Graduate Texts in Mathematics 14 Managing Editors: P. R. Halmos C. C. Moore M: Golubitsky V. Guillemin Stable Mappings and Their Singularities Springer-Verlag New York·Heidelberg·Berlin Martin Golubitsky Queens College Department of Mathematics Flushing, New York 11367 Managing Editors P. R. Halmos Indiana University Department of Mathematics Swain Hall East Bloomington, Indiana 47401 Victor Guillemin Massachusetts Institute of Technology Department of Mathematics Cambridge, Mass. 02139 C. C. Moore University of California at Berkeley Department of Mathematics Berkeley, Ca. 94720 AMS Subject Classification (1973) Primary: 57D45, 58C25, 57D35, 57D40 Secondary: 57D50-57D70, 58A05, 58C15, 58D05, 58D15, 57E99 Library of Congress Cataloging in Publication Data Golubitsky, M 1945Stable mappings and their singularities. (Graduate texts in mathematics, 14) 1. Differentiable mappings. 2. Singularities (Mathematics) 3. Manifolds (Mathematics) I. Guillemin, V., 1937- joint author. II. Title. III. Series. QA613.64.G64 516.36 73-18276 ISBN-13: 978-0-387-90073-5 e-ISBN-13: 978-1-4615-7904-5 DOl: 10.1007/978-1-4615-7904-5 All rights reserved. No part of this book may be translated or reproduced in any form without written permission from Springer-Verlag. @) 1973 by Springer-Verlag New York Inc. PREFACE This book aims to present to first and second year graduate students a beautiful and relatively accessible field of mathematics-the theory of singularities of stable differentiable mappings. The study of stable singularities is based on the now classical theories of Hassler Whitney, who determined the generic singularities (or lack of them) for mappings of Rn ~ Rm (m ~ 2n - 1) and R2 ~ R2, and Marston Morse, who studied these singularities for Rn ~ R. It was Rene Thorn who noticed (in the late '50's) that all of these results could be incorporated into one theory. The 1960 Bonn notes of Thom and Harold Levine (reprinted in [42]) gave the first general exposition ofthis theory. However, these notes preceded the work of Bernard Malgrange [23] on what is now known as the Malgrange Preparation Theorem-which allows the relatively easy computation of normal forms of stable singularities as well as the proof of the main theorem in the subject-and the definitive work of John Mather. More recently, two survey articles have appeared, by Arnold [4] and Wall [53], which have done much to codify the new material; still there is no totally accessible description of this subject for the beginning student. We hope that these notes will partially fill this gap. In writing this manuscript, we have repeatedly cribbed from the sources mentioned above-in particular, the Thom-Levine notes and the six basic papers by Mather. This is one of those cases where the hackneyed phrase "if it were not for the efforts of ... , this work would not have been possible" applies without qualification. A few words about our approach to this material: We have avoided (although our students may not always have believed us) doing proofs in the greatest generality possible. For example, we assume in many places that certain manifolds are compact and that, in general, manifolds have no boundaries, in 0rder to reduce the technical details. Also, we have tried to give an abundance of low-dimensional examples, particularly in the later chapters. For those topics that we do cover, we have attempted to "fill in all the details," realizing, as our personal experiences have shown, that this phrase has a different interpretation from author to author, from chapter to chapter, and-as we strongly suspect-from authors to readers. Finally, we are aware that there are blocks of material which have been incluaed for completeness' sake and which only a diehard perfectionist would slog through -especially on the first reading although probably on the last as well. Conversely, there are sections which we consider to be right at the ••heart of the matter." These considerations have led us to include a Reader's Guide to the various sections. Chapter I: This is elementary manifold theory. The more sophisticated reader will have seen most of this material already but is advised to glance through it in order to become familiar with the notational conventions used elsewhere in the book. For the reader who has had some manifold theory before, v vi Preface Chapter I can be used as a source of standard facts which he may have forgotten. Chapter II: The main results on stability proved in the later chapters depend on two deep theorems from analysis: Sard's theorem and the Malgrange preparation theorem. This chapter deals with Sard's theorem in its various forms. In §l is proved the classical Sard's theorem. Sections 2-4 give a reformulation of it which is particularly convenient for applications to differentiable maps: the Thom transversality theorem. These sections are essential for what follows, but there are technical details that the reader is well-advised to skip on the first reading. We suggest that the reader absorb the notion of k-jets in §2, look over the first part of §3 (through Proposition 3.5) but assume, without going through the proofs, the material in the last half of this section. (The results in the second half of §3 would be easier to prove if the domain X were a compact manifold. Unfortunately, even if we were only to work with compact domains, the stability problem leads us to consider certain noncompact domains like X x X - LiX.) In §4, the reader should probably skip the details of the proof of the multijet transversality theorem (Theorem 4.13). It is here that the difficulties with X x X - LiX make their first appearance. Sections 5 and 6 include typical applications of the transversality theorem. The tubular neighborhood theorem, §7, is a technical result inserted here because it is easy to deduce from the Whitney embedding theorem in §5. Chapter III: We recommend this chapter be read carefully, as it contains in embryo the main ideas of the stability theory. The first section gives an incorrect but heuristically useful "proof" of the Mather stability theorem: the equivalence of stability and infinitesimal stability. (The theorem is actually proved in Chapter V.) For motivational reasons we discuss some facts about infinite dimensional manifolds. These facts are used nowhere in the subsequent chapters, so the reader should not be disturbed that they are only sketchily developed. In the remaining three sections, we give all the elementary examples of stable mappings. The proofs depend on the material in Chapter II and the yet to be proved Mather criterion for stability. Chapter IV gives the second main result from analysis needed for the stability theory: the Malgrange preparation theorem. Like Chapter II, this chapter is a little technical. We have provided a way for the reader to get through it without getting bogged down in details: in the first section, we discuss the classical Weierstrass preparation theorem-the holomorphic version of the Malgrange theorem. The proof given is fairly easy to understand, and has the virtue that the adaptation of it to a proof of the Malgrange preparation theorem requires only one additional fact, namely, the Nirenberg extension lemma (Proposition 2.4). The proof of this lemma can probably be skipped by the reader on a first reading as it is hard and technical. In the third section, the form of the preparation theorem we will be using in subsequent chapters is given. The reader should take some pains to under- Preface vii stand it (particularly if his background in algebra is a little shaky, as it is couched in the language of rings and modules). Chapter V contains the proof of Mather's fundamental theorem on stability. The chapter is divided into two halves; §§1-4 contain the proof that infinitesimal stability implies stability and §§5 and 6 give the converse. In the process of proving the equivalence between these two forms of stability we prove their equivalence with other types of stability as well. For the reader who is confused by the maze of implications we provide in §7 a short summary of our line of argument. It should be noted that in these arguments we assume the domain X is compact and without boundary. These assumptions could be weakened but at the expense of making the proof more complicated. One pleasant feature of the proof given here is that it avoids Banach manifolds and the global Mather division theorem. Chapters VI and VII provide two classification schemes for stable singularities. The one discussed in Chapter VI is due to Thom [46] and Boardman [6]. The second scheme, due to Mather and presented in the last chapter, is based on the" local ring" of a map. One of the main results of these two chapters is a complete classification of all equidimensional stable maps and their singularities in dimensions :s; 4. (See VII, §6.) The reader should be warned that the derivation ofthe "normal forms" for some stable singularities (VII, §§4 and 5) tend to be tedious and repetitive. Finally, the Appendix contains, for completeness, a proof of all the facts about Lie groups needed for the proofs of Theorems in Chapters V and VI. This book is intended for first and second year graduate students who have limited-or no-experience dealing with manifolds. We have assumed throughout that the reader has a reasonable background in undergraduate linear algebra, advanced calculus, point set topology, and algebra, and some knowledge of the theory of functions of one complex variable and ordinary differential equations. Our implementation of this assumption-i.e., the decisions on which details to include in the text and which to omit-varied according to which undergraduate courses we happened to be teaching, the time of day, the tides, and possibly the economy. On the other hand, we are reasonably confident that this type of background will be sufficient for someone to read through the volume. Of course, we realize that a healthy dose of that cure-all called "mathematical sophistication" and a previous exposure to the general theory of manifolds would do wonders in helping the reader through the preliminaries and into the more interesting material of the later chapters. Finally, we note that we have made no attempt to create an encyclopedia of known facts about stable mappings and their singularities, but rather to present what we consider to be basic to understanding the volumes of material that have been produced on the subject by many authors in the past few years. For the reader who is interested in more advanced material, we viii Preface recommend perusing the volumes of the" Proceedings of Liverpool Singularities" [42,43], Thorn's basic philosophical work, "Stabilite Structurelle et Morphogenese" [47], Tougeron's work, "Ideaux de Fonctions Differentiabies" [50], Mather's forthcoming book, and the articles referred to above. There were many people who were involved in one way or another with the writing of this book. The person to whom we are most indebted is John Mather, whose papers [26-31] contain almost all the fundamental results of stability theory, and with whom we were fortunately able to consult frequently. We are also indebted to Harold Levine for having introduced us to Mather's work, and, for support and inspiration, to Shlomo Sternberg, Dave Schaeffer, Rob Kirby, and John Guckenheimer. For help with the editing of the manuscript we are grateful to Fred Kochman and Jim Damon. For help with some of the figures we thank Molly Scheffe. Finally, our thanks to Marni E1ci, Phyllis Ruby, and Kathy Ramos for typing the manuscript and, in particular, to Marni for helping to correct our execrable prose. Cambridge, Mass. August, 1973 Martin Golubitsky Victor W. Guillemin TABLE OF CONTENTS Preface Chapter I: Preliminaries on Manifolds §l. Manifolds §2. Differentiable Mappings and Submanifolds §3. Tangent Spaces §4. Partitions of Unity §5. Vector Bundles §6. Integration of Vector Fields Chapter II: Transversality §l. Sard's Theorem §2. Jet Bundles . §3. The Whitney Coo Topology §4. Transversality §5. The Whitney Embedding Theorem §6. Morse Theory §7. The Tubular Neighborhood Theorem Chapter III: Stable Mappings §l. Stable and Infinitesimally Stable Mappings §2. Examples §3. Immersions with Normal Crossings §4. Submersions with Folds . Chapter IV: The Malgrange Preparation Theorem §l. The Weierstrass Preparation Theorem §2. The Malgrange Preparation Theorem §3. The Generalized Malgrange Preparation Theorem Chapter V: Various Equivalent Notions of Stability §l. Another Formulation of Infinitesimal Stability §2. Stability Under Deformations . §3. A Characterization of Trivial Deformations §4. Infinitesimal Stability => Stability §5. Local Transverse Stability §6. Transverse Stability §7. Summary ix v 1 6 12 15 18 27 30 37 42 50 59 63 69 72 78 82 87 91 94 103 111 118 123 127 131 138 141 x Table of Contents Chapter VI: Classification of Singularities, Part I: The Thorn-Boardman Invariants §1. The Sr Classification §2. The Whitney Theory for Generic Mappings between 2-Manifolds . §3. The Intrinsic Derivative §4. The Sr,s Singularities §5. The Thorn-Boardman Stratification §6. Stable Maps Are Not Dense . Chapter VII: Classification of Singularities, Part II: The Local Ring ofa Singularity §1. Introduction. §2. Finite Mappings §3. Contact Classes and Morin Singularities §4. Canonical Forms for Morin Singularities §5. Umbilics §6. Stable Mappings in Low Dimensions Appendix §A. Lie Groups . Bibliography Symbol Index Index 143 145 149 152 156 160 165 167 170 177 182 191 194 200 205 207 Stable Mappings and Their Singularities Chapter I Preliminaries on Manifolds §1. Manifolds Let R denote the real numbers and Rn denote n-dimensional Euclidean space. Points of Rn will be denoted by n-tuples of real numbers (Xl' ... , xn) and Rn will always be topologized in the standard way. Let U be subset of Rn. Then denote by V the closure of U, and by lnt (U) the interior of U. Let U be an open set, f: U -l> R, and X E U. Denote by (8ff8xi)(X) the partial derivative off with respect to the ith variable Xi at x. To denote a higher order mixed partial derivative, we will use multi-indices, i.e., let a = (al , ..• , an) be an n-tuple of non-negative integers. Then 81al 81al c;af = 8 ct 8 ct 8 ct f where lal = al+ ... +anuX Xl 1 X2 2· •• Xn n andf: U -l> R is k-times differentiable (or of class C", or Ck ) if (8Ict IJf8xct )(x) exists and is continuous for every n-tuple of non-negative integers a with lal ~ k. (Note that when a = (0, ... ,0), 8ctff8xct is defined to be f.) f is real analytic on U if the Taylor series offabout each point in U converges to fin a neighbourhood (nbhd) of that point. Suppose Rm where U is an open subset of Rn andf is some realvalued function defined in the range of Rm, U an open subset ofRn. (a)

is differentiable of class Ck • (c) 1> is real analytic if the pull-back by 1> of any real analytic real-valued function defined on the range of1> is real analytic. Let Rm be Cl differentiable in U and Xo a point in U. Then by Taylor's theorem there exists a unique linear map (d1»xo: Rn -l> Rm and a function p: U -l> Rm such that f(x) = f(xo) + (d at Xo; it is given with respect to the coordinates Xl> ••. , Xn on Rn and Yl, ... , Ym on Rm by the m x n matrix where 4>i : Rm -+- R(l :$ i :$ m) are the m coordinate functions defining 4>. The chain rule holds, of course. That is, if 4>: U -+- Rm and if;: V-+- RP are both C1 differentiable where U c Rn and V c Rm are open and V:::::> 4>(U), then d(if;·4»xo = (dif;)",(xo)·(d4»xo for every Xo in U. Theorem 1.2. (Inverse Function Theorem). Let U c Rn be open andp be a point in U. Let 4>: U -+- Rn be a C" differentiable mapping. Assume that (d4»p: Rn -+- Rn is invertible. Then there exists an open set V in Rn contained in the range of 4> and a mapping if;: V-+- U, differentiable of class C", such that 4>.if;(x) = xfor every x in V, and if;.4>(x) = xfor every x in if;(V). Proof See appendix of Sternberg; or Lang. 0 Definition 1.3. A local homeomorphism of Rn is a homeomorphism of some open subset ofRn onto another. (So the domain ofa local homeomorphism need not be all ofRn.) Let 4> be a mapping. Denote by dom 4> the domain of 4>. Also, if U c dom 4> denote by 4> IU the restriction of 4> to U. If X is a set, then idx : X -+- X denotes the identity mapping on X. Definition 1.4. A pseudogroup on Rn is a collection r of local homeomorphisms on Rn with the following properties: (a) idR n is in r, (b) if4> and if; are in r with dom if; = range of4> then if;.4> is in r, i.e., r is closed under compositionfor allpairs ofelementsfor which this operation makes sense. (c) if4> is in r, then 4> -1 is in r (where 4> -1 denotes the inversefunction of4» (d) if 4> is in rand U is an open subset of dom 4>, then 4>1 U is in r, and (e) if{UoJaEI (I some index set) is a collection ofopen subsets ofRn, 4> is a local homeomorphism ofRn defined on U = UaEI Ua, and 4>1 Ua is in r for every ex in I, then 4> is in r. Some examples of pseudogroups are: (a) (diff)" = the set ofall local homeomorphisms on Rn (n fixed) which are differentiable of class C". (b) (diff)CO = the set oflocal homeomorphisms ofRn (n fixed) which are smooth. (c) (diff)W = the set of all local homeomorphisms of Rn (n fixed) which are real analytic. §1. Manifolds 3 To show that (a) and (b) satisfy the conditions of the definition you need to use only the chain rule, the inverse function theorem, and the local character of differentiability. For (c) you need the strengthened versions of the above theorems for analytic functions. A more general class of pseudogroups can be given as follows: (d) Let G be a group of linear mappings of Rn -»- Rn. Then the pseudogroup r Gk is the set {c/> E (diff)k IVx E dom C/>, (dc/»x E G} (i) G = all linear maps on Rn with positive determinant. Then r Gk = (diff)~ consists of orientation preserving Ck mappings. (ii) G = all linear maps on Rn with determinant equal to 1. Then r Gk consists of all volume preserving Ck mappings. (iii) Let ( , ) be an inner product on Rn. Let G be the group of orthogonal matrices relative to ( , ); namely, A E G iff (x, y) = (Ax, Ay) for every x, y in Rn. Then r Gk consists of all Ck isometries in Rn. Definition 1.5. Let r be a pseudogroup on Rn and X a Hausdorff topological space which satisfies the second axiom ofcountability. Let A be a subset of all local homeomorphisms of X into Rn, i.e., homeomorphisms which are defined on an open subset of X and whose range is an open subset ofRn. Then (i) A is a r-atlas on X if (a) X = Ut>eA dom c/> (b) if c/>,.j; are in A, then .j;.c/>-Ilc/>(dom c/> n dom.j;) is in r. (ii) The elements of A are called charts on X. (iii) Two r-atlases Al and A2 on X are compatible if .j;.c/>-Ilc/>(dom c/> n dom .j;) is in r whenever c/> is in Al and.j; is in A2. (iv) A Hausdorff space X together with an equivalence class of compatible r -atlases is called a r -structure on X. Note. If Xhas a r-structure, then Xis locally compact, since it is locally Euclidean. Definition 1.6. Let X have a r-structure. (a) Ifr = (diff)kandk > O,thenXisadifferentiablemanifoldofclassCk. (b) If r = (diff)D, then X is a topological manifold. (c) If r = (diff) 00, then X is a smooth manifold or a manifold ofclass COO. (d) If r = (diff)W, then X is a real analytic manifold. (e) Ifr = (diff)~ and k > °then X is an oriented Ck differentiable manifold. Any differentiable manifold which has a (diff)6 structure in which the charts are elements of the original (diff)1 structure is orientable. Examples (1) sn-I = {x = (xI, ... ,Xn)ERn L~ Xi2 = I}. Let N = (1,0, ... ,0) and S = (-1,0, ... ,0). 4 Preliminaries on Manifolds Let ~N: {sn-1 - {N}} -+ Rn-1 be stereographic projection via N, i.e., ~N(X1"'" xn) = (1/(1 - X1))(X2, ... , xn) and ~s: {sn-1 - {S}} -+ Rn-1 be stereographic projection via S, i.e., ~S(X1' ... , xn) = (1/(1 + Xl)) (X2' ... , xn)· Then ~S'~N -1: Rn-1 - {O} -+ Rn-1 - {O} is given by y -+ y/lyl2 for all Y in Rn-1 - {O}. Since (~S'~N -l)'(~S'~N -1) = id we see that det (d~S'~N -l)y = ±1. Evaluate at Y = (1,0, ... ,0) to see that, in fact, det (d~S'~N -1) = -1. To show that sn-1 is an oriented analytic manifold we can change the last coordinate of ~N to - xn/(1 - Xl) thus changing the determinant to +1. (2) pn = real projective n-space. To define pn we introduce the equivalence relation ~ on Rn+1 - {O}: (xo, ... , xn) ~ (x~, ... , x~) iff there is a real constant e such that X, = ex; for all i. pn = Rn + 1 - {O}/~ is the set of these equivalence classes. Let 7r: Rn+1 - {O} -+ pn be the canonical projection. pn is given the standard decomposition space topology and note that with this topology 7r is an open mapping. To show that pn has a manifold structure it is necessary to produce local homeomorphisms of pn into Rn which overlap properly. Let Vi = Rn+1 - {hyperplane Xi = O} for 0 .:0; i .:0; n. Vi is open in Rn+1 - {O}, hence 7r(Vi) = Ui is open in pn. Clearly pn = U1 u· .. U Un. Define ~i: Ui -+ Rn by (-IY A ~i(P) = -- (Xo, ... , X;, ... , xn) where P = 7T(XO' ... , Xn) Xi and A indicates that coordinate is to be omitted. Using the equivalence relation defining pn and the fact that p is in U;, one sees that ~i is a well-defined homeomorphism onto Rn. ~i( Ui n Uj ) = Rn - {hyperplane Yi = O} ~l Ui n Uj) = Rn - {hyperplane Yj + 1 = O} (i > j) (i .:0; j) where we assume 11, ... , Yn are the coordinates on Rn. So for i < j ~i'~j -1: Rn - {hyperplane Yi+1 = O} -+ Rn - {hyperplane Yj = O}. A computation yields for i < j -1 (_I)i+j ~i'~j (Y1,···, Yn) = (11,···, Yi> Yi+2,···, Yj' I, Yj+l>"" Yn) Yi+1 which is a real analytic mapping so pn becomes a real analytic manifold. Another computation yields det (d~.•~.-l) = (__I_)n+\ _lyn-1li+(n+1lj , J (Yl, .. ·,Ynl Yi+1 from which we see that real projective space in any odd dimension (p2n+1, n ~ 0) is orientable. It can be proved that p2n is not orientable. (3) G/c,n = Grassmannian space of k-planes through the origin in Rn. = set of all k-dimensional subspaces of Euclidean n-space. §1. Manifolds 5 Note that Gl,n + 1 = pn. We will give Gk,n a decomposition space topology. Let W = all ordered k-tuples P = (Pl , ... , Pk) of k linearly independent vectors in Rn. W is an open subset of Rn Efj ... Efj Rn. '-----v---' k-times Define an equivalence relation ~ on Was follows: span the same k-dimensional subspace of Rn. Clearly G",n can be identified with WI ~ as sets so we may give G",n the topology induced by this identification. We now give G",n an analytic structure. Equip Rn with an inner product ( , ). Then given a subspace V of Rn, there is an orthogonal projection TTV of Rn onto V. Suppose V is a k-dimensional subspace of Rn. Let 7Tu,V = restriction of TTV to U. Let Wv = {U E G",n I7Tu,V is a bijection onto V}. Let Vl. = the orthogonal complement of V in Rn. Define Pv: Wv -?- Hom (V, Vl.) as follows: Let UE Wv. Thenpv(U) = 7TU,Vl.·7Tu,ly EHom (V, Vl.). We leave it to the reader to check that Pv is a homeomorphism. Now make the identification Hom (V, Vl.) ~ Rk(n-"l, to get a chart is bijective, and the rank of is n at every point of X, then is a diffeomorphism. (e) if : X ---0>- Y is an immersion and a homeomorphism (into), then it is an embedding. (f) if there exists a diffeomorphism of X ---0>- Y, then X and Yare diffeo- morphic. Note. If: X ---0>- Y is a diffeomorphism, then -1: Y ---0>- X is well-defined and is as differentiable as is by the inverse function theorem (Theorem 1.2.) We will show that locally immersions" look like" linear injections, submersions "look like" projections, and diffeomorphisms "look like" the identity mapping. (The notion of "looks like" will be made precise in 2.5 and 2.6.) To do this we need the implicit function theorem. Let Vl c Rk and V2 c Rl be open sets. Let : Vl x V2 ---0>- Rl be differentiable. Define (dyxo)YO where Xo in Vb Yo in V2and xo : V2---o>Rl is given by xo(Y) = (xo, y) for all y in V2. Theorem 2.4. (Implicit Function Theorem). Suppose : Vl x V2---0>- Rl is CS-differentiable and (xo, Yo) = Yo. If(dy- V2 such that (x, .f(x, y)) = y for every x in V{ and y in V~. Moreover .f can be chosen so that .f(xo, Yo) = Yo. Proof Define $: Vl x V2-)- Rk X Rl to be the graph of, i.e., $(x, y) = (x, (x, y)) for all x E Vl , Y E V2. In the standard coordinates xI. ... , Xk on Rk andy!, .. ',Yl on Rl where I}e is the k x k identity matrix. The assumption on (dy(.f1(X, y), .f2(X, y))) Hence .fl(X, y) = x and y = (x, .f2(X, y)). Take .f = .f2' 0 Corollary 2.5. Let VeRn be open, Xo in V, and : V ---0>- Rm an immersion at Xo. Then there exists an open set V' in V with Xo in V', an open set V in Rm with (V') c V, and a map T: V ---0>- Rm which is a diffeomorphism onto its image so that A = T' is the standard injection of Rn ---0>- Rn x Rm-n restricted to V. (Thus by a change ofcoordinates in the range, can be linearized locally.) 8 Preliminaries on Manifolds Proof Since (do/)xo has rank n, there is an n x n minor which is nonsingular. Let 0/1> ... , o/m be the coordinate functions defined by 0/. Then C 00/1 Of)OX1 OX2 OXm (do/)xo = Oo/m Oo/m °o/m OX1 OX2 OXm The appropriate minor is determined by n columns i1> ... , in. Let T1 be a linear isomorphism of Rm which maps BiJ f--+ Bj (1 :::; j :::; n) where Bj is the unit vector along the jth coordinate. Then T1·0/ has the property that (dT1 ·o/)xo has rank n and the appropriate n x n minor which is nonsingular is given by the first n-columns. By including T1 in the definition of T we assume that 0/ has this property. Write Rm = Rn x Rl where I = m - nand Rn is given by the first n-coordinates Xl, ... , xn and Rl by the last I-coordinates Y1>···, Yl. 0/: V --+ Rn x Rl is given by 0/ = 0/1 + 0/2 where 0/1 : V --+ Rn, 0/2: V --+ Rl, and (do/lLo has rank n. Since V is in Rn, we may construct f,: V x Rl --+ Rn x Rl given by (x, y) f--+ o/(x) + (0, y) where x is in V and y is in Rl. Then which has rank n. By the inverse function theorem, there exists a differentiable inverse T to f, on a nbhd of (xo, 0). Let A(X) = T"f,(x, 0) = (x, 0). Then A: Rn --+ Rn x Rl = Rm is given by A(X) = (x, 0) which is a linear map of rank n. 0 Corollary 2.6. Let VeRn be open, Xo a point in V, and 0/: V --+ Rm a submersion at Xo. Then there exists a nbhd V' of Xo in V, a diffeomorphism a: V' --+ Rn (onto its image), and Aa linear mapping ofrank m so that 0/ = A·a on V'. (Infact, Acan be taken to be the standard projection ofRm x Rn-m--+ Rm. Thus by a change ofcoordinates in the domain, 0/ can be linearized.) Proof Let Rn = Rm x Rl with coordinates Xl, ... , Xm on Rm and Y1, ... , Yl on Rl. By an appropriate choice of bases on Rn, this decomposition can be done so that (dxo/)xo has rank m. Define f,: V --+ Rm x Rl by f,(x, y) = (o/(x), y). Then §2. Differentiable Mappings and Submanifolds 9 which has rank n. By the inverse function theorem ~ is locally a diffeomorphism. Let a = ~ and '\: Rm x R! ~ Rm be given by '\(x, y) = x. Then '\·~(x, y) = '\((x), y) = (x). 0 Definition 2.7. Let X be a Ck-manifold ofdimension n. Let Y be a subset of X. Then Y is a submanifold of X of dimension m iffor every point p in Y, there exists a chart : dom ~ Rn of the differentiable structure on X so that -l(V) = Y n dom where V = {(Xl' ... , xn) E Rn IXm+1 = ... = Xn = O} and Xl, ... , Xn are the canonical coordinates on Rn. Note. If Y is a submanifold of a Ck-differentiable manifold, then it itself is a Ck-differentiable manifold. Give Y the induced topology from X. (Warning: There are weaker definitions of submanifold in which Y does not bear the subspace topology. See Definition 2.9.) For each p in Y, let P be the chart on X, given in the definition of submanifold. Y n dom is an open set of Yand pl Y: Y n dom ~ Rm is a local homeomorphism. The set of mappings {pl Y}PEY give Ya Ck-differentiable structure of dimension m. Theorem 2.8. Let X and Y be Ck-differentiable manifolds of dimensions nand m respectively with n > m. Let : X ~ Y be a Ck-mapping. Then (1) If is a submersion, then (X) is an open subset of Y. In fact, is an open mapping. (2) Let Z be a submanifoldof Y. If is a submersion at each point in -l(Z), then -l(Z) is a Ck submanifold of X with codim -l(Z) = codim Z where codim Z = dim Y - dim Z. Proof (1) Let U be an open set in X and Van open set in Y with (U) c V and Yo in V. Let.p: U ~ Rn and p: V ~ Rm be charts. Choose Xo in Un -l(yO)' All of this is possible since is continuous. Now p. •.p-1: U' ~ Rm is a submersion where U' = .p(U) is open in Rn. By Corollary 2.6 there exists a nbhd U" of .p(xo) in U' and a diffeomorphism a: U" ~ a(U") c Rn and a linear mapping ,\ of rank m so that p. •.p-1 = '\.a on U". Let .p' = a •.p. .p is a chart on X with Xo in dom.p' and p..(.p')-l = ,\. Since'\: Rn ~ Rm has rank m, it maps open sets to open sets. Choose Wan open nbhd of Xo in X so that .p'(W) c U". Then '\(f(W)) is open in Rm and p-1(,\(.p'(W))) = (W) is open in Y. So (X) is open in Y. (2) Note that ,\ : Rn x Rn-m ~ Rm can be given by '\(x, y) = x. Let p be a chart which makes Z into a submanifold, i.e., one for which p(Z n dom p) is a hyperplane in Rm. Now ,\..p'(dom .p' n -l(Z)) c p..-1(Z) c p(Z) = hyperplane by the choice of p. Thus .p'(-l(Z) n dom .p') c ,\-l(hyperplane) = hyperplane, since ,\ is linear. Thus .p' is a chart near Xo making -l(Z) into a Ck-submanifold of codimension = codim Z. 0 Example. Let : Rn ~ R be given by (xb .•. , xn) = X12 + ... + xn2 • This is a submersion on sn-1 = -1(1). Thus sn-1 is an n - 1 dimensional submanifold of Rn. 10 Preliminaries on Manifolds Note. Let X be a differentiable manifold, Y a set, and f: X -+ Y a bijection. Then there is a natural way to make Y into a differentiable manifold. First declare that the topology on Y is the one which makes f a homeomorphism. Then define the charts on Y, to be the pull-backs viaf- 1 of the charts on X. Definition 2.9. The image ofa 1-1 immersion, made into a manifold in the manner just described, is an immersed submanifold. (Warning: this definition of immersed submanifold is not the same, in general, as that of a submanifold. In particular, the topology of the immersed submanifold need not be the same as the induced topology from the range.) Proposition 2.10. Let : X -+ Y be an immersion. Thenfor every p in X, there exists a nbhd U ofp in X such that (1) \ U: U -+ (U) is a homeomorphism where (U) is given the induced topology from Yand (2) (U) is a submanifold of Y. Proof Given p in X, there exist open nbhds U ofp in X and V of (p) in Y with (U) c V, charts p: U -+ Rn and 7: V -+ Rm, and a linear map '\: Rn -+ Rm of rank n so that the diagram commutes. This is possible by Corollary 2.5. Now (1) follows since ,\ : Rn -+ Rm is a homeomorphism onto its image. For 7((U)) is homeomorphic to (U) with the induced topology since 7 is a local homeomorphism defined on V. 7((U)) c Im'\ since the diagram commutes, thus ,\ -1(7((U))) is homeomorphic to (U) with the induced topology from Y. Finally p -1(,\ -1(7((U)))) = -1(U) = U is homeomorphic to (U) with the induced topology from Y. To see that (U) is a submanifold, use the chart 7. Decompose Rm into '\(Rn) x Rm-n. Then 7\(U): (U) -+ Rn x {O}. 0 Notes. (1) Proposition 2.10 is only a local result since not every immersion is 1 :1. For instance, the mapping of R -+ R2 given pictorially by is an immersion (when drawn smoothly enough!). §2. Differentiable Mappings and Submanifolds 11 (2) The image of an immersion need not be a submanifold even if the immersion is 1: 1. For example, consider oP where P = Limt --+ oo 4>(t). The induced topology on 4>(R) from R2 is not the same (near P) as the induced manifold topology on 4>(R). The following corollary is left as an exercise. Corollary 2.11. Let 4> : X --0>- Y be an immersion. Then (1) For every yin Y, 4>-l(y) is a discrete subset of X. (2) 4>(X) is a submanifold of Y if.! the topology induced on 4>(X) from its inclusion in Y is the same as its topology as an immersed submanifold. Clearly, in the second example above, open nbhds of 4>(P) in the two relevant topologies on 4>(R) are different. Definition 2.12. Let X and Y be topological spaces with 4>: X --0>- Y continuous. Then 4> is proper iffor every compact subset K in Y, 4>-l(K) is a compact subset of X. Theorem 2.13. Let 4> : X --0>- Y be a 1: 1proper immersion. Then 4>(X) is a submanifold of Y. Proof Using Corollary 2.11 (2) we see that 4>(X) is a submanifold iff 4> : X --0>- 4>(X) is a homeomorphism where 4>(X) is given the topology induced from Y. Clearly 4> : X --0>- 4>(X) is continuous and bijective, so we need only show that 4>-1 is continuous. Let 11, Y2, ... be a sequence in 4>(X) converging to y in 4>(X). Let Xi = 4>-l(Yi) and X = 4>-l(y). It is enough to show that Limi--+ 00 Xi = x. Let K be a compact nbhd ofyin Y. Since 4>(X) has the topology induced from Y, K II 4>(X) is a nbhd of y in 4>(X) and we may assume, without loss of generality, that each Yi is in K. Since 4> is proper, 4>-l(K) is compact and 4>14>-l(K): 4>-l(K) --0>- 4>(X) II K is a homeomorphism. Thus Limi--+ oo Xi = X by the continuity of 4>-l l4>(X) II K. D Note. A I: I immersion can be a submanifold even if the immersion is not proper. Consider the spiral of R+ --0>- R2 given pictorially by and analytically by fer) = (r cos (1/r), r sin (1/r)). Clearly,Jis a 1-1 immersion andfis not proper sincef-1(B1) = [1,00) where B1 is the closed disk of radius I centered at the origin. But the two possible topologies onf(R+) are the same so feR +) is a submanifold of R2. 12 Preliminaries on Manifolds Exercises: (1) Letf: Rn _ R2 be defined by (Xh ... ,Xn)I-+(XI2 + ... + Xn2,X12 - (X22 + ... + xn2)) (a) For which x in Rn isfa submersion at x? (b) Let fl and f2 be the coordinate functions of f For which r, s in R is fl -1(r) nf2-1(S) a smooth submanifold of Rn. (2) Let Mn be the set ofn x n real matrices. Let Mnkbe the set ofmatrices in Mn of rank k. Prove that Mnk is a submanifold of Mn and compute its dimension. (Hint: Let S = (~ ~) be in Mn where A E Mkk. Show that SEMnk iff D - CA-IB = 0.) §3. Tangent Spaces Definition 3.1. Let X be a differentiable n-manifold. (1) Let c: R - X be differentiable with c(O) = p. Then c is a curve on X based atp. (2) Let Cl and C2 be curves on X based at p. Then Cl is tangent to C2 at p if for every chart c/> on X with p in dom C/>, (*) (dc/>·Cl)O = (dc/>.C2)O' This makes sense since c/>.Cl and c/>.C2 are mappings of open nbhds of 0 in R into Rn.) Lemma 3.2. If (*) holds for one chart C/>, then it holds for every chart. Proof Let ifi be another chart defined near p. Then (difi·Cl)O = (dific/>-Ic/>.C1)O = (difi·c/>-I),p(p)(dc/>.Cl)O = (d!fo.c/>-I),p(p)(dc/>.C2)O = (difi·C2)O o Definition 3.3. Let SpX denote the set ofall curves on X based at p, p a point in X. Let Ch C2 E SpX. Cl ~ C2 ifCl is tangent to C2 at p. ~ is clearly an equivalence relation. The set TpX == SpX/ ~ is called the tangent space to X at p. If Cl is in SpX, let c1 denote the equivalence class ofC1 in TpX. Let c/> be a chart on X withpin dom c/>. Note that cv(t) = c/>-l(c/>(p) + tv) is a curve on X based at p where v is some vector in Rn. Define A,pP: Rn _ TpX by A,pP(V) = cV' Lemma 3.4. Let X be a differentiable n-manifold, p a point in X. Let c/> be a chart on X near p. Then A,pP: Rn _ TpX is bijective. Proof (a) A,pP is 1: 1. Let VI> V2 ERn and A,pP(Vl) = A,pP(V2)' Then CUI and CV2 are tangent at p; i.e., (dc/>.cv)o = (dc/>.cV2)o. Now (dc/>.cV1)o = (dc/>.c/>-I(c/>(p) + tVl))O = (d(c/>(p) + tVl))O §3. Tangent Spaces 13 but t 1-+ 4>(p) + tV1 has derivative at t = 0 equal to v1. Similarly for v2, so V1 = V2' (b) Aq,P is onto. Let ct be in TpX. Let c be a curve representing the equivalence class ct. Let v = (d4>'c)o be a vector in Rn. By the calculation in part (a), (d4>,cJo = v so (d4>.cv)o = (d4>'c)o which implies that c and Cv are tangent at p. Stated differently, Aq,P(V) = Cv = e = ct. 0 Proposition 3.5. There exists a unique vector space structure on TpX such that for every chart 4> on X with p in dom 4>, the mapping A",P : Rn -i>- TpX is a linear isomorphism. Proof Let 4>, if be charts with p in dom 4> n dom if. Then (*) Assuming this formula, it is clear that if A",P is linear for some chart 4>, then AI// is linear for any other chart if. Let the vector space structure on TpX be the one induced by A",P from Rn, i.e., if ct and f3 are in TpX, then ct + f3 = A",P[(A",P)-l(ct) + (A",P)-l(f3)] We now prove the formula (*). Let v be in Rn and let A = (dif·4>-l)tf;(p). Then (d4>·cv)o = (d4>(p) + tv)o = v = A -1Av = (d4»p .d(if-1(if(p) + tAv))o Therefore Atf;P(V) = A,l(Av), which is what was to be shown. 0 Definition 3.6. Let f: X -i>- Y be a differentiable mapping with p in X and q = f(p). Then f induces a linear map (df)p: TpX -i>- TqY called the Jacobian off at p as follows: Let c be in SpX; then f· c is in Sq Y. To induce a map from TpX -i>- Tq Y we need to know that ifC1 ::: C2 in SpX, thenf,c1 ::: f,c2 in Sq Y. Let 4> be a chart on X near p and if a chart on Y near q. Then C1 ::: C2 implies that (d4>'c1)o = (d4>·c2)o. Hence (dif·f·c1)o = (dif·f·4>-1)tf;(pM4>·c1)o = (dif·f·4>-1)tf;(p)(d4>'C2)O = (dif·f·C2)O using the chain rule. So by definition, f,c1 ::: f'C2' This defines (df)p: TpX-i>Tq Y. To check that (df)p is linear, we have the following formula: (**) (df)p = Al/fq(dif·f·4>-l)",(plAtf;P)-l Let ebe in TpX. Then we may take c(t) = 4>-l(4)(p) + tv) for some v in Rn. Now Al/fq(dif·f·4>-l)q,(p)(Atf;P) -le = Al/fq(dif·f·4> -l)tf;(plv) which is equal to the equivalence class of the curve c1(t) = if-1(if(q) + t(dif·f·4>-l)tf;(p)(v)). Thus (dfMe) is the equivalence class of the curve c2(t) = f·4>-l(4)(p) + tv). 14 Preliminaries on Manifolds To see that CI and C2 are tangent at q, we compute and Remark. Using (**) and the fact that lIl/fq and IItPP are isomorphisms we have that f is an immersion at p if rank (df)p = dim X and that f is a submersion at p if rank (df)p = dim Y. Definition 3.7. Let X be a differentiable manifold. Then TX = U TpX = tangent bundle to X peX Let n: TX -+ X denote the natural projection. Proposition 3.8. Let X be a Ck-differentiable n-manifold (k > 0). Then TX has, in a natural way, the structure of a Ck-I manifold of dimension 2n. Proof Let p be a point in X, U an open nbhd of p in X, and 1> a chart with domain U. Let TuX=n-I(U). Define .f:TuX-+1>(U)xRn by .f(a) = (1)'n(a),(v,(a))-I(a)) for every a in TuX. .f is bijective. We claim that if {1>a} is an atlas on X, then TX can be topologized so that {.fa} is an atlas on TX. Note that .f.~-I(a, v) = (1) •.p-I(a), (IItPq)-IIll/fq(V)) = (1) •.p-I(a), (d1> •.p-IMv)) where q = .p-I(a), by using the formula (*) in Proposition 3.5. Now 1> ..p-I: Rn -+ Rn is Ck-differentiable and (d.p'1>-I): U x Rn -+ U x Rn is Ck -I-differentiable since it is given by a matrix whose coefficients are first partial derivatives of .p'1>-1 on U. Define the topology on TX so that all the .fa are homeomorphisms. Then TX has the structure of Ck-I-differentiable manifold. 0 Notes. (1) Let V be a (finite dimensional) vector space with p in V. It is obvious that there is a canonical identification of V with Tp V given by v 1->- cwhere c(t) = p + tv. (2) Let V be a vector space and let G(k, V) be the Grassmann manifold of k-dimensional subspaces of V. Let W be in G(k, V). (We shall view W both as a point in G(k, V) and a subspace of v.) We show that there is a canonical identification of TwG(k, V) with Hom (W, V j W). Choose a complementary subspace S to W in V. Let C(t) be a curve in G(k, V) based at W. Define At: W -+ VjWby At(w) = neSt) where n: V -+ VjWis the obvious projection and 11' = St + Ct where Ct EO C(t) and St EO S. (Note that for t small, writing 11' = St + Ct is always possible.) First show that if C(t) and e'(t) are two curves on G(k, V) tangent at W, then dAt (11') I = dA; (11') Idt t~o dt t~O §4. Partitions of Unity 15 as mappings of W -+ V j W. Thus we have a linear mapping ,p : TwG(k, V)-+ Hom (W, VjW) given by dC (0) f-+ dAt (w) Idt dt t=o Next show that ,p is, in fact, an isomorphism. Finally show that ,p is independent of the choice S. Hint: Let S' be another complementary subspace to W in V. Then St - s; = c; - Ct is in C(t). Thus there is an at in C(t) such that St - s; = tat. Now show that d(At - A;) (w) I = o· dt t=o §4. Partitions of Unity Manifolds are geometric objects that locally "look like" Euclidean space. It would then be convenient to be able to do whatever analysis or calculus that we have to do locally; i.e., in Euclidean space. The use of partitions of unity is the technique to accomplish this goal. Definition 4.1. Let X be a topological space. (1) {V"}"EI (I some index set) is a covering of X if each V" is contained in X and X = UaEI Va. (2) Let {VaLEI and{Vp}pEJ be coverings ofX. Then {Vp}PEJ is a refinement of {V"LEI iffor every 13 in J, there is an a in I so that Vp eVa. (3) Let {Vp}pEJ be a covering of X. Then {Vp}PEJ is locally finite iffor every p in X, there is a nbhd V ofp in X so that V n Vp = 0 for all but a finite number of13's in J. (4) X is paracompact ifevery open covering of X has a locally finite refine- ment. Proposition 4.2. Let X be a topological space which is locally compact and satisfies the second axiom of countability. Then X is paracompact. In particular, all manifolds are paracompact. (Recall that X satisfies the second axiom of countability if the topology on X has a countable base.) Proof We first construct a sequence of compact sets Kb K2 , ••• such that (1) Ki c Int (Ki+ 1) for all i, and 00 (2) X= UKi· i= 1 Since X is locally compact and second countable, we may choose a sequence of open sets Nb N2 , .•• each of which has compact closure and such that the N;'s cover X. Let MIc = U~=l Ni• Let Kl = M1 • Since Kl is compact there exists Mil' ... , MiT so that Kl eMil U· .. U MiT' Let K2 = Mil u· .. U MiT' Thus K2 is compact and Kl c Int (K2)' Proceed inductively. 16 Preliminaries on Manifolds Now let {Wa}aEI be an open covering of X. We construct a locally finite refinement, For each i, let Wa/, .... , W~k' be a finite subcovering of the compact subset Ki - Int (Ki- l ). Let V/ = Wa/ (\ lnt (Ki+l - Ki- 2). Then the collection {V/} is a locally finite refinement of the covering {Wa}. 0 Corollary 4.3. Let X be a differentiable manifold and let {Ua}aEI be an open covering of X. Then there is a locally finite refinement {Vp}pEJ of{Ua}aEI such that (a) for every f3 in J, there is a chart 0 for s < 1. If y exists, then consider p: Rn --+ R defined by p(x) = y(lx - xoI 2 jr2). Clearly p is smooth and has the desired properties. Now just define if s ~ 1 ifs < 1 We leave it to the reader to check that y is indeed smooth. 0 §4. Partitions of Unity 17 Theorem 4.6. Let X be a Ck differentiable manifold and let {Ua}aeI be an open covering ofX. Then there exists a partition ofunity {Po}oeJ on X subordinate to the covering {Ua}ae]' Moreover, ifI is countable, then we may let J = I and assume that supp Pa C Uafor all (X in I. Proof Let {Vo}oeJ be the locally finite refinement of {Ua}ae] whose existence is guaranteed by Corollary 4.4. Define go: X ~ R by ( ) _ {Y(4)oCP)) go p - o ifp E V otherwise where y: Rn ~ R is a smooth function which is positive on B2 and zero off B2 (using Lemma 4.5). Let h(p) = LoeJ goCp). Then h is well-defined (i.e., the sum is finite), and Ck since {VoheJ is a locally finite covering for X. Also h(p) > 0 for all p. Let pp = (l/h)go. Then {PoheJ is a partition of unity subordinate to the cover {U,,}ae]' For the moreover part, let Ul> U2, ••• be the covering and let j; = Loe]! Po where f3 is in Ji if supp Po C Ui and supp Po ¢ Uj for j < i. 0 Corollary 4.7. Let X be a C k differentiable manifold. Let U and V be open subsets of X with [J c V. Then there is a Ck differentiable functionf: X ~ R such that f(x) = 0{ I o~ f(x) ~ 1 ifxE U ifx¢V otherwise. Proof Let {f1,f2} be a partition of unity subordinate to the cover {V, X - [J} given by Theorem 4.6. Take f = fl' Certainly supp f c V and f == 1 on U since f2 == 0 on U. 0 We present the following Proposition just to indicate the great number of smooth functions which exist (as compared to, say, analytic functions). Proposition 4.8. Let C be a closed subset ofRn; then there exists a smooth functionf: Rn ~ R such that f;:: 0 everywhere and C = f-1(0). Proof Cover Rn - C by a countable sequence of open balls Bb B2, ... each contained in Rn - C. Let j; be a smooth function zero off B! and positive on B!. (Use Lemma 4.5.) Let (ola~)MI = sup ---;} . lal:<:1 ox (M! is well-defined since each ola~/oxa is compactly supported.) Let 00 j; f="2-'.1=12!MI The choice of the M;'s guarantees that for each (X the series ~ _l_ola~ 1=1 2iM! OXa 18 Preliminaries on Manifolds converges uniformly. Using a standard theorem from advanced calculus (see Dieudonne, Foundations of Modern Analysis, 8.6.3, p. 157), 81"1j18x" exists and is continuous for each a, so that f is smooth. Thus f is the desired function. 0 Exercise: Let X be a smooth compact manifold. Show that there exists a 1: 1 immersion of X into some Euclidean space, and thus conclude that any compact manifold can be realized as a submanifold of RN for some large N. §5o Vector Bundles Definition 5.1. (l) Let E and X be smooth manifolds and 7T : E -+ X a submersion. Let Eu = 7T- 1(U)for any subset U of X. Then E is a family of vector spaces over X of dimension k iffor every p in X, Ep is a real vector space ofdimension k whose operations (addition and scalar multiplication) are compatible with the topology on Ep inducedfrom E. Let k be denoted by dimx E. (2) A section of E is a smooth mapping s: X -+ E such that 7T. s = idx. C"'(E) denotes the space ofsmooth sections ofE. (3) Let 7TE: E -+ X and TTF : F -+ X be families of vector spaces over X. Then cfo : E -+ F is a homomorphism from E to F if (a) 7TF°cfo = 7TE (b) cfo is smooth (c) For every p E X, cfo: Ep -+ Fp is a linear map. cfo is an isomorphism ifcfo is a diffeomorphism and a homomorphism. Example. Let V be a vector space (finite dimensional), X a smooth manifold, and E = X x V. Let 7T : E -+ X be a projection on the first factor. Then E ~ X is a family of vector spaces known as a product family. A family of vector spaces F over X is trivial if it is isomorphic to some product family. Definition 5.2. Let E ~ X be a family of vector spaces over X. E is a vector bundle over X is every point p in X has an open nbhd Up so that the family of vector spaces Eu" is trivial (i.e., a vector bundle is a locally trivial family ofvector spaces). Note that dim E = dim X + dimx E. Example. Let X be a smooth manifold. Then TX (the tangent bundle over X) is a vector bundle with dimx TX = dim X. The charts that were constructed in Proposition 3.8 to show that TX is a manifold also show that it is a locally trivial family of vector spaces. When working with a vector space V, it is often useful to consider certain associated spaces such as the dual space V*, the space S2(V*) of symmetric bilinear forms on V, etc. In a similar fashion, when given a vector bundle E §5. Vector Bundles 19 over X, it is sometimes useful to construct associated vector bundles over X. For instance, one should be able to replace Ep by E: (the dual space to Ep) for each p in X and make the new set into a vector bundle. One could also replace Ep by S2(Et) (the space of symmetric bilinear forms on Ep), etc. The following will show how to formalize such a process to yield new vector bundles. Let T be a covariant functor which takes (finite dimensional) vector spaces into (finite dimensional) vector spaces, (i.e., T: Vector spaces --+ Vector spaces and if V and Ware vector spaces, then T: Hom (V, W) --+ Hom (T(V), T(W». This latter map has the property that if J: V --+ Wand g: W --+ Z, then T(g.J) = T(g).T(f). Note that Hom (V, W) denotes the set of linear mappings from V to Wand is vector space isomorphic to Rmon where n =. dim V and m = dim W.) Definition 5.3. T is smooth ifJor every pair oj vector spaces V and W, the mapping T: Hom (V, W) --+ Hom (T(V), T(W» is smooth. (Note that the above isomorphism oj Hom (V, W) with Rm'n gives Hom (V, W) the structure oja smooth manifold.) Proposition 5.4. Let E be a vector bundle over X and T be a smooth covariant Junctor defined on (finite dimensional) vector spaces. Then T(E) = Upex T(Ep) (disjoint union) has the structure oj a vector bundle over X. Proof Let E be a set, X a smooth manifold, and 1T: E --+ X a map. Assume that Ep is a vector space for each p in X. To put a vector bundle structure on E is to make E into a smooth manifold so that E becomes vector bundle over X with projection map 7T. Suppose F is a vector bundle and cfo: E --+ Fis a bijection which is linear on the fibers and for which 7T = 1TF·cfo. Then there is a unique way to put a manifold structure on E so that E becomes a vector bundle and cfo an isomorphism. (1) We note that if cfo: E --+ F is a homomorphism, then there is a map T(cfo) : T(E) --+ T(F) which is linear on fibers. T(cfo)(e) = T(cfop)(e) where cfop = cfolEp: Ep --+ Fpforp E Xande E Ep.tSoT(cfo): UPEX T(Ep) --+ UpeX T(Fp). (2) Suppose that E = X x V is a product family where V is some vector space. Then T(E) = UPEX T(V) = X x T(V), the last equality being an obvious bijection. Give T(E) a vector bundle structure by making this identification an isomorphism. (3) Next assume that E is a trivial bundle. Then there exists an isomorphism cfo: E --+ X x V = F. As noted in (1), T(cfo) : T(E) --+ T(F). Since T(F) is a vector bundle (by (2», we can give T(E) a vector bundle structure so that T(cfo) is an isomorphism. It is necessary to check that this vector bundle structure is independent of the choice of cfo. So let !ft :E --+ G be an isomorphism where G = X x W. 20 Preliminaries on Manifolds Then T(F) 1T( X has a metric. Proof Let E = X x V be a product bundle. Then S2(E*) = X x S2(V*) is also a product bundle. Now let B be any positive definite symmetric bilinear form on V. Then define s: X -?> X X S2(V*) by s(p) = (p, B). s is smooth and a metric on S2(E*). If E is a trivial bundle then there exists an isomorphism 4>: E -?> X X V (for some product bundle X x V). 4> induces an isomorphism 4>(2) : S2«X x V)*) -?> S2(E*). If s is a metric on X x V, !hen 4>(2)·S is a metric on E. Finally, let E be an arbitrary vector bundle. For each P in X, choose an open nbhd Up of p so that Eu is trivial. Let {Vi}i~l be a countable locally finite refinement of {Up}PEX' Let {Pih~ 1 be a partition of unity subordinate to the cover {Vi}i~l of X. (See Theorem 4.5.) Let Si be a metric on Ev;. Define Si: X -?> S2(E*) by _( ) _ {Pi(P)Si(P) Si P - 0 for all P E Vi otherwise; then Si is a smooth section. Let S = L.i~ 1 Si' This sum makes sense since for each P in X, only finitely many Pt(p) are not zero. Let v E Epo Then s(p)(v, v) = LPi(P)Si(P)(V, v) > Pt(p)St(p)(v, v) where i is chosen so that Pi(P) > O. Thus s(p) is positive definite since Si(P)(V, v) #0, and S is a metric on E. 0 Given a Riemannian metric s: X -?> S2(T* X) where X is a connected manifold, then there is a natural way to define a metric d: X x X -?> R so that (X, d) is a metric space. (There is, unfortunately, no way to change the fact that the word "metric" has two different though related meanings!) We show how to define d. LetP and q be points in X and c : R -?> X a (continuous, piecewise smooth) curve with c(O) = P and c(l) = q. By piecewise smooth, we mean that the curve is infinitely differentiable except at a finite number of points. Let (dldt)lto be the tangent vector in TtoR defined by the curve t f-+ to + t of R -?> R. Then t -?> (dldt)lt is the canonical vector field on R. DefineJ: R -?> R by Jis a piecewise smooth function and c = f~J(t) dt makes sense. Note that c is just the arc-length of the curve c relative to the Riemannian metric s. Define d(p, q) to be the infimum of cwhere c ranges over all piecewise smooth curves connecting P to q. It should be noted that d(p, q) is always defined and finite. Define an equivalence relation", on X by P '" q if there exists a piecewise smooth curve offinite length connecting P to q. Since X is locally Euclidean, the equivalence classes are open. Since X is connected there is only one nonempty equivalence class. All steps in showing that d is a metric are easy except showing that §5. Vector Bundles 23 if d(p, q) = 0, then p = q. This will be proved later. In any case, d is a pseudometric. Example. Let Xl, ... , Xn be the standard coordinates ofRn. s = Lf=l dX.2 is a metric and induces the standard metric on Rn. Let c : R -+ Rn be a curve with c(t) = (c1(t), ... , cnU». Note that (dC)t( d I) = ~ OCi~.. dt t .=1 ot OX, Hence So c = f J(0;;)2 + ... + (0;;)2 dt which is just the standard arc-length in Rn. As is known from Euclidean geometry of Rn the shortest distance between two points is the straight line distance, so this metric on Rn is just the standard one. Lemma 5.7. Let s = Lf,j=l aijdxidxj be a Riemannian metric on Rn. Let d be the induced pseudo-metric on Rn. Then on a given compact set K there exist positive constants Land M so that Md(p, q) ;:::: d(p, q) ;:::: Ld(p, q) for every p and q in K where d is the standard metric on Rn. Proof Let c : R -+ Rn be a curve. As noted above ( d I) ~ OCI 0(dc)t - = L --. dt t i=l dt OXi Thus St((dC)i(~I}(dC)t(~ It)) = i'~l ati ~t ~i' Let v = (oc1 10t, ... , ocnlot) and A = (aij). Then s = vtAv where vt is the transpose of v. Now t 1-+ IA(t)1 is a continuous function and hence is bounded above by a constant M on the compact set K. Thus s = vl(Av) ::s; IAlvlv ::s; Mvlv = Mlvl 2 • Thus c ::s; (length of c in the standard metric) x M which implies that d(p, q) ::s; Md(p, q). Since A is a positive definite, symmetric matrix at each point we also have that vlAv ;:::: Llvl 2 and the rest of the proof follows as above. 0 We can now prove the following: Proposition 5.8. Let s be a Riemannian metric on a connected manifold X. Let d be the corresponding pseudo-metric on X. Then d is a metric and the topology induced by d on X is the same as the original topology on X. Proof Fix p in X. For this proof we will call an open set in the topology induced by d on X, d-open. Then, it is sufficient to prove that every open nbhd ofp contains ad-open nbhd ofp and, conversely, that every d-open nbhd of p contains an open nbhd ofp. 24 Preliminaries on Manifolds Let V be an open nbhd ofp. Choose V', also an open nbhd ofp, so that (a) The closure of V' is compact and is contained in V, and (b) S2(T* X)I V' is locally trivial. Hence there exists a chart :V' ---7- Rn and a bundle homomorphism so that the diagram S2(T* X)I V' f> (V') x S2(Rn)------>- 1 1 v'

(V') commutes, where n = dim X. Let B be an open ball in (V') of radius r centered at (p). (Note B is open in Rn.) Let c: R ---7- X be a curve centered atp, i.e., c(O) = p. Suppose that c([O, 1]) does not lie entirely within -l(B). We claim that this curve has length at least N for some constant N not depending on c. If this last statement is true, then B(p, N) = {q E XI d(p, q) < N} is contained in V'. Also this statement completes the proof that d is a metric. For if p, q E X and p i= q, then take V' small enough so that p E V' and q 1= V'. Then the length of any curve connecting p to q is greater than N and d(p, q) i= 0. To prove the claim, we note that C- 1(-1(B)) is open in R, so that there exists a smallest t in (0,1) for which c(t) 1= -l(B). c= length of c is ;:::: length of c([O, t]). Let s' = f>.s· -1. Then s' is a Riemannian metric on (V') c Rn and the length of c([O, tn under the metric s is the same as the length of (.c)([O, t]) under the metric s'. Using Lemma 5.7, we see that for some constant L the length of (.c)([O, t]) under s' is ;:::: L x length of (.c)([O, t]) using the standard metric on Rn, since ( V') is compact. Now we note that since X is Hausdorff c(t) is in V'. For if c(t) 1= V', then there exists an open subset V of X such that c(t) E V and V n -l(B) = 0. Also c- 1(V) is open and contains t. Hence c-1(V) n C- 1(-1(B)) i= 0, a contradiction. Thus . c[O, t] is a curve connecting p with some point outside of B. Hence the length of ·c[O, t] is ;::::r, but length of .c[O, t) = length of .c[O, t]. Thus length of cis ;::::L-r = N. For the converse we suppose that V is some d-open nbhd ofp. Make the same construction as above for V', , f>, s', r, and B. Let (q) E B. Then the straight line, c, from (p) to (q) has length < r. By Lemma 5.7, the length of c in the metric s' is :0; Mr where M is a constant depending only on (V'). Thus the length of -1. c, a curve connecting p to q in X is < M· rand q E B(p, Mr) = the ball in X of radius Mr about p. By choosing r small enough B(p, Mr) c V since V is d-open and B(p, Mr) is a basic d-open set. But the above says that -l(B) c B(p, Mr) and -l(B) is an open nbhd of p. 0 Lemma 5.9. Any differentiable manifold is metrizable. §S. Vector Bundles 2S Proof Let Xl, X2 , ••• be the components of X and let di be a metric on XI making Xi into a metric space. (Use Proposition 5.8.) Define d: X x X ~ R by if X,YE Xm if x E Xi> Y E Xi, and i of j. Then d is a metric on X compatible with the original topology. 0 We need the following two results to show that X can be made into a complete metric space. Lemma 5.10. Let (X, d) be a metric space andf: X ~ R be a continuous proper function. Then d': X x X ~ R defined by d'(p, q) = d(p, q) + If{p) - f(q) Ifor every p, q in X is a complete metric on X which is compatible with the given topology on x. Proof That d' is a metric is clear. Let T be the topology on X induced by d and T' the topology induced by d'. Since d' is continuous in the topology T, we have that T' c T. Conversely, let U be in T and let p be in U. Choose e so that B'(p, e) = {x E X Id'(x, p) < e} c B{p, e) c U. Thus U is in T' and T = T. Finally we show that d' is complete. Let {Xn}:'=l be a Cauchy sequence in the d' metric. Thus there exists a constant L > 0 so that d'{Xh xn) < L for all positive integers n. Hence If(xl) - f(xn) I < L for all n, and {Xn}:'=l c f-I(Lf(xl) - L,f(xl) + L]). Sincefis proper this later set is compact and the sequence {Xn}:'=l has a limit point in X and thus converges. 0 Proposition 5.11. There always exists a smooth proper function on a smooth manifold X. In particular, any differentiable manifold can be made into a complete metric space. Proof Let Kl, K2 , ••• be a sequence of compact subsets of X such that KI c Int (Ki + l ) for i = 1,2, ... and X = Ur;l K1• Let LI = KI - Int (Kt - l) with Ll = Kl. Then Ut"~ 1 LI = Ut;,1 Ki = X. Define smooth functions Pi : X ~ R such that { I p. = 0 , O:S;p:s;1 onLI on KI _ 2 U (X - Ki+1) onX. Then letf = L:1";1 ipl' This sum is locally finite and hence is a smooth function. We claim thatfis, in fact, proper. First note that ifp ELj, then i :s; f(P) :s; 3i since f(p) = (i - I)Pi-I(P) + ipi(P) + (i + I)Pi+l(P). Then to show thatfis proper we need only show thatf-I([A, B]) is compact where A, BE R. Given a p in Li such that f{p) E [A, B), we have the inequalities i :s; Band 3i ~ A, or that i E [AJ3, B). Thus f-I[A, B) c Ui LI where i E [AJ3, B] and is then a closed subset of a compact set and hence compact. The last assertion of the proposition follows immediately from Lemma 3.2. 0 26 Preliminaries on Manifolds We now define what we mean by subbundles of a vector bundle and give one way to construct them. Definition 5.12. Let E be a vector bundle over X with projection 'fT. F is a subbundle of E ifF is a smooth submanifold of E and 'fTIF: F -7 X is a vector bundle where, for each x in X, Fx has the vector space structure induced from Ex. Definition 5.13. Let E -7 X and F -7 Y be vector bundles. cP: E -7 F is a homomorphism if (1) there exists a smooth function f: X -7 Y called the base mapping, so that 1 1lX -----'----+ Y commutes. (2) cP is smooth. (3) cPp: Ep -7 Ff(p) is linear where cPP = cPIEp. Example. Let f: X -7 Y be a smooth mapping. The Jacobian off is a map (df)p: TpX -7 Tf(p) Y for each p E X. So (df): TX -7 TY defined by (df)ITpX = (df)p is a mapping which is linear on the fibers. Locally, TX and TY are trivial and via trivializations are just U x Rn and V x Rm where n = dim X, m = dim Y and U eRn, V c Rm are open. (df): U x Rn-7 V x Rm is given by (p, v) H>- (f(p), (dfp)v) which is a smooth mapping. So (df) is smooth and with this extended definition of a homomorphism between vector bundles, (df) is a homomorphism. Proposition 5.14. Let E ~ X and F -7 Y be vector bundles and cP : E ~ F be a homomorphism with base mapping f Suppose cPP has constant rank for all p in X. Then Ker cP = UPEX Ker cPP is a subbundle of E. Proof The problem of showing that Ker cP is a smooth manifold is a local one. By using trivializations we may assume that U c Rn and V c Rm are open subset with f: U ~ V, cP: U x RS ~ V X Rt and with the appropriate diagram commuting where s = dimx E and t = dimy F. Fix p in U and choose W, a vector space complement to Ker cPP in RS. Note that cPp: W -7 f(p) X Rt is 1:1. Since cP is continuous and dim (Ker cPq) is constant throughout U, there exists an open nbhd U' of p on which cPq: q x W ~ f(q) X Rt is 1: 1for all q in U', i.e., W is a vector space complement to Ker cPq, for all q in U'. Let W be a vector space complement to cPp(W) c f(p) X Rt. We can then restrict U' to U", also an open nbhd ofp on which W is a vector space complement to cPiW) in Rt for all q in U". Let a: V x Rt -7 Rt be projection on the second factor and T: RI -7 Z = RtIW be the natural §6. Integration of Vector Fields 27 projection. Then g = T.a.¢: U' x RS --';>- Z is a smooth mapping. Note that T'a.¢(q, v) = T'a(f(q), ¢iv)) = T(¢iv)) and that T(¢iv)) = 0 iff ¢q(v) = 0 iff (q, v) E Ker ¢q. So g-l(O) = Ker (¢I U"). Ifwe show that g is a submersion, then by Theorem 2.8 Ker ¢ n (U" x RS) is a submanifold of U" x RS and thus a submanifold of E. To show that g is a submersion, it is sufficient to show that if (q, v) in U" x RS, r = g(q, v), and c: R --';>- Z is a (smooth) curve based at r, then there is a (smooth) curve c: R --';>- U" x RS based at (q, v) with g·c = c. Let c be such a curve. Note that T: ¢iW) --';>- Z, a :f(q) x ¢q(W) --';>- ¢q(W), and ¢q: W --';>- ¢q(W) are isomorphisms so that ¢q-l'a-1'T-1.C: R--,;>-q x Wis a smooth curve. Define cby c(t) = (T.a'¢q)-l.c(t) + V - (T.a'¢q)-l.c(O) c:R--,;>-q x We U" x Wisasmoothcurvebasedat(q,v)andg·c=c. D Proposition 5.15. Let F be a subbundle of E. Then there exists another subbundle G ofE }vith FEB G = E. G is called a complementary subbundle to F. Proof Choose a metric s: X --';>- F as given by Proposition 5.6. Let 7Ts: E --';>- E be given by orthogonal projection onto Fusing s, i.e., on each fiber (7Ts)p: Ep --';>- Fp is orthogonal projection. 7Ts is a homomorphism and G = Ker 7Ts is a subbundle of Eby Proposition 5.11. At eachp in X, Fp EB Gp = Ep so E = FEB G. D §6. Integration of Vector Fields There is a close relationship between vector fields and smoothly parametrized families of curves which we shall explore now. Definition 6.1. A one parameter group on X is a smooth mapping ¢: X x R --';>- X satisfying ¢o = idx and ¢s+t = ¢s'¢t for all s, t in R where ¢t(x) = ¢(x, t). Notes. (1) Let ¢ be a one-parameter group on X. Then ¢t is a diffeomorphism on X for each t. Tn fact, ¢-t = (¢t)-l. (2) Let ~p be the tangent vector at t = 0 to the curve p f--7 ¢t(p). Then the mapping p f--7 ~p defines a vector field on X called the infinitesimal generator of ¢. (The joint smoothness of ¢ in p and t guarantees that ~ is a smooth section.) We call a curve c: Ie --';>- X an integral curve for ~ if (dc)rC(d/dt)lr) = ~c(r) for all r. The following lemma shows that the infinitesimal generator of¢ .is the vector field for which the curves t f--7 ¢t(p) are integral curves. (Note Ie = (-e, e) e R.) Lemma 6.2. Let ~ be a vector field on a manifold X with p in X. Then there is a nbhd U of p in X, an e > 0, and a unique smooth function ¢: U x Ie --';>- X satisfying; (a) The curves t f--7 ¢t(q) are integral curves of ~for all q in U; 28 Preliminaries on Manifolds (b) h' 0, and a unique smooth function 1 :s; i :s; n}. Define the volume of C(a, b) to be vol [C(a, b)] = (b1 - a1)· ... •(bn - an) Definition 1.1. (1) Let S be a subset of Rn. Then S has measure zero iffor every e > 0, there is a covering of S by a countable number of open cubes C1, C2 , ••• so that Lf=1 vol [Ca < e. (2) Let X be a differentiable n-manifold and let S be a subset of X. Then S is of measure zero if there exists a countable open covering U1, U2, ••• of S and charts CPI: UI -+ Rn so that CPI(U n S) is of measure zero in Rn. To see that "measure zero" is well-defined on a manifold, we need the following two results: Lemma 1.2. A countable union ofsets ofmeasure zero in Rn is ofmeasure zero. Proof Let Sl, S2, ... be sets of measure zero in Rn. Given e <0, cover each SI by open cubes whose total volume is less than (e/21+ 1). Then the union of all of these cubes covers S = Uf=1 SI and has total volume less than e. 0 Recall that if A : Rn -+ Rn is a linear map, then IAI = sup IAvl. veRn_{O) Ivl Also, if lx,y denotes the line between two points x and y in Rn, then for any C1-differentiable functionf: Rn -+ Rm If(x) - f(y)1 :s; Ix - yl sup I(df)pl· PElx •y (This is just a corollary to the Mean Value Theorem.) Proposition 1.3. Let f: Rn -+ Rn be C 1-differentiable and let S be a measure zero subset ofRn. Thenf(S) has measure zero. 30 §1. Sard's Theorem 31 Proof Without loss of generality, S can be assumed to be contained in some large open cube. On this cube I(df)p Iis bounded by some constant K, so that if x, YES, then If(x) - f(y) I < Klx - yl. Given e > 0, cover S by open cubes Ci whose total volume is less than e/(Yn K)n. We note that f(Ci) is contained in a cube whose volume is (Yn K)n vol (C) using the above inequality. (To see this assume Cj has equal length sides with length a. Let p be the center of Ci• Then f(Ci) is contained in the sphere of radius (KYn/2)a centered atf(p) which is, in turn, contained in a cube centered at f(p) all of whose sides have length KYn.a.) Thus the total volume of cubes containing f(S) is less than e. 0 This generalizes immediately to a statement of manifolds. Corollary 1.4. Let X and Y be differentiable n-manifolds, let f: X ---i> Y be a C1-differentiable, and let Z be a measure zero subset of X. Thenf(Z) has measure zero in Y. Proof Let", be a chart on Y with domain V. Cover f-1( V) by a countable open covering Ub U2 , • .• each of which is the domain for a chart ~,: Ui ---i> Rn and for whichf(Ui) is contained in V. Since Z is of measure zero in Rn, ~i(Z II Ui) has measure zero in Rn. Now ""f'~' -1 is C1 on its domain in Rn. By Proposition 1.3 ""f'~i-1'~i(Z II U,) = "'(f(Z) II Ui) has measure zero in Rn. Hence U~ 1 "'(f(Z II Ui)) = "'(f(Z) II V) is of measure zero in Rn. So feZ) has measure zero in Y. 0 Lemma 1.5. Let X be an n-dimensional submanifold of a differentiable m-manifold Y with n < m. Then X is of measure zero in Y. Proof We first claim that an n-dimensional plane, Rn, in Rm is of measure zero. Rn can be subdivided into a countable number of unit n-cubes so it is sufficient to show that the unit n-cube in Rm is of measure zero. Let e > °be given. The unit n-cube can be covered by (2/e)n cubes each of volume em. Then the total volume of the cubes is em(2/e)n = 2nem- n which converges to zero as e f---+ 0 since m > n. Since X is a submanifold of Y, there exists a countable covering Ub U2 , • •. of Y with charts "'i: Ui ---i> Rm such that "'i(Ui II X) is contained in a fixed n-plane in Rm. Hence ",;CUi II X) has measure zero in Rm and X has measure zero in Y. 0 Proposition 1.6. Let X and Y be differentiable manifolds of dimensions n and m respectively with n < m. Let f: X ---i> Y be C1-differentiable, then f(X) has measure zero in Y. Proof Let s = m - n. Define l: X x RS ---i> Y by l(p, a) = f(p) for every p in X and a in RS. X x {O} is a submanifold of X x RS and, by Lemma 1.5, has measure zero in X x RS. By Corollary 1.4 l(X x {O}) = f(X) has measure zero in Y. 0 We need one more result before coming to Sard's Theorem, namely Fubini's Theorem for measure zero sets. 32 Transversality Let ia: Rn-l -+ R x Rn-l = Rn be the embedding given by ia(x) = (a, x) where a is in R. Theorem 1.7. Let A be a compact subset of Rn. Suppose that for every a E R, ia -leA) has measure zero in Rn -1. Then A is of measure zero in Rn. Let I be a closed interval in R. Suppose I is covered by subintervals [alo bl], ... , [am' bm]. Then the cover is minimal if the covering minus anyone element of the covering is no longer a covering. Lemma 1.8. Let I = [a, b] be a closed interval in R. Then the sum of the lengths of any minimal covering of I (by closed intervals in 1) is less than 2(b-a). Proof Order the intervals of a minimal covering [ai' bd, ..., [am' bm] so that al ~ a2 ~ ... ~ am' Then the minimality implies that bl ~ b2 ~ ... ~ bm. Moreover, [ak, bk] n [ak+2, bk+2] = 0 for 1 ~ k ~ m - 2. Otherwise ak+2 ~ bk and [ak+1o bk+d c [ak' bk] U [ak+2, bk+2] since ak ~ ak+l and bk+l ~ bk+2. Hence the sum ofthe lengths of [alo bl], [a3' ba], [a5, b5], . . : is less than b - a. Similarly for [a2' b2], [a4' b4], • • •• 0 Lemma 1.9. Suppose the set ia-leA) is covered by open sets {Ul> ..., Uk} ofRn-l. Then there exists an open intervalla about a such that {Ul, ... , Uk} covers it -leA) for every t in Ia. Proof If there were no such interval, then there would exist a sequence {ta~l of real numbers with Liml.... "" tl = a and a point XI E itt-leA) such that XI is in the complement of Ul u· .. U Uk' Since (tlo XI) is in A and A is compact, there exists a subsequence of the x/s which converges to some point x in Rn-l and for which (a, x) is in A. Since U~=l UI is open, x ¢:. U~=l Ui• But (a, x) E A implies that x E ia-leA) and the fact that {Ul, ... , Uk} covers ia-leA) gives a contradiction. 0 ProofofTheorem 1.7. Since A is compact and hence bounded, there is a closed interval I such that A c I x Rn-l. By hypothesis ia -leA) has measure zero for each a in I. Thus, given e > 0, there is a cover of ia -leA) by open cubes in Rn-\ {Cia, ... , CN"a} such that Lf!l vol (Cia) < e. By Lemma 1;9, there exists an open interval Ia in I about a so that Cia, ... , CN"a covers it-leA) for every tin Ia. Hence the collection of open sets {Ia x Cia} covers A. Thus there is a finite subcover {Ia x C,an:J:SN" where B is some finite set. Let Ja = ia. The finite collection {Ja}aeB covers I and can be assumed to form a minimal covering of I. Then Lf!l vol [Ja x Cia] ~ e vol [Ja]. Hence N" L: L: vol [Ja x Cia] ~ e L: length (Ja) < 2e length (I). aeB '=1 aeB Since vol [Ia x Cia] = vol [Ja X Cia], the total volume of the covering of A by {Ia X C,an:J:SN" can be made arbitrarily small, A has measure zero in Rn. 0 §1. Sard's Theorem 33 Since all of the results given in this section have been about measure zero sets, it is instructive, perhaps, to show at this time the following obvious but surprisingly complicated result. Proposition 1.10. Let S be a nonempty open subset of Rn. Then S is not ofmeasure zero. Proof Every open set S contains a nonempty open cube C whose closure is contained in S. Let {C!}~ 1 be an open covering of S by open cubes. Since C is compact in Rn, there is a finite subcover of C by Cb ••• , Cm. We claim that vol [C] ::; I~=l vol [Cal. If this is true then we are done since I~l vol [Cd;::: II"=l vol [Ci] ;::: vol [C] > O. So the sums of the volumes ofcubes in a covering of S are bounded away from zero and S does not have measure zero. To prove the claim, let Na = number of integer lattice points of Rn (Le., points of Rn all of whose coordinates are integers) which are contained in Ca. Now Ca = C(aa, bIZ) where aa, ba ERn. Let aa = (ala, ... , ana) and ba = (b1a, ... , bna). Then for each j there are at most bt" - at" + 1 and at least It" = max {bt" - at" - 1, O} integers in [at", bt"]. Hence n n TI It" ::; Na ::; TI (bt" - at" + 1). j=l j=l Similarly let N = number of integer lattice points in C = C(a, b) and obtain similar bounds on N. Certainly N ::; II"=l Ni since {Ca}~=l covers C. Hence n m n TI Ij::; L: TI (bt" - at" + 1). j=l a=lj=l For Ain R sufficiently large, let CA = C(Aa, Ab) and CaA(Aaa, Aba). Apply the above argument to CA and CaAto obtain n m n TI (Abj - Aaj - 1) ::; L: TI CAbt" - Aat" + 1) j=l «=lj=l Hence TI bj - aj - - ::; L: TI bt" - at" + - .n ( 1) m n ( 1)j=l A a=l f=l A Taking the limits of both sides as A-+ O. Denote by C[f], the set ofcritical points of! 34 (3) a point q E Y is a critical value off ifq Ef(C [f]). (4) a point p E X is a regular point offifp rt C[f]. Transversality (5) a point q E Y is a regular value offif it is not a critical value off. So, in particular, a point not in Image f is a regular value. Theorem 1.12. (Sard's Theorem.) Let X and Y be smooth manifolds. Let f: X -»- Y be a smooth mapping. Then the set of critical values off has measure zero in Y. Notes. (1) Sard's Theorem can be generalized as follows: Assume that k > max (dim Y - dim X,O). Iff is a Ck-differentiable mapping, then the measure of the set of critical values is zero. Since we will be using only smooth mappings in later chapters we will prove only the more restricted version here. (2) If dim X < dim Y then Sard's Theorem follows directly from Proposition 1.6 and the fact that the image of a subset of measure zero has measure zero. Sard's Theorem is in reality a local theorem and follows from: Proposition 1.13. Letf: U -»- Rm be smooth where U is an open set in Rn. Then the set of critical values off is of measure zero in Rm. The proof of Theorem 1.12 proceeds from Proposition 1.13 precisely as the proof of Corollary 1.4 proceeded from Proposition 1.3. The details are left for the reader. The proof of Proposition 1.13 will be done by induction on n. Start the induction at n = O. RO is, by convention, just a point and the proposition is trivial in this case. By induction, we assume that Sard's Theorem holds for all smooth mappings of Rn-l -»- Rm, where m is arbitrary. Lemma A. Let f: U -»- Rm be smooth, where U is an open subset of Rn. Let fh ...,fm: U -»- R be the coordinate functions given by f. Assume that fl(Xb ... , xn) = Xl for all (Xl' ... , xn) E U. Let C = critical point set off. Thenf(C) has measure zero in Rm. Proof. The proposition is trivial for n = 1, so we may assume n > 1. GivenaER, recall thalia: Rn-l-»-Rn by ia(x) = (a, .\')where.\' = (X2,"" xn). Define gaCx) = (f2(a, x), ... ,Jm(a, x)). Then the following diagram com- mutes. u l. ) where Ua = ia-l(U). §1. Sard's Theorem 35 Note that Hence rank (df)(a,x) = rank (dga)x + 1; i.e., x is a critical point of ga iff (a, x) is a critical point off So the critical point set of ga is ia-l(C). By the induction hypothesis gaCia -l(C)) is of measure zero in Rm-l. Since ia -l(f(C)) = gaCia -l(C)) we may conclude by Theorem 1.7 thatf(C) has measure zero. (Note that C is a closed set, which is a countable union of compact sets. Thus f(C) is a countable union of compact sets so that 1.7 applies.) D Let f: U -+ Rm be smooth. Let C = C[f] be the critical point set off Denote by C; = {p E C I~;~fz(p) = 0 whenever 0 < lal :::; i and 1:::; I :::; m}- (i=I,2,oo.) The outline of the rest of the proof of Sard's Theorem is: Lemma B. fCC - Cl ) has measure zero. Lemma C. f(C; - Ci +1) has measure zero for i ~ 1. Lemma D. For some i, f(C;) has measure zero. Proof ofLemma B. Let P be in C - Cl. Then there exists some partial derivative of fat P which is not zero. Assume that (8flj8xl)(p) i= O. Let h: U -+ Rn be defined by hex!> ... , Xn) = (fl(X1 , ••• , Xn), X2 , ..• , xn). Then atp which is invertible. By the Inverse Function Theorem, there exists open sets U' c U and VeRn so that h: U' -+ V is a diffeomorphism. Let g: V -+ Rm be given by g = f·h-1, thenf(C[f] (J U') = g(C[g]). Now gl(Yb"" Yn) = fl·h-l(Yl,' .. , Yn) = YI. So we can apply Lemma A to g, and get that g(C[g]) has measure zero in Rm. D Proof ofLemma C. On Ci - Ci + 1 all ith partial derivatives vanish but not some (i + l)st partial derivative. We may assume that g is an appropriate ith partial derivative so that (8gj8xl)(P) i= O. Let h: U -+ Rn be defined by hex) = (g(x), X2 , ••• , xn). Then (dh)p is non-singular, so that It restricted to 36 Transversality u~ c U is a diffeomorphism, where U~ is an open nbhd ofp. Let V = h(U;). By definition g(Ci) = 0, so h(Ci n U;) c {O} X Rn-l in Rn. Let k: Rn-l--+Rm be defined by f·h- 1 restricted to V n ({O} x Rn-l). Finally we note that f(Ci n U;) c k(C[kD and that by the induction hypothesis k(C[k]) has measure zero. Hence for eachp in Ci - Ci+1, there is a nbhd U; of p for which f(C! n U;) has measure zero. We can choose a countable number of the U;'s to cover Ci - Ci+l' So f(Ci - Ci+l) has measure zero in Rm. Proof of Lemma D. Without loss of generality, we may assume that U is an open cube with sides of length b, since U may be covered by a countable union of such sets, and thatfis defined on a nbhd of V. By Taylor's Theorem, if x E Ck, and y E U, then (*) !f(y) - f(x)! :;:; K!x - y!k+l where K is some constant independent of y. Let r be a large integer. Subdivide U into subcubes with sides oflength blr denoted by Bb ... , BN where N = rn. Nowf(Ck nBs) is contained in a ball D of radius K(blr)k+l using (*), so the circumscribed cube has volume (2K(blr)k+l)m. Thus f(Ck) is contained in the union of cubes whose volume is m(~) m(k +1) _ (2K)mbm(k +1) • N(2K) - mk+m n r r When k > (nlm) - 1, mk + m - n > O. Therefore, as r --+- 00, the volume of the cubes containing f(C,J --+- O. So f(Ck ) has measure zero in Rm. 0 CO/'ollary 1.14. (Brown). The set of regular values of a smooth mapping f: X--+- Y is dense in Y. (Recall from 1.11 that a point in Y which is not in Imf is a regular value off) Proof Points of Yare either critical values or regular values forf If the set of regular values is not dense, then there is a nonempty open set in Y consisting entirely of critical values. We have shown in Proposition 1.10 an open set of Rn does not have measure zero; this clearly extends to nonempty open subsets of Y, by using charts. Thus the set ofcritical values offdoes not have measure zero, a contradiction to Sard's Theorem. Hence the regular values off are dense in Y. 0 Exercises (1) Let f: X--+- Rm be a 1: 1 immersion and let n = dim X. Let v i= 0 be in Rm and let 7Tv : Rm --+- Rm-l be the orthogonal projection whose kernel is the subspace (v). Show that if m > 2n + 1, then there exists a vector v so that 7Tv ·f: X--+- Rm-l is a 1: 1 immersion. Hint: Consider the composite mapping g defined by TX - {O-section} (dt) TRm = Rm x Rm ~ Rm _ {O} ~ pm-l where 71"2 is projection on 2nd factor and if1 is the standard projection of Rm onto projective (m - I)-space. Show thatg is well-defined; i.e., 0 rf= 1m 7T2·(df) §2. Jet Bundles 37 and that 7rv ·f is an immersion iff vrt 1m g where v is the point in pm-l corresponding to the subspace (v) in Rm. Next consider the composite mapping h defined by x x X - ~X ~ Rm - {O} --+ pm-l where/(p, q) = f(p) - f(q). Show that trv 0 fis 1: 1 iff vrt 1m h. (2) Use Exercise (1) above and the Exercise of I, §4 to conclude that any compact n-manifold can be realized as a submanifold of R2n+l. (3) Observe that the immersion part of the proof of Exercise 1 is valid when m > 2n. Thus show that there exists an immersion of any compact n-manifold into R2n. (4) Does there exist a smooth functionf: Rn --7 Rn such thatf-l(a) is an uncountable set for each a in Rn? §2. Jet Bundles Definition 2.1. Let X and Y be smooth manifolds, and p in X. Suppose f, g: X --7 Yare smooth maps withf(p) = g(p) = q. (1) f has first order contact with g at p if (d!)p = (dg)p as mapping of TpX --7 TqY. (2) f has kth order contact with g at p if (d!) : TX --7 TY has (k - l)st order contact with (dg) at ,every point in TpX. This is written as f ~ Ie g at p. (k is a positive integer.) (3) Let JIe(X, y)p,q denote the set ofequivalence classes under "~Ie at p" of mappings f:X --7 Y where f(p) = q. (4) Let JIe(X, Y) = U(P,q)EXXYJIc(X, Y)p,q (disjoint union). An element a in JIe(X, Y) is called a k-jet of mappings (or just a k-jet) from X to Y. (5) Let a be a k-jet, then there exist p in X and q in Y for which a is in JIc(X, Y)p,q. p is called the source of a and q is called the target of a. The mapping a: JIc(X, Y) --7 X given by a f-+ (source of a) is the source map and the mapping f3: JIe(X, Y) --7 Y given by a f-+ (target of a) is the target map. Note that given a smooth mapping f: X --7 Y there is a canonically defined mappingjlcf: X --7 JIc(X, Y) called the k-jet offdefined by j'1(p) = equivalence class offin JIc(X, Y)p.f(P) for every p in X. We will also show that j1cf(p) is just an invariant way of describing the Taylor expansion off at p up to order k and that Piis a smooth mapping. Note thatJO(X, Y) = X x Y, sofhas ~o contact withg atp ifff(P) = g(p), andjOf(p) = (p,j(p)) is just the graph off Lemma 2.2 Let U be an open subset of Rn and p be a point in U. Let f, g: U --7 Rm be smooth mappings. Then f ~ Ie g at p iff 38 Transversality for every multi-index a with lal :::; k and 1 :::; i :::; m where j; and gi are the coordinate functions determined by f and g, respectively and Xl> ••• , Xn are coordinates on U. Proof We proceed by induction on k. For k = 1'/"'1 g iff (df)p = (dg)p iff the first partial derivatives of f at p are identical with the first partial derivatives of g at p. Assume the Lemma is true for k - 1. Let Yl> •.• , Yn be the coordinates of Rn in U x Rn = TU. Then (df): U x Rn -+ Rm x Rm = TRm is given by (x, y) r-+ (f(X),]l(Y), ... ,]m(Y)) where - ~ oj; j;(x, y) = L.. ox (x)YJ. j=l j Similarly for (dg). By assumption (df) '" k-1 (dg) at every point (p, v) E {p} X Rn. By induction, the partial derivatives of (df) at points (p, v) E {p} X Rn are equal to the partial derivatives of (dg) at these same points. Let a be an n-tuple of non-negative integers with lal :::; k - 1, then olay; olalg, oxa (p, v) = oxa (p, v). Evaluate at v = (0, ... , 1, ..., 0) with the 1 in the jth coordinate. Then we have that Olal oj; olal Ogi OXaOXj (p) = Oxa OXj (p). Clearly all partial derivatives offand g of order :::; k are obtained this way. To obtain the converse, just note that the partial derivatives of(df) oforder :::; k - 1are determined by knowing the partial derivatives offoforder :::; k. 0 Corollary 2.3. f and g: U -+ Rm have kth order contact at p iffthe Taylor expansions offand g up to (and including) order k are identical at p. Lemma 2.4. Let U be an open subset ofRn and Van open subset ofRm. Let f1'};: U -+ V and gl, g2: V -+ Rl be smooth mappings so that gl·f1 and g2 ·f2 are defined. Let p E U and suppose that f1 '" kf2 at P and gl '" k g2 at q = f1(P) = f2(P)' Then gl ·f1 '" k g2 ·f2 at p. Proof Again proceed by induction. For k = 1, this is just the chain rule, i.e., d(gl·f1)P = (dg1MdfI)p = (dg2)q(df2)p = d(g2·f2)p. Assume true for k - 1. Then again apply the chain rule, using the inductive assumption that at all (p, v) in {p} x Rn. 0 Proposition 2.5. Let X, Y, Z, and W be smooth manifolds. §2. Jet Bundles 39 (1) Let h: Y --+ Z be smooth; then h induces a mapping h* : J"(X, Y)--+ J"(X, Z) defined as follows; Let 0' be in Jk(X, Y)P.q and letf: X --+ Y represent o'. Then h*(O') = the equivalence class ofh·f in Jk(X, Y)P,h(q)' (2) Let a:Z--+W be smooth. Then a*·h*=(a·h)* as mappings of J"(X, Y) --+ Jk(X, W) and (idy)* = idJk(x,y). Thus if h is a diffeomorphism, h* is a bijection. (3) Let g: Z --+ X be a smooth diffeomorphism; then g induces a mapping g* : J"(X, Y) --+ J"(Z, Y) defined as follows: let T be in J"(X, Y)p,q and let f: X --+' Y represent T. Then g*(T) = equivalence class off·g in Jk(X, Z)g -l(p),q' (4) Let a: W --+ Z be a smooth diffeomorphism. Then a*g* = (g.a)* as mappings ofJk(X, Y) --+ J"(W, Y) and (idx)* = idJk(x,y) so that g* is a bijec- tion. Proof A simple application of Lemma 2.4 shows that h* and g* are well-defined mappings. The rest of the proposition is equally easy. 0 Let An" be the vector space of polynomials in n-variables of degree :s; k which have their constant term equal to zero. Choose as coordinates for An" the coefficients ofthe polynomials. Then An" is isomorphic to some Euclidean space and is, in this way, a smooth manifold. Let B~,m = 81;"=1 An". B~,m is also a smooth manifold. Let Ube an open set in Rn andf: U --+ R be smooth. Define T"f: U --+ An" by T,,(f)(xo) is the polynomial of degree k given by the first k terms of the Taylor series offat Xo after the constant term. Let V be an open subset of Rm. Then there is a canonical bijection Tu,v:J"(U, V)--+ U x V x B~,m given by Tu,v(O') = (xo,Yo, T"fl(XO)' ... , Tdm(xo» where and Xo = a(O') = source of 0', Yo = fi(O') = target of 0', f: U --+ V is smooth and represents 0', It: U --+ R (1 :s; i :s; m) are the coordinate functions associated to f By Corollary 2.3, Tu.v is well-defined; i.e., independent of the choice off, and injective. That Tu,v is onto is clear. 0 Lemma 2.6. Let U and U' be open subsets ofRn and let Vand V' be open subsets of Rm. Suppose h: V --+ V' and g: U --+ U' are smooth mappings with g a diffeomorphism. Then Tu'.v,(g-l)*h*Tij}: U x V x B~,m --+ U' X V' x B~,m is a smooth mapping. Proof Let D = (xo, YO,fl(X), . . .,fm(x» with It E An" (1:s; i :s; m). Define f: U --+ Rm by f(x) = Yo +(fl(X - xo), ... ,fm(X - Xo». Thenf(xo) = 40 Transversality Yo and let a = equivalence class of fin Jk(U, V)(xo.Yo)' Tu.v(a) = D. Now (g-l)*h*(a) = r(h.f.g-l)(xo). So Tu'.v,(g-l)*h*Tu}(D) = TU'.V·(jk(h·f·g-l)(XO» = (g(xo), h(yo), Tk((h·f·g- 1)1)(g(XO»,"" Ti(h·f·g-l)m)(g(x») where (h·f·g-l)i: U' -7 R are the coordinate functions ofh·f·g-l: U' -7Rm. To show that this mapping is smooth we need only show that the mapping of U x V x B~.m -7 Ank given by D f--7>- Tk((h·f·g-1)i)(g(xo» is smooth. Let c{> = h·f·g-1. Then To show that D f--7>- Tk(c{>i)(g(XO»is smooth it is enough to show that D f--7>(olalC{>doxa)(g(xo» mapping U x V x B~,m -7 R is smooth for each multiindex a for which lal ~ k. This is done by the chain rule and induction on· I = lal. In fact, one can show by induction that (olalC{>doxlX)(g(xo» is sums and products of terms of the form of. (0) ox; , a -1 ~~. (g(xo» J where Yb ... , Ym are coordinates on Rm and hb gi are the coordinate functions determined by hand g respectively. Each of these terms vary smoothly with D; hence (olalC{>dox")(g(xo» varies smoothly with D. 0 Theorem 2.7. Let X and Y be smooth manifolds with n = dim X and m = dim Y. Then (1) ]k(X, Y) is a smooth manifold with dim Jk(X, Y) = m + n + dim (B~,m)' (2) a: Jk(X, Y) -7 X, f3 : Jk(X, Y) -7 Y, and a X f3: ]k(X, Y) -7 X X Y are submersions, (3) If h: Y -7 Z is smooth, then h*: JI'(X, Y) -7 Jk(X, Z) is smooth. If g : X -7 Y is a diffeomorphism, then g* : Jk( Y, Z) -7 ]k(X, Z) is a diffeo- morphism. (4) Ifg : X -7 Y is smooth, then rg : X -7 Jk(X, Y) is smooth. Proof (1) Let U be the domain for a chart c{> on X and V be the domain for a chart f on Y. Let U' = c{>(U) and V' = f(V). Then (c{>-l)*f*: ]k(U, V)-7 ]k(U', V') and TU,V == TU',V,·(c{>-l)*f*: ]k(U, V) -7 U' X V' x B~,m' Give Jk(X, Y) the manifold structure induced by declaring that TU.V is a chart. To see that this structure is well-defined we need only check to see what hap- §2. Jet Bundles 41 pens on overlaps. Let ,pl, .p1> U1> Vl , U~, V~ be the data for another chart TU1,Vl' Then note that TU1,Vl .(TU,V) -1 = T U1',V1-C,pl-l)*(.pl)*(.p*)-l,p*Tu'!v' = TU1',V1-C,pl-l·,p)*·(.pl·.p-l)*Tu,!v' since lower *'s and upper *'s commute. This last mapping is smooth by Lemma 2.6. (2) In local coordinates a has the form ,p.a'Tij.\r(D) = ,p.a.(.p*)-l.,p*Tu'!v.(D) = ,p.a·r(.p-l·f·,p)·,p-l(xo) wherefis defined in Lemma 2.6. Thus ,p.a'Tij.\r(D) = Xo since a.peg = idx for any mapping g. So a is a smooth mapping and a submersion. Similarly .p'{3'Tij.\r(D) = .p·{3·r(.p-l·f·,p)·,p-l(xo) = .p•.p-l·f·,p·,p-l(XO) = f(xo) = Yo since {3·rg = g for any mapping g. Thus {3 is also a smooth mapping and a submersion. Since T(p,qlX x Y) ~ TpX E8 Tq Y, a X {3: Jk(X, Y) -+ X X Y is a submersion. (3) is obvious from the calculations in (1). (4) rg: X -+Jk(X, Y). Suppose g: Rn -+ Rm. Thenrg: Rn -+Jk(Rn, Rm) = Rn x Rm x B~.m and is given by jkg(XO) = (xo, g(xo), (Tkgl)(XO), ... , (Tkgm)(XO)) where gl, ... , gm are the coordinate functions of g. Now Tkgj is a smooth function being only the sum of partial derivatives of the gt's. So in the local situationrg is a smooth function. With the standard use of the charts given above, one can see that rg is smooth as a mapping of X -+ ]k(X, Y). D Remarks. (1) Jk(X, Y) is, in general, not a vector bundle since there is no natural addition in]k(X, Y)p,q' However, if Y = Rm, then Jk(X, Y) is a vector bundle over X x Rm where the addition ofjets in]k(X, Rm)p,q is given by the addition of functions representing these jets. (2) ]leX, Y) is canonically isomorphic to Hom (TX, TY) where the isomorphism .p is given as follows: Let a be a I-jet with source p and target q, and letf: X -+ Y represent a. Then .p(a) = (df)p in Hom (TpX, TqY). As an exercise show that .p is well-defined and a diffeomorphism. Also note that a X {3 = 7T'.p where 7T is the projection which makes Hom (TX, TY) into a vector bundle over X x Y. Using this identification we can think of]leX, Y) as a vector bundle over X x Y. (3) Although ]k(X, Y) is not a vector bundle, .it does have more structure than just the fact that it is a manifold would indicate, We isolate that structure with the following Definition. Definition 2.B. Let E, X, and F be smooth manifolds and let 7T : E -+ X be a submersion. Then E is a fiber bundle over X with fiber F and projection 7T 42 Transversality iffor every p in X, there exists a nbhd U ofp and a diffeomorphism CPu : Eu --+ U x F where Eu = 'IT-I(U) such that the diagram commutes CPu Eu ----+ U x F u where 'lTu is the obvious projection. Notes. (1) Ep is diffeomorphic (under CPu) with F for all p in X. (2) Clearly every vector bundle of dimension n over X is a fiber bundle with fiber F = Rn. But not every fiber bundle with a Euclidean space as fiber is a vector bundle. (Consider Jk(X, Y) --+ X x Y!) Exercises (1) There is an obvious canonical projection 'lTk.!: Jk(X, Y) --+JZ(X, Y) for k > I defined by forgetting the jet information of order > I. Show that Jk(X, Y) is a fiber bundle over JZ(X, Y) with projection 'lTk.! and identify the fiber. (2) Let P(X, Rh.o be the set of alII-jets whose target is O. (a) Show that P(X, Rh.o is a vector bundle over X whose projection is the source mapping. (b) Show that P(X, Rh.o is canonically isomorphic (as vector bundles) with T*X. §3. The Whitney Coo Topology Definition 3.1. Let X and Y be smooth manifolds. (i) Denote by C"'(X, Y), the set ofsmooth mappings from X to Y. (ii) Fix a non-negative integer k. Let U be a subset of Jk(X, Y). Then denote by M(U) the set {fE C"'(X, Y) Irf(X) c U}. Note that M(U) n M(V) = M(U n V). (iii) Thefamily ofsets {M(U)} where U is an open subset ofJk(X, Y)form a basis for a topology on COO(X, Y). This topology is called the Whitney Ck topology. Denote by Wk the set ofopen subsets ofC 00 (X, Y) in the Whitney Ck topology. (iv) The Whitney COO topology on COO(X, Y) is the topology whose basis is W = Uk'=o Wk' This is a well-defined basis since Wk c Wz whenever k ::;; I. To see this use the canonical mapping 'lTk! : JZ(X, Y) --+ Jk(X, Y) which assigns to a in JZ(X, Y) the equivalence class off in Jk(X, Y) where f represents a. Then M(U) = M«'lTk!)-I(U))for every open set U in Jk(X, Y). §3. The Whitney Coo Topology 43 In order to develop a feeling for these topologies we will describe a nbhd basis in the Whitney Ck topology for a function f in C "'(X, Y). Choose a metric don Jk(X, Y) compatible with its topology. This is possible since all manifolds are metrizable by (1,5.9). Define BoCf) == {g E C"'(X, Y) IVx EX, d(jkf(x),jkg(x)) < o(x)} where 0: X ~ R + is a continuous mapping. We claim that BoCf) is an open set for every such O. For consider the continuous mapping,:l: Jk(X, Y) ~ R defined by a 1---+ o(a(a)) - d(jkf(a(a)), a). Let U = ,:l-l(O, 00). Then Uis open in Jk(X, Y)andBoC/) = M(U).Nowlet Wbean open nbhd oflin C"'(X, Y), let V be an open set in Jk(X, Y) so that fE M(V) c W, and let m(x) = imf {d(a,jkl(x)) Ia E a-lex) n (Jk(X, Y) - V)}. Note that m(x) = 00 if a-lex) c V. Let 0: X ~ R+ be any continuous function such that o(x) O. Choose ml so that Xl is in Kml and let 0 = al on Km1. Assume inductively that we have chosen functions fz1,...,.fzs with II < ... < Is; a compact set Kms; a continuous positive-valued function 0 defined on Kms ; and points Xl> ... , Xs in Kms so that for every i ~ s d(rfz,(Xi),jkf(Xi)) > o(xj ). 44 Transversality Now choosefzs+l where IS+1 > Is so thatfzs+l =1= f off Km,,+l' Let Xs+I be a point not in Kms+1where d{Jkfzs+/xs+ I),rf(xs+1)) = as+1 > O. Then choose ms+1 so that Xs+I is in KmS+ 1 ' Extend 8 to be a continuous positive-valued function on KmS+l which is =as+1 on KmS+l - Kms+l ' In this way we construct a subsequencefz1 ,fz2 , ••• and a continuous positive-valued function 8 defined on X so that for every j,fzJ rf= Blf). Thus fn does not converge to f and we have a contradiction. Finally we note that for a noncompact manifold X, C 00 (X, Y) in the Whitney Ck topology does not satisfy the first axiom of countability. To see this, let WI. W2 , ••• be a countable nbhd basis off in C 00 (X, Y). Then choose for each m a continuous function 8m : X-+- R + so that Blim(f) C Wm and a sequence of points Xl, X 2, ... with no limit point. Now construct a continuous function 8 so that 8(xm) < 8m(xm) for every m. Since WI' W2 , ••• is a nbhd basis off, there is an m such that Wm C Bo(f) which implies that Bom(f) C BIi(f) which is a contradiction. Thus we see that there is a great qualitative difference in the Whitney Ck topology on C"'(X, Y) depending on whether or not the domain X is compact. If X is compact then we get a standard type of topology. If, on the other hand, X is not compact we have defined a very fine topology on Coo(X, y), one with many open sets. In either case, though, a theorem which asserts that a given set is dense in C 00 (X, Y) is saying that this set is indeed quite large and is a rather strong result. Definition 3.2. Let F be a topological space. Then (a) A subset G of F is residual if it is the countable intersection of open dense subsets ofF. (b) F is a Baire space if every residual set is dense. Proposition 3.3. Let X and Y be smooth manifolds. Then Coo(X, Y) is a Baire space in the Whitney Coo topology. Proof For each integer k choose a metric dk on Jk(X, Y) which makes Jk(X, Y) into a complete metric space. Let UI , U2 , ••• be a countable sequence ofopen dense subsets of C 00 (X, Y) and let V be another open subset of Coo(X, Y). We must show that V n n~l Ui =1= 0. Since V is open in the Whitney COO topology, there is an open subset Win Jko(X, Y) such that M(W) C V and M(W) =1= 0. It is clearly enough to show that M (W) n n~ I Ui =1= 0. To do this we inductively choose a sequence of functions fl,12, ... in Coo(X, Y); a sequence of integers kI. k2 , ••• ; and for each i an open subset Wi in Jkt(X, Y) satisfying: (Ai) j; E M(W) n nj:i M(Wj) n Ui' (Bi) M(Wi) s:; Ui andj; E M(W1) (Ci) (i > 1) ds{J"f.(x);j"f.-I(x)) < 1/2i for all X in X and 1 :s; s :s; i. We first show that by choosing the above data we can prove the theorem. Define gS(x) = Limi_ooj"f.(x). This makes sense since ds is a complete metric and for each X the sequence j'fl(x),Pf2(X), . .. is a Cauchy sequence in §3. The Whitney Coo Topology 4S ]s(X, Y) by (e). Note that jDj;(x) = (x,};(x», so we can define g: X ~ Y with gO(x) = (x, g(x». We claim that g is smooth. If so, we are done. Indeed, each}; is in M(W) by (A) and thus g = Limi --+ oo }; is in M(W). Now, by (B), Ws was chosen so that M(Ws) C Us and by (A) each}; for i > s was chosen to be in M(Ws)' Thus g = Limi --+ oo }; is in M(Ws)' Since s is arbitrary g E M(W) II nS"'.:l Us and we are done. We will now show that g is smooth. This is a local question, so choose x in X and compact nbhds K of x and L ofg(x) with g(K) C L. By choosing K and L small enough we may assume that they are contained in chart nbhds and then via these charts that K and L are subsets of Rn and Rm respectively. Since the metric ds is compatible with the topology on ]s(X, Y), conditions (C) translate, in the local situation, to the fact thatj~ converge uniformly to gS on K. Using local coordinates we see that the coordinate functions of Ph are just 8liJ l};/8xiJ for /,8/ :s; s. Thus locally 8liJ l};/8xiJ converges uniformly on K. Using a classical theorem (Dieudonne 8.6.3, p. 157), 81iJig 8xiJ on K for all /,8/ :s; s. Since s is arbitrary all partial derivatives ofg exist at x, in fact gS(x) = pg(x) and g is smooth. Finally we will show that one can choose the};, ki' and TV; inductively satisfying (Ai), (Bi ), and (Ci). Choose/l in M(W) II Ul . This is possible since M(W) is open and nonempty while Ul is dense. Thus (Al) is satisfied. Since Ul is open and/is in Ul we may choose kl and an open set Wl in J k l(X, Y) so that/l E M(WI) and M(WI) CUI' Thus (BI) is satisfied. (el) is vacuous. Now assume inductively that the data is chosen for all j :s; i-I. We will choose}; satisfying (Ai) and (ei) and then we can easily choose Wi and ki so that (Bi) holds. Consider the set D, = {gEeOO(X, Y) / ds(j"g(X),j~_l(X» < ~ for 1:s; s :s; i and for all x in X}. If Di is open, then Ei = M(W) II n}-:i M(Wj) II D; is open. It is easy to check that};_l is in E; using the inductive hypotheses (Ai-I) and (Ei - l) and the definition of Di• Since Ui is dense and E; is open and nonempty we may choose}; in Ui II Ei• By the definition of E{, (Ai) is satisfied, and by the definition of D;, (ei) is satisfied. So the proof of the Theorem reduces to showing that Di is open in eOO(X, Y). Let Fs = {g E COO(X, Y) / ds(j"g(X),j~_l(X» < ~ Vx EX}Since D; = Fl II· .. II F;, it is enough to show that Fs is open in eOO(X, Y). Now define Ex = a-lex) II E(1/2i,j~_l(X» where a: ]s(X, Y) ~ X is the source mapping and B(~,j~_l(X») = {UE]S(X, Y) / ds(U,j~-l(X» <~}- 46 Transversality Let G = UxEX Bx. It is easy to see that Fs = M(G), so that we need only show that G is an open subset of ]s(X, Y). Let a be a point in G and x = a(a). Note that the mapping 'Y: X --+ R defined by q H> ds{j"j;-l(q),j"j;-l(X» is continuous. Thus H = a- llY-l(-812, (12) is an open subset of ]s(X, Y) where 8 = 1/2i - d.(a,j"j;_l(x». (Note that 8 > 0 since a is in G.) Clearly H n B(812, a) is open and contains a so that if H n B(812, a) c G, we are done. Let T E H n B(812, a). To show that T E G, we need to show that ds(T,j"j;_l(a(T») < 1/2i. But d.(T,j"j;_l(a(T») ::; dsCT, a) + ds(a,j"j;_l(x) + ds(j"j;_l(X),j"j;_l(a(a») < ~ + (~ - 8) + ~ =~. 0 Proposition 3.4. Let X and Y be smooth manifolds. The mapping jk: C"'(X, Y) --+ C"'(X, Jk(X, Y» defined by fH> jkf is continuous in the Whitney C'" topology. Proof Let V be an open subset of Jl(X, Jk(X, Y». Then M(V) is a basic open set in C"'(X, Jk(X, Y». It is sufficient to show that (r)-l(M(V)) is an open subset of C"'(X, Y). First we define a mapping ak,l: Jk+I(X, Y) --+Jl(X, Jk(X, Y» as follows: let a be a (k + I)-jet in Jk+I(X, Y) with source x and letf: X --+ Y represent a. By Theorem 2.6 (4), jkf: X --+Jk(X, Y) is a smooth mapping. Define ak,zCa) = PV'f)(x). That ak'l(a) is well-defined, i.e., does not depend on the choice of representative f, can be seen from Corollary 2.3, and the fact that p(rf)(x) depends only on the partial derivative off at x of order ::; k + I. For the same reasons, it is clear that ak,l is a smooth mapping. (In fact, it is an embedding.) Thus ak}(V) is an open subset of Jk+I(X, Y). We claim that M(ak.t(V») = (r)-l(M(V» and thus (jk)-l(M(V» is an open subset of C"'(X, Y). The claim follows trivially from the fact that ak,l·r + If = P ·rf as mappings of X --+Jl(Jk(X, Y». 0 Proposition 3.5. Let X, Y, and Z be smooth manifolds. Let *: Jk(X, Y) --+Jk(X, Z) defined by a = jkj(x) H>r(4)·f)(x). Thus * is continuous. 0 We shall now investigate the properties of C"'(X, R) which is a vector space over R. It would be nice if C"'(X, R) were a topological vector space, but alas, scalar multiplication viewed as a mapping of R x C"'(X, R)--+ C"'(X, R) is not continuous unless X is compact. For if X is not compact and f: X --+ R is a smooth function with noncompact support, then y!: R--+ §3. The Whitney Coo Topology 47 Coo(X, R) defined by r H>- rf would be continuous. Thus Limn~'" fin = O. But this cannot happen unless there is some compact set K off of which each function fin is zero which contradicts the assumption on the support of f Addition and multiplication of functions fare better. Proposition 3.6. Let X, Y, and Z be smooth manifolds. Then C "'(X, Y) x Coo(X, Z) is homeomorphic (in the COO topology) with Coo(X, Y x Z) by using the standard identification U; g) H>- f x g where f x g(x) = (I(x), g(x». To prove this proposition we need the following Lemma on topological spaces. Lemma 3.7. Let A, B, and P be Hausdorff spaces. Suppose that P is locally compact andparacompact. Let 7TA: A --0>- P andTTB : B --0>- P be continuous. Set A x p B = {(a, b) E A x B I7TA(a) = 7Tib)} and give A x p B the topology induced from A x B. Let K c A and L c B be subsets such that 7TAIK and 7TBIL are proper. Let U be an open nbhd ofK x p L in A x p B. Then there exists a nbhd V of K in A and a nbhd W ofL in B such that V x p W S; U. Proof First note that if X and Yare Hausdorff spaces with Y locally compact and iff: X --0>- Y is continuous and proper, thenfis a closed mapping. For let Z be a closed subset of X and y be a point inf(Z). Let Yb Y2, ... be a sequence of points inf(Z) with Limi~oo Yi = y. Since Y is locally compact, there is a compact nbhd V of y. We may assume that Yi in V for all i. Choose Xb X2, ... so that f(Xi) = Yi' Since f is proper f-l(V) is compact. Thus by restriction to a subsequence we may assume that the sequence Xl, X2, . .. converges. Suppose Limi~ ro Xi = x. Then X is in Z since Z is closed and by the continuity ofJ,f(x) = y. So Y is inf(Z) andf(Z) is closed. Now consider 7TA X 7TB: A x B --0>- P X P. Note that /:::..p, the diagonal of P x P, is closed and that A x p B = (7TA x 7TB)-l(/:::..p). So E = A x B A x p B is open. For eachp inP, letKp= K () (7T A)-l(p) andLp= L () (7TB)-l(p). Note that Kp x Lp = Kp xpLp c U and that Uu E is open in A x B. Since 7TAIK and 7TBI L are proper, Kp and Lp are compact. Thus there is an open nbhd Vp of Kp in A and Wp of Lp in B such that Vp x Wp c U u E. To see this, choose for each (k, 1) in Kp x Lp open nbhds Vk,! of k in A and Wk,! of 1in B such that Vic,! x Wk,! C U u E. For a fixed k, the collection {Wk'!heLp is an open covering of Lpo Since Lp is compact, there is a finite subcovering Wk'!l, ... , Wk'!m. Set Vk = Vk'!l () ... () Vk'!m and Wk = Wk'!l u· .. U Wk'!m and note that Vk x Wk C U u E. The collection {VkheKp is an open covering of Kpo So, by compactness, there is a finite subcovering V"l, ... , V"n. Set Wp = Wkl () ... () W"n and Vp = Vkl u· .. U Vkn • To continue with the proof of the lemma, note that 7TAIK and 7TBI L satisfy the hypotheses of the first paragraph; thus 7TA(K - Vp) and 7TB(L - Wp) are closed in P and Pp = P - 7TA(K - Vp) - 7TiL - Wp) is open. Moreover, p is in Pp. Thus the collection {Pp}pep is an open covering of P and since P is 48 Transversality paracompact there is a locally finite refinement {PIX}' For each cx choose cx(p), a point in P, so that PIX c p"(P). Let V" = V,,(P) U TTA -l(P - PIX) and W" = W,,(P) U TTB -l(P - Pa). Let V = nIX Va and W = nIX W". To complete the proof, we need to show that K c V, LeW, V and Ware open, and VxpWc U. (a) K c V. It is clearly enough to show that K c V" for each cx. So assuming that k is in K, we must show that k E V"(P) or k E TTA -l(P - PIX)' But this is trivial, since if k rf= V"(P)' then 7TA(k) rf= p"(P)· (b) LeW. Just the same as (a). (c) V is open. Let v be a point in V. Since {P,,} is locally finite, there exists a nbhd V of 7TA(V) and finitely many PIX'S; namely, P"l'" "P"r' for which Vnp", # 0. Let V' = 7TA- 1(V) n V"l n···n V"r and note that V' is an open nbhd of v. If V n PIX = 0, then V' c 7T- 1(P - PIX) c V". If V n PIX # 0, then cx = CXi for some i and V' c V". So V' c V and V is open. (d) W is open. Just the same as (c). (e) V x p We V. Choose (v, w) E V X p Wand set p = 7Tiv) = 7Tiw). Choose an cx such that p is in PIX' Thus v E V" - TTA -l(P - PIX) c V"(P). Similarly w is in W,,(P)' Hence (v, w) E V,,(P) X W,,(P) c VuE. Since 7TA(V) == 7Tiw) (u, v) ~ E, thus V x pW c V. 0 Proof of Proposition 3.6. The projections 7Ty: Y x Z -+ Y and 7TZ: Y x Z -+ Z induce continuous mappings (7Ty)*: C"'(X, Y x Z)-+ C "'(X, Y) and (7TZ)*: C"'(X, Y x Z) -+ C "'(X, Z) by Proposition 3.5. Since the identification of C"'(X, Y x Z) with C"'(X, Y) x C"'(X, Z) is given by (7Ty)* x (7Tz)*, it is continuous. To show that the identification is a homeomorphism we need only show that it is an open mapping. To do this we let (I, g) be in C"'(X, Y x Z). Choose an open set Win Jk(X, Y x Z) so that (I, g) is in M(W). Now notice that J"(X, Y x Z) = Jk(X, Y) X x J"(X, Z) where A = P(X, Y), B = P(X, Z), P = X, and 7TA: P(X, Y) -+ X and 7TB: Jk(X, Z) -+ X are the respective source mappings. Applying Lemma 3.7, there are open sets V in Jk(X, Y) and V in Jk(X, Z) such that V x xV c W. Finally we note that M(V) x M(V) c (7Tyh x (7Tz)*(M(W)), so that this identification is an open mapping. 0 Corollary 3.8. Addition and multiplication of smooth functions are continuous operations in the C'" topology, i.e., C"'(X, R) x C"'(X, R)-+ C"'(X, R) given by (I, g) 1-+ f + g or (I, g) -+ f·g is continuous. Proof +: R x R -+ R given by (x, y) 1-+ X + y is continuous so (+)*: C"'(X, R x R) -+ C"'(X, R) is continuous by Proposition 3.5. Thus via the homeomorphism of C"'(X, R x R) with C"'(X, R) x C"'(X, R) given by Proposition 3.6 (I, g) 1-+f + g is continuous. The proof for multiplication is similar. 0 For completeness sake, we make some further remarks about the Whitney C'" topology on C"'(X, Y). §3. The Whitney Coo Topology 49 Proposition 3.9. Let X, Y, and Z be smooth manifolds with X compact. Then the mapping of Coo(X, Y) x COO(Y, Z) ---J>- COO(X, Z) given by (J, g) f---7> g·f is continuous. Remark. This proposition is not true if X is not compact although if we replace COO(X, Y) by the open subset of proper mappings of X into Y, then the conclusion is still valid. Proof Let D be the fiber product Jk(X, Y) xyJk(Y, Z) described in Lemma 3.7, where A = Jk(X, Y), B = Jk(Y, Z), P = Y,7I'A = {3 (the target mapping), and 71'B = a (the source mapping). The mapping y: D ---J>-Jk(X, Z) defined by (a, T) f---7> T'a is continuous. (Note Toa = jk(g.f)(a(a)) wherefrepresents a in Jk(X, Y) and g represents Tin Jk(Y, Z)). To prove the proposition it is enough to show that iff is in C "'(X, Y), if g is in C"'(Y, Z), and if S s Jk(X, Z) is open with g·fin M(S), then there are open sets V c Jk(X, Y) and We Jk(Y, Z) with y(V Xy W) c S. Then iff' is in M(V) and ifg' is in M(W), g'.f' will be in M(S). Thus composition will be a continuous mapping in the Whitney Ck topology for arbitrary k and thus continuous in the Coo topology. First we note thatr(g·f)(X) = yUkf(X) xyrg(Y)) i.e., We now apply Lemma 3.7 with K = jkf(X), L = jkg( Y), and U = y-l(S) to show the existence of the desired Vand W. That Lemma 3.7 is applicable follows from the facts that U is open (since y is continuous), 71'A!K is compact (since X and thus K are compact), and 71'B! L is proper (since 71'B·jkg = idy ). 0 Notes. (1) Let f: X ---J>- Y be smooth; then 71' induces f* : COO( Y, Z)---J>COO(X, Z) given by g f---7> g·f The Remark after this last proposition shows that iff is not a proper mapping then this" nice" functorially defined mapping is not continuous. In particular, if X is an open subset of Y and f is just given by inclusion then 71'* is not continuous; i.e., the restriction mapping of COO(Y, Z) ---J>- C"'(X, Z) given by g f---7>g!X is not continuous. (2) An easy consequence of the proof of Proposition 3.9 is that f* is continuous iff is proper. The only use that was made of the compactness of X was to show that 71' A! K is proper; this statement is still true iff is proper. For future reference we make one last comment about the continuity of these types of functorial mappings. If X is a set, let Xl = X X ... x X. Proposition 3.10. Let X and Y be smooth manifolds. The mappings i)l: COO(X, YY ---J>- COO(XI, yl) given by (fl>'" ,/z) f---7> fl x ... x fz where (fl x··· X/z)(Xl>"" Xl) = (fl(Xl), .. .,fz(xl)) is continuous. Proof We assume that I = 2 as the proof for general I is essentially the same. First we claim that the mapping y : Jk(X, Y) x Jk(X, Y) ---J>- Jk(X2, y2) given by (a, T) ---J>- a X T (where a(a X T) = (a(a), a(T)), fi(a X T) = ({3(a), (3(T)), and if f represents a and if g represents T, thenf x g represents a X T) is continuous. To see that this claim is sufficient, let Wbe an open nbhd inJk(X2, y2) 50 Transversality withf x gin M(W). Since y-l(W) is open in Jk(X, Y)2 and containsi1(X) x jkg(X), there exist open sets U and V in Jk(X, Y) so thatj'1(X) x PCg(X) c U x V C y-l(W). Thus f x g E M(U) x M(V) C 82 -l(M(W)) and 82 is continuous. To see that y is continuous (in fact, a smooth embedding), look in local coordinates. In these coordinates for fixed sources and targets y is just a linear injection which varies smoothly with X x X x Y x Y. 0 §4. Transversality Definition 4.1. Let X and Y be smooth manifolds and f: X --7 Y be a smooth mapping. Let W be a submanifold of Y and x a point in X. Then f intersects W transversally at x (denoted by Fm W at x) if either (a) f(x) rJ W, or (b) f(X)E Wand T[(x)Y = Tf(x)W + (df)xCTxX). If A is a subset of X, thenfintersects W transversally on A (denoted by fm Won A) iffm W at x for all x E A. Finally, f intersects W transversally (denoted by fm W) if fm Won X. Examples. (1) Let X = R = W, Y = R2, and f(x) = (x, x 2). Then fm W at all nonzero x. Y W Notice thatf can be perturbed ever so slightly to be transversal to W; e.g., -------------+------------- W §4. Transversality 51 ) ----------~~+-~----------- W (2) Suppose that X, Y, Ware as in (1) and thatfis given by the graph Thenf does not intersect W transversally on the segments within the brackets and does elsewhere. (3) If X = R = Wand Y = R3, then iffis any mapping of X --+ Y, it is transversal to Wonly if f(X) n W = 0. Notice that here too a nontransversal mapping can be approximated closely by a transversal mapping, since in 3-spacefcan avoid Wby just "going around" it. Moreover,fdoesn't have to move far to accomplish this task. This will be made precise shortly. In any case, it becomes apparent quickly that the relative dimensions of X, Y, and W play an important part in determining what transversality means in a particular instance. Also, for any trio X, Y, and W, the set of transversal mappings is quite large. In fact, the Thorn Transversality Theorem is just this observation formalized. Before discussing this theorem, we will give some consequences of the property that a mapping is transversal. Proposition 4.2. Let X and Y be smooth manifolds, W c Ya submanifold. Suppose dim W + dim X < dim Y (i.e., dim X < codim W). Let f: X --+ Y be smooth and suppose that fm w. Then f(X) n W = 0. Proof Suppose f(x) E W. Then dim (Tf(x)W + (df)xCTxX)) ::::; dim Tf(x)W + dim TxX = dim W + dim X < dim Y = dim Tf(x)Y' 52 Transversality So it is impossible for Tf(x) W + (df)iTxX) = Tf(x) Y. Hence if F mWat x, f(x) f/= w. 0 Lemma 4.3. Let X, Y be smooth manifolds, W c Y a submanifold, and f: X --+ Y smooth. Let p E X andf(p) E W. Suppose there is a nbhd U off(p) in Yand a submersion rp: U --+ Rk(k = codim W) such that W n U = rp-1(0). Then fm W at p iff rp·f is a submersion at p. Remark. Such a nbhd U always exists. For there exists a chart nbhd U of f(p), a chart a: U --+ Rm (m = dim Y) and a decomposition of Rm = Rk X Rm-k so that W n U = a- 1(0 X Rm-k). Let TT: Rm --+ Rk be projection on the first factor, then let rp = TT·a. Proof. One can show easily that Ker (drp)f(P) = Tf(p) W. So fm W at p iff Tf(p) Y = Tf(p) W + (dfMTpX) iff Tf(p) Y = Ker (drp)f(P) + (dfMTpX) Since (drp)f(P) is onto we see that (drp .f)p is onto iff this last equality holds. Thus rp·f is a submersion at p ifffm W at p. 0 Theorem 4.4. Let X and Y be smooth manifolds, W a submanifold of Y. Letf: X --+ Y be smooth and assume thatfm W. Thenf-1(W) is a submanifold of X. Also codimf-1(W) = codim W. In particular, if dim X = codim W, then f-1( W) consists only of isolated points. Proof. It is sufficient to show that for every point p Ef- 1(W), there exists an open nbhd V ofp in X so that V n f- 1(W) is a submanifold. Choose U and rp as in Lemma 4.3. Choose Va nbhd ofp so thatf(V) c U. By Lemma 4.3 rp·fis a submersion at p. Thus, by contracting V if necessary, we assume that rp·f is a submersion on V. Thus f-1(W) n V = (rp·(fl V»-l(O) is a submanifold, by (1,2.8). 0 Proposition 4.5. Let X and Y be smooth manifolds with Wa submanifold of Y. Let Tw = {fE COO(X, Y) Ifm W}. Then Tw is an open subset of COO(X, Y) (in the Whitney Ct, and thus, Coo, topology) if W is a closed submanifold of Y. Proof. Define a subset U of P(X, Y) as follows: let 0' be a I-jet with source x and target y and let f: X --+ Y represent 0'. Then 0' E U iff either (i) y ¢ Wor (ii) y E Wand TyY = TyW + (dfMTxX). Recall that M(U) = {fE COO(X, Y) ICj1f)(X) c U}. It is clear that Tw = M(U), so that if we can show that U is open, then so is Tw. We show that V = P(X, Y) - U is closed. Let 0'1, 0'2, ... be a convergent sequence of I-jets with O'j in V for all i. Let 0' = Limj_ 00 0'1> we will show that 0' is in V. Let p = source 0' and q = target of 0'. Since the targets of O'j are in W for all i and W is closed, q is in W. Let f: X --+ Y represent 0'. Choose coordinate nbhds U ofp in X and V ofq in Y so thatf(U) c V. Assume that the chart defined on V takes W onto a subspace of dimension k. Via these charts we may assume that X = Rn, x = 0, Y = Rm, and W = Rk C Rm. §4. Transversality S3 Let cp: Rm --+ RmjRk = Rm-k be projection. By applying Lemma 4.3 we see thatfm Wat 0 iff cp·fis a submersion at 0 iff cp.(df)o rf: F where F = {A E Hom (R", Rm-k) Irank A < m - k}. Consider the mapping R" x W x Hom (R", Rm) c P(R", Rm) -4- Hom (R", Rm-k) given by (x, w, B) 1--+ cp.B. Since F is closed and TJ is continuous TJ-1(F) is closed in R" x W x Hom (R", Rm) which, in turn, is a closed subset of P(R", Rm). Now V = TJ-1(F) since T = (x, y, (dg)x) is in V iff y is in Wand g does not intersect W transversally at 0 iff TJ(T) is in F. Since V is closed in this local situation u is in V. 0 Note that Proposition 4.5 does not hold if W is not assumed to be closed. As an example take X = S1, Y = R3, and W = {(t, 0, 0) I0 < t < I}. Transversality in these dimensions means thatf(X) n W = 0. Letf: S1--+ R3 be given by f(x, y) = (x + 1, y, 0) where S1 c R2 is thought of as the unit circle centered at the origin. Thenfm W but arbitrarily small perturbations off given byJ.(x, y) = (x + 1 - e, y, 0) intersect Wand are thus not transversal to W. Lemma 4.6. Let X, B, and Z be smooth manifolds with W a submanifold of z. Let j: B --+ C "'(X, Y) be a mapping (not necessarily continuous) and define : X x B --+ Y by (x, b) = j(b)(x). Assume that is smooth and that mw. Then the set {b E B Ij(b) mW} is dense in B. Proof Let W<1) = -1(W). Since m W, W<1) is a submanifold of X x B. (Apply Theorem 4.4.) Let 7T be the restriction to W<1) of the projection of X x B --+ B. First note that if b rf: 1m 7T, then j(b)(X) n W = 0 so j(b) m W. Now if dim W<1) < dim B, then 7T(W<1) has measure zero in B by Proposition 1.6 and for a dense set of b in B; namely B - 1m 7T, j(b) m w. Thus, in this case, the Lemma is true and we may assume that dim W<1) ~ dim B. We claim that ifb is a regular value for 7T, thenj(b) m W. If this claim is true, then the lemma is proved for we may apply Sard's Theorem (actually Corollary 1.14) to 7T. To prove the claim let b be a regular value for 7T and let x be in X. If (x, b) ¢ W<1), thenj(b)(x) rf: Wandj(b) m Wat x. So we may assume that (x, b) is in W<1)' Since b is a regular value for 7T and dim W<1) ~ dim B, we have that T(X.b)X x B = T(x,b)W<1) + T(x,b)X x {b}. Apply (d m W so T<1)(x,b) Y = T<1)(X,b) W + (d = G. Assume G rh W, then the set {b E B \ Gbrh W} is dense in B. This remark is the basic fact about transversality; that is, if a parametrized family of mapping intersects a given submanifold transversely, then for a dense set of parameters the individual mappings also intersect this submanifold transversely. In the same spirit, we have the following: Corollary 4.7. Let G: X x B -» Y be a smooth mapping. Let (x, b) = j"Gb(x) Assume that rh W where W is a submanifold ofJk(X, Y). Then the set {b E B \jkGbrh W} is dense in B. Proof. Define j: B -» C 00 (X, Jk(X, Y)) by b 1-+ jkGb and apply Lemma 4.6. 0 Definition 4.8. Let X and Y be smooth manifolds withf: X -» Y a smooth mapping. Let W be a submanifold of Y with W' a subset of W. Thenfrh Won w' iffor every x in X withf(x) in W',frh Wat x. Theorem 4.9. (Thom Transversality Theorem). Let X and Y be smooth manifolds and Wa submanifold ofJk(X, Y). Let Tw = {fE COO(X, Y) \j'1"rh W}. Then Tw is a residual subset of C 00 (X, Y) in the Coo topology. Proof. We need to show that Tw is the countable intersection of open dense subsets. To construct the sets which will go into this countable intersection, we first choose a countable covering of Wby open subsets WI, W2 , • •• such that each Wr satisfies (a) the closure of Wr in Jk(X, Y) is contained in W, (b) Wr is compact, (c) there exist coordinate nbhds Ur in X and Vr in Y such that 7T(Wr) C Ur X Vr where 7T: Jk(X, Y) -» X x Y is the projection mapping, and (d) Dr is compact. This choice is possible since around each point w in W, we may choose an open set Ww satisfying (a), (b), (c), and (d), since W is a submanifold of Jk(X, Y). Since W is second countable we may extract a countable subcovering from {WW}WEW' Let TWr = {fE COO(X, Y) \frh Won Wr}. It is clear that Tw = n,:"= I TWT ' Thus the proof reduces to showing that each TWr is open and dense in COO(X, Y). Define Tr = {g E COO(X, Jk(X, Y)) \ g rh Won Wr}. The proof of Proposition 4.5 can be easily adapted to show that Tr is open since Wr is closed and contained in W. Since P': COO(X, Y) -» COO(X, Jk(X, Y)) is continuous (by Proposition 3.4), TWr = (P') -1(Tr) is open. We now continue with the harder part of the Theorem, that is, to show §4. Transversality 55 that Tw, is dense. Choose charts ¢;: UT --+ Rn and Tj : VT --+ Rm and smooth functions p: Rn --+ [0, 1] c Rand p': Rm --+ [0, 1] c R such that p = {~ and , {Ip = o on a nbhd of ¢;.a(Wr) off ¢;(Ur) on a nbhd of Tj.f3(WT) off Tj(Vr) where n = dim X, m = dim Y, a is the source map, and f3 is the target mapping. The choice of p and p' are possible since Wr is compact. We will use Corollary 4.8 to show thatf can be perturbed slightly to be transversal to Wr • The perturbation will be accomplished locally using the data defined in the previous paragraph. Let B' be the space of polynomial mappings of Rn --+ Rm of degree k. For b in B', define gb : X --+ Y by { f(X) if x ¢ UT orf(x) ¢ Vr gb(X) = Tj-l(p(¢;(x))p'(rd(x))b(¢;(x)) + Tjf(x)) otherwise. The choice of p and p' guarantees that gb is a smooth function from X to Y and is just a polynomial perturbation off done locally and smoothed out so that it is equal to f off the domain of interest. Define G: X x B' --+ Y by G(x, b) = gb(X). By inspection of the formula defining gb, one sees that Gis smooth. Define : X x B -'r Jk(X, Y) is locally a diffeomorphism near (x, b). For let a be in Jk(X, Y) near <1>(x, b), let x' = a(a), and let b' be the unique polynomial mapping of degree ::;, k such that a = i'(T)(b'.if; + T) ·f))(x'). Then a 1-+ (b', x') is a smooth mapping and is the inverse of <1>. 0 Corollary 4.10. For each integer i, let Wi be a submanifold ofJk!(X, Y). Then the set of smooth mappings f: X -'r Y for which j k1m Wi is dense in cro(X, Y). Ifthe number of W;'s isfinite and each Wi is closed, then this set is open as well. Proof Follows immediately from Theorem 4.9 and the fact that cro(X, Y) is a Baire space. 0 Corollary 4.11. Let X and Y be smooth manifolds and Wa submanifold ofJk(X, Y) such that a(W) is contained in an open subset U of X. Letf: X -'r Y be a smooth mapping and Van open nbhd offin cro(X, Y). Then there exists a smooth mapping g in V such that i'g m Wand g = f off u. Proof This is really a corollary to the proof of the Thorn Transversality Theorem. Under the assumptions of the corollary it is clear that for each Wn a(Wr) C U. Thus we can choose Ur so that Ur C U. Then note that the perturbation gb = f off Un and thus off U, for each b in B. So the constructed transversal mapping does in fact equal f off U. 0 Corollary 4.12. (Elementary Transversality Theorem). Let X and Y be smooth manifolds with Wa submanifold of Y. Then (a) the set ofsmooth mappings of X to Y which intersect W transversally is dense in cro(X, Y) and if W is closed, then this set is also open. (b) let UI and U2 be open subsets of X with VI C U2 • Letfbe in cro(X, Y) and V be an open nbhd off in cro(X, Y). Then there is a smooth mapping g: X -'r Y in V such that g = f on UI and g m W off U2 • Proof (a) Note that JO(X, Y) = X x Y and rf(x) = (x,J(x)). The projection [3: X x Y -'r Y is a submersion so [3-I(W) is a submanifold of X x Y. If rfm [3-I(W) at x, thenfm Wat x. For eitherjOf(xH [3-I(W) in which case f(x) ¢ W or jOf(x) E [3-I(W) and T(x.[(X»[3-I(W) + (djOf)xTxX = T(x.[(x))(X x Y). Apply (d[3)(x.[(X)) to each side to obtain Tf(x) W + (df)x(TxX) = Tf(x) Y. Thusfm Wat x. Since the set of transversal mappings to W contains the set {fE cro(X, Y) IjOfm [3-I(W)} which is dense by Theorem 4.9, we are done. Note that the last part of (a) is just Proposition 4.5. For (b) note that W' = [3-I(W) n (X x Y - a- I(V2)) is a submanifold of X x Y since X x Y - a- I(V2) is an open subset of X x Y. Also a(W') is contained in the open set X - VI so by Corollary 4.11, there exists g: X -i>. Y in V such that g = f on UI and jOg m W'. This latter condition is the same as jOg m [3-I(W) off U2 • Thus g m Woff U2 as in (a). 0 §4. Transversality 57 We now present a generalization of transversality in jet spaces to transversality in multijet spaces which is useful for studying the injectivity, or alternately, the self-intersections of smooth mappings. Let X and Y be smooth manifolds. Define XS = X x ... x X (s-times) and X, Z = J/,(X, Y) and j: B --+ coo(xs,Jsk(X, Y)) be given by j(b) = j/,Gb. Apply Lemma 4.6. 0 Exercises (1) (a) Let X and Y be smooth manifolds with W a submanifold of Y. Assume that dim X = codim W. Let p be in X and letf: X --+ Y be smooth. Assume thatf(p) in W andfffi W at p. Then there exists a nbhd N of fin COO(X, Y) (in the Cl topology) and an open nbhd U ofp in X such that if g is in N, then g-l(W) ('\ U consists of one point q and g ffi W at q. (Hint: Use Lemma 4.3 and note the similarity with Theorem 4.4.) (b) Assume that Xis compact. Letj':X--+ Yffi W. Showthatthereis an open nbhd N of fin COO(X, Y) such that the number of points inf-l(W) is equal to the number of points in g-l(W)" for any g in N. (2) Let f: X --+ X be smooth with p in X a fixed point for f Definition: p is a non-degenerative fixed point iff (df)p : TpX --+ TpX does not have 1 as an eigenvalue. Let Diff (X) be the group of smooth diffeomorphisms on X and give Diff (X) the relative topology as a subset of COO(X, X). (a) Show that {IE Diff (X) Ifixed points off are nondegenerate} is open and dense in Diff (X), and (b) Show that nondegenerate fixed points are isolated. Hint: Consider what it means for rfffi ~x at p in X where ~X is the diagonal in X x X = JO(X, X). (3) Let {Wa}aeI be a family of submanifolds of Jk(X, Y) where I is some index set. Let Tw = {IE COO(X, Y) IVa E I,jkjffi Wa}. Suppose that UaeI Wa is closed in Jk(X, Y). Then show that Tw is an open subset of COO(X, Y). §5. The Whitney Embedding Theorem Let X and Y be smooth manifolds. We want to show that if dim Y is large enough relative to dim X, then the set of immersions of X into Y is dense in COO(X, Y). The idea of this proof of the Whitney Immersion Theorem will be to translate the fact thatfis an immersion into (a finite number of) transversality conditions. 60 Transversality Let a be in P(X, Y); then a defines a unique linear mapping of TpX---+ Tq Y where p is the source of a and q is the target of a. Letfbe a representative of a in COO(X, Y). Then (df)p is that linear mapping. Define rank a = rank (df)p and corank a = q - rank a where q = min (dim X, dim Y). Let Sr = {a E P(X, Y) Icorank a = r}. We will show that Sr is a submanifold of P(X, Y). The significance of the submanifolds Sr is illustrated by the following obvious lemma. Lemma 5.1. f: X ---+ Y is an immersion iffPf(X) n (Un'o Sr) = 0. One also observes the following: Lemma 5.2. Let S be an m x n matrix where S = (~ ~) where A is a k x k invertible square matrix. Then rank S = k iff D - CA -1B = O. Proof The matrix is an m x m invertible matrix. So rank S = rank TS = rank (~ D _ ~A-1B) . Clearly this latter matrix has rank = k iff D - CA -1B = O. 0 Let V and W be vector spaces of dimension nand m respectively. Let q = min {n, m}. Let S: V ---+ W be linear, then define corank (S) = q rank (S). Define V(V, W) = {S E Hom (V, W) Icorank S = r}. Proposition 5.3. V(V, W) is a submanifold of Hom (V, W) with codimV(V, W) = (m - q + r)(n - q + r). Proof Let S be in Lr( V, W) and let k = q - r = rank (S). Choose bases of V and W so that the matrix of S = (~ ~) where A is a k x k invertible matrix. Choose an open nbhd U of S in Hom (V, W) so that if S' is in U and S' = (~', ~,) then A' is a k x k invertible matrix. Consider the smooth mapping f: U ---+ Hom (Rn-k, Rm-k) given by f(S') = D' C'(A')-lB'.fis a submersion since if we fix A, B, C, then g: Hom (Rn-I<:, Rm-k) ---+ Hom (Rn-I<:, Rm-k) given by g(D) = f (~ ~) = D - CA -1B is a diffeomorphism. (In particular (dg)v = identity, so (df)s is surjective.) By Lemma 5.2f-1(0) = V(V, W) n U which is a submanifold sincefis a submersion. Moreover codim V(V, W) = dim Hom (Rn-k, Rm-k) = (n - k)(m - k). 0 5. The Whitney Embedding Theorem 61 Theorem 5.4. Sr is a submanifold of Jl(X, Y) with codim Sr = (n - q + r)(m - q + r). In fact, Sr is a subfiber-bundle of Jl(X, Y) with fiber Lr(Rn, Rm). Proof Let U c X and V c Y be coordinate charts. Then Jl(X, Y)u x v ~ U x V x Hom (Rn, Rm) and under this isomorphism Sr ~ U X V x veRn, Rm). Applying Proposition 5.3, Sr is then a submanifold. 0 Let 1m (X, Y) be the subset of immersions in C 2·dim X = dim X(2) we have that j2°flt W iff j2°f(X(2» n W = 0. So f is 1: 1 iff j2°flt Wand the result follows from the multijet transversality theorem. 0 Proposition 5.8. Let X and Y be smooth manifolds with X compact. Then the set of 1: 1 immersions is open in C: U --+ Rn a chart. Assume that U is small enough so that there exists a chart nbhd V off(p) with chart if;: V --+ Rm such that f( U) c Vand if;·f·4> -1 is a linear injection. This is possible since f is an immersion at p. Choose an open nbhd Up of p so that Up is compact and contained in U. Suppose that g: X --+ Y so that g(Up ) c V. Then define 62 Transversality g = ifIogocp-l: cp«(]p) ---+ Rm. Let p' = cp(p) and let M = infIX\;l l(dj)p,(x)l. Note that M > 0 since (dJ)p' is 1: 1. Let WI = {g E C"'(X, Y) Ig(Up ) C V and I(dj)p' - (dg)xl < ~'r/x E cp(Up)}WI is an open nbhd in C"'(X, Y) since the first condition is a Co open one and the second one is (:'1 open. Using the triangle inequality in Rm we have that l(dj)p,(Xl) - (dj)P,(X2)I ::; Ig(Xl) - g(X2)I + I«dj)p' - g)(xl) - «dj)p' - g)(x2)1. Now if g is in W" then I«dj)p' - g)(Xl) - «dj)p - g)(X2)I ::; (d«dj)p' - g))xl IXl - x21 ::; ~ IXl - x21 where the first inequality is given by the Mean Value Theorem for some x in cp(Up). Thus for g in W" Ig(Xl) - g(X2)I~ Mlxl - x21 - ~ IXl - x21 = ~ IXl - x21 and g is 1: 1. Since the set of immersions of X into Yis open we are done. 0 Proof of Proposition 5.B. Let f: X ---+ Y be a 1: 1 immersion. First we show that there is a nbhd Z of t::..x in X x X and an open nbhd W' off in C"'(X, Y) such that for every g in W' g is an immersion and g2(Z - t::..X) n t::.. Y = 0 where g2 : X2 ---+ y2 is given by g x g. For every p in X, choose Up and W/ as guaranteed by Lemma A. Since X is compact we can choose Pl> ... ,ps so that UPl'· ..' Ups cover X. Let W' = nf=l Wl' and Z = Uf=l (Up, x Up,). Let g be in W'. Then glUp, is an immersion for each i, so g is an immersion. Also if (p, q) is in Z - t::..X then (p, q) E Up, X Up, for some i and thus g(p) =F g(q) since glUp, is 1: 1. Now suppose that there does not exist an open nbhd offconsisting entirely of 1:1 immersions. Since the set of immersions is open, there does not exist a nbhd off consisting of 1: 1 mappings. Since X is compact and C"'(X, Y) satisfies the first axiom of countability, there is a sequence of functions fn converging to f each of which is not 1: 1. We may assume that each fn is in W'. Let (Pn, qn) be in X x X - t::..X such that fn(Pn) = fnCqn). The pair (Pn, qn) exists since fn is not 1: 1. By the choice of W', (Pm qn) is not in Z. X x X - Z is compact so we may assume that Limn _", (Pm qn) exists and is (p,q). Since (p,q) rtz,P =F q. Also Limn_",fn(Pn) =f(p) and Limn_",fn(qn) = f(q). Since with an appropriate choice of metric on Yfn converges uniformly to J, f(p) = f(q) contradicting the fact thatfis 1: 1. 0 Proposition 509. (Whitney Embedding Theorem). Let X be a smooth manifold ofdimension n. Then there exists an embedding f of X into R2n+l. §6. Morse Theory 63 Proof By Theorem 5.7 the set of 1: 1 immersions of X -+ R2n + 1 is dense. We showed in the first paragraph of Lemma 3.7 that the image under a continuous proper mapping is closed. Thus if there is a 1:1 proper immersion of X -+ R2n+1, the result is proyed. The following lemma is thus sufficient. Lemma 5.10. Let X be a smooth manifold. Then the space of smooth proper mappings of X -+ Rm is nonempty and open in COO(X, Rm) (in the Co topology). Proof By 1,5.11 there exists a proper mapping of X -+ R. Compose this mapping with any linear injection of R -+ Rm to obtain a proper mapping of X -+ Rm. To show that the set of proper mappings is open, let f: X -+ Rm be proper and let Vx = {y E Rm Id(y,f(x» < I}. Let V = UXEX Vx in JO(X, Rm) = X x Rm. The continuity of f guarantees that V is open and clearly fis in M(V). Now let g be in M(V), then g is proper for g-l(BrCy» c f-1(Br +1(y». Sincefis proper g-l(Br(Y» is a closed subset of a compact set and thus compact. 0 §6. Morse Theory In the last section we used transyersality to analyze what" most" mappings of a manifold X into some high dimensional manifold Y look like. We now use the same technique to analyze the other extreme: that is, to determine the structure of most mappings of X into R. In particular, the Whitney Theorem shows that a generic mapping is as nice as possible in the differentiable sense, namely, the Jacobian always has the maximum rank possible. With Morse Theory we will show that, in general, the Jacobian is not of maximal rank; in other words, the mapping has singularities, but these singularities must be of a particular type. Since the dimension of R is 1, the only non-empty submanifolds of ]leX, R) of the type Sr (see §5) are So and Sl. Thus p in X is a singularity (or critical point) for f: X -+ RiffPf(p) is in Sl. Definition 6.1. (a) Let p be a Singularity off: X -+ R. pis nondegenerate ifPfffi Sl at p. (b) f is a Morse function if all of the singularities off are nondegenerate. Theorem 6.2. Let X be a smooth manifold. Then the set of Morse functions is an open dense subset of C 00 (X, R). Proof Apply the Thorn Transyersality Theorem. 0 Proposition 6.3. Letf: X -+ R be a smooth function with a nondegenerate critical point p. Then there exists a nbhd V ofp in X such that no other critical points off are in V, i.e., nondegenerate critical points are isolated. Proof Note that codim Sl = dim X (see 5.4) and apply Theorem 4.4 (or, more precisely, Exercise (I)(a) of S4t n 64 Transversality Now that we know that nondegenerate critical points are isolated and that they represent a generic situation for functions, we wish to analyze their character locally. Proposition 6.4. Let U be an open subset ofRn andf: U -+ R be smooth. Then f has a nondegenerate critical point at p iff the Hessian of f at p = «fPf/oxi OXj)(p)) is nonsingular. Proof P(U, R) ~ U x R x Hom (Rn, R). Note that the projection 7T: P(U, R) -+ Hom (Rn, R) is a submersion and 7T- 1(O) = S1' Now apply Lemma 4.3; that is, jIfffi S1 at p iff 7T·jIfis a submersion at p. But 7T·jlJat x is ( Of of) (df)x = OX1 (x), ... , OXn (x) in the standard coordinates on Hom (Rn, R). Thus 7T·jIfis a submersion at p iff the mapping of Rn -+ Rn given by X-+ (:~ (x), ... , o~ (X)) is a submersion at p iff «o2J/oxl OXj)(p)) is nonsingular. 0 We shall now give an invariant definition for the Hessian of a function at a critical point. To do so we need the concept of intrinsic derivative. Let E be a vector bundle over X. E always has a distinguished global section called the zero section io : X -+ E which is defined by io(p) = 0 in Ep. Let 7T : E -+ X be the projection mapping. Then it is clear that Ker (d7T)io(P) n (dioMTpX) = {O} and that Tio(p)Ep = Ker (d7T)io(P) so that Tio(p)E ~ Tio(p)Ep EB TpX. Now T.o(p)Ep ~ Ep in a canonical way since Ep is a vector space. Finally let a : Tio(p)E -+ Ep be the obvious projection. Definition 6.5. (a) Let.p: X -+ E be a section such that .p(p) = O. Then define (D.p)p: TpX -+ Ep by (D.p)iv) = a·(d.p)iv). (D.pp is called the intrinsic derivative of.p at the zero p. (b) Let f: X -+ R be smooth with a critical point at p. The I-form df: X -+ T* X has a zero at p. Define (d2fMv, w) = «D(df))pv, w) where v, WE TpX. Then (d2J)p is a bilinear form on TpX called the Hessian off at the critical point p. We leave the following two Lemmas as exercises. (A) Let U be an open subset of Rn and f: U -+ R be smooth with a critical point at p. Then with the standard identifications 2 (0 I 0 I) 02J(d 'f)p OXi p' OXj p = OXi OXj (p). §6. Morse Theory 65 (B) Let 4> : X --+ Y and f: Y --+ R be smooth. Suppose that q is a critical point forfandp E 4>-I(q). Thenp is a critical point for 4> *fand (d24> *f)p = (rp * (d2f))p where rp * (d2fMv, w) = (d2fM(drp)pv, (d4»pw) Vv, WE TpX. (Hint: Compute both lemmas by using local coordinates.) Proposition 6.6. Let f: X --+ R be smooth with a critical point at p. Then (d2f)p is a symmetric bilinear form on TpX and is nondegenerate iffp is a nondegenerate critical point off. Proof. Let U be a nbhd of p in X and rp: U --+ Rn a chart centered at p. Let U' = rp(U) and g = f·rp-I. By Lemma (B) above 0 is a critical point of. g and (d2f)p = (d2rp *g)p = rp *(d2g)o. By Lemma (A), (d2g)o is symmetric. Hence (d2f)p is symmetric. By Proposition 6.4, 0 is a nondegenerate critical point of g iff (d2g)o is a nondegenerate bilinear form iff (d2f)p is a nondegenerate bilinear form since (drp)p: TpX --+ ToRn is an isomorphism. Finally the diagram Jl(U, R) (rp*)-l , ]leU', R) Pfl 4> IjIg U ' U' commutes and rp * (SI n ]leU', R)) = SI n ]leU, R) so that jIg mSl at 0 iff jIfm SI at p. Thus 0 is a nondegenerate critical point of g iff p is a nondegenerate critical point off. 0 We recall the following proposition from Linear Algebra. Proposition 6.7. Let B be a symmetric, nondegenerate, bilinear form on a real vector space V of dimension n. Then there is an integer k :s: n and a basis VI, ... , Vn of V such that B(v;, v)) = SiDt) where k is called the index of B. ifi :s: k ifi > k. Definition 6.S. Let p be a nondegenerate critical point off: X --+ R. Then the index off at p is the index of(d2f)p. Theorem 6.9. (Morse Theorem). Let g: Rn --+ R be given by where c is some constant. Then (a) g has a nondegenerate critical point of index k at the origin and has no other singularities. (b) Let f: X --+ R be smooth with a nondegenerate critical point ofindex k 66 TransversaIity at p. Then there is a nbhd U ofp and a chart a: U -+ Rn centered at p such that commutes where c = f(p). A simple calculation gives the proof of (a). We shall need a sequence of Lemmas to prove (b). Note that in the coordinates al>"" an defined bya,f has the "normal form" f(x) = f(p) - (a12(X) + ... + a/c2(x» + a~+l(x) + ... + an2(x) for all x in a nbhd of p. Thus the behavior of a function in the nbhd of a non-degenerate critical point is determined. Lemma 6.10. Let U be an open convex subset ofRn, a E U, andf: U -+ R smooth. Then there exist gl> ... , gn: U -+ R all smooth so that for every x in U n f(x) = f(a) + Lgt(x)(Xt - at). i=l Moreover gt(a) = (of/oxt)(a). Proof Fix x in U and let cfo(t) = f(a + t(x - a». This is well-defined for t in [0,1 ] by convexity. Then f(x) - f(a) = cfo(l) - cfo(O) = {Ia;dt and dcfo (t) = i of (a + t(x - a»(xt - at) dt t=10Xt by the chain rule. Let gt(x) = II ~f (a + t(x - a» dt. o uXj o Lemma 6.11. Let f be a smooth function on an open convex subset U of Rn. Let a be a critical point for f Then there are smooth functions gtj: U -+ R (1 :::; i,j :::; n) such that (a) gtj = gji; n (b) f(x) = f(a) + L gt;(x)(Xt - at)(xj - aj) "Ix E U; t,j=l and 1 02f (c) gj;{a) = 2- ~ (a). uXi uXj Proof From Lemma 6.10 there are functions gl,"" gn such that f(x) = f(a) + L:f=l gt(x)(xt - ai) and gt(a) = (of/8xj)(a) = 0 since a is a critical point off Now apply Lemma 6.10 to each gt(x) to insure the existence §6. Morse Theory 67 of smooth functions hlj: U -+ R so that g;(x) = 2.7=1 hilx)(Xj - aj). This is possible since gi(a) = O. Let gij = !(hij + hii). Then we have written n f(x) = f(a) + 2: glj(x)(Xj - ai)(xj - aj) "Ix E U ;,i=1 and gij = gj;' To see that g!j(a) = !(o2flox; oxj)(a) differentiate both sides of the above equation with fJ2loXi OXj and observe that the only terms on the right-hand side which do not disappear when evaluated at a are gila) and gi!(a). 0 Lemma 6.12. Let U be an open convex nbhd of0 in Rn. Letf: U -+ R be a smooth function with a non-degenerate critical point at O. Assume that (o2flox; OXj)(O) = ±Sij' Then there exists a nbhd V of 0 contained in U and smooth functions hi: V -+ R (1 ::;; i ::;; n) satisfying oh· (a) h;(O) = 0 and ~ (0) = ± aii ; uXj (b) f(x) = f(a) + (±h12(X) ± ... ± hn2(x)) "Ix E V. Proof. The Lemma will be proved by induction on r. The induction hypothesis for each r with 0 ::;; r ::;; n is that there exist smooth functions hi: Vr-+R (1 ::;; i::;; n) and gij: Vr-+R (r + 1 ::;; i,j::;; n) where Vr is a nbhd of 0 contained in V, satisfying (a') g;j = gjl; , 1 o2J (b) g!j(O) = 2ax; OXj (0) oh· (c') h;(0) = 0 and ~ (0) = ±a;i; uXj and (d')f(x) = f(a) + (±h12(X) ± ... ± hr2(x)) ± i gilx)h;(x)h;(x). ;,i=r+1 The Lemma is proved by taking r = n. For r = 0, let h;(x) = x; and use Lemma 6.11 to obtain the gi/S which satisfy (a') through (d') on Vo = U. Assume that the induction hypothesis is true for r - 1, giving the existence of smooth functions Ui (1 ::;; i ::;; n) and Vii (r ::;; i, j ::;; n) of Vr -1 into R. By (b'), vr,(O) = t(02Jloxr2)(0) =f O. Thus there is a nbhd Vr of 0 contained in Vr- 1 on which Vrr is nowhere zero. Let hi = ud Vr for i =f r and define hr is well-defined and smooth on Vr. A straight-forward calculation will show that (c') holds. Now let a= sign (vr,(O)). From the induction hypothesis (d') 68 Transversality we know that f(x) = f(O) + (±h12(X) ± ... ± hr _ 12(x» + SeX) wheres(x) = 2:f.j=r Vij(x)ui(x)ulx). Letgij = Vij - VrjVri/vrrfor r + 1 ~ i,j ~ n. Then (a') and (b') hold. Finally to see that (d') is true, compute and n (ii) L gijhjhj = Mr2 + S. l,j=r+ I Then f(x) =f(a) + (±hI2(X) ± ... ± h~_l(X» - ohr2(x) + i gijhi(x)hlx). 0 ;,j=r+ 1 ProofofTheorem6.9 (b). Let Ube a coordinate nbhd ofp and",: U -+ Rn a chart. Then (d2(f."'-I» is a symmetric, nondegenerate, bilinear form on ToRn. Choose a matrix A which diagonalizes this form, i.e. d2(f.",-I.A) = (-Ik 0)o I n - k where Is is the s x s identity matrix. Let TJ = A-I.",. TJ is also a chart on U. Now g(x) = f(TJ-I(X» - f(p) satisfies the conditions of Lemma 6.12 and g(O) = O. Thus there exist smooth functions hl' ... , hn defined on a nbhd V of 0 in Rn so that g(x) = ±hl2(X) ± ... ± hn2(x) where h;(O) = 0 and (8h;/8xj)(0) = ±oij' Now define H: V -+ Rn by H(x) = (hl(x), ... , hn(x». H is a diffeomorphism on a nbhd of O. Let a = H· TJ be a chart defined on a nbhd U' ofp in X. Then define g(x) = ± X l 2 ± ... ± Xn2 where the signs in the definition of g are the same as those in the representation of g above. Hence g·H = f·TJ-I - f(p) and g.a = f - f(p) or f(q) = f(p) + g(a(q». Let al,"" an be the coordinates with the chart a, then f(q) = f(p) + (± aI2(q) ± ... ± an2(q». Finally Since (d2f)p has index k, so does (d2g)o. Thus by a simple reordering of the a;'s we may assume thatf(q) = g(a(q». 0 Having analyzed the structure of a function in the nbhd of a nondegenerate critical point, we can now make a statement about the critical values. Proposition 6.13. Let X be a smooth manifold. The set ofMorsefunctions all of whose critical values are distinct form a residual set in C "'(X, R). Proof Let S = (Sl x Sl) ('\ Jl(X, R) ('\ (B2)-I(LlR). We claim S is a submanifold of the multijet bundle J2 l(X, R). Let U be an open coordinate §7. The Tubular Neighborhood Theorem 69 nbhd in X diffeomorphic to Rn. In these local coordinates J12(V, R) ~ (Rn x Rn - ~Rn) x (R x R) x Hom (Rn, R)2 and S ~ (Rn x Rn - ~Rn) x (~R) x (0,0) which is clearly a submanifold. Moreover codim S = 2n + 1 where n = dim X. Apply the Multijet Transversality Theorem (4.13) to conclude that the set of mappings f: X ~ R for which j21frF1 S is residual. Transversality in these relative dimensions means that j21f(X x X - ~X) n S = 0. Thus if p and q are critical points ofj(Pf(p),Pf(q» E Sl x Sl n J21(X, R). The fact that Nf(p, q) ¢= S means that f(p) =f. f(q), i.e., f has distinct critical values. Thus the proposition is proved. 0 §7. The Tubular Neighborhood Theorem Definition 7.1. Let X be a submanifold of the smooth manifold Y. A tubular neighborhood of X in Y is an open subset Z of Y together with a submersion 7T : Z ~ X such that (a) Z ~ X is a vector bundle, and (b) Xc Z is the zero section of this vector bundle. Theorem 7.2. Let X be a submanifold of Y, then there exists a tubular nbhd of X in Y. We prove some preliminary results first. Proposition 7.3. Let Y and Y' be smooth manifolds with Xc Y and X' c Y' submanifolds. Let f: Y ~ Y' be smooth and satisfy: (a) fl X: X ~ X' is a diffeomorphism. (b) (df)x: Tx Y ~ T,(x) Y' is an isomorphism for every x in X. Then there is an open nbhd V of X in Y such that f( V) is open in Y' and fl V is a diffeomorphism. Proof If X is a point, then Proposition 7.3 is just the Inverse Function Theorem. In any case, since X is second countable, there is a countable covering of X by open sets Vb V2, . .. in V so that}; = fl Vi is a diffeomorphism; i.e., f is a diffeomorphism on a nbhd of Vi' Moreover, since X' is paracompact, there is a locally finite covering of X' by open sets W1, W2, ••• in Y which is a refinement of the open covering f1(V1)'/2(V2), ... By replacing Vi with h-1(Wi), where Wi c Vj, we may assume that Wi =};(Vj). Let W = Ut~l Wi and F = {y E W Iif y E Wt n Wj, then };-l(y) = h-l(y)} clearly contains X. Moreover, we claim that Fcontains an open nbhd G of Xin Y. For each x in X, there is an open nbhd Gx which intersects onlyfinitely many W;'s, say Wi1' ... , Wts' since the covering {Wi} is locally finite. By making Gx even smaller we may assume that x is in Wi1 n· .. n Wis' Now};t -1 is a local inverse for f near x. Thus there is an open nbhd of x, H, such that 70 TransversaIity A-11H = ... = ];.-11H using the uniqueness of inverse functions and the fact that];r -l(X) = ... =];.-l(X). Then (ix = H n Gx is an open nbhd of x in Yand (ix c F. Let G = UXEX (ix' Now define g: G --+ Y by g(y) =]; -l(y) if y is in Wi. g is well-defined since G c F and smooth since locally g =];-1 for some i. Also f· g = idG so g is a diffeomorphism. Let V = g(G). 0 Lemma 7.4. Let E ~ X be a vector bundle. Let V be an open nbhd of the zero section in E. Then there exists a diffeomorphism h : E --+ V (into) such that 1T·h = 11'. Proof. Choose a metric t on E. The sets BxCa) = {v E Ex It(x)(v, v) < a} form a nbhd basis of 0 in Ex where a is in R+. Since V n Ex is an open nbhd of 0 in Ex, there is an a > 0 for which BxCa) c V n Ex. Since t is smooth we may choose, by a partition of unity argument, a smooth function 8: X --+ (0, I) such that BxC8(x» c V for all x in X. The mapping h(v) = 8(x)vj(1 + t(x)(v, V»1/2 where x = 1T(V) is a diffeomorphism of E into V whose inverse is given by w 1-+ 8(x)wj(82(x) - t(x)(w, W»1/2 where x = 1T(W). 0 Combining the last two results we have: Proposition 7.5. Let E ~ X be a vector bundle and let U be an open nbhd of the zero section Xo in E. Suppose that X is a submanifold ofa smooth manifold Yand thatf: U --+ Y is smooth and satisfies: (a) fl Xo : Xo --+ X is a diffeomorphism. (b) (df)x: TxE --+ T,(x)Y is an isomorphismfor every x in Xo. Then there is a tubular nbhd of X in Y. Proof. By Proposition 7.3, there is an open nbhd V of Xo such that fl V: V --+ Y is a diffeomorphism into. Let h : E --+ V be a diffeomorphism guaranteed by Lemma 7.4. Then Z = f·h(E) is a tubular nbhd of X in Y with projection mapping 11" = f·h.1T·(j·h)-1. 0 Lemma 7.6. Let X be a n-submanifold ofRP, then there is a tubular nbhd of Xin RP. Proof. Equip RP with the standard inner product. Let Ex be the (p - n)plane of normal vectors to X at x. The plane through the origin Ex - x has a vector space structure which we give to Ex. Let E = UXEX Ex. Then E is a vector bundle over X, since, in effect, E is just the complementary subbundle to TX in TxRP = X x RP. (See 1,5.12). The explicit construction of E gives a smooth mapping f: E --+ RP such that flXo = fiX = idx. To show the existence of a tubular nbhd we will apply Proposition 7.5. So we must show that (df)x: Tx E --+ T,(x)RP is an isomorphism for all x in E. Since dim TxE = dim TxRP we need only show that Ker (df)x = O. Let v E Ker (df)x. Since TxE ~ TxX E8 TxEx, v = V1 + V2 where V1 E TxX and V2 E TxEx. (df)xv = 0 implies that V1 = 0 sinceflX = idx. Represent V2 by a curve c in Ex.f·c is then just the curve c in Ex when Ex is thought of as in RP. Thus (df)xCV2) = 0 implies that V2 = O. 0 §7. The Tubular Neighborhood Theorem 71 ProofofTheorem 7.2. Using the Whitney Embedding Theorem (Proposition 5.9), we may assume that Y c RP for p = 2 dim Y + 1. Thus X is a submanifold of Rk and by Lemma 7.6 there is a tubular nbhd of X in RP which we call Z'. Let Ex be the set of vectors normal to X and tangent to Y at x for x in X and let E = Uxex Ex. E is a vector bundle over X since it is just the complementary subbundle to TX in Tx Y. As in Lemma 7.6, this explicit construction of E gives a smooth mapping f: E --+ RP such that flXo = fiX = idx. Let Z" be a tubular nbhd of Y in RP with projection map 'IT". Then U = f-1(Z' n Z") is an open nbhd of Xo in E. Consider 'IT" ·f: U --+ Y which is a smooth mapping. 'IT" ·flX = idx since Xc Yand 'IT"I Y = idy • We wish to apply Proposition 7.5 to obtain the desired result. In order to do so we need to know that (d'IT"'f)x : TxE --+T[(x) Y is an isomorphism for all x in X. Since dim E = dim Y we need only show that Ker (d'IT"'f)x = {O}. Now TxE = TxX EB TxEx. Since 'IT" ·fl X = idx we have to show that if v is in TxEx and (d'IT" ·f)xCv) = 0, then v = O. Let e be a curve in Ex representing v, then f·e is a curve in RP whose tangent at 0 is tangent to Y at x. Since 'IT"I Y = idy, 'IT" ·f·e is a curve in Y whose tangent at 0 is the same as the tangent to f·e at O. Thus (d'IT" ·f)xCv) = 0 implies that v = O. 0 Remark. If one traces through the proof of the Tubular Nbhd Theorem, one sees that the tubular nbhd Z is always isomorphic (as vector bundles) to a complementary subbundle of TX in TYI X. Such a complementary subbundle is always isomorphic to the quotient bundle N where Nx = Tx Y /TxX for each x in X. N is called the normal bundle to X in Yand Nx is called the normal spaee to X in Y at x. Thus a tubular nbhd is a realization of the normal bundle as an open nbhd of X in Y. Chapter III Stable Mappings §1. Stable and Infinitesimally Stable Mappings Definition 1.1. (a) Letfand/, be elements of COO(X, Y). Thenfis equivalent tof' if there exist diffeomorphisms g : X ---J>- X and h: Y ---J>- Y such that the diagram /'X~Y commutes. (b) Let f be in COO(X, Y). Then f is stable if there is a nbhd Wf off in COO(X, Y) such that each/, in Wf is equivalent to f In other words f is stable if every nearby mapping /' is identical to f, after suitable changes of coordinates, both in the domain and the range of/,. We now describe an alternative formulation of stability of mappings. Recall that a group G acts on a set A if there is a function G x A ---J>- A, written as (g, a) 1-+ g·a, with the properties that (gg').a = g.(g' ·a) and e·a = a for every g, g' E G and a E A where e is the identity in G. The orbit of a in A is the set G·a == {b E A Ib = g·a for some gin G}. In the case at hand, let G = Diff(X) x Diff( Y) where Diff(X) (resp. Diff( Y)) is the group of all diffeomorphisms on the manifold X (resp. Y) and let A = C"'(X, Y). Then there is a natural action of G on A defined by (g, h)·f = h·f·g-l (i.e., change of coordinates) where g E Diff(X), hE Diff(Y), andfE COO(X, Y). Lemma 1.2. Let f be in COO(X, Y). Then f is stable iff the orbit off in C 00 (X, Y) under the action of Diff(X) x Diff( Y) is an open subset. We recall the following: Lemma 1.3. Let X and Y be smooth manifolds with g : X ---J>- X a diffeomorphism. Then g* : COO(X, Y) ---J>- COO(X, Y) given by fl-+ f·g is continuous. Proof Since g is proper, Note (2) after II,3.9 applies. 0 72 §1. Stable and Infinitesimally Stable Mappings 73 Proof of Lemma 1.2. Let g be in Diff(X) and h be in Diff(Y). Let Y(h,g): C"'(X, Y) -+ C"'(X, Y) be induced by the action of Diff(X) x Diff( Y) on C"'(X, Y). Thus Y(g,h) = (h)*(g-l)* and is continuous by the above Proposition and II,3.5. Moreover Y(g,h) is a homeomorphism since Y(g-l,h- 1 )·Y(g,h) = ide"'(x,y). Now observe that f' is in the orbit off ifff' is equivalent to f Also the orbit off is open iff there is an open nbhd of C"'(X, Y) contained in the orbit off(since any such nbhd can be translated by an element of Diff(X) x Diff( Y); i.e., Y(g,h), to an open nbhd around any other point in the orbit). These two facts taken together immediately yield the proof. 0 This definition of stability proves difficult to apply in practice. However, using a criterion suggested by Rene Thorn, John Mather has produced a theorem ~hich provides a truly computable method for determining whether or not a mapping is stable. We now present that criterion. Definition 1.4. Let f: X -+ Y be smooth. (a) Let 7Ty: TY -+ Y be the canonical projection, and let w: X -+ TY be smooth. Then w is a vector field along f if the diagram TY commutes. Let Cr(X, TY) denote the set of vector fields along f (b) f is infinitesimally stable iffor every w, a vector field alongf, there is a vector field s on X and a vector field t on Y such that (*) w = (df)·s + t·f where (df) : TX -+ TY is the Jacobian mapping off Remark. Vector fields alongfcan be identified with sections of a certain vector bundle. Let E be a vector bundle over Y. Define f* E = UVEX Ef(p) (disjoint union) and let 7T : f* E -+ X be the obvious projection. We claim that f* E can be made into a vector bundle over X, the pull-back bundle ofE byf, as follows. Let V be an open nbhd of Y such that EI V is trivial, say EI V ~ V X Rk. Then make f* E into a smooth manifold by demanding that f*Elf-l(V) ~ f-l(V) x Rk. That the transition functions are smooth follows from the fact that E is a smooth vector bundle over Y. Now let s be a section off*(TY). Then sex) fJ*(TY)x = Tf(x) Y so s may be thought of as a smooth mapping of X -+ TY such that 7Ty·S = f, (i.e., can be identified with a vector field along f). The converse is also clear. Thus we can identify Cr(X, TY) with C "'(f*(TY)). Theorem 1.5. (Mather). Let X be a compact manifold and f: X -+ Y be smooth. Then f is stable ifff is infinitesimally stable. 74 Stable Mappings Note. It is sufficient to assume that f is a proper mapping and to drop the assumption on X. The proof of this theorem will be given in Chapter V. For the moment we will content ourselves with explaining what originally motivated the introduction of the concept of infinitesimal stability. This will require a sketchy development of the theory of Frechet manifolds. Definition 1.6. Let V be a topological vector space, i.e., a vector space with a topology in which addition and scalar multiplication are continuous. Let 1·1 : V -+ R be continuous and satisfy for all x and y in V: (a) Ixl ~ 0 (b) Ixl = oiffx = 0 (c) Ix + yl :::; Ixl + Iyl (d) Ixl = I-xl (not IAxI = l"llxl for arbitrary" E R). We can now define a metric d on V by setting d(x, y) = Ix - y I. If V is complete with respect to this metric, then it is a Frechet space. Notes. (1) The norm on a Banach space satisfies axioms (a) through (d), so every Banach space is a Frechet space. . (2) Let VI and V2 be Frechet spaces. Then L(Vl' V2), the set of all continuous linear mappings of VI into V2, is a Frechet space if we define IfI = SUPlxl = 1 If(x) I. (This is well-defined by the linearity and the continuity of 1·1 and f.) Definition 1.7. Let VI and V2 be Frechet spaces and U an open subset of VI' Let f: U -+ V2 be continuous and p be a point in U. Then f is differentiable at p iff there is a linear mapping Ap: VI -+ V2 such that Lim If(p + tv) - f(p) - tAp(v)I = 0 Itl->O It I for every v in VI' Note that the linear mapping Ap is unique ifit exists. Thus we may define (df)p = Ap when f is differentiable at p. f is differentiable on U iff is differentiable at p for every p in U. f is ktimes differentiable if (df): U -+ L(VI> V2) defined by (df)(p) = (df)p is (k - I)-times differentiable.Jis smooth iffis k-times differentiable for every k. We note that the chain rule holds for differentiation on Frechet spaces. Definition 1.8. Let X be a Hausdorff topological space. Then X is a Frechet manifold if (a) there is a covering {U,,}aeI of X by open sets. (b) for each 0: in I there exists a homeomorphism ha: Ua-+ Va with Va an open subset ofsome Frechet space. (The ha's are called charts.) (c) for every 0:, f3 E I, ha·h/J -1 is smooth where defined. If Va is contained in a Banach space, then X is called a Banach manifold. §1. Stable and Infinitesimally Stable Mappings 75 As in the ordinary manifold case, it is possible to define the tangent space to a point p on a Frechet manifold X. Let Sp(X) be the set of smooth curves c: R -+ X such that c(O) = p. (Note that smooth mappings between Frechet manifolds are defined exactly as in the finite dimensional case.) Define C1 is tangent to C2 at p iffor every chart ha, (dh".c1)o = (dha·c2)o. Since the chain rule is valid, "is tangent to" is a well-defined concept. Finally, let TpX = set of equivalence classes of Sp(X) under the equivalence relation" is tangent to." As in the finite dimensional case TpX is a vector space. Note that a smooth mapf: X -+ Ybetween Frechet manifolds induces a well-defined linear mapping (df)p : TpX -+ Tf(p) Y just as in the familiar case of Chapter I. The following proposition will present our basic examples of Frechet spaces. Proposition 1.9. Let X be a compact finite dimensional manifold. Then (a) C""(X, R) is a Frechet space, (b) if E is a vector bundle over X, then C""(E) is a Frechet space, and (c) let f: X -+ Y be smooth, then CT(X, TY)-the set of vector fields alongf-is a Frechet space. It is understood that in each case the topology induced by the Frechet "norm" is the Whitney C "" topology. Proof (a) Cover X be a finite number of open sets V" where Va is contained in a coordinate patch. This is possible since X is compact. Let g: X -+ R be smooth. Define Then define Iglk = L" IglkUa and, finally, define ~ 1 Igllc Igl = Ic~ 2k 1+Iglk· It is now an easy exercise to check that C""(X, R) is a Frechet space with 1·1 defined in this way. (b) Choose the Va's so that EI Va is trivial. Then let the Frechet norm 1·lkU. be the supremum over the Frechet norm of the coordinate functions. Then continue as in (a). (c) Since C}"(X, TY) can be identified with C""(f*TY), the result follows from (b). (See the Remark after 1.4.) 0 We note that one gets different metrics on Cro(E) for different choices of Va but the underlying topology is the same, and that this underlying topology isjust the one induced on C""(E) from the Whitney C'" topology on C"'(X, E). Proposition 1.10. Let X be a smooth manifold. Then Diff(X) is an open subset of C "'(X, X) and hence a Frechet manifold. (For simplicity in the proof, we assume that X is compact.) 76 Stable Mappings Proof By II, Proposition 5.8 there is a nbhd of f which consists of 1:1 immersions. Let g: X ~ X be a 1:1 immersion and Xo a connected component of X. g(XO) is closed since XO is compact, while it is open since g is a submersion. So g(XO) is a connected component of X. We claim that g: XO ~ g(XO) is a diffeomorphism; this is enough to prove the proposition. Now (gl XO) -1 exists since g is 1:1. Also, at each point of g(XO), (gl XO) -1 is smooth by the Inverse Function Theorem. So g is a diffeomorphism. D The following theorem shows our basic reason for studying Frechet manifolds. Theorem 1.11. Let X and Y be smooth manifolds with X compact. Then COO(X, Y) is a Frechet manifold. Proof It is easy to check that COO(X, Y) is a Hausdorff space. Letfbe in COO(X, Y). We wish to produce an open nbhd off, Uf, which is homeomorphic to an open subset Vf' of it Frechet space. We do this via tubular nbhds. Let Xf = graph (f) c X x Y. Xf is a submanifold of X x Y so we may apply the tubular nbhd theorem (II, 7.2) to find a tubular nbhd Z of Xf with projection 7T. Vf will be an open nbhd of the zero section in COO(Z): Since 7T: Z ~ Xf is smooth 7T*: COO(X, Z) ~ COO(X, Xf) is continuous (II, Proposition 3.5), so that (7T*)-l(Diff(X, Xf)) is an open subset ofCOO(X, Z). (Apply Proposition 1.10.) Since COO(X, Z) is an open subset of COO(X, X x Y) = COO(X, jO(X, Y)) we see that Uf = (j0)-l(7T*)-l(Diff(X, Xf)) is an open nbhd offin COO(X, Y). To define Vf we let 7TX: X x Y ~ X be projection. The mapping (7Txh: COO(Z) ~ COO(X" X) given by s f-+ 7Tx'S is continuous since the topology on COO(Z) is the topology induced from COO(Xf' Z). (Again apply II, Proposition 3.5.) Thus Vf = (7Tx)* -l(Diff(Xf' X)) is an open nbhd of the O-section in COO(Z). Next we define hf : Uf ~ Vf by hlg)(x,J(x)) = jOg'(7T.jDg)-l(X,J(X)). Note that 7T.jDg: X ~ Xf is a diffeomorphism since g is in Uf so that hlg) : Xf ~ Z is well-defined. Since 7T·hlg) = idx" hf(g) is actually a section of the vector bundle Z. Finally 7TX·hf(g) = (7T.jOg)-l which is in Diff(Xf' X) so that hlg) is in Vf' To see that hf is a bijection, define kf : Vf ~ Uf by kls)(x) = 7Ty,s'(7TX'S)-l(X) where 7Ty: X x Y ~ Y is projection. Since 7T.jDkf(s) = (7TX'S)-l we see that kls) is in Ufo It is an easy exercise to show that kf = hf -1. Since our reason for introducing Frechet manifolds was just to motivate the criterion ofinfinitesimal stability, we will leave the details of showing that hf is a homeomorphism and that hf·hg -1 is smooth (where defined) to the " interested" reader. D Now suppose that the Implicit Function Theorem were true for smooth functions between Frechet manifolds. Then consider the mapping Yf: Diff(X) x Diff(Y) ~ COO(X, Y) given by (h, g) f-+ g·f·h- 1• 1m Yf is the orbit of fin COO(X, Y) under the action of Diff(X) x Diff(Y), so a reasonable criterion for the stability off §1. Stable and Infinitesimally Stable Mappings 77 would be that Yf be a submersion. (Using Lemma 1.2.) In fact we would need to know only that Yf is a submersion at the identity e = (idx , idy), since then 1m Yf would contain an open set and, as we saw in the proof of Lemma 1.2, this would imply that 1m Yf is itself open. We claim that f is infinitesimally stable iff (dYf). is onto. To see this we need to identify the spaces Te(Diff(X) x Diff(Y)) and TfC"'(X, Y). Lemma 1.12. Let t f-+ ft and t f-+ gt be smooth curves in C"'(X, Y) with fo = go· Then ft and gt are tangent at t = 0 ifffor each p in X, the curves in Y t f-+ ft(p) and t f-+ glp) are tangent at t = o. Proof Let h = fo = go and Xh = graph (h) c X x Y. Let Z be a tubular nbhd of Xh in X x Y. As we saw in the Proof of Theorem 1.11, nearby functions tofcan be identified with sections in C"'(Z). So for t small we can think of t f-+ ft as a smooth curve of sections in C"'(Z). The Frechet derivative of t f-+ ft at t = 0 is given by (dfMl) = (Lim 1ft - fol)(p) = Lim Ift(p) - fo(p)l. t-+o ItI t-+o ItI So (dfMl) = (dft/dt(P))lt=o. Writing the same for g concludes the proof of the Lemma. 0 Proposition 1.13. (a) TfC"'(X, Y) ~ Cr(X, TY) (b) T!dx Diff(X) ~ C"'(TX). Proof (b) follows immediately from (a) since ~dx Diff(X) = ~dxC"'(X, X) ~ C;"JiX, TY) = C"'(TX). To prove (a) let w be in TfC '"(X, Y) and let t --+ft be a curve representing w. Define w': X --+ TY by w'(p) = (dft/dt(p))lt=o. By the last Lemma, this definition is independent of the choice of curves ft. Using the identification of functions ft with sections of C"'(Z) we see that 1Ty'W' = f, so w' is in Cr(X, TY).The smoothness of w' is left as an exercise. Lemma 1.12 also shows that w f-+ w' is l: 1. To show that this map is onto, it is sufficient to work locally. Let w' be in Cr(X, TY). For each p in X choose a curve t --+ft(p) which represents w'(p) and do this so thatft(p) is jointly smooth in p and t. (This can be done by integrating the vector field. See I, §6.) The curve t f-+ ft generates a section win TfC"'(X, Y). Linearity of this mapping of Cr(X, TY) --+ TfC"'(X, Y) follows from Lemma 1.12 and the pointwise linearity of vectors in TY. 0 Thus (dYf)e: Te(Oiff(X) x Diff(Y)) --+ TfC"'(X, Y) can be identified with a mapping af EB f3f: C"'(TX) EB COO(TY) --+ C,(X, TY). Theorem 1.14. In the notation above, af: COO(TX) --+ Cr(X, TY) is given by s f-+ -(df)·s and f3f: C"'(TY)--+ C,(X, TY) is given by t f-+ t·f Thus infinitesimal stability reduces to the following criterion: for every vectorfield w along g, there exist vector fields s on X and ton Ysuch that w = (df)·s + t·f 78 Stable Mappings Proof Let s be in COO(TX). Let ii: Diff(X) -»- COO(X, Y) be given by h f-+ f·h- 1; so (dii)idx = af. Let t f--7 ht be a curve in Diff(X) based at idx representing s. Then af(s) = (dii)idx(S) is represented by the curve t f--7 f·ht- 1. Thus d dh -11 dt (f.ht-1)!t=o = (df)·+ t=o by the chain rule and the pointwise definition of dht -ljdt given by Lemma 1.12. To compute (dht-ljdt)!t=o, we consider the mappings tl: R -»- R2 given by t f--7 (t, t) and r: R2 -»- X given by r(u, v) = hJlv -1(p) where p is some fixed point in X. Clearly r.tl(t) = p for all t. Using the chain rule we see that °= or (0, 0) + or (0, 0) = dht (p) I + dht-1 (p) Iou ov dt t=o dt t=o since ho = ho -1 = idx. Thus (dht-ljdt)!t=o = -s and a/s) = -(df)·s. The computation of f3f is similar and is left as an exercise. 0 To summarize, Mather's criterion for stability of a mapping, namely, infinitesimal stability, arises naturally when C"" Xn so that fl Up is given by c + E1X 12 + ... + EnXn2 where El = ... = En = ± 1. (See II, Theorem 6.9.) The collection {Up}PEX forms an open covering of X. Since X is compact, there exists a finite subcovering Ul>"" Um corresponding to Pl, .•. , Pm' Let Pl, .•. , Pm be a partition of unity subordinate to this covering. Choose vector fields Si on X (1 :::; i :::; m) as follows: (a) ifPi is a regular point, then let (b) ifPi is a critical point, then W(Pi) = 0 and pjW = 21=1 hjXj for selected smooth functions hj in the coordinates on Ui given above. (See II, Lemma 6.10.) Moreover, the h/s are compactly supported functions in Ui since PiW is. Let ~ Ejhj 0 Si = L. - - on Uj i=l 2 OXj and extend it to be zero off Ui• Finally, define s = Sl + ... + Sm' Then s[f] = sl[f] + ... + sn[fl· In case Pi is a regular point, since Pi is zero off Ui• In case Pi is a critical point, = PiW, on Ui = PiW off Ui on Ui off Ui Therefore s[f] = P1W + ... + Pmw = W 2']'=1 Pi = W. 0 Notes. (1) By definition stable mappings in COO(X, Y) always form an open subset; Proposition 2.2 tells us that in the case of C I,PS) (h LlYs. It is easy to see that the two examples at the end of the last section are not immersions with normal crossings, while the small perturbations are. Proposition 3.2. The set of mappings of X into Y with normal crossings is dense in C"'(X, Y). Proof Let f: X -+ Y be an immersion. Let (3": JsO(X, Y) -+ ys be the multijet target mapping. Since (3" is a submersion WS = (ftS) -l(Ll P) is a submanifold of JsO(X, Y). It is easy to check thatfis a mapping with normal crossings iffjsof(h Ws. Sofis a mapping with normal crossings ifffsatisfies a countable number of multijet transversality conditions. Applying the multijet transversality theorem and the fact that C"'(X, Y) is a Baire space, we have the result. 0 §3. Immersions with Normal Crossings 83 Corollary 3.3. Immersions with normal crossings are dense in the set ofall immersions. Proposition 3.4. Iff: X -+ Y is an immersion which is stable, thenfis an immersion with normal crossings. Proof Any mapping equivalent to an immersion with normal crossing is an immersion with normal crossings. D We shall now proceed to prove the converse of this proposition, but first we need some preliminaries. Definition 3.5. Let V be a vector space and let HI, ... , Hr be subspaces of V. Then Hh ... , Hr are said to be in general position iffor every sequence of integers ih •.• , is with 1 ::; il < ... < is ::; r. codim (Hll n· .. n His) = codim (HI) + ... + codim (Hi'). Note. In the case r = 2, then HI and H2 are in general position iff HI + H2 = V. For dim (HI + H2) = dim HI + dim H2 - dim (HI n H2) = n - (codim HI + codim H2 - codim (HI n H2)). So dim (HI + H 2) = n iff codim (HI n H2) = codim (HI) + codim (H2). Lemma 3.6. Letf: X -+ Y be an immersion with normal crossings. Choose q in Y. Let f-l(q) = {PI> ... , Pr} all distinct points. (I-l(q) isjinite sincef is an immersion and X is compact.) Then the spaces (df)Pl(Tp1X), ... , (df)Pr(TPrX) are in general position as subspaces of Tq Y. Proof Choose a sequence of integers iI, ... , is such that 1 ::; il < ... < is ::; r. Let H, = (df)p,;(Tp'l X ) (1 ::; j ::; s). Now ft = (Pll' ... , PI,) is a point in X(s) and f(ft) = if = (q, ... , q) is in D.. ys. By the transversality condition of normal crossings off, we have that Tq ys = (dfYs)TpX(s) + TqD.. Y' = HI EB· .. EB Hs + TijD.. P. Therefore s.dim Y = dim TqYs = dim(HlEB···EBHs) + dim Y- dim(HlEB···EBHsnTqD..P) So codim HI + ... + codim Hs = codim (HI EB· .. EB Hs n TqD.. P). But HI EB···EB Hs n TqD..Y· = HI n··· n H.. D Lemma 3.7. Let HI> ... , Hr be subspaces of V in general position. Let D = HI n· .. n Hr. Then there exist subspaces Flo ... , Fr of V satisfying (a) V = DEB Fl EB···EB Fr (b) HI = DEB Lu.I Fj (c) V = FI EB HI. 84 Stable Mappings Proof. Let DI = ni¢1 Hj. Choose a complementary subspace FI to D in Dj (1 :s;. i :s; r). First note that dim FI = codim HI> since dim FI = dim DI - dim D = dim V - codim (n Hj) - dim V + codim (HI n· ..n Hr) 1'#1 = codim(Hj ) since the H/s are in general position. So dim D + dim FI + ... + dim Fr = dim D + codim HI + ... + codim Hr = dim D + codim (HI n· .. n Hr) (by general position) = dim V (since D = HI n· .. n Hr). Thus to show (a) we need only show that the sum D + FI + ... + Fr is direct. Suppose d + f2 + ... +f, is in FI wherefj in Fj for 2 :s; j :s; r. Note that eachfj is in Dj - D; so fj is in HI for i =? j. Hence d + f2 + ... +f, is in HI. Now Fl C Dl - D = H2 n···n Hr - HI n···n Hr. Hence FI n HI = 0 and thus d + f2 + ... +f, = o. Suppose d + fl + ... +f, = 0; then d + f2 + ... +f, E Fl and by the above equals o. By a simple induction argument we know that each fi = 0 (1 :s; i :s; r) and thus that d = O. To show (b) we just note that D cHi and that Fj c HI for i =? j. So Hi :::> DEB Lj>Fi Fj. But codim (D EB Li>Fj Fj) = dim Fj = codim Hj • (c) follows from (a) and (b). 0 We need two more preparatory lemmas, but first some definitions: let Xo be a submanifold of X. (a) An X vector field s on Xo is a section of TX\ Xo. (b) s is tangent to Xo if for every pin Xo s(p) E TpXo c TpX. Lemma 3.B. Let Hlo ... , Hr be subspaces of Rn in general position. Regard each Hi as a submanifold of Rn. Let tj be an Rn vector field on Hi (1 :s; j :s; r). Then there is a vector field t on Rn such that for every j, t - ti (restricted to Hi) is tangent to Hi. Proof. Choose Flo ..., Fr as in Lemma 3.7. Choose an inner product on Rn so that D, Fl, ... , Fr are mutually orthogonal subspaces. Define TTi: Rn ~ Hi to be orthogonal projection. View ti as a map of Hi into Rn. Let ij be the vector field on Rn given by ii = ti·TTj. Let ~i = ii - TTj.ii (1 :s; j :s; r). ~i is the normal component of ii to Hi. Indeed, TTj~j = TTij - TT/ij = o. since TTj2 = TTj. So 1m (~j) is in Fj. Define t = ~l + ... + ~r. We claim that tlHj - tj is tangent to Hi. It is sufficient to show that TTlt - tj) = t - tj on Hj. Note that TT;(~I) = ~I for i =? j since 1m (~I) c F, c Hj and TTi\Hi = idHj" Hence on Hi' TT;(t - tj) = t - ~i - TTjtj = t - ii + TTij - TTiti = t - ti §3. Immersions with Normal Crossings 85 Definition 3.9. Let Y be a manifold with submanifolds Y1,···, YT' Suppose q E Y1 n· .. (1 YT' Then Yh ••• , YT are in general position at q if Tq Y1, ... , Tq YTare in general position in Tq Y. Lemma 3.10. Assume that the submanifolds Yh . .. , YTof Yare in general position at a point q. Then there exists a nbhd W ofq in Y, a chart if>: W ~ Rn and subspaces Hl>' .. , HT such that 1'; (1 W = if>-l(Hj) (1 ~ i ~ r) where n=dimY. Note. This lemma says that in a nbhd of a point of general position, the submanifolds Yh ... , Yr can be simultaneously linearized. Proof. Let mj = codim (Yj) in Y. There is a nbhd WI of q and functions !t..h' ..,!t..m, such that 1'; n WI = {p E WI Ifijp) = ... = !t..m,(P) = O} since Yj is a submanifold of Y. Let W = nr=l WI. By general position, we know that codim (TqY1 (1 ••• (1 Tq Yr) = codim (Tq Y1) + ... + codim (TqYT) = ml + ... + mT ~ n. Let 1= n - ml _ ... - mT• Consider the functions {fi.j}(lSISr)(lS'Sml)' The number of such functions is ml + ... + mr • We claim that (dfij)q form a linearly independent set in T: Y. Clearly the subspace of Tq Y annihilated by all the (dfi.;)q is just Tq Y1 (1 ••. (1 Tq Yr which has dimension ml + ... + mr• Since there are exactly that many (d!t..,)q, they must be linearly independent. Now choose functions gl, ... , gl defined on V so that (dg1)q, ... , (dgl)q, (dfi.;)q form a basis of T: Y. Consider if>: W ~ Rn given by By construction the Jacobian of this map is nonsingular at q. Choose Wa nbhd of q, We Won which if> is a diffeomorphism. In terms of this chart Yj (1 W is given by the linear equations ft., = 0 (1 ~ j ~ mj). 0 Theorem 3.11. Let f: X ~ Y be an immersion. Then f is stable ifff has only normal crossings. Proof. We assume that f has only normal crossings and prove that f is infinitesimally stable. The converse has already been shown in Proposition 3.4. Let q E Y and {Ph' .. , Pr} = f-l(q). We claim that there exists a nbhd Wq of q in Yand nbhds Uj of Pt in X (1 ~ i ~r) satisfying (a) Uj (1 Uj = 0 1 ~ i < j ~ r. (b) fl Uj is a 1: 1 proper immersion. (c) f(Ut) c Wq• (d) f-l(Wq) = Ur=l UI • (e) Wq can be chosen as small as desired. 86 Stable Mappings It is easy to choose Vi satisfying (a) and (b). Also, there is a nbhd Wq of q such that f-1( Wq) c Uf= 1 Vi. Wq can be chosen as small as wished, for if not, there is a sequence of points Xl, X2, ... in X - Uf=l Vi such thatf(xi) converges to q. Since X - Uf= 1 Vi is compact we can assume that {Xj} converges to some point p not in f-1(q). The continuity of f guarantees that f(p) = q, a contradiction. Let Vi = VI nf-1(Wq). Now Yi = f(Vi) (1 :c; i :c; r) are submanifolds of Y sincefl Vi is a 1: 1 proper immersion. By Lemma 3.6 Y1, ••• , Yr are in general position at q sincef is assumed to have only normal crossings. Choose Wq small enough so that it satisfies the conditions of Lemma 3.10, i.e., Y1 , .•. , Yr are simultaneously linearized in Wq • The collection {Wq}qEY form an open covering of Y. Hencef-1(Wq) is an open covering of X. By the compactness of X, there is a finite subcover of X, given by f-1( Wq ), ••• ,f-1(Wqm). Choose a partition of unity Pi (1 :c; i :c; m) subordinate to this covering. Let w be a vector field alongf Then PiW is also in Ct'(X, TY). Since w = 2:l"= 1 Piw, it is sufficient to show that the criterion for infinitesimal stability holds for vector fields along f whose support lies in a givenf-1(Wq). So let wE Ct'(X, TY) with supp we f-1(Wq) for some fixed q. In Wq, we have the submanifolds Yi = f(Vi). Define ti a vector field on Yi by ti = w·(fl V;)-l. This is possible since fl Vi: Vi -7 Yi is a diffeomorphism. Moreover each ti is compactly supported. By the general position of Y1, •.• , Yr at q and Lemmas 3.8 and 3.10, there exists a vector field ton Wq (which is compactly supported since each tj is compactly supported) such that tlf(Vj) - tj is tangent to Yj • Extend t to a vector field on Y by setting t == 0 off Wqo Consider w' = w - t·f w' has the property that for every p in Vb w'(p) = w(p) - t(f(p» is tangent tof(Vj) atf(p). So there exists a unique vector field Sj on Vi such that (df).sj = w' sincefis a 1: 1 immersion from Vi -7 f(V;), Moreover Sj is compactly supported so there exists a vector field S = Si on Vi and S = 0 offf- 1(Wq). By construction w = (df)·s + t·f 0 Proposition 3.12. Let dim Y = 2 dim X, X compact. Then f: X -7 Y is stable ifff is an immersion with normal crossings. Proof By the Whitney Immersion Theorem (Theorem II, 5.6) the immersions of X -7 Yare open and dense, so stable mappings must be immersions. Using Corollary 3.3 any stable map is an immersion with normal crossings. The converse is given by Theorem 3.11. 0 Having settled the question of stability for mappings without singularities (i.e., submersions and immersions) we will now focus our attention on mappings with singularities. As our study of Morse functions suggested, in order to understand singularities it is useful to describe the behavior of a function in a nbhd of a given singularity by a fixed" normal form." In the next section of this chapter we will investigate the stability of a class of mappings similar to Morse functions and in doing so again demonstrate the usefulness of normal forms. §4. Submersions with Folds 87 Exercise. Interpret what it means for an immersion of X -+ Y to have normal crossings when dim Y = 2 dim X in terms of the number of self- crossings. In general show that iff: X -+ Y is a stable immersion then the number of points inf- 1(q)-for any q in Y-is bounded by dim Y/(dim Y - dim X). 4. Submersions with Folds Let X and Y be smooth manifolds with dim X ~ dim Y. Let k = dim X - dim Y. Letf: X -+ Ysatisfy PfrFl SI where SI is the submanifold ofJ1(X, Y) ofjets of corank I as defined in II, §5. Applying II, Theorem 4.4, we see that SI(f) = (Pf)-I(SI) is a submanifold of X with codim SI(!) = codim SI = k + I (II, Theorem 5.4). Note that at a point x in SI(!), dim Ker (df)x = k + I; that is, the tangent space to SI(f) and the kernel of (df)x have complementary dimensions. Definition 4.1. Let f: X -+ Y satisfy PfrFl SI' Then x in SI(f) is a fold point if TxS1(!) + Ker (df)x = TxX. Definition 4.2. (a) A mappingf: X -+ Yis a submersion with folds ifthe only singularities off are fold points. (Note that a submersion with folds, J, satisfiesPfrFl SI') (b) Letf: X -+ Y be a submersion withfolds. Then SI(!) is called the fold locus of! Example. In the case Y = R, the set of submersions with folds is precisely the set of Morse functions on X. The first obvious fact to observe about submersions with folds is: Lemma 4.3. Let f: X -+ Y be a submersion with folds, then f restricted to its fold locus is an immersion. The main theorem for this section describes a simple criterion to determine when a submersion with folds is stable. Theorem 4.4. Let f: X -+ Y be a submersion with folds. Then f is stable ifffISl(f) is an immersion with normal crossings. Notes. (1) The criterion thatfrestricted to its fold locus is an immersion with normal crossings in the case that f is a Morse function is just the criterion thatf has distinct critical values. (2) We shall actually prove Theorem 4.4 with "stable" replaced by "infinitesimally stable" and appeal to the as yet unproved Theorem 1.5 of Mather to obtain the desired result. 88 Stable Mappings Example. The following is an example of the necessity of this criterion S I I I I I I I I I I I I I I I I TC~~ S ""----~ Figure 1 I I I I I I I I I I I I I I I I I I I II III I II I I i II II I I II I[ I i~: il I T~~T' X is a sphere dented at the equator, Y is R2, and f: X ~ Y is the projection ofR3 onto R2, restricted to X. The singular set is a curve offold points consisting of the three circles S, S', and S" while the image forms two circles T and T". f is obviously not stable. For example, we can perturb the picture above so that the image of Sand S' intersect transversely at isolated points. ProofofTheorem 4.4. Necessity. We know thatfISl(!) is an immersion and-with the assumption that f is infinitesimally stable-we show that fISl(f) has normal crossings. Applying Theorem 3.11 it is enough to show thatfISl(f) is infinitesimally stable. Let T be a vector field along g = fISl(f). Extend T to a vector field T along f (Check that this can be done locally and use a partition of unity argument.) Since f is infinitesimally stable there exist vector fields' on X and 7J on Y so that T = (df)(D + 7J • f On Sl(f), we have T = (dg)('ISl(f)) + 7J • g. Now 'ISl(!) = ~ + r where ~ is a vector field on Sl(f) and ~' is in Ker (dg) since by the definition offold points TXISl(!) = T Sl(f) EB Ker (dg). Thus T = (dg)(~) + 7J • g and g = fISl(f) is infinitesimally stable. 0 Before proving the sufficiency part of Theorem 4.4 we shall need a normal form theorem for the local behavior of a submersion with folds near a fold point similar to II, Theorem 6.9 for Morse functions. Theorem 4.5. Let f: X ~ Y be a submersion with folds and let p be in Sl(f). Then there exist coordinates Xl> ... , Xn centered at p and Yl> ... , Ym centered at f(p) so that in these coordinates f is given by (Xl, ... , Xn) 1-+ (Xl' ... ' Xm-l> Xm2 ±... ± Xn2). §4. Submersions with Folds 89 Remark. From this local normal form we see the reason for the nomenclature "fold point." Take the particularly simple example of 2-manifolds. In this case the normal form is given by (Xl> X2) 1--7>- (Xl' X22). This map is depicted in Figure 2. We first map the (Xl' x2)-plane onto the parabolic cylinder X3 = X22 in (Xl' X2, X3) space by the map (Xl> X2) 1--7>- (Xl, X2, X22) and then follow this by the projection onto the (Xl' X3) plane. Figure 2 Proof of Theorem 4.5. First choose coordinates YI, ... , Ym centered at f(p) so that the image of SIC!) under f is described by the equation Ym = O. Since f!SIC!) is an immersion the image is locally an m - I-dimensional manifold so this choice of y's is possible. (See I, Theorem 2.10.) Since f: Sl(!) ----»- {Ym = O} is a diffeomorphism locally near p, we can choose coordinates Xl, ... , Xn near P so that X, = y, • f for I ::;; i ::;; m - I and SIC!) is defined near p by the equations Xm = ... = Xn = O. In those coordinates f has the form (Xl' ... , xm) 1--7>- (Xl' ... , xm-l,fm(X)), Of course SIC!) is also described by the equations afm/ax, = 0 (m ::;; i ::;; n). So these partial derivatives vanish when Xm = ... = Xn = O. Now fm itself vanishes when Xm = ... = Xn = 0 since the equation Ym = 0 describes the image of SI(!)' These two facts imply that fm(x) = .L h1lx)x,Xj i,j';2:m where h!j(x) are smooth functions. (This is similar to II, Lemma 6.10). Suppose that {h,lO)} is a nonsingular (n - m + 1) x (n - m + 1) matrix. The rest of the proof is similar to the proof of II, Theorem 6.9 about Morse functions. In particular, using arguments as in II, Lemma 6.11 show that by a change of coordinates in (xm' ... , xn) we can assume that fm has the form xm2 ±... ± Xn2• 0 We prove the supposition in the folIowing lemma and in doing so emphasize that point in the proof where the transversality hypothesis is used. Lemma 4.6. The matrix h,lO) is nonsingular. Proof Suppose it were singular. Then we could make a linear change 90 Stable Mappings of coordinates in the variables Xm, ••. , Xn so that this matrix is in diagonal form with entries ± 1 and 0 along the diagonal and at least one of the diagonal entries is zefD. Thus we could assume thatfhas the same 2-jet at 0 as the map (*) where r < n. Now the condition thatjIfffi Sl at p is a condition onpf(p) and (djIf)p' i.e., on the 2-jet off at p. Thus if the transversality condition is satisfied by jIf at p it is satisfied by any other map with the same 2-jet at p as f For the mapping (*) the set Sl(*) is given by the equations Xm = ... = Xr = 0 and thus has codimension r - m + 1 < n - m + 1. If the transversality condition were satisfied it would have codimension n - m + 1; so we have a contradiction and the matrix is nonsingular. 0 We now isolate the main step in the pfDofofthe sufficiency part ofTheorem 4.4 with the foIIowing: Lemma 4.7. Let p be a fold point off and let T be a vector field along f defined on some nbhd U ofp such that TI(Sl(f) n U) = o. Then there exists a vector field ~ defined on a nbhd V c U ofp such that T = (df)(~) on V. Proof Choose coordinates Xl, ... , xn centered at p and coordinates Yl, ... , Ym centered at f(p) satisfying Theorem 4.5. In these coordinates T is just an m-tuple of smooth functions T = (Tl' ... , T m) (that is T(X) = 2r~ 1 T;(x)(8/8y;) and a vector field ~ is given by an n-tuple of functions (~1' ... , ~n). Given the normal form of f in these coordinates solving the equation T = (df)(~) is equivalent to solving the system T; = ~i l:s;i:s;m-l and The first equations are trivially solvable and the last equation is solvable providing that T m = 0 on the points Xm = ... = Xn = O. But these equations describe the fold locus SI(f) n U and by assumption Tm == 0 on this set. 0 Proof of Theorem 4.4. Sufficiency. We assume that f is a submersion with folds and thatfISl(f) is an immersion with normal crossings. We will show thatfis infinitesimally stable. To do this let T be a vector field alongf; we must find vector fields ~ in X and TJ on Y so that T = (df)(~) + TJ • f Since g = fISl(f) is an immersion with normal crossings, g is infinitesimaIIy stable so there exist vector fields' on SI(f) and TJ on Y so that TIS1(f) = (dg)(,) + TJ • g. Extend, to a vector field ~ on X and consider a new T = T - (df)(O - TJ • f This new T has the property that TIS1(f) = o. Applying Lemma 4.7 we can assume that around each point pin Sl(f) there is a nbhd V of p and a vector field ~ on V so that T = (df)(~) on V. At points p not in Sl(f) there exists a nbhd ofp such thatfl V is a submersion. On these nbhds the equation T = (df)(~) is clearly solvable. Using a partition of unity argument we obtain a global solution by patching. 0 Chapter IV The Malgrange Preparation Theorem §1. The Weierstrass Preparation Theorem In this chapter we will prove a technical theorem about smooth functions which will be used to prove Mather's Theorem about stable mappings and to establish the existence of normal forms for singularities of certain stable mappings. In order to make the theorem palatable, we first state and prove the corresponding but less complicated result for analytic functions of several complex variables. Theorem 1.1. (Weierstrass Preparation Theorem). Let F be a complexvalued holomorphic function defined on a nbhd of 0 in e x en satisfying: (a) F(w, 0) = wleg(w) where (w, 0) E e x en and g is a holomorphic function of one variable in some nbhd of 0 in e, and (b) g(O) # O. Then there exists a complex-valued holomorphic function q defined on a nbhd of 0 in e x en and complex-valued holomorphic functions Ao, ... , A,e-l defined on a nbhd of 0 in en such that (i) (qF)(w, z) = wle + :2J':-Ol A;(Z)Wi for all (w, z) in some nbhd of 0 in ex en, and (ii) q(O) # o. Remark. The reader may well ask what such a theorem is good for. Before we proceed we point out one trivial consequence. Given a nonzero holomorphic function F of n + 1 complex variables, we may assume (by a linear change of coordinates) that F = F(w, z) is in the form above. Then the Weierstrass Preparation Theorem states that the zero set of F equals the zero set of the function Ie-I Wlc + :2 Ai(Z)Wi i= 0 which is just a "branched covering surface" over the z hyperplane. We will actually prove a more general result. Theorem 1.2. (Weierstrass Division Theorem). Let F, g, and k be as above and let G be any complex-valued holomorphic function defined on a nbhd of 0 in e x en. Then there exist complex-valued holomorphic functions q and r defined on a nbhd of 0 in e x en such that (i) G = qF + r, and (ii) r(w, z) = :2J':-Ol ri(z)wi for all (w, z) in some nbhd of 0 in e x en where each r, is a holomorphicfunction defined on a nbhd of 0 in en. (iii) q and r are unique (on some nbhd of 0). 91 92 The Malgrange Preparation Theorem Proof that Theorem 1.2 => Theorem 1.1. Let G(w, z) = wi' and apply Theorem 1.2. Setting At equal to rl proves (i). To show that q(O) i= 0, consider wk = q(w, O)F(w, 0) + r(w, 0) k-1 = q(w, O)wkg(w) + 2: rl(O)wl. 1=0 Since both sides are analytic functions of w, we may use power series techniques to conclude that q(O)g(O) = 1 and thus q(O) i= O. 0 Proof ofTheorem 1.2 (iii). Uniqueness. Suppose G = qF + r = q1F + r1' Then (q - q1)F = r1 - r. Fix z in cn; then r1 - r is a polynomial of degree :s;k - 1 in wand has at most k - 1 roots (including multiplicity). We shall show that there is a nbhd of 0 in cn such that for every z in this nbhd (q - q1)F has at least k zeroes when viewed as a function of w. Thus we can conclude that r1 = rand q = q1 (since F ¥: 0 on a nbhd of 0). It is clearly enough to show that F(·, z) has k zeroes. Let'F(w) = F(w, 0). Since the zeroes of a nonzero complex analytic function of one variable are' isolated, and F(O) = 0, there is a constant S > 0 for which F(w) i= 0 whenever 0 < Iwl :s; S. Let e = inf1wl=o IF(w)l. Since F is continuous there is a constant a > 0 for which IF(w, z) - F(w) I < e whenever IZjl < a for j = 1, ... , n where z = (Zl"'" zn) and Iwl = S. Choose such a z and let hew) = F(w, z). Since Ih(w) - F(w) I < e :s; !F(w)I when Iwl = S we can apply Rouche's Theorem [see Ahlfors, Complex Analysis, p. 152] and conclude that hand F have the same number of zeroes in the disk Iwl < S. Since F(w) = wkg(w) on a nbhd of 0, we know that F has k-zeroes (counting multiplicity) in this disk. 0 Definition 1.3. Let Pk : C X cn X Ck -7 C be the polynomial Piw, z, A) = wk + L~~l AiWi where A= (Ao, ••• , Ak- 1). The heart ofthe proof ofthe Division Theorem lies in proving the tlJ.eorem for the polynomials Pk • Theorem 1.4. (Polynomial Division Theorem). Let G(w, z) be holomorphic on a nbhd of 0 in C x cn. Then there exist holomorphic functions q(w, z, A) and r(w, z, A) defined on a nbhd of0 in C x cn X Ck satisfying: (i) G(w, z) = q(w, z, A)Pk(W, A) + r(w, z, A), and (ii) r(w, z, A) = wk + L~;;l ri(z, A)Wi where each rl is a holomorphic function defined on a nbhd of0 in cn X Ck. Proof that Theorem 1.4 => Theorem 1.2. Let F and G be as in the hypotheses of the Division Theorem. Using Theorem 1.4 choose holomorphic function qF, rF, qG, and rG satisfying (*) §1. The Weierstrass Preparation Theorem 93 AlsorF(w, z, A) = :2}:l r/(z, A)W1• First we note that r{(O) = OandqF(O) i= 0; for wkg(w) = F(w,O) = q F(W, O)Pk(w, 0) + rF(w, O) k-l = qF(W, O)wk + L r/(0)w1• 1=0 Now apply a simple power series argument to both sides, recalling that g(O) i= O. Next let f.(A) = r{(O, A). We claim that det «of./oAj)(O)) i= o. For let z = 0, then wkg(w) = F(w,O) = qFPk + rF = qF(W, 0,1.)(wk + ~~ Aiwi) + ~~ f.(A)wi. Differentiate both sides with respect to Aj and evaluate at A = 0 to obtain _ OqF ( 0) k (0) j k~l of. (0) i o- 01. w, w + qF W, W + L.., 01. w. j i=O j Matching coefficient on Wi for i < j we see that (of./oAj)(O) = 0 and also (oiJ/oAj)(O) = -qF(O). So the matrix «of./oAj)(O)) is lower triangular and has determinant equal to (-I)kqiO) i= 0 (shown above). We now apply the Implicit Function Theorem (for holomorphic functions) [10, p. 17] to insure the existence of holomorphic functions 8;(z) (0 :-:; i :-:; k - 1) satisfying (a) rjF(z, 8(z)) == 0 where 8(z) = (80(z), ... , 8k_1(Z)), and (b) 8(0) = 0 (since r/(O) = 0). Define (j(w, z) = qF(W, z, 8(z)), and pew, z) = Pk(w, 8(z)). Then F(w, z) = qiw, z, 8(z))P(w, 8(z)) + riw, z, 8(z)) = (j(w, z)P(w, z). Moreover (j(0) = qF(O) i= 0, so pew, z) = F(w, z)/(j(w, z) on a nbhd of 0 in e x en. From the second equation in (*) we obtain where and G(w, z) = qG(w, z, 8(z))P iz, 8(z)) + rG(w, z, 8(z)) = q(w, z)F(w, z) + r(w, z) ( ) _ qG(w, z, 8(z)) q w, z - ( )q w,z k-l r(w, z) = rG(w, z, 8(z)) = L riG(z, 8(z))w1• i=O Finally let r;(z) = riG(z, 8(z)). 0 94 The Malgrange Preparation Theorem Proof of Theorem 1.4. Given a holomorphic function G(w, z) we must produce q and r so that G(w, z) = q(w, z, .\.)Pk(w, .\.) + r(w, z, .\.) where r is of the form wk + L~':Ol ri(z, .\.)wi. We recall the following form of the Cauchy Integral Formula: 1 f G(l), z) G(w, z) = 2-' ( ) d"') 7Tl Y "') - w where y is a simple closed curve in the complex plane with w in the interior of y. Now note that for appropriately defined holomorphic functions Si(W, .\.), k-1 Pk ("') , .\.) - Pk(w, .\.) = ("') - w) 2: Si("'), .\.)wi, i=O or (*) Pk("'), .\.) _ Pk(w, .\.) k.;;;;l ( \) i - + L ~"'),AW. "') - w "') - W i=O Thus (Using (*» if the appropriate integrals are in fact well-defined. Thus we should like to set 1 f G(",), z) q(w, z, .\.) = 27Ti yPi"'), .\.)("') _ w) d"') and 1 j' G(",), z) ri(z, .\.) = -2' P ( \) SiC"'), .\.) d"'). 7Tl y k"'), A But these integrals give well-defined functions as long as the zeroes ofPk ("') , .\.) do not occur on the curve y for.\. near 0 in Ck. Such a y is easily chosen. 0 2. The Malgrange Preparation Theorem The proof given in §1 of the Weierstrass Preparation Theorem can be adapted to a corresponding theorem about real smooth functions, the difficulties in the adaptation appearing in the Polynomial Division Theorem (l.4). Our proof follows Nirenberg [41]. §2. The Malgrange Preparation Theorem 95 Them'em 2.1. (Mather Division Theorem). Let F be a smooth real-valued function defined on a nbhd of 0 in R x Rn such that F(t, 0) = g(t)tk where g(O) #- 0 and g is smooth on some nbhd of 0 in R. Then given any smooth real-valued function G defined on a nbhd of 0 in R x Rn, there exist smooth functions q and r such that (i) G = qF + r on a nbhd of 0 in R x Rn, and (ii) ret, x) = "L7,;l rj(x)ti for (t, x) E R x Rn near O. Notes. (1) The Malgrange Preparation Theorem which states that there exists a smooth q with q(O) #- 0 such that (qF)(t, x) = tk + "L7,;l '\lx)t follows from 2.1 in precisely the same way that Theorem 1.1 follows from Theorem 1.2. (2) In the complex analytic theorem q and r are unique; this is not necessarily true in the real Coo case. As an example, let F(t, x) = t2 - x and G(t, x) == O. Then q1 = 0 = r1 and are two pairs ofq and r which satisfy the conclusions of the Division Theorem. This is not surprising when one realizes that the proof of the uniqueness part of the Weierstrass Division Theorem used methods that depended crucially on complex variable theory (of one variable). It is possible to state a division theorem for formal power series algebras and in this setting uniqueness also holds. [See Zariski and Samuel; Commutative Algebra, Vol. II, p. 139]. (3) For the case when k = 1, however, q and r are unique; in fact, the Mather Division Theorem follows from the Implicit Function Theorem. By the Implicit Function Theorem (I, 2.4) there exists a unique real-valued smooth function 1jJ(t, x) such that F(ljJ(t, x), x) = t and 1jJ(0) = O. Suppose that G = qF + r, then G(ljJ(t, x), x) = q(ljJ(t, x), x)t + rex). Setting t = 0, we see that rex) = G(ljJ(O, x), x) is uniquely determined. Now suppose that G = q1F + r also, then (q - q1)F == 0 and, in particular, H(ljJ(t, x), x)t == 0 where H = q - q1' Now (t, x) ~ (ljJ(t, x), x) maps a nbhd of 0 in R x Rn onto a nbhd of 0 in R x Rn so that H == 0 on a nbhd of 0 and q is also uniquely defined. It is now easy to see how to prove the Malgrange Theorem in this special case. Choose 1jJ as above and define rex) = G(ljJ(O, x), x) and q = (G - r)jF. We leave the verification that q is a smooth function as an exercise. (4) For the case of one variable t (n = 0) the Malgrange Theorem is trivial and q and r are uniquely defined. This is left as an exercise-use Taylor expansions of order k. (5) The proof of the Division Theorem given by Mather [26] yields a somewhat more general result; namely, the choice of q and r can be made to depend linearly and continuously (in the Whitney Coo topology) on F and G. 96 The Malgrange Preparation Theorem This proof can be used to modify and extend Nirenberg's proof [32] to obtain these extra results. For our purposes the Division Theorem as stated is sufficient. (6) Mather also proves a global division theorem not just a local one [26]. As in the complex analytic case, the crucial theorem is the following: Theorem 2.2 (Polynomial Division Theorem). Let G(t, x) be a complexvaluedfunction defined and smooth on a nbhd of0 in R x Rn. Then there exist smooth, complex-valuedfunctions q(t, x, A) and r(t, x, A) defined on a nbhd of0 in R x Rn x Rk satisfying: (i) G(t, x) = q(t, x, A)Pit, A) + ret, x, A), and (ii) ret, x, A) = tk + 2:r,;l ri(x, A)ti where each ri is a smooth function defined on a nbhd of0 in Rn x Rk. Moreover if G is real-valued then q and r may be chosen to be real-valued. Note. The" moreover" part is obtained by equating the real parts of both sides of equation (i) since Pk is real-valued. Also each ri is easily seen to be real-valued. Proof that Theorem 2.2 => Theorem 2.1. This proof is word for word the same as the proof that Theorem 1.4 => Theorem 1.2 with the single exception that smooth is substituted for holomorphic throughout. 0 We shall use the same idea to prove 2.2 as we used to prove 1.4 but first we need an analogue of the Cauchy Integral Formula. This is provided by Green's Theorem from Advanced Calculus. Let z = x + iy be a complex coordinate on R2. Then we can solve for x and y in terms of z and z (= x - iy); namely, x = -t(z + z) and y = (Ij2i)(z - z). Let f: C ~ R and define ofjoz so that the chain rule holds; that is, of = of ox + of oy = ! (of + i of) . oz ox oz oy OZ 2 ox oy Now suppose that F: C ~ C is given by f + ig where f, g: C ~ R. Then (*) of = of .og =! ({of _ Og) .(Of og)). oz oz + I OZ 2 \ ox oy + I oy + ox Thus dFjoz == 0 iff F satisfies the Cauchy-Riemann Equations iff F is holomorphic. It is easy to check that the standard rules of differentiation work when differentiating with respect to Z. We also make the convention that dz 1\ dz = -2i dx 1\ dy. Lemma 2.3. Let F: C ~ C be a smooth function (when viewed as a mapping ofR2 ~ R2 ). Let y be a simple closed curve in C whose interior is D. Then for w in D F(w) = _1. r F(z) dz + ~ff o~ (z) dz 1\ dz 2m. r z - W 2wl OZ Z - W D §2, The Malgrange Preparation Theorem 97 Note. If Fis holomorphic in D, then this formula reduces to the Cauchy Integral Formula. Proof Let w be in D and choose s less than the distance from w to y. Let D. = D - (disk of radius s about w) and y. = boundary De. Now recall Green's Theorem for R2. Let M, N: D. -+ R be smooth and continuous on y., then ,. "J (ON OM)Jy M dx + N dy =.1 ox - By dx /\ dy. o Do Note that the formula still holds if M and N are complex-valued since we integrate the real and imaginary parts separately. Apply Green's Theorem and (*) above to F = f + ig and obtain (**) tF dz = L(f + ig)(dx + i dy) = 2i.fJ a:;dx /\ dy Do Ij of -= -.I ozdz /\ dz Finally, apply (**) to F(z)j(z - w). Since Ij(z - w) is holomorphic on D., 0_ ( F(z) ) = (oF joz)(z). oz z - w z - w Thus (***) -Jfo~(z) dz /\ dz = J" F(z) dz = /' F(z) dz - f F(z) dz OZ z - W Yo Z - W • y z - w So Z - W Do where S. is the circle of radius s about w. Using polar coordinates centered at w, one obtains f F(z) dz = J2" F(w + sei9)i dO. So z - w 0 So letting s -+ 0, we see that the RHS of (***) goes to I' F(z) . - - dz - 2mF(w) • y z - w while the LHS converges to -Jfo~ (z) dz /\ d!. oz z - W D (Note this last limit exists since oFjoz is bounded on D and Ij(z - w) is integrable over D.) Thus taking (***) to the limit proves the lemma. 0 Proof of Theorem 2.2. Let G(t, x) be a smooth complex-valued function defined on a nbhd of °in R x Rn; we need to show that for appropriate choices of q and r, G = qPlc + r. Let G(z, x, .\) be a smooth function defined on a nbhd of °in C x Rn x C lc so that G is an extension of G, that is, 98 The Ma1grange Preparation Theorem G(t, x, .\) = G(t, x) for all real t. Then G = qPk + r (on C x Rn x Ck) where (. .\) = _1 J' G(TJ, x, .\) dTJ _1 fJ" (oG/oz)(TJ, x, .\) dTJ /\ dYj q If, x, 27ri y Pk(TJ,.\) (TJ - w) + 27ri PiTJ, .\) (TJ - w) D and if these integrals are well-defined and yield smooth functions. (This follows from Lemma 2.3 as the calculations in Theorem 1.4 follow from the Cauchy Integral Formula.) The first integral in the definition of both q and r is welldefined and smooth for the same reasons as for the corresponding integral in Theorem 1.4. 0 The problem lies with the second integrals since D may contain zeroes of Pk' But if it is possible to choose a smooth extension Gof G so that oG/oz vanishes on the zeroes of Pk and for real z, then we will have q and r welldefined. Yet this is not enough, since q and r must be smooth functions. Since the integrands are bounded we can differentiate under the integral sign and then quickly see that an appropriate condition for insuring that q and rare smooth is the existence of a smooth extension Gof G such that oG/oz vanishes to infinite order on the zeroes of Pic and for z real. Thus the last detail-in fact, the crucial detail-is to show the existence of such an extension. This we shall now do. Proposition 2.4. (Nirenberg Extension Lemma). Let G(t, x) be a smooth complex-valuedfunction defined on a nbhd of0 in R x Rn. Then there exists a smooth complex-valuedfunction G(z, x, .\) defined on a nbhd of0 in C x Rn x C Ic satisfying (1) G(t, x, .\) = G(t, x) for all real t, (2) oG/oz vanishes to infinite order on {1m z = O}, and (3) oG/oz vanishes to infinite order on {Piz, .\) = O}. As a first step in proving the Nirenberg Extension Lemma we recall a more elementary extension lemma due to Emile Borel. Lemma 2.5. Let fO(X),fl(X), ... be a sequence of smooth funCtions defined on a given nbhd of0 in Rn. Then there is a smooth function F(t, x) defined on a nbhd of 0 in R x Rn such that (oIF/otl)(x, 0) = fz(x) for aliI. Set (*) Proof Let p : R -)- R be a smooth function such that pet) = {~ ItI :c:; t ItI ~ l. co t l F(t, x) = 2: Ii P(fLlt)fz(x) 1=0 • §2. The Malgrange Preparation Theorem 99 where the fLz'S are an increasing sequence of real numbers such that Lim/_ oo fLl = 00. We will choose the fLz'S so that F(t, x) is smooth on a nbhd of 0 in R x Rn. First observe that the RHS of (*) is well-defined for all t (since for any given t only finitely many terms are nonzero) and is a smooth function of t when t =1= 0 (since for t =1= 0 only finitely many terms are not identically zero on a nbhd of t). Next choose a compact nbhd K of 0 in Rn contained in the common domain of the fz's and let MI = SUpxEK Ifz(x)l. Now differentiate the terms on the RHS of (*) s times with respect to t; the resulting series is dominated in K by 00 I t ll - 8 C8/~ (1- s)! P(fLlt)fL/MI where C8 is some constant depending on P and its first s derivatives. Since P(fLlt) = 0 as soon as 1t 1 > 11fLl' this series is itself dominated by C ~ M p + 8 8 L p " p=o fLp+8P· which will converge for all s if the fLz'S tend toward infinity rapidly enough. This shows that if the RHS of (*) is differentiated with respect to t termwise (s-times) the resulting series converges uniformly on R x K. The corresponding result obtained by differentiating with respect to the x-variables is clear. Suppose that you want to show that the series for (olotS)(olox,,)f converges uniformly. Let Mz" = sup __"I (x)1 01"1Jz I XEK ox and proceed as before. Finally to do this for all possible mixed partials simultaneously we use the diagonal trick of I, Proposition 4.8. That is, let Ml = sup __"I (x) .I01"1Jz IXEK oXlalsl Proceed as above choosing the fLl'S and note that now the series for each partial derivative converges uniformly on K and that this f is smooth on a nbhd ofO. 0 Note. One can use Lemma 2.5 to show that for any power series about oin Rn there exists a smooth COO real-valued functions whose Taylor series expansion at 0 is this given power series. We will need another elementary extension lemma which is, in fact, a special case of the Whitney Extension Theorem. [59] Lemma 2.6. Let V and W be subspaces of Rn such that V + W = Rn. Let g and h be smooth functions defined on a nbhd of0 in Rn. Assume that for all multi-indices ex Olalg 01((111 oxa (x) = ox" (x) for all x in V n W. 100 The Malgrange Preparation Theorem Then there exists a smooth function F defined on a nbhd ofO in R" such that for all a lo,a,g olalF X _ oxa (x) oxa ( ) - olalh oxa (x) ijx is in V ijxisin W Proof We observe first that it is sufficient to prove the lemma for h == 0. For let F1 be the extension for g - hand 0, then F = F1 + h is the required extension for g and h. So we do assume that h == 0. Next choose coordinates Y1, ..., Y" on R" so that V is defined by the equations Y1 = ... = Yi = °and W is defined by the equations Y1+1 = ... = Yle = 0. This is possible since V + W = R". Then set co ya olalg ( i 2)F(y) = 2: "I 0''« (O""'O'Y1+b""Y")p p.lal 2:Yi lal=O a.)' 1=1 a =(al•.. .,al' 0, ... ,0) where p is the same smooth functions which appeared in the proof of the lastlemma and the sequence {p.z}i= 0 is increasing to infinity. As in the last lemma the p.z's can be chosen to increase rapidly enough to infinity to insure that F is a smooth function on a nbhd of °in R". We need only check that F has the desired properties. IfY = (ll, ... , Y,,) is in W, then Yi+1 = ... = Yle = °and every term of (oIPIF/oyP)(y) contains a factor of the form (olrlg/fJyr)(o, ..., 0, YIe+1, ... , y,,). Since (0, ..., 0, Ylc+b ... , Y,,) is in V n W that factor equals zero by assumption. Thus (oIPIF/oyP)(y) = 0. On the other hand, if Y is in V, then Y1 = ... = Yi = °and Thus OIPIF ~ 01131 (ya olalg ) I-13 (y) = L.. -13 ,-a(y) . oY lal=O oY a. oY Yl =...=y,=o Let f3 = (b1, ••• , b,,). It is easy to see that if bl # al for some i :::; j, then the given term in the series is 0. In fact, the only nonzero term is (oIPlg/oyP)(y). So F is the desired extension. 0 Finally we need to solve formally an "initial value" problem for certain partial differential equations. Lemma 2.7. Let f(x) be a smooth complex-valued function defined on a nbhd of °in Rn and let X be a vector field on Rn with complex coefficients. Then there exists a smooth complex-valuedfunction F defined on a nbhd ofO in R x Rn such that (a) F(O, x) = f(x) for all x in R", and (b) of/at agrees to infinite order with XF at all points (0, x) in R x Rn. §2. The Malgrange Preparation Theorem 101 Proof. An obvious candidate for such a solution is the formal expression In fact, by differentiating the LHS of this equation term by term and evaluating at t = 0 we see that (b) holds. Clearly (a) holds. The only problem is that Fneed not be smooth. Now by the Borel Theorem (Lemma 2.5) we may choose a smooth function F of the form having the same power series expansion as F at t = O. This F will solve our "initial value" problem. D Proof of Proposition 2.4. The proof will be done by induction on k. When k = 0, Pk(z) == 1, so we need only show that there exists a smooth function G(z, x) such that G(t, x) = G(t, x) for real t and (oG/oi)(t, x) vanishes to infinite order for real t. Let z = s + it. Then .. 0 0 .0 1 oi = ot + 1 os and the existence of such a Gfollows from the last lemma by taking X = -i(%s). We now assume that the case k - 1 has been proved and attempt to prove the proposition for k. In particular, we will show that there exist smooth functions E(z, x, A) and F(z, x, A) satisfying (i) E and F agree to infinite order on the set {Pk(z, A) = O} (ii) F is an extension of G (iii) of/oi vanishes to infinite order on {1m z = O} (iv) Let M = F I{Piz, A) = O}. Then oM/oivanishes to infinite order on {oPk/oz)(z, A) = O} and (v) oE/oi vanishes to infinite order on {Pk(z, A) = O}. First we show that the existence of E and F is sufficient to prove the proposition. Set u = P(z, A) == Pk(z, A), and let A' = (AI" .., Ak- 1). Then consider the change of coordinates (z, Ao, .\') ~ (z, u, A') on C x C X Ck-l. Recalling that P(z, A) = Zk + J.r::01 A1Zi we see that OU/OAo == 1 so that this is a legitimate change of coordinates. In these new coordinates the hypersurface {P k(Z, A) = O} is given by the simple equation U = O. The coup de grace is then administered by Lemma 2.6. By this lemma there exists a function Gwhich agrees to infinite order with E on U = 0 and to infinite order with F on 1m z = O. (Note that u = 0 and 1m z = 0 are subspaces of R2k+2 which intersect transversely. Along with (i) this guarantees that Lemma 2.6 is applicable.) Properties (ii), (iii), and (v) then imply that Gis the desired extension of G. Now to show the existence of E and F. First we assume the existence of F and construct E. Consider again the coordinates (z, u, A') and notice that in 102 The Malgrange Preparation Theorem these coordinates the vector field 8/8z has the form 8/8z + (8P /8z)(8/8u). Similarly, 8/8i becomes 8/8i + (8P/8z)(8/8ii). So our problem in these coordinates is to find E(z, x, u, ,\') such that (a) E = F to infinite order on {u = O}, and (b) (8/8i + (8P/8z)(8/8U)E) = 0 to infinite order on the set {u = O}. Let X == -(8P/8z)-1(8/8i). (We will deal with the problem of the zeroes of 8P/8z in a moment.) Then this problem can be reformulated in a form analogous to Lemma 2.7, i.e. find a smooth function E satisfying (a) and (b') 8E/8i = XE to infinite order on {u = O}. This admits the same sort of formal solution as Lemma 2.7; namely, (*) where p is the same bump function used before. Now since 8M/8i = 0 to infinite order on the set {(8P /8z)(z, '\') = O} (assumption (iv)) we see that XIM is a smooth function in z, x, and '\' for all I. Hence we can choose fLz'S which increase to infinity rapidly enough to guarantee that the RHS of (*) is a smooth function of z, x, u, and '\. Thus the zeroes of 8P/8z cause no problem and this E is then the desired function. Thus to complete the proof we need to construct a smooth function F satisfying (in the z, x, u, '\' coordinates) (ii) F(t, x, u, '\') = G(t, x) for all real t, (iii) 8F/8i = XF to infinite order on {1m z = O}, and (iv) Let M = F I{u = O}. Then 8M/8i = 0 to infinite order on {8Pk/8z = O}. Consider the hyperplane u = 0 and the change of coordinates '\' = ('\1"'" '\,e-1) 1---7- ('\1/1, ... , '\k-1/k - 1) = ,\". These conditions translate to finding a smooth function M(z, x, '\") (which will be F restricted to {u = O}) satisfying (1) M(t, x, '\") = G(t, x) for all real t, (2) 8M/8i vanishes to infinite order on {1m z = O}, and (3) 8M/8i vanishes to infinite order on {Pk- 1(Z, '\") = O}. By our induction hypothesis such a smooth function M of the variables z, x, and '\" exists and we can view M as a smooth function of z, x, and,\'o Finally we define Fusing M and (*); that is, Again since 8M/8i vanishes to infinite order when (8P/8z)(z,,\') = 0 = Pk - 1(Z, /.."), X1M is smooth in z, x, and '\'. Thus the fLz'S can be chosen so that F is a smooth function satisfying (ii) and (iii). Also on {u = O} F = M so 8F/ai vanishes to infinite order and F satisfies (iv). 0 Thus we have proved the Nirenberg Extension Lemma and the Malgrange Preparation Theorem. §3. The Generalized Malgrange Preparation Theorem 103 Exercise. Let f: R --+ R be a smooth even function. Show that there exists a smooth function g: R --+ R such that f(x) = g(x2 ). Hint: Use the trick used to prove Lemma 2.5. §3. The Generalized Malgrange Preparation Theorem Our purpose is to generalize the Malgrange Preparation Theorem to a statement about certain local rings. We will only discuss rings which are commutative and have a mUltiplicative identity. Definition 3.1. Let X be a smooth manifold and let p be a point in X. (a) Two smooth real-valued functions f and g, defined on nbhds of p, are equivalent near p iff = g on some nbhd ofp. (b) Let f: U --+ R be a smooth function where U is some nbhd ofp. Then [f]p == germ of f at p is the equivalence class off in the equivalence relation defined in (a). Let Cp""(X) be the set of all germs of smooth, real-valued functions defined on a nbhd ofp. (c) A local ring is a ring with a unique maximal ideal. Lemma 3.2. Cp""(X) is a local ring if the ring operations are given by [f]p + [g]p = [f + g]p and [f]p' [g]p = [fg]p where f + g andfg are assumed to be defined on domf n dom g which is a nbhd of p. Let ,Ap(X) = ([f]p E C;(X) If(p) = O}. Then ,Ap(X) is the unique maximal ideal. Proof It is easy to convince oneself that these operations are welldefined and that C;(X) is a commutative ring with multiplicative identity. It is also a trivial exercise to see that ,Ap(X) is an ideal in Cp""(X). As for unique maximality, let ~ be any other ideal in Cp""(X). Suppose [f]p E,A ,Ap(X). Then [ljf]p is defined since f(O) "# 0 and therefore [ljf]p·[f]p = [l]p E,A. SO u#t = Cp""(X). Thus U#tvCX) is the unique maximal ideal in Cp""(X). 0 Note. We will sometimes omit the brackets when discussing germs, and use the same symbols as for functions. The context should allay any possible confusion. Lemma 3.3. Let ~p2(X) be the ideal generated by germs of the form fg where!, g E ,Ap(X). Then u#tp(X)j,Ap2(X) is a vector space (over R) canonically isomorphic with T: x. This isomorphism is induced by the mapping if; : ,Ap(X) --+ T: X given by [f]p H> (df)(p). Proof The facts that U#tp(X)j,Ap2(X) is a vector space and that if; is welldefined and linear are easy to verify. Equally easy to see is that if; is onto. For let Xl> ... , Xn be local coordinates on X based at p, and V be a cotangent vector in T; X, so that v = (v1dx1 + ... + vndxn)lx=p. If we let f(x) = L:f= 1 ViXi which is defined on a nbhd of p, then (df)(p) = v. Finally we show that Ker if; = u#tp2(X). It is an easy calculation to show that,Ap2(X) c Ker if;. 104 The Malgrange Preparation Theorem So let [f]p E Ker l/;. Since f(O) = 0, f(x) = L:f= 1 Xij;(X) where j;(0) = (ofloXI)(O) by II, Lemma 6.l0. Since (df)(p) = 0, j;(0) = O. Thus [f]p is in .4p2(X). 0 Lemma 3.4. Let : X ---+ Y be a smooth mapping with q = (p). Then induces a ring homomorphism *: C;'(Y) ---+ C;(X) given by [f]q f-+ [f·]p. Moreover is locally (near p) a diffeomorphism iff* is an isomorphism. Proof It is easy to see that * is well-defined. To show that * is a ring homomorphism and that if is a local diffeomorphism then * is an isomorphism is also easy. (Note that (*)-1 = (-1)*.) So we assume that * is an isomorphism. Clearly * induces an isomorphism of .4i Y)j.4q2( Y) ---+ .4vCX)j.4p2(X) so that by Lemma 3.3 dim Y = dim X. Choose local coordinates Xl> ... , Xn on X based at p corresponding to the chart 7]. Let [xtlp = *[h;]q for some smooth functions hi. Define H: (dom h1 n· .. n dom hn) ---+ Rn by H(y) = (h1(y), ... , hn(y)). H is smooth and 7] = H. on a small nbhd of p. Applying the chain rule, we have (d7])p = (dH)id is a local diffeomorphism. 0 Let f!lI be a ring (commutative with identity) and A an abelian group (with the group operation denoted by +). Then we recall that A is an fltmodule if there is a mapping of flt into the set of homomorphisms on A. We denote the action of r in flt on a in A by ra and demand that the following relations hold for all r1, r2 in flt and a in A: (r1 + r2)a = r1a + r2a, (r1r2)a = r1(r2a), r(a1 + a2) = ra1 + ra2, and l·a = a. Note that if f!lI is a field, A is simply a vector space over flt. Recall also that an f!lI-module A is finitely generated over f!lI if there is a finite number of elements a1, ... , an in A such that each element a in A can be written as a linear combination a = r1a1 + ... + rnan for some r;'s in flt. (Warning: In an arbitrary module the linear combination need not be unique, even if the generating set {al> ... , an} is minimal.) We assume in what follows that the reader is familiar with such elementary notions as submodules, quotient modules, module homomorphism, etc. We will need two lemmas about flt-modules. Lemma 3.4. (Nakayama). Let flt be a commutative local ring with identity and let .4 be the maximal ideal in flt. Let A be an flt-module. Assume that (i) A is finitely generated, and (ii) A = .4A (= the set of sums of elements of the form ra with r E .4 and a EA.) Then A = {O}. Proof Let el> ... , en be a finite set of generators for A over flt. We will show that each e" = o. First we may write e" = m1a1 + ... + msas where §3. The Generalized Malgrange Preparation Theorem 105 each ?ni is in .A since A = .AA. But since el, ... , en are a set of generators we may also write ai = If=l rijej. Thus ek = If=l skjej where Skj = If=l ?nirij is in.A. Using the Kronecker delta we have that If=l (8kj - Skj)ej = 0 for each k. This is a system ofn linear equations in n unknowns el, ... , en' Now note that if the matrix (8ij - Sij) is invertible then the system of equations has only the trivial solution el = ... = en = O. Now a matrix D (over a commutative ring with unit) is invertible iff det (D) is invertible in the ring. (To review the theory of determinants of matrices over a commutative ring with identity, see Chapter V of Hoffman and Kunze, Linear Algebra.) Now by using the standard expansion of det (8ij - Sij) by permutations it is easy to see that det (8ij - SIj) = 1 + S where S is in .A. Also, in a local ring, the maximal ideal is precisely the set of noninvertible elements: Suppose t is in Jlt; then t is not invertible. For if t were invertible then 1 = tt- l would be in .A. Conversely if t is not in .A, then t is invertible, for the ideal generated by t is not contained in .A so it must be all of B? Thus there is t' in [J£ such that tt' = 1. From this we may conclude that 1 + S (which is not in .A) is invertible and, therefore, el = ... = en = O. 0 Remark. A more sophisticated way of formulating the last half of the above argument is that the quotient ring B?j.A is afield. Corollary 3.5. Let A be a finitely generated B?-module. Then Aj.AA is a finite dimensional vector space over the field B?j.A. Let 4> : A --'>- Aj.AA be the naturalprojection and Vb ••• , Vn a basisfor this vector space. Choose eb ••• , en in A so that 4>(ei) = Vi' Then eb ... , en form a set ofgenerators of A over B? Proof Since the action of B? on A clearly induces an action of B?j.A on Aj.AA we see that Aj.AA is a module over the field B?jj/; i.e. a vector space. To show that dim9f!/.4't Aj.AA < 00, let al, ... , ak be a set of generators of A over B? and let v be in Aj.AA. Choose a in A so that 4>(a) = v and choose ri in B? so that a = rIal + ... + rkak. Then v = [rd4>(al) + ... + [rk]4>(ak) where [r;] denotes the equivalence class of ri in B?j.A. Thus 4>(al), ... , 4>(ak) form a set of generators of AjJitA. Conversely let Vb ••• , Vn be a basis for Aj.AA with el, ... , en chosen as in the statement of the corollary. Let B be the submodule of A generated by eb ... , en and let C be the quotient module AjB. Since A is finitely generated over [J£, C is finitely generated over B? Now A = B + .AA. For if a is in A, then 4>(a) = [rdvI + ... + [rn]vn. So a = rIel + ... + rnen + s where s is in ",itA. Thus C = AjB = (B + .AA)jB = j/(AjB) =.Ac. Use arguments about cosets to check these equalities. Finally apply Nakayama's Lemma to show that C = 0 and thus A = B. 0 Returning to our local ring of interest, suppose that A is a Cp""(X) module and that 4> : X --'>- Y with q = 4>(p) is a smooth mapping. The induced ring homomorphism 4>* allows us to view A as a Cq""( Y) module. More specifically if a is in A and [f]q is in Cq""( Y), then we define [f]qa == 4>*[f]qa. We now state the local ring generalization of the Malgrange Preparation Theorem. 106 The Malgrange Preparation Theorem Theorem 3.6. (Generalized Malgrange Preparation Theorem). Let X and Y be smooth manifolds and

Y be a submersion with dim X = n = dim Y + 1 and q = 7T(p). Let A be a finitely generated C;(X)-module.lf V = AjV«q{ Y)A is afinite dimensional vector space over R then A is afinitely generated Cq"'( Y)module (via 7T*). Proof. Since this is a local result, we may assume, with a proper choice of charts, that X = Rn, Y = Rn-\ p = °=.q, and 7T: Rn-l>Rn-l is given by (Xl" .. , Xn) H>- (X2' ... , xn). Let .p: A -l> V be the canonical projection and choose el, ... , en in A so that {.p(el), ... , .p(en)} is a basis for V. Step I: eb ... , en generate A as a Co"'(Rn)-module. To see this note that V«o(Rn-l) c V«o(Rn) so that there is a natural surjection YJ: AjV«o(Rn-l)A-l> AjV«o(Rn)A. Thus YJ •.p(el), ... , YJ' .p(en) is a set of generators for AjV«o(Rn)A and Step I follows from Corollary 3.5. Step II: All elements of A have the form 2:7=1 (Ciei + hei) where Ci is a scalar in R and.!; is in V«o(Rn -l)Co"'(Rn). Since .p(el), ... , .p(en) is a basis for V, we see that if a is in A, then a = 2:r=1 Ciei + ewhere eis in V«o(Rn-l)A. Thus e = 2:7=1 gjaj where gj is in V«o(Rn-l) and aj is in A. But by Step I aj = 2:r=1 hiei where hi is in Co"'(Rn). Thus e = 2:r=1 (2:1'= I gihj)ej and by letting j; = 2:1'= I g/lj. Step II is proved. Now we prove the Lemma. By Step II, xlei = 2:1=1 (Cij +hj)ej with Cij and hj in the appropriate places. Using the Kronecker delta we obtain the n-linear equations in n-unknowns el , ... , en (*) n 2: (XIOij - Cij - /ij)ej = 0. j=l Let P(Xb ... , xn) be the determinant of the matrix (XIOij - Cij - j;J. By Cramer's Rule Pei = °for each i. Now note that /ij(xl , 0, ... , 0) = °since hj is in V«o(Rn-I)Co"'(Rn). Therefore, P(Xl' 0, ... ,0) = det (XlOij - cij) which is a polynomial in Xl of degree ~n. Hence there exists k ~ n such that P(Xb 0, ... , 0) = xr"g(xl) and g(O) #- 0. By Step II, if a is in A, then a = 2:r=1(Ciei +hei)' Apply the Malgrange Preparation Theorem to hand P to obtain Ie-I h = QiP + 2: Rilx2 , ••• , Xn)x/. j=O Since Pei = 0, we have thathei = 2:J~J Rijx/ei and a = i (ciei + leII Rijx/ei). i=l j=l Thus A is generated by the nk elements eb ... , en, xlel , . .. , xlem ... , xr"en as a module over Co"'(Rn-l) since Rij is in CO'(Rn-I). 0 Lemma 3.8. Let rj>: X -l> Y be an immersion with q = rj>(p). Let A be a finitely generated Cp"'(X)-module. Then A is a finitely generated C;:(Y)- module. 108 The Malgrange Preparation Theorem Proof 4>*: Cq""( Y) --+ C;(X) is onto, since there is a nbhd U of p in X such that 4>1 U is a 1: 1 immersion and 4>(U) is a submanifold of Y. Clearly, using the definition of submanifold, every smooth function on 4>(U) can be extended to be a smooth function on a nbhd of 4>(U) in Y. (Since we are working locally we can assume that 4>(U) is small enough to be made a kplane in Rm by one chart on Y.) The surjectivity of 4>* implies the stated result. 0 Proof of Theorem 3.6. Define q;: X --+ X x Y by q;(x) = (x, 4>(x)). Using charts, we may assume that X = Rn and that p = O. Let 71"i : Ri x Y--+ Ri-l x Ybegivenby(xl, ... ,x;,Y)H>-(x2, ... ,x;,y).Then4> = 71"l.···.71"n.q; (locally). Since q; is an immersion Lemma 3.8 applies and A is a finitely generated C<,~),q)(Rn x Y)-module. Now assume that A/vfti Y)A is a finite dimensional vector space. Since ~o.qlRn -1 X Y)A ::> vfti Y)A, there is a natural surjection of A/Jli Y)A --+ A/vft(O,q)(Rn -1 X Y)A so that the latter space is finite dimensional. Thus the hypotheses of Lemma 3.7 are satisfied for 71"n and we may conclude that A is a finitely generated q~,q)(Rn-l x Y)-module. A simple induction argument implies that A is a finitely generated Cq""( Y)-module. 0 Examples (A) Letf: R --+ R be a smooth even function; then there exists a smooth function g: R --+ R satisfying f(x) = g(x2 ). This is easy to prove if f is assumed real or complex analytic near 0 (using Taylor series) but is not quite so obvious in the stated case. (This fact was first proved by Hassler WhitneyDuke Journal of Mathematics, volume 10, 1943.) Proof Let p = 0 in the domain and q = 0 in the range. Let A = CpOO(R) which is clearly a finitely generated module over C;(R). Let 4>(x) = x2 • Then via 4>, Cp""(R) becomes a CqOO(R)-module, the module action being given by (ba)(x) = b(x2 )a(x) where a is in CpOO(R) and b is in Cq""(R). Observe that vftiR)C;(R) = (x2) and that the images of 1 and x span the vector space C;(R)/vftq(R)CpOO(R). Apply Theorem 3.6 and Corollary 3.5 to see that f(x) = g(x2) + xh(x2). Since f is even, h(x2) == 0 and f(x) = g(x2) on a nbhd of O. It is easy to see how to make this equality a global one. 0 (B) Let g!> ... , gn be the n elementary symmetric polynomials in nvariables; that is, gl(X!> ... , xn) = Xl + ... + Xn g2(Xl, ... , xn) = X1X2 + ... + X1Xn + ... + Xn-1Xn gnCxl, ... , xn) = Xl' .. Xno Let f(x!> ... , xn) be any smooth symmetric function; that is if 0' is any permutation on n-Ietters, then f(x!> ... , xn) = f(Xa(l), ... , xa(n»)' Then there exists a smooth function h : Rn --+ R satisfying f(x) = h(gl(X), ... , gn(X)). This global result was proved originally by Glaeser [9]. We shall only prove the corresponding local result. Define g: Rn --+ Rn by g(x) = §3. The Generalized Malgrange Preparation Theorem 109 (gl(X), ... , gn(x». Using the same convention concerning p and q in (A) we see that via g Cp"'(Rn) is a Cq"'(Rn)-module. Let B be the set of multi-indices 13 = (131, ... , f3n-l, 0) where f3i < n. Then the set of nomomials {xil I13 E B} is a generating set for the vector space C;'(Rn)jvllq(Rn)cp'" (Rn). (To see tJ"c note that (x - Xl)· .. (X - Xn) = Xn + gl(Xl, ... , Xn)Xn- l + ... + gn(X1 , ••• , xn). Substituting Xi into this polynomial we have that Xin is in the submodule J/{iRn)cp'" (Rn). Also Xn = -Xl - ... - Xn- l modulo vIIiRn)C;'(Rn).) Applying Theorem 3.6 and Corollary 3.5 we see that f(x) = h(g(x» + L hP(g(x»xB. IlEB Since f is symmetric and Xn is not a factor of any of the monomials we see that each ha(g(x» == 0 and thatf(x) = h(g(x». 0 In all of our applications of the Malgrange Theorem we shall be dealing with modules of smooth functions. The most obvious problem in dealing with such functions (as distinct from analytic functions) is that the Taylor series about a point does not necessarily converge to the given function. Thus, in order to show that a module is finitely generated it would be nice to show that the prospective generators need only generate the module in question up to some finite order, thus eliminating the problem of what happens to the smooth functions" at the tail". We now show that this is, in fact, the case. Define inductively a sequence of ideals vIIp"(X) in C;'(X) by letting vIIpleX) be vIIP(X), and vIIpk(X) be the vector space generated by germs of the formfg wherefis in vIIP(X) and g is in vIIpk-l(X). Lemma 3.9. vIIo"(Rn) consists precisely of germs of smooth functions f whose Taylor series at 0 begin with terms of degree k, i.e., 8afj8xa(0) = 0 for lal ::; k - 1. Thus Co"'(Rn)/vllo"(Rn) can be identified with the vector space of polynomials in n variables of degree ::; k - 1. The proof of this Lemma is a simple induction argument based on II, Lemma 6.10 and is left to the reader. Theorem 3.10. Let A be ajinitely generated Cp"'(X)-module. Let 4>: X---+ Y be smooth with q = 4>(p) and let el, ... , e" be elements ofA. Then el, ... , e" generate A as a Cq"'( Y)-module iff7)(el), ... , 7)(e,,) generate Ajvllp,,+l(X)A as a Cq"'(Y)-module where 7): A ---+ Ajvllpk+l(X)A is the obvious projection. Proof The forward implication is obvious, so assume that 7)(e1), ••• , 7)(e,,) generate A/vllp k+ l(X)A as a Cq"'( Y)-module. Let B = A/(vllp"+l(X)A + vIIiY)A). Note that vIIq( Y) acts trivially on B; thus we may consider B as a module over C;:( Y)jvllq{ Y) = R, i.e., a real vector space. Since the images of eb ... , e" generate B, dimB B ::; k. Consider the sequence of vector spaces B::;, 110 The Malgrange Preparation Theorem vff/(X)B :::::J ••• :::::J vff/+l(X)B = O. There are k + 2 vector spaces in this decreasing sequence. Applying the" pigeon-hold principle" we see that there must be i ::; k such that vffpi(X)B = J/{pi + l(X)B. Thus J/{pi(X)A + vffiY)A = vffpi+l(X)A + vffiy)A, since vff/+l(X) c vffpi+l(X) C vffpi(X). Finally consider the CpOO(X)-module C = Ajvffg{ Y)A and note that vftpi(X)C = vffpi+l(X)C = vffiX)vffpi(X)C. Thus we may apply Nakayama's Lemma and deduce that vffpi(X)C = O. (Note that as a CpOO(X)-module J/tpi(X)C is finitely generated. This follows since vffpi(X) is a finitely generated ideal in CpOO(X) and C is a finitely generated module.) But C = Ajvffi Y)A, so vffpi(X)A c vffi Y)A for some i ::; k. Since the images of eb ... , ek generate AjJ/tpk+l(X)A, 'I)(e1), ... ,'I)(e/c) must generate Ajvffq{Y)A as a CqOO(y)_ module and thus as a CqOO( Y)jvffq{ Y) = R-module. Said differently, dimR AjJ/tq{Y)A ::; k. We can now apply the generalized Malgrange Preparation Theorem to conclude that A is a finitely generated CqOO ( Y)-module and Corollary 3.5 to conclude that eb ... , e/c generate A. 0 The usefulness of Theorem 3.10 is illustrated by the following: Corollary 3.11. If the projections of e1, . .. , ek form a spanning set of vectors in the vector space Aj(vffv";+l(X)A + vltq(Y)A), then e1, ... , ekform a set ofgenerators for A as a CqOO( Y)-module. Chapter V Various Equivalent Notions of Stability §1. Another Formulation of Infinitesimal Stability In this section we have three objectives: to show that (1) Infinitesimal stability is locally a condition of finite order; i.e., if the equations can be solved locally to order dim Y then they can be solved for smooth data. (2) Infinitesimal stability is globally equivalent to a multijet version of local infinitesimal stability. (3) Infinitesimally stable mappings form an open set. We won't be able to achieve our last objective just yet; but, at least, we shall be able to give a sufficient condition for the existence of a neighborhood of infinitesimally stable mappings around a given infinitesimally stable mapping. Let X and Y be smooth manifolds with p in X and q in Y. Denote by COO(X, Y)P.Q the germs at p of mappings of X -7 Y which also map p to q. Recall that a germ at p is an equivalence class of mappings where two mappings are equivalent if they agree on a neighborhood of p. (We shall use the symbol [f]p to indicate the germ off: X -7 Y at p-at least at those times when pedagogy overwhelms natural instincts.) Let E be a vector bundle over X. Denote by C 00 (E)p the germs of smooth sections of E at p. Since sections are mappings of X -7 E this makes sense according to the above prescription. In particular, we can speak of germs of vector fields alongfas germs of sections off*(TY) using the identification of Ct(X, TY) with coo(f*(TY)) discussed in III, §I after Definition 1.4. Definition 1.1. Let f: X -7 Y, let p be in X, and let q = f(p) in Y. (a) the germ [f]p is infinitesimally stable iffor every germ ofa vector field along f, [TJp, there exist germs of vector fields mp in C 00 (TX)p and h]q in COO(TY)q so that (*) (b) f is locally infinitesimally stable at p if [f]p is infinitesimally stable. It is clear that if f: X -7 Y is infinitesimally stable, then f is locally infinitesimally stable. Choose coordinates Xl> .•. , Xn on X based at p and coordinate Yl, ... , Ym on Y based at q. We will compute equation (*) in these coordinates. If T is a vector field along ,f, then we can write T(X) = 2~ 1 Ti(X)(OjoyJ So equation (*) becomes (**) 111 112 Various Equivalent Notions of Stability where n 0 ~ = L ~j-' j=1 OXj m 0 7J = i~ 7Ji oy/ andfh" .,fm are the coordinate functions off Equations (**) can be solved to order k if for each set of germs Th ... , T m in CO'(Rn), there exists germs ~1"'" ~n in CO'(Rn) and germs 7J1,·· ., 7Jm in CO'(Rm) so that Theorem 1.2. Letf: X ---+ Y with p in X, q = f(p) in Y, and m = dim Y. Then [f]p is infinitesimally stable iffequations (**) can be solved to order m. Proof We shall use the Generalized Malgrange Preparation Theorem. First note that coo(f*TY)p = EB\"=1 C;'(X) since T(X) = L\"= 1Ti(X)(OjOYi) (locally). Thus COO(f*TY)p is a finitely generated C;'(X) module. Let A = {(df)(~) I~ E COO(TX)p}. A is a submodule of coo(f*TY)p so that Ml = coo(f*(TY))pjA is a finitely generated CpOO(X)-module. Via f* we can view Ml as a CqOO( Y)-module. Let ei be the projection off*(ojoYi) in Ml. We first observe that [f]p is infinitesimally stable iff eh ... , em generate Ml as a C:'( Y)-module. Recall that Jlp m+ leX) consists of germs of functions at p whose Taylor series start with terms of order m + 1 or greater (IV, Lemma 3.9). Now apply IV, Theorem 3.10 to see that [f]p is infinitesimally stable iff the module Ml/.;Itpm+l(X)M/ is generated over CqOO(Y) by the projections of e1, ... , em. This last statement is easily seen to be equivalent to solving equations (**) to order m, for if [T]p is in coo(f*TY)p, then T = L\"=l (7Ji·f)ej + (df)(~) + g where ~ and 7J are defined as usual and g is in .;Itpm+l(X)COO(TY)q; i.e., g is O(lxlm+1). 0 Theorem 1.2 makes it clear that whether or not f is infinitesimally stable at p is determined by r+Y(p). We shall formalize this notion. Let E be a vector bundle over X and let Jk(E) = {a EJk(X, E) Ia is represented by a section of E} = k-jet bundle ofsections ofE. Let 7T : E ---+ X be the projection. Then 7T* : Jk(X, E) ---+ Jk(X, X) is a submersion. Let I be the submanifold of Jk(X, X) given by {a EI Ia is represented by idx}. Then Jk(E) = (7T*)-1(I) and is thus a submanifold of Jk(X, E). The source map a: Jk(X, E) ---+ X restricts to a map ofJk(E) ---+ X. It is not hard to see that this is a fiber map. Let Jk(E)p = fiber of Jk(E) at p. This has a natural vector space structure. In fact, given two elements ofJk(E)p we can find sections that represent them. Add these sections and take the k-jet of the sum. We let the reader check that this operation is well-defined; i.e. independent of the choice of sections and that this gives Jk(E) a vector bundle structure over X. Hint: Do this first for the trivial bundle whose sections are just maps of X ---+ Rn. The following is just a restatement of Theorem 1.2. §1. Another Formulation of Infinitesimal Stability 113 Corollary 1.3. Letf: X -+ Y with p in X, q = f(p) in Y, and m = dim Y. Then f is locally infinitesimally stable at p iff Jm(f*TY)p = (dj)pJm(TX)p + f* Jm(TY)q where (df)p and f* are the obvious mappings into Jm(f*TY)p induced by the action of(df) andf* on vector fields. We wish to obtain conditions analogous to those in Corollary 1.3 which will be equivalent to infinitesimal stability. We have local conditions but these are not sufficient; for what happens at the self intersections of a function is not taken into account. In particular, the choice of 7] in Jm(TY)q might be forced in two different ways at two different pre-image points. We shall extend our results to take care of these cases. Letf: X -+ Ybe smooth and let q be in Y. Let S = {Pl, ... , Pk} C f-l(q). Define C;'(X) = ffif=l Cp":(X) and note that Cs""(X) is a ring where the operations are done coordinatewise. Since f induces a ring homomorphism f*: Cq""(Y) -+ Cp":(X) for each i, it induces a ring homomorphism of Cq""(Y) -+ C;'(X) which we also denote by f*. So if A is a C;'(X)-module, then via f* A becomes a Cq""(Y)-module. We reformulate the Generalized Malgrange Preparation Theorem so that it is applicable to these modules. Lemma 1.4. Let Ai (1 :0:; i :0:; k) be a finitely generated C;:(X) module. Then A = Al EB···EB Ak is a finitely generated C;'(X) module (where the action of Cp":(X) on Aj for i =f j is zero). Let elo ... , em be in A so that the projections of the e/s in AJAq( Y)A span this vector space. Then el, ... , em generate A as a Cq- Y be smooth and S = {PI' ... ,Pk} C f-I(q). Then f is simultaneously locally infinitesimally stable at PI>···, Pk iff Jm(f*TY)s = (df)Jm(TXh + f*Jm(TY)q. Proof For S consisting of a single point this result is given by Corollary 1.3. The proof for general S is exactly as in the single point case except that we substitute Lemma 1.4 and the subsequent remark for the Generalized Malgrange Preparation Theorem. In particular choose coordinates on X at PI, ... , Pk (with disjoint domains) and coordinates on Y at q and write down the equations generalizing (**) to solve the local infinitesimal stability condition simultaneously to order m at PI> ... ,Pk. Let M/ = ffi~=1 M/i. Continue as before-except for the substitution of Lemma 1.4 and its subsequent remark. 0 Theorem 1.6. Let f: X ->- Y be smooth. Then f is infinitesimally stable iff (t) for every q in Yand every finite subset S off-I(q) with no more than (m + 1) points The necessity part of this theorem is obvious. Before proving the sufficiency we need some preparatory lemmas. Lemma 1.7. Let HI, ... , Hk be subspaces of a finite dimensional vector space V. Then HI, ... , Hie are in general position (see III, Definition 3.5) (If (*) given VI> ••. , Vk in V, there exists hi in Hi and Z in V such that Vi = Z + hi for all i. Proof Let l7i: V -i>- V jHi be the natural projection and let 17: V-i>- VjHIEB···EB VjHk be given by l7(V) = (l7I(V), ... , l7k(V)). Clearly Ker 17 = HI n· .. n Hie so that the sequence o-i>- HI n··· n Hk -i>- V ~ VjHI EB· .·EB VjHk is exact. Now 17 is onto iff codim (HI n· .. n Hie) = 2~=1 dim VjHi iff HI, ... , Hie are in general position. But clearly 17 is onto iff given VI, ... , Vk in V, there exists z in V such that l7i(Z) = Vi> i.e. there exists hi in Hi so that Vi = Z + hi. So 17 is onto iff condition (*) holds. 0 Lemma 1.S. Letf: X -i>- Y satisfy (t) and let S = {PI> ... , Pie} C f-I(q). Let Hi = (df)p,(Tp,X) for 1 ::::; i ::::; k. Then HI, ... , Hk are in general position as subspaces of Tq Y. Proof Let ZI, ... , Zk be in TqY. By Lemma 1.7 we must show that there exist hi in Hi and y in Tq Y so that Zi = hi + y for all i. Choose a vector field r alongf so that r(pi) = Zi. By (t) choose vector fields ~ on X and 7J on Y so §1. Another Formulation of Infinitesimal Stability 115 that T = (df)(~) + f*Tj on a neighborhood of S. (Use Proposition 1.5.) Let hi = (df)pl~i) and y = Tjq. 0 For our current purposes we shall call p in X a critical point off: X --+ Y if (df)p : TpX --+ Tf(p) Y is not onto. Thus a critical point is either a singularity in case dim Y;::: dim X or an arbitrary point if dim X < dim Y. Lemma 1.9. Letf: X --+ Y satisfy (t), let m = dim Y, and let q be in Y. Then the number of critical points in f-l(q) is -:s; m. Proof We shall argue by contradiction. Suppose S = {Pb ... , Pm + l} consists of distinct critical points off in f-l(q). Let Hi = (df)p,(Tp,X). The last lemma states that Hl, .. ., Hm+l are in general position as subspaces of TqY. Thus m ;::: codim (Hl (\ ... (\ Hm+l) = L:f;} codim Hi ;::: m + 1. The last inequality holds since if P is a critical point codim (dfMTpX) ;::: 1. 0 Proof of Theorem 1.6. Sufficiency. We assume that f satisfies (t). Let I: be the critical point set of f in X and let I:q = I: (\f-l(q). By the last lemma I:q is a finite set with -:s; m points. Let T be a vector field along X. To prove that f is infinitesimally stable we need to show that there exist vector fields ~ on X and Tj on Y so that T = (df)(~) + f*Tj. We first show that this equation can be solved on a neighborhood of I:. We claim that there exist open sets Ul, ... , UN in X, Vb ... , VNin Y, and Wl, ... , WNin Y; and vector fields ~i on Ui and Tji on Vi satisfying (a) f(I:) c Uf"=1 w; (b) f(Ui) C Vi (c) T = (df)(~i) + f*Tj on Ui (d) f-l(Wi) (\ I: C Ui> and (e) Wi C Vi. One need only construct U, V, W, ~, and Tj for a given q inf(I:) (i.e., W must be a neighborhood ofq) satisfying (b)-(e). Sincef(I:) is compact, the necessary N will exist. By Proposition 1.5 we may choose open neighborhoods U of I:q in X and V of q in Y, and vector fields ~ on U and Tj on V so that (b) and (c) are satisfied. Next we choose W satisfying (d). Since I:q = f-l(q) (\ I: C Ui> there is a small open neighborhood W so that f-l( W) (\ I: CUi. If not there would exist a sequence of critical points Xl, X2, ... with Limi~ '" f(Xi) = q and Xi ~ Ui for all i. Since X is compact we may asume that Xi --+ x. By continuity f(x) = q and X is a critical point. Thus X is in I:qo A contradiction since X is in X - U and (X - U) (\ I:q = 0. By shrinking Wwe may assume thatf-l(W) (\ I: c U and that Wi c V. Next choose a partition of unity Pl, ... , PN on W = Uf"=1 Wi with supp Pi C Wi. Choose an open neighborhoodZ ofI: such thatf-l(Wi) (\ Z C Ui. (This is possible. Otherwise there exists a sequence Xl, x2 , ••• converging to X in I: with Xi in f-l(Wi) (\ (X - U). Since both f-l(Wi) and X - U are closed X is inf-l(Wi) and X is not in U. Contradiction.) Choose a smooth function P: X --+ R such that supp P C Z and P == 1 on a neighborhood ofI:. Let ~ = L:f"=1 pf*(Pigi. This is well-defined on X since supp Pf*(Pl) C 116 Various Equivalent Notions of Stability i-l(Wi) n Z CUi' Next let 7) = 2:f=l Pi7)i' Then calculate that T = (df)(O + f*7) on a neighborhood of~. Thus we may assume that T == 0 on a neighborhood U of ~ in X. If dim X < dim Y, then we are finished for ~ = X. In case dim Y ::::: dim X, thenfis a submersion on X - ~. Thus there exists a vector field ~ on X - ~ so that (df)(~) = T. Use III, Proposition 2.1 which states that all submersions are infinitesimally stable. (Note the fact that X is compact was not used in that proof.) Choose a smooth function P: X.......,.. R which is zero on ~ and 1 off U. Then p~ is globally defined on X and (df)(pO = T since supp T C X- U. 0 We now attack the last of our three objectives. Proposition 1.10. Letf: X.......,.. Y be infinitesimally stable. Then there exists a neighborhood W off such that every g in W is locally infinitesimally stable. Proof Let p be in X and let q = f(p). Since f is infinitesimally stable Jm(f*TY)p = (df)Jm(TX)p +1*Jm(TY)q. Now consider the mapping 1: Jm(TX)p (£; Jm(TY)q""""" Jm(f*TY)p given by (df) +1*. This is just a linear mapping between vector spaces. In particular, if we choose chart neighborhoods U of p and V of q such that f( U) C V, then in these local coordinates jis a linear mapping of B':.n (£; B:::.m.......,.. B':.m where B~.m is the vector space of polynomials ofdegree::::; k from Rn.......,.. Rm. Now clearlyjdepends continuously on p and f (in fact on jm+ If). Thus there is an open neighborhood Up of p and an open neighborhood Wp off such that if g is in Wp then g(U) c V (a Co condition) and ifp' is in Up, then g is onto at p'. Thus g is infinitesimally stable at p' using Corollary 1.3. Since X is compact there is a finite covering of Xby Up's. The intersection ofthe corresponding Wp's is an open neighborhood off with the desired properties. 0 We would like to prove a corresponding theorem for (global) infinitesimal stability; unfortunately there are some difficulties. We have the global result given in Theorem 1.6 to use instead of Corollary 1.3 but the proof above will not work for X(s) is not compact (even though X is compact). To avoid these difficulties for the moment we make the following definition. Definition 1.11. An infinitesimally stable mapping f: X.......,.. Y satisfies condition (!) iffor every p in X, there exists a neighborhood Up of p and a neighborhood Wp off such that if g is in Wp and if S = {Pl, ... , Ps} c Up n g-l(q), then Jm(g*TY)s = (dg)Jm(TXh + g*Jm(TY)q. The object of this definition is to finesse-for the moment-the question of what happens in X(s) near the generalized diagonal. This is shown by the following. Lemma 1.12. Iff: X.......,.. Y is an infinitesimally stable mapping satisfying condition (!), then there exists an open neighborhood W off which consists of infinitesimally stable mappings. §1. Another Formulation of Infinitesimal Stability 117 Proof We will prove the lemma by applying Theorem 1.6. In order to make the notation a little easier to follow we shall prove that there is a neighborhood W2 off such that each g in W2 satisfies (t) for every finite subset S consisting of two points. A similar proof will yield an open neighborhood Wr of f consisting of mappings satisfying (t) for all sets S consisting of r points. The desired W is then given by nr=V Wr where W1is given by Proposition 1.10. Let (p, t) be in X x X. We will choose neighborhoods Up of p, Vt of t, and Wp,t off such that if g is in Wp,t, x is in Up, and y is in Vt with g(x) = g(y) = q, then Jm(g*TY){x,y} = (dg)Jm(TX){x,y) + g*Jm(TY)q. If p = t, then the relevant data (with Vt = Up) is given by condition @. If p i= t and f(p) i= f(t), then we may choose Up, Vto and Wp,t so that g(Up) n g(i7t) = 0 for every g in Wp,t. Finally if p i= t and f(p) = f(t) = q, we may choose disjoint neighborhoods Up and Vt and a neighborhood Wp,t so that TXI Up and TXI Vt are trivial and g(Up) U g(i!;) C Zq where Zq is an open neighborhood ofq in Ywith TYIZq trivial for every g in Wp,t. Sincefis infinitesimally stablel: Jm(TX)\p,t} EEl Jm(TY)q -)- Jm(f*Ty){p,t} is onto where! is the linear mapping between these finite dimensional vector spaces given by (d!) + f*. Since TXI Up, TXI Vto and TYI Zq are trivial we may identify!with a mapping of B':,n EEl B':,n EEl B;;:,m -?o- B':,m which depends continuously on p, rand f as long as the perturbation off is in Wp,t. Now the sets Up x Vt cover X x X which is compact. So there exists a finite subcover given by the Up x Vt's. Intersecting the corresponding Wp,t's gives the desired open neighborhood W2 off 0 Exercises (1) Consider the mapping f: R2 -?o- R2 defined by (x, y) 1--+ (x, xy _ y3). Try showing thatf is infinitesimally stable using only the definition of infinitesimal stability. In doing so you should get to a functional equation which is rather difficult to solve on a nbhd of the origin, i.e., for every pair of smooth functions 71, 72 : R2 -?o- R there exist smooth functions g10 g2, 7)10 and 7)2: R2 -?oR such that (*) [ 71 = gl + 7)1 • f 72 = yg1 + (x - 3y2)g2 + 7)2 • f Use Theorem 1.2 to show that [flo is infinitesimally stable by solving equations (*) to order 2. Try showing that f is infinitesimally stable by applying Theorem 1.6. (Sincefis proper the theorem is still valid even though X = R2 is not compact.) (2) Letf: R2 -?o- R3 be given byf(x, y) = (x, xy, y2). Show thatfis locally infinitesimally stable at O. (3) It is possible to reduce even further the calculations needed to compute local infinitesimal stability using the following observation due to Arnold. Let f: Rn -?o- Rm and assume that f(O) = O. Say that f satisfies con- 118 Various Equivalent Notions of Stability dition (Z) if for every germ

condition (Z). Let TI = (0, ... , (**), let T = (Tl, •.. , Tn,) and solve TZ = (df)H + K 0 f Let ti = 2.'J'= 1 YJij and YJi = 2.'J'=1 kij where H = (hiJ and K = (kij). Then let t = (tl>"" tn) and YJ = (YJl>"" YJm)· (b) Show that if condition (Z) can be solved for - Y be smooth and let p be in X. Choose coordinates Xl, ... , Xn on X at p. Then f is locally infinitesimally stable at p iff equations (**) are solvable to order m for TZk = (0, ... , Xb ... , 0) where the X k appears in the lth position for 1 ~ I ~ m and 1 ~ k ~ n. (4) Let f:Rn-J>-Rn be given by (Xl>""Xn)I-+(X1"",Xn-l,XIXn + X 2X n2 + ... + Xn_1Xnn-1 + xnn+1). Show that f is locally infinitesimally stable at 0. §2. Stability Under Deformations Our intention in this chapter is to prove (under the assumption that the domain is a compact manifold) that stability is equivalent to infinitesimal stability. To accomplish this task it seems necessary to introduce several other notions of stability-all of which are, in fact, equivalent to the ones just mentioned. The most natural of these is the concept of stability under deformations introduced by Thorn and Levine. Definition 2.1. Let f: X -J>- Y be smooth and let Is = (- e, e). Then (a) let F: X x Is -J>- Y X Is be smooth. F is a deformation off if (i) for each s in (- e, e), F: X x {s} -J>- Y X {s}. Denote by Fs the mapping of X 1-+ Y defined by F(x, s) = (FsCx), s). (ii) Fa = f (b) Let F: X x Is -J>- Y X Is be a deformation off Then F is trivial if there exist diffeomorphisms G: X x 10 -J>- X x 10 and H: Y x 10 -J>- Y x 10 §2. Stability Under Deformations 119 (where 0 < I) :::; e) such that G and H are deformations of idx and idy, respective/y, and such that the diagram x X 10 F I Y X 10 Gl lH X x 16 f X id10 I Y X 16 commutes. (c) f is stable under deformations (or homotopically stable) if every deformation off is trivial. Remarks. (1) In a trivial deformation F of f each Ft is equivalent to f Also, iff is stable then for a given deformation F each Ft (for t small enough) is equivalent to f Unfortunately, this is not enough to show that F is trivial, since the conjugating maps need not vary smoothly. (2) By viewing a deformation of f as a mapping of Ie 1-+ cro(X, Y) we can equate deformations offwith curves in cro(X, Y) based atf Recall that in Chapter III, §1 we motivated the definition of infinitesimal stability in terms of Frechet manifolds. In particular, we showed thatf is infinitesimally stable iff (d')lr)id is onto where ')Ir: Diff(X) x Diff( Y) -+ cro(X, Y) is defined by ')IrCg, h) = h·f·g-1. Certainly (d')lr)id is onto if for every curve t 1-+ Ft in cro(X, Y) where Fo = f, there is a curve t 1-+ (Gb Ht) in Diff(X) x Diff(Y) so that ')Ir(Gb Ht) = Ft for all small t; i.e., for small t the diagram commutes. But this is just the condition that the deformation F be trivial. So it should come as no surprise that we will show that homotopic stability implies infinitesimal stability. In fact, they are equivalent notions and this also will be shown later. (3) As for the relationship between stability under deformations and stability, the only fact which can be immediately proved is the following: Lemma 2.2. Let f: X -+ Y be stable under deformations. Suppose there exists an open nbhd W off in C ro (X, Y) such that each g in W is stable under deformations, then f is stable. Proof By shrinking, if necessary, we can assume that W is "arc-wise connected" ; that is, for each g in W, there is a deformation F: X x [- 1, 1] -+ Y x [-1, 1] offso that F1 = g and Ft is in W for all tin [-1, 1]. (In III, 120 Various Equivalent Notions of Stability Theorem 1.12 we identified functions in a nbhd U of f with sections of a tubular nbhd of graph (I) in X x Y. For functions in U the deformation is obvious. Thus by "shrinking" we mean replacing W by W n U.) Now let g be in Wand choose such an F. Consider the equivalence relation on [-1, 1] defined as follows: s ~ t if Fs is equivalent to Ft as mappings of X -+ Y. The assumption that each mapping in W is homotopically stable along with Remark (1) implies that each equivalence class is open. Since [-1, 1] is connected there is only one equivalence class and thus g is equivalent to f. 0 So if we know that infinitesimally stable maps form an open set and that infinitesimal stability is equivalent to homotopic stability, then we would know that infinitesimally stable mappings are stable. Recall that in the last section we showed (in Lemma 1.12) that if every infinitesimally stable mapping satisfies condition @ then the set of infinitesimally stable mappings is open. We now generalize the concept of homotopic stability and show that a mapping which is both infinitesimally stable and satisfies this generalized homotopic stability criterion also satisfies condition @. Definition 2.3. Let f: X -+ Y be smooth and let U be a nbhd of 0 in Rk. (a) Let F: X x U -+ Y x U be smooth. F is a k-deformation off if (i) for each v in U, F: X x {v} -+ Y x {v}. Denote by Fv the mapping of X -+ Y defined by F(x, v) = (Fv(x), v). (ii) Fo = f (b) Let F: X x U -+ Y x U be a k-deformation off Then F is trivial if there exist diffeomorphisms G: X x V -+ X x V and H: Y x V -+ Y x V where V is an open nbhd of0 contained in U such that G and H are deformations of idx and idy respectively, and such that the diagram Xx V F ) Y x V Gl IH XX V ) Yx V f x idv commutes. (c) f is stable under k-deformations if every k-deformation off is trivial. Remarks. (1) Stability under I-deformations = homotopic stability. (2) Iffis stable under k-deformations, thenfis stable under I deformations for I s k. In particular, f is homotopically stable. Before giving our discussion of condition @ we make some preparatory lemmas. Ifg: Rn -+ R is smooth and K is a compact subset of Rn, then define Ialai IJJgJJsK = max a!(x) . XEK X O:s:lal:S:s §2. Stability Under Deformations 121 Lemma 2.4. Let p be in a convex compact subset K of Rn and let g: Rn -+ R be smooth. Let g(x) = L a..(x - p)a + Osla!ST L gp(x)(x - p)/3 IPI =T+ I be the Taylor expansion with remainder term oforder r + 1. Then if \\g\\/ < s, then \\gp\\:f-T-I < sfor r < sand \.8\ = r + 1. Proof Without loss of generality we may assume that p = O. We proceed by induction on r. For r = 0, g(x) = g(O) + :Lf=l xig!(x) where gi(X) = f~ (8gj8xi)(tx) dt. (See II, Lemma 6.10.) Thus I8~:!! (x) I~ fI~;~ :!!(tx) Idt < e for \ex\ ~ S - 1 since \ex\ + 1 ~ sand \\g\\sK < e. (Note that tx is in K since K is convex.) So \\gi\W-l < e. For the general case just note that if g(x) = L aaxa + L hp(x)xP, OslalsT-I IPI=T then by expanding hp(x) = hP(O) + :Lf= 1 gp,;(X)Xi we get the Taylor expansion of g to order r. Apply induction and the r = 0 case to obtain the desired result. 0 Let Anl be the vector space of polynomials of Rn -+ R of degree ~ I. Lemma 2.5. Let r ~ 0 and s > 0 be integers and let K be a compact convex nbhd of 0 in Rn. Let Z be an open nbhd of 0 in Anl where I = (r + I)'. Then there exists an e > 0 so that ifPI, ... , Ps are in K and if g: Rn -+ R is smooth and satisfies \\g\\fcr+l) < e, then there exists a polynomial v in Zsuch that Proof We prove the lemma by induction on s. Let s = 1 and let p = Pl' By Taylor's Theorem g(x) = L a..(x - py + L (x - p)agaCx) OSlalST lal=T+l where each ga is a smooth function. Let v = :Lo sial ... , Ps be distinct points in K. Again apply Taylor's Theorem to g and obtain g(x) = L aa(x - ps)a + L g..(x)(x - ps)a. OslalSr lal=r+l 122 Various Equivalent Notions of Stability IfIIg!lf ..., P. are assumed to be in K, and the coefficients of Va are bounded by a constant times e. Thus by choosing e small enough we can guarantee that V is in Z. Finally, we note that 811llg 8181 ( '" '" 8xll (Pi) = 8x/3 L. aaCx - p.)a + L. O,,;lal,,;r lal=r+l 811llv = 8Xll (pj) for 1 ~ i ~ sand 0 ~ 1,81 ~ r since the middle term of this equality depends only onp!> ... ,p., aa (lal ~ r), and (8IYlga/8xY)(Pi) (y ~ ,8) and the RHS of the equation depends in exactly the same way on these parameters. Thus the induction is proved. 0 Proposition 2.6. Suppose that f is infinitesimally stable and that f is stable under k-deformationsfor k large. Thenfsatisfies condition (!). Proof Let p be in X. Choose coordinate nbhds U of p in X and V of f(p) in Y such that f( U) c V. Let W be a Co open nbhd offsuch that if g is in W, then g(U) c V. Choose an open nbhd Up ofp such that Up is convex, compact, and contained in U. (By convex, we mean convex in the coordinates chosen on U.) Note that ifg is in W, then gl U can be thought of as a mapping of Rn --+ Rm. Define IIglV'Jp = maXl,,;t,,;m IIgiliaup where gl>' .. , gm are the coordinate functions of g. Let p : X --+ R be a smooth function which is 1 on a nbhd of Up and 0 off U. Let rand s of Lemma 2.5 both equal m + 1 so that I = (m + 2)m+l. Let B~,m be the polynomial functions of Rn --+ Rm; i.e., B~,m = EEl;"=l Ani. Let k = dim B~,m' We now define a k-deformation of F. For v in B~,m' let FvCx) = f(x) + p(x)v(x). Certainly F: X x B~,m --+ Y x B~,m is smooth and Fo = f so that F is a k-deformation. Sincefis stable under k-deformations there is a nbhd Z of oin B~,m on which F is trivial. Let W. = {g E W Ilig - f!l~+l)(m+2) < e}. First note that g - f makes sense since g(U) c V. Next note that W. is an open nbhd offin the cm2 +3m+2 topology on Coo(X, Y) and thus open in the Coo topology. Now we choose e > 0 by using Lemma 2.5 as follows: Choose e so that if P!> ..., P. are s distinct points in Up where s ~ m + 1 and if g is in W., then there exists a v in Z for whichr+l(g - f)(Pi) = r+lv(pj) for 1 ~ i ~ s. §3. A Characterization of Trivial Deformations 123 Finally we shall show that if g is in We> then g satisfies (t); that is, if S = {Pl, ... , Ps} c Up 11 g-l(q), then Jm(g*TYh = (dg)Jm(TXh + g*Jm(TY)q. Note that if this statement is true then f satisfies condition (!) for the choices Up and We. So let g, P1, ... , p., and q be given satisfying the conditions of (t). Since g is in We, there exists a v in Z so thatr+1g(Pi) = jm+1(f + v)(pj) for 1 :0:; i :0:; s. Since p == 1 on a nbhd of Up, jm+1(f + V)(Pi) = r+ 1Fv(Pi) for 1 :0:; i :0:; s. Since F is trivial on Z, Fv is equivalent (as mappings of X -+ Y) with f and since f is infinitesimally stable so is Fv. Thus (t) is satisfied by Fv at the points P1, ... , P., q. Now the equations in (t) depend only onjm + 1 Fv(Pi) (1 :0:; i :0:; s) so these same equations must be satisfied by g sincer+ 1FvCPi) = r+ 1g(Pi). 0 To summarize this discussion we have: Proposition 2.7. If infinitesimal stability is equivalent to stability under k-deformations for k large (e.g., k = dim Bit.m where 1= (m + 2)m+1), then infinitesimal stability implies stability. Proof By Proposition 2.6 all infinitesimally stable mappings satisfy condition (!). By Lemma 1.12 the set of infinitesimally stable mappings is an open set. Apply Lemma 2.2 and the hypothesis of this Proposition to see that infinitesimal stability implies stability. 0 §3. A Characterization of Trivial Deformations Let V be a nbhd of 0 in Rk and let t1, ... , tk be the standard coordinates on Rk. Definition 3.1. Let f: X -+ Y be smooth and let F: X x V -+ Y x V be a k-deformation off Define the vector field along F 7/ = (dF)(!) - F*(!) where 8/8ti is a vector field on X x Vor Y x Vas required. We now establish some notation. Let 7T: X x V -+ V and p: X x V -+ V be the obvious projections. Then T(X x V) = 7T*(TX) EEl p*(TV). So any vector ~ in T(X x V) can be written uniquely as ~ = ~x + ~v where ~x is in 7T*(TX) and ~v is in p*(TV). We call ~x the X-component of ~ and ~v the Rk_ component of ~ and denote ~x by 7T(~) and ~v by pm. Lemma 3.2. Let F be a k-deformation off Then F = f x idv iff 7Fi == 0 for 1 :0:; i :0:; k. In particular F is independent of ti if 7F i == o. Proof If F = f x idv, then (dF)(x,v)(!i I(x.v») = !iIF(X'V) 124 Various Equivalent Notions of Stability So that 7Fi = O. So assume that 7Fi = 0 for all i. Fix (xo, va) in X x V and choose coordinates Xl> ... , Xn near Xo in X and Yl, ... , Ym near Fvo(xo, va) in Y. In these coordinates we may write F(x, v) = (Fl(x, v), ... , Fm(x, v), v). Then (dF)(x,v)(: I )= (~Fl (x, v), ..., o!'m (x, V)) + : I .uti (x,V) uti uti uti F(x,v) Thus 7/ == 0 implies that of)ot;Cx, v) == 0 for 1 ::;; j ::;; m. Thus Fix, v) is independent of ti for all i and Fix, v) = Fix) = fix) where f = (fl> ... ,1m) in these coordinates. Since (xo, va) is arbitrary we find that globally F(x, v) = f(x). 0 Theorem 3.3 (Thom-Levine). Let/: X --+ Y be smooth and let F: X x V --+ Y x V be a k-deformation off Then F is trivial iffthere exists an open nbhd U of 0 in V and vector fields ~i on X x U and r/ on Y x U (for 1 ::;; i ::;; k) satisfying (a) pW) = 0 = p(r/), and (b) 7Fi = (dF)W) + F*(r/) on X x U. Proof Necessity. Assume that F is trivial, then there exists a nbhd U and diffeomorphisms G: X x U --+ X x U and H: Y x U --+ Y x U satisfying Definition 2.3(b). First we note that for any deformation K the Ric-component of (dK) (·~I )-~Ip otj p - oti K(p) since p. K = p. Thus (dK)p ( ~j \J = ~j IK(P) + 7T(dK)p (~iIJ. Now, by assumption, F = H·(f X idu)·G-l where all the mappings are k-deformations. Let p be in X x U and let r = (f x idu)·G-l(p). Then compute (*) (dF)p(~j IJ = ~i Ip + 7T(dH)r(~J) + (dHMdf X idu)a-l(p)7T(dG-l)p(~j IJ· Let ~pi = (dG)a-1(p)7T(dG-lM(%ti)lp)' Thus ~i is a vector field on X x U and pW) = p.7T(dG-l)(O/ot j) = 0 since G is a deformation and p.7T = O. Now insert (dG)p -l.(dG)a-\p) before 7T in the last term of the RHS of(*) to obtain (dF)p(OO I)= 00 I + 7T(dH)r(ool ) + (dFM~pi).ti p ti F(p) ti p Thus (**) 7Fi(p) = (dF)p(: I)-00 I = 7T(dH)r(oo I) + (dFM~pi).uti p ti F(p) ti r §3. A Characterization of Trivial Deformations 125 Define TJql = 7T(dH)H-1(Q)«0/otl)IH-1(Q» where q is in Y x U. Clearly TJI is a vector field of Y x U and the Rk-component of TJI is zero. Substituting TJ in (**) we see that 'TFI(p) = (dF.M~pl) + TJk(r)' But H(r) = H.(j·idu)·G-l(p) = F(p). SO 'TFI = (dF)(~I) + F*(TJt). 0 Before proving the sufficiency part of the theorem we make some preparatory calculations. Lemma 3.4. Let ~ be a compactly supported vector field on X x Rk such that the Rk-component of~ is zero. Then there is a di./feomorphismg: X x Rk-+ X X Rk which is a deformation of idx satisfying (*) Proof. Since ~ is compactly supported and %tk has a globally defined one parameter group, I, Corollary 6.5 guarantees that ~ + %tk has a globally defined one parameter group ~ : (X x Rk) X R -+ X X Rk. Let ek = (0, ... ,0,1) in Rk. We claim that~.: X x {v} -+ X x {v + sek}' Let p = (x, v) be in X x {v}. Then (dp) ~ + - =-(( 0 ) 1 ) 0 1P otk p- otk p(p) since the Rk-component of ~ is zero. Since ~ + (a/atk) is the infinitesimal generator of ~, the curve s I-l>- ~s(p) is an integral curve for ~ + o/atk and thus represents the vector (~ + %tk)l",s(p) for each s. Thus (d/ds)p(~s(p» = ek and p(~s(p» = sek + p(~o(p» = v + sek' This proves the claim. Next, define g: X x Rk -+ X X Rk by g(x, v) = ~Vk(X, V - Vkek) where v = (Vb' .. , Vk)' Then g: X x Rk -+ X X Rk is a smooth mapping Since g: X x {v} -+ X x {v} and g(x, 0) = ~o(x, 0) = (x, 0), g is a deformation of idx. Note that glX x {v} is a diffeomorphism since hex, v) = ~-vJx, v + Vkek) is the smooth inverse ofglX x {v}. To see that g is a diffeomorphism we need only show that Ker (dg) n p*(TRk) = {O}. But this is clear since g is a deformation; i.e., (dg)(o/at!) = o/at! + 7T(dg)(o/att) :1= O. Finally we compute (*). The curve s I-l>- (x, V + sek) represents (a/otk)I(x,V) so the curve s I-l>g(x, V + sek) = ~Vk+'(X, V - Vkek) = ~.(~vix, v - Vkek» = ~.(g(x, v» repre- sents (dg)(x,v) - = ~ + - .( a 1 ) ( a)1otk (x,V) otk g(x,V) Lemma 3.5. Using the same notation as in Lemma 3.4, we have that (i) ~ = 7T(dg)(g-I)*(O/oti)' and (ii) ~ = -(dg)7T(dg-1)(a/ott). o Proof. (i) follows trivially from Lemma 3.4 since 7T(O/Ott) = O. Applying (dg)p -1 to both sides of Lemma 3.4 we have that 126 Various Equivalent Notions of Stability Thus Since the Rk-component of ~p = 0 so does the Ric-component of (dg-l)p(~p). So 7T(dg)p -l(~p) = (dg)p -l(~p). Apply (dg)g-l(p) to obtain (ii). 0 Proof of Theorem 3.3. Sufficiency. Let F: X x V --+ Y x V be a kdeformation of f and let ~i and 1]i be vector fields on X x V and Y x V respectively, such that the Ric-components of ~i and 1]i are zero and T/ = (dF)W) + F*(1]i) on X x V. We must show that F is trivial. Since X is compact ~i is trivially compactly supported. By shrinking V we may assume that V is compact and that T/ = (df)(~i) + F*1]i on X x V. We can then damp 1]i to zero off a compact nbhd of F(X x V) and assume that 1]i is compactly supported. Apply Lemmas 3.4 and 3.5 to show the existence of diffeomorphisms G: X x V -;. X x V and H: Y x V --+ Y x V so that and 1]k = -(dH)7T(dH-l)(8~J' Let M = H-l·F·G and let p be in X x V with q = G(p) and r = F·G(p). Then (dM)p(8~JJ = 8~JM(P) + 7T(dH-l)r(8~J) + (dH-l)r7T(dF)q(8~k IJ + (dH-IMdF)q7T(dG)p(8~1cIJ using the fact that H, F, and G are deformations. So (dM)p(8~k IJ = 8~1c IM(P) + (dH-1)r( - 1]/ + 7T(dF)q(8~1c IJ -(dFM~qk)) Now TFk(q) = (dFM~qlc) + 1]/ by assumption. Hence TMk(p) = (dM)p(8~" IJ -M*(8~J = (dH-1)r( - TFk(q) + 7T(dF)q(8~k IJ) =0 since §4. Infinitesimal Stability => Stability 127 Applying Lemma 3.2 we see that M is, in reality, a (k - 1) deformation trivially extended to a k-deformation. Thus if we can show that 7Mi (1 :0; i :0; k - 1) can be written in the form 7Mi = (dM)(~i) + M*(iji) where ~i and iji are vector fields whose Rk-components are zero, then we will be able to use induction to conclude that F is trivial. Now note that (*) 7Mi ·G = (dH)7pi - (dM)7T(dG)(8/8t). For (dH)(7pi) = (dH)7T(dF)(8/8ti) = 7T(dH)(dF)(8/8ti) since Hand Fare deformations. So (dH)(7pi) = 7T(dM)(dG)(~) = 7T(dM)(~i la) + (dG)(~) = 7T(dM)(~i la) + (dM)7T(dG)(~} This proves (*) since 7T(dM)(8/8ti)1 a) = 7Mi .G. From (*) we see that to show that 7Mi has the desired form it is sufficient to show that (dH)7/ has the desired form. Finally we compute (dH)(7pi) = (dH)(dF)W) + (dH)(7]ilp), Define ~a(p) = (dGM~pj) and ij/ = (dH)H-1(Q)(7]k-1(Q)' Then ~ and ij are vector fields whose Rk-components equal zero. Moreover, o Exercises It is possible to use the Thorn-Levine Theorem to prove that certain mappings are homotopically stable. For example, show that: (1) Submersions are stable under deformations and (2) 1: 1 immersions are stable under deformations. §4. Infinitesimal Stability => Stability Proposition 4.1. Let /: X -+ Y be stable under k-de/ormations,' then / is infinitesimally stable. Proof. Since/ is stable under k-deformations, / is homotopically stable. To show that/is infinitesimally stable we must produce for each vector field 7 along/vector fields ~ on X and 7] on Y so that 7 = (d/)(O + /*7]. Consider Xf = graph/in X x Y. We can view 7 as a vector field on Xf pointing in the Y-direction as follows: T(P,f(P)) = 7p in {O} EEl Tf(p) Y C T(p.f(p))(X x Y). Extend T to a compactly supported vector field on X x Y. (This is possible since Xf is closed in X x Yand thus has a tubular nbhd Z. Trivially translate T(p.f(p)) along the vector space fiber of Z at (p,f(p» and damp-out off a compact nbhd of 0 in this vector space. This can clearly be done smoothly.) 128 Various Equivalent Notions of Stability Next let CPt: X x Y ~ X x Ybe the one parameter group whose infinitesimal generator is T. Finally define F: X x R ~ Y x R by F(x, t) = (7Ty·CPt(X,J(x», t) where 7Ty: X x Y ~ Y is the obvious projection. Clearly F is smooth and is a deformation of f since F(x, 0) = (7Ty(X,J(X», O) = (f(x), 0). Since f is stable under deformations there exist vector fields { on X x 16 and Tj on Y x 16 (where I) > 0) whose R-components are zero satisfying TF = (dF)({) + F*Tj on X x 16 by Theorem 3.3. Restrict this equation to X x {O} to obtain TFlxx{o} = (df)W +f*'TJ where ~p = {(P,O) and 'TJq = Tj(q,O) define vector fields on X and Y respectively since the R-components of { and Tj are zero. Finally we compute TFI(p,o) = 7T(dF)(p,o)«8/8t)l(p,o»' The curve t ~ (p, t) represents (8/8t)l(p,o) so that t ~ 7Ty·cpt(P,J(p» represents TFI(p,o)' Now ~7Ty·cpt(p,J(P»lt=o = (d7TY)(P,f(P»(TI(p,f(P») = Tp since TI(P,f(P» points in the Y-direction. Thus TFI(p,o) = Tp. 0 The proofofthe fact that stability under deformations implies infinitesimal stability is a calculation involving nothing deeper than the global integration of certain vector fields. This is not true for the converse statement. As we shall see, the proofofthis implication uses the generalized Malgrange Preparation Theorem and is quite similar in spirit to the proof of the formulation of infinitesimal stability given in §1. Theorem 4.2. Let f: X ~ Y be infinitesimally stable, then f is stable under k-deformations for all k. Let F: X x V ~ Y x V be a k-deformation off We need to show that F is trivial. By applying Theorem 3.3 we see that it is enough to find a nbhd U of 0 with U C V and vector fields ~ on X x U and 'TJ on Y x U such that the Rk-components of ~ and 'TJ are zero and TF = (dF)W + F*'TJ on X x V. First we prove that ~ and'TJ exist locally. Proposition 4.3. Iff is infinitesimally stable at p, then there exists germs ofvector fields ~ and'TJ with Rk-components equal to zero such that [TF](P,O) = (dF)m(p,o) + ['TJlf(p),o). Proof Let NFP == N == {germs of vector fields T: X x Rk ~ T(Y X Rk) along Fat (p, 0) IRk-component of T = O}, and let AFP == A == NjK where K = {(df)[np,o) I~ is a vector field on X x Rk with Rk component = O}. There is an obvious action of q;;,O)(X x Rk) on N given by multiplication. Thus N is a module over C(';;,O)(X x Rk) and is finitely generated. For if we choose coordinates Xl>"" Xn based at p on X and Y1,' •• , Ym based at f(p) on Y, then every vector field along F whose Rk-component is zero can be written as L: TtCX, t) - .m 81 1=1 8YI F(X,t) §4. Infinitesimal Stability => Stability 129 Thus the vector fields along F, F*(8/8Yi) are generators of the module N. Thus A is a finitely generated module over q;,O)(X x Rk). Finally we note that via F*, A is a module over q:,O)( Y x Rk) where q = f(p). We claim that A is also a finitely generated Ct:.,o,( Y x Rk) module with a set of generators given by ei = projection of F*(8/8YI) in A. It is in proving the claim that we shall use the fact thatfis infinitesimally stable at p. First we show that the claim is sufficient to prove the Proposition. In A, [TF](V,O) = i ['TJiF*(88y )] . 1=1 1 (v,O) Thus in N, [TFlv,o) = (dF)[nv,o) + F*[i 'TJI c;,8y ] i= 1 (J 1 (V,O) where t has Rk-component equal to zero since (dF)[nv,o) is in K. Now 'TJ = Lf'=l 'TJi(8/8YI) clearly has Rk-component equal to zero so TF = (dF)(t) + F*'TJ on the germ level near (p, 0). Apply the obvious local form of Theorem 3.3 to prove the proposition. To prove the claim we shall use the Malgrange Preparation Theorem (IV, Theorem 3.6). Using Taylor's Theorem write k (*) T(X, t) = TO(X) + 2: tITI(X, t). 1=1 Since T is a vector field along F, TO is a vector field alongf. Sincefis infinitesimally stable, there exist vector fields t on X and 'TJ on Y such that TO = (df)(0 +f*'TJ. Extend t and 'TJ trivially to be vector fields on X x Rk and Y x Rk and apply Taylor's Theorem again to obtain k (**) TO(X) - «dF)(0 + F*'TJ)(x, t) = 2: tIT;(X, t). 1=1 Substituting (**) in (*) we obtain k (***) T(X, t) = [df(t) + F*'TJ](x, t) + 2: tIT~(X, t). 1=1 Next we consider the vector space A/(tl, ... , tk)A. The equivalence class of T in A/(tl , ••• , tk)A is F*'TJ. Now F*'TJ = Lf'=l ('TJI'f)(8/8YI)IF since'TJ is the trivial extension of a vector field on Y to Y X Rk. So the projections of F*(8/8YI), ... , F*(8/oYm) generate the vector space A/(tl> ..., tk)A. Finally consider the vector space A/.Ac.q,o,(Y x Rk)A. Since (tI. ... , tk) C .Ac.q,O)( Y x Rk) there is a natural projection of A/(tl, ..., tk)A onto A/.Ac.q,O)(Y x Rk)A so that e1 , ••• , em generate this last vector space. Now apply the Malgrange Theorem to obtain the desired result. 0 Corollary 4.4. Let f be infinitesimally stable and let F: X x V -+ Y x V be a k-deformation off. Let S = {Pl> ... , P.} C f-l(q). Then there exists a nbhd U ofS x {O} in X x Rk and vector fields t on X x Rk and'TJ on Y x Rk such that the Rk-components of t and'TJ are zero and TF = (dF)(t) + F*'TJ on Xx U. 130 Various Equivalent Notions of Stability Proof In the case that S is a single point, this is just Proposition 4.3. The proof proceeds precisely as the proof of Proposition 4.3 substituting Aps = AFP1 EEl···EEl AFPk for AFP and using Lemma 1.4 rather than the Malgrange Preparation Theorem. 0 Our next step is to show that TF can be written in the desired form on a nbhd of the critical point set off Let ~q denote the critical points of fin f-l(q) and let ~ = UqEY ~q denote the critical point set off Note. p is a critical point offif (df)p : TpX --+ Tf(p) Y is not onto. Recall Lemma 1.8 which showed that ~q is a finite set. Proposition 4.5. Let X be compact, letf: X --+ Y be infinitesimally stable, and let F be a k-deformation off Then there exist vector fields ~ on X x Rk and 7J on Y x Rk with the Rk-component of ~ and 7J equal to zero on a nbhd B of~ x {O} in X x Rk such that TF = (dF)W + F*7J on B. Proof We claim that there exist open sets Ul, ... , UN in X; VI' ...' VN in Y; and WI' ... , WNin Y; and e > 0 satisfying (a) f(~) c Uf=l Wi (b) Wi C Vi (c) for all v E Rk with Ivl < e, Fv -l(Wi) n ~ C Ui where F(x, v) = (Fv(x), v) (d) for Ivl < e, Ui C Fv -leVi) (e) there exist vector fields ~i on Ui X Be and 7Ji on Vi X Be with Rk_ components equal to zero (where Be = {v E Rk I Ivl < e} so that TF = (dF)(~i) + F*7Ji on Ui x Be. We need only verify the choices at each point q inf(~). The finiteness then follows sincef(~) is compaet. Since ~q is a finite set we may apply Corollary 4.4 for S = ~q and gain the existence of U, V, ~, 7J, and e satisfying (e). Shrink U so that feU) C V. Now choose W satisfying (b) and (c) for f = Fo. By taking e smaller if necessary we may assume that (c) and (d) hold. Next choose a partition of unity PI, ... , PN on Uf=1 Wi with supp Pi C Wi and extend Pi to be =0 off Wi. Now there is a nbhd U of ~ such that f-l( Wi) n U C Ui for each i since ~ is compact and Ui is open. Choose a smooth function P on X such that supp pC U and P =1 on.a nbhd of~. Let ~ = 2:f=1 pF*Pi~i. (This is globally defined and smooth since supp (pF*Pi) C supp Pn supp F*Pi C Un F-l(Wi X Be) C Ui X Be. Hence pF*Pi~i can be extended to all of X x Rk trivially.) Let 7J = 2:f=l Pi7Jl. (7J is globally defined on all of Y x Rk.) Then calculate (dF)(~) + F*7J = (dF)(~ pF*Pi~i) + i~ F*(Pi7Jl) N = LF*Pi[(dF)(~i) + F*7J;] on a nbhd of~ x {O} 1=1 = TF· Let B be that nbhd. 0 §5. Local Transverse Stability 131 Proof of Theorem 4.2. If dim X < dim Y, then ~ = X and the last proposition proves the theorem. So assume dim X ;::: dim Y and let B, " and 'YJ be given by Proposition 4.5. Choose a nbhd Z of ~ x {O} so that Z c B. Let a = Tp - (df)W - F*'YJ. Then a is a vector field along F whose R"-component is zero and which is zero on B. Now fl(X - Z) is a submersion. Thus for v in R" small enough Fvl(X - Z) is a submersion. Since F is a deformation, F is a submersion on a nbhd D of X x R" - Z in X x R". Thus T(X x R") = TR" EB Ker (dF) EB G on D where G is some complementary subbundle. Moreover (dF): G -+ TY is an isomorphism on D. Since a is a vector field along F whose R"-component is zero, there exists a vector field r on D so that (dF)(O = a on D. Moreover, we can assume that " is a section of G so that the R"-component of " is zero. Since a =0 on B, we can extend r to be =0 near the boundary of D and thus extend " to a vector field on X x R" whose R"-component is zero. Then Tp = (dF)(, + 0 + F*'YJ on a nbhd of X x {O}. 0 Theorem 4.6. Suppose f: X -+ Y is smooth and X is compact. Iff is infinitesimally stable, then f is stable. Proof A trivial consequence of Proposition 4.1, Theorem 4.2, and Proposition 2.6. 0 Q.E.D. §5. Local Transverse Stability We will show that local infinitesimal stability (see Definition 1.1) is equivalent to a certain transversality condition. First we must construct the submanifolds which will appear in this transversality statement. To do this, consider the action of Diff(X) x Diff(Y) on J"(X, Y) given by (g, h).a = j"h(q).a.j"(g-l)(g(p» where a is in J"(X, Y)p,q. Let!?d" be the orbit of the action thru the k-jet a. It is true that !?d" is a submanifold of J"(X, Y) but, for our purposes, we shall not need this fact. We shall only prove the follow- ing. Theorem 5.1. !?d" is an immersed submanifold ofJ"(X, Y). Before proving this result, we need some facts about extending diffeomorphisms and damping translations. Lemma 5.2. Let 'YJ: Rn -+ Rn be an immersion such that 'YJ is a diffeomorphism outside ofsome compact set K. Then 'YJ is a diffeomorphism. Proof We need only show that 'YJ is 1: 1 and onto as the Inverse Function Theorem will imply the result. To show that 'YJ is onto we note first that 'YJ is a submersion and so 1m 'YJ is open. Let L be a compact set with K c Int L. Then Im'YJ = 'YJ(L) U 'YJ(Rn - Int L). Both sets in the union are closed so Im'YJ is closed. Thus 1m 'YJ = Rn. To show that 'YJ is 1: 1, define S = {x ERn I3y ERn, y t= x, with 'YJ(y) = 'YJ(x)}. Since 'YJ is a diffeomorphism off K, Rn - S t= 0. Thus it is enough to show that S is both open and closed; for then S = 0 and 'YJ is 1: 1. Let x 132 Various Equivalent Notions of Stability be in Sand y be in Rn - {x} such that 7){x) = 7){y) = q. Choose nbhds U ofx, V ofy, and W ofq so that U () V = 0 and 7)IU: U --+ Wand 7)IV: V--+ Ware diffeomorphisms. (This is possible since 7) is an immersion.) Then U c S for if a is in U, then b = (7)1 V)-l·{7)1 U){a) satisfies b i' a and 7){b) = 7){a). Thus S is open. To see that S is closed let Xl> X2, ... be a sequence of points in S converging to x. Choose an open nbhd W of x so that'l)l W is a diffeomorphism. We may assume that each Xi is in W. Choose Yt i' Xi so that 'I){Yt) = 'I){Xi)' Clearly the y;'s are not in W since '1)1 W is 1: 1. Also, the y;'s are contained in the compact set 7)-l{'I){K)). Thus we may assume that the y;'s converge to y in 'I)-l{'I){K)) - W. Clearly y i' x and the continuity of 'I) guarantees that 'I){x) = 'I){y). Thus x is in Sand S is closed. 0 Proposition 5.3. Let Ta : Rn --+ Rn be translation by a in Rn-i.e., Ta{x) = x + a. Given an open set B in Rn, there exists a diffeomorphism 7) : Rn --+ Rn such that 7) = Ta on Band 7) = idR R outside ofsome compact set. Proof Choose a smooth function u : Rn --+ R which is 1 on a ball centered at 0 containing B and which has compact support. Let p(x) = u{tx) for some t. Choose t so small that Idpl = t Idul < 1/14 By also demanding that t ::; 1 we see that p == 1 on B. Now consider 'I){x) = x + p(x)a and observe that 7) = Ta on Band 7) = idR R off of some compact set. By applying Lemma 5.2 it is enough to show that 'I) is an immersion in order to show that 'I) is a diffeomorphism. Now (d,). ~ I, + c:)(::,....,::.) where a ~ (a... .•, aJ· (A short computation is necessary here.) Thus for v i' 0, I(d'l))xCv)I ;::: Ivl - lal·ldpl·lvl > 0 by the choice of p. Hence 'I) is an immersion. 0 The following is left as an exercise. Lemma 5.4. The connected component ofthe identity in GL{n, R) is the set ofmatrices with positive determinant. Proposition 5.5. Let 1> be a local diffeomorphism on Rn defined near 0 satisfying 1>(0) = 0 and det (d1»o > O. Then there is a diffeomorphism 'I) : Rn --+ Rn such that 'I) = idR n outside of some compact set K and 'I) = 1> on some nbhd ofO. Proof Since 1>(0) = 0,1> = (d1»o + f3 where f3 is O(lx12) near O. We first show that {3 can be damped out off K. Let p : Rn --+ R be a smooth function such that p == 1.on a nbhd of 0 and p == 0 off K. Consider T = (d1»o + p{3. Clearly T = 1> on a nbhd of0 and T = {d1»o offK. We wish to choose p so that T will be a diffeomorphism. By Lemma 5.2 it is enough to show that T is an immersion. Now for v in TxRn = Rn, I(dT)xCv) I ;::: Id{d1>)O(v)I - l{dp(3)xl·lvl ;::: (c - I{dp (3)xDlvl where c = I(d1» 0 -11. Thus if we choose p so that I{dp (3)1 < c, then T will be an immersion. Choose u: Rn --+ R such that u == 1 on a nbhd of 0 and supp u c B(l) = ball of radius 1 centered at the §5. Local Transverse Stability 133 origin. Let p(x) = a(rx) for some positive constant r. Then p == 1 on a nbhd of 0 and supp p c: B(1/r) = ball of radius l/r centered at O. By choosing r large enough supp p c: K. Let M = sup Idal. Then rM ~ Idpl. Since f3 is O(lx12) and supp p c: B(1/r) there exist constants e and f so that 1f3(x)I :::; elxl2 and I(df3)xI :::; flxl on supp p. Thus I(dpf3)I :::; Idpl·If31 + Ipl·ldf31 :::; (eM +/)/r. Choose r large enough so that (eM + f)/r < c. Next we show that given a linear map a with det a > 0 there exists a diffeomorphism g so that g = a on a nbhd of 0 and g = idR n outside of K. Ifg exists, then 1) = a-1.g'T is the desired diffeomorphism where a = (d(/»o' (We use the hypothesis that det (dcP)o > 0 here.) Moreover, it is sufficient to show that there exists a 0 > 0 so that g exists whenever la - Inl < o. For we may choose a curve c: R -? Gl(n, R) so that c(O) = In and c(1) = a using Lemma 5.4. Also, since [0, 1] is compact there exist points to = 0 < tl < ... < tk = 1 such that Ic(ti)·c(tt_l)-l - Inl < 0 for 1 :::; i:::; k. Let gj be the diffeomorphism associated with c(ti)'C(tt-l)-t, then g = gk'" . 'gl is the desired diffeomorphism. Let p : Rn -? R be a smooth function such that p == 1 on a nbhd of 0 and p == 0 off K, then consider g = In + pea - In). Clearly g = a near 0 and g = In off K. Using Lemma 5.2 again, we need only show that g is an immersion to see that g is a diffeomorphism. Indeed I(dg)xCv)I ~ Ivl - (I(dp)xl + Ip(x)IHa - Inl·lvl· Thus if we choose 0 < l/sup (Idpl + Ipl), then whenever la - Inl < 0, the associated g will be an immersion. 0 At this point we shall need some standard facts about Lie groups. We refer the reader who is not familiar with this topic to the appendix where we give definitions, examples, and sketch the results that are used here. Now let p be in X and q be in Y. Let Gk(X)p and Gk( Y)q be the invertible k-jets in ]k(X, X)p,p and J"( Y, Y)q,q respectively and let G = Gk(X)p X Gk( Y)q. Then G is a Lie group (see Example (4) after Definition A. I) and there is an obvious action ofG onJ"(X, Y)p,q given as follows: (a, f3)(a) = f3'a'a- 1 where (a, f3) is in G and a is in J"(X, Y)p,q. Let (!}(1 be the orbit in ]k(X, Y)p,q thru the k-jet a. Applying Theorem A.13 we see that (!}(1 is an immersed submanifold. (In fact, (!}(1 is a submanifold as it is the orbit of an algebraic group acting algebraically on a manifold. See Borel, Linear Algebraic Groups-Proposition 6, 7, p. 180. For our purposes we shall not need this fact. We also note that the knowledge that (!}(1 is a submanifold would be enough to prove that !l)(1 is actually a submanifold of J"(X, Y).) Now let &(1 be the connected component of(!}(1 containing a. Clearly &(1 is also an immersed submanifold of ]k(X, Y)p,q. ProofofTheorem 5.1. Suppose that dim X = n and dim Y = m. Choose p in X and q in Y with chart nbhds U of p and V of q. Via charts we may identify U with Rn, V with Rm, p with 0, and q with O. Consider T: U x V X ]k(U, V)p,q-?J"(U, V)definedby(p',q', T)'t-+j"Tq"T.j"(Tp,-l) where Tc is translation by the vector c. (This makes sense using the identifications above.) T is a diffeomorphism as it is essentially the inverse of a chart 134 Various Equivalent Notions of Stability in the manifold structure ofJk(X, Y). (Identify the domain with Rn x Rm x ]"(Rn, Rm)o,o and see II, Theorem 2.6.) We claim that T(U x V x @(,) = ~r;.v == connected component of~(J n ]"(U, V) containing a. This will imply the Theorem as @(J is an immersed submanifold since T is a diffeomorphism; as long as we also show that ~(J n ]"(U, V) has at most a countable number of components. First we show that T(U x V x @(J) C ~r;'v. Since U x V X @(J is connected, it is enough to show that T(U x V x @(J) C ~(J n ]"(U, V). Let (p', q', r) be in U x V X {g(J' Now @(J = Goa (using Lemma A.14) and G= G"(X)p X G"(Y)q where Gk(X)p = {a EO Gk(X)p Idet (da)p > O}. Thus r = f3 oa oa- 1 where a EO Gk(X)p and f3 EO Gk(Y)q. Now we can represent a and f3 by mappings a: X ---»- X and i3: Y ---»- Y which are diffeomorphism nbhds of p and q respectively. Using Proposition 5.5 we can insure that a and i3 are globally defined diffeomorphisms. So r = j"(i3)(q) 0 aojk(a-1)(p). Now T(p', q', r) = j"(Tq.)(q) 0 7"j"(Tp' -l)(p'). By Proposition 5.3, we may assume that Tq.: Y ---»- Y and Tp .: X ---»- X are globally defined diffeomorphisms. Thus T(p', q', r) = (Tp.oa, Tq.oi3)oa EO~(J' For the reverse inclusion, let r be in ~r;.v. Let the source of r be p' and the target be q'. Consider p = j"(Tq.-l)(q') 0 roj"(Tp.)(p). Since Tp. and Tq. are in Diff(X) and Diff(Y) respectively (Proposition 5.3 again), we see that p is still in ~(J' Thus there exist (y, 8) in Diff(X) x Diff(Y) such that p = jk(8)(q)oaoj"(y-l)(p) and so p is in @(J' So we have shown that ~r;,v C T(U x V x @(J)' Also we have shown that ~(J n Jk(U, V) C T( U x V x @(J) so that ~(J n]k( U, V) has at most a countable number of components. Since T is a diffeomorphism and ~r;.v is connectedwemusthavethat~r;'v C T(U x V x (connectedcomponentof@,,)). But certainly a is in ~r;'v n T(U x V x @(J) so that ~r;.v C T(U x V x @(J)' Thus the components of ~" n ]k( U, V) are a subset of the components of T(U x V x @(J) and this last set is at most countable. 0 Definition 5060 Letf: X ---»- Y be smooth and let p be in X. Let m = dim Y and let a = rf(p). Then f is locally transverse stable at p ifrrm ~(J at p. Notes. (1) The concept of intersecting transversely an immersed submanifold makes sense in an obvious way, since the tangent space to an immersed §ubmanifold at a point is well-defined. More precisely, let W be an immersed submanifold of Y and let f: X ---»- Y. Then rm W at p if either f(p) rt W or f(p) EO Wand Tf(p) Y = (dfMTpX) + Tf(p) W. (2) Another way of phrasing this definition is that an immersed submanifold, W, is the countable union of submanifolds WI> W2 , ••• which are open subsets of W. Thus fm W iff fm Wi for each i. (3) Applying (2) we see that the Thorn Transversality Theorem still applies to immersed submanifolds; i.e., if W is an immersed submanifold of ]k(X, Y), then TW = {fEO C ""(X, Y) [jkfm W} is a residual subset of C""(X, Y). (Trivial, since TW = nj;1 TWi where TWi is defined in the obvious fashion.) A corollary of all this is the following: Lemma 5070 Let f: X ---»- Y be stable, then f is locally transverse stable at p for all p in X. §5. Local Transverse Stability 135 Proof. Choose an open nbhd U offin C"'(X, Y) such that all g in U are equivalent to f Let a = rf(p), By the Thorn Transversality Theorem (and Note (3) above), there is a g in U such that rg m£»'1" Choose (a, (3) in Diff(X) x Diff(Y) such that f = f3.g.a- l • Then jmfm £»(J at p since jmg m£»(J at a(p). 0 Let f: X ~ Y be smooth with p in X and a = j' ht) in Diff(X) x Diff( Y) such that e(t) = rht(q)·a.jkgt -l(gtCp)) then v is in the image (dY2) EB (dY1)' Let ~p = ~; (P)lt=o and TJq = ~t (q)lt=o. Then ~ and TJ are vector fields on X and Y respectively. (Since we are only interested in the germ of TJ at q we may assume that TJ has compact support.) Let 4>t and ifit be the one parameter groups associated with, and TJ respectively. Then the curve e(t) = rifit(q)·a.jk4>-MtCp)) satisfies del del dt t=o = dt t=o = v since (defdt)lt=o only depends on (dgtfdt)lt=o and (dhtfdt)lt=o but dgt I = d4>t I and dht I = difit Idt t=o dt t=O dt t=o dt t=o' Thus Recall that when we proved that filJ" is an immersed submanifold (see the proof of Proposition 5.1) we showed that filJ;:'v = connected component of filJ" n Jk(U, V) containing a is equal to T(U x V x (g,,) where @" is the orbit thm a of the action of the Lie group G= Gk(X)p X Gk( Y)q on Jk(X, Y)p,q. Since T"filJ" = T"filJ;:'v we may assume that the curve e(t) is in filJ;:·v. Since T is a diffeomorphism there is a curve aCt) = (p(t), q(t), T(t)) in U x V x @" 138 Various Equivalent Notions of Stability such that T(a(t» = c(t). Thus c(t) = jk(Tq(t)' T(t) .jk(T;(t\). So if we can show that T(t) = jk(h(t».a·/«g(t)-I) then we will be finished by the last paragraph. But T(t) is a curve in @a. Thus there is a curve (g(t), h(t» in Gk(X)p x Gk(Y)q such that h(t).a.g(f)-1 = T(t) since Gis a Lie group. In local coordinates (;k(X) is just polynomial mappings of degree -::;,k on Rn which are diffeomorphisms on a nbhd of o. Let get) be the unique polynomial of degree -::;, k such that jkg(t) = get). Similarly for h(t) and h(f). Since det (dgo)p > 0 and det (dho)q > 0 we may assume that gt and ht are globally defined diffeomorphisms on X and Y respectively. (Apply Proposition 5.5.) Thus we obtain the desired gt and ht· 0 We now state and prove the main Theorem of this section. Theorem 5.13. Let f: X ~ Y be smooth and let m =dim Y. If f is locally transverse stable at p, then f is locally infinitesimally stable at p. Proof Assume that f is locally transverse stable at p. By Corollary 1.3 it is sufficient to show thatfsatisfies the conditions of infinitesimal stability to order m. In particular, we need to show that if T is in Jm(f*TY)p, then there is a VI in Jm(TX)p and a V2 in Jm(TY)q so that T = (df)(Vl) + j*(V2). Consider ,\(T) in TaJm(X, Y). Since f is locally transverse stable at p there exists w in Ta~a and v in TpXso that '\(T) = w + (drfMv). By Lemma 5.12 there exists ~ in COO(TX)p and T) in C"'(TY)q so that w = -(dY2)(O + (dYl)(T). Let VI = 7T;:::m and V2 = 7T;:::(T). Applying Propositions 5.10 and 5.11 we have that w = ,\·f*(V2) - :t""(Vl). Thus '\(T) = ,\·(df)(Vl) - (drf)p' 7Tom(V1) + ,\·j*(V2) + (drfMv). Apply (da)a to both sides and apply Proposition 5.9 to obtain SO '\(T) = ,\.(df)(v1) + ,\.j*(V2). But ,\ is injective (Proposition 5.8), so T = (df)(v1) + f*v2. 0 §6. Transverse Stability The problem of transferring the result that local transverse stability implies local infinitesimal stability to a global result is, as usual, at the selfintersections of the mapping in question. Multijet transversality is again the tool used to solve the problem. Let ~a' be the orbit through the s-fold multijet a under the action of Diff(X) x Diff(Y) on Jsm(x, Y). Proposition 6.1. ~as is an immersed submanifold of Jsm(x, Y) for each s-fold multijet a. Remarks. (1) ~(Js is, in fact, a submanifold for the same reasons that ~(J is a submanifold of Jk(X, Y). §6. Transverse Stability 139 (2) We are only interested in s-fold m-jets a = (aI' ... , as) where target al = ... = target as since these jets reflect the problem of self-intersection. Such multijets are called diagonal elements. Proof The proof splits into two cases: namely, whether or not a is a diagonal element. Suppose a = (aI> ... , as) is a diagonal element. Recall from Proposition 6.1 the diffeomorphism T: U x V X @a ~J'C(U, V). Now we extend this to an immersion as follows: Let Ub ... , Us be disjoint chart nbhds of source (al) = PI>"" source (as) = Ps respectively and Va chart nbhd of q = target (al) = ... = target (as). Define T: UI x ... x Us x V X @al x··· X @as ~Jk(UI' V) X ... x Jk(Us> V) c J/(X, Y) as follows: T(x1o ... , x s, y, T10 •.• , Ts) = (T(x1o y, TI), ••• , T(xs> y, Ts»). Define ~a"(U1o"" Us> V) = connected component of a in ~as n (Jk(UI, V) x ... X Jk(Us, V). With arguments similar to those in Proposition 5.1 one shows that 1m T = ~a"( UI, ... , Us> V) so that ~/ is an immersed submanifold. The only catch is that if we have s diffeomorphisms one each defined on Ui that there is a global diffeomorphism on X which is equal to each on a nbhd of ai' This is possible as long as the diffeomorphisms are given by translations or have a fixed point where the diffeomorphism has a Jacobian with positive determinant. But these are the only diffeomorphisms that are considered in the proof. In other words, the diffeomorphisms on Ui can be damped near the boundary to extend smoothly to the identity off Ui' Now suppose a is not a diagonal element. For simplicity suppose that s = 2 so that a = (aI' a2) and target al =1= target a2' Then we claim that ~a2 = (~al X ~(2) n Z where Z is the open set {(aI' a2) EJ2m(x, Y) Itarget al =1= target a2} (at least locally). Ifthis is true, then certainly ~a2 is an immersed submanifold. We shall leave the proof of this claim to the reader as it is not difficult and we shall not make further use of this fact. Note the case for general s is similar to s = 2; the details of this observation are also left as an exercise. 0 Definition 6.2. Let f: X ~ Y be smooth and let m = dim Y. Then f is transverse stable iffor every s with 1 :s; s :s; m + 1 and diagonal element a, js"1m ~/. Lemma 6.3. Let f: X ~ Y be smooth. Iff is stable, then f is transverse stable. Proof The proof is the same as the proof of Lemma 5.7 except that we substitute the Multijet Transversality Theorem for the Thom Transversality Theorem. 0 The main result is the following: Theorem 6.4. Letf: X ~ Y be smooth. Iff is transverse stable, thenfis infinitesimally stable. 140 Various Equivalent Notions of Stability The proof of Theorem 6.4 is almost identical to that of Theorem 5.13. We will just give a sketch as the important ideas have already been presented. Sketch afProof We assume thatfis transverse stable; we need only show that condition (t) of Theorem 1.6 is satisfied by f So let S = {Ph' .., P.} C f-l(q), with 1 :s; s :s; m + 1. Define AS: Jm(f*TY)s -+ T,,J.m(x, Y) as follows: A: Jm(f*TY)p! -+ Ta/m(X, Y) has already been defined in the discussion before Proposition 5.8. Since Jm(f*TY)s = EBf=l Jm(f*TY)v! and T"J.m(x, Y) = EBf=l T,,/m(x, Y) it makes sense to define AS = EBf=l A. Proposition 5.8 still applies so that AS is a linear injection. Let 0:.(,): J.m(x, Y) -+ X(·) be the source map. Just as in Proposition 5.9 the sequence m(f* ) AS m( ) (du(·»" (.) 0O~J TYs~TaJ. X'Y-T(Vl .....V.)X ~ is exact. Next let TJ be a vector field on Y with compact support represent [TJ]q and let cfot be the associated one parameter group. Define (dyl)(TJ) by (de/dt)lt=o where e(t) = j.kcfotCq)·a. Then (dylS) is a mapping of COO(TY)q -+ T~a". Just as in Proposition 5.10 the diagram l~: * rAS Jm(TY)q --.l- Jm(f*TY)s commutes. Similarly, we can define (dyl): COO(TX)s -+ Taflj,,·· Let ['l]Vl"'" [,.]v, be germs of vector fields on X. Since Pl, ..., P. are all distinct, there is one vector field' on X such that l'lV! = ['i]V!' Let ifJt be the one parameter group associated with ,. Then define (dyl)(['l]Pl"'" ['.]p.) = (de/dt)lt=o where e(t) = a.j.mifJt-lifJt(S). Next define :/('S = -As·(df) + (djs'T'f)s'~om where ~om : Jm(TX)s -+ TsX(') is the obvious projection. Then just as in Proposition 5.11 the diagram COO(TX)s (dYl») Taflja" l~~~ commutes where ~: is the obvious projection. Finally we note that (dy2S) EEl (dY1S): COO(TX)s EEl COO(TY)q -+ Taflja" is onto as in Lemma 5.12. The calculations to show that condition (t) holds for the set S proceed in an entirely analogous way as the calculations in the proof of Theorem 5.13. 0 §7. Summary 141 §7. Summary The following summarizes the work of this Chapter. Theorem 7.1. Let f: X --+ Y be smooth and let X be compact. Then the following are equivalent: (a) fis stable; (b) f is transverse stable; (c) f is infinitesimally stable; (d) f is homotopically stable; (e) f is stable under k-parameter families ofdeformations. Proof (a) => (b) Lemma 6.3; (b) => (c) Theorem 6.4; (c) -¢> (d) -¢> (e) Theorem 4.2 and Proposition 4.1. (d) => (a) Proposition 2.6. 0 At this point a few comments about the heuristics of stable mappings seem in order. It should be clear that "generic" properties have something to do with stable mappings. This can be made precise as follows. A property P of smooth mappings of X --+ Y is generic if it satisfies the following two conditions. Let Wp = {fE C"'(X, Y) ifsatisfies property P}: (1) Wp contains a residual subset of C"'(X, Y). (Preferably Wp is open and dense.) (2) Iff is in Wp, then any mapping equivalent to f is in Wp; that is, Wp is an invariant subset under the action of Diff(X) x Diff( Y) on C "'(X, Y). A simple argument shows that with this definition of generic a stable mapping does satisfy every generic property. Examples of generic properties were developed in Chapters II and III. (e.g., Morse functions with distinct critical values, 1: 1 immersions when 2 dim X < dim Y, and immersions with normal crossings when 2 dim X = dim Y). In these cases we used the Thorn Transversality and Multijet Transversality Theorems to show that the property in question is valid for a residual set of mappings and is thus a generic property. John Mather's Theorem which states that infinitesimally stable mappings are stable enabled us to show (in Chapter III) that (for the relative dimensions of X and Y under consideration) there are no other interesting generic properties; i.e., if there were other generic properties they would be satisfied automatically ifthe function in question satisfied the generic properties listed above. The subsequent chapters will be devoted to finding generic properties for mappings between manifolds of arbitrary dimensions. One might hope for a list of "interesting" generic properties along with a result which states that if a mapping satisfies the properties on this list then it is stable. This turns out not to be possible! The proof that infinitesimal stability implies stability-in particular, the notion of transverse stability-allows us to describe what all of the generic properties are (at least those generic properties which depend only on the m-jet of the function for some m). Since the m-jet of a function determines 142 Various Equivalent Notions of Stability whether or not the function is stable (Theorem 1.6 where m = dim Y), this means that we can describe all of the "interesting" generic properties. Said more precisely-to each orbit, (!), in JJ::+ l(X, Y) under the action ofDiff(X) x Diff(Y) there is associated a generic property, P(f) of mappings of X -+ Y; namely, the property that the m-jet extension of the function intersects (!) transversely. Theorem 7.1 states that a mapping is stable iff it satisfies all generic properties P(f) constructed in this fashion. Since the number of orbits is uncountably infinite it is by no means clear that the set of stable mappings is dense. Quite the reverse, it seems remarkable that a large subset ofmappings could satisfy all of these properties simultaneously. The problem with the generic properties P(f) described in the last paragraph is that they do not translate easily into more familiar properties of smooth mappings such as the structure of the singular sets. In the next chapter we will show how to construct submanifolds of Jk(X, Y) (which are unions of orbits) and which do translate into nice geometric notions. In doing so, we shall also be able to show that certain of these properties are contradictory so that stable mappings are not always dense. In fact, there exist manifolds X and Y for which there are no stable mappings in C"'(X, Y). Chapter VI Classification of Singularities Part I: The Thorn-Boardman Invariants §1. The Sr Classification For a mappingf: X -+ Y we can make the following rudimentary classification of singularities. We say that f has a singularity of type Sf at x in X if (df)x drops rank by r; i.e., if rank (df)x = min (dim X, dim Y) - r. Denote by SrCf) the singularities offof type Sr. Recall that in the proof of the Whitney Immersion Theorem we introduced the submanifolds Sr of ]leX, Y) consisting of jets of corank r. (See II, Theorem 5.4.) Clearly SrCf) = (Pf)-l(Sr). To prove the Whitney Theorem we showed that if X and Y have the" right" relative dimensions then generically Sr(f) = 0 (r > 0) and f has no singularities; i.e., f is an immersion. Without restricting the relative dimensions of X and Ywe can still say that generically SrCf) is a submanifold of X and codim Sr(f) = codim Sr = r2 + er where e = [dim X - dim yr. This statement follows immediately from the Thorn Transversality Theorem and II, Theorem 4.4. In particular, the set of mappings for which Pf(fl Sr (for all r) is residual. Besides the Transversality Theorem, the main fact used in the proof of this statement is that Sr is actually a submanifold of ]leX, Y). We shall sketch a different proof of this fact in order to motivate the material in §4. Given a pair of vector spaces V and W, let U( V, W) be the set of linear maps of V into W which drop rank by r. The main fact needed is: Proposition 1.1. U(V, W) is a submanifold of Hom (V, W) of codimension r2 + er l",here e = [dim V - dim WI. This is just II, Proposition 5.3, which we shall reprove here using a trick involving Grassmann manifolds. Let s = r + max (0, dim V - dim W) and let G(s, V) be the Grassmannian of s planes in V. Let E be the canonical bundle over G(s, V). (See Example (4) after I, Proposition 5.4.) We will denote by Q the vector bundle with fiber VjEp at p E G(s, V) and by Hom (Q, W) the vector bundle over G(s, V) whose fiber at p is Hom (Qp, W). (The construction is functorial and thus yields a smooth vector bundle. See I, Proposition 5.4.) This fiber contains L°(Qp, W) as an open subset, and LO(Q, W) = Up U(Qp, W) is an open submanifold of Hom (Q, W). Now we claim there is a natural identification (as sets) (1.2) LO(Q, W) ~ U(V, W). 143 144 Classification of Singularities Part I In fact, for each p E G(s, V) we have a projection 7T: V --+ Qp and this induces a transpose map backwards. (1.3) 7T*: Hom (Q, W) --+ Hom (V, W). The image of this map is the set of elements in Hom (V, W) of corank '2::.r. Moreover, 7T* maps LO(Q, W) bijectively onto the elements in Hom (V, W) which are precisely of corank r; so 7T* gives us the identification (1.2) whose existence we asserted. We now use (1.2) to provide D(V, W) with a manifold structure. We will let the reader compute the dimension of this manifold and check that it gives the same answer as our computation in Chapter II. (Hint: It is the same as the dimension of Hom (Q, W) regarded as a manifold.) This does not yet show that D(V, W) is a submanifold of Hom (V, W). Let 7T* : Hom (Q, W) --+ Hom (V, W) be the map described above and let M be the subset of Hom (V, W) consisting of elements of corank > r. To conclude the proof one has to show Proposition 1.4. The map 7T*: P(Q, W) --+ Hom (V, W) - M is a 1:'1 proper immersion. The proofis straightforward but a little tedious. We will not include it here. (See exercises.) Anyway, given Proposition 1.4, it is easy to prove our assertion about SrCf). Namely, we first observe that Proposition 1.1 is true for vector bundles as well as vector spaces. Given two vector bundles E --+ X and F --+ Y denote by D(E, F) the fiber bundle over X x Y with typical fiber D(Ex, Fy). Then by Proposition 1.1 D(E, F) is a subfiber-bundle of Hom (E, F) of codimension r2 + er. Recall now the canonical identification P(X, Y) = Hom (TX, TY). (See Remark (2) after II, Theorem 2.7.) Just note that Sr is the subfiberbundle of ]leX, Y) corresponding to D(TX, TY) and we have the desired conclusion that Sr is a submanifold of ]leX, Y). Definition 1.5. We will say that a mapping f: X --+ Y is one generic if Pfffl Srfor all r. From now on, we will assume all maps are one-generic. Exercises (The purpose of these exercises is to supply the reader with an outline of the proof of Proposition 1.4.) (1) Show that the map (1.2) is a homeomorphism; i.e., give D(V, W) the induced topology and show that the inverse of the map (1.2) is continuous. (2) Let Z 4 X be a fiber bundle and let p: Z --+ Y be a map. Let Z E Z, X = a(z), and y = p(z). Show that §2. The Whitney Theorem for Generic Mappings between 2-Manifolds 145 is exact, and deduce the existence of a mapping (1.6) TxX -+ TyYj(dpMTzZx)· (3) Apply Exercise (2) with the following data: X = G(s, V), x = an s dimensional subspace K of V, Y = Hom (V, W), y = a linear map A in Hom (V, W) with Ker A = K, Z = LO(Q, W), and z = A'7T where 7T is as in Proposition 104. (Note that V/K is the fiber of Q above x and LO(V/K, W) is the fiber of Z above x.) Let p = 7T*: LO(Q, W) -+ Hom(V, W) and let a: LO(Q, W) -+ G(s, V) be the canonical projection. Note that Ty Y = TA Hom (V, W) ~ Hom (V, W) and TzZx = TA." Hom (V/K, W) ~ Hom (VjK, W) since Hom's are vector spaces. (i) Show that (dp)z: TzZx -+ Ty Y is an injection. In particular show that the image is all maps containing K in their kernels. (ii) From (i) conclude that (dp)iTzZx) ~ Hom (V/K, W). (iii) Finally conclude that Ty Y/(dp)zCTzZx) ~_ Hom (K, W). Moreover, show all these identifications are canonical. (4) Let x E G(s, V) and let K be the s-dimensional subspace it represents in V. Recall there is a canonical identification TxG(s, V) ~ Hom (K, VjK). (See Note (2) at the end of!, §3.) Show that the map (1.6) of Exercise 2 is just the map A* : Hom (K, V/K) -+ Hom (K, W) given by composition on the last factor where A is given in Exercise 3. (5) From 2-4 deduce that the map 7T* in Proposition 104 is an immersion. (6) Using Exercise 4, prove that NA = TA Hom (V, W)jTAL'(V, W) the normal space to LT(V, W) in Hom (V, W) at A is canonically isomorphic to Hom (Ker A, coker A). (7) Let f: Rn -+ Rn be given by f(Xh ... , xn) = (fl(X), ... ,!rex), Xr+h ... , Xn). Suppose thatf(O) = 0 and thatfhas an ST singularity at O. Show that PI(fl Sf at 0 iff the r2 vectors (d~J°where 1 ~ i,j ~ rare all linearly independent. Hint: Use the proof of II, Proposition 5.3 to identify T<1L'(Rn, Rn) in coordinates where a = PI(O). §2. The Whitney Theorem for Generic Mappings between 2-Manifolds The Thorn-Boardman Theory has to do with the behavior of maps restricted to their singular sets; i.e., if I: X -+ Y is one-generic, Sr(f) is a submanifold, so IISr(f) is again a map between manifolds; and we can, for example, ask whether it has singularities generically. It is this type ofquestion to which the Thorn-Boardman Theory addresses itself. Before we outline this theory, we will discuss one example in detail-the Whitney Theory for maps between 2-manifolds. Let X and Y be 2-dimensional manifolds and let I: X -+ Y be a onegeneric mapping. By our computation in §l Sl(f) is of codimension 1 in X 146 Classification of Singularities Part I and S2(f) does not occur since it would have to be of codimension 4. Let p be in Sl(f) and q = f(p). One of the following two situations can occur. (2.1) { (a) (b) TpS1(f) EB Ker (df)p = TpX; Tp S1(f) = Ker (df)po Note that ifp is a Sl singularity satisfying (a), then p is a fold point. (See III, Definition 4.1.) The first theorem ofWhitney for maps between 2-manifold gives the normal form for fold points. Theorem 2.2. If (a) occurs then one can choose a system of coordinates (Xl' X 2) centered at p and (Y1, Y2) centered at q such that f is the map (*) This theorem is just a special case of the normal form that we derived for submersions with folds. (See III, Theorem 4.5.) Now we will suppose that condition (b) holds; i.e., ker (d!)p = TpS1(f). This situation is considerably more complicated. Let us choose a smooth nonvanishing vector field, g, along Sl(f) such that at each point of Sl(!) gis in the kernel of (df). (Locally this is always possible.) By assumption, g is tangent to Sl(!) at p. The nature of the singularity at p obviously depends on what order ofcontact ghas with Sl(f) at p. Let us make this statement more precise. Let k be a smooth function on X, such that k = °on Sl(f) and (dk)p i= 0. Consider (dk)W as a function on Sl(f). By assumption this has a zero at p. We let the reader check as an exercise that the order of this zero does not depend on the choice of g or k. (Hint: for another choice (f, k') show that f and k' are nonzero multiples of gand k.) Definition 2.3. We will say p is a simple cusp if this zero is a simple zero. The second main theorem of Whitney states Theorem 2.4. Ifp is a simple cusp then one can find coordinates (Xl' X 2) centered at p and (Y1, Y2) centered at q such that { J*Y1 = Xl J*12 = X 1X 2 + X23. A picture of this map is sketched in Figure 3. Let X be the graph of X3 = X1X2 + X23. This graph can be viewed as a family of cubic curves in (X2' x3) depending on the parameter Xl' For Xl positive these curves are without critical points. For Xl = °there is a critical point which is a point of inflection and for Xl < °there are max and min's. Letf: X -+ R2 be the projection of X onto the x1x3-plane. There is a natural set of global coordinates on X given by (Xl. x2) H>- (Xl. X2, X 1X2 + X23). In these coordinates on X f has the form of Theorem 2.4. The fold curve is the locus of extrema and the cusp is the inflection point. Note that Sl(f) is a parabola twisted in R3 so that any vector tangent to Sl(f) at (0,0,0) is killed by f Also note that the image of Sl(f) under fis the cusped plane curve t H>- (-3t 2 , -2t3). §2. The Whitney Theorem for Generic Mappings between 2-Manifolds 147 X3 ----:::::~---__ Xl II/ Figure 3: The Simple Cusp SIU) (in coordinates) ={xl=-3x2 2 } The proof we will give of Theorem 2.4 is due to Morin and uses the Malgrange Preparation Theorem. Whitney gives a more elementary but slightly more complicated proof in [58]. Proof Let us choose coordinates (Xl' X 2) centered at p and (11, Y2) centered at q such that1 has the form 1*11 = Xl 1*Y2 = h(X1> X2)' Since 1 has rank I at p this is possible. We can also assume that at the origin (df)a = [b ~] in this coordinate system; i.e., (Oh/OXl)(O) = (oh/ox2)(0) = O. We note, however, d(oh/ox2)o =P 0, otherwise 1 would not be one-generic. (Proof Suppose 148 Classification of Singularities Part I at O. Let a = t(a2hjax12)(0) and compare fwith the map (tt) (Xl' X2) f--+ (Xl' aX12). They have the same 2-jet at 0, but (tt) is of rank 1 everywhere, so it is not one-generic.) The set Slf) is defined by the equation ahjax2 = 0; so at each point of Sl(f) the kernel of (df) is spanned by aj8X2. This means that we can take ahj8x2 to be the function k and ajax2to be the vector field gof Definition 2.3. The condition for the origin to be a cusp is that (82hj8x22)(0) = 0; and for it to be a simple cusp, (83hj8x23)(0) #- O. Therefore, at the origin, we have 8h a2h a3h h = -a = -a2 = 0 and -a3 #- O. X2 X2 X2 The Generalized Ma1grange Preparation Theorem allows us to write X23 = 3aixl, h)X22 + al(Xl> h)X2 + ao(xl, h) where ao, aI, and a2 are smooth functions ofYl and Y2 vanishing at O. (To see this recall that f is given by f(xl> x2) = (Xl' h(Xl> X2». Then via f CO"(R2) becomes a module over itself; i.e., a.b(xl, x2) = a(f(xl, x2))b(xl> X2) where a is in the ring CO"(R2) and b is in the module C0"(R2). By the Malgrange Theorem (IV, Corollary 3.11) this module is generated by 1, X 2 , and X 22 if the vector space CO"(R2)j«Xl' h) + JtO(R2)4) is generated by 1, X2, X22. The assumptions on h guarantee that this is so.) Now the equation above can be written in the form (*) (with a = a2, and band c new functions of (Yl> Yz) vanishing at 0.) If we set Xl = 0 in (*) we see that the left hand side is of the form X 23 + ..., the dots indicating terms of order> 3 in X2. Since h(O, X2) = X23 + ..., the right and left hand sides of (*) can be equal only if(8ejaY2)(Yl> Y2) #- 0 at O. The leading term in the Taylor series of h is a nonzero multiple of XI X2; so, comparing the linear and quadratic terms on the right and left hand sides of (*), one easily sees that 8ej8h = 0 and abj8Yl #- 0 at the origin. This means that the following are legitimate coordinate changes: { Xl = b(xl , h) X2 = X2 - a(xl' h) { Yl = b(Yl> Y2) Y2 = c(Yl> Y2). In these coordinates we have f*Yl = Xl f*Y2 = X23 + XIX2 which is Whitney's canonical form. 0 Finally, Whitney proved that the singularities described above are generically the only singularities that can occur for maps between 2-manifolds. §3. The Intrinsic Derivative 149 Theorem 2.5. Let X and Y be 2-manifolds. Then there is a residual set in C 00 (X, Y) such that iff belongs to this set, its only singularities are folds and simple cusps. This is not hard to show directly, but we prefer to deduce it from a more general result. We defer the proof to §4 below. (See §4, Exercise 4.) §3. The Intrinsic Derivative In this section we will develop a technique due to Porteous for differentiating maps between vector bundles. This is the intrinsic derivative, and it plays a rather important role in the Thorn-Boardman theory (a special case of the intrinsic derivative was introduced in Chapter II, Definition 6.5). We will first of all give a pedestrian (and uninvariant) definition of this notion, then later a more sophisticated (invariant) definition. Let X be a manifold and let E = X X Ric and F = X X RI be product bundles over X. Let p: E ~ F be a vector bundle homomorphism. We may view p as a mapping of X ~ Hom (Ric, Rl). Then for p in X, (dp)p: TpX ~ Tp(p) Hom (R\ RI) = Hom (Ric, RI) makes sense. Let Kp = Ker pep) and let Lp = Coker pep). Then we define the intrinsic derivative, (Dp)p, in this local situation by the composite map TpX ~ Hom (R\ Rl) ~ Hom (Kp, Lp) where this second arrow is given by "restricting and projecting". We claim that the intrinsic derivative does not depend on which choices of trivializations of E and F are made. More precisely, let A: E ~ E and B: F ~ Fbe vector bundle isomorphisms. We may view A and B as mappings of X ~ Hom (R\ Ric) and X ~ Hom (Rl, Rl) respectively. With these trivializations p has the form p where p(x) = B(x).p(x).A(x)-l. Let Kp = Ker pep) and Lp = Coker pep). Clearly A(p) and B(p) induce isomorphisms of Kp ~ Kp and Lp ~ Lp respectively. Thus the different trivializations give a natural mapping ••• , Xn in Kx and the linear coordinates YS+l,"" Ym in Lx).) (2) Show that the intrinsic derivative D(df)x is determined by the 2-jet off at x (Hint: Use Exercise 1.) (3) Show that the map (3.9) when restricted to Kx Q9 Kx is in fact a symmetric map; i.e., D(dfMkl' k2) = D(dfMk2' k 1). Thus D(df)x induces a mapping (3.10) (where Kx 0 Kx denotes the symmetric product of Kx with itself). (Hint: Use Exercise 1.) (4) Letfbe a real valued function and let x be a critical point off Then Kx = TxX and Lx = R. Show that is just the Hessian off at x. (In Chapter II, Definition 6.5.) §4. The Sr,s Singularities Letf: X -+ Y be one-generic. We will denote by Sr,s(f) the set of points where the mapf: Slf) -+ Y drops rank s. Note that codim Sr(f) > dim X dim Y by Proposition 1.1, so dim Slf) < dim Y. Therefore, x E Sr,.(f) if and only if x E Sr(f) and the kernel of (df)x intersects the tangent space to Slf) in an s dimensional subspace. For example, for maps between 2-manifolds the points Sl,O(f) are fold points and the points Sl,l(f) are cusps. (See §2.) Our goal in this section is to show that the Sr,sCf) are generically manifolds (just like the sets Slf) and to compute their dimensions. The idea of the proof will be to construct universal Sr,s's in J2(X, Y) (analogous to the Sr's described in §1) such that x E Sr.s(f) -¢> j2f(x) E Sr,s' To begin, recall the identification: Sr;:;; D(TX, TY). Given a E Sr with source at x and target at Y we now can attach to a the vector spaces K" = ker a and L" = coker a in TxX and Ty Y respectively. This defines for us vector bundles on Sr which we denote by K and L. As we saw in §3, the normal bundle to Sr in P(X, Y) is canonically isomorphic to Hom (K, L). §4. The ST•• Singularities 153 Now let S,<2) be the pre-image of ST in J2(X, Y). By Exercise 2 at the end of §3 the intrinsic derivative gives us a map of fiber bundles: (4.1) S,<2) --+ Hom (K 0 K, L) ST Moreover, the top arrow,is surjective. (This is clear from Exercise 1 in §3. The last two terms on the RHS of (3.8) are completely arbitrary except for the symmetry condition.) We will construct our universal Sr.:s from the diagram (4.1). The main step in the construction is a theorem about vector spaces similar to Proposition 1.1 of §1. Let V and W be vector spaces, let V 0 V be the symmetric product of V with itself, and let V 1\ V be the space of alternating tensors. Recall the standard algebraic fact that V 0 V = (V 0 V) EEl (V 1\ V), so that there is a canonical projection TT: V 0 V ~ V 0 V whose kernel is V 1\ V. Consider the map (4.2) Hom(Vo V, W)~Hom(V0 V, W)~Hom(V,Hom.(V, W» where the first arrow is given by A ~ A'TT and the second arrow is given by B ~ q,B where q,a(v)(v') = B(v ® v'). Let Hom (V 0 V, W). be the pre-image under (4.2) of U(V, Hom (V, W». (Note that Hom (V 0 V, W). is not the same as U(V 0 V, W).) Proposition 4.3. Hom (V 0 V, W). is a submanifold of Hom (V 0 V, W) of codimension (4.4) 1 I - k(k + 1) - - (k - s)(k - s + 1) - s(k - s) 2 2 where k = dim V and 1 = dim W. Proof We use the "Grassmannian trick" used to prove Proposition 1.1. Let G(s, V) be the Grassmannian of s planes in V. Let E be the canonical bundle over G(s, V) and let Q be the vector bundle over G(s, V) whose fiber at pis V/Ep. Let Hom (Q 0 Q, W) be the vector bundle over G(s, V) whose fiber at p is Hom (Qp 0 Qp, W). The set Hom (Qp 0 Qp, W)o is an open subset of this fiber (being the inverse image under the continuous map (4.2) of an open set), and Hom (Q 0 Q, W)o = Up Hom (Qp 0 Qp, W)O is an open subfiber-bundle of Hom (Q 0 Q, W). The map TTp: V ~ Qp induces a map TTp ® TTp: V ® V ~ Qp 0 Qp. It is easy to see that TTp 0 7Tp: V 0 V ~ Qp 0 Qp and is onto. Just as in §1, this map induces a transpose map TT*: Hom (Q 0 Q, W) ~ Hom (Vo V, W) whose image is the set Ut;'8 Hom (Vo V, W)t. Moreover, its restriction to Hom (Q 0 Q, W)o maps this set bijectively onto Hom (Vo V, W).; so there is a canonical isomorphism (of sets) (4.5) Hom (Q 0 Q, W)o ~ Hom (V 0 V, W)•. 154 Classification of Singularities Part I The left hand side is a manifold; so we can define a manifold structure on Hom (V 0 V, W)s by requiring (4.5) to be a diffeomorphism. To prove that Hom (Vo V, W)s is a submanifold of Hom (V 0 V, W) requires a little more work. We will need, in fact, the following proposition. Proposition 4.6. Let Ms = Ut>s Hom (Vo V, W)t. The map 71"*: Hom (Q 0 Q, W)o -+ Hom (Vo V, W) - Ms is a 1: 1 proper immersion. The proof of this is almost identical with the proof of Proposition 1.2. For the details of that proof see the Exercises at the end of §1. Finally we have to compute the dimension of Hom (V 0 V, W)s' This is the same as the dimension of Hom (Q 0 Q, W). The fiber dimension of Q is k - s; so the fiber dimension of Hom (Q 0 Q, W) is (k - s)(k - s + 1)// 2. The dimension of the base space (i.e., G(s, V)) is s(k - s); therefore, the total dimension is (k - s)(k - s + 1)//2 + s(k - s); and the codimension is as asserted in Proposition 4.3. 0 Proposition 4.3 is valid for vector bundles as well as vector spaces. Given two vector bundles E -+ X and F -+ X let Hom (E 0 E, F)s be the fiber bundle whose fiber at p in X is Hom (Ep 0 Ep, Fp)s. Then by Proposition 4.3 Hom (E 0 E, F). is a fiber subbundle of Hom (E 0 E, F), and its codimension is given by (4.4) with k the fiber dimension ofE and I the fiber dimension of F. Let us now go back to the map (4.1) described earlier in this section. Hom (K 0 K, L). is a submanifold of Hom (K 0 K, L), so its pre-image is a submanifold (of the same codimension) in S,'2) (since (4.1) is a submersion). We will denote this manifold by ST,•. Our main result of this section is Theorem 4.7. Let f: X -+ Y be one-generic. Then x E ST,S(f) -¢> Pf(x) E ST,S' Proof Let Pf(x) = a in Sr. The normal space to ST in J1(X, Y) at a is Hom (K", L,,); and the map (dPf)x : TxX -+ T"J1(X, y) induces a map: (4.8) which is, as we saw in the last section, the intrinsic derivative of (df). This map is surjective since PfrFi ST by assumption, and its kernel is the tangent space to ST(f) at x. If x is in ST,.(f) the kernel of (4.8) intersects the kernel, K", of (df)x in an s dimensional subspace; that is, the restriction of (4.8) to K" has an s dimensional kernel. This means Pf(x) is in Hom (K" 0 K", L,,)•. The converse is equally easy to see, and this proves Theorem 4,7. 0 Corollary 4.9. Let f: X -+ Y be smooth. IfFfrFi ST,., then ST,.(f) is a submanifold of ST(f) whose codimension is given by the formula (4.4) where I = dim Y - dim X + k and k = r + max (dim X - dim Y, 0). §4. The ST,s Singularities ISS Thus ST,S(f) is a submanifold of X and dim ST,SCf) = dim X - r2 - er (codim ST,sCf) in ST(f)) where e = Idim X - dim YI. The Transversality Theorem says that the condition Pf?1 ST.S is satisfied by a residual set of mappings. A mapping for which this condition is satisfied for all r, s will be called 2-generic. Exercises (1) Show that the condition PFm Sr.s at x is a condition on the three jet offat x. (2) Let X and Ybe 2-manifolds andf: X -+ Ya 2-generic map. Show that dim SI(f) = 1 and dim Sl,l(f) = O. (3) Let X and Y be two manifolds and let f: X -+ Y be one-generic. Let x in X be a cusp point. Show that if x is not a simple cusp there exist coordinates (Xl> x2) centered at x, coordinates (Yl> Y2) centered at Y = f(x), and a mapping of the form (4.10) with the same 3-jet at 0 asf Hint: Use the coordinates in the proof of Theorem 2.2. In these coordinatesfis of the form (Xl> x2) -+ (Xl> h(x1' x2)); and at a cusp point which is not simple (4) Let X and Y be 2-manifolds andf: X -+ Y be a 2-generic mapping. Show that its only singularities are folds and simple cusps. (Hint: Show that for the map (4.10) Sl.l is the whole X 2 axis. Now use Exercise 1 and Exercise 2.) (5) Show that for 2-generic maps of n manifolds into n manifolds S2.1 occurs for the first time in dimension 7 and S2,2 occurs for the first time in dimension 10. (6) Let A be an element of Hom (V 0 V, W)s' Using (4.2) A is associated with a linear mapping A of V -+ Hom (V, W) of corank s. Let K = ker A and let L = coker A. Since K c V, there is a natural mapping (4.11) K* ® V* -+ K* ® K* given by restnctlOn; i.e., k* ® v* -+k* ® (v*IK). Since Hom (V, W) is naturally identified with V* ® W, we can regard L as a quotient space of V* ® Wand obtain a natural map (4.12) K* ® V* ® W -+ K* ® L given by id ® 7T where 7T is the obvious projection of V* ® W -+ L. Let K* 0 V* be the pre-image of K* 0 K* with respect to (4.11), and let K* 0 L be the image of (K* 0 V*) ® W with respect to (4.12). Finally let NA be the 156 Classification of Singularities Part I normal space of the point A to Hom (V 0 V, W)s in Hom (V 0 V, W). Show that there is a canonical identification (Hint: Use the same argument as in Exercise 6 of §I.) (7) Letf: X -7 R2 be a 2-generic mapping where X is a compact manifold. Show that S1C!) is a disjoint union of a finite number of circles and that S1.1C!) is a finite collection of points. Also show that no other types of singularities aside from fold points and simple cusp points occur. (8) Let f: X -7 X be 2-generic. How large must dim X be to allow the existence of an S3 singularity? Similarly for an S3.1 singularity? §5. The Thorn-Boardman Stratification It is clear, in principle anyway, how to define higher order versions of the Si./S. Iff: X -7 Y is 2-generic, Si.j.if) is defined to be the set of point~ in Si,tCf) where the map drops rank by k. This definition makes sense because, as we know from the previous section, Si,if) is a submanifold of X. If, by chance, Si.j.IcC!) turns out to be a manifold, we can define Si.j.Ic.lf) similarly. Thorn conjectured that for a residual set of maps this process could be continued indefinitely. This conjecture was proved by Boardman in his I.H.E.S. paper [6]. Specifically what Boardman proved is the following. Theorem 5.1. For every sequence of integers r1 ;::: r2 ;::: ... ;::: ric ;::: 0 one can define a fiber subbundle, ST1.....Tk of Jk(X, Y) (relative to the fibration Jk(X, Y) -7 X X Y) such that ifif is transversal to all the manifolds SI1 •...•I, where I < k, then ST1•....Tk(f) is well-defined and x E ST1.....Tk(f) - Y is a submersion. By the transversality theorem there is a residual set of maps, f: X -3>- Y, such that jkfm Tj for j = I, ... , s. For the moment just consider such maps, and let Tj(f) = (r.f) -l(Tj). (The transversality assumption assures that the T/f)'s are submanifolds of x.) Consider the following normal crossing condition, relative to Tr, ... , Ts: Let Xl, ... , Xs be distinct points of X with Xj in T/f) and f(xl) = ... = f(xs) = y. Let Hj be the tangent space to T/f) at Xj' Then (5.5) are in general position in Ty Y. We will prove: Proposition 5.6. The above normal crossing condition relative to the T/s is satisfied for a residual set of maps f: X -3>- Y. (Note that Theorem 5.2 is a corollary of this proposition: just take the T/s to be the Sf/S.) Proof The target map f3: Tl x ... x Ts -3>- Y X ... x Y is a submersion, so the pre-image ,8-l(L~Y) where ~Y is the diagonal in Y x··· x Y is a submanifold of Tl x ... x Ts. By the multi-jet transversality theorem there exists a residual set of maps f: X -3>- Y such that the multi-jet extension jskfis transversal to f3-l(~ Y). It is easy to see that if the usual k jet extension jkfis transversal to Tj for j = 1, ... , s thenjs"fis transversal to Tl x ... x Ts and vice-versa. Suppose both these transversality conditions are satisfied. The pre-image of Tl x ... x Ts with respect to js'tis just the set T1(f) x ... x Ts(f) with the generalized diagonal X(S) deleted. Call this set W. By Lemma 5.3 the map: j/f: W -3>- Tl X ..• x Ts is transversal to f3-1(~Y); so, by Lemma 5.4, the map f x ... x f: W -3>- Y X .•• x Y is transversal to ~ Y. We showed, however, in Chapter III, §3 that this is the same as the normal crossing condition. 0 Let's now go back to the conjecture we made earlier (with the normal crossing condition added). Conjecture. f: X -3>- Y is stable -¢;> f is a Boardman map satisfying NC. §5. The Thorn-Boardman Stratification 159 We will see in the next section that this conjecture is false for maps between 9-manifolds. (In fact it is false for maps between n manifolds of dim > 3 though in dimensions 4-7 it is "nearly right".) Nevertheless, let's attempt to give a proof of it. Iff: X -?>- Y is a Boardman map we can partition X into a disjoint union of subsets consisting of the nonsingular points: X - U Si (f) and the Boardman sets ST1.....Tk(f) with rk = O. The map is an immersion or a submersion depending on whether dim X :s; dim Yor dim X ~ dim Y; and if rk = 0 the map f: ST1.....Tk(f) -?>- Y is an immersion. This partition of X that we have just described is called the Thom-Boardman stratification. It has the property that f, restricted to each "stratum," is a particularly simple kind of stable map (either a submersion or an immersion with normal crossings). How do the various strata fit together, i.e., how do the closures of the higher dimensional strata intersect the lower dimensional strata? Obviously the story is quite complicated; but because of the transversality theorem, they might be expected to fit together in the same way that the universal strata ST1....,Tk fit together in Jk(X, Y). Hence, if we perturb f, the Thom-Boardman stratification of the perturbed map should look like the Thom-Boardman stratification of the unperturbed map. This suggests a way to prove the conjecture: construct an isotopy of X carrying the first stratification into the second, and then adjust it so that it conjugates the first mapping into the second. This" proof" is unfortunately based on an erroneous assumption, namely that if we know the stratification and know that on each stratum f is either an immersion or submersion, then we have enough data at our disposal to describefin the large. In fact, this data doesn't even describe the Co structure of the mapping; e.g., compare the two Morse functions (x, y) 1--7 x2 + y2 and (x, y) 1--7 x2 _ y2. One might suppose that if we know f, the Thom-Boardman data give us enough information to determine the structure of small perturbations off; but even this isn't true as we will see in the next section. Our "proof" does however have an intriguing air of plausibility about it, and we might ask whether some refined conjecture is true. It turns out that if we just restrict ourselves to the Co stability problem (two maps being equivalent if they can be conjugated, one into the other, by homeomorphisms of the source space and target space) then there is a finer stratification of the jet space than the Thom-Boardman stratification for which the "proof" above can be made rigorous. Thom is able to conclude from this that for all X and Y the Co stable maps form a residual subset of COO(X, Y). A careful proof of this can be found in a forthcoming book of John Mather [33]. See also [48]. In the next section we will see that the usual stable maps don't always form a residual subset of COO(X, Y) or even a dense subset. 160 Classification of Singularities Part I §6. Stable Maps Are Not Dense We have seen that immersions with normal crossings are stable; so stable maps are dense in C(X, Y) if dim Y ~ 2 dim X. Stable maps are also dense in C(X, Y) if dim Y = 1 (Morse theory) and if dim X = dim Y = 2 (the Whitney theory sketched in §2). For a time it was conjectured that stable maps are always dense in C(X, Y). Thorn and Levine proved in [18] that this is not the case. In fact, they showed that for maps of 9 manifolds into 9 manifolds, stable maps are not dense. In this section we will give their demonstration of this fact. We will first prove: Proposition 6.1. Let X and Y be manifolds of dimension n2• Then there exists a one-generic map f: X -+ Y such that Sn(f) is nonempty. Proof It is enough to prove that a map f exists taking on an Sn singularity transversely at a single point, say xo, because we can always find a nearby mapping which is one-generic. If this mapping is close enough to f it must also take on an Sn singularity at a point close to Xo by Exercise I of II, §4. Let N = dim X = dim Y. With choices of coordinate on Xand Ywe can identify C(X, Y)P.q with C(RN, RN)o.o; so to exhibit a map of X into Y taking on an Sn singularity transversely it is enough to construct a map germ f: (RN, 0) -+ (RN, 0) with this property. To do this we will need the following lemma. Lemma 6.2 Let V and W be vector spaces and let K be a subspace of V. If dim V ~ dim K·dim W, there exists an element A in (V* 0 K*) @ W such that the map (6.3) V-+K*@ W associated to A is onto. (For notation, see Exercise 6 of§4.) Proof The requirement that A be in (v* 0 K*) @ W means simply that A, regarded as a map of K @ K into W, is symmetric, otherwise A can be arbitrary. In particular, if H is a complement to K in V we can define A so that (6.3) is completely arbitrary on H. Since dim H ~ dim (K* @ W) dim K it is enough to show that there exists a B in (K* 0 K*) @ W such that the map (6.3), K-+K*@ W associated to B is injective. Let Wl be a one dimensional subspace of W. One can already construct a B in (K* 0 K*) @ W such that the induced map K -+ K* ® WI is injective. (Just take a nondegenerate bilinear form.) 0 To prove Proposition 6.1, choosefas in Exercise 1 of §3 and identify A with the last two terms of expression (3.8). If A is chosen as in the lemma, then Pf(f) Sn at 0 by Proposition 3.7. 0 We will now prove §6. Stable Maps Are Not Dense 161 Proposition 6.4. Let X and Y be manifolds ofdimension n2 • Letf: X --+ Y be one-generic./f Sn(f) is nonempty and n > 2 thenfis not stable. Combined with Proposition 6.1 this shows that stable maps are not dense in COO(X, Y) when dim X = dim Y = n2, n > 2. For the proof of Proposition 6.4 we will need Lemma 6.5. Let X and Y be manifolds and Wan open subset ofC 00 (X, Y). Then the set Aw = {O' E Jk(X, Y) I3fE Wand x E X with 0' = rf(x)} is open in Jk(X, Y). Proof. Let 0' = rf(x) be in Aw. Choose coordinate nbhds U and V about x andf(x) such thatf(i7) C V and let p be a function which is 1 near x and has support in U. Consider the set of maps g defined by g=ff+ph inU V inX- U where h is a polynomial map of degree :$; k. If the coefficients of h are sufficiently small this is well-defined (i.e., g( U) c V) and is in W. The setjkg(y) for y near x and h with coefficients small define an open nbhd of Jk(X, Y) in Aw. 0 Now let us go back to the diagram (4.1) of §4. S/2) --+ Hom (K 0 K, L) ~/ Sr The groups Diff(X) and Diff( Y) act on Sr and on Sr(2) in obvious ways as subsets of Jl(X, Y) and J2(X, Y). They also act on the bundles K and Lin the following simple way. If u in Sr has source at x and target at y we can think of u as an element of Hom (TxX, Ty Y). Then by definition K" = kernel u and L" = cokernel u. If g is in Diff( Y) then g acts on u by sending it to u' = (dg)y'u; so (dg)y maps the cokernel of u onto the cokernel of u' and leaves the kernels fixed. Thus Diff( Y) acts on K trivially and on L by the tangent bundle action. Diff(X) acts similarly; i.e., by the tangent bundle action on K and trivially on L. Lemma 6.7. The mappings in the diagram 6.6 commute with the respective actions of Diff(X) and Diff(Y). A proof is sketched in the exercises. We will now prove Proposition 6.4. Letf: X --+ Ybe one-generic. By the dimension formula in Proposition 1.1 dim Sn(f) = 0, so Sn(f) consists of a countable number of isolated points, say xo, Xl, X2, •••• Let aO = j2f(xo), al = Pf(Xl), etc. Now supposefis stable. Then the orbit, W" offin COO(X, Y) is open; so by Lemma 6.5, the set (6.8) AWl = {u E Jk(X, Y) I3g E W, and x E X with u = rg(x)} is open in J2(X, Y). If a is in the set AWl and is also in Sn(2) then it must 162 Classification of Singularities Part I be conjugate to one of the (Xi'S. This means that the open set Awr II Sn(2) in Sn(2) is covered by a countable number of orbits of the action of the group Diff(Z) x Diff( Y). Let a = j1f(xo). Since the top arrow in (6.6) is surjective, this means there is an open set in Hom (Ka 0 Ka, La) covered by a countable number of orbits of the group GL(Ka) x GL(La)' Suppose all these orbits were of dimension less than the dimension of Hom (Ka 0 KI1 , L(1)' Since they are immersed submanifolds they would have to be of measure zero by II, Lemma l.5; and so would their union. We conclude that one of the orbits has to be open. We wiIl show this is impossible by showing: Lemma 6.9. Let V and W be vector spaces of dimension n. Let G be the group GL(V) x GL(W) acting on the space Hom (V 0 V, W) by its standard representation. (The action is given as follows: let C be in Hom (V 0 V, W) and let (A, B) be in G, then (A, B).C(u) = C·B·A-l(u)for u in Vo V. Note that A(vl <29 V2) = AVl <29 AV2 and V 0 V is an invariant subspace of V <29 V under this representation.) Then G has an open orbit only when n < 3. Proof. The dimension of Hom (V 0 V, W) is n2(n + 1)/2 and the dimension of Gis 2n2 • Note that an orbit of G is diffeomorphic to some homogeneous space GIH. SO the dim (orbit) = dim G - dim H ~ dim G. (See Theorem A.I3). Clearly n2(n + I)/2 ~ 2n2 only when n ~ 3; so if n > 3, the dimension of G is less than the dimension of Hom (V 0 V, W), and the assertion is trivial. When n = 3 the dimensions are equal; but G contains a one-dimensional subgroup which acts trivially on Hom (V 0 V, W) namely {(c idv, c2 idw)}, c a nonzero real number; so the orbits are at least one dimension less than dim G. 0 This concludes the proof of Proposition 6.4. One can prove a slightly stronger result by the same technique; namely, one can show that stable maps are not dense in C: :! .:::::r::: ::::: .. :' . .. . . ... , ' ..... , .... . :::}:....:).::<::,/::::::::.::::.;.:::':'::::;.:.::'i::i:::::::::/::/:':':':::::'::;:;::: . 11/ = 7 1 1 1 ! I 1. 1 I , .. " r I , I • l ~ -t-.~.L-..~' 5 10 15 20 25 30 11 = dim X Figure 4: The region where stable maps are not dense is the shaded region including the boundary. 164 Classification of Singularities Part I Exercises (1) Let E, F, and G be vector bundles. Let p: E --7- F be a vector bundle morphism, and let T: E --7- G be a vector bundle isomorphism. Prove the chain rule formula for the intrinsic derivative: D(rop)x = ~.(Dp)x where T: Hom (ker p, coker p) --7- Hom (ker p, coker Tp) is the obvious map induced by T. (2) Determine the corresponding formula for a vector bundle isomorphism acting on the left. (3) Let f: X --7- Y be a smooth map, and g: Y --7- Y a diffeomorphism. Show: 82(g·f) = (dg)f(X)·(82f)x. (For notation, see Exercise 3 in §3.) (4) Derive a similar formula for composition by a diffeomorphism on the left. (5) Prove that the mappings in the diagram (6.6) commute with the action of Diff(X) x Diff( Y). Chapter VII Classification of Singularities Part II: The Local Ring of a Singularity §1. Introduction The Thorn-Boardman theory gives us a way of breaking up a map into simple constituent pieces; however, from the Thorn-Boardman data alone we usually cannot reassemble the constituent pieces and see what the map itself looks like. Consider for example the maps and g: R2 -+ R, (Xl> x2) f-+ X1 2 - X22. fand g have isolated Sl singularities at the origin and are regular everywhere else. However, their map germs at the origin are not equivalent, even under homeomorphisms of R2 and R, sincefhas an extremum at 0 and g does not. From the Thorn-Boardman data alone there is no way of computing the Hessian off at 0; and, of course, it is the signature of the Hessian which distinguishesf from g. (See II, Theorem 6.9.) In this chapter, we will be concerned with a more subtle invariant of a singularity-its local ring. To define this we recall some notation from Chapter IV, §3. If X is a point in a manifold X then C: = C:(X) denotes the ring of germs of smooth functions at x. This is a local ring and its maximal ideal, vllx = vltx(X), is the ideal of germs of functions vanishing at x. A map germ f: (X, x) -+ (Y, y) induces a map f* : Cy'" -+ C: by pull-back; and this is a morphism of local rings. Definition 1.1. Let f: (X, x) -+ (Y, y) be a map germ. The local ring off is the quotient ring: C:/C:f*vlly • This ring will be denoted :!Ilf and its maximal ideal ilif. Iff: X -+ Y is a map, not just a map germ, then at each point x in X we get the local ring of the germ off which we will denote :!Ilf(x) and call the local ring off at x. Example 1 (Immersions). Iff: (X, x) -+ (Y, y) is the germ of an immersion, we can choose coordinates centered at x and y so thatf is the immersion f*y, = x, f*y, = 0 i=l, ... ,n i=n+l, ... ,m where n = dim X and m = dim Y. Then vllx = f*vlly and :!Ill = R. Example 2 (Submersions). If f: (X, x) -+ (Y, y) is the germ of a submersion we can choose coordinates so that f is the canonical submersion i=i, ... ,m. 165 166 Classification of Singularities Part IT Then C':!*Jlty is the ideal generated by (x1o "" xm) and Pllf is the ring of germs of smooth functions in the remaining variables Xm + 10 ..• , X n• Example 3 (Morse functions). If f: X ---+ R is a Morse function with critical point at x, the product operation in the local ring Pllf induces a linear map (1.2) The kernel of the map (1.2) turns out to be one-dimensional, and is, in fact, all scalar multiples of the Hessian. (See Exercise 5 below.) Therefore up to scalar multiple the Hessian can be computed from Pllf • Example 4 (Generic maps between 2 manifolds). At a fold, we can write such a map in the form (Xl' X2) 1-7 (X1o X22). If we divide C[{' by the ideal (Xb X22) we get the truncated polynomial ring R[x2]/(X22), so f?Jtf ~ R[t]/(t2 ). At a simple cusp we can choose coordinates so that the mapping is (Xl' X 2) 1-7 (Xl' Xl X2 + X23). If we divide Co'" by the ideal (Xl' Xl X2 + X 23). (which is just the ideal (Xl' X 23)), we get the truncated polynomial ring R[X2]f(X23). Hence Pllr ~ R[t ]/(t3). We recommend that you try a few other examples on your own. The ring f?Jtr has been around a long time in algebraic geometry. For example, if f: X ---+ Y is a morphism between schemest, Pllf is in a natural sense the "fiber" of fat x. From the algebraic geometer's point of view, f?Jtf(X) is a much more natural invariant to attach to a singularity than, say, its Boardman data. I ts use in differential geometry is fairly recent, however; and is mainly due to Malgrange, Mather, and Tougeron. Its importance is indicated by the following theorem of Mather. (See [29].) Theorem 1.3. Let f, g: (X, x) ---+ (Y, y) be germs of stable maps. Then f and g are equimlent ifand ollly if2llr and!?Jlg are isomorphic as rings. Note that f and g are equivalerit if there exist germs ofdiffeomorphisms h: (X, x) ---+ ex, x) and k: ( Y, y) ---+ ( Y, y) such that g = k I h - 1 near x. The proof of this theorem is beyond the scope of this book. We will, however, see it corroborated by the simple examples we are going to discuss in the following sections. Exercises (1) Let f: (X, x) ---+ (Y, y) be map germs. Then f induces a map f*:C:/jt~+l---+C,:/J!t~+l for each k. Let g:(X,x)->-(Y,y) be another map germ. Show that j~f(x) = }"g(x) iff!* = g*. Thus!* is the k-jet off at x in algebraic disguise. (Hint: Let (Yl' ... , Ym) be coordinates centered at Y and .h, ... ,Ym be the associated elements of the quotient ring. Show that !*Yt is the k-jet of the ith coordinate function off) t Whatever that means. §2. Finite Mappings 167 (2) Let y: C;:lAyIe +1---» C; jJlxIe +1 be a ring homomorphism. Show that y = f* for some map germf. (3) From Exercises 1 and 2, deduce an isomorphism (of sets) Jk(X, Y)x,y;;:; Hom (Cy"'jJI/+\ C;jJlxle+l) (Hom meaning homomorphisms of rings). (4) Identify JlyjAy2 with the cotangent space of Y at y. Compare with IV, Lemma 3.3. Show that f* : AyjJly2 ---» JlxjJ("2 is just the transpose of (df)x: TxX ---» Ty Y. (5) Letf: Rn ---» R be the Morse function given by (xb ... , xn) 1-+ X12 ± X 2 2 ± ... ± xn2 • Show that at the origin ~;zrl?Jz? is just the space of linear functions in Xl, ... , Xn; and mf2fm/ is the space of quadratic functions with X12 ± X 22 ± ... ± xn2 identified to zero. Conclude that the kernel of the map (1.2) is the one-dimensional space spanned by Xl ® Xl ± ... ± Xn ® Xn(6) Verify that nonconjugate map germs can have the same local ring. (Hint: Try (Xl' X2) 1-+ (Xl' XIX2 + X23) and (Xl' X 2) 1-+ (Xl' X 23).) Why doesn't this contradict Mather's theorem? (7) Prove that the dimension of mf(x)/mf2(x) is equal to the dimension of Ker (df)x: TxX ---» Tf(x) Y. (Hint: Use Exercise 4.) §2. Finite Mappings Letf: X ---» Y be a smooth map. We say thatfisfinite at X if: (2.1) f is finite if it is finite at every point. Note that if dim X > dim Y, then dimR~rCx) = 00, (i.e., even formally the functionsfb" .,fn cannot generate an ideal of finite codimension in the formal power series ring R[[Xb ... , xm]] if m > n. See [61].) Therefore, in talking about finite maps, we are implicitly assuming dim X :S; dim Y. "Finiteness" implies among other things that the map f is locally" finite-to-one" at x. In fact, we have Proposition 2.2. Iff is finite at x, and a = f(x) then there exists an open nbhd U of X such that X is the only point in U mapping onto a. (In particular, iff is finite, it is "finite-to-one" on compact subsets of X.) Proof We can assume that X and Yare Rn and Rm respectively and that X = a = O. Let fb .. .,fm be the coordinate functions off The assumption (2.1) means that some power of the maximal ideal in Co'" is contained in the ideal (f1, ... ,fm). Therefore, there exists an open set U and an integer N such that on U, XiN = 2']'= 1 aijjj, the ai/s being smooth functions on U. Therefore, if the f,.'s vanish on U so do the coordinate functions Xi' In other words, zero is the only pre-image of zero. 0 Remark. The converse is not true. The map f: R ---» R, given by t---» exp (-1/t 2) is "finite-to-one" but dim ~f(O) = 00. 168 Classification of Singularities Part IT Supposef is finite at x. Let Pl> ... , Pk be elements of C:' projecting onto a basis of ~f(X) over R. Then the Malgrange Preparation Theorem says that Pl> ... , Pk generate C:' as a module over Cyoo. Conversely, if Pl> ... , Pk generate C:' as a module over C;: it is clear that they project onto a spanning set of vectors for ~r ... , Ym a coordinate system centered at the image point. We can assume that the functions PI, ... , Pk above are polynomials in xt's of degree < N. We will show that there exists a fixed open set, U, about X such that on U every polynomial in the x;'s can be written as a linear combination of the Pi'S with smooth functions of the y;'s as coefficients. First note that this statement is true for some U and for all polynomials of degree :5: N. (Note that for any monomial we can find an open set U since the statement is true for germs at x. Thus we can find an open set U which works for all monomials of degree :5: N. By linearity U works for all polynomials of degree :5: N). Now consider a polynomial of the form XfXp where deg P = N. XfXp can be written as a linear combination of the XfXp;'S and, hence, by induction as a linear combination of the l's themselves. Thus U works for all polynomials of degree> N as well. This proves that for all x' in U, C; is formally generated by the Pt's as a module over C;. Therefore, by the Malgrange Preparation Theorem, it is actually generated by them. 0 In some sense, dimR ~f(X) measures the multiplicity ofthe point X as a root of the equation f(x) = a. Over the complex numbers this vague statement can be made precise (see Remark 1 below), but over the reals we have to content ourselves with: Proposition 2.4. Let dimR ~f(X) = k. Then there exists a neighborhood, U, ofx such that every Y sufficiently close to f(x) has at most k pre-image points in U. Proof Choose U and Pl, ... , Pk as in the proof of Proposition 2.3. Let Xl> ... , Xr be pre-image points ofY in U, and let S = {Xl' ... , xr}. Let C; = EtX=l C;; be germs of COO functions on S, and let ~f(S) be the quotient ring C;IC;f*.Ay • Csoo is a finitely generated module over C;: with generators Pl, ... , Pk (by the same reasoning as in the proof ofProposition 2.3), so ~f(S) is a finite dimensional vector space over R with the images of Pl> ... , Pk as spanning vectors. On the other hand, the restriction map ~r X2) -+ (Xl> X1X2 + tX23) of R2 into R2. Given a in R2 what values can the sum (2.7) .2: dimR~f(p), pef- 1(a) take on? (2) Consider the analogous problem for the complex Whitney map I: (Xl> X2) -+ (Xl' X 1X 2 + tX23) of C2 into C 2. (3) Call dimR ~tCx) the multiplicity of the point X (with respect to the mapping I: X -+ Y). Prove Proposition 2.4, counting pre-image points with multiplicity. (Hint: Confirm that we did prove this stronger result in the text.) (4) Prove Tougeron's Theorem 2.6 for maps between two-manifolds using the Whitney theory. (5) Let X and Y be manifolds of dimension n and let I: (X, x) -+ (Y, y) be a map exhibiting an Si singularity. Show that dimR ~f > i(i + 1)/2. (Hint: Choose coordinates Xl> ..• , Xn centered at X and Yl> .•• , Yn centered at Y such thatI has the form I*Yi = iJ j .$ i I*Yi = Xi j> i where the leading term ofiJ is quadratic. Show that ml/m/ is the space of all quadratic polynomials in Xl, • •• , Xi with the quadratic terms of 11, ...,It identified to zero.) 170 Classification of Singularities Part II §3. Contact Classes and Morin Singularities We shall now attempt to give some insight into the geometric content of the local ring !?4?f(P) defined in §2. This will be accomplished through the notion of contact equivalence. Let Z be a manifold with p in Z. There is an obvious way to define a submanifold germ of Z near p; namely, two submanifolds are equivalent if they are identical in a nbhd of p and a submanifold germ is one of these equivalence classes. (Clearly we may think of a submanifold germ as a small piece of a submanifold.) Definition 3.1. Let A, Bl, and B2 be equidimensional submanifold germs ofZ near p. Then Bl and B2 have the same contact with A ifthere exists a germ of a diffeomorphism (bb"" bl)' Conversely, suppose that f: R'" x RI --+ R vanishes on graph b. Then we may write f(x, y) = (b1(x) - Yl)fl(X, y) + ... + (Mx) - YI)fz(X, y) where each}; is a smooth function. (To see this, let g(t) = f(x, (1 - t)y + tb(x)) for t in R. Then f(x, y) = g(O) - g(1) = f~ dg/dt(t) dt. Expanding dg/dt by the chain rule gives the desired result.) Thus Mkx (Od)(x) = b1(x)fl(X,0) + ... + Mx)fz(x, O) and is in the ideal (b1 , ••• , bl)' 0 Proof of Theorem 3.3. As noted above, it is easy to see that contact equivalent submanifold germs give rise to identical local rings. So we assume that :Jlt,4 , B, = :JltA •B2• Choose a tubular nbhd U = Rk X RI of A = Rk X {O} as in Lemma 3.4. The transversality assumptions imply that we can find smooth maps b1 and b2 : Rk --+ Rl so that Bi = graph (bi) (near p, of course). Let b1i, ... , bli be the coordinate functions of bi. Then ~(A, Bi) = (b1i, ..• , b/) by Lemma 3.6. Now the assumption that :JltA •B, = :JltA ' B2 implies that Jp(A, Bl ) = Jp(A, B2). Thus the calculation that ~(A, Bi) = (bI i , ... , b/) implies that there exist smooth functions gaP and hpy where 1 ::::; ex, (3, y ::::; I so that I I b,/ = 2: g"pbp2 and bp2 = 2: hi3yb/. P=1 y=l Let G and H denote the matrices (gan) and (hi3Y)' We claim that we may choose the h{3/s so that H(x) is invertible for all x near O. Using Lemma 3.5, there exists a matrix F whose entries are smooth functions (smoothness is obtained by looking at the proof of Lemma 3.5 and noting that the choices which are made can be made smoothly) such that the matrix K = F(I - GH) + His invertible for all x near O. A simple computation shows that b{32 = 2:~=1 kpyb/ so we may replace H by K. Now define cp: U = R'" X RI--+ U by cp(x, y) = H(x)y. By (*), cp: BI --+ B2 • Since cp is linear on fibers of U, cpiA = cpiRk x {O} = idR k x {O}. Finally, cp is a diffeomorphism on a nbhd ofp since H(O) is invertible. The mapping cp shows that BI has the same contact with A as B2 • 0 We now specialize this construction of contact equivalence to obtain results on 2l2f (p) where f: (X, p) --+ (Y, q). Lemma 3.7. Mf(p)=cJ)*Plxx{q}.graPhf \l'here j:X--+graphf is the canonical map and the ambient manifold Z is taken to be X x Y. Proof Choose coordinates Xl, ... , Xn near p and h, ... , Ym near q. Then Lemma 3.6 states that ~P.q> ... ,fm), 172 Classification of Singularities Part IT where fI. .. .,fm are the coordinate functions off (having identified X with X x {q}). Since/*(fI. ... ,fm) = (fI. ... ,fm) and rJilf(p) = Cp00 (Rn)/(fl, ... ,fm) the lemma is proved. 0 We now make the obvious definition. Definition 3.8. Two map germs J, g: (X, p) -l>- (Y, q) are contact equivalent ifgraph f and graph g have the same contact with X x {q} as submanifold germs of X x Y near (p, q). Notes. (1) Theorem 3.3 says thatfand g are contact equivalent as germs near p iff rJilf(p) = rJilg(p). We thus have a geometric interpretation of what it means for two map germs at a point to have the same local ring. (2) We would like to generalize the definition of contact equivalence so that we can interpret what it means for two map germs to have isomorphic local rings. For example, we could weaken the definition of contact equivalence to allow germs of diffeomorphisms 4> which leave A invariant and not demand that 4>IA = idA' It is then easy to see that if graphf is "contact equivalent" to graph g with respect to X x {q} with this new definition, then rJilrCp) is isomorphic to rJilg(p). The problem is in the converse statement. Suppose that rJilf(p) ~ rJilg(p), then what we would like to know is whether there exists a germ of a smooth diffeomorphism 4>: (Z, p) -l>- (Z, p) such that 4>* induces the isomorphism between rJilf(p) and rJilg{p). The thrust of the proof of Theorem 3.3 is, of course, the construction of such a 4>. Let us spend a moment to reflect on the problem. Suppose that there is no obstacle to lifting the isomorphism of rJilf(p) -l>- rJilg{p) to an isomorphism tjJ: C;'(X)-l>C;'(X). With a choice of coordinates we may CO'(Rn). Question: Is tjJ = 4>*? Since we know what tjJ(Xl)' ... , tjJ(xn) are, there is only one possibility for 4>, namely, 4>(x) = (tjJ(Xl)' ... , tjJ(xn». It is easy to show that 4>* = tjJ on any analytic function. The problem is that tjJ is not uniquely defined on flat functions, so 4>* does not have to equal tjJ. But we are saved by the Malgrange Preparation Theorem and Mather's Theorem (V, Theorem 1.2) that a stable map germ is determined by only a finite portion of its Taylor series. Thus to circumvent our problem we need only "jetify" our result and work with the local ring rJilrCp)/~pk(X) (for some appropriate k). It should be clear that any isomorphism between these finite dimensional local rings is induced by a smooth mapping. We introduce some terminology. A k-jet of a submanifold at p of a manifold Z is an equivalence class of submanifold germs at p where two germs A and B are equivalent if every smooth function of Z -l>- R which when restricted to A vanishes to kth order at p also, when restricted to B, vanishes to kth order at p. To analyze submanifold k-jets at p, we can assume by choosing coordinates that Z = Rn and p = O. Let H{;.! be the set of k-jets of I-dimensional submanifolds of Rn at O. First we claim that two equidimensional submanifold germs at 0, A and B, are in the same I-jet equivalence class iff ToA = ToB eRn. This is clear since a function tjJ: Z -l>- R vanishes to first order at p in the A (respectively, B) directions iff (dtjJMToA) = 0 (respectively, §3. Contact Classes and Morin Singularities 173 (dif!MToB) = 0). Thus we may identify Hr.l with the Grassmann manifold of I-planes in n-space by assigning, to each A in Hr.!> ToA in Gl.n • In this way Hr.l is a manifold. In fact, we have the following: Lemma 3.9. HT:.1 is a smooth manifold. Moreover, HT:.1 ~ Gl.n given by peA) = ToA makes HT:.1 into afiber bundle over Gl.n whose typical fiber is the fiber F of the bundle P'(RI, Rn-l)o.o --+ peRl, Rn-l)o.o. Proof Recall how the manifold structure of Gl •n is obtained. We assume that Rn comes equipped with an inner product. Then a nbhd of V in Gl.n is given by @y = {W E Gl •n l7Ty: W --+ V is a bijection} where 7Ty: Rn --+ V is just orthogonal projection. We then identify Hom (V, Vol) with @v by the mapping A f-+ graph A. We claim that we can identify p-l(@y) with @y x F (where V = Rl in the definition of F). Given a submanifold germ A, there exists a unique germ of a smooth map f: ToA --+ (ToA)ol such that A = graphf Now just note that two submanifold germs A and B which are tangent at 0 yield the same submanifold k-jet iff the corresponding f and g satisfy jkf(O) = jkg(O). (Note that since A = graphf and B = graph g, Pf(O) = pg(O) = 0.) The "iff" is clear since if!: Rn --+ R vanishes to kth order at 0 when restricted to A iffr(if!·f)(O) = 0, butr(if!·f)(O) = r(if!·g)(O) iffrf(O) = rg(O). Now let A be in p-l(@y). Then the maps 7Ty : ToA --+ Vand 7Ty.l: (ToA)ol --+ Vol are bijections. The map a: p-l(@v) --+ @v x F given by A f-+ (ToA, projection of r(7Tv.l·f·7Ty -1)(0) into F) where A = graphf is a bijection. We topologize HT:.1 by demanding that all such a be homeomorphisms and give HT:.1 a manifold structure by demanding that all such a are diffeomorphisms. We leave it to the reader to check that everything works right on overlaps. 0 Definition 3.2'. Let A, Bl , and B2 be equidimensional k-jet submanifold germs of Z near p. Then Bl and B2 are contact equivalent with respect to A ifthere exists a germ ofa diffeomorphism : (Z, p) --+ (Z, p) such that (A) = A and (Bl) = B2. Let ~lB = C;(A)/(~(A, B) + .A~+l(A)). Then we have Theorem 3.3'. The k-jets of submanifolds Bl and B2 are contact equivalent with respect to A iff ~~.Bl is isomorphic with ~~.B2' Definition 3.7'. Two k-jets of maps jkj(p) andrg(p), both with target q, are contact equivalent iff graphf and graph g are contact equivalent with respect to X x {q} as k-jet submanifolds of X x Y at (p, q). Let ~/(p) = ~f(P)/.A~+l(X). If a is a k-jet with source atp, then define ~~ = ~/(p) where rf(p) = a. Then the following is immediate. Proposition 3.10. Let a and T be in Jk(X, Y)p,q. Then a and T are contact equivalent iff ~~ ~ ~,. Definition 3.11. Let ~ be a local ring. The contact class Sge c Jk(X, Y) is given by Sge = {a EJk(X, Y) I~~ ~ ~}. 174 Classification of Singularities Part IT Remarks. (1) Let a and T be equivalent k-jets, then clearly £!il(J ~ £!il. so a and T are contact equivalent. Thus the contact class of a singularity is an invariant of that singularity type. (2) John Mather has proved that two stable k-jets are equivalent (with k = dim Y) iff they are contact equivalent. This result clearly demonstrates the importance of contact classes. (For the details see [29].) Tn the Thom-Boardman Theory one of the crucial technical tools is given by showing that the sets Si.i....,k are in fact submanifolds of the appropriate jet bundle. The same is true for contact classes. Theorem 3.12. Let £!il be a local ring. The contact class SGl is an immersed submanifold ofJk(X, Y). Note. In fact, Sge is a submanifold but, as before, we shall not need this fact. Proof As in previous theorems of this type, the crucial part of the proof is in showing that Sge n Jk(X, Y)P.q is a submanifold since SGl is clearly a subfiber bundle of Jk(X, Y) over X x Y. For this purpose, we may choose coordinates so that X = Rn, Y = Rm, and p = q = O. Now there is a natural action of the Lie group Gk(Rn)o (=invertible kjets on Rn at 0) on H~.l' Let (r(/»o be in Gk(Rn)o and let A be in H~.l' Then ... , xn) 1-+ (f2(X), ... , fn(x)) having the property that dXI, df2, ... , dfn are linearly independent at O. This is an open subset ofJk(Rn, Rn-l)o,o containing the image of ao. To each a in U we can associate an invertible k-jet {j in Jk(Rn, Rn)o,o-namely the k-jet of the map (Xl' ... , xn) 1-+ (Xl' f2, ... , fn). Let 2; be the vector space of all polynomials of degree ::; k in Xl, ... , Xn with zero constant term. 2; is the" typical fiber" of the fibration 7T*. We will show that the fibration 7T* is trivial over U by constructing an explicit trivialization Given p in 2; let kp be the k-jet of the map (Xl' ... , xn) 1-+ (p, X2, ... , xn). Then we define T(p, a) == kv' {j for p in 2; and a in U. It is easy to see that this is a diffeomorphism between 2; x U and (7T*) -l(U). Let 2;l k be the set of pin 2; satisfying :p (0) = ... = :kPk (0) = O. These uXI uXI conditions are independent so 2;l k is a codimension k subspace of 2;. We let the reader check that T gives us a smooth identification 2;lk X U ~ (7T*)-I(U) II Slk' -+ Hint: use (*) 0 Finally we shall show that the Morin singularities are actually singularities of the type studied in Chapter VI. We recall that for a map f: (X, p) -+ (Y, q), P is in Sl(f) ifffhas corank 1 at p and that p is in Sl.l(f) ifff!SI(f) has corank 1 at p (assuming-naturally-that Sl(f) is a submanifold). We can k continue this inductive construction of SOC!) as long as at each stage k-l SOC!) is a submanifold of X. The Boardman Theorem states that this is the case for a residual set off We will prove this theorem in the special case of the Morin singularities by proving the following. 176 Classification of Singularities Part IT Proposition 3.15. If jkf(f) Slk for all k ::;; dim X + 1, then Slk(f) = k SO(f)· Proof We need only prove this locally. Let p be in Slk(f). By choosing coordinates on X based at p and on Y based at f(p) we may assume that p = f(p) = 0 and thatf(x1, ... , xn) = (h(x), X2, ... , xn) where 8sh/8x1S(0) = ofor s ::;; k. In fact, SIJf) is given by the equations 8h/8xI = ... = 8I ... , Xn centered at Xo and a coordinate system Yl, ... , Yn centered at f(xo) such that f has the form (4.2) f*Yl = X2XI +... + X"XI"-l + Xl"+l f*Y2 = X2 The result is due to B. Morin (See [39]). Note that in dimension 2 it gives both Whitney canonical forms of Chapter VI, §2. Proof We can choose coordinates Xl> ..., xncenteredatxoandYl>" .,Yn centered at f(xo), so that f(xl> ... , xn) = (h(x), X2, ... , xn). Since Xo is an Slk singularity the local ring 8flxo) is generated as a vector space over R by 1, Xl, ... , Xl'" By the Malgrange preparation theorem every germ of a 178 Classification of Singularities Part II function at Xo can be written as a linear combination of 1, Xb ... , Xllc with smooth functions of the y's as coefficients; so in particular, we can write (4.3) Xllc + l = f*al + (f*a2)xI + ... + (f*alc+l)xllc or (4.3)' f*al = -(f*a2)xI - ... - (f*alc+l)xllc + Xlk+l the a;'s being smooth functions of y. Furthermore we can assume alc+l = 0 (Proof: replace Xl by Xl + (l!k)f*alc+l and leave X2,"" Xk fixed.) By comparing the two sets of (4.3) we see that al(O) = a2(0) = ... = aiO) = O. Let us now set X 2 = ... = Xn = Y2 = ... = J'n = 0 in (4.3) and expand both sides in powers of Xl' By assumption f*h = h(xl' 0, ... ,0) = CXllc+ I + .. " C being a nonzero constant and the dots indicating terms of degree> k + 1 in Xl' Therefore, if the Xllc+l terms in (4.3) are to be equal, we must have al(h, 0, ... ,0) = (l!C)YI + .. '. In particular oal!oh of. O. This means that the map: (YI. ... , Yn) f-+ (al(Y), Y2, ... , Yn) is a legitimate coordinate change in Y space. In other words we can assume to begin with that our x;'sand Y;'s satisfy: f*YI = f*a2xI + ... + f*alcxllc-1 + x/+ l (4.4) i = 2, ... , n. (We have changed at to -ai to make the first line more visually appealing.) Since the map f has the form of Exercise (4) of §3, the transversality condition says that ( of*Y) (olcf*Y)d OXI (0), ... , d ox/ (0) are linearly independent. This implies that are linearly independent. So this means that the differentials of the functions a2(0, Y2, ... , Yn), ... , alc(O, Y2, ... , Yn) are linearly independent at O. Permuting, if necessary, the y;'s we can assume the matrix ( oai (0») °Yi 2 ::; i,j ::; k is nonsingular. Therefore the mappings (Xl"", Xn) f-+ (xI.f*a2, ... ,f*alc, Xk+l,"" Xn) and §4. Canonical Forms for the Morin Singularities 179 are legitimate coordinate changes. With respect to the new coordinatesfhas the form (4.2). 0 Morin obtained canonical forms for all the" Morin" singularities of §3, not just the equidimensional ones. Here we shall confine ourselves to discussing one nonequidimensional example, the "cross-cap" for maps of n manifolds into 2n - 1 manifolds. This canonical form is due to Whitney [57], though the derivation we will give of it is due to Morin. First of all we need: Definition 4.5. Let X be an n man(fold, Ya 2n - 1manifold, andf: X -+ Y a smooth map. A point Xo in SI(!) is called a cross-cap ifPf(f1 SI at Xo. Note that for these dimensions, codim SI = n (See VI, Proposition 1.1) so cross-caps occur as isolated points of X. Whitney's result is: Theorem 4.6. Iff: X -+ Y has a cross-cap at x o, there exists a coordinate system Xl, ... , Xn centered at Xo and Yb ... , Y2n -1 centered at f(xo) such that f has the form: (4.7) f*Y1 = X12 f*Yi = Xi f*Yn+i = XlXi ;=2, ... ,n j= 1, ... ,n - 1. Proof We can choose coordinates Xl, ... , xn centered at Xo and Yb ... , Y2n -1 centered at f(xo) such that f*Yi = Xi for i = 2, ..., n. The set Sl(f) is the locus of points for which (4.8) alI = ofn+l = ... = Of2n-l = 0 oXI OXI OXI where}; = f*Yi' We let you check that the transversality assertion is equivalent to the assertion that (4.9) are linearly independent at the points where (4.8) holds. (Hint: Use the " D - BA -Ie" lemma of Chapter II, Lemma 5.2. See also Exercise (7) of VI, §l.) This means in particular that one of the differentials (4.9) must be nonzero when evaluated on OjOXI' By a linear transformation of the Y coordinates we can arrange: (4.10) It is clear from (4.10) that the local ring 8?tCxo) is generated by 1 and Xl' By the Malgrange preparation theorem we can write (4.11) 180 Classification of Singularities Part II a1 and az being smooth functions of the Y variables. If we make a change of x variables, substituting Xl + f*az/2 for Xl and leaving the other x/s fixed, we can make a2 = 0 in (4.11); that is, we can assume (4.11), Now let us set X2 = ... = Xn = Y2 = ... = Yn = 0 and expand the left hand side of (4.11)' in powers of Xl' By (4.10), f1 = f*Y1 = CX12 + ." and !s(x1, 0, ... ,0) = 0(X13) for s > n, with C #- 0, so we must have a1(Yl> 0, ... ,0) = (l/c)Y1. In particular, oa1/oY1 #- 0, so the map (Yl>"" Y2n-1) ~ (a1(y), Y2,···, Y2n-1) is a legitimate coordinate change. Replacing the old Y coordinates by the new Y coordinates, we have f*Y1 = X12 and we continue to have f*Yi = Xi for 2 :s; i :s; n. The remaining coordinate functions ,J; = f*Yi' i > n, can be written in the form (using the Malgrange Theorem) If we replace Yi by YI - gi(Yb"" Yn) for i = n + I, ... , 2n - 1 and leave the other y/s as before, we obtain the system of equations f*Y1 = X12 f*YI = Xi i = 2, ... , n f*Yn+j = x1hn+lx12, X2,···, Xn) j=l, ... ,n-l. This is almost the form we want. In fact if we can show that the following are legitimate changes of coordinates: (Xl>"" xn) ~ (Xl' hn+1(X12, ... , Xn),·.·, h2n_1(X12, ... , Xn», and (Yl>"" Y2n-1) ~ (Yl, hn+1(y), ... , h2n - 1(y), Yn+l>"" Y2n-l) then we will have precisely the set of equations (4.7). To show this we must go back to the transversality condition (4.9). At Xl = 0 this reduces to the condition that be linearly independent at 0, or, in other words, that the matrix ( ohl )n+l':IS2n-l OXj 2.:1sn be nonsingular. This however is precisely what is needed to make the changes of coordinates above legitimate. 0 Remark. We showed in Chapter II that for every n manifold X we can find an immersionf: X -+ R2n. Cross caps arise as obstructions to lowering the dimension of this immersion by 1. In (60] Whitney proved that every n manifold can be immersed in R2n-1, the idea of the proof being to delete cross caps, two at a time, from a generic mapping. Pictures of cross-caps in §4. Canonical Forms for the Morin Singularities 181 R3 can be found in classical books on topology in connection with the problem of immersing p2 topologically in R3 (See exercise 3.) Exercises (1) Let f: X -+ R2n be an immersion of the n manifold X. Let v be a regular value of the induced map (df) : TX -+ R2n and let 7T: R2n -+ R2n-1 be a surjective linear map with v in its kernel. Show that 7T·f has no singularities except cross-caps (Compare with II, §1, Exercise 1.) (2) Describe the image of the map (Xl> X2) ~ (XI2, X2, XIX2). What are the images of the curves Xl = const and X2 = const. Show that this map is 1-1 except along the "double line" X2 = 0 (See Figure 6). L Figure 6: The Cross Cap \ \ I (3) Construct a topological immersion of p2 into R3 whose image is a "cross cap". (See Figure 7.) Ob b a Figure 7: The Topological Cross Cap 182 Classification of Singularities Part IT §5. Umbilics From Exercise 5 of §2 we know that S2 singularities cannot occur with multiplicity 1, 2, or 3. An S2 singularity which occurs with multiplicity 4 is called an umbilict. Umbilics are the simplest Sz singularities, and are the only ones that can occur stably in dimensions < 6. We will begin our study of them by showing that for singularities of multiplicity 4, the only local rings that are allowable are those on the following list: (5.1) (5.2) (5.3) (5.4) (5.5) R[t]J(t4) R[x, y]J(x2 - y2, xy) R[x, y]J(x2, y2) R[x, y]J(x2, xy) R[x, y, z]J(x2, y2, Z2, xy, xz, yz). In fact we will prove Proposition 5.6. Let fJJl be a local ring over R with dimR f!il = 4. Then fJJl is isomorphic to one of the rings on the above list. Proof Let j/ be the maximal ideal of fJJl. Suppose first that dimR j(/j(2 = 1. Let t be an element of j( - j(2. j(i = {cti} + j(i+l, so the mapping of R[t] into fJJl is onto. The kernel is generated by a polynomial p(t), which we can write as tkq(t), q(t) having a nonzero constant term. The image of q(t) in fJJl is invertible, so the kernel is also generated by tk. Hence fJJl is isomorphic to R[t]/Ctk). Since dimR fJJl = 4, k = 4. Next suppose dimR j(Jj(2 = 2. We note that (5.7) by Nakayama's lemma. (If j(i = j(i+l then j(l = 0.) Therefore if fJJl is 4 dimensional, dim jf2/jf3 = 1 and jfl = 0 for i > 2. Consider now the bilinear map jf/jf2 ® jfJjf2 -+ jf2 induced by the product operation on the local ring. We will, for the moment, fix a basis vector in jf2 and regard this as a map: (5.8) i.e., as a symmetric bilinear form on jf/j{2. This form cannot be identically zero otherwise jf2 = 0; therefore there are three possibilities for it: it can be nondegenerate and definite, nondegenerate and indefinite, or degenerate. We will show that if the first is the case then fJJl is isomorphic to the ring (5.2). In fact if (5.8) is definite we can find a basis for jf/j{2 such that X·X = t Because the "umbilical points" of a surface in R3 are the points where its normal bundle map exhibits this kind of singularity. §5. Umbilics 183 y.y and x·y = O. Therefore, [Jf is isomorphic to the polynomial ring in two variables divided by the ideal of relations (X2 - y2, xy). A similar argument shows that if (5.8) is nondegenerate and indefinite then [Jf is the ring (5.3) and if (5.8) is degenerate, [Jf is the ring (5.4). Finally if dim jtjjt2 = 3, j/2 = 0 by (5.7) and [Jf is the ring (5.5). 0 (5.1) is the local ring of an Sl3 singularity (a "swallow's tail"). In the equidimensional case, which is the only case that we shall consider here, the local ring (5.5) cannot occur (by Exercise 5 of §2.) (5.2) (5.3) and (5.4) are all possible candidates for umbilics. Definition 5.9. An umbilic is called hyperbolic if its local ring is (5.3), elliptic if its local ring is (5.2) and parabolic if its local ring is (5.4). So the name for an umbilic is given by the name ofthe associated quadraticform (5.8.) Our main theorem about umbilics will be that two generic umbilics are equivalent (as map germs) if and only if they are of the same type. For simplicity we shall just prove this for the elliptic and hyperbolic umbilics. (The parabolic case will be treated in the exercises.) Specifically, we will prove: Theorem 5.10. Let X and Y be n dimensional manifolds, n ~ 4. Let f: X -+ Y be a smooth map exhibiting either an elliptic or hyperbolic umbilic at Xo in X. Suppose jIf7fl S2 at Xo. Then we can find coordinates, Xl, ... , Xn centered at Xo andYI, ... , Yn centered at f(xo) such that f has one of the follOlving two canonical forms: (5.11) (5.12) Hyperbolic case f*YI = Xl2 + X3X 2 f*Y2 = X22 + X4Xl f*Ya = X3 Elliptic case f*Yl = X12 - X22 + X3Xl + X4X2 f*Y2 = XIX2 + X4Xl - X3 X2 f*Y3 = X3 f*Yn = Xn Note that the assumption n ~ 4 is essential. The transversality condition cannot be satisfied in dimensions < 4. In the proof of (5.11) we will need the following Lemma 5.13. Let (5.14) 184 Classification of Singularities Part IT be a quadratic form whose coefficients are smooth functions ofa set ofparameters: Z = (Zl' .•• ' Zm). Suppose that Ivhen Z = 0 (5.14) is equal to X12 - X22. Then one can find a rotation, whose coefficients are smooth functions ofz, such that S(O) is the identity, and such that in the rotated coordinate system (Xl> X2) (i.e., Xl = SllXl + S12X2, X2 = S21Xl + X22X2) (5.14) has the form OX12 + 5x? where 0(0) = 1 and 5(0) = -1. Proof Solve for the eigenvalues of the matrix A = [: !]. Since A is close to [~ ~1] for Z small the eigenvalues are distinct and are close to 1 and -1. Moreover they depend smoothly on z. For the eigenvalue close to 1 we can find an eigenvector (1, T) with Ta smooth function of z and T(O) = o. Since A is symmetric the other eigenvector will be (- T, 1). Let S be the rotation vI ~ T2 (~T ;). Note that 0 and 5 are the eigenvalues of A. 0 We will now derive the normal form (5.11). Since Xo is in S2 we can choose a coordinate system Xl> ••• , Xn centered at Xo and Yl, ... , Yn centered at f(xo) such thatfhas the form (5.15) (Xl> ..., xn) ~ (J;.(X),f2(X), X3, ..., xn) where the linear terms in f1 and f2 vanish and the quadratic terms are of the form (5.16) fl = Xl2 + .. . f2 = X22 + .. . the dots indicating quadratic terms like X3Xl, X32, etc. and higher order terms in all the x's. Therefore the local ring will be generated over R by 1, Xl, X2, and X1X2. By the Malgrange preparation theorem every germ of a function at ocan be written as a linear combination of 1, Xl> X2 and X1X2 with smooth functions of Y as coefficients. In particular we can write: (5.17) Xl2 - X22 = f*al + f*blXI +f*CIX2 +f*dIXIX2, and Xl2 + X22 = f*a2 + f*b2xI +f*C2X2 + f*d2XIX2 where the a's, b's etc. are smooth functions ofY vanishing at.y = O. Replacing Xl by Xl + f*d2/2 and leaving the other coordinate fixed we can arrange that d2 == o. Applying the lemma to Xl2 - X22 - f*dlXIX2 we can also arrange that dl == O. Note that since the change of coordinates (Xl> X2) -+ (Xl> X2) is given by a rotation we have that Xl2 + X22 = Xl2 + X22. Dropping the §5. Umbilics 185 -'s on j\ and X2 and using the fact that il(O) = I and 5(0) = -1 we may solve for X12 and X22 to get and X22 = J*a2 +J*(32X1 +J*Y2X2 the a's, (3's and Y's being functions of y. One last simplification is possible. Replacing Xl by Xl +J*(31/2 and X2 by X2 +J*Y2/2 we can assume (31 and Y2 are zero. Therefore, with Y = -Y1 and (3 = -(32' we have J*a1 = Xl2 + J*YX2 J*a2 = X22 +J*(3Xl (5.18) We continue, of course, to haveJ*Yi = Xi for i > 2 since we have not made any changes in these coordinates. Now we will set X3 = Y3 = ... = Xn = Yn = oin (S.18). Comparing the quadratic terms on both sides and using (S.16) we see that: al(YI. Y2, 0, ... ,0) = Yl + .. . a2(YI. Y2, 0, ... , 0) = Y2 + .. . the dots indicating terms of degree> I in Yl and Y2' (Now in the coordinate changes above Xl = Xl + ... and x2 = X2 + ... where· .. stands for higher order terms so that (S.16) is still applicable.) This implies that the map (YI."" Yn) f-+ (al(Y), a2(Y), Y3,"" Yn) is a legitimate coordinate change. In the new coordinates we have (5.19) J*Yl = Xl2 +J*YX2 J*Y2 = X22 +J*(3Xl J*Y3 = X3 This is nearly the canonical form we want. In fact if we can show that and are legitimate coordinate changes, then in the new coordinates (S.19) will have the form (S.lI). We will show that these coordinate changes are allowable precisely because of the transversality hypotheses. In fact letting hl(x) and h2(X) denote the right hand terms on the first two lines of (S.19), the set S2(f) is defined by the set of 4 equations (5.20) 8hl = 8h2 = 8hl = 8h2 = 0 8Xl 8Xl 8X2 8X2 . 186 Classification of Singularities Part IT Use exercise 7, VI, §1 to show that the transversality hypothesis reduces to the assertion that on the set defined by (5.20) the differentials (5.21) are linearly independent. When we replace hl and h2 by X 12 +f*YX2 and X22 +f*f3xl this condition reduces to the condition that dXb dx2, d(f*f3), d(f*rx) be linearly independent at 0 which is precisely what is needed to legitimatize the above coordinate changes. This concludes our proof for the hyperbolic case of Theorem 5.10. The derivation of the canonical form (5.12) for the elliptic case is similar. We will indicate what changes need to be made in the proof above: (1) Lemma 5.13 has to be replaced by the following "elliptic" analogue: Lemma 5.22. Let (5.23) be a quadratic form whose coefficients are smooth functions ofa set ofparameters, z. Suppose that for z = 0 (5.23) is just the form X 12 - X22. Then there exists a function ,\ depending smoothly on z such that '\(0) = I and such that with respect to the coordinates Xl = ,\xl' X2 = (l/,\)x2 (5.23) has the form a(x12 - X22). Proof This is much easier than Lemma 5.13. Just take ,\ to be (aJ(a - b))1/4 (2) Now choose the coordinates (5.15) such that fl = X12 - X22 +... f2 = X1X2 +... and show that the local ring is generated over R by Xl, x2 and X22. (3) Using the Generalized Malgrange Preparation Theorem show that X1 2 - X22 = f*al + f*blXl +f*C1X2 + f*d1X22 and X1X2 = f*a2 + f*b2Xl + f*C2X2 + f*d2X22 where the a;'s, b;'s etc. are smooth functions of y vanishing at O. Replace Xl by Xl + f*d2X2 to make d2 = 0 and apply the lemma to make dl = O. (4) By a linear change of coordinates makes CI = b2 and bi = C2' The rest of the proof is as before. 0 Exercises (1) Show that over the complex numbers there are just 4 local rings with dime:Jf = 4. §s. Umbilics 187 (2) Let z and w be complex numbers. Show that the map of C 2 into C 2 defined by (5.24) (z, w) 1-+ (Z2 + ZW, w) has an elliptic umbilic at 0 when viewed as a map of R4 --+ R4. (Hint: Compare with the canonical form (5.12).) For w fixed and real sketch the curve: u = Z2 + ZW, Izl = canst., in the u plane. Show that it has the appearance indicated in Figure 8. Izl = canst < M Izi = lcl Izl lcl= canst> 2 2 2 (a) (b) (c) Figure 8 (3) Show that for the map (5.24) the image of Sl in the 3 dimensional plane: 1m w = 0 has the appearance of the cusped surface depicted in Figure 9: The elliptic Umbilic 188 Classification of Singularities Part IT Figure 9. (Hint: Show that Sl is given by the equation Izl = Iw1/2. Use the identity ( OU1 0U2) OXl oX2 lou 12 lou \2det = - - --:: OU10U2 OZ OZ OX2 0X2 where U = Ul + iU2 and Z = Xl + iX2. Now look at Figure 8(b). (4) Let/: R4 -+ R4 be the mapping (5.25) (Compare with 5.11.) Sketch the images of some of the lines Xl = const., X3 = const., X4 = const., and of some of the lines X2 = canst., X3 = canst., X4 = const. (5) Show that for the map (5.25) the image of Sl in the 3 dimensional plane Y4 = 0 has the appearance of the surface depicted in Figure 10. Figure 10: The Hyperbolic Umbilic §S. Umbilics 189 (6) Letf: Rn -+ Rn be the mapping (Xl> ... , xn) 'r-+ (fl(x),f2(X), X3, ..•, xn) where and f2 = a2(Xl> X2) +... the dots indicating quadratic terms in XIX3, X32, etc. plus higher order terms in Xl> ... , Xn• al and a2 are assumed to be homogeneous quadratic polynomials in Xl and X2 alone. Show (a) 0 is a parabolic umbilic -¢> al and a2 have a common linear factor. (b) 0 E S2,l => at = Ct(SlXl + S2X2)2, i = 1,2. (c) 0 E S2,2 => al = a2 = O. (7) Let Q be the vector space consisting of all pairs (ai, a2) where al and a2 are homogeneous quadratic polynomials in (Xl> x2). Let P be the subset of Q consisting of all (ai, a2) for which al and a2 have a common linear factor. Let U be the subset of P consisting of all (al> a2) for which al and a2 have a common quadratic factor (i.e., are constant mUltiples of each other.) Let W be the subset U consisting of all (ai' a2) for which Prove: i = 1,2. (a) P - U is a submanifold of Q of codimension 1. (b) U - W is a submanifold of Q of codimension 2. (c) W - {O} is a submanifold of Q of codimension 3, (An elegant way to do this exercise is to define these sets using the resultant, R(al> <;(2), of the polynomials al and a2' See van der Waerden, [52], Vol. 1, Chapter IV, §27.) (8) Let PA be the local ring: R[xl> X2]/(X12, xl, X12X2, X1X22). If X and Yare n dimensional manifolds show that the contact class, SUI, in J2(X, Y) is a submanifold ofcodimension 7. Iff: X -+ Y has the property: Pfm S2, show that X E S2,l(f) -¢> j2f(x) E SUI. (Hint: Use Exercises 6 and 7, and the same kind of trick as in the proof of Proposition 3.l3.) (9) Let PA be the local ring: R[Xl, x21/(x13, X12X2, X1X22, X23). If X and Yare n dimensional manifolds show that the contact class SUI in J2(X, Y) is a submanifold of codimension to. Iff: X -+ Y has the property: Pf(fi S2, show that X E S2.2(f) -¢> Pf(x) E SUI. Hint: Use exercises 6 and 7, and the same kind of trick as in the proof of Proposition 3.13. (to) Show that the map f: R4 -+ R4 by 190 Classification of Singularities Part II has an S2,O singularity at the origin and satisfies the transversality condition Pfm S2' Show it is not an umbilic. Hint. Use Exercise 7, VI, §l. Also show by an example that a small perturbation off will yield an umbilic at O. (11) Let X and Y be n manifolds. Show that if n < 6, the set of maps, f: X -+ Y, which exhibit only Morin singularities and umbilics, is residual. (Hint: Show that the phenomenon illustrated by exercise 10 occurs generically only in dimensions ;:::6. Use part (b) of Exercise 7, and the same sort of trick as in the proof of Proposition 3.13.) (12) Let SE, SH and Sp be the contact classes in J2(X, Y) associated with the rings (5.2), (5.3) and (5.4) respectively. Show that the codimension of the submanifolds SH and SE in J2(X, Y) is 4 and that Sp is a submanifold of codimension 5. (Hint: For Sp, use part (a) of Exercise 7.) (13) Prove the following: Theorem 5.26. Iff: X -+ Y has a parabolic umbilic at Xo and Pfm Sp at Xo, then there exist a system of coordinates XI. ... , Xn centered at Xo and YI. ... , Yn centered at f(xo) such that f has the canonicalform: (5.27) f*Y1 = X12 + X2XS + X22X4 f*Y2 = X1X2 + X2XS f*ys = Xs f*Yn = Xn• (Note that for the transversality condition to hold the dimension of X must be ~5.) Hint: Assumefis in the form (5.15) with f1 = X12 + .. . f2 = X1X2 +.. . the dots indicating terms in XSX1' X4X2, XS2 etc. and higher order terms. Show that the local ring is generated over R by 1, XI. X2, and X22. Using the Malgrange Preparation Theorem show that X12 = f*a1 +f*b1X1 +f*C1X2 + f*d1x22 and X1X2 = f*a2 +f*b2X1 +f*C2X2 + f*d2x22 the a's, b's etc. being smooth functions of Y vanishing at O. Use algebraic tricks to make b1 = b2 = d2 = 0, and finally make a coordinate change in Y1. Y2 so that f has the form f*Y1 = X12 +f*f3X2 +f* 8x22 f*Y2 = XIX2 +f*YX2 f*ys = Xs f*YrI = Xn• Finally use the transversality condition to show that f3, 8 and y can be introduced as new coordinates in place ofYs. Y4 and Ys. §6. Stable Mappings in Low Dimensions 191 §6. Stable Mappings in Low Dimensions Using the results of this chapter it is not hard to get a picture of what stable mappings look like in low dimensions. For simplicity we will restrict ourselves to the equidimensional case: dim X = dim Y. We will first make a list of the kinds of "nonremovable" singularities that can occur in dimensions ::;4. (By "nonremovable" we mean they can occur transversely, and therefore cannot be eliminated by small perturbations.) (6.1) dim 1 dim 2 dim 3 dim 4 S1 (folds) Sl, S12 (folds, cusps) S!> S12, S13 { S1' S12' S13' S14 elliptic and hyperbolic umbilics (S2,0) That the Slk singularities which we have listed are the only ones that can occur is clear from Corollary 3.14. In §5, exercises 11 and 12, we indicated how to prove analogous results for the S2 singularities on the list above. S3 singularities are, of course, removable as long as dim X < 9. In particular, for dimensions ::; 4, a stable map can only exhibit the above singularities; and, being stable, it must exhibit them transversely. Summarizing we have: Proposition 6.2. In dimensions ::; 4, for a map germ to be stable it is necessary and sufficient that it exhibit only singularities on the list (6.1), and that it exhibit these Singularities transversely. Remark. It is instructive to verify directly from Mather's criterion that the canonical forms described for the singularities in the list (6.1) which are given in Theorems 4.1 and 5.10 do indeed represent infinitesimally stable germs. For example, we verify this criterion for the Morin singularities where f(x!> ... , xn) = (f1(X), X2, ... , xn) and f1(X!> ... , xn) = X2X1 + ... + XkX/-1 + X1k+1. Locally the equation T = (df)a) + YJ ·ftranslates to the system of functional equations [ T1 = (X2 + 2X1X3 + ... + (k - 1)X1k-2Xk + (k + 1)X1k)~1 + X1~2 + ... + X1k-1~k + YJ1 • f (*) T2 = ~2 + YJ2 • f Tn = ~n + YJn • f where we must solve for the functions ~i(X) and YJi(Y) given the functions Tj(X). By V, Theorem 1.2, we need only solve the equations (*) to order n and by Arnold's criterion (V, Proposition 1.13) we need only solve equations (*) when TZ = Xs and Ti = 0 for i f= I. When I > 1, let ~z = xs, ~i = 0 i f= I, and YJi = 0 for all i to solve (*). So we assume that T1 = xs, and Tj = 0 for i > 1. If T1 = Xl, then let ~2 = 1, YJ2 = -1, ~1 = ~3 = ... = ~n = 0, and YJ1 = YJ3 = ... = YJn = O. If T1 = Xs for s > 1 then let YJ1(y) = Ys (so that YJ1 ·f(x) = xs), 192 Classification of Singularities Part II ~l = ... = ~" = 0, and 7]2 = ... = 7]" = O. We leave it to the reader to check that umbilics are infinitesimally stability. We will now try to find global criteria for stability. We made a tentative investigation of this problem in §5 of Chapter VI. We will apply our conclusions there to maps in low dimensions. There are at most six distinct types of singularities on the list (6.1); so for a map f: X -+ Y, with dim X = dim Y :s; 4, we can partition the set of singularities of f into six disjoint subsets: Xl>"" Xe. We showed in VI, §5 that for f to be stable it must satisfy the following "normal crossing" condition. Condition NC. Given distinct points XI, in Xf,. r = 1, ..., k such that f(x!) = ... = f(x!k;) = y then the subspaces (df)x"(Tx,,X!,) for r = 1, ..., k of Ty Yare in general position. (See VI, Proposition 5.2.) Inter alia, this condition implies that J, restricted to each "stratum" Xi> is an immersion with normal crossings, and that the images of these strata intersect transversely. For example it implies. that a point cannot simultaneously be the image of an umbilic and of an Sl singularity. Theorem 6.3. Let dim X = dim Y:s; 4 and letf: X -+ Y be a map which exhibits only the singularities on the list (6.1) and exhibits these transversely. Then a necessary and sufficient condition for f to be stable is that it satisfy the condition NC described above. We will deduce this from a slightly more general result. Theorem 6.4. Let X and Y be n dimensional manifolds, andf: X -+ Ya map which is of rank ;::: n - 1 everywhere. Then f is stable if and only if it satisfies the transversality conditions of Morin: k=I, ... ,n+l and, in addition, satisfies the condition NC,Jor the stratification, Xk = Slk;(f), k = 1, ..., n + 1. Proof The necessity is obvious. To prove the sufficiency we only have to show that Mather's criterion for infinitesimal stability is true on the multijet level. (V, Theorem 1.6.) The specific result we are going to prove is the following "canonical form" lemma. Lemma 6.5. Letf: X -+ Y be a map satisfying the hypotheses ofTheorem 6.4. Let Pl, ... , P. be points of X such that P! is in Sl,,(f) for i = 1, ... , s and such thatf(Pl) = ... = f(p.) = q. Then we can choose a coordinate system Yl>"" y" centered at q and coordinate system Xl(l), ••• , x,,(t) centered at each of the p;'s such that f has the canonical form (4.2) Simultaneously in each of these coordinate systems. §6. Stable Mappings in Low Dimensions 193 Proof For simplicity we will just consider the case s = 2 (to spare the reader rather than the authors; the general case is not any harder, but there are more indices to keep track of.) Let P = PI and P' = P2; and for the moment let us just considerf in the vicinity ofp. Ifp is an S:f.k singularity we can choose coordinate systems centered at p and at q such that f has the canonical form (4.2). Let Xl be the first coordinate function in the coordinate system at p, and Yl, ... , Yk the first k coordinate functions in the coordinate system about q. Then the tangent space to Slk(f) at p is characterized by the equations dXI = df*Y2 = .•• = df*Yk = 0 (Compare with §3, Exercise 4) and the image space by the equations (6.6) dYI = dY2 = ... = dYk = O. If we make a similar choice of coordinates relative to p' and q, then the tangent space to Sk,(f) at p' is characterized by the equations dx~ = df*y; = ... = df*Y~' = 0 and the image space by the equations (6.6), dy~ = ... = dy~ = O. By assumption the subspaces (6.6) and (6.6)' are in general position. This means that the differentials dYb ... , dYk, dy~, ... , dy~, are linearly independent at q, and, therefore, that Yb ... , Yk, y~, ... , y~, can be introduced as the first k + k' coordinate functions of some coordinate system. In this coordinate system we will simultaneously have the Morin canonical form for an Sl1< singularity at p and the Morin canonical form for an Sik singularity atp'. 0 From this it is easy to prove Theorem 6.4. One merely checks that the multijet conditions of (t) in V, Theorem 1.6 are satisfied using the multijet canonical forms given in Lemma 6.5. The details are similar to the argument in the remark following proposition 6.2 and we leave them to the reader. To prove Theorem 6.3 we observe that since the condition NC is satisfied no multi-germ can have a source consisting ofan umbilical point and another singular point since the umbilics occur as isolated points (in dimension 4) or not at all (in dimension < 4). Therefore, to verify the multi-jet criterion for stability we only have to verify it for multi-jets involving singular points of type Slk; so we are back in the situation of Theorem 6.4. 0 Appendix §A. Lie Groups The theorem that we need states that the orbits of a Lie group action are immersed submanifolds. First we define and sketch some facts about Lie groups. Definition A.I. Let G be both a group and a smooth manifold. Then G is a Lie group if the mapping of G x G --+ G given by (a, b) 1-+ ab- 1 is smooth. Examples (a) Rn where the group operation is addition. (b) Sl where the group operation is addition of angles. More generally Tn = Sl X ... X Sl = n-torus is a Lie group where the group operation is coordinate-wise addition. Note. The only abelian connected Lie groups are Rn x Tm (no proofl). (c) All matrix groups. For example GL(n, R) = group of n x n invertible real matrices, O(n) = group of n x n orthogonal matrices, and SL(n, R) = group of n x n real matrices with determinant equal to 1. All of these groups are submanifolds of Rn2. (See Theorem A.7.) Also GL(n, C) = group ofn x n invertible matrices with complex entries. Here we view GL(n, C) as a submanifold of Rn2 EEl Rn2. (d) Let X be a manifold with p in X. Let a be a k-jet (k > 0) in Jk(X, X)p,p. Then a is invertible if any representative of a is a diffeomorphism on a nbhd of p. The invertible k-jets form a group under composition and a manifold since they are an open subset of Jk(X, X)PoP' We shall denote the set of invertible k-jets at p by Gk(X)po To see that Gk(X)p is a Lie group, we choose coordinates near p and inspect Gk(Rn)o. Further we may identify Jk(Rn, Rn)o,o with polynomial functions from Rn --+ Rn of degree::; k mapping o to O. Under this identification Gk(Rn)o is the open subset of polynomial mappings f for which (df)o is nonsingular. Here we see that the group operation is given by composition of the polynomial mappings but throwing away all terms in the composition of degree> k. This is clearly a smooth operation. It is also not hard to see that the mapping b 1-+ b- 1 in G"(Rn)o is a smooth operation. The tangent space to a point in a smooth manifold is always locally diffeomorphic to the manifold (using chart mappings). On a Lie group, G, there is a naturally defined identification exp: TeG --+ G which is a diffeomorphism on a nbhd of O. We shall construct this mapping. Let v be in TeG. Then v along with the group action defines a vector field on G. For a in G let La: G --+ G be defined by LaCg) = a·g. Clearly La is a 194 §A. Lie Groups 195 diffeomorphism. Define ~av = (dLaMv). The smoothness of the group action guarantees that~V is a vector field on G. Also ~v satisfies (dLa)b~bV = ~~.b for all a, bin G. Definition A.2. A vector field ~ satisfying (dLa)~ = ~ for all a in G is called left invariant. Lemma A.3. Let ~ be a left-invariant vector field on G. Then ~ is the infinitesimal generator of a globally defined one parameter group. Proof Let o/t: U -+ G (It I < e, U a nbhd of e in G) be a locally defined one parameter group near e in G as given by I, Lemma 6.2. As we saw in the case that G is compact (I, Theorem 6.5), the trick in showing that there is a globally defined one-parameter group is to show that o/t is globally defined on G for ItI < e. Now since ~ is left-invariant (dLg)~ = ~. Thus (d/dt)go/t(a) = (d/dt)o/tCga) for all g, a, and ga in U. Hence (*) go/tea) = o/t(ga). In particular, o/t(g) = go/tCe) for g in U. Thus we can clearly extend o/t to be globally defined and smooth on G. The left invariance of ~ guarantees that o/t (It I < e) is still a one-parameter group for ~ on all of G. 0 Remark. Given a vector v in TeG, let ~v be the left invariant vector field that it generates. Let o/V be the globally defined one parameter group whose existence is assured by the last Lemma. We can think of 0/ as a mapping of (TeG) x G x R -+ G given by (v, g, t) f-+ o/tV(g). Thus we have the following: Proposition A.4. The mapping 0/ is smooth and satisfies (1) o/tV(ga) = go/tV(a), and (2) ifJs!V = o/tsv• Proof For fixed v, o/V is just a one-parameter group and is thus smooth. Varying v just varies the initial conditions to the first order system of ODE's which define o/v. Since solutions to such a system vary smoothly with the initial conditions, 0/ is a smooth mapping. Note that (1) isjust a restatement of (*) in the proof of Lemma A.3. For (2), note that o/stv and o/st are both oneparameter groups on G for fixed t and v. Now o//v has infinitesimal generator ~tv and !!.. o/stV(g) I = t!!'" o/rV(g) I = t~gV = ~gtv.ds 8=0 dr r=O Thus the infinitesimal generator of o/stVis also ~tv. Applying the fact that oneparameter groups are unique we have o//v = o/stv. 0 Define exp: TeG -+ G by exp (v) = o/lv Theorem A.5. exp : TeG -+ G is smooth and is a diffeomorphism on a nbhd ofO. Infact (d exp)o = identity. (Note: we identify To(TeG) with TeG.) Proof Clearly exp is smooth. Using (2) in the last Proposition, we have (d expMv) = dd (exp tv) \ = dd o/ltV(e) \ = dd o/tV(e) \ = ~/ = v. t t=O t t=o t t=o So (d exp)o = idTGc ' 0 196 Appendix Exercise: We may identify TeGL(n, R) with M(n, R) = vector space of n x n real matrices. Using this identification show that exp A = Exp A where A is in N(n, R) and co AI ExpA = 2:"7['i=O 1. Corollary A.6. Let Vand W be subspaces ofTeG such that V EB W = TeG. Define y: TeG --+ G by y(v, w) = exp v·exp w for v in V and win W. Then if> is a diffeomorphism on a nbhd ofO in TeG with a nbhd of e in G. Proof Certainly y is smooth. Moreover (dY)ol V = (d exp)ol V = idv by Theorem A.5. Similarly for W. So (dy)o is invertible. 0 Definition AS. Let G be a Lie group. Then He G is a Lie subgroup if (i) H is a subgroup of G; (ii) H is an immersed submanifold of G; and (iii) H is a Lie group with the group operation assumed in (i) and the manifold structure assumed in (ii). Note. Lie subgroups are not, in general, submanifolds. For example, let G = T2 viewed as the decomposition space R21 Z2 where Z2 = the subgroup of integer lattice points in R2. Let H' be a line in R2 through the origin with irrational slope and let H = 7T(H') where 7T: R2 --+ T2 is the obvious projection. Then 7TIH' is a 1: 1 immersion so that H is an immersed submanifold and a Lie subgroup. But H is not a submanifold of T2 since H is dense. One of the more interesting facts about Lie subgroups which indicates the strong connection between the geometry and algebra on a Lie group is the following. Theorem A.7. Let H be a subgroup of the Lie group G which is a topologically closed subset. Then H is a Lie subgroup. Remark. The content of this theorem is that any closed subgroup of a Lie group is an immersed submanifold and thus a submanifold. First some lemmas. Lemma A.B. Let II be a norm on TeG. Suppose that Vb V2, ... is a sequence ofnonzero vectors in TeG such that Liml-> h2' . .. of points in H with Limt.... Y, 42 germs of maps from (X, p) -> (Y, q), 111 germs of functions of X~, Rat p, 103, 165 sections of the vector bundle E, 18 vector fields along f, 73 codimension cokernel, Jacobian offatp, 2,13 Hessian off at p, 64, 65 intrinsic derivative of pat p, 64, 150 group of smooth diffeomorphisms on X, 72 various pseudogroups, 2, 3 domain off, 2 131 138 dimension pull-back function or homomorphism via f, 1, 104 pull-back bundle of E via f, 73 f'IX',57 Gk,n or G(k, V) Grassmann Manifolds, 4, 14 GL(n, R) or GL(V) general linear group, 194 Gk(X)p invertible k-jets on X at p, 194 Hom(V, W) Hom( V 0 V, W), Jk(X, Y)p,q Jk(X, Y) jkf J/(X, Y) j,kf Jk(E) Jk(E)p linear maps of V ->- W, 19 153 k-jets of mappings of (X, p) -)- (Y, q), 37 k-jet bundle over X x Y, 37 k-jet extension off, 37 s-fold k-multijet bundle, 57 s-fold k-multijet extension off, 57 k-jet bundle of sections of E, 112 fiber of Jk(E) at p, 112 205 206 Ker or ker V(V, W) M(U) m,(x) .,#P(X) or .,#p nbhd 0" pn supp(f) Sr Sr(f) Sr.• or Sr.....! Sr.s(f) Slk Slk(f) S~ S2(V) or VO V V VO Vor S2(V) XS X(S) kernel linear maps of V --+ W of corank r, 60, 143 basic open set in Coo topology, 42 maximal ideal in ~f(X), 165 maximal ideal in C;(X), 103, 165 neighborhood 133 real projective n-space, 4 real n-space, 1 local ring off at x, 165 local ring of the k-jet a, 173 support off, 16,28 universal set of singularities of corank r, 60 singularities off of type Sr, 87, 143 Symbol Index universal singular sets of Boardman type, 154, 156 singularities of type Sr.s off, 152 contact classes for Morin Singularities, 174 Morin singularities off, 174 contact class of type ~, 173 symmetric 2-tensors on V, 20, 153 tangent space to X at p, 12 tangent bundle to X, 14 cotangent bundle of X, 20 dual space to V, 20 symmetric 2-tensors on V,20, 153 57 57 Baire space, 44 Boardman map, 157 Borel Extension Lemma, 98 Canonical bundle, 21 Chart, 3, 74 Ck topology, 42 Condition NC, 157 Condition (!J, 116 Contact class, 173 Contact equivalence germs, 172 k-jet submanifold germs, 173 k-jets, 173 Coordinate neighborhood,S Corank, 33, 60 Cotangent bundle, 20 Covering, 15 Critical point, 33, 115 non-degenerate, 63 critical value, 34 Cross cap, 179, 181 Curve, 12 Cusp, 146 simple, 146, 147 Deformation, 118, 120 Diffeomorphism, 7 Differentiable, 1, 6 class Ck, 1,6 Embedding, 7 Equivalent mappings, 72 exp, 195 Family of vector spaces, 18 Fiber bundle, 41 Finite mapping, 167 Fold locus, 87 Fold point, 87, 146 Frechet space, 74 r-atlas, 3 r -structure, 3 INDEX Generic property, 141 Germs function, 103 mapping, 111 submanifold, 170 Generalized Malgrange Preparation Theorem, 106 General position, 83, 85 Grassman manifold, 4 Hessian, 64, 132 Homogeneous space, 198 Homotopic stability, 119 Immersed submanifold, 10 Immersion, 6 Implicit Function Theorem, 7 Infinitesimal generator, 27 Infinitesimal stability local, 111 mappings, 73 germs, 111 Index bilinear form, 65 non-degenerate critical point, 65 Intrinsic derivative, 64, 150, 151 Inverse Function Theorem, 2 Isotropy subgroup, 198 Jacobian, 2, 13 Jet bundle, 37 of sections, 112 Jets mappings, 37 submanifolds, 172 k-deformation, 120 kth order contact, 37 Lie group, 194 action, 198 subgroup, 196 207 208 Local coordinates, 5 Local homeomorphism, 2 Local infinitesimal stability, 111 Local ring, 103 of contact, 170 of germs, 173 of k-jets, 173 of mappings, 165 Local transverse stability, 134 Manifold,3 Banach,74 class Coo, 3 differentiable, 3 Frechet,74 orientable, 3 real analytic, 3 smooth,3 topological, 3 Mather Division Theorem, 95 Measure zero, 30 Metric, 21 Riemannian, 21 Module, 104 finitely generated, 104 Morin singularity, 174 Morse function, 63 Multijet bundle, 57 Multijet Transversality Theorem, 57 Multiplicity, 168, 169 Nakayama's Lemma, 104 Nirenberg Extension Lemma, 98 Non-degenerate fixed point, 59 Normal bundle, 71 Normal crossings, 82, 157 Normal sp,ace, 71 I-form, 21 I-generic, 144 One parameter group, 27 Orbit, 72, 198 Paracompact, 15 Partition of unity, 16 Product family, 18 Proper mapping, 11, 25 Pseudogroup, 2 Pull-back bundle, 73 Pull-back function, 1 Rank, 6 Refinement, 15 locally finite, 15 Regular point, 34 Regular vallie, 34 Residual, 44 Sard's Theorem, 34 Section, 18, 21 s-fold k-jet bundle, 57 Singularity, 143 contact class, 173 critical point, 33 cross cap, 179, 181 cusp, 146 elliptic umbilic, 183, 187 fold, 87, 88, 146 hyperbolic umbilic, 183, 188 Morin, 174 Morse, 63 multiplicity, 168, 169 non-degenerate, 63 parabolic umbilic, 183, 190 simple cusp, 146, 147 swallow's tail, 176, 177 type ST, 143 type ST,S' 152 umbilic, 182 Smooth function, 1, 6 functor, 19 manifold,3 Source, 37 Stability, 72 homotopic, 119 infinitesimal, 73 local infinitesimal, 111 local transverse, 134 mappings, 72 transverse, 139 under deformations, 119 under k-deformations, 120 Subbundle, 26 complementary, 27 Index Index Submanifold, 9 immersed, 10 Submersion, 6 with folds, 87, 88 Support functions, 16 vector fields, 28 Symmetric function, 108 2-generic, 155 Tangent, 12, 75 bundle, 14 space, 12 Target, 37 Thorn-Boardman stratification, 159 Thorn Transversality Theorem, 54 Transversality, 50, 54 Transverse stability, 139 Trivial deformation, 118, 120 family of vector spaces, 18 k-deformation, 120, 124 vector bundle, 18 Tubular neighborhood, 69 Tubular Neighborhood Theorem, 69 Umbilic, 182 elliptic, 183, 187 hyperbolic, 183, 188 parabolic, 183, 190 Vector bundle, 18 base mapping, 26 canonical bundle, 21 cotangent bundle, 20 homomorphism, 18, 26 isomorphism, 18 jet bundle, 41 normal bundle, 71 pull-back bundle, 73 tangent bundle; 12, 75 trivial bundle, 18 tubular neighborhood, 69 Vector field, 21 along!, 73 compactly supported, 28 left invariant, 195 209 Weierstrass Division Theorem, 91 Weierstrass Preparation Theorem, 91 Whitney Ck Topology, 42 COO Topology, 42 Embedding Theorem, 62 Immersion Theorem, 61 sum, 20