where

f1i =]-4, 4[ x cR2,Vl = identity,

f12=]1,1( X ]-E,E[U [-1,-2[ XP2 =

/

identity on ] 2, 1 [ x

,p2(x, b) _ (x + 2, -y) on [-1, -' [ x - E, E[,


(x, y are coordinates on R2). Then

i21f1S22-]-41[ x ]-E,E[U]2,d[ x ]-E,E[CR2.

Let (xl, yl) and (x2r y2) be the considered coordinate systems on f1l andrespectively.

On I", I [ x E, e [ the change of coordinates is x2 = X1, y2 = yl, andon ] - 4, [ x ] - E, E[ it is x2 = xl + 2, y2 = -yl. On the first open set thechange of coordinate chart has positive Jacobian, but on the second it hasnegative Jacobian.

In the proof of Theorem 2.28 we saw that, if a manifold is orientable,from an atlas A we can construct an atlas A all of whose changes of chartsare positive only by the eventual change of coordinate x2 -p -x2. Here ifwe do that, we always have one positive Jacobian and one negative Jacobianon the two open sets of fl, fl 51,. Thus the Mobius band is not orientable.

The Klein bottle is the compact version of the Mobius band. It is in R°.Consider a cylinder with end circles C1 and C2. Let ABCD be four points(in that order) on C1, and A'B'C'D' four points (again, in order) on C2.We identify C1 and G. We get a torus if we identify A with A', B with B',C with C', and D with D. On the other hand, we get a Klein bottle if weidentify A with A', B with D', C with C', and D with B.

The cylinder is a two-sheeted covering manifold of the Mobius band.The torus is a two-sheeted covering manifold of the Klein bottle.

2.30. Theorem. If M is nonorientable, M has a two-sheeted ori.entablecovering manrold M. If M is simply connected, then M is orientable.

For the proof see Narasimhan [10].M is said to be simply connected if any closed curve in M may be reduced

to one point by a continuous deformation. More precisely,

2.31. Definition. Let C be a circle, f and g two continuous (respectivelydifferentiable) maps of C into a manifold M. Let f (C) and g(C) be closed(respectively differentiable) curves of M. f (C) is said to be homotopic tog(C) if there exists a continuous map F(s, t) of C x [0,1] into M such thatF(x, 0) = f (x) and F(x,1) = g(x). If f and g are C', we require thatF(x, t) be C' in x on C x [0,1].

2.32. Definition. The manifold M is said to be simply connected if anyclosed curve f (C) in M is homotopic to one point; that is, there existsF(x, t) as above with g(C) reduced to one point. f (C) homotopic to g(C) isan equivalence relation. The equivalence classes are called homotopy classes.

2.33. Definition. Let M be a differentiable orientable manifold. We de-fine the integral of w, a differential n-form with compact support, as follows:

58 2. Tangent Space

Let (fZ, A)iEI be an atlas compatible with the orientation chosen, and{ai}iEi a partition of unity subordinate to the covering {Sli};EI. On Sti, wis equal to fi (x)dx, A dx? A dx; . By definition,

JM_ J (ai(x)fi(x)] o Bpi

1dx' n dx2 n ... n dxn.iEI -0'(a)

One may verify that the definition makes sense. The integral does notdepend on the partition of unity, and the sum is finite.

Manifolds with Boundary

2.34. Let E be the half-space xl > 0 of R", XI being the first coordinateof Rn. Consider E C R" with the induced topology. We identify the hyper-plane n of R", x1 = 0, with R"-1. Letting fl and 8 be two open sets of E,and ' : 0 0 a homeomorphism, it is possible to prove that the restrictionof tp to St n II is a homeomorphism of fl n II onto 0 n II. So the boundaryII of the manifold with boundary E is preserved by homeomorphism. E isthe standard manifold with boundary, as R" is the standard manifold.

LI

IP

2.35. Definition. A separated topological space M" is a manifold withboundary if each point of M" has a neighbourhood homeomorphic to anopen set of E.

The points of Mn which have a neighbourhood homeomorphic to R" arecalled interior points. They form the interior of Mn. The other points arecalled boundary points. We denote the set of boundary points by W. andcall it the boundary of M.

As in 1.6, we define a Ck-differentiable manifold with boundary. Bydefinition, a function is Ck-differentiable on E if it is the restriction to E ofa Ck-differentiable function on R".

2.36. Theorem. Let M. be a (Ck-differentiable) manifold with boundary.If 8M is not empty, then aM is a (C"-differentiable) manifold of dimensionn - 1, without boundary: O(OM) _ 0.

Manifolds with Boundary 59

Proof. If Q E 8M, there exists a neighbourhood n of Q homeomorphicby W to an open set 9 C . The restriction Yf of Sp to ft = ii fl 8Mis a homieomorphism of the neighbourhood 12 of Q E OM onto an openset 8 = e n n of Rn-1. Thus OM is a manifold (without boundary) ofdimension n - 1 (Definition 1.1). If M is Ck-differentiable, let (S2i, {p,);EIbe a Ck-atlas. Clearly, (57,, cp;);El turns out to be a Ck-atlas for OM.

2.37. Theorem. If Mn is a Ck-differentiable oriented manifold with bound-ary, then 8M is orientable. An orientation of Mn induces a natutl omen-tation of OM.

Proof. Let (12 j, SO3 )jE! be an admissible atlas with the orientation of Mn,and (S2 j, oj) jE I the corresponding atlas of OM, as above. Let i : OM Mbe the canonical imbedding of 8M into M. We identify Q with i(Q), andX E TQ(OM) with 1.(X) E TQ(M). Given Q E OM, pick el E TQ(M),el V TQ(&M), el being oriented to the outside, namely, ei(f) > 0 for alldifferentiable functioos on a neighbourhood of Q which satisfy f < 0 in Mnand f (Q) = 0. We choose a basis {e2, e3, , en} of TQ(8M) such that thebasis {el, e2, eg, , en} of TQ(N) belongs to the positive orientation givenon Mn. Then {e2, e3, , en} is a positive basis for Tq(8M).

This procedure defines a canonical orientation on OM, as one can see.

2.38. Stokes' Formula. Let M be a Ck-differentiable oriented compactmanifold with boundary, and w a differential (n -1)-form on Mn; then

1I dw

= JamW,

,u aM

where OM is oriented according to rthe previow theorem. For convenience

we have written Jam w instead of J i*w (for the definition of i* see 2.78M 8A/

and 2.25).

Proof. Let (12{, cp,);EJ be a finite atlas compatible with the orientation ofMn; such an atlas exists, because Mn is compact. Set e, = w(Sl;). Consider{a;}, a Ck-partition of unity subordinate to {1'4}. By definition,

d(a,w).f dw = Ej41 iEj ;

Thus we have only to prove that

Jd(a;w) = j(o.w).

60 2. Tangent Space

we recall that f4 = (Zi fl am and we have set 6i = w,(SZj) = O1 fl fI. In((Z;, gyp;) we have

na,w=Efj(x)dx1A...AddjA...Ade,

j=1

where the fj(x) are CA-differentiable functions with compact support in-cluded in dxj means this term is missing. Now,

d(aiw) = dfi(x)Adx'Adx2A...AdxjA...ndx"

j=1

[(_1p_10)1ds1A2A...hdx

by 2.25. According to Fubini's theorem,

J6d(aaw) = jf1(x)dx2 A dx3 A ... A Can

j*(Cw),

where j is the inclusion R E. Indeed,

a a aj*(aiw) (5121... i a;w (j. ax fi(x),

and we identify j*a/axk with a/axk. Observe that

W1)*w An-1(lI)

So, we have (gyp 1)* o i1' = j* o 60-T.


2.39. Exercise. Let M and W be two differentiable manifolds, and f adiffeomorphism of M onto W. Let P E M, and set Q = f (P). ConsiderY E TQ(W) and ,'y E AQ(W), q > 1. Express f*i(Y)%r in terms of ry = f*j.

2.40. Exercise. Let (Z be a bounded connected open set of R3 such thatSt is a differentiable manifold with boundary. On R3, endowed with anorthonormal coordinate system (x1, x2, x3), we consider a vector field X ofcomponents Xi(x):

3

(x) = (x)axi.i=1


Is the boundary 80 orientable? At P E 852, let v (P) be the normalunit vector of 80 oriented to the outside of Q.

Prove the well-known formula

18X dE = fan v . X do,ax Jest

where dE is the volume element on R3, do, the area element on 8f2, andv . X the scalar product of v and Y. Hint. Proceed as follows:

a) Find a differential 2-form w on R3 such that

s ;

dw = dx' A dx2 A dx3.;

c=1

b) Let (u, v) be a coordinate system on a neighbourhood 0 C OIL of P,orthonormal at P and such that (v ,, a) is a positive basis inTp(R3). We set

3 349

U a andava

= v ax;.

Compute w(&, 8) in terms of the components of and .

c) Use Stokes' formula.

2.41. Exercise. Consider on R3, endowed with a coordinate system(x, y, z), the following three vector fields:

X = 1(1-x2y2-zz)ax+(-El-z) +(xz+y) z

Y = (xy+z) +2(1-x2+y2-z2)ay+(yz-x)1,

Z = (xz - y) + (yz+x)19 +2(1-x2-y2+z2)z.

a) Verify that the three vectors X(m), Y(m), Z(m) form an orthogonalbasis of R3 at each point m E R3 (the components of the Euclideanmetric are in the coordinate system ej = 6,J).

b) Compute the three Lie brackets [X, Y], [Y, Z], [Z, X] and express themin the basis (X, Y, Z).

c) Let 0 = R3 - {0} and consider the map cp of t2 into fl, (x, y, z)(u, v, w), defined by

xU

(x2 + y2 + z2)'

62 2. Tangent Space

w=(x2 + y2 + x2).

The restrictions of X, Y, Z to fl are still denoted X, Y, Z. Verify thatV is a diffeomorphism, and compute 9#X, cp*Y and p.Z.

d) Using an atlas with two charts on S3, deduce from c) the existenceof three vector fields on S3 forming a basis of the tangent space ateach point of S3. Notice that p is the diffeomorphism of change ofcharts for an atlas with two charts on S3 constructed by stereographicprojection.

2.42. Exercise. Let W be the set of real 3 x 3 matrices whose determinantis equal to 1.

a) Exhibit a differentiable manifold structure on W. Show that W is ahypersurface of R", and specify n.

b) Identify to a set of matrices the tangent space Tg(W) of W at 1, theidentity matrix.

2.43. Problem. In R3, we consider a compact and connected differentiablesubmanifold M of dimension 2.

a) Show that 11 = R3\M has at most two connected components.b) We admit that $I has at least two connected components. Prove that

one of them is bounded. We call it W.c) Construct on M a continuous field v(P) of unit vectors in R3 such that

v(P) is orthogonal to Tp(M) at each point P E M. The norm of thevectors comes from the scalar product (.,.) defined by the Euclideanmetric on R3 endowed with a coordinate system {xi}, i = 1, 2, 3.

d) Deduce that M is orientable.e) On R3, consider the differential 3-form w = dx' A dx2 A dx3 and a

vector field X. Verify that3 8Xk

d[i(X )w] = (divX)w, where divX =8xk

k=1

f) Prove that

VV = (x2

+ y2 +z2),

z

(divX)w = JM ,)j*[i(v)wJ1

for an orientation on M. Here j is the inclusion M C W.g) Let f be a C2 function defined on R3. We suppose that f satisfies

i8;;f =O on W.If f I M = 0, or if 8 f = 0 on M, prove that f is constant on W.


h) When M is the get of zeros of a C1 function on R3 such that R3 9 x -f (x) is of rank 1 at any point x E M, show that Cl has at least twoconnected components.

2.44. Problem. In this problem R" is endowed with the Euclidean metricE, {x1} is a coordinate system denoted (x, y, z) on R3 , and S1 is the setof the points of R" satisfying a1(x')Z = 1. T is the inclusion Sn_1 c R".

a) On R3 - {O}, consider the differential 2-form

w=(xdyAdz-ydxAdz+zdxAdy)(x2+y2+z2)-.

Verify thatrw is closed.

b) Compute J **&.,. For lack of a better method, one can use spherical

coordinates.

c) Is w homologous to zero on R3 - {0}? (See 5.18 for the definition.)

d) On R" - {0}, consider the differential 1-formn n

I

Compute *a (the adjoint of a, see 5.16 for the definition; here t712..n= 1) and prove that *a is closed.

e) What is the value of fs +Il*(*a)? Is *a homologous to zero onR"-{0}?

2.45. Exercise. Let 0(n) be the set of the n x n matrices M such thattMM = I, the identity matrix.

a) Show that 0(n) is a manifold. What is its dimension? Hint. If youwant, first consider the problem in a neighbourhood of I.

b) Identify Tj(O(n)) with a set S of n x n matrices.c) Show that if A E S and M E 0(n), then MA E TM(O(n)). Verify

that the map Is: (M, A) ---+ (M, MA) is a diffeomorphism of 0(n) xS onto T(O(n)).

d) Deduce from this result that 0(n) is parallelisable.

2.46. Problem. In this problem w is a C°° differential 1-form on Rn+1which is not zero at 0 E Rn+1

a) Show that if there exist two C°° functions f and g, defined on aneighbourhood V of 0 E Rn+1 with f (0) 0, such that w = f dgon V, then there exists a differential 1-form 9 in a neighbourhood of0 E R"+1 such that dw = 0 A w.

64 2. Tangent Space

b) Exhibit an expression of 0 if w = yzdx + xzdy + dz, (x, y, z) beinga coordinate system on R3. Show that we can choose f = e-11'.Compute g.

c) Ifw=dz - ydx - dy,do f andgexist?d) Next, put w in the form w = fdg. In the general case, show that the

problem may be reduced to the case where, in a neighbourhood of 0,

w=dz-A,(x,z)dx'.i=1

Here z E R, x = (xl, X2'.. , x") E R", and the Ai are Coo functions.e) a = (a', a2, , a") E R" and c E R being given, consider the differ-

ential equation

dz"

dt= Ai(at,z)a', z(0) = c.

Show that there exists a C°° map F : (t, a, v) -' F(t, a, v) of I x W xJ into R with I, W, J neighbourhoods respectively of 0 E R, 0 E R"and c E R, such that

OF "_ Ai (at, F)ai, F(0, a, c) = c.i=1

Verify that F(t, a, v) = F(1, at, v) if one of these terms exists.f) Prove that u and v defined by u = x and F(1, u, v) = z form a

coordinate system in a neighbourhood of (0, c) E R" x R. Hint. Showthat (8F/8v) (t, a, v) does not vanish in a neighbourhood of (0, 0, c).In this chart w = 1 Pi (u, v)dui + B(u, v)dv. Show that

Pi(at, v)a' = 0.i=1

g) LetWbethemapofRxR"xRinto R"xRdefined by1@ (t, a, v) _ (ta, v) = (u, v).

Compute 41*w. Denote by Ri(t, a, v) the coefficient of dai in 4' w.h) We suppose that dw = 0 Aw, 0 being a differential 1-form in a neigh-

bourhood of 0 E R"}1. Show that 8Ri/8t = HRi, with H(t, a, v) thecoefficient of dt in IIY*B.

Deduce from this result that there exists a neighbourhood of0 E Rn+1 where w = Bdv.

2.47. Exercise. Let M be a C°° differentiable manifold of dimension n.Consider its cotangent bundle T*(M), and denote by H the canonical pro-jection T*(M) -i M.


a) (9, gyp) being a local chart on M and {C'} the corresponding coordinatesystem, on II-1(9) consider the coordinate system (x', x2, ... , xn,y1, y2, . , yn) with x' _ V o II and y' the components of the 1-formin the basis {dl;'} (i = 1, 2, , n). Let a E 7= (M), and define alinear form on TQ (7" (M)) by

T.(T*(M)) D u -p (II*u, a) = (u, W or).

Show that a - I o defines a differential 1-form w on M. What isits expression?

n timesb) Compute n = dw A dw A - . A dw. Deduce from the result that T* (M)

is orientable.

2.48. Exercise. Prove that for any C°° differentiable manifold M the tan-gent space T(M) is orientable.

2.49. Problem.

a) Describe an atlas of the projective space P2(R) with three local charts(f2i,Oi), i = 1,2,3.

b) Compute the changes of charts, and deduce that P2 is not orientable.

c) Consider the cylinder H = [-1,11 x C, where C is the circle quotientof R by the equivalence relation in R: 9 - 9 + 2kx (k E Z). Verifythat the Mobius band M may be identified with the quotient of Hby the equivalence relation in H: (t, 9) - (-t, 9 + x) . Prove that theboundary of M is diffeomorphic to a circle C : C 3 0 - ' 0(9) E 8M.

d) Let (r, w) be a polar coordinate system on the disk

D={xER2IIxI< 1}.Show that P2(R) may be identified with the quotient of D U M bythe equivalence relation: (1, 0) E D is equivalent to ¢(9) E M.

2.50. Problem. On a C°° differentiable manifold M of dimension 2n, wesuppose that there exists a closed 2-form fZ of rank 2n. That is to say that

n timesf2 A A .. A 76 0 everywhere on M.

a) To a C°° vector field X E r(m) we associate the 1-form wX = i(X)S2.Verify that wX is closed if and only if CXSl = 0.

b) Show that the map h : X - wx is an isomorphism of r(m) on/l' (M)

c) a and Q being two 1-forms, we set (a, Q) = h([X,,, X,6]) with X,, =h-1(a) and X0 = h-1(13). If a is closed, prove that Cx013 = (a,#).Deduce that (a,,S) is homologous to zero if a and j3 are closed.

66 2. Tangent Space

d) Let f, g be two CO° functions on M, and set

(f, 9) = fl(Xdq, XdI).

Show that (d f, dg) = d(f, g). Deduce that f is constant along theintegral curves of Xd, if and only if g is constant along the integralcurves of X.

e) Assume that locally there exists a coordinate system (x1 , ... , x",y1, ... , yn) such that 12 =

1 dx' A V. Compute the local ex-pression of (f, g) in this coordinate system.

2.51. Exercise.

a) Let f be a differentiable map of Rn into R of maximal rank every-where. Show that f-1(0) is an orientable manifold.

b) Prove that the manifold that is the product of two C°° differentiablemanifolds M and W is orientable if and only if M and W are ori-entable.

2.52. Exercise. Let w be a differential 1-form on a C°° differentiable man-n times

ifold M of dimension 2n + 1. When wA dw A dw A . . A is never zero onM, we say that w is a C form.

a) Verify that wo = dx2i+1 + E1 x2d-ldz&2i is a C-form on withcoordinates {x?} (j = 1, 2, ... , 2n + 1).

b) Let (6k, ^-)LEK be an atlas and -y a differential 1-form such that oneach Sk we have y = fk'pkwp, with fk a differentiable function whichdoes not vanish. Show that 7 is a C-form.

c) Conversely, 7 being a C-form, find an atlas (0k,''k)kEK such that7 = on each Bk. For simplicity, do the proof when n = 3.

Solutions to Exercises and Problems

Solution to Exercise 2.39.f*i(Y)7 is a (q - I)-form on M. Let Xi (1 < i < q -1) be q -1 vectors

in Tp(M). We have

[f*i(y)i](Xl, X2, ... , Xq-1) = i(Y)7(Yi, ... , Yq-i) if Y = f.Xi,i(Y)?'(Y1,... ,Yq-1) = i(Y,Yi,Y2,... ,Yq-1) = f"7(X,Xl,X2,... ,Xq-i)

Thus f*i(Y). = i(X)7 with X = f,,-'Y.Solution to Exercise 2.40.0 C W is orientable; thus 811 is orieltabie.


a) Set w = X1 dx& A dx3 + X2dx3 A dx1 + X3dx1 A dx2, and

3

dw =1: ;X'dxl A dx2 A dx3.i=1

b) w(J, &) = Xl(u2v3 - u3v2) + X2(u3v1 - v3u1) + X3(t'v2 - u2v1).c) Stokes' formula gives

i*w,Iona =1 on

where i is the inclusion a52 Q. Let us compute i*w at P:

(8 a)duAdv=X. v do.

Indeed, du A dv is the area element and

v = (u2v3 - u3v2)8

+ (u3v1 - v3u1) + (4t1v2 - u2v1)8

.ax ay az

We can verify that v is orthogonal to and to , and that

Il v ll =1Solution to Exercise 2.41.

a) fransposing x y -+ z -p x, we obtain X --i Y -+ Z -. X.Thus if we prove that X does not vanish and that X is orthogonalto Y, the three vectors form an orthogonal basis of R3 at each pointm E R3. X does not vanish. Indeed, in that case z = xy, y = -xzand 1 + x2 = y2 + z2; thus z = -x2z and y = -x2y. We would havey = z = 0, which is impossible since y2 + z2 > 1. We verify that thescalar product (X, Y) = 0.

b) A straighforward computation yields [X, YJ = -2Z Thus [Y, Z) =-2X and [Z, X] = -2Y.

c) On 0, , o jP = Identity; thus W is invertible. It is a diffeomorphismsince V and io-1 are differentiable. Using the fact that cp* = Dip, acomputation leads to

rp,X=-Z(1+u2-v2-w2) a -(w+uv) -(uw - v) .

We obtain tp*Y and o*Z by transposing.

d) We saw in 1.8 that V is the diffeomorphism of change of charts for anatlas with two charts on S3 constructed by stereographic projectionfrom the north and south poles P and P, respectively.

rp*X extends into a vector field U on R3, and V,X at zero is equalto Likewise for p*Y and V,Z in V and W.

68 2. Tangent Space

Thus the vector field on S3 equal to X in one chart (on 53 - {P}for instance) extends by U in the other chart.

Solution to Exercise 2.42.a) The set T(3,3) of real 3 x 3 matrices M is in bijection with R9. If

M = ((aii) ), 1 < i, j < 3, and {xk } is a coordinate system on R9,the bijection +Il : M _ x = {xk} may be defined by xk = a j withk = 3(i - 1) + j. W is the subset of T(3,3) such that det IMO =1.

The map $: T(3,3) --i detIMI of R9 in R is of rank 1 on W.Indeed, OW /Bail = Mi3, the minor of ai f in det IM 1. Now all minorscannot vanish on W, since det I.MI = Ls1 a1jMlj = 1 on W.

Thus according to Theorem 1.19, W is a submanifold of dimension8inR9.

b) Let t -- W(t) be a map of a neighbourhood of zero in R into W suchthat V(O) = I.

If W(t) _ ((aj,(t))), then ail(0) = 6, 3, the Kronecker symbol. Writ-ing det 0 leads to t a'a(0) = 0, and the tangent vectorto the curve t -+ v(t) at t = 0 is the matrix ((a'1(0))). Thus the tan-gent space Tr (W) may be identified with the set of real 3 x 3 matricesof zero trace.

Solution to Problem 2.43.a) At P E M, there is a local chart (V, V) of R3 (cp(V) a ball in R3)

such that cp(V n M) is a disk D of R2. V n Sl has two connectedcomponents 01 and 02. 01 is included in a connected component Wof i2. Then 8W n v = v-1(D) is an open set of M.

Thus 8W n m is an open set in M, and it is nonempty. But it isalso compact (OW is closed and included in M, which is compact).Hence 8W = M, since M is connected. At the most there are twoconnected components: one contains Bl and the other contains 92.


b) M is compact, so M is included in a ball B. R3\B is connected; thusit is included in a connected component of f), while W is included inB.

c) At each point P, we denote by 92 the open set which belongs to theconnected component of it which is not bounded. We choose the unitvector v(P) orthogonal to Tp(M) oriented to 92. It is well definedand continuous.

d) On each local chart on M (whose coordinates are X1,1;2) we arrangethat (v, 8/81, 8/82) is a positive basis in R3 (this means replace 11

by -Cl, if necessary).Thus we exhibit an atlas for M, all of whose changes of charts

have positive Jacobian.

e) We have

i(X )w = X ldx2 A dx3 - X2dx1 A dx3 + X 3dx1 A dx2,3

49X kd[i(X)w]

= E w.k=1

f) According to Stokes' formula, for the canonical orientation on M (see2.37) we have

fW div Xw = fMIi(X)w],

[i(X)w](e1, e2) = w(X, el, e2) = w((X, v)v, el, e2)(X, v)[i(v)w](el, e2),

where el = 8/8t1 and e2 = 8/82. Thus

fdivX1J= JMN)1g)

We choose X = fV f (on R", the gradient V f of f is the vectorV f = X18, f 8/8x' ), and get

3 3 3

f Eai(fb4f)w= f (Of)2W+ff B fW.W W i=1 W f=1

Thus

JM

3

f [i(v)w] = f (af)2w - 0Wi=1

according to the result above and the hypothesis 18;; f = 0 in W.If f6 f = 0 on M, we find that I V f I = 0 in W. Thus f is

constant.

70 2. hngent Space

h) The proof is by contradiction. Suppose Il is connected, eo that theexists an are y starting from 01 which goes to 02 without cuttingacross M (M n y = 0). Now as IVfI(P) # 0 at each point P of M,we have for instance f (x) < f (P) = 0 for x E 01 and f (z) > f (P) = 0for x E 02. Thus f I.y has the value f (P) somewhere, and y cuts acrossM.

Solution to Problem 2.44.

a) We have

dw=3(z2+y2+z2)-hdxAdyAdz

- 3(x2 + y2 + z2)-z(x2 + y2 + z2)dx A 4 Adz = 0,

and w is closed.

b) Let (0,,p) be spherical coordinates,

x = coo V Cos 0,y = coo W sin 8,z = sin W.

Thus

dx = -(sin <p cos 84 + cos W sin Ode),dy = cos cp c o s 8d8 - sin rp sin 0dcp,dz = cos cp4,

*w = [cosg co c os2 B + cos3 <p sin2 8

+ sin2 W(cos2 B cos W + cos cp sin2 B)]dB A dye

Cos Vd9 A d(p.

If v' is the unit exterior normal vector at a point of S2, the basis(v 5S, 4) is positive. Thus

rj 4c*w = / cos 04P f d8 = 49r.

c) w is not homologous to zero on l3 - {0}. Otherwise we would havew = dy and qf*w = d'k*y. Thus according to Stokes' formula wewould have fS2 % w = 0.

d) (*a)23..,, = at, (*a)ls-..n = -a^2r ak (k meansthat k is omitted), since the metric is the Euclidean metric £.c, = b(the Kronecker symbol).


Let us compute d(*a):

d(*a) _ E(-1)k-1 a (*a)12...&...ndX1 A ... A dxnk=1

ndx' A ... A dxn

8Ck

Since a; = 2'[j-=1(X02]-j' we havenn F r

[11

= (xj)2J - n(xr)2I'E(xf)2]

Thus we verify `that

`l

n Oaka-rk -0.E

k=1

e) Integrate aldx2 A - - A dxn on the half sphere where x1 > 0, withx2, X3'... , xn as coordinates:

Al= aidx2A...AdxnJ_, n{s' >0}

j1_E(x*)2dx2...dx,En- t=2

where Bn-i is the unit ball in Rn-'. We have dx2 A ... A dxn =dz' dx", since (v , 8/8x2, , 8/8sn) is a positive basis if v isthe unit exterior normal vector at a point of Sn-l n {x1 > 01. SetP = Z7-2(XI)2; we find that

jAl = n-2 Pn-2 p2dp,

where 0n-2 is the volume of Sn-2, the unit sphere in RI-I. ThusAl = j vol(BB). When we integrate. dx2 A ... A dxn = -dx2 dxnon the half sphere x1 < 0, we find the same value, since then x1 =-. Thus

aldx2A...Adxn=vol(Bn)gn-1

and

'Y* (*a) = n volBB = on_i.

*a cannot be homologous to zero, since otherwise we would havefS.-, q`*(*a) = 0.

72 2. Tangent Space

There are two alternative short proofs without computation. Let

then

i=1

J '@*(*a) = J 'I`*(*7) = J d(*y) = nvolB" = an-1.n-1(1) n-i(1) (1)

Or, if we consider the unit normal vector v = x'8 at x E S,1i_1, then*ry = i(v)dx1 Adx2 A. Adx" and W*(*ry) =area of Sn-1(1) = Qn-1-

Solution to Exercise 2.45.a) We saw that the set T (n, n) of n x n matrices M is in bijection with

Rn' . If M = ((a, j)) and {xk } is a coordinate system in Rn', thebijection maybe defined by xk = a,j with k = n(i - 1) + j. ThenM E O(n) if and only if

n n

Bj = j(a,1)2 = 1 and Bjk = > aijaik = Oi=1 i=1

for all jandailk>j.There are n n+1 equations. If we prove that the map

r : M - (B1, ... , B., B12, ... , B1n, B23, ... , B(n-1)n)

is of rank n n+l on 0(n), Theorem 1.19 will imply that 0(n) is asubmanifold of Rn' of dimension n n-1

We have13Bj = 2a,

aBjk - OBjk =j,

aik, a,j.

At I we obtainaB

-aB;k - aBjk = 1.

aajj 2' aajk - 1' Oakj

The other derivatives vanish.If we suitably arrange the Bj and Bjk, there is in Dr a diagonal

matrix with 2 and 1 on the diagonal. Thus 1' is of rank 24R at I.Since I' is CO°, r is of rank n n+1 in a neighbourhood of I in 0(n).So this neighbourhood is a submanifold of Rn' of dimension 21V-11

Now consider the question in a neighbourhood V of A E 0(n).The map of V into T(n, n) defined by M -- p tAM is a homeomor-phism of V onto a neighbourhood of I. Thus V fl O(n) is a sub-manifold of T(n, n) of dimension As A is arbitrary, 0(n) is a

n-1submanifold of T(n, n) of dimension !!n-1-

,Y= 2y = [

n

n (xi)2]n/2

Or;


We can give an alternative proof. Let W be the map of GL(R')into the set E of symmetric matrices, defined by co : M -'MM.GL(Rn) is an open set of T(n, n), thus a submanifold, and it is easyto see that E is also a submanifold of T(n, n). We have (DSp)M(H) ='MH + tHM. The rank of cp at M E O(n) is equal to 24a, thedimension of E, since (Dcp)M is surjective. Indeed, let B E E; thenJMB satisfies (D'p)M('MB) = 2tMMB + LBtMM = B. ThusO(n) = 9-1(1) is a submanifold of GL(R") of dimension n n-1

= n2 - 14U, since dimGL(Rn) = n2.

b) Let t - M(t) be a differentiable curve in O(n) such that M(O) = I.Set M(t) = ((aij(t))). aij(0) = S;, the Kronecker symbol. We have

(dB(O)a o

2a(0) and akj(O) +a,k(0)

Since Bj(t) and Bjk(t) are constants, it follows that ajj(0) = 0and akj(0) + ajk(0) = 0 for all j and k 96 j. Thus dM(t)/dt is atangent vector of O(n) at I. This vector may be identified with theantisynunetric matrix ((a;,(0))). Thus Tj(O(n)) may be identifiedwith the set S of antisymmetric n x n matrices.

c) Let u -- M(u) be a differentiable curve in O(n) such that M(O) =M. Then u -' M(u) = tMM(u) is a differentiable curve in O(n)through I, M(u) = MM(u), and dM/du = MdM/du. The resultfollows.

We saw just above that * is bijective. Moreover 1Y is differen-tiable. %-1 is the map of T(O(n)) onto O(n) x S : (M, B) -+(M, tMB), and this map is differentiable.

d) T(O(n)) is trivial since it is diffeomorphic to O(n) x Rn s ' . Indeed,

S may be identified to R"('2-1), since S is defined by n n+i linearequations in T(n,n) identified to Rn2. Moreover, W-1 is linear oneach fiber. Thus T(O(n)) is parallelisable.

Solution to Problem 2.46.a) We have dw =4f A dg = d log if I A w = dO n w, with O= log l f l in a

neighbourhood where f does not vanish.

b) We have dw = dz A (ydx + xdy). With f = e-'Y,

d log f A w = -(xdy + ydx) A [(ydx + xdy)z + dz]= dz A (ydz + xdy).

We have indeed dw = d log a-n A w. So

dg = exd(yzdx + xzdy + dz) = d(ze`y),

74 Z. Tangent Space

and we can choose g = zen.c) If f and g exist, we can write dw = B A w. Now here dw = dx A dy

does not have a term with dz like 0 A w with B not zero. So f and gdo not exist.

d) Since w(O) 96 0, there is a coefficient of the 1-form w which does notvanish on a neighbourhood V of 0 E Rs+1: w = B(x, z)dz + withB(0, 0) 36 0. Set w = w/B(x, z). If we can write w = f dg, thenw = fdz with f = B(x, z) f, and w has a simpler expression.

e) According to Cauchy's theorem, F(t, a, v) exists, and depends differ-entially on t and on the parameter (a, c).

Consider b = Aa with A E R, and let F(u, b, v) be a solution ofthe equation

OF= E A,(aAu, F)Aa', F(0, b, c) = c.

i=1

Setting t = Au yields

OF °cat

= A,(at, F)a', F(0, aA, c) = c.i=1

Thus, since the solution of the equation is unique,

F(Au, a, v) = F(u, Aa, v).

Pick u = 1, A = t. This gives us the result.f) Since F(0, a, c) = c in J, OF (0, a, c) = 1.

Thus (t, a, v) 36 0 in a neighbourhood of (0, a, c). Applying theinverse function theorem, we can express v in terms of u and z. (u, v)form a coordinate system in a neighbourhood of (0, c). If dv = 0 (cis given) and u = at with a = constant, then w = 0 according to thedefinition of F. Thus Pi(at, v)a' = 0.

g) *w = L1 P,(at,v)tda' + B(at,v)dv according to the result above.Thus R, (t, a, v) = tP;(at, v).

h) We have d(l!*w) = *dw = TV A W*w. Thus

dR, Adai + dB A dv = Vk*0 A Rda'+Bdv).i=1

Since the coefficient of dt A da' must be the same on both sides,OR,8t

=HR, and R,(0,a,v)=0.The solution of this equation is unique according to Cauchy's theorem.Thus R, 0 and w = B(u, v)dv.


Solution to Exercise 2.47.

a) Letu= 1(u'O+v'8). Then 14u u4 Z'FO.

Let a = EL-1 a;dx`. Then (II*u, a) _ 1 a;u'.Thus w = E! 1 y'dC, since y' = a; at a.

b) We have dw = E7 1 dy n de and Q = n1 dy1 A dCI A dya n ... A dC',which is nonzero everywhere. Hence 7'"(M) is orientable.

Chapter 3

Integration of VectorFields and DifferentialForms

The first part of this chapter concerns the integration of vector fields. Asa vector field is, by definition, a section of T(M), its components in a localchart are differentiable functions. So, by the Cauchy theorem, we provethat a vector field X is integrable. It defines a one-parameter local group oflocal diffeotnorphisms. That allows us to define Ex, the Lie derivative withrespect to the vector field X, of a vector field Y or a differential form w.

Instead of considering a vector field X (x) (1-direction field), we can con-sider p-direction fields H. (1 < p < n) and ask the question: Do there existintegral manifolds W of dimension p? That is, do there exist submanifoldsWsuch that Tx(W)=H., for all xE W?

Frobenius' theorem states that, if a certain necessary and sufficient con-dition is satisfied, there exists, through a given point xo, a unique integralmanifold of dimension p.

Integration of Vector Fields

3.1. Definition. A differentiable map (t, P) -+ cp(t, P) = Wt(P) of R x Minto a differentiable manifold M is called a one parameter group of diffeo-morphisms on M if

a) W(0, P) = P, Wo = identity, and

b) for all s and t belonging to R, Va+t = P. o Ot

77

78 3. Integration of Vector Fields and Differential Forts

cpt is indeed a diffeomorphism, since cpt o cp-t = cP-t o 9t = cpo

A one parameter group of diffeomorphisms on M defines a vector fieldon M by Xp = [dpt(P)/dt)t=o. It is the tangent vector at t = 0 to thetrajectory ,y of P : t ---+,y(t) = cpt (P).

The tangent vector at cp,(P) to ry is

141,P)l= [dy.+u(P)] _ {d<pu[P,(P)I1 - = Xv

dt t-, du -o L du J U=0

Moreover, since cw,+u(P) = co3[cPu(P))1, we have

dcp, Ldu(P)J = (w,)XP-du u-0

3.2. Definition. A differentiable map (t, P) - (t, P) = Et(P) of 11, anopen neighbourhood of 0 x M C R X M, into M is called a one parameterlocal group of local difeomorphisms i;t of M if

a) t(0, P) = P, to = identity, andb) for all P E M and all t and s in R, {(t+s, P) = £[s, t(t, P)] whenever

(t, P), (t + s, P) and [s, l; (t, P)] are in 11.

Let us verify that £ is indeed a local diffeomorphism. Let t 0 0 be a realnumber such that (ut, P) exists for all u E [0,1). P is fixed. Consider acompact neighbourhood 0 of the curve C = f t(ut, P) I u E [0, 1] }, and e > 0such that) - e, e[ x R C Q. We verify easily that B and a exist, since C is acompact set.

Let q E N be such that Iti < qe. We can write

&(P) _ (: ore 0...o£4)(P),4 9 9

the product of q factors £ j, which is invertible. Thus C1 is everywhere of9 9

rank n on C. Consequently, 6 is of rank n at P, it is locally invertible atP, and

4_f o t;_f o ... o c-t (q times).4 4

A one parameter local group of local diffeomorphisms £t defines a vectorfield Xp = [det(P)/dt)t o. The converse is the goal of

3.3. Theorem. A C'''1 vector field X on M, r > 1, defines a one parameterlocal group of local difeomorphisms obtained by integrating the followingdifferential equation on M.

d<pt(P)

dt= X (p).

An integral curve of X is a differentiable curve 7(t) such thatdy{t} = X

dt y(

Lie Derivative

L e t (0, *) be a local c h a r t with coordinate system xl, x2, ,

let P E 0. The equation is, denoting (P)] by xt(t),

dx' n).=X`(x1,x2,... ,x

79

The Cauchy theorem asserts that there exist a neighbourhood I x Uof (0, P) in R x 0 and a unique differentiable map p(t, Q) of I x U into 0satisfying the equation above and V(0, Q) = Q for all Q in U. Moreover,since dp..+t/dt = dp.+t/d(s+t) = Xw.+t, uniqueness implies that p.+t(P) _Spt[V.(P)1-

Thus we exhibit a cover of M by open sets U., j E J, such that on I j x Ujthere is a map tp j with the properties defined above. Set f l = U J EJ(If x Uf )(Cl is an open neighbourhood of 0 x M in R x M), and set op(t, P) = pj(t, P)if (t, P) E I j x Uj. I 'his definition of co makes sense because of the uniquenessof the solution.

Lie Derivative

3.4. Definition. Let X and Y be two C2 vector fields on a differentiablemanifold M, and apt the local group of local diffeomorphisms related to X.The vector field

Cx(Y) =tlim

tYP]

is called the Lie derivative of the vector field Y with respect to the vectorfield X. Likewise we define the Lie derivative of contravariant tensor fields.The Lie derivative, with respect to the vector field X, of a function f is

Gx(f)=t mt1

(f Pt -f),and that of a differential form w is

Cx(w) = tli 0 t [(apt)* &' (P) - ''!PI

Let us verify that Cx (f) = X (f ). Cx (f) is the differential at t = 0 off o pt Now

(d(fo))=dfo(±Pt) =X(f)

dt t=o

In a neighbourhood of P we saw that for t small enough cpt and Sp-texist. (w-t)* at P is a map of Ti(p) onto Tp(M). So in the definitionof Cx (Y) we have the difference of two vectors at P in the bracket. Fordifferential forms we must use (spt)* to have the difference of two forms atP in the bracket.

3.5. Propoattlon. Lx (Y) = [X, Y1.

3. Integration of Vector Fields and Differential Forms

Proof. Let f be a C2 function in a neighbourhood Cl of P. Compute[Gx (Y)] p(f ). For t small, we can write

[(v-t)*Yv'(P)](f) - YP(.f)= [YwrP](f O P-) - y

(p)](f) + [Y49 (P)](f) - YP(f)The result follows. Indeed,

rt 0 t (f)]P} = X [Y(f)]P

and

o t {[Yv,(P)](f o v-t) - -tli o LY(Jo

ot - f )J tP)_ -Y[X(t)Jp.

t is small enough so that Spt(P) and W_t(P) belong to Cl. [Y(p)](f) is thevalue of Y(f) at VI(P).

3.6. Proposition. a) If u and v are two contrnvariant tensor fields,then

,Cx (u ®v) = Lx (u) ®v + u (3 Lx (v).

b) If w and w' are two differential forms, then

Gx(wAw') =Gx(w) Aw'+wACx(w').On differential forms,

d,Cx = Gxd and Gx = i(X)d+di(X).

The proof of (a) is similar to the proof of the first part of (b). Since0*(w A w) _ (cp*w) A we can write

(wc) (w A w ),, (P) -- wp A wp

_ (V*war(p)) A (tptwsvt(P) - wP] + [V*wp,(P) - wp] Awp.

The first result in (b) follows. For the second we use Proposition 2.26:p*d = dip*. Thus

Ex(dw) = tl t ((jpt)*(dw)s.,(P) - (dw)p]

tll t d[(dot)*(w) (P) - "P]= dCxw.

And now let us prove the last result. It is true for functions: i(X) f = 0and i(X)df = X (f) = Lx(f)

For differential 1-forms, such as 4f,

d[i(X)df] = dX(f) = dCx(f) =Gx(df) and [i(X)d]4f = 0.

The Flobenius Theorem 81

Since GX is linear, we have to prove the result only for p-forms such asw = fdx" A A dx":

,CXw = Lx! dx" A ... A dx'"P

+ f E dx'' A ... A dx"-' A GXdx'J A dx'J+t A ... A dxi'.1=1

Observe that Cxdx' = dCxx' = dX(x') = dX', where {X'} are the com-ponents of the vector X in the basis {8/8x'}. Moreover,

i(X)w = f t(-1Y^1dx" A ... A dx'J-- A X'dx'J+' A ... A dx'",

j=1

P

di(X)w = tV A E(-1)j-1dx" A ... A dx'J-' A Xjdx'J+' A ... A dx'Pj=1P

+fEdx" AdXjAdx'J+'

A A

A E(-1)jdx'' A ... A dx'J ' A X3dx'f}1 A ... Aj=1

Thus Lxw = [i(X)d + di(X )]w.

The Frobenius Theorem

'p.

3.7. Definition. A p-direction field H= on M is defined by prescribing, ateach point x of M,,, a vector subspace H,, of dimension p of TZ(M) whichsatisfies a differentiability condition. We can write this condition in one ofthe following forms:

a) Each Xo E M has a neighbourhood V where there exist p differen-tiable vector fields X1, X2i - , XP such that Xi(x), X2(x), , XP(x)form a basis of HS for all x E V.

b) Each xo E M. has a neighbourhood V where there exist q = n - pdifferential 1-forma wl, w2,- , wq such that

XEHr4-*wl(X)=w2(X)=...=wq(X)=0.3.8. The F robenius Theorem. In order that, for each point xo E M,,,there exist a submanifold 9 xo of dimension p (called the integral man-ifold through xo) of a neighbourhood V of xo, tangent to H. in each pointx E WW,, it is necessary and sufficient that [Xi, Xj]. E Hx for all i, j and

82 3. Integration of Vector Fields and Differential Fbrms

x (1 < i < j < p) if we consider condition (a). On the other hand, if weconsider condition (b), a necessary and sufficient condition for the existenceof an integral manifold through any given point is that

foralli(1<i<q).

We will give below a proof of Frobenius' theorem and some developmentsof Pfaff systems.

If p = 1, we have to integrate vector fields. We saw that this is al-ways possible. We can verify that the conditions of Frobenius' theorem aresatisfied when p = 1. In this case H,, has dimension 1. Let X # 0 be avector field such that X (X) E Hr; when x E V, we have (X, X} , = 0 E Hr.The condition written with differential 1-forms is also trivially satisfied. Asq = n - 1, dwi Awl A w2 n . n w,a_1 is a differential (n + 1)-form whichvanishes.

But when p > 1, intuitively there is no integral manifold in general (forexample, when p = 2 and n = 3). Let X, Y be two vector fields generatingH2 for xE V.

zo

From xo, we integrate X, then Y. We get Co and Co. Let C1 be theintegral curve of X from a point xl of Co, and let C2 be the integral curveof Y from a point x2 of Co. There is no reason that C2 should intersect Cl.But if there is an integral surface S through xo, then C1 meets C2 since theyare included in S.

3.9. Definition. A system of exterior differential equations on M is a seto f N equations {wQ = 0}, a = 1, 2, .. , N, where wQ are exterior differentialp.-forms. The solutions of the system are vector subspaces T such thatw- (XI, X2, ... , Xp.) = 0 for all a whenever X1i X2, , Xp. belong to T.

The Frobenius Theorem 83

The pair (Wp,') consisting of a manifold and of a differentiable map 4of Wp in M. is an integral manifold of the system if $*w° = 0 for all a.

Consider the ideal I of the exterior algebra A generated by the w°:

w E 14=4-w = w° n 0° with O° E A.1<°<x

3.10. Proposition. Every solution of the system {w° = 0) is a zero ofeach form of I, and conversely.

3.11. Definition. Two systems of exterior differential equations are saidto be algebraically equivalent, {w° = 01 {wO = 01, if they generate thesame ideal. Then each element 06 = F,° w° A 01, and vice versa.

According to Proposition 3.10, the two systems have the same solutions.

3.12. Definition. The differential system {dw° = 0, ° = 01 is called theclosure of the system {w° = 0}.

3.13. Proposition. A system of exterior differential equations and its clo-sure have the same integral manifolds.

Indeed, Vw° - 0 implies 4*dw° = d(4*w°) - 0, since d and $* com-mute.

If two systems are algebraically equivalent, so are their closures. Indeed,if a)P = E. w° A 0°, then &A = E, dw° A 0° + E° (-1)p°w° A d0°.

3.14. Definition. A system is said to be closed if it is algebraically equiv-alent to its closure: dI C I. The closure of a system is closed (d2 = 0).

3.15. Definition. A system of exterior differential equations is called aPfaff system if the w° are all differential 1-forms. The equations {w° = 0}determine at one point a vector subspace which contains the tangent space ofall integral submanifolds W p C M,,. A Pfaff system of rank q, in an open setIl, is said to be completely integrable if there exist q differentiable functionsy# (0 = 1, 2, - , q) such that the system {w° = 0} (a = 1, 2, , N) isalgebraically equivalent to the systems {y# = 0} (j3 = 1, 2, - - - , q). That isto say, w° = aady'. The YO are called first integrals of the Pfaff system.

In writing w° we omit the sign F,. This is the Einstein convention.The Einstein Convention. When the same index (say 0) is above

and below, summation over this index is assumed (here from 1 to q). 6 iscalled a dummy index. Henceforth we will use the Einstein convention tosimplify the writing.

3.16. Theorem. If a Pfaff f system of rank q in 11 C M is completelyintegrable in f3, then through each point xO E fl there is exactly one integralsubmanifold of dimension p = n - q.

3. Integration of Vector Fields and Differential Forms

Proof. The equations ys(x) = yO(xo) (1 <,3:5 q) define a differentiablesubmanifold W of dimension p = n - q through x0. Indeed, the system{w" = 0} being of rank q, the system {dys = 0} is of rank q, there are qfunctions ys. We verify easily that (W, i), i being the inclusion, is an integralmanifold of the system, the unique integral manifold of dimension p throughXO.

3.17. The Frobenius Theorem (second version). A necessary and suffi-cient condition for a Pfaf system {w°1 = 0}, of rank q in fl, to be completelyintegrable in 1, is that the system is closed in f2.

Proof. Necessity. Let {V} be the q first integrals, w" = a°dy,6, whereA = ((a')) is an invertible matrix. Let B((b4)) be its inverse. Then

dw0 -doeAdye-b.Oda,3Aw7.

Thus, if I is the ideal generated by the w°, then dI C I.Sufficiency. The proof proceeds by induction on the dimension n. First,

in a space of dimension q, a system of rank q is completely integrable. Indeed,since a, dx= with i, j = 1, 2, ... , q, the system {wj = 0} is equivalent tothe system {dx' = 0}, the matrix A = ((A.,)) being invertible by hypothesis.

In the general case, when the dimension is greater than q, suppose wehave proved the theorem up to dimension n -1 > q. More precisely, if a Pfaffsystem {w° = 0}, of rank q in fl, is closed, the 411" depending differentiablyon n - 1 variables x1, X2'. .. , xn-1 and on some parameters xn, x"+1, ,

then there exist q first integrals z" which depend differentiably on the n -1variables and on the parameters.

Consider the Pfaff system of rank q,

n{w _ a°dr = 0}

i=1

(a=1,2,...,q),

where the as depend differentiably on n variables and some parameters.

Without loss of generality we can suppose that {w" = E;_i a, dx` = 0}is of rank q. The forms iv°are differential 1-forms on Rn-1 depending onxn (at this stage, xn is a parameter), and on some other parameters. ddesignates differentiation in Rn-1 and d differentiation in R":

dw" - dbp + 00 A dx", 00 E A1(t').

Since by hypothesis dw" E I, the ideal generated by the w° in A(fl), itfollows that d/' E I the ideal generated by the w" in A(SZ n 7r) (7r being thehyperplane x' = Const). Indeed, if dw" - =1 ry,9 A wO, we can write dw"

Integrability Criteria 85

in the formq

dw° E ry0* A w + 9° A dxn=1

with E Al (SZ n ir) and dm = nw150=1Thus there exist q functions z°, depending differentiably on the variables

and the parameters, such that w° = A°dz,6, ((A°)) being an invertiblematrix (((B4)) its inverse) whose elements A' depend differentiably on thevariables and on the parameters. The system {w° = 0} is equivalent tothe system JUPO = BQw° = 0}, where w6 is nothing else but = dz' +BQandxn = dz# + b1dxn. Since this system is closed (cam E I),

dba n dxn = 9. n (dz° + b°dx")

with 0,0, a differential 1-form which is a linear combination of the dz° anddx" only. Indeed, in the coordinate system zi, z2, ... , zq, xq+r, ... , xn, 0.0cannot have a term in dxi with q < i < n, as otherwise we would have inthe right hand side a term in dx' A dz°, but no such term in the left handside. Consequently there is no term in dxt A dx" in the right hand side, andb-8 is a function of the z°, of xn and of the parameters only, 8b9/8xt = 0 forq < i < n. Thus the system {00 = 0} is an ordinary system of differentialequations

615 ydxn

= -b (f , z, ... zq, x") ( 1< 0<0 ,

which locally has a solution according to Cauchy's theorem. Consider q inde-pendent first integrals y°(zl, Z2,. -. , zq, xn) (1 < a < q) in a neighbourhoodof the considered point. The y° depend differentiably on the parameters.The system {w° = 0} is equivalent to the system {dy° = 0}-it is completelyintegrable.

Integrability Criteria.

3.18. If w1, w2, .. , wq are q independent Pfaff forms, a necessary and suf-ficient condition for the p form 0 to belong to the ideal I generated by thewQ (1<a<q) is that 0Awlnw2A...Awq=0.

Obviously the condition is necessary; if 0 = F,a_10° A w°, 0 satisfiesthis condition. To prove the sufficiency, consider a basis of A'(Sl) withw1, w2, , wq completed with n - q 1-forms wq+i, w:

9= E ai,,12,...,ipw'1 n w 2 n ... A w".it<is<<iy

86 3. Integration of Vector Fields and Differential Forms

Then 0 A w l A w2 A . A wa = 0 implies that a;1,12; .. ,j, = 0 whenever i t > q.Thus 0EI.

The necessary and sufficient condition of the Frobenius theorem is

dwaAwI for 1 <a <q.

3.19. Remark. We saw already that in a space of dimension n, a Pfaffsystem of rank n is always integrable. The criterion above shows that thisis also the case if the rank of the system is equal ton -1 (an (n+ 1)-form isidentically zero). We can prove this result directly. If the wa = F 1 qd-T'(a = 1, 2, . , n - 1) are independent, we can suppose that the matrix((aq)) (/3 = 1, 2, , n - 1) is invertible. The system {E,"_1 a°dx` = 0}is then equivalent to a system of the type {dx=/dzn = A=(x2,x")} withi, j = 1, 2, , n - 1. The Cauchy theorem gives us the first integrals.

3.20. Example. The Pfaff equation in R3,

w - a1dx1 + a2dx2 + a3dx3 = 0

is completely integrable in a neighbourhood Il of a point xo if al, a2 and a3do not vanish simultaneously and if w A dw = 0-that is, if rank w = 1 andif

Bag 8a3(-Gla2

aals1

8aa3

x2 - 8x3) + a2aOnX3ai

- axl) + a3 x1 - 9X2 = 0.

This can be written as A rot A = 0 with A the vector of componentsa;. Under this condition, in a neighbourhood of xo, w = 0 is equivalentto dcp = 0 for some function cp, and through xo there is one and only oneintegral manifold of dimension 2. Its equation is W(x) = W(xo).

3.21. Example. The Pfaff equation in R2,

('J - x1dx2 - x2dx1 = 0,

is completely integrable in P2 - {0}. In R2 - {0} the rank of w is I (it is zeroat 0). The integral manifolds are straight lines x2 = kx' (k E R). Througheach point xo 96 0 there is one and only one integral manifold.

3.22. Proposition. If a p-direction field on M is defined by p vector fieldsXi(x), i = 1, 2, . , p (see Definition 3.7), the necessary and sufficient con-dition of the F vbenius theorem is [X;, Xi]z E H, for all i, j, and x (1 < ij:5 p)-

Indeed, consider q differential 1-forms wa (a = 1, , q = n - p) suchthat wa (X) = 0 for all a is equivalent to X E H1. The necessary andsufficient condition for the Pfaff system to be completely integrable is that


dwa E I, the ideal generated by the wa, and, according to the definition ofthe differential d,

dw(Xi,Xj) = X,[w(Xj)] - Xj[w(X.)] - w([X{,Xj]).If dw E I, then dw(X1, X j) = 0 and we have w([Xi, Xj]) = 0; thus [Xi, Xl]rE H.

If [Xi, X f]: E H=, we have dw(Xi, Xj) = 0, which implies dw E I. To seethis we consider a basis of A' (fl) with wl, w2, , w4 completed with n -q = p differential 1-fag w¢+1, ... , wQ+v such that w(Xj) = b (the Kro-necker symbol). If dw = E1<k< <n aktwk Awl, then dw(Xi, Xj) = aq+i,q+j ifi<j. We must have ail=Owlieneveri>q. That is,dwEI.


3.23. Problem. Let M and W be two COD differentiable manifolds of thesame dimension, and let (u, x) --+ Gu (x) be a C°° differentiable map of]a, b[ x M into W (]a, b[ C R) such that, for fixed u, Gu (x) is a diffeomor-phism of M onto C W. We denote by Xu(y) the tangent vector aty = G. (x) to the differentiable curve ]a, b[ E) t -- Gt(x).

a) Let fl be a differential 2-form on W (Il E A2(W)). Let {xi} bea coordinate system in a neighbourhood U of xo E M and {y} acoordinate system in a neighbourhood 0 of W = Gu(xo). Express onU the components of Gu I in terms of the components flap of fl on8.

b) Compute GuCx Q.c) Prove that &(GGSZ) = GuCx.fl.d) From now on M = W = Bo(1) = B, the ball of center 0 and radius 1

in Rn. We consider for u E ]0, 1] the homothety Gu (Gu(x) = ux).On which open set is X,(y) a vector field? What are its compo-

nents?

e) Henceforth Cl is closed (dC2 = 0). For a, b E ]0, 1], prove that GaIZ -Gafl is the differential of a differential 1-form ry(a, b).

f) What is the limit of Galwheng) Show that there exists -, E A'(B) such that fZ = dry.

3.24. Problem. Let M be a C°° differentiable manifold of dimension n =2m, and let Cl E A2(M). We assume that the rank of fl is 2m, that is to

m timess a y , that, atanyypoint xEM,w= Aft Q A - - - - - - .l 96 0, a n d

The differential forms are assumed to be C°°.

a) Show that for any X 76 0, i(X)Sl is nonzero.


b) Exhibit a basis e; (i = 1, 2, , 2m) of xo fixed, such thatSa(e2k_1i elk) = 1 for k = 1, 2,- , m and fl(ei, ei) = 0 for the otherpairs e;, ej with i < j. What is the expression of fk(xo) in a co-ordinate system {x` } associated to the local chart (V, III) at xo, if(g/ax`)"o = e:?

We suppose that *(V) is a ball of R".

c) On V, consider the constant differential 2-form fl such that fl = faat xo. F o r t E [0,11, set fat = fa + t(Sl - fl). Verify that there isa neighbourhood W C V of xo on which fat is of rank 2m for allt E [0,1], and that there exists a differential 1-form ti on V such thatry(xo) = 0 and dry = fl - Cl (use Problem 3.23).

d) Show that the map X i(X)fl is an isomorphism of r(M), thespace of C°O differentiable vector fields, on Al (M). Denote by Xtthe vector field defined by i(Xt)f4 = -y. We admit that the map(t, x) ---' Xt(x) of [0, 1] x M in T(M) is C°°.

e) Prove that there is, on a neighbourhood Vo of (0, xo) in [0, 1] x M, aC°° differentiable map ft(x) such that Oft(x)l& = Xt(x), fo(x) = Id.Prove that & (f t Sat) = 0 (use the result of Problem 3.23).

f) From the previously posed questions, deduce the existence of a coor-dinate system {yJ } on a neighbourhood of xo such that

Sa = dy1 A dye + dy3 A dy4 + ... + dye"-1 A dy2m.

We will admit that there is W such that [0,1] x W C V0.

3.25. Problem. Let (x, y, z) be a coordinate system on R3 where we con-sider the differential 1-form

w = (1 - yz)dx - x(z + x)dy + (1 + xy)dz

a) Is the equation w = 0 completely integrable? What is its rank? Isthe set M of points where w = 0 a submanifold of R3?

b) Find a nonzero vector field Y such that w(Y) = 0, Y being of theform Y = ai + c with a and c polynomials in x, y, z (its secondcomponent vanishes). Integrate the vector field Y. What are theintegral curves?

c) Exhibit two independent first integrals y and u of the equations con-sidered in (b).

Write w in the coordinate system (x, y, u). Was the result ob-tained foreseeable?

d) What are the integral manifolds of the equation w = 0? Verify thatthey are indeed submanifolds of R3. How many integral manifoldsare there through a given point of R3?


e) Beginning by integrating a vector field X whose first component iszero, prove again the previous result.

3.26. Problem. Let E be a vector space of dimension n. Set F = E x R.The pair consisting of a point (x, z) E E x R and of a hyperplane H throughit is called a contact element of F. If the hyperplane is not parallel to theOz axis, we say that the contact element is regular.

a) Show that the set of regular contact elements may be identified toFxE`.

b) For a submanifold W of F of dimension p (p < n), we say thatthe contact element (x, z, H) is tangent to W if (x, z) E W andT(r,z)(W) C H.

Prove that the set of regular contact elements tangent to W formsa submanifold W of F x E* of dimension n.

Hint. Consider q equations f, (x, z) = 0 which define W, withft r= Cl such that df, (1 < i < q) are independent when (x, z) E W.

c) We denote by {x.' } (1 5 j < n) the coordinates of E and by { ay }(1 < j < n) those of E*. On F x E* we consider the differential formw = dz - a,dx'.

Prove that W is an integral manifold of the Pfaff equation w = 0.

d) Is it possible that the equation w = 0 has integral manifolds of di-mension d > n?

3.27. Problem. On R4, endowed with the coordinate system (x, y, z, t),we consider two vector fields,

X =tax by

+z -yaz 'Nand

a a a aY=-y8x+xay-taz+z&.

a) On which submanifold M of R4 do these two vector fields define atwo-plane field? Is this two-plane field integrable on M?

b) We consider on R4 a third vector field49Z = -z +t +xax - yat.

Integrate this vector field. Do the integral paths have any particularproperty?

c) Show that X, Y, Z define at each point P E M a 3-hyperplane fieldHp. Is this field Hp integrable?

d) Find a differential 1-form w such that for all P E M and 1: E Tp(M),w(t)=0 atEHp.


e) If they exist, what are the integral manifolds of the ques-,ions askedin (a) and (c).

3.28. Problem. Consider in 1R4 the Pfaff system S:

{wl = [x3 - (xl )2]dxl + x1 x2dx2 + dx3 = 0,

l w2 = fdx1+gdx2=O,

where f and g are functions of the four coordinates xl, x2, x3, x4.

a) Find functions f and g for which the system S is completely inte-grable. Show that it is then equivalent to a system of rank 2 in R3.

b) Find the integral manifolds of dimension 2 of the system S whenf=9=1.

c) Consider the differential equation in 1R3

w == [x3 - (x1)2 - xIx2]dxl A dx2 - dx2 A dx3 + dx3 A dxl = 0.

Write the integral manifold through the line

x1 = alv, x2 = a2v, x3 = a3v.

3.29. Problem. Let S = {w"}"EA be a family of exterior forms on thevector space R" each of degree q" > 0, and I the ideal generated by theseforms in A(R"). Denote by Q the smallest subspace of Al (R" I such thatS C A(Q). Q is associated to S.

a) Prove the existence and uniqueness of Q. Its dimension is called therank of S.

b) Establish that the 1-forms of S belong to Q.

c) Show that a necessary and sufficient condition for an exterior informw to be a monomial (that is to say, w is the exterior product of linearforms) is that the rank of S = {w} is p.

Verify that the rank of a nonvanishing 2-form on R3 is always 2.

d) Denote by H the subspace of Wt such that ixO = 9(X) = 0 for all0EQwhen XEH. Show that ixw" = 0 for all a E A.

Hint. Consider a basis of Al (R") formed by a basis of Q completedwith a basis dual ro a basis of H.

e) Prove that XEHifandonly ifixwEIforallwEI.Hint. For sufficiency, proceed by induction on the rank of the

forms.

f) Let S = {wl, w2} with wl = adx + bdy + cdz and w2 = Cdx A dy +Bdz A dx + Ady A dz on R3. According to question (e), wri-,e the threeconditions for the vector X of components (x, y, z) to belong to H.Here a, b, c, A, B and C are CO0 functions on 1R3.


g) Find three 1-forms (ryf } (j = 1,2,3) such that {ry) (X) = 0 (j =1, 2,3)1 is equivalent to X E H. Find sufficient conditions for S ={ryl, rya, rys} to be of rank 2.

Henceforth E = {w* = 0}QEA is a system of exterior differentialforma on an open set Q of a differentiable manifold M. Each pointx E fl is associated to a subspace Q. of 2 (M) corresponding toS = {w°(x)}QEA. We assume that there exist p differential 1-formsyi on 12 forming at each point x a basis of Qy. P = { = 0}1<i<p issaid to be associated to E.

h) Show that an integral manifold of the Pfaff system P = {ryi = 01is an integral manifold for the system E.

i) Consider on R3 two systems El = {wl = 0,w2 = 0} and E2 = {w2= 0} with wl = dx + 2dy and w2 = dx A dy + 2dz A dz + 4dy A dz.What are the Pfaff systems P1 and P2 associated respectively to Eland E2?

j) Integrate the Pfaff systems PI and P2, and find the integral manifolds.

k) What are the integral manifolds of the system El and those of E2 ?

Problem 5.30 (see Chapter 5) is a beautiful application of the Frobeniustheorem.

3.30. Exercise. On R2 we consider the following vector field:

X' Y,

y' _ -x + (1 - x2 - y2)y.

Prove that there is only one closed integral curve.

3.31. Exercise. On R2 - {0} find the integral curves of the vector field

J x' = (x2 + U2)-3/2(12x2y2 - 3x4 - y4),j lr = (10xy3 - 6x3y)(x2 + lf2)-3/2.

3.32. Exercise. We consider in R4 the Pfaff system S:

W 1 = [z3 - (xl )2]dx1 + xlx2dx2 + dx3 = 0,1 w2 fdxl+gdxa=0,

where (z'} is a coordinate system on R4 and f, g two functions on R4.

a) Find the functions f and g for which the system S is completelyintegrable. In this case, show that it is equivalent to a system of rank2inR3.

b) What are the integral manifolds of dimension 2 of the system S whenf=g=1?


c) Let us consider the following equation in R3:

W = {x3 - (z1)2 - x'x2}dx' A dx2 - dx2 A dx3 + dx3 A dx' = 0.

Write the integral manifold through the straight line l(v) defined by

x1 = av, x2 = by, x3 = CV,

a, b, c being real numbers.

3.33. Exercise. On R2 we consider the vector field defined by

R2 (x, y) ---+ X (x, 1J) _ x + y(1 x2 - y22)

a) Show that for any (xo, yo) E R2 \ (0, 0) this vector field has a uniqueintegral curve Ot (xo, yo), defined for all t E R, such that 00(xo, yo) _(xe,YO) -

b) According to the value of zo + 14 > 0, determine whether the imageof R by the map i : t - ot (zo, yo) is a submanifold of R2, andwhether ti is an injective immersion or even an imbedding in R2.

3.34. Problem. On R4 we consider the following vector field:0 .Y= e-2x1 1 - e-2.,' 0+ e-2,,.,4

a) Write the Pfaff system (1) associated to Y. Find three independentfirst integrals of system (1): y1, y2, y3.

b) Write the equation Y(f) = 0 in a new coordinate system {y*}(i = 1, 2, 3,4); y4 is to be chosen. Deduce from this the solutionsofY(f)=0.

c) Let Z be the vector field

Z=5j1Write the Pfaff system (2) corresponding to the 2-plane field definedby Y and Z (A and p are C1 functions on R4).

Hint. Integrate an arbitrary vector field which is a linear combi-nation of Y and Z.

d) What is the Frobenius condition for system (2) to be completely in-tegrable. Find functions .1 and p satisfying this condition.

Hint. Write system (2) in the form

w1 = A (e2s'dxl + e22dz2) - (e2=' +e2_2 ) dx3 = 0.

W2 = !ice (e2h' dx1 + e2xadx2)

+ e2(x1+12) (dx1 - dx2) - (e2x1 + 212) e2x`dx4 = 0


e) Verify that this condition is satisfied if we choose

,\(xt)= O(x3)'41,

/I e2''' + e2x2 1

where 0 and 4, are two arbitrary C1 functions of one variable with ¢never zero, and

P(x'.) =e2(r'-r2)

or µ(x') = 2(x - -_l ).

In these two cases, find the first integrals of system (2). Use thesefirst integrals to solve the system Y(f) = Z(f) = 0.

f) Let IV = 8/8x3. We consider the Pfaff system (3) corresponding tothe 3-plane field defined by Y, Z, W. What is the Frobenius conditionfor which system (3) is completely integrable?

h) We choose p(x') as previously. Integrate system (3) in the two cases.Find the solutions of the system Y(f) = Z(f) = IV (f) = 0.



a) Set A?(u,x) = 8Gu(x)/8x. Then (G* 11);j (x) =b) According to Proposition 3.6,

Lx.,IZ = Xu(fl.j)dya A dya + naf3(dXu A dyO + dya A dXu ),

since Lxd = dLx and LXya = Xa. Thus

Lx., SZ = (XU-'t 8.yQ,,3 + SZ yI88X'y + f2aryarjXu) dy" A dyf3,

and

GuLx., I = A;'A j (X,'O. S2ryi8aX; + Sta. Of3X.u) dx' A dxJ.

c) Since

a (Gun)>; = a -(AaA' )Q,,, + A A; X'Y8S1au,

the result follows. Indeed, 8; = A '?O,,.

d) Xu(y) is a vector field on G.u(B) = Bo(u). We have

Xu(y) =8Gu(x)

=x _ Y;

49U u'its components are XU (y) = ya/u.

e) According to b) and c),b

G11- =jb

(G1)du = fa

= dfbGL[i(Xu(y))1]du = d'Y(a,b).

a


f) We have A;' = OGu(x)/ax' = u6 and (GafZ)ij = a2S;'8gi2Q,3(y)Thus Ga11-+0when

g) Gi = Id; hence f2 = d-y with y = f0

Solution to Problem 3.24.a) If X ,- 0, there exist Y1, Y2, , Yn_1 such that (X, Y1, Y2. ,Yn_1)

is a basis of T;,(M). Thus w(X,Y1,Y2i ,Yn_1) is a sun of termseach having a factor Q(X,Y) for some i. If SZ(X, Y) = 0 for all Y, wewill have w(X, Y1i Y2, , 1) = 0, which is in contradiction withthe hypothesis.

b) In T0(M) we choose el 96 0, then e2 such that 1l(el. e;) = 1. e2exists according to question a).

Let El = {X E T,,, (M) I fl(e1,X) = 0 and 52(e2rX) _= 0}. Thendim El = 2(m - 1). Then in El we choose e3 # 0 and e., such thatSZ(e3, e4) = 1. By induction we exhibit the basis e1. Then

S2(xo) = dx1 A dx2 + dx3 A dx4 + + dx2ni-1

A dx2"'

c) w = P(SZij)dx1 A dx2 A ... A dxn with P a polynomial in the Stij.By hypothesis P(Rj) 54 0 on V; we can suppose that V is compactand that the coordinate system exists on V. Since P is continuouson the f2ij, there exists f > 0 such that jaijI < e for all (i, j) impliesP(S2ij + aij) 54 0 on V. According to the continuity of Q. there is aneighbourhood W ofxn such that for x E W we have

//1152(x) - f2(x0))ijI < e

for all (i, j). Then the rank of Q1 is 2m on W fort E 10, 11. Moreover,since d(f) -!5) = 0, there exists -y E A1(V) such that y(ap) = 0 anddy = SZ - fl (see 3.23).

d) We saw in a) that the kernel of g : X - i(X)SZ is reduced to {0}.Thus g is an isomorphism of Tx(V) onto T(V). We can solve thesystem Xjftij = yj, j = 1, 2,- , n (yj being the comporents of thegiven differential 1-form y), at each point x E V. The solution isa C°O function of the yj and S2ij, since the determinant MR AI isnowhere zero. So toy E A' (V) there corresponds X E F(V ). We dothe same on the local charts of an atlas. The map y X is linear.

e) The existence and uniqueness of fi(x) are given by the Cauchy theo-rem. Moreover,

N(Put) = atfjf2 +t5ift 0 -fl)+fi*0 -Cl).According to 3.23c we find that

aa t (ft* Q0 = f 11xrI1t + Cl - f2].


Since d(It = 0, it follows that Cx,f4t = d[i(Xt)flt] = dry. Thus(f;nt)=0.

f) We have f152 = fOM = ft according to the previous equality. In thecoordinate system corresponding to the local chart (f1(W),* o (fi ') ),n is the constant 2-form fl.


a) We have dw = -2xdx n dy - 2ydz A dx + 2xdy A dz, and

w A dw = [(1- yz)2x - x(z + x)(-2y) + (1 + xy)(-2x)]dx A dy A dz = 0.

Thus the equation w = 0 is completely integrable. Its rank is equalto 1 except on M. Indeed, its rank is zero when yz = 1, x(z+x) = 0,and xy = -1. The second equation gives z = -x, since x = 0 isimpossible according to the third equation. So the rank is zero on thedifferential curve M defined by the equations z = -x and yz = 1.

If the rank of the map W : (x, y, z) (fi, f2) with fi = z + xand f2 = yz - 1 is two on M, then M is a submanifold of Rs. Nowthe rank ofD4c=(os is two on M, since z=Dandy=0donothappen simultaneously on M.

b) w(Y) = a(1-yz)+c(1+xy). We can choose a = 1+xy and c = yz-1.Now we have to consider the system

1 = 1 + xy, = 0, d zy - 1.

Integrating yields y = yo, log 11 + xyo yot + Constant andlog jzlp -11 = yot+Constant. So if yb 0, the integral curves of thevector field Y are given by

x = (Ae" - 1), y = yo, z = (aew + 1),

A = (xoyo + 1) and A = (yozo - 1). They are straight lines in theplanes y = yo. If yo = 0, the integral curves are straight lines

I x+zy

where A,,u, v are constants.

c) We have

dx _ dz1+xyo zyo-1

Thus the second first integral may bezy - 1uxy+f

96 3. Integration of Vector Fields and Differential Fbrms

Putting z = ux + (u + 1)/y in the expression of w yields

= d- (1+ - [ +u+I dw u x x x+uxxy) y

+(1+xy) [udx+xdu+ u -(u+1)!]

w =

,

(1 +xy)2y-2[-(u+ 1)dy+ydu].

We could expect this result. As u and y are first integrals, we havedy(Y) = 0 and du(Y) = 0.

So w(Y) = 0 and dx(Y) 9k 0 imply that dx cannot appear in theexpression of w in the coordinate system (x, y, u).

d) We can write w = (1 + xy)(z + x)[du/(1 + u) - dy/y].If (1 + xy) (z + x) 0 0, then w = 0 is equivalent to

1du

d-yw = 0.+ u y

Cv = 0 leads to 1 + xy = k(z + x), with k E R. They are the integralmanifolds of the equation w = 0. Indeed, we verify that 1 + xy = 0(k = 0) is an integral manifold (when x = -y ,w = 0). The planez + x = 0 is also an integral manifold. Fork E R - {0}, we have toconsider the map

cp : (x, y, z) -i 1 + xy - k(z + x),

Dip= (y-k,x,-k),which is of rank 1, except for k = 0 if x = y = 0.Hence the integral manifolds are submanifolds of R.

Through a point (xo, bb, zo) of R3 - M, there is a unique integralmanifold, the one with k = (1 + xoyo)/(zo + xo).

Through a point of M, there are two integral manifolds: the planex+z=0and the surface 1+xy=0.

e) If X is of the form X = bO + c with b = 1 + xy and c = x(z + x),then w(x) = 0. Integrating the equation

xdy _ dz1+xy z+x

with x = xo gives at once the first integral

Vz+X



a) A hyperplane H is defined by a vector orthogonal to H. Let {af}(1 s i : n + 1) be its components. If the hyperplane H is not parallelto Oz, then 96 0. Then we choose, as vector orthogonal to H,the vector having a bijection between thehyperplanes which are not parallel to Oz and the set {al, a2, . ,

which may be considered as the coordinates of a point of E*.

b) The c; = (8f,/8xj)dx' +(Of;/8z)dz (1 < i < q) are independent. Aregular contact element will be tangent to W if 8 = a;dx' + dz is alinear combination of the df;.

Suppose that the determinant IDI = 18f1 /Oxj I (1 : i < q, 1 < j< q) does not vanish. Then, for every k > q,

Dk =

D of8xk

at ... aq ak al

D=0.

W belongs to the space F x E*, which has dimension 2n + 1.But among the n + 1 first coordinates only p = n + 1- q are inde-pendent: (x, z) E W. For the n last coordinates, we have p inde-pendent conditions. Indeed, ODk/Oak = IDI # 0, and at (k > q)appears only once in the determinants Dj (when j = k). Moreover,b = [DI + via' = 0.

Since IDI 0, at least one of the A' (1 < i < q) is not zero, sayAj 36 0. In the Jacobian matrix of

(at,a2,... ,a,.) -* (Dq+t,... ,D.,D),

we can exhibit the determinant

IDI 0

IDI

IDI

0

ODk

Oaj,

which does not vanish.The dimension of W is (p + n) - p = n.

c) When p = n, at a point (x, z) E W there is only one contact hyper-plane, defined by aj = Oz/8xj. Thus dz = ajdza. When p < n, W isdefined by z = ff(x) (1 < i < q) at least when the contact element isregular. And aj = AzOfjl8zi for some A with E a' = 1.

Thus we verify that w = 0 on W.

= 0 and b =


d) The answer is no. Otherwise the integral manifold depends on dvariables. For instance we would have z = f (xl, - - , z", ail, - - , aaf ),and dz would be expressed as a function of the dad but also as afunction of some daj, which is not the case.

Chapter 4

Linear Connections

In R", there is a natural notion: the parallel transport of a geometric pic-ture. This allows one to compare two vectors in R' which do not have thesame origin. On a manifold, that notion does not exist-it is impossible tocompare two vectors which do not belong to the same tangent space. Closelyrelated to this is the fact that if we consider a vector field Y, we do not knowwhat it means to differentiate Y at a point in a direction. Henceforth, weassume that the manifold is C°° differentiable.

First Definitions

4.1. In the preceding chapter, thanks to the linear tangent map (V_t)*, wetransported the vector Y,,e(p) into the tangent space Tp(M), and we definedCXY. But, as we saw, CXY = [X, Y] depends not only on Xp E Tp(M),but also on the vector field X.

This is the reason for introducing a connection on a manifold. We definebelow the derivative of a vector field (then, more generally, of a tensor field)at a point in a direction.

4.2. Defnftion. A connection on a differentiable manifold M is a mappingD (called the covariant derivative) of T(M) x r(M) into T(M) which hasthe following properties:

a) If X E Tp(M), then D(X, Y) (denoted by DXY) is in Tp(M).b) For any P E M the restriction of D to Tp(M) x F(M) is bilinear.c) If f is a differentiable function, then

Dx(fY) = X(f)Y + fDxY.

99

100 4. Linear Connections

d) If X and Y belong to f(M), X of class C' and Y of class C''+1, thenDxY is in r(M) and is of class C".

Recall that l'(M) denotes the space of differential vector fields (2.14).A natural question arises: On a given C°O differentiable manifold, does

there exist a connection? The answer is yes, there does. We will prove inChapter 5 that a particular connection, the Riemannian connection, exists.So we will be able to differentiate a vector field with respect to a givenvector. Then, applying Proposition 4.5, we have all connections.

Let us write the covariant derivative DxY of a vector field Y with respectto a vector X E Tp(M) in a local coordinate system {x'} corresponding toa local chart (fl, gyp) with P E Q:

X = X' axi and Y = Yl 19axi

{8/ax'} (i = 1, 2, . . , n) are n vector fields on f2 which form at each pointQ E f2 a basis of TQ(M), as we saw in Chapter 2. {Xt} (1 < i < n) and{Yi} (1 < j < n) are the components of X E Tp(M) and Y E F(0).

According to (c) and the bilinearity of D,

DxY=XID;Y=X'(OYj) +X'YJD;

where we denote Dal&: by Di and 9f /8xt by &J, , to simplify the notation.

According to (d), Di(8/8xi) is a vector field on fl. Writing in the basis{a/axk},

D. (ate;) = I';`; Oxk.

Christoffel Symbols

4.3. Definition. T are called Christofel symbols of the connection D withrespect to the local coordinate system x1, x2, , x". They are COO functionsin 11, according to d) (the manifold is assumed to be C°°). They define thelocal expression of the connection in the local chart (f), g'). Conversely, if forall pairs (i, j) we are given Di(8/axi) = I ;FHB/8xk, then a unique connectionD is defined in A.

4.4. Definition. VY is the differential (1,1)-tensor which in the local chart(Il, cp) has (D;Y)j as components. (D;YY is the jrn component of the vectorfield D;Y.

To simplify we write V;YJ instead of (VY);. According to the definitionabove, (D;Y)' is equal to V;Y1:

V;Yj = 4-Y' + l' Yk

Torsion and Curvature 101

4.5. Proposition. T h e C h r i s t o ff e l symbols r of a connection D are notthe components of a tensor field. If I' are the Christofel symbols of anotherconnection b, then Ckj = r s - q. are the components of a (2,1)-tensor,twice covariant and once contravariant.

Conversely, if r k are the Christoffel symbols of a connection D and Cikjthe components of a (2,1)-tensor field. then ]: = Ck, +r are the Christoffeisymbols of a connection.

Proof. Let (9, 0) be another local chart at P, {y°} the associated coordi-nate system. Let Ai' = 8}I°/8xi and B336 = 8z'/8V. If X = Xi8/8x' =Xa8/8y°', Y = Y.8/8xi = Y08/O*J, and w = wkdxk = wadyA is a differ-ential 1-form, then

Xi(V,Yj)wj = Xa(V"Y3)w'3X`Aa[8Q(YkAk) +F AkYk[B'@wj.

Xd(ViYj)wj is a real function, so it is the same in any chart. Moreover,we can write X° = ArV, YO = AkYk and w'9 = Biwj. Since the equalityabove is true for any X and w,

8jYi + rJikYk = 4. [(BQYk)AL$ + Yka,,Ak + ro.AAkYkIBB.

As 0,YJ = As BaYJ, A'O '' = (= 0 if k j; = 1 if k = j) and since,9 Vk

the equality is true for any Y, we get

rjk = BJOA-A"k

+ A°AkB'5r0a.

Christaffel symbols I`iik are not the components of a tensor; in a change of lo-cal chart they do not transform like the components of a tensor, whereas 01,are the components of a tensor. Indeed, since 1 = B/S Ak + Ai° Ak B10,I4.a,

we get G°ik = Aj

conversely, in SZ, the family of coo functions c9 + r k defines a connec-tion D. Indeed, the mapping

D : (X, Y) -+ DXY + C,'kX'Yk Sri

satisfies conditions (a), (b), (c) and (d) of Definition 4.2.

Torsion and Curvature

4.6. Definition. The torsion of a connection D is the map of r x r into rdefined by

(X, Y) -' T (X, Y) = DXY - DyX - [X, Y].


Let us verify that the value of T(X, Y) at P depends only on the val-ues of X and Y at P, and does not depend on the first derivatives of thecomponents of X and Y. We have

[T(X,Y)]' = XiVjY' -YjVjX' - (X'83Y' -Yj8jX')= (I'jk - r=kj)X;Yk = T9kXjyk.

The operator of Tp(M) x Tp(M) into Tp(M) defined by

T :11 p

is represented by 7?k(P), which are the components of a (2,1)-tensor sincecontraction with two vectors yields a vector. In fl, the 7lk are C°° functions.

It is possible to give an alternative proof. Since 8 Ak = 8ikyO = 06e _according to the formula in 4.5 with r and 1' we get

7;k = A,°AkB30TQa.

So the T;k are the components of a (2,1)-tensor.

4.7. Definition. The curvature of a connection D is a 2-form with valuesin Hom(r, r) defined by

(X, Y) -+ R(X, Y) = DxDy - DyDx - D[x,y].

For the definition we suppose that the vector fields are at least C2. Oneverifies that R(X,Y)Z at P depends only on the values of X,Y and Z atP. In a local chart.

[R(X,Y)Z]k = X'DI[Y)(DjZ)k] - Y'DI[X1(DDZ)k]

- (X'BiYj - Y'BoXj)(D,,Z)k

= ()0OYj - Y'B Xj - [X,Y]j)(D,Z)k+ X*Yj[8i(BjZk + rj1Z1) +rki(DjZ)']

- X'Yj [8j (8 Zk + I'k,j Z') + I4j1(Di Z )t].

The terms in (D3Z)k, aiZ' and BjZ' vanish. Thus we get,,1[R(X, Y)Z]k = X'YjZ-(air - ajr ,n + r l jm - rjtl im).

Denote by Ri j the k'h component of R(8/8x', 8/8x) 0&. We have

Rj,,jj = &r - ajr + r,r - rk,,,r .

R,ij are the components of a (3,1)-tensor on 1, since after contraction withthree vectors we get the components of a vector R 4ZLX'YI = [R(X, Y)Z]k.

Parallel Transport. Geodesics 103

According to the definition, R(X, Y) _ -R(Y, X); thus R , = -R jkj,.Since [8/84 8/8z1] = 0, if we choose X = 8/8x' and Y = 8/exi in il, weget

a (a al az = , jz'&[(DjDjZ)k - (DjDoZ)k]&,k = R axi, azi

Parallel Transport. Geodesics

4.8. Definition. A vector field X is said to be parallel along a differentiablecurve C if its covariant derivative in the direction of the tangent vector toC is zero. Letting X(t) __ X(C(t)), we get

D fX(t) = (t)vjx(t) =*(t)

[,%..j(t) + r k(C(t))xk(t)l-a-s = o.

Thus X(t) is a parallel vector field along C if, in a local chart,

fi= 0.

dj(t) +Nik(C(t))X (t) Wi(t)

One verifies that if two vector fields are equal to each other at each pointof C, then one is parallel along C if and only if the other is. Thus, for avector field X to be parallel along C depends only on the values of X alongC. This is why we will talk about parallel vector fields along C, even if theyare defined only along C.

4.9. Definition. Let P and Q be two points of M, C(t) a differentiablecurve from P to Q (C(a) = P, C(b) = Q), and Xo a vector of Tp(M).According to Cauchy's theorem, the initial value problem X(a) = Xo of theequation above has a unique solution X (t) defined for all t E [a, bJ, sincethe equation is linear. The vector X (b) of this parallel vector field along C(with X (a) = Xo) is called the parallel translated vector of Xo from P to Qalong C.

The parallel translate along a piecewise differentiable curve C = ( 1 Ciis defined in a natural way. X1 is the parallel translated vector of X0 alongthe differentiable curve C1, and by induction X, is the parallel translatedvector of Xi_1 along the differentiable curve Cg. X. is called the paralleltranslated vector of X0 along C. The definition of the parallel translate of atensor along C is similar, once we define the covariant derivative of a tensorfield.

The parallel translate at Q of Xo along C is unique according to Cauchy'stheorem, but it depends on C. That is not the case in R. For instance, onS,,, consider a unit tangent vector Xo at the north pole P, and let C thefollowing piecewise differentiable curve: half meridian tangent to Xo, then


a piece of equator AB, and finally another half meridian back to P. Anygiven unit tangent vector at P may be the parallel translated vector of X0along C; it depends on B. It is possible to verify this result after Chapter 5(Exercises 5.36 and 5.51).

The parallel translated vector of Xn along C (resp. PA) is X3 (resp.X1); X2 is the parallel translated vector of Xn along AB.

4.10. Definition. A differentiable curve C(t) of class C2 is a geodesic if itsfield of tangent vectors is parallel along C(t). That is to say, Ddeldtw = 0.Writing in a local chart, we find that C(t) is a geodesic if and only if

d2C?(t) + rj. (C(t)) dC'(t) dCk(t) _ 0dt2 ik dt dt

This is the geodesic equation. It is obtained from the parallel translationequation (see 4.8) with X(t) = dC(t)/dt. But this equation is not linear-itis much more complicated. We will study it in detail in Chapter 5. It is anequation of the second order. According to Cauchy's theorem,

4.11. Proposition. Given P E M and X E Tp(M), there exists a uniquegeodesic, starting at P, such that X is its tangent vector at P. This geodesicdepends smoothly on the initial conditions at P and X. If X = 0, the geodesicis reduced to the point P.

4.12. Example. Connections on an open set 9 C R' (recall t:liat fl is amanifold). There are atlases with one chart, for instance (St, l.d). If on(fl, Id) we choose arbitrary Christoffel symbols, they define a connection.When there is more than one chart in the atlas, this procedure cannot beapplied according to Proposition 4.5.

If on (St, Id) we choose r = 0, the curvature vanishes. The par-allel transport is the usual one. Indeed, the equation (see 4..3) is then

Covariant Derivative 105

dXJ(t)/dt = 0, so the components of X(t) are constant. The geodesicequation is then d2Ci(t)/dt2 = 0, and so geodesics are straight lines (eachCi(t) is a linear function oft).

Covariant Derivative

4.13. Definition. The definition of covariant derivative extends to differ-entiable tensor fields as follows:

a) For functions, Dx f = X (f ).b) Dx preserves the type of the tensor.c) Dx (u ®v) = (Dxu) ®v + u ®(Dx v ), where u and v are tensor fields.

d) Dx commutes with the contraction.

Let us show how we compute the covariant derivative of a differential1-form w in a local chart. Let X = X48/0x', Y = Yj8/8x' and w = wkdxk.

According to (a)

DX(wkYk) = X 8,(WkYk),

according to (c)

Dx(w ®Y) _ (Dxw) ®Y + w 0 (DxY),

and according to (d)

Dx(wkYk) = Dx[W(Y)] = (Dxw) (Y) +w(DxY).Thus

(DXw)kYk = r(8iwk)Yk + X`WkO Yk - wk`V,Y't.

Setting X = 8/8x' leads to

(Diw)kYk = (aiWk)Yk - wkrijY9.

Interchanging the dummy indices k and j in the last term yields, since theequality is true for any Y,

(Diw)k = 8iwk -

By definition, Vw is the twice covariant tensor having components (Vw)ikwritten Viwk equal to (DiW)k:

ViWk = aiwk - 1NkWj.

In a similar way we compute the covariant derivative Vi of an (r, s)-tensorfield u.

The rule is the following: Viu is the sum of the partial derivative8iu, of a terms with 1;k where j is equal successively to the values of thes contravariant indices and k is a dummy index, and r terms with -N?


where j is a dummy index and k is equal successively to the values of the rcovariant indices.

4.14. Examples. Let g be a (2, 0)-tensor field. We have

Vi9jk = a9jk - rij9lk - IlkYjl

Let Rlkij be the components of the curvature tensor. Its covariancederivative is

VmRtkij = $mRlkij - W&kij + IkR4° - I'°Rtkaj - I'OmjRtk;Q.

4.15. Exercise. Prove the following formulas (commuting of the covariantderivatives):

andV,VjZk - VjVZk =1 {'Zi - T,jV,Zk

ViVjwk - VjViwk = -R1 jw1- lijVtwk.

The first formula looks strange when we notice that, by Definition 4.7,(D;DJZ)k - (DjD{Z)k = 1 ijZ'.

The difference comes from the fact that DjZ is a vector field whereasVZ is a (1,1)-tensor field,

(DiD?Z)k = &(DDZ)k + I'(DfZ)t,

whereasViVjZk = 8c(V,Zk)+IkVjZl - IjV,Zk.

Thus

V{VjZk - VVViZk = (DiDjZ)k - (DjDiZ)k - TijV1Zk,

and the first result follows. For the second result, we can compute A =V*Vj(Zkwk) -VjVi(Zkwk) in two different ways:

A = &; (Zkwk) - (f81(Zkwk) - ;i(Zkwk) + I'ji8t(ZkuJk)

= -7 jVt(Zkwk),since for functions Oi f = V 1 f . And if we develop A as

A = wk(V;VjZk -VjVjZk) + Zk(ViVjwk - VjViwk),

the second formula follows.

4.16. Proposition. identities are

nijk = (VAN - 7 '1 k)a(ij,k) o(ij,k)

and

E Vj "ki = E 7 j,o(iJ,k) a(i,i,k)


where Ea(i,j,k) means the sum on the circular permutations of i,j and k.

When the connection is torsion free, the Bianchi identities are simple.The first identity is E0(i,j,k) R;jk = 0 and the second Ea(i,J,k) OjR1% = 0.We will prove them in Chapter 5.


4.17. Exercise. Consider on R2 the connection defined by 1'11 = xlir12 = 1, r222 = 2x2, the other Christoffel symbols vanishing. Let c bethe arc on R2 defined by C'(t) = t, C2(t) = 0, t E (0, 1].

a) Compute the parallel translate along C of the tangent vector 8/8x2at 0.

b) Write the geodesic equation of this connection.

4.18. Exercise. Consider a differentiable manifold endowed with a con-nection.

a) Give, in local coordinates, the expression of D010 (dx') and of DXT,where T is a (2,1)-tensor field, by means of the Christoffel symbols.

b) Write the differential equation satisfied by a (1, 1)-tensor field parallelalong a differentiable curve C.

4.19. Problem. Let Mn (n > 1) be a C°° differentiable manifold endowedwith a linear connection (1' are the Christoffel symbols in a coordinatesystem).

a) We say that two vector fields that are never zero have the same di-rection, if at each point of M the vectors of these vector fields havethe same direction. So we define an equivalence relation in the set ofvector fields that do not vanish.

A differentiable curve C(s) being given, we say that the vectorfield X preserves a parallel direction along C if there exists in theequivalence class of X a parallel vector field along C.

Prove that this property is equivalent to the condition that

[4:1k(s) k dCi(s)1 8+l a;-] 8xkand X

have the same direction.Write the differential system satisfied by the curve C whose tan-

gent vectors preserve a parallel direction along C. Show that thesecurves C are geodesics up to a change of parameters.

b) We consider on M a second linear connection whose Christoffel sym-bols are and we set T j = ft -1' . Determine the Tk, so thatthe two connections define the same parallelism: for any differentiable


curve, if a vector field preserves a parallel direction for one connection,it does the same for the other.

Deduce from the above result that the necessary and sufficientcondition we are looking for is _ pjb , where the µ are the cony

83 .7

ponents of a differentiable 1-form.

c) Given a linear connection on M,,, is it possible to exhibit a connectionwithout torsion which defines the same parallelism?

4.20. Exercise. Let M be a differentiable manifold and let D, D be twoconnections on M. For two vector fields X and Y we set

A(X, Y) = DXY - DXY and B(X, Y) = A(X, Y) + A(Y, X ).

a) Show that a necessary and sufficient condition for D and b to havethe same geodesics is that B(X, Y) = 0.

b) Verify that D and D are identical if the two connections have thesame geodesics and torsion.

c) Prove that for D and b to have the same geodesics up to a changeof parameterization, it is necessary and sufficient that that A(X, X)be proportional to X.

d) In this case exhibit a I-form w such that

B(X, Y) = w(X)Y + w(Y)X.

Solutions to Exercises


a) Let X`(t) = X(C(t)) be the components of the parallel vector fieldX along C which is equal to C7/8x2 at 0. Then

1 9 1

+ r,k(C(t)) xk = +X IXI + X2 = 0,

dX2dt

0.

Thus X2 = 1 and (X 1), + x1X 1 + 1 = 0 with XI (0) = 0.The general solution is

r z

0

=

X1(t) = -e-cJ

e$du+ ke-4.

Also, X1(0) implies k = 0. So

r1 z

57X-1 9X2


b) We have1 1 1 2 1 dC2+C(dCd2

dt=0,2C2 + 2C2

(a)2

= o.dt2

Solution to Exercise 4.18.a) Delft: (w), with w = dx', is a differential 1-form. Also, w = wkdxk,

with w; = i and wk = 0 if k j. Thus Da18:.(&) = _r? dxkFinally,

DxT = Xi[8,T - rmTm; - rj,R ]dx= 0 dx' ® C)Xk

b) The equation is DdcldtZ = 0 with Z = ZJ dx` ®818xj. That is, forany (i,j),

Chapter 5

Riemannian Manifolds

In this chapter we will apply what we learned in the previous chapters tothe study of Riemannian manifolds. When we have to study a manifold,it is convenient to consider a Riemannian metric on it. We can do thiswithout loss of generality, since there exists a C°O Riemannian metric gon a C°O paracompact differentiable manifold M. We define a distance on(M,9), so (M, 9) is a metric space. Then we introduce normal coordinatesystems at a given point. In these systems the computations are easier. Wecontinue with the exponential mapping. Then we define some operators onthe differential forms, and we conclude with the proofs of some theorems,using global notation. There are a lot of exercises and problems. Most ofthem extend the course itself, which is confined to the main topics.

Some Definitions

S.I. Definition. A CO° Riemannian manifold of dimension n is a pair(M,,, g), where M. is a CO0 differentiable manifold and g a Riemannianmetric. A Riemannian metric is a twice-covariant tensor field g (that is tosay, a section of T*(M) ®T*(M)) such that, at each point P E M, gp is apositive definite bilinear symmetric form:

gp(X, Y) = gp(Y, X) for all X, Y,

gp(X,X)>0 if Xt0.

In the sequel, (Ms, g) will always be a connected C°° Riemannian mani-fold endowed with the Riemannian connection (Definition 5.5). If the mani-fold is not connected, we study each of its connected components separately.If the manifold is CI, we consider a C' atlas C'-equivalent to the C' atlas.

ill

112 5. Riernannian Manifolds

5.2. Theorem. On a paraonnpact C°° differentiable manifold M, there ex-ists a C°° Riemannian metric g.

Proof : Let (Vi,'pi),EJ be an atlas of M and {ai} a partition of unitysubordinate to the covering {Y }. This partition exists, since the manifoldis paracompact (Theorem 1.12). E = ((4k)) being the Euclidean metricon R" (in an orthonormal basis EJk = bJ ), we consider the tensor field9EEla+4(E)

Let us verify that g is a Riemannian metric on M. Indeed,

g(x, Y) = E aie(Spi.x, Vi*Y) = g(Y, X),iEl

according to the definition of the linear cotangent mapping and since C issymmetric.

Moreover, g(X, X) = iE1 0 if X 34 0.

Indeed, at a point P, at least one a, does not vanish. Let aio(P) # 0.Then X # 0 implies (+pia.) pX 96 0. since Vi,),, is invertible. Thus

gp(x,x) >_ ai'(P)0Piofx, .X) > 0.

By using the Withney Theorem 1.22, we can also define a Riemannianmetric on Mn: the imbedding metric. If h is an imbedding M - R2n+', theimbedding metric g is g = W. In a local chart (0, gyp) at a point P E Mlet us compute the components of g. If {xi} is the coordinate system on Rcorresponding to (12, cp), let {y°} (a = 1, 2, , 2n + 1) be coordinates in(R2n+1 e)

The imbedding h is defined on 11 by 2n + 1 C-functions

e(xi x2,... ,xn), a= 1,2,... ,2n+1.

The components of the imbedding metric tensor are given on Il by

ft- w 2n+'fta Slta

$xi 8xj _ E 8xi 8rjor-- I

S.S. Definition. The length of a differentiable curve C : R D [a, b] -+ Mnof lass C' is defined to be

L(C) _jb

gc(t) ( dt.

If a curve is piecewise differentiable, its length is the sum of the lengthsof the pieces.

Some Definitions 113

We verify that the definition makes sense. It does not depend on theatlas or on the parameterization. If t = t(s) with dt/ds > 0, then -y(s) _C(t(s)) and C(t) have the same length.

A connected manifold is path connected (Proposition 1.3). For givenP, Q in a C°O differentiable manifold M, there is a COO differentiable arcfrom P to Q. Indeed, the continuous path ry from P to Q is compact, andis coed by a finite set of local charts (St;, Bpi), i = 1, 2,- .. , m, such that9i(Sli) is a ball of R. Let Pe be in Sli (1 R-+ I. Instead of the are 7 from Pto Pl in il1i we consider a C°O differentiable are Cl in Sl1 from P to A, forinstance w where means the segment in R". And soon, until the are C. from P,n_1 to Q. We get a piecewise C°O differentiablearc from P to Q, the union of the arcs Ck (1 < k < m). Then it is possibleto smooth the corners at the points P, to obtain a COO differentiable arcfrom P to Q.

5.4. Theorem. Set d(P, Q) = inf ET-1 L(Cj), when the infimwn is over-all piecewise C1-diferentiable paths C from P to Q, C = LrQ1 Ci. Then(P, Q) -+ d(P, Q) defines a distance on Mn, and the topology determined bythis distance is equivalent to the initial topology on M.

Obviously d(P, Q) = d(Q, P) and d(P, Q) < d(P, R) + d(R, Q), sincethe inf of the length of the piecewise differentiable arcs from P to Q is lessthan or equal to the length of the piecewise differentiable arcs from P to Qthrough R. It remains to prove that d(P, Q) = 0 implies P = Q; we do itby contradiction.

Let (S2, gyp) be a local chart at P with q(P)_= 0. If Q 54 P, there is a ballBr C R", of radius r with center 0, such that Br C V(fl) and Q f tp-1(Br).

Let us consider the map rY defined by

E-1 x 3 (t,x) E R,

where F"_1 is the unit sphere of dimension n - 1 and radius 1 in R", andW;' (e). Since 34 0, it follows that g.(1;, 1:) > 0.

Moreover, 0 is continuous (in fact C) and K = x is acompact set; thus 4' achieves its minimum A2 and its maximum µ2 on Kwhich satisfy 0 < A < p < oo and, for any f E R" and x E

A2 If MI2 <_ µ2IKII2

Qbeing the Euclidean norm.Let r be the connected component of P on the are C from P to Q in

The end points oft are P and R, C(a) = P, C(b) = R. We have

L(G')2L(r)°r. 9c" ( 4f ')dt Aj 11d( dtlldt>Ar.

114

Indeed,

5. Riemannian Manifolds

dC _ d(V oC)dt dt

and the length of V(1:) in &k" is at least r. Thus d(P, Q) > 0 if P # Q.Setting Sr = {Q E M I d(P, Q) E r}, we have, according to the proof

above, SA,. C p-1(Br). Likewise we prove that. 1(90 C SNr Indeed, wehave

d(P, R) < L(1') < µ b I) d( dt C) II dt

for any arc r from P to R. Picking V o C to be the straight lint from 0 toV(R), we find that d(P, R) < µr. Hence the topology defined by the metricis the initial topology.

Riemannian Connection

5.5. Definition. The Riemannian connection is the unique connectionwith vanishing torsion tensor, for which the covariant derivative of the metrictensor is zero (Vg = 0).

Let us compute the expression of the Christoffel symbols in a local coor-dinate system. The computation gives a proof of the existence aad unique-ness of the Riemannian connection.

The connection has no torsion; thus r 1 = r i. Moreover, let us writethat Vk9i; = Vi9;k = V;9ki = 0:

ak9i3 - rki91; - Iik;9, = 0,t

a;9ki - rjk91, - r,9k, = 0.

Riemannian Connection 115

Taking the sum of the last two equalities minus the first one, we find that

Ili = 2 [8igkj + 8j9ki - 8k9ijI9M,

where the gke are, by definition, the components of the inverse matrix of thematrix ((gii)) : giigkj = 6 (the Kronecker symbol).

5.6. With the metric tensor of components gij, the indices go down by con-traction. For example, if 4j (1 < j < n) are the components of a vectorfield, then ai = giji are the components of a 1-form. If Rj are the com-ponents of the curvature tensor (see below), then Rklij = gk,,,R7ji are thecomponents of the 4-covariant curvature tensor.

For the tensor of components gii, the indices go up by contraction. Forexample, if ai are the components of a 1-form, then i = giiai are thecomponents of a vector, and R1'i = gk.n

Rkii 'j.

5.7. Definition. A normal coordinate system at P E M,, is a local coor-dinate system for which the components of the metric tensor at P satisfygi (P) = 6,1 and 8kgij(P) = 0 for all i, j, k. (4gii(P) = 0 is equivalent to1'h(P) = 0; see the proof above in 5.5.)

Let us prove that at each point P, there exists a normal coordinatesystem. Let (0, W) be a local chart at P with cp(P) = 0, and {x} thecorresponding coordinate system. First, by a linear transformation, we maychoose a frame in R" so that 9,j(P) = 4. Then consider the change ofcoordinates defined by

xk-yk=-1F (P)yiy12 %J

{yi} is a coordinate system in a neighbourhood of P according to the inversefunction theorem. Since the Jacobian matrix ((8xk/8yi)) is invertible at P,it is the identity matrix.

In the coordinate system {yi}, the components of the metric tensor are

iii(y) = 9ki(x)[ak -

since 8xk/8yi = &k - q p)%'. x and y represent the same point Q in a93

neighbourhood of P. The first order term in y°L of 9ij(y) - 9ii(x) is

-[1`'im(P)+1m(P)lym=-(a2M)pym -( )Pym.

Hence (8gii /8y'") p = 0 and all the Christoffel symbols t(P) are zero.The value of a normal coordinate system will be obvious when we consider,for instance, the proof of the Bianchi identities.

116 5. Riemannian Manifolds

5.8. The curvature tensor.

Consider the 4-covariant tensor R(X, Y, Z, T) = g[R(X, Y)T, Z]; itscomponents are Rijw = gi,,,Rm . It has the properties Rijjd = -Rijlk (bydefinition), Rijkl = Rklij, and

Rijkl + Riljk + &U., = 0,VmRijkl + VkRijtm + VIRij,,,k = 0.

The last two equalities are the Bianchi identities. Let us prove them. Con-sider at P a normal coordinate system. According to the expression of thecomponents of the curvature tensor,

I4 (P) _ (8ejk)P - (8 L'tik)P,Rijk(P) = (Sjl1ki)P - (8kfji)P,

Rjki(P) _ (8kl ij)P - (8irkj)P.Taking the sum of these three equalities, we get zero. This is the first Bianchiidentity (a tensor equality is proved when it is proved in some coordinatesystem). Differentiating the components of the curvature tensor at P, wefind that

(VmRjkl)P = ((8kws j)P - (BImI"kj)P,

(VkRjjlm)P = (8tk mj)P - (8mk1' )P,

(VIRmj,,,k)P = (8lmAkj)P - (8lk1"m,)P

Taking the sum of these three equalities, we find that

VmRjkl + Vkim + VI1r,mk = 0,which is the second Bianchi identity. We put it into the form written aboveby multiplying it by gih. Since Vg = 0,

9ihVmRjkl = Vm(gihRfk!) = VmRhjkl

So we find that

VmRijkl + VkRijlm + VIRij,,,k = 0.

The equality remaining to be proved is R{jW = Rjjj. In a normal coordinatesystem at P we have

Rklij(P) =2

[Sjkga + 8it9jk - 8ki9jl - 8jl9ik]P.

In this expression we see that Rijkl = R. Let {tk} be the components ofa vector field, and recall the following equality (see 4.15 with zero torsion):

Vivitk - V,Vi£k = Ri{jS.

Exponential Mapping 117

5.9. Sectional curvature. Ricci curvature. Scalar curvature.

By definition, o ,(X, Y) = R(X, Y, X, Y) is the sectional curvature of the2-dimensional subspace of T(M) defined by the vectors X and Y, X andY being orthonormal (i.e., g(X, X) = 1, g(Y, Y) = 1, and g(X, Y) = 0).Otherwise o(X,Y) = R(X, Y, X, Y)/[g(X, X)g(Y, Y) - (g(X, y))2].

From the curvature tensor is it possible to obtain, by contraction, othertensors? We obtain only one nonzero tensor (or its negative); it is called theRicci tensor. Its components are Rij = Rkk_,. The Ricci tensor is symmetric:

Rtj = gkI Rtikj = g"Rkjti = gtkRkjl, = Rj,

according to the properties of the metric and curvature tensors. The Riccicurvature in the direction of the unit tangent vector X = {e} is RijThe contraction of the Ricci tensor R = Rijgij is called the scalar curvature.

Exponential Mapping

5.10. Let (U, gyp) be a local chart related to a normal coordinate system {x'}at P; we suppose W(P) = 0. Also, let X = {Ci} 54 0 be a tangent vectorof Tp(M). Let C'(t) be the coordinates of the point C(t) belonging to thegeodesic defined by the initial conditions C(O) = P and (dC/dt)t=o = X.C(t) is defined for the values oft satisfying 0 < t < 0 (i3 given by the Cauchytheorem).

Then the quantity

gij(C(t))WidC- (t) dC (t)

is constant along C. Indeed, the covariant derivative along C of each of thethree terms is zero: Dg = 0 and Dd l 4 = 0. Hence s, the parameter ofare length, is proportional to t (s = IIXllt):

gij(C(t))Wi(t)

(t) =gij(C(0))(dC*(t))

i=o

(dCdtt)Jt -11X112.

The C'(t) are C°° functions not only of t, but also of the initial conditions.We may consider C'(t, Q, X) = C`(t, xl, x2, ... , xn, £1, ... , V), where the{x'} are the coordinates of Q E U and {C'} the components of X in thebasis {8/8x'}. According to the Cauchy Theorem 0.37, /3 may be chosenvalid for initial conditions in an open set, for instance for Q E V- I (B,.) andIIXII < a (B,. C v(U) being a ball centered at 0 of radius r > 0 and a > 0).As we will consider only geodesics C(t, P, X) starting from P, we will writeC(t, X) for C(t, P, X).


Let us verify that C(t, AX) = C(At, X) for all A, when one of the twonumbers exists. Set -y(u) = C(Au, X). Then

dry _ A- and d2ry A2 d2Cdu dt du2 dt2

Thus, since C is a geodesic, ^y satisfies the geodesic equation (4.10). ry is ageodesic starting from P such that (dry/du)u, o = AX; it is C(u, AX). Hencein all cases, if a is small enough, we may assume /3 > 1 without loss ofgenerality.

5.11. Theorem. The exponential mapping expp(X), defined by

R"DO X - C(1,P,X)EMn,is a difeomorphism of 0 (a neighbourhood of zero, where the mapping isdefined) onto a neighbourhood fl of P, f = expp 0. By definition, expp(O) _P, and the identification of R" with Tp(M) is made by means of gyp,:

X = (W.1)pX.

Proof. We saw in 5.10 that expp(X) is a COO map from a neighbourhoodof 0 E R" into M" (Q may chosen greater than 1). At 0 the Jacobian of thismap is 1; then, according to the inverse function theorem, the exponentialmapping is locally a diffeomorphism (on 0): f1, f2'... , £n can be expressedas functions of C', C2, . , C". To compute the Jacobian matrix at 0, wecompute the derivative in a given direction X:

(fexPpAx) _ (c(1.xt)) _ (c(..t))A=O a=o A=O

=X=(rp;l)pX.

Let Bo(r(P)) be the greatest ball with center 0 and radius r(P) in 0.r(P) is called the injectivity radius at P, and r; = infpEM r(P) is called theigjectivity radius of the manifold.

5.12. Corollary. There exists a neighbourhood f2 of P such that every pointQ E fl can be joined to P by a unique geodesic entirely included in f2.(ft, expel) is a local chart, and the corresponding coordinate system is calleda normal geodesic coordinate system.

It remains to be proved that this coordinate system is normal at P,since by 5.11 if we have Q (that is to say C', C2, , C"), then we havet 1, e, , r the components of k such that C(1, X) = Q. In this newcoordinate system, {!;'} are the coordinates of Q E Q.

Let C(t) _ {C`(t)} be the geodesic from P to Q lying in Q. We verifythat C'(t) = t k' for t E [0,1]-this comes from the equality C(t, X) _C(1, tX ).

Exponential Mapping 119

Q_C(1,k)-a,prx-(ti)

M-C(t,k)-C(1,tk)=.xpPix -(t')

k

P

The arc length s = II X IIt = II X 11 t, son

s9ij(Q)tto = F(S )2 = IIX112,i=1

and the length of the geodesic from P to Q is IIXII.Since C(t) satisfies the geodesic equation (4.10), d2Ck/dt2 = 0 implies

1'tJ[C(t)]C'41 = 0. Letting t - 0 gives 1' (P)s:'t:3 = 0 for all (el. ThenI,=I%f implies 1'k, (P)=0 foralli,j,k.

5.13. Proposition. Every geodesic starting from P is perpendicular toEp(r), the image by expp of the sphere of center 0 and radius r in Rnwhen r is small enough.

Ep(r) Is the subset of the points Q E f2 satisfying 1(f`)2 = r2, wheret"' are the geodesic coordinates of Q.

Choose an orthonormal frame of RI such that the geodesic coordinatesof Q E Ep(r) are CI = r and t;2 = t;3 = _ = 0. The desired result willbe established if we prove that glt(Q) = bl for all i, because a vector Y inTQ(M) tangent to Ep(r) has a zero first component:

n j( L C 'if gl,(Q) = 0 for i > 1.

1-2

Indeed, if y(u) is a differentiable curve in Ep(r) through Q, then we haveELl[, {(u))2 = r3 and E7_1 y'(u)d7'(u)/du = 0 by differentiation. Thus,since y(u) = 0 at Q for i > 1, that implies dyl(u)/du = 0 at Q.

We saw that g j(t;)t;'t;j _ 3 (i)2 t; = {t;'}. At Q that gives gii(Q)= 1, and differentiation with respect to yields

8kgi., (Q)eCj + 29gk(Q)C' = 24k.

Hence, at Q, if k 11,

r8k911(r) + 291k(r) = 0,

120 5. Riemannian Manifolcks

where g,j(r) are the components of g at the point with coordinates £1 = r,e = 0 for i > 1. Moreover, I jj(r)E = 0 leads to 28rglk(r) = Okgll(r).Thus we get glk(r) +r8,.glk(r) = 0; that is, N[rglk(r)] = 0 and rglk(r) isconstant along the geodesic from P to Q. It is zero at P; hence glk(Q) = 0fork96 1.

5.14. Theorem. A C2 differentiable curve which minimizes the distanceis a geodesic. If HX N is small enough, the distance from P to Q = expp Xis IIXII, which is the length of the unique geodesic from P to Q given byCorollary 5.12.

Proof. Let C be a minimizing C2 differentiable curve from P to Q parame-trized by the arc length t ([0, r] 3 t -' C(t) E C), which lies in a chart(R, cp), and let {x'} be the associated coordinate system. We will prove thatit is a geodesic. Consider a family {Ca} of C2 differentiable curves fromP to Q (Ca C 11 for A E ] - e, c[), defined by Ca(t) = &(t) + V(t) withC'(0) = Ci(r) = 0 for all i, t fi(t) being Coo functions on [0,r].

Since the quantity

r ( ' )dt0

attains its minimum at A = 0, we have

(dLdAa)) - 2 f [+2g,1(C(t)) -x-o= 0.

As we have

d dCJ 11 dCk dCjt [9<j(C(t)) & } = &2 dt dt

integration by parts of the second term gives

dC'2

28k9t4(C(t))]dt - 2gct(C(t))

Pd2C az

} dt =0

for all t vanishing at 0 and T. Hence we obtain

i dC*dt

+F k(C(t))dt dt =0.

This is the Euler equation of the problem: "Minimize L(r) for all C2differentiable curves I' from P to Q." It is also the geodesic equation (4.10).Thus C is a geodesic.

Some Operators on Differential Forms 121

We choose V = expel. If the geodesic coordinates of Q are {l;' }, we sawthat the equation of the geodesic from P to Q is {C' (t) = t(.' for 1 < i < n}(t E [0,1]), and its length is r = i=1(Z;')

Consider ry : [0,1] 3 t -y(t) E M, a differentiable curve from P to Qlying in 12 (-y(0) = P, ry(1) = Q). Let us prove that the length of y is greaterthan or equal to r. If (p,0) are geodesic polar coordinates (0 E thesphere of dimension n - 1 and radius 1), Proposition 5.13 implies that

with p(-I(t)) = i=1(ry=(t)) . Thus

10'r

d'y' d7' j1 dp(7(t))L(7) = V9., f7(t)] dt dt

dt > dt = p(7(1)) = r.

Hence d(P,Q), the distance from P to Q, is equal to r, the length of thegeodesic from P to Q.

Some Operators on Differential Forms

5.15. Definition. Let (M,,, g) be an oriented Riemannian manifold and Aan atlas compatible with the orientation (all changes of charts have positiveJacobian). In the coordinate system {x'} corresponding to (S2, gyp) E A, thedifferential n-form q is by definition

q = 19(x)Idxl A dxz A ... A dxn,

where Ig(x)I is the determinant of the metric matrix ((9,J)). q is a globaldifferentiable n-form, called an oriented volume n -farm, and it is nowherezero.

Indeed, in another chart (0, E A such that snit # 0, with coordinatesystem {y°}, consider the differentiable n-form it = I9(y)Idyl A dy2 A . Adyn.

On f2 n 0, let us consider A? = 8y°/8x', Bp = 8x)/8ya. The Jacobianmatrix A = ((A°)) E GL(Rn)+, the subgroup of GL(Rn) consisting of thosematrices A for which det A = I AI > 0. We set B = ((Bk)). On f2 n 0 wehave g.0(y) = and hence I9(y)I = IBI2I9(x)I. Moreover,

That yields ij = I BI (9(x)IIAIdx1 A dxz A ... A dxn. Since I B I > 0 andIAIIBI= 1, we have ,' on fin 0.

On an oriented Riemannian manifold Mn, we define the following oper-ators on the differentiable forms.

122 5. Riermannian Maoibide

5.16. Adjolnt operator *. The operator * associates to a p-form a an(n-p)-form *a, called the acyoint of a, defined as Mows. In a chart (52, p),the components of *a are

(*a)v+= 'Pp

e

1 ag u.t .. g p"

1,A2,... ,ay

See in 5.6 how the indices go up and down. Let us verify that

*1=17, *,7=1, **a = (-1)a("-P)a

and

a A (*0) = (C"MV,

where 3 is a p-form; here (a, $9) denotes the scalar product of a and $9:

(a,3)

We will prove the formula above at a point P. We choose a system ofnormal coordinates at P. Thus t7(P) = dxl A dx2 A ... A dx".

Since * is linear, we can suppose without loss of generality thata = Q1.2,--- ,p&' A dx2 A .. A dxp. *a has only one component which isnot zero, namely (*a(P))(a+i),(a+2),...,n = Also,

(* * a(P))1,2,...,p = rl(p+1),(F}2),..' ",1,2,... y(*o(P))(p+l),(P+2),...,n

= (-1"-p)(a(P))1,2,...,p.

Since the exterior product is bilinear, we can suppose thata = a1,2,... ,pdx1 A dx2 A .. A dxP. Thus

(a A (*$))p = (a(P))1,2; ..,pd 1 A dx2 A ... A dxeA[($(P))1.2,...,pdxx+1 A dx"+2 A ... A ds"l

_ (a,fA)prl

Note that the adjoint operator is an isomorphism between the spaces AP(M)and A--P(M).

5.17. Co-differential J. Let a E AP(M). We define Sa by its componentsin a local chart as follows:

On functions, S vanishes: b f = 0. For the definition of 6, the manifold neednot be orientable.

Some Operators on Differential Fbrms 123

The differential (p -1)-form is called the co-dj femntial of a and has thefollowing properties: on the p-forms

6 = (_1)P *-1 d*, 66 = 0.

The last result is a consequence of the first: bd = - *-1 dd* = 0.Th prove the first result at P, we choose a system of normal coordinates

at P. Then

1d(*a)] ,a,. 1 1fpJ+P+1: ,v,. ,,...,lAP 8

In9'._

`` V1,... , v9 isa permutation of pi, I3, N; according to theµt,lrsr~ ,I+.T...

sip of the permutation f"'P9 = ±1. We saw that * 1 = 1(P-1)(n-p+i)*

It1.... JAq { )

on an (n - p + 1)-form; hence(-1)P+(p-1)(n-p+l)

(&t)n,...,Vj1 - p!(n -J" p)r(n - p + 1)!XfPPP-rl,...,An 1'-%AP

Since

and

714. ,).ft :.. dtn - ,W,y,,... ,VP-1

Mti..,I8P Jy. 1i...,l.n - 1(n-p)(P-1)fl+lr...,l'yI µP+1,' Jf I - PVl *,,,& -1,

it follows that

eov .~

5.18. The Laplacian operator 0 is defined by

0 = d5 + 8d.

If a is a differential p-form, then Aa is a differential p-form and theLaplacian commutes with the adjoint operator: A* = *A.

Indeed, on differential p-forms

*dd = (-1)P+'d*d= (-1)P+i+P(n-P)d*d**

_ (-1)l+p(n-p+l}d[{-1}(p+1)('a-P-i) * ' d*] *

_ db * .

For a function f ,

of = sdf = -vvv,,f,a is said to be closed if da = 0, co-dosed if Sa = 0, and harmonic if Lea = 0.a is said to be enact or homologous to zero if there exists a differential formf such that a = dfi. a is said to be co-exact or cohomologous to zero if


there exists a differential form y such that a = by. Two differential p-formsare homologous if their difference is exact.

5.19. Global scalar product. On a compact orientable Riemannian man-ifold, the global scalar product (a, /3) of two differential p-forms a and /3 isdefined as follows:

(a,13) = fM(a,A)n-

Recall that (a, 8) = v '4. The name of the operatorb comes from the formula

(da,y) = (a, 6-y)

for all differential (p + 1)-forms y. 6 is the adjoint operator of d for theglobal scalar product. Let us prove this formula. According to 5.16

f(aM)i= f aA*by=(-1)x''1 faAd*y.

But d(a A *y) = da A *y + (-1)Pa A d * y, and so

(a, by) = fda A *-y - f d(a A *i) = (da, y)

according to Stokes' formula. Using this result yields

(Aa, /3) = (ba, b$) + (da, d,3).

A is an elliptic self-adjoint differential operator. If f E C2(M), then

(Af,f) = (4,df) = r(MVLfV-f'7.

5.20. Proposition. On a compact oriented Riemannian manifold M anyharmonic form is closed and co-closed. A harmonic function is constant.

This follows from the equality above with 0 = a:

(Da, a) = (Sa, be) + (da, da).

Da=0implies be=0andda=0. Ifba=0andda=0, then &a=0by definition. A harmonic function satisfies df = 0; thus the function f isconstant.

5.21. Remark. If M is not orientable, the statement remains true. Weconsider a two-sheeted covering manifold !tit of M such that M is orientableOr : V --. V), and g = x*g (Theorem 2.30). We work on a = ,r*a, which isharmonic. Wehave da=0andba=0;thus da=0andba=0.

Spectrum of a Manifold 125

5.22. Hodge Decomposition Theorem. Let M,, be a compact and ori-entable Riemannian manifold. A differential p-form a may be decomposedinto the scam of three differential p forms:

da = dA + Sµ + Ha.

where Ha is a harmonic differential p-form, dA is exact and 8p co-exact.The decom;x9sition is unique.

For the proof, see de R.harn [111.

Uniqueness comes from the orthogonality of the three spaces for theglobal scalar product:

(a, a) = (dA, dA) + (tp, bp) + (Ha, Ha).

The dimension by of the space Hp(M,,) of harmonic p-forms is called theih Betti number of M. It is finite.

Since A* = *0 (see 5.18), * defines an isomorphism between the spacesH,(M.) and HR-p(Mn). Hence bp(MM) = bn-p(Mn). The number

XWO = E(-1)pbp(Mn)P=O

is called the Euler-Poincarl characteristic. Clearly, bn(Mn) = 1(Proposition 5.20). H0(M,,) is the space of constant functions and Hn(Mn)is the space of differential n-forms proportional to the oriented volume n-fiorm rl. Indeed, r! is harmonic since q = *1; or we can see directly thatdrl = 0 and bq = 0. There is no nonzero (n + 1)-form, and, since Vg = 0,

- I E1,2,--. ,n V" V 191 = o.

For a connected manifold, bn = 1 or bn = 0 according to whether it isorientable or not. In general, by is the number of connected components.

Spectrum of a Manifold

5.23. The Lebesgue Integral. Let (Mn, g) be a Riemannian manifoldand (11, cp) a local chart, with {xi} the associated coordinate system. For acontinuous function f on M with cco;compact support lying in It, we set

.

JM fdV= 191f) o V-1dx1dx2 ... dx".

For a continuous function f on M with compact support, we set

f fdv = J aifdV,M iEI M

126 5. Rienannian Manifolds

where {aj}iE1 is a partition of unity subordinate to the covering {Ri}m,(IZj,w,)iE/ being an atlas.

In the sum, only a finite number of terms are nonzero, since K fl supp arjis empty except for a finite set of indices i if K is compact.

We have to prove that this definition makes sense. It depends neitheron the local chart nor on the partition of unity. Indeed, let (0, tai) be anotherchart, {ya} the associated coordinate system. Suppose supp f c SA fl 8; setas usual i/8x' = A? and 8xj/8yJ = B'jq (see 5.15). Then

( Ie(x)If) o w-ldxldx' ... dzn

4PrA

( Ig(y)1f)o0-'dyldy2... ^,

since Ig(v)I = IB1alg(x)I and dy'dy2...dy"' = IIAiIdx'dx2...dx".

Consider another atlas {8j, Oj } jEJ and a partition of unity {,3j }jEJ sub-ordinate to the covering (8j}jEJ. We have

ail = JM(at[Nj)dV OjfdV,iE1 M iEt JEJ jEJ M

since the sums are finite.

Hence f - fm fdV defines a positive Radon measure, and the theory ofthe Lebesgue integral can be applied. We call dV = (x) I dx'dx2 ... dx"the Riemannian volume element.

5.24. Proposition. Let M,, be a compact Riemannian manifold, and w adiferential 1 form. Then fjW bwdV = 0. In particular, if f E C2(M), thenfmAfdV =0.

If M is orientable, by choosing the correct orientation we have

IMdwdV = f &ori = (6w, 1) = (w, dl) = 0.

If M is nonorientable, we consider an orientable two-sheeted Riemanniancovering manifold M of M (Theorem 2.30). Let 7r be the covering mapM-+ M, and letg=7r*gandw=7r*w.

Since M is orientable, fM &JdV = 0. !Moreover,

I &"adV = 2 J & .'dV.M

Spectrum of a Manifold 127

Indeed, let {ai}1<i<m be a partition of unity subordinate to the covering{f2P}PEM, 1p being a neighbourhood of P such that r-1(Up) is diffeomor-phic to ftp x F, where F consists of twofpoints. We have

f 2 J b(a;w)dV,M M

whence the result follows. If f E C2(M), then fm 8(df )dV = 0.

5.25. Theorem. Let M be a compact Riemannian manifold with strictlypositive Ricci curvature. Then bl(M), the first Betti number of M, is zero.

Let a be a harmonic 1-form. Then, by Proposition 5.20 and Remark5.21,da=0 and &a=0. That is to say, we have Ojaj=Ojajand viai=0in a local coordinate system where a = aidxl. But C7iaj = Ojai impliesviaj=Aaj--rilak=aai-rjiak=V;ai

Contracting the equality at the end of 5.8 (i = k) with l k = ak gives

(*) Rijai = V (Vja`) - vj(viai).

Then multiplying by or) and d integrating over M lead to

IM JRljaWdV = / V [&(V ai)]dV -1M (Viai)(O1a`)dV.

since via; = 0. Moreover, according to Proposition 5.24,

[aj(Vja)Jd'V = 0.IM

V1

Hence, if a does not vanish everywhere, the left hand side will be strictlypositive, while the right hand side is < 0, since Viaj = V jai.

5.26. Let M be a compact Riemannian manifold. The spectrum of M isSS(M) = {A E R J there exists f E C2(M), f 0- 0, satisfying Af = Af}.

A is called an eigenvalue of the Laplacian, and f an eigenfunction. IfA E SS(M), then A > 0, because

Al f2dV=fofdV=V,JVVdV>0.M 'Al IM

The eigenvalues of the Laplacian form an infinite sequence 0 = Ao <Al < A2 < - . - going to +oo. And for each eigenvalue Ai, the set of corre-sponding eigenfunctions forms a vector space of finite dimension (Fredholm'stheorem). Fbr AO the vector space is one-dimensional.

5.27. Lichnerowics's theorem. Let M be a compact Riemannian man-ifoid. If its Ricci curvature is greater than or equal to k > 0, then Al >nk/(n - 1).


Let f be an eigenfunction: A f = A f with A > 0. Multiplying formula(*) used in 5.25, with a = c(f,, by Vj f , and integrating over Mn lead to

J RijV`fVJfdV=1M VifV`(VjVif)dV+A JM VJfVjfdV.f

According to the hypothesis, the 2-tensor Rij - kgij is non-negative. ThusRij V' f Vi f > kV' f Vi f . Integrating the second integral by parts givesjf(if)=

f JMV`[V'fV3Vif]dV - ju V'VJ fvjVifdV

= - / VVJJ .Vjf r

M

thanks to Proposition 5.24 and the equality Vi V i f = V j V i f . At once weget A > k. But we have more. Since

(ViVif + Af94i/n)(V4VJf +,Afg'J/n) > 0,it follows that fV1VjfV'Vjf > (.f)'/n, becauseg;ig#j =n. Hence

al V,ffVifdV -1

fAfdV >_ k f VfVafdV.Af n M M

Since

V'(fVjf)dV+I V'fVifdVIM ffdV=-1M Ar

_ ( V'fVifdV 96 0,M

the result follows.

5.28. Remark. The inequality of Theorem 5.27 is the best possible. In-deed, for the sphere of dimension n and radius 1, the sectional curva-ture is 1. Thus the Ricci curvature is n -1. If k = n -1, we get Aj > n, andwe will see that n is an eigenvalue for the Laplacian on Sn(l) (see Exercise5.36 or Problem 5.38).


5.29. Exercise. An Einstein metric is a metric for which the Ricci tensorand the metric tensor are proportional at each point.

If the metric is Einstein, prove that the scalar curvature R of the Rie-mannian manifold is constant when the dimension n > 2.

5.30. Exercise. Let (M,,, g) be a compact, oriented Riemannian manifoldof dimension n > 1. When X is a C°° vector field on Mn, we associate toX the 1-form a such that a(Y) = g(X, Y) for all vectors Y E T(M).

a) For an oriented volume n-form rt, express Lxq in terms of ba and g.


b) In a local coordinate system, express the components of Aa by meansof the components a; of a, VjVja,, and the components of the Riccitensor.

c) Show that a necessary and sufficient condition for LXg = 0 is thatViaj+Vjaj=0 for all (i,j).

d) Prove that Gxg = 0 if and only if dda = 0 and Act = 2RjXjdx',where the X? are the components of X.

e) Let G be the one parameter local group of local diffeamorphismsVt(X) associated to X.

Verify thatDeduce from this equality that pig = g if LXg = 0.

f) Let w be a harmonic p-form. Show that Exw = 0 if GXg = 0.Hint. If Gxg = 0, then £x6 = 6LX.

5.31. Problem. Let (M,,, g) be a C°° Riemannian manifold with zero cur-vature (Ri;M = 0).

Prove that there exists an atlas whose local charts ((I, cp) are such thatW*E = g, where E is the Euclidean metric on R'2.

Hint. Let (9,10) be a local chart at P and {x'} the corresponding coor-dinate system. We have to exhibit a local chart (Il, w) at P with coordinatesystem {ya} such that V*E = g. On 9 set 41 = I' ikjdx', and in a neighbour-hood of P set A? = 8y°/8x'.

a) Express d4 in terms of 4.

b) Which system of differential equations must the 4 satisfy?

c) Represent this system as a Pfaff system of order n 2 on a space ofdimension p (p to be specified) and use the Frobenius theorem.

d) Solve the problem.

5.32. Problem. Let M be a C°° submanifold of RP endowed with theEuclidean metric 6, and let b be the Riemannian connection on (RP, E).Denote by 4' the inclusion M C RP, and consider on M. the imbeddingmetric g = 4*E. Let {x'} be the coordinate system in a neighbourhood ofQ E M,, associated with the chart (11, gyp), and {ya} the coordinates on RP.

a) Recall the expression of the components gij (x) of g as functions ofthe coordinates ya(x) of the point 4'(x), x E Q.

By 4'* we identify a tangent vector of M to a tangent vector ofRP (T(M,) C T(RP)). Let Y C T(RP) be a vector field defined onM,,, and let X E

Show that DXY is well defined.


b) Let irQ be the orthogonal projection of vectors of TQ(RP) ontoTQ(M,,). If Y C is a vector field on M,,, prove that DxY =1rQDXY defines a connection on M which is the Riemannian con-nection of (M,,, g).

c) For two vector fields X and Y on M,,, we set H(X, Y) = DXY-DXY.Verify that H(X,Y) = H(Y,X).

d) If p = n + 1 and if M,t is orientable, show that there exists on 4b(M" )a differentiable vector field N C T(R"+i) of unit Euclidean norm,orthogonal to TQ(M,,) at each point Q E M,,. In this case verify thatwe can write H(X, Y) = h(X, Y)N, h being a symmetric bilinearform satisfying h(X, Y) = -E(Y, DXN).

5.33. Problem. Let (M", g) be a connected and complete COD Riemannianmanifold of sectional curvature greater than or equal to k2 > 0. We considera closed geodesic y; that is, y is a differentiable map of the circle into M.We assume the following result: any pair (P, Q) of points of M can be joinedby a geodesic arc whose length is equal to d(P, Q).

a) Let x f y be a point of M. and set d(x, y) = infy d(x, y).Show that there is a point yo of -y with yo = y(uo), such that

d(x, 4Jo) = d(x, y). We set d(x,7) = a.

b) Let us consider a geodesic C from x to yo of length L(C) = a. Let[0, a] 3 t , C(t) E M with C(O) = x and C(a) = yo. Do we haved(z, yo) = d(z, y) when z E C?

Ebwill be thesetofpoints QEMoftheform Q=C(1,z,X)(see 5.10 for the notations) with g,(X,X) = b2. If b is small enough,prove that d(z, Q) = b when Q E Eb.

c) Show that Y = is orthogonal to (dC/dt)t--a.Hint. Consider a local chart (St, exp T') at z E C close enough to

yo so that yoE0.d) Let e(t) be the parallel vector field along C such that e(a) = Y. Verify

that 9C(c)(e(t), ) = 0 and that gC(e)(e(t), e(t)) = g ,,(Y, Y).

e) We consider the family of curves Ca defined for A E ] - e, e[ (e > 0small) by

[0, a] 3t-'C,,(t)=C (Asin C(t)ie(t) J.

Verify that Ca is a differentiable curve with endpoints x and a pointy(ul) of y. What is the value of ul?

f) We will admit the existence of a neighbourhood 9 of C and of a localchart (9, cp) such that the corresponding coordinate system is normal


at any point of C. We set f (A) = L(CA). Prove that

f (0) = T. gciii(v(t), )dt

for some vector field v. Verify that f'(0) = 0.

g) From the fact that f (A) is minimum at A = 0, deduce that d(x, y) <x/2k.

5.34. Exercise. Let (M, 9) be a complete C°° Riemannian manifold andC a geodesic through P: [0, a[ 9 t --r C(t) with C(0) = P.

a) For an increasing sequence (t, } c R converging to a, show that thesequence C(tj) converges to a point Q E M.

b) Consider the map expQ. It is a diffeomorphism of a ball 0 = Bo(r) inR' onto a neighbourhood of Q in M. Verify that there exists an iosuch that C(t,) E 11 = expQ 0 for i > io, and prove that the geodesicC from P to Q can be extended beyond Q (for t E [a, a + e) withf > 0).

c) Deduce from b) that on a complete Riemannian manifold all geodesicsare infinitely extendable.

6.35. Exercise. Let (M,,, g) be a compact and oriented C°° Riemannianmanifold.

a) Let ry E AP(M). Show that as = -y has a solution Cr E AP(M) if andonly if y E Ay, the set of the differential p-forms which are orthogonalto ff. for the global scalar product. Here H. is the set of harmonicdifferential p-forms.

b) Exhibit a map G : Ap -, A. such that AG = GA = Identity on A,,.G is defined for each p (0 p < n).

c) Verify that dG = Gd, OG = Gb and G* = *G.d) For two closed differential forms y E AP(M) and E A P(M), we

denote by y (respectively ) the set of differential p-forms homologousto y (respectively 0). When y' E y and 0' E 0, show that f y' A 0'depends only on y' and ¢. Let (y, O) = f -y A 0.

e) Which harmonic forms p satisfy (cp, *cp) = 0?

5.36. Exercise. (x, y, z) is a coordinate system on R3 endowed with theEuclidean metric. S2 is the unit sphere centered at 0 in R3 and 4' is theinclusion of S2 in R3.

Let P be the point of S2 for which z = 1. When M = (x, y, z) E S2.0 will denote the angle (Ox,OM -z ), and r the length of the are PM ofmeridian.


a) What is the greatest open set 11 C S2 where (0, r) is a polar coordinatesystem?

Express the coordinates (x, y, z) of 4'(M) in terms of the coordi-nates (O,r) of ME Il.

From that, deduce the components of the metric g = 4*E on fl.

b) (S2, g) being endowed with the R.iemannian connection, compute thecomponents of the curvature tensor and of the Ricci tensor, and thevalue of the scalar curvature. Is the Ricci tensor proportional to themetric tensor?

c) For which values of 0 are the curves 0 = Constant geodesics? Fbrwhich values of r are the curves r = Constant geodesics? What arethe geodesics of the sphere?

d) Prove that the function f = cos r is an eigenfunction of the Laplacian.What is the corresponding eigenvalue? Is it the first (the smallest)nonzero eigenvalue of the Laplacian?

5.37. Exercise. a) Let (f2, 4') be a local chart of a C0° Riemannianmanifold (M,g). On Sl, compute r, in terms of the determinant I9Iof the matrix ((gu)).

b) Consider a C°° function V on fZ such that cp(Q) = f (r) for all pointsQ E fl with r = d(P, Q), P being a point of Q. Find a simpleexpression of App.

Hint. Consider a polar geodesic coordinate system at P, andcompute Vi(ghivjw)

5.38. Problem. Let S be the unit sphere in Rn+1 endowed with the Eu-clidean metric S, 0 its center. We consider on S, the imbedded metricg = 4'*E, 4' being the inclusion S C Rr+1. Let P be the point with on-ordinates xJ = 0 (1 < i < n), xi+1 = 1 (the x' are the coordinates onRn+' ), and let Q be the point opposite to P on the sphere. To a pointx E fl = we associate the point y, the intersection of the straightline Px with the plane it with equation xn+1 = 0: y = 4'(x). We will de.note by {x'} (1 < j c n + 1) the coordinates of x E Ri+1, and by {yk}(1 < k < n) the coordinates of y in 7r.

a) Is {yk} a coordinate system on 52 corresponding to a local chart C?

b) Express the coordinates {r} of x E S) in terms of the coordinates{yk}ofyE7r.

Hint. Consider the arc or =Qx and compute in polar coordinateson ?r,

c) What are the components of the metric g in the local chart C?


d) If £,r is the Euclidean metric on 7r, verify that g = f (p)£,, f being apositive C100 function of p = [En_, (yk)2] .

e) The straight lines of x through 0 are geodesics for £,r. Are theygeodesics for the metric g? (Take care of the parametrization.)

f) What are the geodesics of the sphere Sn?g) What is the distance r from Q to x on (Sn,g)?h) What are the components on D of the metric g in a polar geodesic

coordinate system (r,9) with 0 E Sn_1?

i) Let +p be the trace on Sn of the coordinate function xi±1 : V(x) =x"'1 Express rp as a function of r.

j) Show that i+p = aV for a real number A to be determined. Thus 'pis an eigenfunction for the Laplacian A on (Sn, g).

k) Does there exist an eigenfunction for A on (Sn, g) equal to cost r + kfor some constant k?

For j) and k), the result of Exercise 5.37 is useful.

5.39. Problem. Let M and W be two connected, compact and orienteddifferentiable manifolds of dimension n, and f a differentiable map of Minto W.

a) Prove that there is a real number k such that, for all differentialn-forms w E An (W),

four=k w.

Hint. Use the Hodge decomposition theorem.

b) If f is not onto, show that there exists an open set 0 C W such thatf-1(e) = 0. Deduce that k = 0 under this hypothesis.

c) For the rest of this problem we suppose f is onto. Let Q E W be aregular value of f (the rank of (f,,) p is n if Q = f (P)).

Prove that f-1(Q) is a finite set and that there exists an openneighbourhood 0 of Q in W such that f -1(8) is the disjoint union ofopen sets 1- 4- M.

d) Prove that k is an integer.Hint. Consider a differential n-form w with suppw C 0, and

compare fw w and fll, f"w for each i.

e) Let Sn be the set of unit vectors in R"t1. We endow Sn C IRn+1 withthe imbedding metric. Suppose there is on Sn a differentiable field Xof unit tangent vectors (g(X, X) = £(X, X) = 1).

Let X}(x) be the components of X in R"+1. We consider the map

Ft of Sn into IZn+1 : x - y of coordinates yt = xi cos in+Xl (x) sin zrt,


where the xt are the coordinates of x E R"+1. Verify that Ft isa differentiable map of S" into S. We denote by k(t) the numberdefined in a) for the map Ft. Show that t - k(t) is continuous.

f) What are the values of k(0) and k(1)? When n is even, prove thatthere does not exist a vector field on S" which is nowhere zero.

5.40. Exercise. Let (M,,, g) be a C°° Riemannian manifold and P E M.For two vectors X, Y of R", show that for any e > 0, there exists it > 0 suchthat if0<t<rl,then

d(expptX,expptY) < (1 +F)IIX - Y11t.

5.41. Problem. Let (Mn, g) be a complete C°O Riemannian manifold ofdimension n > 2. We admit that for every pair of points P and P of Mthere exists at least one geodesic C through P and P whose length is d(P, P).So let C be a geodesic from P to P whose length

L(C) = d(P, P) = a: R D [0,a] Dt-C(t)EM.

P = C(a), and t is the arc length parameter. At P = C(0), we consider anorthonormal basis {e,} (i = 1, 2, ... , n) of Tp(M) such that el = ( )t o.We denote by e;(t) the parallel translated vector of ei from P to C(t) alongC.

a) Verify that the vectors {el (t), e2(t), - , en(t)} form an orthonormalbasis of TC(t)(M).

b) In the sequel we consider, on a neighbourhood SI of C, a coordinatesystem {x`} such that C(t) has (t, 0, 0, , 0) for coordinates and suchthat (8/8xr)c(c) = e,(t) for all t and j.

What can we deduce from this property for the componentsg,,i(C(t)) of g and for the Christoffel symbols I'}j(C(t))?

c) Let B be the unit ball in R"-1, and consider the map f of [0, a] x Binto M defined by

(t, ) - f (t,expC(t)1

' (t)J ,awhere {`} (2 _< i < n) are the components of the vector t; E R"-and where e;(t) = (8/8xt)o, {xti} being the coordinates of R'. Whatis the differential of f at (t, 0)? Is it invertible?

d) For a given point Q of M. verify that there exists at least one pointQ E C such that d(Q,Q) = d(Q,C), where by definition

d(Q, C) = inf {d(Q, R) I R E C}.


e) Prove that if Q, one of the points found in d), is neither P nor P, thena geodesic y from Q to Q, of length L(y) = d(Q,Q), is orthogonal toCat Q.

Hint. For two nonorthogonal vectors X, Y E T4 (M) (gg(X, Z)36 0), observe that we can choose a vector Y proportional to Z sothat 9§(X - Y, X - Y) < g(X, X). Use the result of Exercise 5.40.

f) Prove that there is e > 0 such that, if d(Q, C) < e, then there existsa unique point Q of C such that d(Q, Q) = d(Q, C).

g) Deduce from f) that (t, t;l, t a, .. , C") is a coordinate system on 9 =f 00, a( xBB), B. being the open ball of radius a in R".

h) In this coordinate system, compute the Christoffel symbols at C(t).

5.42. Problem. Let (x, y, z) be a coordinate system for R3 and (u, v) acoordinate system for R2. For two real numbers a and b with a > b > 0 wedefine the map 90 of R2 into R3 as follows:

I x = (a + boos v) cos u,y = (a+bcosv)sinu,z = bsinv.

a) What is the rank of ti?b) We identify C x C (C the circle of radius 1) to the quotient of R2 by

the equivalence relation

J u - u = 2kir, kEZ,(u, v} {u, v) - l v - v = 2hir, h E Z.

Show that we can define a map of C x C into R3 from the map,0.Is ib an imbedding?

c) Find a polynomial P of the fourth degree in x, y, z which vanishes onM =,O(R2) and only on M.

d) Prove that M is a submanifold of R3. Is M diffeomorphic to C x C?

e) Determine the Riemannian metric g on M induced by the Euclideanmetric on R3. Verify that the nonzero components of g (in the coor-dinate system u, v) are equal to b2 and (a + b oos v)2.

Compute the Christoffel symbols of the Riemannian connection.

f) Compute the scalar curvature R of (M, g). Note that it has the samesign as cosv. What is the value of fm RdV?

g) Write the differential equations that arcs of geodesics t (u(t), v(t))satisfy on (M, g).

Are the intersections of M with the planes through the axis Ozarcs of geodesics?


Find all geodesics which are included in a plane perpendicular toOz.

h) Show that, if at a point of the are of geodesic ry, does not vanish,then is not zero at any point of 'y.

Hint. Use Cauchy's theorem.i) Deduce from the result above that we can substitute the parameter t

for u, except for some geodesics 9 that one will identify.What is the differential equation (E), where the unknown is v(u),

satisfied by the differentiable curves which are arcs of some geodesicof M different from the geodesic S found above.

Verify that ddv/dus always has the same sign as - sin v.j) We study the geodesics from P = tP(0,0). What is the geodesic for

which (J')p = 0? Which symmetry permutes the geodesic for whichd') p = a > 0 with the one for which (d) p = -a?( 94

We denote by v,, the solution of equation (E) with va.(0) = 0 ande"'(0) = a.

k) Prove that if a is small enough (0 < a << 1), then dv,,/du vanishesfor the first time at a value u,, < a/2 when u increases from zero.What is the sign of va(u,,)?

1) Show that vo(2ua) = 0. Deduce that there are at least three geodesicsfrom P to Q = tt'(2ua, 0). Compare their lengths to it b(a + b) and(a + b)Tr.

5.43. Problem. Let M be a COD differentiable submanifold of R"+*(k > 1). We denote by F the inclusion M -, R"+k.

By F we identify X E T (M) to F*X E T(R"+k); thus T(M) C T(R"+k).

Let £ be the Euclidean metric on R"+k, and set g = F"£ 1

X, Y and v will be three vector fields defined on M, with X and Yincluded in T(M).

a) Consider three vector fields X, k and v in R"+k which are equal toX, Y and v on M. For the Riemannian connection b on (R"+k,£),verify that DX f,/M depends only on X and v.

b) For Q E M, set (DxY)Q = trQ(DgY)Q, where 7rQ is the orthogonalprojection of TQ(R"+k) on TQ(M).

Compare (9, fl /M with [X, YJ.Verify that (X, Y) DxY is a connection on M, which is the

Riemannian connection of (M, g).

c) In a neighbourhood 0 of P E M, we consider k orthonormal vectorfi e l d s v ; (i = 1, 2, - , k) orthogonal to M (£(v;, vi) = b; , and ifQE 8,, then trQuu = 0). Set 4(X, Y) = E(Dxv,,Y).


Establish the following formula:k

DxY = DxY - 1.(X,Y)vi.

Is tt(X,Y) a symmetric bilinear form on TQ(M)?

d) Compute bxbyZ, and then show thatk

£(DxbyZ,T) = g(DxDYZ,T) - £,(X,T)1i(Y, Z)

for two vector fields Z < T E M included in T(M).e) Deduce from this formula that the curvature tensor of (M, g) is given

by

k

R(X,Y, Z, T) = E(fi(X, Z)t1(Y,T) - fi(X,T)e!i(Y, Z)J.i=i

HinL Use the fact that the curvature tensor of the Euclidean metricvanishes.

f) Suppose k = 1, and let A (j = 1, 2, . , n) be the eigenvalues of tjat Q.

If n = 2, compute the scalar curvature of (M, g) at Q in terms ofthe Ai. If n > 2, prove that the sectional curvature of (M, g) at Qcannot be negative with respect to all 2-planes of TQ(M).

5.44. Problem. Let (M, g) be a CO0 compact Riemannian manifold of di-mension n and let X, Y be C°° vector fields on M. We consider the oneparameter local group G of local diffeomorphisms Vi (x) corresponding to X.

a) For a covariant tensor field h, verify that d(pih)/dt = pitxh. De-duce from this that a necessary and sufficient condition for h to beinvariant under G is Cx h = 0.

By definition, h is invariant under G if V *h = h when V *h exists.

b) A C°° differentiable map t' of M into M is called a conformal trans-formation of (M, g) if Tp*g = efg, f being a C°° function on M. Seth = JgJ` y g, where 191 is the determinant of ((gig)). Compute Ivp*gJand t h.

Show that Cxh = 0 is a necessary and sufficient condition for thediffeomorphisms of G to be conformal transformations of (M,g).

c) Let P be a point where Y(P) 0. Consider a local chart (fZ, y) at P,with y(fl) a ball centered at y(P) = 0 E R" . We choose (0, y) suchthat Y # 0 on Q. {xi} being the associated coordinate system, wesuppose Y" 96 0 on is (Y = Y'8/8x'). Let iA be the one parametergroup of local diffeomorphisms corresponding to Y and let ao be a


point in fZ such that xo = {x1, x2,. - -, xr-1, 0}. If it exists in ft, weset x = (xo). Prove that x', . . , x"-1, t form a coordinate systemin a neighbourhood of P.

d) Prove that Cyh = 191n [Cyg - 2V,Y'g]_Hint. Use the coordinate system constructed above.

5.45. Problem. Let (M, g) be a connected complete CO° Riemannian man-ifold of dimension n.

If W is a submanifold of M, we consider the tangent vectors to W astangent vectors of M (each X E TQ(W), Q E W, is identified to (i,,)QX,where i : W -+ M is the inclusion). Moreover, W is endowed with themetric g = i*g.

a) Since the manifold is complete, prove that expp X exists for anyXER' and any P E M.

Hint. Consider a sequence {t,} C R, which converges to r, suchthat Q, = exp tX exists in M for any i. Prove that the sequence {Q, }in M converges to a point Q E M. Then apply at Q the theorem ofthe exponential mapping (5.11).

b) The compact submanifold W is said to be totally geodesic if anygeodesic of W is a geodesic of M. If W has this property, provethat a geodesic C of M is included in W if at one of its points C(t),(J)C(t) ET(W)-

c) We admit that there exists a coordinate system {x'} in a neighbour-hood of a geodesic arc C : [0,,r] D t - C(t) E M, such that thecoordinates of the points C(t) are x1 = t, x' = 0 for i > 1, and suchthat this coordinate system is normal at each point C(t) of C. Provethat the vector fields (8/&) are parallel along C.

d) Let W, and Vq be two compact totally geodesic submanifolds of di-mension respectively p and q such that W. n Vq = 0. Prove thatthere exist Po E W and Qo E V such that d(Fb, Qo) = d(W, V). Bydefinition r = d(W, V) = inf d(x, y) for all x E W and y E V.

e) We admit that there exist a geodesic C from PO to Qo (Po, Qo as ind)) of length L(C) = r, and a coordinate system {x'} as in c). SetFb = C(O) and Qo = C(r). Let H be the set of the vectors of TQo (M)obtained from the vectors of Tpo (W) by parallel transport along C.

For the rest of this problem we suppose p + q > n. Prove thatH n TQ° (V) is not reduced to the zero vector. Let Y E H n TQ, (V)be of norm 1 (gQ(,(Y, Y) = 1).


f) Consider the vector field X(t) parallel along C such that X(r) = 1'.For a chosen A E R. we define the are -)a by

[0, T] t - -Y,\ (t) = eXJ)(-(1) AX (t).

Verify that y,\ is a C" differentiable are whose endpoints are apoint of W and a point, of Al. What can we say about f (A) = L(ya)?

g) Compute first f'(0). then f"(0). in terms of the sectional curvaturea-(X(t). ). We recall that the curvature tensor is given by

RAtij (C(t)) =1

(a,2 ,g k - c ,xg t - 8j,g,k + akg )Cct)

in a coordinate system which is normal at. C(t).

h) Deduce from g) that the intersection of WVp and V. is not empty if thesectional curvature is everywhere strictly positive.

5.46. Problem. Let. (Al, 9) be a compact Riemannian manifold. and con-sider a Riemannian cover (IV, g) of (M, g) such that TV is simply connected.If a : TV , Al is the projection of W on M, then g = lr*g. We suppose thatIV has more than one sheet.- that is to say, (Al, g) is not simply connected.

a) Let C be a differentiable arc of W. C = ir(C) its projection. Showthat L(C) > L(C) and that the projection of a geodesic are is ageodesic are (L denotes the length).

b) Verify that. (W, g) is complete.

c) Let C : [a, b] Al be a differentiable closed curve of Al which isnot homotopic to zero. Set P = C(a) = C(b) and choose a pointP E 7r-'(P). Show that we can construct by continuity a differen-tiable curve e C TV, locally diffeomorphic to C by 7r/C. such that.C(a) = P. Why is C(b) = P not the point P?

(1) If CO is a curve from P to P, homotopic to C, prove that we find thesame endpoint P = Co(b). C0 being constructed from Cc, as C fromC with Co(a) = P. We set f (P) = d(P, P'). So f (P) depends on Pand on the chosen homotopy class.

e) For a point R of Al, we consider a differentiable curve y : [a, a] - Alhomotopic to C and such that y(a) = y(i3) = R. For instance. if r isa differentiable curve from R to P. y may be a regularization of thefollowing curve: r, then C. then r from P to R.

From a point R E it-1(R). we construct (as in c) and d)) acurve y C TV locally diffeomorphic to -y by it/y` with y(a) = R.Set R' = ry(j3) and f (R) = d(R, R').

Prove that Al D R - f (R) E IlP is a strictly positive continuousmap which achieves its minimum at least at one point. Q E Al.


f) As above, we construct the point Q' from a point Q E 7r-1(Q). Thenwe consider a minimizing geodesic ry" from Q to Q'.

Show that 7r(5) is a geodesic from Q to Q, then a closed geodesic(the curve is also a geodesic at Q). What is its length?

We assume that any pair of points (P, Q) of a complete Riemann-ian manifold (M, g) may be joined by a minimizing geodesic -y (thatis to say, L(-y) = d(P,Q)).

5.47. Exercise. Let (M, g) be a compact Riemannian manifold. Considerthe metric g' on M defined by g' = of g, where f is a C°O function on M.

a) Show that (M, g') is a Riemannian manifold.

b) In a normal coordinate system at P E M, compute the Christoffelsymbols I'; (P) of (M, g`) in terms of f.

c) In a neighbourhood of P, deduce the expression of c ilk = I ; , - I'';kwhere V. are the Christoffel symbols of (M, g).

d) In terms of f , compute Rik - Rt i where RI 'k Rik,, are respectivelyii 4the components of the curvature tensors of (M, g') and of (M, g).

e) Deduce, from the result above, the expression of R;t - RJi, the differ-ence of the components of the Ricci tensors of (M, g') and (M, g).

f) Express R' as a function of R and f, R' and R being the scalarcurvatures of (M, g') and (M, g).

5.48. Problem. Let (M, g) be a Riemannian manifold with nonpositivesectional curvature (v(X, Y) < 0). Given a geodesic C : R D [a, b] 3 t -,C(t) E M, w e consider an orthonormal f r a m e el, ea, , e,, of Tp(M)(P = C(a)) such that el = (Vi)a. Then at Q = C(t) we consider thevectors ei(t) obtained from e; by parallel translation along C. We supposethat t is the are length.

a) Do the vectors ei(t) (i = 1, 2, . , n) form a basis of TQ(M)?

b) Let x1, x2, , x" be a coordinate system in a neighbourhood of Csuch that (8/8x')Q = ei(t). What are the values of the componentsof the metric tensor at Q, and what can we say about the Christoffelsymbols at Q?

c) We suppose that there exists a coordinate system in a neighbourhoodSl of C such that at any point Q of C we have gij (Q) = S` andq- (Q) = 0. Along C, we consider a vector field X whose componentsX'(t) written for X'(C(t)) satisfy

c2 '(t) _ -R;,1(C(t))X'(t).dt2


Shaw that such vector fields J exist, and that they form a vectorspace E. What is the dimension of E?

d) Verify that there exist vector fields J which are orthogonal at eachpoint to the geodesic, and that they form a vector subspace of E.

e) Prove that a nonzero vector field J may be zero only once.

f) If X and Y are two vector fields J along C vanishing at P, show thatg(Y, X') = g(Y', X), where X' _.

g) Prove that there exists one and only one vector field J such thatX(a) = 0 and X(b) = Xot a given vector of Tc(b)(M).

h) Consider n-1 vector fields J which satisfy Ye(a) = 0 and Y (b) = ej (b)(i = 2, 3, , n). Verify that n - 1 differentiable functions fi(t) on]a, b] are associated to a differentiable vector field Z along C vanishingat P and orthogonal to C such that Z(t) = E a f;(t)Y(t).

i) We define

t.dI(Z) =J {(z' Z') - g [R(dt'Z) Z, d]I

Set X(t) = E,n2 fs(b)Y(t), and show that I(X)=g(X'(b),X(b)).

j) Prove that I(Z) > I(X).Hint_ Consider W (t) = En

i=2 f; (t)Y=(t), and establish the equality

I(Z) _ f g(W(t),W(t))dt+g(X'(b),X(b)).0

5.49. Problem. Let (Mn, g) be a connected complete C°O Riemannianmanifold of dimension n, and f a map of M into M which preserves the

,Q) E M x M.distance: d(P, Q) = d(f (P), f (Q)) for all pairs (P., Q) can be joined by a geodesic.We assume that any pair of points (P.

a) Show that the image by f of a geodesic arc is a geodesic arc. Recallthat if d(P, Q) < .1(P), the injectivity radius at P, then there existsa unique geodesic arc from P to Q whose length is d(P, Q).

b) If X and Y belong to W', prove that

li m[d(expp tX, expp tY)t-1] = IIX - 9.

c) Prove that any geodesic arc through P can be extended infinitely(that is to say, expp X exists for any X E R").

d) The image by f of a geodesic through P (X is its tangent vector atP) is a geodesic through P' = f(P). We denote by X' = +p(X) itstangent vector at P'.

Show that V is a linear map of Tp(M) onto Tp,(M).


e) Establish that f is differentiable and that f*(g o f) = g. f is calledan isometry.

f) Prove that two isometries f and h are identical if there exists a pointP E M such that f (P) = h(P) and (ff)p = (h*) p.

g) Define a map of I(M), the set of the isometries of M, into M x 0,,.What can we say when M is the sphere S,,?

5.50. Problem. Let (M., g) be a compact Riemannian manifold of dimen-sion n, and X, Y two C°° vector fields on M. Consider the differential1-forms and a associated to X and Y; gijXj and ai = gijYj are theircomponents in a local chart.

We set h = lgI -g, where I I is the determinant of ((gij)).

a) Show that, in a neighbourhood of a point where Y is nonzero, thereexists a coordinate system {x'} such that [Y, 8/8x'] = 0 for all i.

b) Deduce from a) that Gyh = jgj_n [Gyg - 2D;Y'g].

c) Set t(a) _ IginGYh. Show that the components of t(a) are t(a)ij =Viaj + Vjai + -abagii.

d) Verify that (Aa)j = -O'Viaj+Rija', where Rij are the componentsof the Ricci tensor.

e) Prove that O't(a)ij = 2R;ja' - (da)j - (1 - 2)(d6a)j.f) Consider the differential 1-form u(a) defined by u(a)i = ajt(a)ij.

Compute 8u(a) in terms of a1V't(a)ij and (t(a), t(a)).g) Show that Lxh = 0 if and only if

(n-1)+ ` 1- n) 64 =

where S(k)i = nRij£j.h) Prove that Lxg = 0 if and only if

V2Vitj+RijX° =0 andi) Compute Deduce f om the result that if GXh = 0 and if the

scalar curvature R is constant, then

(n-1)&5 =R5C.

5.51. Exercise. Let us consider the unit sphere S C R"+1 endowed withthe canonical metric g = i'E (i the inclusion), and let C be the geodesicof S through x E S with initial condition (x, v), v belonging to TT(S ).Consider the orthogonal symmetry u with respect to the 2-plane P definedby x and v in R'.

a) Prove that C is included in P.


b) Deduce that the geodesics of the sphere are the great circles.c) What are the geodesics of the quotient P,,(R) of S. by the antipo-

dal map? Show that they are periodic with period r if they areparametrized by are length.

5.52. Exercise. Let us consider the unit sphere Sea+1 C Cn+1 endowedwith the canonical metric g. The complex projective space P,, (C) is the quo-tient of by the equivalence relation R (see 1.41); let q: Pri(C)be the corresponding projection. Let z E S2n+1 and v E T.(S2,,+1). We con-sider the carve C : (0, x] 3 t -+ z cos t + v sin t.

a) Verify that C C S27 +1 and that q(C) is a smooth closed curve inP.M.

b) Prove that there is a unique Riemannian metric g on Pn(C) such thatq9=9.

c) What is the length of q(C) in (Pn(C), g)?

5.53. Problem. Let M be a C°° Riemannian manifold and let C be ageodesic from P to Q ((0, r] 3 t - C(t) E M). According to Problem 5.41,there exists a coordinate system {x'} on a neighbourhood 0 of C which isnormal at each point of C (the coordinates of C(t) are (t, 0, 0, ..., 0)). ForIAI < E, let us consider a family {Ca} of differentiable curves in 6 definedby the C' maps ]0, r] x (-E, E) 9 (t, A) - xf (t, A) E 0 with xi (0, A) = 0,z'(t,0)=Iandx'(t,0)=0fori> 1.

a) Verify that the first variation of the length integral of Ca is zero atA=0.

b) Prove that the second variation of the length integral of CA at A = 0is

,a)1 =

f{fl()2+ Ri1 (C(t))v`(t)i(t), = 0,

Jwhere

7/(t) = (&x'(t, A)/8A)a=o

5.54. Problem. Prove that a compact orientable Riemannian manifold ofeven dimension with positive sectional curvature is simply connected.

Hint. Use the result of Problem 5.46 and the expression of the secondvariation of the length integral found in the problem above.

5.55. Exercise. Let (MM,g) be a Riemannian manifold, endowed with theRiemannian connection D, and {x'} (i = 1,2,... , n) a normal coordi-nate system on P E M corresponding to the local chart (1Z, tp). In Rn,with coordinates {w'}, we consider the ball B, of center 0 E Rn and ra-dius e > 0, and a C°O map ¢ of BE into i2 (with 0(0) = P) such thatt -+ ¢(t, 0, 0, . , 0) is a geodesic on M. Define z = (ws, w+, ... , W"), and


let Z(t, u, z) = 8,0(t, u, z)/8u be the tangent vector at ¢(t, u, z) to the curveu 0(t, u, z) and T(t, u, z) = 80(t, u, z)/8t the tangent vector to the curvet-+O(t,u,z).

a) Compute the bracket [Z(t, u, z), T(t, u, z)].

b) Set X (t) = Z(t, 0, 0) and Y(t) = T(t, 0, 0). Show that DyDyX =R(Y, X )Y.

c) Prove that g(X (t), Y(t)) = at + b, a and b being two real numbers.

5.56. Exercise. Let M" be a compact Cl* orientable manifold of dimen-sion to > 1. We consider a Riemannian metric g on M" and g E A"(M)defined in a coordinate system by

q= dxlAdx2A...Adx",

where IgI = det((g;,)).

a) Verify that q is a differential reform on M.b) We endow M" with a linear connection whose Christoffel symbols are

Ft,. Show that a necessary and sufficient condition for Vq = 0 is thatNi = 1A 109 191.

c) Does the Riemannian connection satisfy this condition?d) We suppose (here and for e)) that the manifold is endowed with the

Riemannian connection. Let X be a vector field. Show that

Dix' = 18,(X' I9I).

e) Let a = i(X)q. Compute da and the value of fm VZX`q.

5.57. Exercise. Let M,, be a complete Riemannian manifold of dimensionn > 1. For X E S,,-1 we define

p(X) = sup{r E V I d[expp(rX), P] = r},

where {X E R"-1 I IXI = 1} and P is a given point.

a) Show that X -+ u(X) is a continuous map of S,,_1 into I =R+ u {+oo}, endowed with the usual topology.

b) Prove that for M" to be compact, it is necessary and sufficient thatp(X) be finite for all X E S1.

c:) If M,, is compact and has non-positive sectional curvature, show that,given a pair (P, Q) of points of M, there exist only a finite num-ber of minimizing geodesics from P to Q. Then show that the setCp = {expp[p(X)X] I X E S1} is the union of a finite numberof manifolds of dimension n - 1. We will assume (or prove) that,when Q E Cp, if there is only one minimizing geodesic from P to


Q = expp Y then X -. expp X is singular at X = Y. A geodesicfrom P to Q is minimizing if its length is equal to d(P, Q).


Solution to Exercise 5.29.By hypothesis ;j(P) = f(P)gij(P). Contracting this equality gives

R(P) = n f (P) (9ij9ki = dk; if we do i = k = 1, 2,..., n, we find thatgijg'j = n). On the other hand, contracting twice the second Bianchi iden-tity (5.8) yields (we multiply it by gjm, then by gi<)

V'Rijki + VkRi/ - VlRik = 0; then VkR = 2V'Rik.

Indeed, for instance,

9ln'OkRijlm = OkW 'Rijlm) = OkRik

and

g' VkRit = Vk(9'lR;1) = VkR.

The covariant derivative of Rgij = nRij is gijVkR = nVkRij. Contract-ing this equality leads to V2R = nViRij.

We get (n - 2)VjR = 0. Hence when n 36 2 the scalar curvature R mustbe constant. Recall that for functions Vkf = 8k f, and R is a function.


a) If {Y'} are the components of a vector Y in a coordinate system, thenaiY' = gijXjY'; thus the components of a are ai = gijXj. Likewise,Xj = gj'ai, and with our notations we have ai = X'.

Since Lxx' = X(xi) = Xj8jx' = Xi, it follows that

Lxr7 =Xj8j dx' A dx2 A Adxnn

+ Nl'lgldX'i=1

We have

19j V191 = 1M = 29'kaj9ik 191;

then Lxri = ('Xjg'k8j9ik + 81X1t = V,Xig = -bag. Indeed,dXi = 8,X'dxj, and V1X' = ax + I jXj with

r>> _ 2gik(8i9jk + 8j9ik - 8kgij)


For an alternative proof we note that Cx = i(X)d+di(X) on then

forms. Thus Cxil = d[i(X )ri] = d[Xiq. dxl A - - A dx' A .. AR

ndxn; here means that l; is omitted. We obtain

n A

Cxi? = d[EX-i(-1)i-i Wj]dx' Adx2A...A dx' A...Ad^,i=t

Cx'1 = 8(X')dxl Adx2 A... Adx" =ViXif,.b) We have

Aa = 6da + d6a = 6(dai A dx') - d(Vjaj)

and

dai A dx' = 8jaidxi A dx' = V ja{dx' A dx`,

since I'1k, = Pkj implies t ,dxj A dr' = 0. Thus

(Aa)i = -VjVjai + V'Viaj - ViVjaj = -XI - V'Vjai.

c) Cx is distributive and Cxdx' = dXi = 8jX'dxj. Thus

Cx9 = Xkakgidx" ® dx' + gk(dX' 0 dxk + dxi ® dXk)

= (Xk8kgj + 9kj&Xk + gkOjXk)da' ®dxi(Viaj + Vjai)dx' ®dxi.

The result follows.

d) If V iaj + V jai = 0, by contraction we get 6a = -V'ai = 0. Thenmultiplying by Vk gives VkV jak + VkVkaj = 0. Thus, according tob), (.a)i = RjX'+VJViaj = RjXj+V'Viaj-ViVjaj = 2RijXj.

Conversely, d6a = 0 implies

J(ba)2dV = (da, Sa) = (a, dba) = 0.

Hence ba = 0, which implies V,Vja' = Rija'. Moreover,

((Viaj + Vja;)(Viaj +Vja')dV = 2J(V'ai + V'ai)ViajdV.

Then since V'ai = 0, and according to b), integrating by parts leadsto (see 5.27)

r(Viaj + Vjai)(V1& + V'a`)dV = -2 J ai(VzV'aj + Ri ja')dV

= -2 J,1[2RijX' - (tkx)jI dV

= 0

Solutions to Exercise's and Problems 147

e) We haveu

('pi+ug - ;Pi9) = 'pi [ti ( pug - g)] - W*Lxg when u tendsto zero. So Cxg = 0 implies d (Vi g) = 0, and thus apt *g = g.

f) We saw just above that if Cxg = 0, Wi is an isometry. Thus thecovariant derivative commutes with gyp=, e t6 = 6V*. This impliesCxb = bCx. We have dw = 0 and bw = 0, since Aw = 0 (see 5.20).So Cxw is homologus to zero:

Cxw = [i(X)d+ di(X)]w = d[i(X)w].

On the other hand, 5Gxw = Cxbw = 0. Hence Cxw = 0, sinceCxwis homologous to zero (exact) and coclosed.


a) Since Ru,d = 0, according to the expression of R;i,, (see 4.7) we get

dw; = dri`, A dx' = air dx' A dx' = 2 (air; - a,r;`,)dx' A dx'

_ -2(r,;rn,; - r -ij rkW)dx' Adx' = I A dx'.

Then, as rm/ = r, , it follows that dwjk = wjm A wnk, .

b) I17 = r B' BryAk - ByasA; are the Christoffel symbols on (Q, gyp).Recall that ((B,)) is the inverse matrix of ((Aj")). We want thatrg,f, = 0; thus 8,3A, = r B''Ak. Multiplying by 4- yields OiA91 _r kAk, which is

(1)

(u)

{dAj - 0}.

c) We choose p = n + n2; the Ak and the coordinates x' will be thevariables. The system (i) is integrable since it is closed:

dAk n w + Ak dw _ (dAk - Al wk) A wil; .

Moreover, the system is of order n2. This is obvious, since each dA7appears only once in the system.

Thus there are n2 functions Fj of A; and x' which are constant.But since the system {dF) = 0} is equivalent to (i), which is of ordern2 with respect to the A,*, it is the same for the system {dFj = 0}.Hence it may be written oya/ax' = Ajo(x'); that is,

{dy" = A7 (x')dx'}.

d) System (ii) is integrable since dA9 (x) A dxj = 0. Indeed, we have

a- a k&A; - Ak i, = Ak r,i = a,A;


Hence there are n functions ya(x) defined in a neighbourhood f1 ofP which are the coordinates corresponding to the local chart (f2, gyp)which we wanted to find.

Solution to Problem 5.32.a) We have

gg1(x) - Ea0aya av - P fta gya02 i t

Let k be a vector field on RP whose restriction to M is Y:

(DXY)'l = Xa(BQY$ + f$y ') = X'(Ar, + I'si yYy);

we have chosen a coordinate system {ya} in a neighbourhood of Q inRP such that ya = rQ for 1 < a < n. DXY depends on k only by(O,Y')Q and Y(Q). So, DxY is well defined.

b) As X E TA(RP), we have DxY E TQ(RP). Thus fIQDXY EThe restriction to x r (m,,) of (X,Y) IIQDXY is bi-

linear since HQ is linear. Furthermore, if f E C' on a neighbourhoodofQinM,then{Dx(fY)]Q = flQ[X(f)Y+f(Q)DxY] = [X(f)Y+fDxY]Q.But fIQ is C, and thus the differentiability condition holds.

Hence (X, Y) DxY defines a connection with vanishing torsiontensor. Indeed, letting X be a vector field on RP whose restrictionto M is X, we have DAY - DYX = [X, fl, because (RP, e) is aRiemannian manifold. Applying IIQ to this equality yields

DxY - DyX = HQ[X,Y] = [X, Y],

since [X,Y]Q = (Xa8aY,6 - YQBfP)Q = Xi88Y$ - Y'B,X5 (seea)). In fact we have more: as X and Y belong to T(M), B;YS = 0for 3 > n. Moreover, by definition g(X, Y) = E(X, Y). For a vectorfield Z on M, applying Dz to this equality at Q gives

(Dzg)(X, Y) + g(DzX, Y) + g(X, DZY)

= E(DzX, k) + E(X, DzY),

since Dz = Dz on functions. But E(DzX,Y) = E(DzX,Y), becauseY is orthogonal to DzX - DzX. Hence Dz9 = 0.

c) We have

H(X, Y) - H(Y, X) = DXY - DyX - (DxY - DyX)

= [X, Y]Q - [X, Y]Q = 0

(we saw this result in b)).


d) Let (Ilj, cp;);E j be an atlas for M compatible with the orientation,and {x; } the associated coordinate systems. At each point Q E M.we choose N such that the basis (8/8x; , , 8/8x°, N) is positivein R"+1 This definition of N makes sense; it does not depend on i.As E(Y, N) = 0, we have £(DXY, N) + E(Y, DXN) = 0. The resultfollows since £(DXY, N) = 0.


a) y - d(x, y) is a strictly positive and continuous function on a compactset y. Thus there exists yo E y such that d(x, yo) = d(x, y) = a > 0.

b) We have d(z, y) > d(z, yo) for ally E y, as otherwise the are xz U zywould be shorter than the arc ayo. Thus d(z, yo) = d(z, y).

If b is small enough, Eb C f2, the open set of the local chart(11,exps 1). We know that there is a geodesic from z to Q of lengthd(z, Q), and furthermore there is the geodesic of the exponential map

which is included in f2. Its length is b = g=(Jf',X). If the firstgeodesic is not the second, the first geodesic must go outside f2 andits length would be greater than b (see 5.10).

c) Define Sb = {P E M I d(z, P) < b = d(z, yo)}. Then y f1 Sb = 0; thusY = () is tangent to Eb. Moreover, according to Proposition5.13, (4 )t=a is perpendicular to Eb. The result follows.

d) We have Ddc/dtEgc(t)(e(t), )] = 0 and DdC/dt[gc(t)(e(t),e(t))] = 0,since Dg = 0, Da/dt4 = 0 and Ddc/dte(t) = 0.

Thus both expressions are constant along C, so they are equal totheir values at yo.

e) The maps t - Asin , t - C(t) and t -y e(t) are differentiable.Moreover, (u, Q, X) - C(u, Q, X) is differentiable, so the result fol-lows. Also, Ca(0) = C(0, x, e(0)) = x and CC(a) = C(A, yo, Y) _y(ul), a point of y.

Since

dC(Adyo,l')),\_o=1,,A

both geodesics are the same. Thus ul = no + A.

f) We have

= J iloc (dC A dCA)f()

a (:) dt dt00

On C (A = 0) we know that the square root is constant, so it is equalto 1 since f (0) = a. We can differentiate under the integral sign with


respect to A. Moreover, the first derivative of gU on C is zero. Hence

f'(0) =J

9c{:) (v(t), ) dt

with v(t) (30N since we can permute the derivatives

(the function is C2). Now

sinste(t)

aA a=o 2aCOQ\W)

(see 5.10). Hence

g)

v(t) = cos be(t) + sin -DFe(t) = -' cos2a't e(t).

The integral vanishes since e(t) is orthogonal to dc. It is obvious thatf'(0) = 0, according to the definition of yo.

Since f (A) is minimum at A = 0, we have

i , 2a=o

> 0.dal

Now

f"(0) = 2 J dt iii' sine 2a7rt

+ ( 2a)2 /oaco62 -gq(C(t))e`(t)&(t)dt.

But R;k't = -2gij on C, and, moreover,

raCoe2 27rt

a dt = To am2 2a 2 .

Thus

0<f"(0) <[12

2a12-k2]9ya(Y,Y)

and a < it/2k.



a) The parameter s of are length of C is proportional to t (see 5.10).Without loss of generality, we can suppose that t = s. Set xp = C(sp).We have d(xp,xy) < [8p - sqJ. Hence {xp} is a Cauchy sequenceconverging to a point, say Q E M, which does not depend on thesequence {sp}.

b) Let io be such that C(tj) E 1 with tj, > a - r. Since there is aunique geodesic included in it from Q to C(t;a) (Theorem 5.11), itfollows that C(t) for t > t;, belongs to the geodesic from Q to C(tj ).According to 5.10 the geodesic can be extended for all values oft suchthat a<t<a+r.

c) Consequently all geodesics are infinitely extendable.

Solution to Exercise 5.35.Let ti' be a harmonic differential p-form. Then d±b = 0 and 5 ' = 0

according to Proposition 5.20.

a) If Aa = y, then

(y, i) = (dba, ti) + (bda,:l') = (ba, bib) + (da, dio) = 0.

Soy is orthogonal to Hp. Now let us suppose that (y, rli) = 0 for anyharmonic differential Inform 0. Applying de Rham's Theorem 5.22three times, we see that there exist two differential forms B and wsuch that y = d6d/3 + bdbw. Setting a = bw, we get Aa = y.

b) According to (a) we only have to prove that a is unique in Ap. IfAa = 0 then a E H.; thus a = 0 since a E Ap. G exists, a = Gy,and y = Aa = ,4Gy implies AG = Identity on Ap. Let a E Ap; thenAct = dba + bda, and according to (a) a = GAa, GA = Identity onAp.

c) Let A = (,JJp., Ap. We suppose that a linear operator T on A satisfiesT o A = A o T. If a E A, there exists ,3 E A such that a = GS; thusGTO = GT o AGES = GA o TGO = TGQ since AG = GA = Identityon A. As d, b and * are T operators, the result follows.

d) We have 7 = y+ da and ¢' = 0 + dfl for some a and p. Moreover,r(y + da) n (0 + dpi) _

P/ 7 n ¢.

Indeed, since

(da) A (¢ + d$) + y A d# = d[a n (0 + d(3)] + (-1)pd(y A i3),

according to Stokes' formula the integral pn the right hand side van-ishes.


e) Since *,p is harmonic, *(p is closed. Thus (gyp, *ip) makes sense. Also,(gyp, *!p) = f A f (<p, p)rj = 0 implies V - 0. If cp is not zero,then (gyp, *cp) 0.


a) Let Q be the point opposite to P on S2. Then S2 = S2 - {Q} is inbijection with B E R2, where (0, r) is a polar coordinate system, Bbeing the open disk of radius 7r centered at 0 E R2. We write

I x = sinroos0,y = sinrsin0,z = cos r.

We have sin r sin 0, and = sin r cos 0. Thusg e e (

,)2+(5i,Z)2

= Sin2 r.90

Likewise we compute 90r and 9,,.:

60 Orger = + (20 M) + ('2)= sin r [cos r cos 0(-sin 0) + cos r sin 0 cos 0]

= 0,

9rr e CJ2+

()2cos2r+sin2r= 1.

b) 8,-g00 = sin 2r, and the other first derivatives of the components ofthe metric tensor vanish. Thus 17100 j8,.g.. = -

2sin 2r and

rre =2

12 r 8x900 = cotan r.

Moreover, g''" = 1, g0i' = 0 and g00 = sin-2 r (r 34 0).We know that there is only one component to compute for the

curvature tensor:

RB,.O = 8r I'BO - 1'1' 0, 0 = - coe 2r + sin r cos r cotan r =sine r.

We infer from this value that

ROO =sm2 r, RB,. = 0, Rrr = R,.e sg°° =1.

The scalar curvature R = 2, and we have R; = 91Sc) According to the geodesic equation, the curve t -+ [0(t), r(t)] is a

geodesic if

off +I'BOr'0'=0 and r"+r (0')2=0.


The curve 0 = Constant is a geodesic for any value of the con-stant.

The curve r = Constant is a geodesic if 1' = 0, that is, forr=s/2. In this case, 6"=0. Thus 0=at+b.

The meridians and the equator are geodesics. On the sphere, onlythe great circles are geodesics; that is, the intersections of S withthe planes through the center of S C Rri+i

d) Since g''B = 0, when f depends only on r we have

-Af = 9ijvijf = Vrrf + 9"Veef = Vrrf - 9°°rree8rf

and

Acosr = cosr+ mar(-sinrcosr)(-sinr) = 2cosr.

The corresponding eigenvalue is A = 2, and it is the first nonzeroeigenvalue according to Lichnerowicz's Theorem 5.27. Here, withk = 1 and n = 2, we find that Al > 2. Indeed, the Ricci curvature ofthe sphere S2 is 1 (9ijC1 i = 1 implies 1).


a) We have

rilir = 9ij(8i9jk + 8k9ij - 8j9ik) = 2049ij.

Moreover, 8kI9I _ 19194ek2ij, and hence

rsik = 28k log 191 =I

"k 10919 V(1_

b) We have

-A = 9ijViOj+P = Vi(9"w) = 8i(g"O,p) + r{ikgkj8j,p,

since Vg = 0-If cp(Q) = f (r), then, in a geodesic coordinate system, 8jW = 0

when j # r. Since g' = 0 and g''" = 1,

-aV = f" + f'Or log 191.


a) IF : x - y is a bicontinuous bijection of 0 on ir. (S2, *) is a localchart C, and the corresponding coordinates are {yk}.

b) We have e+z = cosa, a.=, (x')2 = sin2a and (y')2 = p2.Since xi = kyi (1 i < n) for some real number k > 0, it follows that


sin2a = k2p2. Thus k = :15. Considering the two triangles POyand PxQ yields

_ 2 sin i _ aP 2sin

- tan2Ee

sin a i _ 2

P 1+p2yfor1<i<n,and

xn+i=_1_p2-1-2

1+p2 1+p2c) We have

INWWIj=8y,8yjEs,,.

Choose polar coordinates on 7r, and cylindrical coordinates on Rnt1(those of r and xn+1). In local coordinates 8j on S,,,-1 we haveO'(x) = Oj(y) (1 < j < n - 1) and

nE(x;)2 = Sin2 a = (2)2

We have

and

gvn=(88pa)2+ (8p

l)2

_ (cos2a+sin2a) r_j2 4\8p (1+p2)2

b .gpgj = 0, ggae, _ e; =Vi sin2 a = (1±p2)2

d) We verify that g = (4/(1 + p2)2)4, so f (p) = 4(1 + p2)-2.

e) In polar coordinates the straight lines have the equationsConstant and p = h(s). They will be geodesics for the metric gif1i,=0,1<i<n-1,andh"(s)+I''°,,(s)[h'(s)]2=0. We easilyverify that I'oP = 0. The arc length satisfies

ds = gppdp2dp+ p2

and so s = 2 Arctan p, p = tan 21, p' = 1(1 + 02), and p" = pp'.Moreover,

_ 1 _ _ 2pImo- 20,logg" 1+p2.

Solutions to Exmrlses and Problems 155

Thus

2=0.hn+r(h,)2P(1+P2)2p

Me)2 1 + p2

f) We found all the geodesics of S throught Q. So the geodesics of thesphere S,+ are the great circles, and only them.

g) r = a = 2 Arctan p.h) Of course 9., = 1, gpo = 0, and by symmetry gtv depends only on r.

Now the length of the horizontal circle through x has two expressions:and 27r sin a. Thus goo = sine r.

i) We have W(x) = - coo r.

= n cos r. Also, AV = no.j) We have -AV = cos r + (n - 1) sin r W-FSo V is a n eigenfunction f o r A on (S , g), and Al = n.

k) Set 10 = cost r + k = h(r). Then h' = - sin 2r and

-Atp _ -2 cos 2r - sin 2r(n - 1)cos rsin r

= 2 - 2(n + 1) cost r

according to the formula of Exercise 5.37.We want that 0& = 2(n + 1)c'. This implies A2 = 2(n + 1) and

(n + 1)k = p -1, k = -1/(n + 1).


a) According to the Hodge decomposition theorem, there exist a realnumber a and a differential (n - 1)-form cp such that w = ail + dip, rlbeing a nonzero harmonic n-form (all harmonic n-forms are propor-tional). We have

JMJMJMJMaccording to Stokes' formula, since f* and d commute. Thus k doesnot depend on w. Since fw w = a fw rf, it follows that

k=fM ,7

/fu'7.

Indeed, fw rl 96 0, since there are n-forms w whose integral is not zero.For instance, if (Sl, cp) is a local chart with coordinates {x'}, choose

hsmooth, h>Oand supphC11.b) Since f is continuous, f (M) is a compact set of W. Thus 0 =

W\f(M) is an open set, not empty if f is not onto. Consider onit C 0, as above, w = h(x)dx' A dx2 A A W. Then fw w # 0 andfm fw = 0. Thus k = 0 in this case.


c) Let Pi E f -1(Q). Since (df) p, is of rank n, f is a diffeomorphism ona neighbourhood SZi of Pi. The proof is by contradiction. Supposef -1(Q) is not finite. There exists a sequence {Pj } C f -1(Q) such thatPj P, which implies by continuity that f (P) = Q. So f would be adiffeomorphism on a neighbourhood f) of P. This is impossible, sincef o r l a r g e j w e would have P j E i2. Hence f -1(Q) = {P1, P2i , P,n).Choose the U disjoint, and set 0 = ( 1 f(fl.) and S?.i = fli fl f -1(B).Then f 1(0) = U'

1SZi.

d) f in, is a diffeomorphism; thus fw w = f fat f *w according to thechosen orientations on M and W. Hence k is an integer (Iki < m).

e) Let f be a unit vector. Then E(s,X(x)) = 0, since X(x) E T=(S,,).Thus

n+1

E (Jj )2 = II cos 7rt2 + sin 7rtX II2 = cos2 7rt + Sine 7rt = 1.j=1

So Ft is a differentiable map of S. into S. Let q be the orientedvolume n-form on S,,. Since the map

RxSnE) (t,x)-iFt(x)ESnis C', it follows that t --i fS. Ft *j? is continuous. Thus t k(t) is alsocontinuous. Then k(t) is constant, since the k(t) are integers.

f) Fo(x) = x and FO = Id; hence k(0) = 1.Also, F, (x) = -x, and F1 is the restriction to Sn of -Id on R1+1.

Set F = -Id on Rn+1. If ij is the oriented volume (n + 1)-form onR"+1 the orientation of S is the natural orientation induced by thatof Rn+1. Also, F"7j = (-I)n+in. Let v the unit normal vector toSn oriented to the outside; E7(x) has {2`' } for components. Moreover,q = i(O) and F*i% = v o F. Hence F*q = i(i1)F*i = (-1)n+lq andk(1) = (-1)n+1. When n is even, k(0) is not equal to k(1). This isin contradiction with k(t) = Constant. So, on Sn when n is even,there does not exist a vector field X which is nowhere zero. We wouldconsider X/II X II

Solution to Exercise 5.40.Let (l, gyp) be a normal coordinate system at P. For any Z > 0 there

exists p such that if IIZII < p (Z E B(p), the ball of radius p in Rn centeredat 0), then

(1- [sj(eXpP I IIFII2(1 +

for any vector E Rn, since gij(P) = b= and (,p1)p = Id. If IItXII < p/2and IItYII < p/2, an are from Q = exptX to Q = exptY is included ine = expp B(p) when its length is smaller than p.


Since d(Q, Q) < d(Q, P) + d(P, Q) < p, all differentiable curves y fromQ to Q that we must consider for the definition of d(Q, Q) are included ine, and so

d(Q, Q) < L(y) < 1 + ELE(v o y),

where Lg is the Euclidean length. This is true for any y. Choosing y =co 1((tX, tYJ), [tX, tYJ being the segment from tX to tY, we find that

d(expp tX, expp tY) < t 1 + E JJX - YH[.


a) We have

D#j [gc(t) (ei (t), ej(t))] = (Dac9)C(t)(ei(t), ej(t))

+ 9C(t) (Dc ei(t), ej (t))

+ 9C(t) [-i(t), DFej(t)).

In the right hand side the three terms vanish, the first since Dg = 0and the others since Ddcldtej(t) = 0 for 1 < j < n. We infer that foralli,j,

9C(t)(ei(t), ei(t)) = 9p(ei, ej) = 5 .

Note that el(t) = dc, since C(t) is a geodesic.

b) We have

9ij(C(t)) = 9C(t) (a ' 8') =8i.

Moreover,

[D4jej(t)Ik = dz(t) +I'ii(C(t))et,(t) = 0.

Thus I'lj(C(t)) = 0 for all pairs j, k, since e,k (t) = 6 . We knownothing about the other Christoffel symbols.

c) According to the properties of the exponential mapping, f is COO.Also, f (t, 0) = C(t), and thus

(L)(t o)

)(t o)

Moreover, fort E Rn-1 ( = 7_2 C`e; (t)) we have

(0f(ttz)) _ Eei(t)

`fit

-i_2


Thus (D f)(t,o) is invertible, and (D f)(t.a) from TC(t)(M) to T0(R') is

n n

i=1 i=1

d) R - d(Q, R) is a continuous function on the compact set C. Thusthe inf is achieved at least at one point Q.

e) The proof is by contradiction. We suppose that X = (dy(u)/du)Qand Z = (dC(t)ldt)Q are not orthogonal vectors. We choose

Y = ZgQ(X, Z)/9Q(Z, Z),

for instance. We have

gQ(X - Y, X - Y) = 9Qi(X, X) - [9Q(X, Z)]2/9Q(Z, Z)< gQ(X, X).

Using the result of Exercise 5.40 yields, for small t,

b=d(expQtX,expQtY) < (1+e)tIIX - Y[] <tIXll

with X = D expel X and Y = D Y. Recall that 9i.1 (Q) _ .Thus the curve formed by the two area of geodesics, a piece of

-y from Q to expQ tX, then the shortest geodesic from expQtX toexpQtY, has a length smaller than the length of y. That is not pos-sible.

f) The proof is by contradiction. Let {Qk} be a sequence of points suchthat there exist Qk, and QA,, Qk, # Qk, with

d(Qk, Qkl) = d(Qk, Qk,) = d(Qk, C) < 1/k.

Since C is a compact set, we can suppose that QA,, and Q10, tend toa point R of C. Thus Qk tends to R also, and, according to c), for klarge

Qk = f (tk, uk) = .f (tk, Uk)

if Qk, = f (tk, 0) and Qa = f (tk, 0) -

Now in a neighbourhood of R, according to the inverse functiontheorem, a point such as Qk, for large k, is the image by f of a uniquepair (t, il). Thus we get the contradiction and the existence off > 0.

g) f is bijective on 10, a[ xBE. Moreover, it admits an inverse functionf -1, which is differentiable according to the inverse function theorem.Thus f is a diffeomorphism of ]0, a[ xBE onto 9, and (9, f -1) is a localchart.


h) We saw that DdC(t)1dcei(t) = 0 implies rii(C(t)) = 0, 1 < ij j < n.Moreover let us write that u f (t, ut;) satisfies the geodesic equation:

82f`(t,u4) 8fj(t,ut)Ofk(t,ul;) - 0.&U2 + rjk(t, U ) 8u 8uNow f 1(t, uu) = t, and f i(t, ut) = uei for i > 1. Thus at u = 0the equation above yields r (C(t))t it j = 0, for all E E Rn-1 andall 1 < k < n. This implies q(C(t)) = 0 when 2 < i, j < n and1 < k < n. So the Christoffel symbols vanish on the geodesic C.


a) The rank of 4, is 2. Indeed,

-(a+bcosv)sinu -bsinvcosuD4,= (a+bcosv)cosu -bsinvsinu

0 bcosv

The 2 x 2 determinants of DO are

D1 = (a + b cos v)b sin v 0 if v 3khir (h E Z);

when v = hr

1)2= -b(a + b cos v) sin u cos v 0 0 if u 54kir (k E Z);

and when u = k7r and v = h7r

D3 = (a+bcosv)bcosvcosu 9k 0.

b) If (u, v) ti (fi., v), then 4'(u, v) = tb(u, v). Thus we can define fromV,. Also, 4, is differentiable of rank 2 like 4,. Moreover, .W is injective.Indeed, suppose 4,(u, v) = 4,(u, v). Then the equality

x2 +y2 = (a+bcosv)2 = (a+bcosb)2

implies a + b cos v = a + b cos v, since a > b. Then cos v = cos v andsin v = sin v imply v = v + 2hir (h E Z). Thus u = u + 2kir (k E Z).Finally, 4 is an imbedding since i/, is proper, C x C being a compactset.

c) We have x2 + y2 = a2 + tab cos v + b2(1 - sine v). On M = 4,(III2)

P(x, y, z) _ (x2 + y2 - a2 - b2 + z2)2 - 4a2(b2 - z2) = 0.

We can write P in the form

P(x,y,z) = (x2 + y2 + z2 + a2 - b2)2 - 4a2(x2 + y2).

If P(x, y, z) = 0, then r2 + z2 + a2 - b2 = tar with x = r cos u andy = r sin u, since a > b. But (r - a)2 = b2 - z2 implies z = b sin v andr= a+ b cos v. Thus M= P-1(0).


d) DP is of rank 1 on M. Indeed, if Q = x2 + y2 + z2 - a2 - b2, thenDP = (4xQ, 4yQ, 4zQ + 8a2z). Since Q = tab cos v, Q = 0 impliesz2 = 0, and then V= 8a2 z 9& 0. If Q 96 0 and z j4 0, then

# 0. As (0, 0, 0) 0 M since P(0, 0, 0) = (a2 - b2)2 > 0, it folknvsthat DP is of rank 1 on M, and M is a submanifold of R3. t' isdifferentiable and injective. To r/' there corresponds 0: C x C -+ M,which is differentiable and bijective. Since D4; is of rank 2, 4i is locallyinvertible and its inverse D4-' is differentiable. M is diffeomorphicto C x C.

e) The components of the metric g in the coordinate system (u, v) are

(,)2+(,4)2

= (a+bcosv)2,

Ox 8x 8y Oy 49z $z -(OX)2 + (,,Z)2 =

b;.

The Christoffel symbols that may be nonzero are those in which Bog,appears (those with one v and two u's):

(a+bcosv) bsinvb sin v and r:.

a + b cos v

f) We have

_ _ 2 cos vR 2g Rte

b(a + boos v)'

because (see 4.7)

bcosv b2sin2v _ 62sin2v bcasva+bcosv + (a+bcosv)2 (a+boosv)2 = a+bcosv'

Since igj = b(a +bcosv),

JM

2xRdV=4zr 1cosvdv=0.

g) The differential equations satisfied by the arcs of geodesics are

d2u = -(I" + I"` )du dv2

b sin v du dvdt2 "e °" dt dt a + b cos v dt dt

and

;2 - -ru (du)2 _ -(a+bbcoev) sine Idu)


The equation of a plane through Oz is u = uo. This implies du = 0,and the equations have a solution u = uo, v = vo + At. The intersec-tions of planes through Oz with M are geodesics.

The equation of a plane orthogonal to Oz is z = zo; thus v = vo.Then u = uo + At is a solution of the first equation, but the secondthen implies sinv = 0. Hence v = kir (k E Z). The geodesics whichare in a plane orthogonal to Oz are those which are in the plane z = 0.

h) If d" = 0 at (uo,ve), then u(t) = uo and v(t) = vo + At is a solutionof the equations. When initial conditions are given, the solution isunique according to Cauchy's theorem. We know the solution withinitial data (uo, vo, 0, A). For any A, whatever t, du = 0. Thus ifdu 96 0 at some point, then du does not vanish.

i) The geodesics whose equations are is = uo, v = vo +,ut are thosewhich are in a plane through Oz. For the other geodesics, du nevervanishes and we can choose is as parameter. Let v' and v" denote 9Zand d respectively. We have

dv du d2u2

b sin vv'

(du) 2d t

dd t ' dtz a + b cos v dt J '

d2 =v"(dt)z+v a2.

Hence we find that

(E)bsinv (v)z

b a+bcosv

v" always has the same sign as - sin v.

j) 0(0, 0) = P is the point of coordinates (a+b, 0, 0) in R3. The geodesicfor which (d')p = 0 has for equation v = 0, is = At; it is the circleof radius a + b centered at O in the plane z = 0. If we substitute -vfor v in (E), (E) remains satisfied. Then (u/u, v/ - v) is a symmetrywith respect to the plane z = 0; it permutes the geodesic for which(d') p = a with the one for which (dv) p = -a.

k) When u-'0we have va(u)Nauand

a+b 2b z

Thus va(u) decreases from a until zero if a is small enough. On)0, ua[, va is increasing, v,, (t&,,) > 0 and va(u) < au. Then va, V'a,v." are of the order of a. Define w = lima.o(va/a); w satisfies theequation w" = -°-w and the initial conditions w(0) = 0, w'(0) = 1.


Thus

Jb +b1-bb

w( u) _ Va+bsin (V u

w' vanishes for the first time at uo = z n+b If a is small enough,

then ua is close to uo and smaller than 7r/2. va(ua) is close to n+b

1) Since we have va(0) = 0 and va(ua) = 0, by symmetry that impliesva(2ua) = 0. There are three geodesics from P to Q: this geodesic,the one with (du) p = -a and the arc of circle in the plane z = 0.They have the same length 1a. We saw that, when a tends to zero,la to = 2(a + b)uo = b(a + b). We saw also that if a is smallenough, then ua exists and la < (a + b)ir.

Solution to Problem 5.43.Questions (a) and (b) are solved in Problem 5.32.

c) We have E(vi, Y) = 0. Since DE = 0,

ei(X, Y) = E(DXvi, Y) = -E(vi, DXY).

Thus

k

DXY = DXY - 2 ti (X, Y)vi

As the Euclidean connection is without torsion, E(v., DXY) _£(v1,DyX) = -4(Y, X) since E(vi, [X, Y]) = 0. Thus f (Y, X) is asymmetric bilinear form.

d) We have

k

DXDyZ = bx[DyZ - ei(Y Z)viJi=1k

= DXDyZ - k £ (X, DyZ)vi

k k

- E fi(Y, Z)DXVi - E[DXti(Y, Z)]vi-i=1 i=1

The result follows since E(vi,T) = 0 for any 1 < i < k.

e) We have

R(X,Y, Z, T) = -g[R(X,Y)Z,T]


with R(X, Y)Z = DX Dy Z - Dy DX Z - DpX,y) Z. Moreover,

e(DXDyZ - DybxZ - D1X,y1Z,T) = -R(X,Y, Z, T) = 0.

The result follows from (d).

f) Ifk=landn=2, thenR(X, Y, X, y) = t(X, X)e(Y, y) - [e(X, y)12.

Choose two orthonormal vectors X, Y such that t is diagonal andR(X, Y, X, Y) = A1A2. Thus R = 2A1A2. If k = 1 and n > 2, choose abasis Xt of TQ(M) such that a is diagonal. Then R(X,, X j, X,, Xj) =A{A f. All these values cannot be negative, since two A; have the samesign or some A; are zero.


a) We havea

(cp h) = Bpi [u

h - h)]. Passing to the limit leadsto d(sh)/dt = O Cxh.

Necessity. If h is invariant under G, the left hand side above iszero by hypothesis; thus Cxh = 0.

Sufficiency. Lx h = 0 implies Ot h = c h = h.

b) We have jO*9I = enf I9I ; thus

O*h = I-G*9I-1/"*9 = ef9/ef I9I'In = h.

Necessity. If 10 is a conformal transformation, h is invariant underG. SO 1Cxh = 0.

Sufficiency. Cxh = 0 implies 'Ge h = h. Consequently, tp*g =I'f'*9I1rn919I-1/n is of the form efg.

c) We have L = Y. Since Y" 0, 0. So we can expresst as a function of the coordinates of x, in a neighbourhood of P,in a unique way according to the implicit function theorem. Thenxe = (x), and x1, x2, , z"`1, t form a coordinate system on thisneighbourhood. Indeed, consider the map

r : (x1,,2,... , -1, t) . (1pi (xo), 1pt (zo), ... 'on (xo))

defined on a neighbourhood of 0 E R. Then

1 0 ... 0 Yl0 1 ... 0 y2

(Dr)o =1

0 0 ... 0 Ynis invertible. Thus r is a diffeomorphism of a neighbourhood 9 of0 E Rn onto a neighbourhood of P.

164 5. Riemannial. Manifolds

d) In this coordinate system,

1

£1 h. = I9I-1/n 11C y9 - n191-1Gy 1919]

= IgI-1/n {c9_ 1

t(9i)GY9ij)9 .

Let us compute gijCygij. Since Y = , we have Yi = 1) for i < nand Y' = 1. Then

ViY' _ .qiY' + rijYj = rin = 29'j [O pnj + an9ij - 8j9, i]

and

91CCygij = 9:3an9ij = 21',n = 2V1Y' since g'2 = gJ'.

The equality is proved in a coordinate system; since it is a tensorequality, it is valid in any coordinate system.


a) Choose X such that IIXII = 1; in this case t is the arc length. Supposethat expp tX is defined for 0 < t < r, -r being the largest real numberhaving this property (r exists according to Theorem 5.11). The proofis by contradiction; we suppose r finite. Let {ti } be an increasingsequence such that ti - r. Then Qi = expp tiX (i E N) is a Cauchysequence. Indeed, d(Qi, Qj) < I ti - t j I, and {ti } is a Cauchy sequence.Since M is complete, there exists Q E M such that d(Q, Qi) -. 0 wheni oc. The geodesic has an end point. Q = exprX.

Now we use the theorem on the exponential mapping .'A Q. Thereis a neighbourhood It of Q where expel is a diffeomorphism. Sinceexpp tX , Q when t - r, there exists an s such that expp tX E I2 fors < t < r. Now we know that there is a unique geodesic y (u -' -y(u))from expp sX to Q included in 0 (y(0) = Q and y(r-s) expp sX).Because of the uniqueness. y(r - t) = expp tX. Let Y = Fu )Q; theny(u) = C(u, Q, Y) exists for any u E ] - e, e[, e > 0 smaL. Thus thegeodesic C(s) extends for r < s < r+e. This is in contradiction withthe definition of r.

b) Let y(u) be the geodesic of W from C(t) such that

Cdulc(c) d dtt))C(t)

Since W is compact, W is complete and -y(u) extends itself infinitely.According to the hypothesis -y(u) is a geodesic of M, but there isonly one geodesic with given initial data C(t), namely (d(.'(t)/dt.)C(t)Thus C and y are the same.


c) We have dC/dt = 8/8x1, so

DA (mss) =0,

since the Christoffel symbols vanish on C. The vector fields 8/8x'are parallel along C.

d) The map W x V 3 ( P , -a d(P,Q) > 0 is continuous on a com-pact set. Thus r is achieved: There exist P,, and Q,, such thatd(P0,Q0)=r>0.

e) Let X (t) be a parallel vector field along C. The norm of X (t) isconstant along C. Indeed, gr(t)(X(t),X(t)) = Constant since thecovariant derivatives D(cfdt of g and X(t) are zero. So the mapT, (M) a X(0) -+ X(T) E TQ®(M) is injective; it is an isomorphism.Thus dim H = p. Since dim TQo (V) = q with p + q > n, it followsthat H f) TO- (V) does not reduce to zero, because p + q - (n -1) > 0.Indeed, ( )QQ is perpendicular to H and to TQo (V) according toProposition 5.13. Thus dim(H +TQ0(V)) < n - 1.

f) t _. X (t) and t _. C(t) are C°° differentiable, and we also know that(H, Z) - expR Z is C°G differentiable. Thus (t, A) -+ rya(t) is C°°differentiable. Also, A -+ expQ0 AY is a geodesic of M, included inV since Y E TQa (V). Likewise A exppe AX(0) is a geodesic of M,included in W since X(0) E Tp,(W). Thus f (A) > r.

g) The function

A f (A) _ 9ai(1'a(t))-'a d2j

dtek- dt00

is C°O differentiable. Since f (A) > f (0), we have f'(0) = 0. ChooseA small enough so that the geodesic 7a lies in the coordinate system.Then yo(t) = C(t); thus ryo(t) = t and yo(t) = 0 for i > 1. However,

(&YA(t)) = X(t) and X(t) = > X' \ a )dA ;=a c(t)

with each X' = Constant, since 9(X (t), dc) = 0 and according to c).We have

8A1

= 0 and 9+i(iu(t))dt dt = 1

A=O

since (8kgtt)c(t) = 0 (the Christoffel symbols vanish on C). SinceA -i 7,,(t) satisfies the geodesic equation, we have

(_ n(t))

(dWi(t)) Cd

Wi(t)) = U,x A=O a=o


since the Christoffel symbols vanish on C. So when we computef"(0), we can get something which is not zero only if we differentiate9ij twice. But as dryo/dt = 0 when 1 > 1, we obtain

f "(O) = 2 fO 911(C(t))Xk(t)X(t)dt.0

Since 497k(C(t)) = 0, it follows that (191t9jk)C(t) = 0. So if l = j =1in the expression of Rk1ij, we get

v (X(t), 1 = Rk111(C(t))Xk(t)X'(t) _ -28ik911X;(t)Xk(t)

since X(t) and dc are unit vectors (gik(C(t))Xi(t)Xk(t) = 1).

h) If o,(X (t), dc) > 0, then

f"(0)=- J a(X(t), )dt<0,o

which is a contradiction. Indeed, f is minimum at 0; thus f"(0) > 0.So in that case Wp n V. 0 0.


a) Let [a, b] 3 t -' C(t). Set C(t) = it o e(t). Locally

dC dC=

(dC dC)9 dt , dt 9 dt , aft

,

since any point P E W admits a neighbourhood diffeomorphic to aneighbourhood of P = ir(P). Thus L(C) _> L(C). We can have strictinequality, since some arcs of e may have the same projection. If Csatisfies the geodesic equation, so does C (it is a local condition).

b) Let {Pi} be a Cauchy sequence in (W, g). According to (a), d(Pi, Pj) >d(Pi, Pj) with Pi = ir(P1). Thus {P1} is a Cauchy sequence in M,which is compact. Let P be the limit of {Pi}. P admits a neigh-bourhood B(p) of radius p centered at P such that ir-1(B(p)) isdiffeomorphic to B(p) x F, with F a discrete space, by definition ofa covering manifold. There exists k such that if i > k and j > k thend(Pi, Pj) < p/2. Pi belongs to an open set 0 of B(p) x F which isdiffeomorphic to B(p) by 7r/S2. Since d(Pi, P5) < p/2, all points Pjfor j > k belong to Sl. Thus there exists P E Cl n it-1(P) such thatPj-+P'when

c) Any point of C admits a neighbourhood 11 such that ir'1(fl) is dif-feomorphic to fl x F. Since C is compact, a finite number of theseneighbourhoods, 01, f12,- , ilk, cover C. P belongs to 01i and is

Solutions to Ekerciges and Problems 167

diffeomorphic to f21 by 7r/521. Let 4)1 be the inverse of 1r/521. Then

('1(Cnf21) is an arc [a,al[ W. We choose

C(al) E 7r-'(%) n t, (C n %).C(al) belongs to 4 and is diffeomorphic to ftra by 7r/52. And soon. We construct an arc e of M which is locally diffeomorphic toC. If C(b) = P, C would be closed, thus homotopic to zero sinceW is simply connected. By 7r, we would have C homotopic to zero,which is in contradiction with the hypothesis. Indeed, let F(t, A) bea continuous map of [a, b] x [0,1] into W such that F(t, 0) = C(t) andF(t,1) = P, for instance. Then (t, A) -+ F(t,.\) = Yr o F(t, A) wouldbe a continuous map of [a, b] x [0,11 into M such that F(t, 0) = C(t)and F(t,1) = P.

d) Since Co is homotopic to C, there exists a continuous map G(t, 1)of [a, b] x [0,1) into M such that G(t, 0) = C(t) and G(t,1) = Co(t).%, ... , lk is a covering of C. For A small enough (A < el ), 01, ... , flkis also a covering of G([a, b] x [0, ell). Set G(t,1) = o G(t,1) for(t,1) E [0, all x [0, ell, 04, 1) _ $s o G(t,1) for (t,1) E jai, aril x [0, ell ,and so on. Then we consider the curve CE, = G(t, el ). From CE,we obtain C2 just as we obtained CE, from C. And so on. Thusthere exists a continuous map G(t, l) of [a, b] x [0,1] into W suchthat 67(t, 0) = C(t) and 0(t, 1) = co(t). Then I --, G(b,1) is acontinuous function in lr-1(P) = F; thus this function is constant,Co(b) = C(b) = P.

e) If we choose R close to P and it close to P, by the constructionabove 1 will be close to C. Thus k is close to P and R - f (R)is continuous. Since f (R) > 0 and M is compact, f (R) achieves itsminimum at least at one point Q E M.

f) We saw (question a)) that 7r(j) is a geodesic from Q to Q, and itslength is f (Q). Moreover ir(7) is also a geodesic at Q, as otherwisethere exists a curve ' close to a(y), thus homotopic to 7r(5), suchthat its length is smaller than f (Q). This is impossible, because thenthe curve ^' constructed above would be strictly smaller than f (Q).

We have proved the existence of a shortest closed geodesic in anyhoumtopy class.

Chapter 6

The Yamabe Problem:An Introduction toResearch

6.1. For leas than forty years, analysis has been used in order to try to solveproblems in geometry. A pioneering work has been completed by Yamabe[161, on the scalar curvature problem. In Chapter 5, we have acquired enoughknowledge to study nonlinear problems in Riemannian geometry. Neverthe-less, for solving then it is necessary to know some theorems in functionalanalysis, at least three of which we will give below. For example, let usstudy the Yamabe problem.

6.2. The Yamabe problem. Yamabe wanted to prove that on a compactRiemannian manifold (M,,, g) of dimension n > 3 there always exists ametric with constant scalar curvature. This is a geometrical problem whichwe will transform into a problem in analysis. Let R be the scalar curvatureof (M,,, g). We suppose R is not constant-otherwise we have nothing todo. If we think of all possible changes of metric, the simplest is a conformalone. Therefore Yamabe considered the conformal metric g' = efg, where fis a C°° function on M.

6.3. Let us compute R' in terms of f. If we are in a local chart, Vilk andi i denote the Christoffel symbols corresponding to 9' and g respectively:

I,rtut - It = 29im(9mki%f + 9miOkf - 9ik8mf ),

according to 5.5.

169

170 6. The Yamabe Problem

To simplify we assume that the coordinates are normal at P for g(gij(P) = S? and rikj (P) = 0) (see 5.7):

R'kij - Rkij = rjk) - aj(rik - rik) + rimrk - rj".rik .

Thus at P

R'kj -Rkj =ai(rk-rk) -a(rik-rik)+rimrk

R'kj -Rkj

1 im= 29 (9mkaij f + 9mjafks,l - 9jkaimf )

rr'mjmik ,

1 im- 29 (9mkaijf +9miOjkf -9ikajmf )

+ 4(8mf +namf -amf)9m`(9(k8jf +9ljakf -9jka,f)

-4

(91m8jf + 9ljamf - 9jmalf)(9gka;f + 9giakf - 9ikagf)9i'9"'q,

R'kj - Rkj = (28kj f + Of9jk) - 2 (n8kjf) + 4 (2akfajf - OV I VVI9jk)

- !(9agajf + bjVg f if )(9gka;f + 9giakf -900j),

R'., - Rkj =-n

2- 2

akjf +121f9jk+

n2akfajf - nV'f01.,fgjk

- 4(n(9jfakf +2akfajf -2O''fO,.fgjk),

and

(1)

n-2 1R'kj - Rkj = -2

VkVj f + 2Of9jk

+ n 4 2vkfvjf - n 4 20" foJ9jk-

We have equality between two tensor fields, so the identity holds in anycoordinate system. Thus, contracting (1) by gkj, we obtain, since 9''j =ef 9'?,

(2) R'ef - R = (n-1)Of - (n-2)(n-1)0°f O,,f

4

On this expression we do not see how to proceed to find a function fsuch that R' is constant. To simplify the equation, Yamabe considered theconformal deformation in the form g' = p4/(n 2).9. Now w must be strictlypositive.

An Introduction to Research 171

Letting f = n422 LogV in (2), we find that

(3) (4(n n-

2)1)Ap

with cP>0.

This is an elliptic equation (the Laplacian operator is elliptic) which is non-linear: the exponent in the right hand side is greater than 1.

6.4. The Yamabe problem is a geometrical one: find a metric with constantscalar curvature. We have proven that if we look for a conformal metric,the problem is equivalent to proving that equation (3) with R' = Constanthas a strictly positive C00 solution. It is a problem of PDEs (PDEs meanspartial differential equations).

It is easy to see that, if there are two solutions of (3), the new scalarcurvatures of them have the same sign. Indeed, assume that g' has constantscalar curvature R' and g = y4/(a-2)g has constant scalar curvature R. Letus compute R' in the metric g. If we set V = yi, this is possible since 'Pand y are strictly positive. Thus g' = 14/(n-2)g and

(4) 4(n - 1)A+ RV, = jeV,(n+2)/(n-2)

(n - 2)

Here -9''(d jV, - T 8ku'). Now let us integrate (4)with respect to the metric

dV(5) k f R' J 0'67since f O./'dV = 0.

Thus R and R' have the same sign (or are both equal to zero).If R = R' = 0, then 0& = 0; hence u,/ = Constant. The solutions of (4)

are proportional. If k = R' < 0 we easily see that (4) has only one solutionrG - 1. Indeed, at a point P where ip is maximum we have (4 )p > 0; thusR'Vi(P).- 2 > Rrp(P), and we find that V,(P)4/(n-2) < 1. Now at a point. Q

where V, is minimum we have(Da')Q < 0; thusR'V,(Q)L(^±2i

< Rt'(Q), andwe find that u'(Q)4/(n-2) > 1. Consequently u'(P) = Ii(Q) = 1.

We have proven that there are three cases, according to whether R' ispositive, negative or zero, and in the negative and zero cases the solutionsof (4) are proportional. In the negative case, if k and R' are constant, then0 = (R/R')(i-2)/4 is the unique solution of (4).

6.5. The variational method. How can we prove the existence of a so-lution of (3) with R' = Constant? There are several methods for solvingnonlinear equations; one of the most powerful is the variational method.We have to consider, on a set A of functions, a functional I boundedfrom below on A such that the Euler equation of the variational problem


is (3). Then, {u;} (i E N) being a minimizing sequence (i.e. {u$} C Aand limi.-,,, I(uj) = p, the inf of 1(u) on A), we try to exhibit a subse-quence {u?} c {u;} which converges to a strictly positive solution of (3)with R' = Constant.

This is roughly the idea of the method. In practice, it is not so easy, aswe will see.

6.6. For some equations, we can imagine several functionals; for of hers thereare. none. Here R' appears as a Lagrange multiplier.

The constraint which Will give j,(n+2)/(n-2) is

K(u) = N J I uI NdV with N = n2n2nf

Indeed, Du[(u2)N/2](v) = 2 (u2) i -l2uv = NIuI4/(ii-2)uv.

The functional which will. give twice the left hand side of (3) is

I(u) _ 4(n - 1) V'uQ;u+ Ru2)dV.- (n - 2)

Indeed,

rDuI(v) = 2

4(n - 1)Jv

(n - 2)V`uViv + Ruv dV.

If we formally perform an integration by parts, we obtain

Du1(v) 2j l (nn- 1) Au + Ru J vdV.l 2)

We have to use the fact that DI(v) is proportional to DuK(v). Thus theEuler equation is

4n-1(6) (n - 2

Du + Ru = lrlul4/(n-2)u

if the constraint is K(u) = 11N. Indeed, p is the Lagrange multiplier:multiplying (6) by u and integrating lead to I(u) = pNK(u).

6.7. The 'Sobolev imbedding theorem. Now we have to choose the setA. Assume that we choose for A the set of CO° strictly positive functions Vsuch that

l/N

IIPIIN = (f Ic INdV) = 1.

It will be very difficult, even impossible, to prove that u, the limit of asubsequence (u,), is a strictly positive C°° function. On the other hand, wecan write the Euler equation only if u E A. For this reason, we introducethe Sobolev space H1.


Let E be the set of C°O functions on M endowed with the norm

IIuIIH; = (IIVu!I2 +IIu1I2)1/2.

The gradient is ]VuJ = (V{uViu)1/2; its L2 norm is (fv V1uV'udV)'/2

The Sobolev apace H1 is the completion of E with respect to the normII IIH2. It is a Hilbert space. Moreover, the Sobolev imbedding theorem (see(2]) asserts that H? C LN for compact Riemannian manifolds, and that theinclusion is continuous (i.e. there exists a constant C such that any u E Hisatisfies nuIIN < Coup). To solve the variational problem, we will chooseA = {u E H1, u >_ 0 1 IIu1IN = 1}, which makes sense according to theSobolev imbedding theorem.

In general, choosing a constraint like u > 0 implies several difficulties(we cannot write the Euler equation when u is zero). These difficulties arenot present here. Indeed, it is a fact that if u E H2 , then Jul E Hl andIVIull = IVul almost everywhere; we have I(u) = I(Jul), and obviouslyK(u) = K(Jul).

Thus the inf of 1(u) on A is equal to the inf of I(u) on A = {u E H1 IIIUIIN = 1}. Therefore, the Euler equation can be written without anytechnical problem, and the limit u will be positive or zero: u = Jul. Henceequation (6) is equation (3) with R' = p.

8.8. We therefore consider the following variational problem: find inf I(u)for U E A. Recall that

I(u) = 4(n - 1) fm IVuJ2dV + J RuadV(rt - 2) M

and A = {uEH?,u>0I IIuIIN=1}.Let us prove that p = infA I(u) is finite. Observe that

1(u) > inf(R,0)IIu0I2 >_ inf(R,0)V2/"IIuIIN = inf(R,0)V2/",

according to Holder's inequality. Here V = fu dV; without loss of generalitywe may suppose that the volume V is equal to 1. Indeed, by a homotheticchange of metric we can set the volume equal to 1, and a homothetic changeof metric is a conformal one. Let g' = kg with k > 0; then

V1 = J dV' = k"/2 J dV = k"/2V.

We have only to choose k = V-2/". Henceforth we assume that the vol-ume is equal to 1. Let us consider a minimizing sequence {i4} C A:1(uj) - p. We can suppose I(ur) < p+1. According to Holder's inequality,IIu D2 < HwlIN =1. Moreover,

<- it + 1- inf (R, 0) IIU,112 < Constant.4(n - 2) O Vu p22 2 -


Thus the sequence {u;} is bounded in H12-

6.9. Weak convergence. The Banach theorem (see [2]) asserts, in par'-ticular, that in a Hilbert space the closed unit ball is weakly sequentiallycompact. This implies, for every bounded sequence {u; } in M12, that thereexist w E H1 and a subsequence {u.) of {u;} such that uj -4 w weakly inHl (i.e., for any u E Hl, (u, uj)Hi - (u, w)Hl as j -' oo). It is easy toverify that convergence in H1 (1 uj - wq, -, 0) implies weak convergence.We say that w is the weak limit in H1 of the sequence {uj}. One can provethat Ilweak l m ll <- in ll II; thus

(7) dwIIH; <- liminf Ilujllfn

6.10. We have used two theorems, the Sobolev theorem and the Banach the-orem; we now have to use a third theorem, the Kondrakov theorem (see [21).It asserts that, on compact Riemannian manifolds, the inclusion H1 C Lq iscompact for 1 < q < N = 2n/(n - 2). This means that, if {u;} is a boundedsequence in H1, there exists a subsequence which converges in Lq.

Returning to our problem, according to the Kondrakov theorem, we canchoose the subsequence {uj} such that uj -. I7 in L2 and also uj 0 a.e.It is not difficult to prove that in fact w = T. Thus we have IIuj112 11w1I2,

I I V w 112 <- lim inf j _.> I l V u j 112 according to (7), and

f Ru3dV - f RwvdV.v v

Indeed,,

RwvdV- f Ruj2dVI <supIRlllw-ujlI2(Ilwll2+IIujII2)- 0-V v

We infer that I (w) < µ = lim 1(u j). Assume that we are able to prove thatIIwIIN = 1. Then w E A and I(w) =,u according to the definition of At (ifw E A. 1(w) > µ). We then write the Euler equation.

6.11. So we have exhibited w E H12, w > 0, w $ 0, which satisfies equation(6) weakly in H12. That is, for any v E H?,

4(n - fm V'wVivdV + f RwvdV = µ f wN-1vdV.(n - 2)M M

The maximum principle then implies w > 0, and some regularity theoremsimply w E C. Thus w satisfies equation (6) in the sense of functions.

Unfortunately it is impossible to prove that (lw11N = 1. This is whyYamabe considered an approximated equation

(8) 4 (n - 1) Au + Ru = icquq-1, u > 0 for 2 < q < N.


For this equation the proof above works. Indeed, now

Aq={uEH?,u>0 Ilupq=l}.According to the Kondrakov theorem, we can choose the subsequence

{ul} such that u1 -+ wq in Lq. Consequently 1 = IIu;IIq -' IIwgIIq. Thuswq E A, and wq satisfies equation (8). Now we have to see what happenswhen q --+ N. This is difficult. Nevertheless, we have succeeded in provingthe following.

6.12. Theorem (Aubin; see [1] or [2]). We always have is < n(nIf s < n(n - 1)w2n/", there exists a strictly positive solution w E COO of (3)with R' =,u and IIwlIN = 1. Here w is the volume of the sphere of radius 1and dimension n. µ is defined in 6.8.

In order to see. that the inequality of Theorem 6.12 is strict, we haveto put test functions in the functional I. If the manifold is not conformalto the sphere, we can prove that p < n(n - 1)w2n/" Thus the Yamabeproblem is solved. There always exists a conformal metric g' such thatR' = Constant (the sphere has constant curvature, hence constant scalarcurvature).

6.13. Perspective in research. On compact Riemannian manifolds, theYamabe problem is solved. One may pose the same problem on completenoncompact manifolds. There are only a few results, and they are differentfrom those on compact manifolds. For instance, it is not clear that thereare three different cases, positive, negative or zero (according to the signof u in the compact case). We can also consider generalised versions ofthe Yamabe problem. For example, let us consider the prescribed scalarcurvature problem:

Let f be a C°° function on the Riemannian manifold (M,,,g). We askthe following question: Does there exist a conformal metric g' for g suchthat R' = f?

In dimension n > 3, this problem is equivalent to solving the followingequation:

(n - 1) .4(n-)Ocp+R = fcp(" s), tp>0.

In dimension n = 2, the equation to solve is (see (2) with n = 2)

AV+R= fe`e.This problem is particularly hard on the sphere, where it is the so-calledNirenberg problem.

Bibliography

[11 Aubin, T.: Nonlinear Analysis on Manifolds. Monge-Ampere Equations (Grundlehren252), Springer-Verlag, New York, 1982.

[21 Aubin, T.: Some Nonlinear Problems in Riemannian Geometry, Springer-Verlag,New York, 1998.

[31 Gallot, S., Hulin, D., and Lafontaine, J.: Riemannian Geometry, Springer-Verlag,New York, 1987.

[4J Helgason, S.: Differential Geometry, Lie Groups and Symmetric Spaces. AcademicPress, New York, 1978.

151 Kobayashi, S., and Nomizu, K.: Foundations of Differential Geometry. I and II,Interscience, New York, 1963.

[61 Lichnerowicz, A.: Ggomktrie des groupes de transformations, Dunod, Paris, 1958.

[7) Malliavin, P.: Gdometrie diffenentielle intrinseque, Hermann, Paris, 1972.

[81 Milnor, J.: Morse Theory (Annals Studies 51), Princeton Univ. Press, 1963.

[91 Milnor, .1.: Topology from the Differentiable Viewpoint, The University Press of Vir-ginia, 1969.

[101 Narasimhan, R.: Analysis on Real and Complex Manifolds, Masson, Paris, and North-Holland, Amsterdam, 1971.

(11] de Rham, G.: Sur la thdorie des formes differentielles harmoniques, Ann. Univ.Grenoble 22 (1946), 135-152.

1121 de Rham, G.: Varietes diffdrentiables, Hermann, Paris, 1955.

[131 Spivak, M.: Differential Geometry (5 volumes), Publish or Perish, Berkeley, 1979.

[141 Sternberg, S.: Lectures on Differential Geometry, Prentice-Hall, Englewood Cliffs,NJ, 1965.

(151 Warner, F.: Foundations of Differentiable Manifolds and Lie Groups, Academic Press,New York, 1971.

[161 Yamabe, H.: On the deformation of Riemannian structures on compact manifolds,Osaka Math J. 12 (1960), 21-37.

177

Subject Index

adjoint operator flJfl conformal metric 03algebraically equivalent (systems) 311 [Aconnected 03arc U connection 4.2arc length &3 RiemannianAseoli's theorem 0.39 contraction (of indices) 0-22, 5.8

contravariant QJAclass coordinates L4d,flnltion L4 geodesic 512equivalent L5 normal

cotangent bundle 23Banach space 0.24 countable at Infinity LiiiBanach'a theorem 5.9 covariant (Index., tensor)basis 23j covenant derivativeBetti numbers 5.25 of a tensor field 413Blanchi's Identities 413, 5.fl of a vector field 4.2boundary 2.35 covering 0.1

2.15 covering manifold LiZbundle 211 thInner Liii

aiticaltheorem point

dependance on Initial condition 0.39chart L4 curvatureChr4.stoffel symbols 4.3. definitIon 4.1closed Ricci &9

form 3.15 scaler 5.2set 03 sectIonal 5.9system 314 tensor 5.8

set 03 diffeomorphiam ILlSsystem 514 differentIable 013

511 functIon Lflcompact set (LB manifold 1.8compatible (vector flelde) 2.11 with boundary 2.35complete 0.24 113completely integrable 2.15 dIfferentIal 2.25

179

Notation

Basic Notation

We use the Einstein summation conv ration.Positive means strictly positive.Nonnegative means positive or zero.Compact manifold means compact manifold without boundary unieas we say otherwise-M is the set of positive integers, n E N.Rn is the Euclidean n-space, n > 2, with points r = (xi, x2, ..., xn), xi E R, the set of realnumbers.When it is not otherwise stated, a coordinate system (mil I 5,:S. in R" (or (z, V, x, t) in R") ischosen to be orthonormal.We often write 8+ for 8/& and Al for 8i8j.Sometimes we write V 1 for V1V1.]a, b( or (a, b) means an open interval in R. (a, b) may also be the point of R2 whose coordinatesare a and b.

Notation Index

Ck , C-, C- difermaiablefunction. 0.23, 0.26manifold. 1.6

CC = nl/(n - P)ipid exterior differential. 2.24, 2.25di dtial off. 0.26dV Rlemannian volume element. 5.23DxY or D(X,Y) 4.2dxj 0.21, 2.25d(P, Q) distance from P to Q. 5.4E={xERn(x1<0) 2.34E is the Euclidean metric 5.2expp(X) 5.11

f'(x) differential of f at x. 0.23

g Riemannian metric. 5.19.j, g`1 the components of g 5.5Cx Lie derivative with respect to X. 3.4M, or Af manifold of dimension n. 1.1(Mn, 9) Riemannian manifold. 5.1O(n) 2.45P,,(R) real projective space. 1.9R(X,Y) 4.7

R 5.9Sn the sphere of dimension n

183

184

Tp(M) 2.3T(M) 2.5T. (M) 2.5T(X,Y) 4.6

p) 1.20T(n,p,k) 1.217';(M) 2.141'(M) space of vector 8ekie on M. 2.14ri`i Christ" symbols. 4.3, 4.5, 5.5A Laplacian operator. S.188M boundary of M. 2.35, 2.38b codifierentiaL 5.17

Kronecker's symbol. 0.14, 5.5'R 5.15

Notation

AD(M) 2.14A(M) 2.22# Lebesgue measure. 028o(X,Y) eectJooal curvature. 5.94. 2.64!" 2.7, 2.23x the 1 Potncar4 characteristic. 5.22

W. the volume of the unit sphere S.Vu 4.13VY = V1Y.abrl ® 8/8si 4.4(8/8x4) p tangent vector at P. 2.3[X, Y) bracket. 2.15* adjoint operator. 5.1

terror product. 0.13

www.ams.org

a course in differential geometry thierry aubin

Documents