Singular Points of Smooth Mappings (Research Notes in Mathematics Series)

Research Notes in Mathematics

C G Gibson

Singular points of smooth mappings

Pitman LONDON SAN FRANCISCO - MELBOURNE

25

C G Gibson University of Liverpool

Singular points of smooth mappings

Pitman LONDON SAN FRANCISCO MELBOURNE

PITMAN PUBLISHING LIMITED 39 Parker Street, London WC2B 5PB FEARON PITMAN PUBLISHERS INC. 6 Davis Drive, Belmont, California 94002, USA Associated Companies Copp Clark Pitman, Toronto Pitman Publishing New Zealand Ltd, Wellington Pitman Publishing Pty Ltd, Melbourne First published 1979

AMS Subject Classification: 57D45

British Library Cataloguing in Publication Data Gibson, C G Singular points of smooth mappings - (Research notes in mathematics; 25). 1. Mappings (Mathematics) 2. Singularities (Mathematics) 1. Title H. Series 515 QA360 ISBN 0 273 08410 0 © C G Gibson 1979

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording and/or otherwise without the prior written permission of the publishers. The paperback edition of this book may not be lent, resold, hired out or otherwise disposed of by way of trade in any form of binding or cover other than that in which it is published, without the prior consent of the publishers.

Reproduced and printed by photolithography in Great Britain at Biddies of Guildford.

TO

DORLE

Contents

Introduction

I

Acknowledgements

7

Chapter I

Chapter II

Smooth Manifolds and Mappings

-

§1.

A Preliminary Review of Some Calculus

§2.

Smooth Manifolds

12

§3-

The Differential of a Smooth Mapping

16

§4.

Vector Fields and Flows

24

§5.

Germs of Smooth Mappings

33

Transversality

-

§1

The Notion of Transversality

38

§2.

The Basic Transversality Lemma

48

An Elementary Transversality Theorem

51

§4.

Thom's Transversality Theorem

53

§5.

First Order Singularity Sets

54

3,

Chapter III

8

-

Unfoldings

The Finite Dimensional Model

§1.

Groups Acting on Sets

61

§2.

Some Geometry of Jets

62

§3.

Smooth Actions of Lie Groups on Smooth

Manifolds

73

Chapter IV

Chapter V

§4.

Transversal Unfoldings

81

§5.

Versal Unfoldings

89

-

Singular Points of Smooth Functions

§1.

Some Basic Geometric Ideas

94

§2.

The Algebra

99

§3.

Determinacy of Germs

§4.

Classification of Germs of Codimension

-

n

116

5 5

122

Stable Singularities of Smooth Mappings

§1.

The Basic Ideas

139

§2.

Contact Equivalence

143

§3.

Deformations under.'-Equivalence

159

§4.

Classification of Stable Germs

168

§5.

Higher Order Singularity Sets

174

§6.

Classifying Germs under,11-Equivalence

191

§7.

Some Examples of Classifying Stable Germs

199

§8.

Singular Points of Stable Mappings

205

Appendix A

-

The Theorem of Sara

215

Appendix B

-

Semialgebraic Group Actions

222

Appendix C

-

Real Algebras

226

Appendix D

-

The Borel Lemma

228

Appendix E

-

Guide to Further Reading

232

Index

237

Introduction

Suppose you take a smooth curved surface

X

made of some transparent material

and imagine it projected downwards onto a plane surface of light from above. on the surface

Think of this as a map

X projected down to a point

Y by shining a beam

f : X - Y

q = f(p)

with every point in the plane

Y.

p

On

the plane you will see the apparent outline of the surface, as it would appear from below.

Here are two simple examples.

Fig. 2

Fig. 1

It is not hard to see which points on X

give rise to the apparent outline

they are precisely the points where the tangent plane to the surface is vertical, the so-called singularities of the mapping

f : X - Y.

Figure 1 repre-

sents the simplest situation one can imagine, with the surface X folding over at

p:

such points

p

are called fold points.

Figure 2 represents a

more complex situation, a curve of fold points on which lies an exceptional point

p

where two folds meet, a so-called cusp point.

cated situation is provided by Figure 3 where

X

A still more compli-

is a torus,

i.e.

a dough-

1

nut - shaped surface.

Here one has again curves of fold points, on which lie

Fig. 3 four cusp points:

but in addition we have two simple crossing points on the

apparent outline, where the curves cross over properly. any point

q

very close to

Thus, if we take

on the apparent outline and look at the nature of the outline we can distinguish just three possibilities.

q,

I

II

III

q

q

q

Common sense, and a certain amount of experimentation, will soon convince one that these are the only essential types of behaviour which can arise, in the sense that any other type of behaviour could be eliminated by the slight-

est change of position of X in space.

For instance our torus might be so

positioned that the apparent outline was as in Figure 4, with the outline touching itself at some point:

2

but clearly we could just nudge

X

slightly

to get back to the previous situation where only possibilities I, II, III can occur.

To make the point even clearer, here are three further types of behaviour

which are all inessential, because they could be eliminated by the slightest change in position of simple crossing points

to yield situations exhibiting only fold, cusp and

X -

as follows.

The broad objective of this book is to introduce the reader to the less technical aspects of a mathematical theory of singularities which seeks to make precise the kind of heuristic reasoning just described.

The basic

3

objects

X, Y are replaced by smooth manifolds, which are natural generaliza-

tions to higher dimensions of the familiar notions of curve and surface.

And

the projection of the surface onto the plane is replaced by an arbitrary smooth mapping

f : X-+ Y.

For such mappings we introduce the general notion of

singularity and begin to list the simplest singularities which can occur,

We shall, for reasons of simplicity, con-

idealizing each type by a model.

cern ourselves principally with the local behaviour of the mapping, behaviour very close to a single point in the domain:

i.e.

its

thus in the situation

discussed above we interest ourselves solely in the fold and cusp points, and neglect simple crossing points which arise from considering what happens close to two distinct points in the domain.

There is nothing particularly new in the notion of a singularity.

Scien-

tists and geometers have recognized them, and appreciated their significance, for a long time now.

But no-one seems to have systematically set about

studying the singularities of smooth mappings till the pioneering work of Hassler Whitney in the mid 1950's.

Around the same time Rend Thom pointed

out the analogy with more finite-dimensional situations and indicated the general lines along which a theory might proceed.

So it was in the 1960's

that a number of mathematicians, principally John Mather, laid the foundations of a general theory.

That was the position in 1967 when Vladimir Arnol'd put

together the bits and wrote his now classic survey paper, a model of lucid descriptive writing.

It was a time of great promise.

Singularity theory

itself threw up a number of provocative problems, and the range of possible applications (both within and without mathematics) added to the excitement. Without question, the intervening years have justified that promise, and singularity theory can hold its own as a flourishing area of mathematics.

I feel that the time has come to provide prospective students with readable introductions to the subject.

4

It is my personal conviction that the way to

get into any area of mathematics is to concentrate on understanding the simplest situations first, so as to build up some intuitive feeling for what is going on, and to leave the deeper matters till later in life.

Singular Points

of Smooth Mappings is the result of following this guiding philosophy.

I have

taken a small number of intuitively appealing ideas and used them to pursue the problem of listing singularity types, one of the goals of the local theory.

It is the kind of book which I would expect a postgraduate student in mathematics to read with little difficulty, and I rather hope that others will find it within their compass as well.

A guide to further reading has been inclu-

ded to help the reader pursue those matters which interest him most. A few words are in order concerning the structure of the book.

In accor-

dance with the philosophy outlined above smooth manifolds are introduced in

Chapter I as subsets of in enjoying certain properties: way everyone should meet them.

I think this is the

Anyone who wants to get to grips with singu-

larity theory should be familiar with the basic ideas of transversality and of unfolding, so these topics provide the subject matter for Chapters II and III respectively.

Here again I have kept to the simplest situations which can

arise, imposing restrictions whenever I felt it was possible to suppress undue technicality:

in particular, unfoldings have been introduced in a finite-

dimensional situation where they are much easier to understand.

Singularity

theory proper is taken up in Chapter IV with the study of functions;

this

enables one to make some distance fairly easily, without getting involved in the subtleties associated to general mappings.

The result is the derivation

of the list of singularities of codimension : 5. case of mappings is taken up:

In Chapter V the general

it is inevitable that one must quote more and

prove less, but I have tried to expose the less technical aspects and give a fairly coherent account of just how one uses the theory to obtain explicit lists of the simplest singularities which turn up.

Of course one can pursue 5

the listing process much further than is indicated in this book, but one soon comes up against much deeper matters which lie beyond the scope of an introductory account.

I have not attempted systematically to attribute results to their authors, mainly on the ground that such a practice is out of place in a book at this level.

In any case, the material of the first four chapters is now pretty

well an established part of the subject.

I should say however that the open-

ing sections on differential topology follow closely the exposition given by John Milnor in his excellent little book "Topology, from the Differentiable The material in the final chapter, basically Mather's classi-

Viewpoint".

fication of stable germs by their local algebras, is not as well-known as it should be.

Here I decided to follow the elegant account of Jean Martinet

(see the Guide to Further Reading) in which the unfolding idea plays the central role.

The key result in this development, namely the characteriza-

tion of versal unfoldings, turns on a real version of the Weierstrass Preparation Theorem which I do not discuss;

I felt it was more important at this

level to place proper emphasis on the underlying geometric ideas, and to leave an exposition of the Preparation Theorem to a volume with more ambitious aims. I decided also to say nothing about the applications of singularity theory, mainly because I feel each area of application is probably worthy of a volume in itself.

For instance, Thom's catastrophe theory is already the subject

matter of several volumes.

Also, the applications within mathematics itself

all seem to be at too early a stage to merit writing-up. that my guide to further reading will prove to be useful.

Liverpool February 1978

6

Here again I hope

Acknowledgements

To Les Lander who drew the pictures,

and helped me find enthusiasm at a time when it was all but lost.

To Peter Giblin who undertook the considerable task of correcting the manuscript,

and whose suggestions have contributed much to the final form of this book.

And to Ann Garfield who did the typing and produced an excellent job in difficult circumstances.

7

I

Smooth manifolds and mappings

A Preliminary Review of Some Calculus

1.

[In this section it will be understood that

]RP, 7Rq Let

U, V, W

are open sets in

Rn

respectively.] f : U --> V be a mapping with components

f1, ..., f

We call

.

f

P

smooth when all the partial derivatives

aaf a

(a = a 1 + ... + an )

a

ax 11...8xnn

exist and are continuous in fice to observe that if

For the purposes of this book it will suf-

U.

f1, ..., f

are all given by polynomials in

P

x1, ..., xn U

onto

then

will be smooth.

f

And we call

V when it is a bijection, and both

Now suppose

linear mapping

f : U -+ V

Daf

:

is smooth.

7Rn - ]RP

f, f-1

f

a diffeomorphism of are smooth.

For every point

there is a

a E U

called the differential of

f

at

a:

it

is precisely the linear mapping whose matrix (relative to the standard bases) is the so-called Jacobian

(af

1

1

of

aa) 1

8

...

of axP- IRk

p, v

x

M.

k

in

Now

v(0) = X.

subset of IRm

x

into IRn

with

(Note that therefore

M, N

respectively.

f(O) = 0

Clearly, one can suppose

whose differential at

0

has rank

By (1.i+) there exists a diffeomorphism

m . n.)

in IRn

k0f(x1, ..., x ) m

=

(x1,

v0k-1

of relatively

is a smooth mapping of an open

o p

0

=

m, n

= (V) 0 M.

v-1

f =

of some open neighbourhood of

and then

having a para-

and O(IRmx 0)

be parametrizations of dimensions

open neighbourhoods of

p(0) = x,

for which

N

Nn C IRk:

onto another for which

..., xm, 0, ..., 0)

has the desired properties. 15

In differential topology (the study of smooth manifolds and smooth mappings between them) one regards two smooth manifolds when they are diffeomorphic.

f1

:

N1 -+ P1,

N1, N2

as being the "same"

Likewise we regard two smooth mappings

as being the "same", formally one calls them

f2 : N2 -+ P2

equivalent, when there exist diffeomorphisms

h, k

for which the following

diagram commutes.

f

P

N1

k

h

f2

N2

> P2

The Differential of a Smooth Mapping

§3.

Our objective in this section is to introduce the concept of "differential" for smooth mappings with domain a smooth manifold, rather than just an open The following definition provides a necessary

set in some Euclidean space. preliminary. :

in

U -+7Rk

Let

be a smooth manifold, let

Nn C IRk

x E N

and let

be a parametrization of a relatively open neighbourhood of

N with 4(u) = x,

say.

We define the tangent space

to be the image of the differential

Du(P

:

IRn ->IRk.

TxN

One pictures

the vector subspace of IRk

parallel to the affine subspace of IRk

x which best approximates

N

16

close to

x.

at

x

x

N

to

TxN through

as

Of course, for the above definition to make sense we must show that it is independent of the choice of parametrization.

So let

J

:

V -*IRk

another parametrization of a relatively open neighbourhood of

0(v) = x, u

say.

Then X =

onto an open neighbourhood

gram

x

in

be

N with

0_1

o 0 V1

of

will map an open neighbourhood V,

U1

of

and we shall have a commuting dia-

of smooth mappings

giving rise to a commuting diagram of differentials

from which it is immediate that

Duc, Dvo

have the same image, as was

required.

17

Let

(3.1)

gent space

Proof

N C IR.k be a smooth manifold, and let

TXN

is a vector subspace of IRk

x E N

of the same dimension

We keep to the notation of the above discussion,

= x.

in

n

as

N.

k

i.e.

is a parametrization of a relatively open neighbourhood of x (u)

then the tan-

:

U ->]R N.

with

Now-1

open set W C IRn with verse image under

(u) -> U -1 =

of

it

is smooth, so (by definition) there is an

on W n 4(U).

Taking

U'

to be the in-

we obtain a commuting diagram of smooth

mappings

U'

(inclusion)

hence a commuting diagram of differentials

It is immediate that

TxN,

the image of

has the required dimension

Duo,

n.

The reader will readily check for himself that the tangent space at a point to an open set in IRk

is precisely IRk:

point to a vector subspace

V

of IRk

also, that the tangent space at a

is precisely

x

is to show that the tangent space at a point is precisely the subspace perpendicular to

x.

V.

A harder exercise

to the n-sphere

Sn C IR

n+1

However for such examples as

this (where the smooth manifold is defined by equations) we shall establish 18

much simpler ways of computing tangent spaces in Chapter II. The importance of the tangent space is that it allows one to introduce the differential at a point of a smooth mapping defined on a smooth manifold, rather than just on an open set.

observation.

f

:

that IRp,

M --> N f

Let Mm C ]RP,

The definition will require a preliminary

Nn C

IRq

be two smooth manifolds, let

be a smooth mapping, let x E N, and put y = f(x).

The fact

is smooth requires that there is an open neighbourhood W of

and a smooth mapping

the differential of claim that

space TyN.

DxF

F

F : W -*]Rq

x

at

with

is a linear mapping

suppose that

D F

To prove this let 0

:

U - IRp,

0(v) = y,

By taking U

say.

q(U) C W,

and that

0

TxM

in

Of course

]Rp -+ 1Rq.

We

into the tangent

V - 7Rq be parametriza-

:

x, y

:

x

necessarily maps the tangent space

tions of relatively open neighbourhoods of

0(u) = x,

on w n M.

f = F

x

in

M, N

respectively with

to be sufficiently small we can

f o 4(U) C O(V),

so

g

:

U --r V

given

0-1

by

g =

o f o 0

is a well-defined smooth mapping.

We now have a com-

muting diagram of smooth mappings

F ]R.q

W

0 I

g

U

1

V

yielding a commuting diagram of differentials D F IRp

x

Ftq

T

D 0 v

D u'p

Dug IRm

It is immediate that

D F

x

maps the image of

n IR

Duo

into the image of

Dv', 19

the tangent space

i.e.

TXM

into the tangent space

The restriction of the linear mapping DXF T f

X

:

T f

TXM

TyN

->

IRp -> 11

and call the differential of

as the best linear approximation to

x

:

is drawn for the case

x = 0

f

at

and y = 0

in lip,

TyN,

f

at

to

as was claimed. TXM

x.

we write One pictures

the illustration below

x:

in IRq.

TXf f

It follows from the very construction of the choice of

T f

x

that it does not depend on

Notice also that in the particular case when

F.

open sets in IRp,

the differential

IRq

as previously defined.

Txf

M, N

is the the differential

are Dxf

It is an easy matter to derive the basic properties

of this more general notion of differential from those mentioned in 1.

We

shall mention two of these, leaving the proofs to the reader.

Take, for instance, the situation when M smooth manifold

x E M TXf

N,

and

the tangent space :

TXM

-+ TXN

f : M - N T M X

is a smooth submanifold of a

is the inclusion mapping.

is a subspace of

is the inclusion.

T N, X

At any point

and the differential

One pictures the situation something

like this.

TXN

20

Likewise, we have the Chain Rule, which we express as follows. A, BY C

Suppose

are smooth manifolds, and we have a commuting triangle of smooth map-

pings

and a point

a E A

with

f(a) = b,

then the corresponding triangle

g(b) = c

of differentials commutes

And as in 1 it follows from these two basic properties that at any point the differential of a diffeomorphism is an isomorphism of vector spaces, so that in particular domain and target have the same dimension.

Our next

result provides us with further examples of smooth manifolds.

First a defi-

nition.

We define the

rash of a mapping

graph f= {(x, y) E N (3.2)

Let

f : N -+ P

is a smooth submanifold of

graph f

at any point

F

:

X P

:

f(x)}.

(x, f(x))

N - graph f

is precisely graph

given by

F(x)

of the smooth projection N x P

-+

then

Also the tangent space to

N x F.

=

tion, indeed a diffeomorphism since its inverse graph f

y

to be the set

be a smooth mapping between smooth manifolds:

graph f

Proof

f : N - P

N.

Txf.

( x, f(x)) F-1

is a smooth bijec-

is the restriction to

It follows immediately that 21

graph f

is a smooth manifold.

tangent space to

graph f

Now

T F

x

is the mapping

will be its image,

(1, T f)

x

graph

i.e.

and the

Txf.

At this point is is convenient to augment (2.1) by the following proposition, determining the tangent space to a product;

the proof is left to the

reader.

Let

(3.3) (x, y)

M, N be smooth manifolds.

to the product

M x N

The tangent space at a point TxM x TyN.

is the product

So far we have talked only about the differential f

at a point

x.

of a smooth mapping

T f

x

We shall conclude this section by mentioning an elegant

globalisation of this notion which is essential to any serious study of differential topology.

The idea is to glue together all the linear mappings

to obtain a single mapping

Formally, we proceed as follows.

Tf.

N C 7Rk be a smooth manifold.

TN

Let

(3.1+)

Let

ly open set in F

:

I (x, v) E N x I R

N,

:

N.

the tangent

2n.

The proposition follows from the observation that -+

(f(u), Duf(x)) is a 2n-

dimensional parametrization of a relatively open set in

f : N --> P

be a smooth mapping with

Ve define the tangent mapping

22

n:

be an n-dimensional parametrization of a relative-

U x IRn - ]Rk x IRk defined by (u, x)

Now let

to

V E TXN1 .

N C 1Rk be a smooth manifold of dimension

f : U -+ IRk

N

i.e.

TN C IR2k is a smooth manifold of dimension

bundle space

Proof

=

Let

9Je define the tangent bundle space to

be the set of all possible tangent vectors to

Txf

Tf

:

TN - TP

N, P

TN.

smooth manifolds.

by the formula

Tf(x, v)

Let

(3.5)

Tf

:

f : N -a P

be a smooth mapping:

Let N C IRs, P C IEtt

open set

U C

TF

]Rs

with

x0 E U

Choose a point

:

-->

IRt x IRt

this mapping is smooth. (x0, v0):

also we have

definition of

M

(x, v)

U X as C

TF = Tf

on

It follows that

Tf.

Notice that if

given by But

and an

(This is possible since f is

Observe that the tangent mapping of

U x IRs

(x0, v0) e TN

for which there exists a smooth mapping

: U -+ IRt with F = f on U (1 N.

smooth.)

the tangent mapping

is likewise smooth.

TN -+ TP

Proof

F

(f(x), Txf(v)).

=

IRs

F

--

is the map

(F(x), DxF(v)) and that

X IRs

is open and contains

(U x IRs) (1 TN,

Tf

going back to the

is smooth.

is a smooth submanifold of a smooth manifold

the tangent mapping of the inclusion

M -+ N

will be the inclusion

N then TM -i TN.

One of the advantages of the tangent mapping is that it allows a simple and elegant formulation of the Chain Rule, namely that a commuting diagram of smooth mappings

gives rise to a commuting diagram of tangent mappings

TB

TA"

TC

23

otherwise expressed,

T(gof)

TgoTf.

=

Notice also that given a smooth manifold N :

TN -+ N

given by the formula

(x, v)

-

there is a smooth projection this mapping is the

x:

1TN

tangent bundle of

N.

It follows that given any smooth mapping

f : N -+ P

between smooth manifolds we have a commuting diagram of smooth mappings.

Tf

f

Vector Fields and Flows

§4.

Let

Nn be a smooth manifold with tangent bundle

v

smooth vector field on N we mean a smooth section of mapping

e

:

N - TN

to each point

x e N

such that

it o e

a tangent vector

=

at

By a

a smooth

i.e.

v,

In other words

1N.

e(x)

TN - N.

:

assigns

x.

For the purpose of illustration take the special case when N C an open set.

In that case

TN

is

N x 7Rn

and

it

:

IRn

is

is pro-

N x IRn -+ N

jection on the first factor, so that a smooth vector field is a smooth map-

ping e

:

N -+ N x IRn of the form x -+ ( x, 9(x)) with

smooth mapping.

Suppose

Bl, ..., 0n

are the components of

vector field does no more than associate with each point (01(x), ..., 6n(x))

24

in IRn.

0

x E N

N -+ IRn

:

B:

a

then our

the vector

Thus, for instance, in the case when

n = 2

we can sketch a vector field by the simple device of drawing at each point

x e N an arrow starting at that point, and in the direction given by the vector

indeed a good exercise is to invent simple formulae for

to check for himself: 6

Here are some examples, which the reader is advised

(61(x), 02(x)).

and sketch the corresponding vector fields.

taken

In all these examples we have

N = 7R2

ay

aY

61 =x: 62=y

61 = y : 02 = -p 2x

61 = x2-y 2 : 0 2 = 2xy

61=x: 02=-y

These pictures have a certain didactic value.

They suggest that if you

start at a point and follow the arrows you will move along a curve, a "flow line" if you like. fold.

We shall make this precise.

By a smooth curve in

is the mid-point of

I.

Nn

N we mean a smooth mapping

an open interval of positive length. 0

Let

be a smooth manif

:

I - N

with

I

For convenience we shall suppose that

And we say that the curve starts at

x0 = f(0).

25

f

I 0

Recall now that we have the following commuting diagram of smooth mappings.

Tf

TI

)TN "N

ITI

f

I Here, of course, TI = I x 1R. field

6

on

i

vector at at

f(t).

at

f(t)

on t,

I

Suppose now that we have a smooth vector

We wish to make precise what we mean by saying that f

N.

a "flow-line" for e. field

N

To this end we introduce a "canonical" smooth vector

by writing and

is

Tf(i(t))

i(t) =

(t, 1).

Thus

i(t)

is a unit tangent

is the corresponding tangent vector to the curve

We want this to be precisely e(f(t)), given by our vector field

i.e.

the tangent vector

In other words we want the follow-

e.

ing diagram of smooth mappings to commute

Tf

-> TN

6

f

-N

Thus we are led to define a flow-line for for which the above diagram commutes.

26

e

to be a smooth curve

f

The composite mapping Tf o i

:

I -> N is

usually written

f':

with that notation the condition for

to be a flow-

f

line is that

f' (t)

=

a(f(t)).

To make this as concrete as possible it might help to go back to the case

when N

and the vector field is given by n

is an open set in ]Rn,

functions

81, ..., Bn

on

A smooth curve

functions

f1, ..., fn

as its components.

N.

smooth

in N will have

f

And the condition

*

n

for

smooth f

to

be a flow-line of the vector field is, written out in full, the following system of

n

simultaneous equations

..., fn(t))

f1(t)

=

01(f1(t),

f' (t)

=

0n(f1(t), ..., fn(t)).

n

Thus the problem of constructing a flow-line n

ving simultaneously the above

f

for

is that of sol-

differential equations for f1, ..., fn.

And the condition for the flow-line to start at the point N

e

(x1, ..., xn)

in

is that the solution should satisfy the "initial conditions"

=

fl.(0)

Example 1

...

x1

fn(O)

Take the smooth vector field on IR2

02 = -p2x with

p I 0 a real number.

=

xz

defined by

01 = y,

(We sketched this one above.)

To

find flow-lines we have to solve the simultaneous differential equations

= 1

f2

f2

=

-p2f 1.

These yield the harmonic equation

27

f

p2f1

+

0

=

having solutions

=

f1(t)

with

p2 x2 + y2

=

f2(t)

2 2

Notice that

c, a E ]R.

the ellipses

c sin(pt + a)

p f

p2c2.

+ f2

=

2

=

p c

plc cos(pt + a)

2

so the flow-lines are

,

(Compare with the sketch given above.)

The most basic fact which it is necessary to know about flow-lines is what we shall term the Local Existence and Uniqueness Theorem.

Let

(4.1)

x0 E N.

(i)

be a smooth vector field on a smooth manifold

e

and let

The following assertions hold.

There exists a flow-line for

(ii) Any two flow-lines for hood of

N,

0

in 3Rn.

e

which starts at

which start at

e

(Existence)

x0.

agree on some neighbour-

x0

(Uniqueness)

The reader will readily check that it suffices to prove the result when N

And in that case the result is just the Local

is an open subset of IRn.

Existence and Uniqueness Theorem for ordinary differential equations, which we assume him to be familiar with from calculus.

In fact we shall need to know rather more than the mere existence of a we need to know that we can simultaneously

single flow-line through a point:

parametrize all the flow-lines through points near to a given one. is captured by the following formal definition. field on a smooth manifold

local flow for U

e

at

x0

N.

x0

is an open neighbourhood of

x0

in

x E U

N,

6

be a smooth vector

be a point in

we mean a smooth mapping

the property that for any point

28

and let

Let

and

I

the mapping

This idea

N.

F : U x I

By a smooth -+

N

(where

is an open interval) with fx

I -> N

defined by

fx(t) = F(x, t)

is a flow-line for

which starts at

x.

One pictures

the situation something like this, with

N F U

the vertical lines in the left-hand picture mapping to the flow-lines in the right-hand picture.

The fact which we need to know is the following result,

which again vie assume the reader to be familiar with from calculus.

(4.2)

Let

xO E N:

there exists a smooth local flow for

be a smooth vector field on a smooth manifold e

at

N,

and let

x0.

The next thing we need to know is that we can "straighten-out" a local flow.

However, for this we need an additional proviso, embodied in the folA critical point for a smooth vector field

lowing definition. smooth manifold

N

is a point

x e N for which

at such a point the flow-line is constant, (Indeed a constant curve through

x

i.e.

fi(x) = 0.

e

on a

Notice that

its image is a point.

is certainly a flow-line, and the Local

Existence and Uniqueness Theorem tells us that it is the only one.)

a critical point

another critical point

a non-critical point

29

We formalise the idea of "straightening out" a vector field as follows. Suppose that

U1, U2

of

N1, N2

in

are smooth-

when there exist relatively open neighbourhoods

x1, x2

x1, x2

on which are

We shall say that 1, e2

defined smooth vector fields i, 2. ly equivalent at

NV N2

are points in smooth manifolds

x1, x2

and a diffeomorphism

:

U1 - U2

for which

the following diagram commutes, and iP(x1) = x2.

TU1

TU2

62

e1

'U2

UI

Here of course we are abusing notation by allowing

their restrictions to UV U2

respectively.

Now let us call a smooth vector field v

constant when there exists a vector x E N.

One pictures the situation

mapping the flow-lines for

something like this with

all

e1, e2 also to denote

e 0

eI

to those for

on a smooth manifold N for which

e(x) =

(x, v)

Of course, such a vector field has no critical points.

for We

contend the following.

(4.3)

Let

e

be a smooth vector field on a smooth manifold N

x0 E N be a non-critical point.

N 30

smoothly equivalent to

e

at

and let

There exists a constant vector field on x0.

e2.

The picture for this result is the same as that above save that one of the vector fields is constant.

Clearly, it is no restriction to suppose that

Proof

bourhood of F

:

U x I

x0 = ->

IRn

0

open interval containing

an open neighbourhood of 0.

stant vector field on IRn (I

v = e(O):

Let

I claim that

critical point.

diffeomorphism

U

is an open neigh-

By (4.2) there exists a smooth local flow

in ]Rn.

with

N

then

v

as

0

is smoothly equivalent at

6

given by

and

in IRn,

I

I

an

is a non-

0

to the con-

0

For this we have to construct a

v.

of some neighbourhood of

0

in IRn

onto another which

satisfies

(1)

(0)

(2)

DX (v)

=

To begin with, note that any vector x e form

x = 7T(x) + t(x)v

where t(x)

0

_

n can be written uniquely in the

is a real number, and

denotes ortho-

v

gonal projection on the orthogonal complement of the line spanned by v. Observe that both

vr,

t

are linear mappings.

that Yi(U) C U, t(U) C I.

We can suppose

Define the smooth mapping

fi(x)

Intuitively, to get to the point

=

:

U

chosen so

U - IRn by

F(7r(x), t(x)).

fi(x)

you start at the point

slide up the flow-line through that point for time

t(x).

7r(x)

and

The pictorial idea

31

is as follows.

Certainly (1) is satisfied because one has For (2) we proceed as follows.

v, t

mind that

Taking differentials in *,

fi(x), t(x)) o

=

with obvious meanings attached to x E U

Thus

and bearing in

aF au ,

t(x)) o t

OF at

The fact that

F(x, 0) = x

represents multiplication of the fixed vector

D0c (x)

=

U

Note

is the identity map on ]Rn.

ir(x) + t(x)v

=

x

so

D0i

v

by a

is the identity map-

It follows from the Inverse Function Theorem that )

ping on ]Rn.

its domain

OF

IT

implies that aU x, 0)

also that at 0, 0) scalar.

x E U.

are linear, one obtains

Dx

for all

F(x, 0) = x for all

will map

is small enough.

diffeomorphically onto its image, provided U

And that completes the proof.

There is one last point which we should clear up, namely that much of what we have said can be extended to an apparently more general situation. In practice one often comes across "time-dependent" vector fields, one-parameter families of smooth vector fields.

vector field on a smooth manifold N with

J

defined by et(x) thinks of

32

e

=

e(x, t)

Formally, a time-dependent

is a smooth mapping

an open interval, such that for each

t E J

6

:

N x J - TN,

the mapping t : N - TN

is a smooth vector field on

as a 1-parameter family

i.e.

N.

Thus one

of smooth vector fields on

N.

To maintain the contrast one can then refer to an ordinary vector field as time-independent.

For a time-dependent vector field

one can mimic the

: N x J -> TN

e

definitions already introduced for time-independent vector fields. flow-line for I

x0 e N

starting at

e

an open interval containing

0,

f'(t)

And a .smooth local flow for (where

U

at

e

is a smooth mapping f

f(0) = x0,

and

=

x0

Let

and let

x0 E N:

x0

in

U x I -+ N

:

and

N,

with the property that for any point

0)

=

is an

I

x E U

F(x, t) is a flow-line for

be a time-dependent vector field on a smooth manifold

e

the

The analogue of (4.2) is

x.

(4.3)

for which

is a smooth mapping F

map x : I - N defined by the rule fx(t) starting at

I - N with

:

*(t), t).

is a relatively open neighbourhood of

open interval containing

Thus a

then there exists a smooth local flow for

6

at

N,

x0.

Germs of Smooth Mappings

§5.

The subject matter of this book is concerned principally with the behaviour of a smooth mapping smooth manifolds. x E N,

point ain

U

N -+ P

and consider the set of all smooth mappings

is a neighbourhood of

we write

f1 - f2

x

in

N.

are

Suppose given a

We can make this precise as follows.

valence relation - .

N, P

close to a point in its domain, where

whose dom-

U --r P

On this set we introduce an equi-

Given two such mappings

fI

when there exists a neighbourhood

f2

: U1 -> P, U

of

x

in

:

U2 -+ P N

33

depending

on

f1

for which the restrictions

f2

and

f2IU

f11U,

coincide

The equivalence classes under this relation are called smooth germs of mappings x, and elements of the equivalence class are called representatives

at

N -> P

Notice that if

of the germ.

f1(x) = f2(x),

then

say, at

y, f

x:

(N, x)

:

->

f1, f2

so all representatives of the germ take the same value

in view of this fact it is usual to adopt the notation

1N

:

x, y respectively the

for the germ, and to call

(P, y)

In the particular case when N = P

source, target of the germ.

notation

are representatives of the same germ

(N, x) - (N, x)

for the germ at

x

we use the

of the identity mapping

N - N. One can handle germs in much the same way as one handles the mappings from which they are derived. g

(P, y) -> (R, z)

:

gof

:

For instance given germs

f(U) C V,

of their composite

g o f

U - P,

which is evidently possible, and then the germ

U - R

:

at

x

can be defined to be the composite

It is a trivial matter to check that this definition does not depend

germ.

on the particular choices of representatives. analogy with mappings to introduce "inverses".

is invertible when there exists a germ

of

and

we can "compose" them to obtain a germ

(N, x) - (R, z): one just chooses representatives f

g : V - R with

has

f : (N, x) --# (P, y)

f o g f.

=

1P,

g o f

=

Further, to a germ

:

x

f

:

(P, y) --> (N, x)

(N, x) --> (P, y)

tial, denoted as one would expect by Txf : TxN the differential at

A germ

and in that case

1N

f

g :

Likewise, one can pursue the

of any representative:

does not depend on the choice of representative.

-'

g

(N, x) -> (P, y)

for which one

is called the inverse

we associate a differenTyP,

and defined to be

once again the definition And we shall leave the

reader to write out in full for himself the obvious version of the Chain Rule for germs.

It is perhaps worth pointing out that the Inverse Mapping Theorem

admits a neat statement in the language of germs, namely that a germ is 34

invertible if and only if its differential is invertible. f

:

(N, x)

(P, y)

-+

dim N

rank equals submersive.

The rank of a germ

is defined to be that of its differential:

the germ is immersive, and when it equals

when the

dim P

it is

Thus a germ will be invertible if and only if it is both immer-

sive and submersive.

A germ which is neither immersive nor submersive is

called singular.

Consider the plane curve

Example 1

Its image is the cuspidal cubic at any point

:

t

defined by

i. -+IR2

pictured below.

is immersive, whilst that at

t j1 0

the only value of

x3 = y2

f

t = 0

t -,

(t2, t3).

Clearly, the germ is singular.

Thus

for which we get a singular germ corresponds to the one

rather exceptional point on the curve, namely the cusp point.

0

There are various equivalence relations under which it is sensible to study smooth germs.

A good starting point from which to develop is the following.

By analogy with the definition of equivalence given for mappings in §2 we call two germs

f1, f2

equivalent when there exist invertible germs

h, k

for

which the following diagram commutes

f1

(Ni, x1) T

h

k f2

(N2, x2)

I

(P2, Y2)

35

Example 2

Consider the germs

(p2,

0)

-

(]R2,

0)

given by the formulae

(x2

f1(x, Y)

=

+ y2, xY)

f2(x) Y)

=

(x2, y2)

These germs are equivalent because in the above diagram we can take

h(x, y)

(x + y, x - y)

=

k(x, y)

(2x + 2y, x - y)

=

which are invertible because the formulae represent vector space isomorphisms, hence diffeomorphisms.

It is maybe worthwhile spelling out one simple consequence of the above definition, namely that any germ some germ

(Mn,

0) -

(IR',

f

:

(N, x) - (P, y)

is equivalent to

It is for this reason that much of this

0).

book is restricted to studying germs of this apparently special type.

Our

broad objective is to introduce the reader to the basic ideas relevant to the problem of classifying germs under the above equivalence relation.

Proposition (1.3) tells us that any submersive germ

Example 3

(Ian, 0)

-+

(]RT, 0)

..., xn )

(x1,

immersive germ

is equivalent to the germ at (x1,

->

..., xP ).

(]Rn, 0)

natural inclusion

->

of the projection

And Proposition (1.4) tells us that any

(]RR, 0)

(x1, ..., xn)

0

-+

is equivalent to the germ at (x1,

of the

..., xn, 0, ..., 0).

This example takes care of the non-singular germs. ones?

0

What of the singular

Here one is up against serious mathematical difficulties, and it will

take us to the final chapter of this book even to outline how one sets about resolving these difficulties.

Finally, we wish to introduce "jets" of smooth mappings, to be thought of as finite approximations to germs of smooth mappings, in the following rough sense:

given a germ we can write down a corresponding Taylor series (of some

representative with respect to appropriate local co-ordinates at source and 36

target) and the jets of the germ correspond to the initial finite segments of the Taylor series. J (n, p)

Let us make this more precise.

we mean the real vector space of all mappings

of whose components is a polynomial of degree 5 k xi, ..., xn

in IRn

be called k-jets. that

By the jet-space

a E IRn.

with zero constant term: Suppose now that

f

:

IRn

If in the Taylor series of

f

:

IRn

in the standard co-ordinates

the elements of --,

each

-, IRp

IRp

will

J (n, p)

is a smooth mapping, and

f(x) - f(a)

ressed in terms of the standard co-ordinates on Mn, IRP)

at the origin

(exp-

we delete all terms

of degree > k the result can be thought of as a k-jet, which one writes jkf(a)

and calls the k-jet of (the germ of)

arrive at a smooth mapping

jkf

called the k-jet extension of

:

f:

IRn

-,

f

at

dk(n, p)

a.

In this way we

given by

a -

jkf(a)

this mapping will play a crucial role in

the final sections of the next chapter.

37

II Transversality

§1.

The Notion of Transversality

The notion of objects intersecting transversally (or in general position) has The simplest situation to

become quite fundamental to singularity theory. look at is two subspaces of a vector space transversally when their vector sum is

transverse

V:

we say that they intersect

V.

not transverse

transverse

The notion is easily extended to smooth submanifolds of a smooth manifold. N1, N2

We say that two smooth submanifolds sect transversally at

x e N1 (1 N2

intersect transversally in

TxN:

when they do so at every point in

of a smooth manifold N

when the tangent spaces and

N1, N2

N1 n N2.

TxN1, T N x 2

intersect transversally in

38

not transverse

N

Probably the best way to under-

stand the idea is to look at a series of pictures.

transverse

inter-

transverse

not transverse

transverse

The idea which we really want to exploit is this. smooth mapping, and let graph

and

f

verse to in

Q

N x P.

N x Q

be a smooth submanifold of

Q

are smooth submanifolds of

and (write

Let

when graph

f (j1 Q)

P.

One pictures it thus. P

be a

Recall that both

We say

N x P.

f, N x Q

f : N -i P

f

is trans-

intersect transversally

graph f

N x Q

N

(1.1)

Let

f

manifold of P. with

y = f(x) E Q

be a smooth mapping, and let

Nn - Pp

An equivalent condition for

N x P points

The condition for graph is that for all points (x, f(x))

be a smooth sub-

is that for all

xEN

we have Txf(TxN) + TyQ

Proof

f (I1 Q

Q

with

to intersect transversally in

f, N x Q

z = (x, y)

f(x) E Q.

TyP .

=

in the intersection,

i,e.

all

we have

Tz(graph f) + Tz(N x Q)

=

Tz(N x P).

In view of (I.3.2) and (1.3.3) this may be re-written as

graph Txf +TxN x TyQ

=

TxN x Ty

which is clearly equivalent to . 39

Some authors use the relation tainly in practice

*

as a definition of

*

is probably easier to work with.

f ( Q, and cerHowever we prefer

the definition given above on the ground that it has more immediate geometric content.

Case 1

There are some special cases well worthy of separate mention.

Take the case when

is submersive.

In that case

p-dimensional vector space submersion

Case 2

f : N -> P

holds (for some

*

if

codim Q > dim N

to the image avoiding

Case 3

Txf (TIN)

is a p-dimensional subspace of the

hence equal to it.

TyP,

its germ at any point

i.e.

We conclude that a

must be transverse to every submanifold

Q C P.

A further particular situation is prompted by the observation that

if

f

is a submersion

f

f(N)

x)

then certainly

then transversality of

being disjoint from

Consequently

codim Q , dim N. f

:

N--+ P

to

Q

is equivalent

or (expressed more vividly) to

Q,

Q.

One last special case is when

diagram above). lar value of

f.

P

A point in A point

f,

the condition for this is that

have rank < n.

to which

x e N

value is a critical point of

Q

comprises a single point (see the is transverse is called a re u-

f

for which and

Txf

f(x)

is a critical value.

should fail to be surjective,

For instance, when N = P =1R

the graph of

f

c

Clearly, i.e.

the critical points are pre-

cisely the points where the derivative vanishes; precisely the real numbers

fails to be a regular

f(x)

for which the line

and the critical values are

y = c

fails to intersect

transversally.

It is maybe worthwhile spelling out the fact that the notion of transversality is invariant under equivalence of smooth mappings, in the following precise sense.

Suppose the smooth mappings

one has a commuting diagram of smooth mappings 40

f1, f2

are equivalent,

i.e.

h

1

k

I

f2

N

P2

2

h, k

with

diffeomorphisms;

respectively corresponding under

P1, P2

submanifolds of

of the invariance of transversality is that only if

is transverse to

f2

are smooth

Q1, Q2

and suppose further that

The statement

k.

is transverse to

f1

Q1

if and

The proof is a straightforward deduction

Q2.

from the definitions, and is left as an exercise for the reader.

Bear this

point in mind when reading the proof of the next proposition, which provides us with a painless procedure for extracting further examples of smooth manifolds.

Let

(1.2)

P

manifold of

be a smooth mapping, and let

f : Nn -a PP

with

having the same codimension as N

in

y = f(x)

with

in

M = f-1Q

then

Q:

f tj

Q

one has

T

N

Further, for any point

x

M = Txf-1 (T Q).

x

For the first proposition it suffices to show that every point x s M

Step I

M

has a neighbourhood whose intersection with dimension

Put

r = p - q.

suppose that

y = f(x).

are open sets in IRn,

N, P

is a smooth manifold of co-

with IRq x

Let

0.

IRp

and that

transversalitv of

f

to

Q

tells us that

7T o f

(x1,

0

in IRn ..., xn)

onto another with -*

(x1,

..., xr).

i

o f o h

on

IRp

has rank

then (I.1.3) tells us that there is a diffeomorphism of

x = 0,

y = 0:

is the intersection of

Q

denote the projection of

7

x, y we can

By taking charts at

indeed (I.2.2) allows us to assume further that P

be a smooth sub-

is a smooth submanifold of

or is empty.

Q,

Qq

k

0 x

r at

IRr.

0,

The and

of a neighbourhood

the projection

The inverse image of

0

under this mapping 41

0 x ,n r,

is an open subset of

M

that

is a smooth manifold of codimension

Step 2

The tangent space

f

h

corresponding under

TxM

to

that shows

M:

r.

The commuting dia-

is computed as follows.

gram of smooth mappings on the left gives rise to the commuting diagram of differentials on the right.

N

inc T M

TQ

f M

T f x

T N

3, P

X

inc

inc

y

inc

T (f IM)

x

TxM

TXM C Txf-1(TyQ).

from which it is clear that

T P

>T

y

Q

To show that these vector

spaces are equal it will suffice to show that their dimensions are equal.

For this one considers the linear mapping

Txf-1 (TyQ)

-->

given by res-

T Q

y tricting T Q

Txf :

Txf(TxN).

(1

y

Txf,

note that it has the same kernel as

and image

The desired equality now follows on using the fact that

the dimension of the domain is the sum of the dimensions of kernel and image, together with the definition of transversality.

Note the special case of this result when

T

xM

f -'(y)

is a smooth manifold of dimension

Let

A = (aid)

Consider the smooth mapping .

In

the conclusion

f;

or empty, and that

n - p,

be an invertible symmetric f

:

IRn -> IR

x

value of

and

f,

is the linear mapping f-1(1),

i.e.

a smooth manifold of dimension

v - 2(Ax.v),

the central quadric (n - 1),

n x n

which is given by

denotes the standard scalar product on IRn.

at the point

42

y.

is the kernel of Txf.

Example I

where

is a single point

y is a regular point of

that case the hypothesis is that is that

Q

or empty.

matrix.

f(x) = Ax.x,

Here the differential so

I

is a regular

a..x.x.

1J 1 J

=

1,

is

And the tangent space

at

x

to this quadric is the kernel of the differential,

Taking A

plane perpendicular to the vector Ax. we see that

is a smooth submanifold of IRn

Sn-1

fact we already knew from Chapter I;

Taking

expect.

the hyper-

to be the identity matrix of dimension

(n - 1),

x,

as we would

we see that ellipsoids, hyperbolic cylinders, ellip-

n = 3

tic cylinders, and hyperboloids are all smooth manifolds of dimension

Consider a smooth mapping

Example 2 rank

I

at every point in

f-1(0)

:

IR.3

f

- {0J -->

IR2

:

---, IR

i.e.

f-1(0)

is then a

a smooth curve in the plane.

given by F(x, y, z)

IR

2.

whose differential has

(1.2) tells us that

:

of dimension 1,

smooth submanifold of IR2

Consider now F

a

moreover the tangent space to this sphere

is precisely the hyperplane perpendicular to

x

at a point

i.e.

=

f(Ix2

+

y2,

z).

This too has the property that the differential has rank 1 at every point in so likewise

F 1(0),

is a smooth submanifold of

F-1(0)

IR3

of dimension 2:

it is just the surface of revolution obtained by rotating the curve f(y, z) = 0, tion let

y > 0

a, b

about the z-axis in

f

Clearly the differential of is the circle radius F 1(0)

a > b > 0,

be real numbers with

f(y, z)

b

By way of specific illustra-

IR3.

and let

f

be given by

(y - a) 2 + z2 - b2

=

has rank I at every point in

centred at

f-1(0),

which

The surface of revolution

(a, 0).

is a torus, and a minor computation will verify that it is given by

the equation

(x2 + y2 + z2 + a2 - b2)2

Example 3 Cc

Suppose

Aa, Bb

C

4a2(x2 + y2).

are smooth submanifolds of a smooth manifold

intersecting transversally in

or a smooth submanifold of

=

C:

the intersection A n B

of dimension

a + b - c.

will be empty,

This follows on 43

observing that the inclusions

A - C, B -4 C

are transverse to

B, A

respec-

tively.

A singular point (or singularity) of a smooth mapping point

x E N where the germ is singular.

f : N - P

(Of course a singular point is a

critical point, though the converse does not necessarily hold.)

E f of

set

is the set of all its singular points:

f

is a

The singular

the image of

Z f is

sometimes called the bifurcation set.

Example 4

f

:

The Whitney cusp mapping of the plane is the smooth mapping

given by (x, y)

]R2 -+ IR2

->

(u, v)

u = x, v =

where

y3

- xy.

The

singular set is the set of points where the Jacobian matrix has rank < 2, i.e.

the parabola

parabola under 4u3 - 27v2

=

x = 3y2.

f ,

And the bifurcation set is the image of this

the cuspidal cubic having the equation

i.e.

0.

f

The Whitney cusp mapping is best understood on a geometric level as the composite of the mapping

g

:

IR2 - IR3 given by

u = x, v = y3 - xy, w = y and the projection (u, v, w) ded surface

(u, v): 5

see the diagram below.

defined by

of dimension 2 by (1.2).

44

v - w3 + uw

=

0,

v

(x, y) :

->

(u, v, w) with given by

IR3 -> IEt2

The image of

g

is the fol-

a smooth submanifold of

IR3

It is maybe worth saying a little more about the projection of a surface onto a plane, as it helps to strengthen one's intuition. surface plane

S C ]R3,

P C R3

tion onto of

P

i.e.

Consider a smooth

a 2-dimensional smooth submanifold of ]R3,

through the origin:

we take

to denote orthogonal projec-

v

and ask for the singular set of the restriction

P,

and a

7rIS.

Think

For

as "horizontal", and the line perpendicular to it as "vertical".

each point

x E S

the commuting diagram of smooth mappings on the left gives

rise to a commuting diagram of differentials on the right.

T S

IT

The condition for x should map T S

x

to be a singular point of

onto a proper subspace of

T S

x

should be "vertical";

be a curve in

S

IT

x

P,

HIS i.e.

is therefore that

that the tangent plane

thus, in general, the singular set of 1rIS

projecting to a curve in

P,

IT

will

the bifurcation set.

4+5

Example 5

Let us return to the model situation presented by the last The folded surface was given by the equation

example.

and was projected onto the plane kernel of the differential of alF

lau' 8v'

i.e.

w),

to

w = 0.

The tangent plane to

(w, 1, u - 3w2):

Thus the singular set of

the equations

u = 3t2,

S

is the

i.e. the plane perpendicular to the vector

F,

and the tangent plane will be

"vertical" when this vector lies in the plane v = -2w.

F = v - w3 + uw = 0,

zrJS

v = -2t3, w = t;

w = 0,

when

i.e.

u = 3w2,

is the "fold curve" parametrized by the geometrically inclined reader

will recognise this as a twisted cubic, projecting onto the cuspidal cubic. Note that one point on the "fold curve", namely that projecting to the cusp. it is the one point where two folds on

differs qualitatively from the rest;

the surface meet, whereas all the others are just points where the surface In some sense therefore we have two types of singular point.

folds over.

Consider the projection

Example 6

the plane y = z.

G

=

Recall that

S

of the torus

S

of Example 2 onto

was given by an equation

(x2 + y2 + z2 + a2 - b2)2

The tangent plane to 46

S

it

-

4a2(x2 + y2)

is the kernel of the differential of

=

(1)

0.

G,

i.e.

the

plane perpendicular to the vector point to be a singular point of plane

y = z,

i.e.

that

ay

raG aG aGll tax, ay, az/:

iris

=

and the condition for a

is that this vector should lie in the

az ,

which computation verifies to be the

condition that

+ y2 + z2 + a2 - b2) + 2a2y

(z - Y X X

The singular set of

iris

=

is given therefore by (1), (2),

0.

i.e.

(2)

it is the

intersection of two surfaces of degrees 4, 3 respectively, hence a space curve of degree 12. set;

And its projection onto the plane

y = z

is the bifurcation

it is what we "see" if we imagine the torus made of glass and viewed

from a distant point on the line perpendicular to the plane y = z.

So much for transversality per se.

The remainder of this chapter is

devoted to the development of a very basic intuition associated with transversality.

Suppose we have a smooth mapping

transverse to a submanifold forced to be transverse to

Q C P. Q

f : N --).P

which fails to be

Common sense tells one that

by arbitrarily small perturbations.

f

can be

The for-

mal development of this intuition leads to a host of results, reaching high degrees of complexity and subtlety, called transversality theorems.

In recent years transversality theorems have assumed a role of increasing importance in differential topology, indeed Singularity Theory could hardly exist without them.

Essentially they all say the same thing, namely that by

an arbitrarily small perturbation a given smooth mapping (or some closely

47

related smooth mapping) can be made transverse to a smooth manifold.

The

object of the next section is to establish a very basic lemma which provides the key to most transversality theorems.

The Basic Transversality Lemma

§2.

F : N x S

Let

P

-4

be a smooth mapping.

family of smooth maps

fs

the elements

Suppose

s e S.

We ask whether

fs (t1 Q

: N -> P

This we think of as a smooth

where with

F (I1 Q

for all parameters

F(x, s)

=

fs(x)

parametrized by

a smooth submanifold of

Q

s?

P.

That the answer can well be

in the negative is shown by

Example 1 F

fs

:

Take

N x S maps

P

-*

IR2

the 2-sphere

N = ]R 2,

S = ]R,

P = IR3

defined by the formula

onto the horizontal plane F

S2.

and consider the smooth family

((x, y), s) in IR3.

z = s

is a diffeomorphism, so

->

F

Q.

And take

the plane

is tangent to

z = s

to be

Q

On the other hand

only provided vie avoid the two exceptional parameters

fs (h Q

Thus

(x, y, s).

s = +1

when

Q.

However, it is clear in this example that any value of the parameter can be approximated as closely as we please by "good" values, for which

Transversality Lemma: Q

for technical reasons we prefer to replace the single

by several smooth manifolds

Q1, ..., Qt :

become clear in the final section of this chapter. Sard's

the reason will

The crucial tool here is

Theorem, one of the truly fundamental results in differential topology,

which we state in the following form.

48

those

That this holds generally is the content of the Basic

fs (f Q.

smooth manifold

i.e.

s

(2.1)

Let

fi

:

Ni

->

P

be a countable family of smooth mappings.

set of common regular values of the

fi

is dense in

The

P.

The proof of Sard's Theorem is quite lengthy, so we have isolated it as Appendix A to this book in order not to break the flow of the text.

The

Basic Transversality Lemma is

(2.2)

Let

F : N x S

P

->

verse to smooth submanifolds of parameters

Proof

B

for which

s

1.2

M.

sider the restriction

= in Ni

be a smooth family of smooth mappings transof

Q1, ..., Qt fs

then there is a dense set

P:

is transverse to all of

F(Q1)

Q1,

..., Qt.

is a smooth submanifold of

of the projection

in

: N x S

->

N x S. S.

Con-

l'Te shall

show that

vjMi

if and only if fs

{sJ

which will prove the result, by Sard's Theorem. ',4'e can, and shall, drop the index

i

41

Qi

Now to the proof of

since it plays no further role.

*.

The

reader is urged to keep the following picture in mind.

N To start with, note that the condition for that for all

z

=

(x, s)

in

N x S

with

TzF(TxN x TSS) + T

w

F

to be transverse to

w = F(z)

Q

=

TWP.

in

Q

Q

is

we have

(1)

49

Consider now the condition for dition that for the same

x

to be transverse to

fs

Q:

as in (1) =

TZF(TXN x 0) + TwQ

TwP.

(2)

Furthermore, the condition for N x (s} to be transverse to x

the same

M = T F-1(TWQ)

T

M

is that for

as in (1)

TXN x 0 + TZM

where

this is the con-

=

TXN x TsS

(3)

If we assume (3) and apply TF to

by (1.2).

z

z

both sides then

(1)

tells us that

And conversely it is a

holds.

(2)

minor exercise in linear algebra to see directly that Thus the condition for

for N x {s}

fs

to be transverse to

to be transverse to

lent to the condition that

7vIM

x

dition says that for the same

implies

is precisely the condition

is transverse to as in

(s}:

indeed this last con-

we have

(1)

TSS

(4)

(3).

By way of explicit illustration consider two smooth manifolds some Euclidean space

(3).

We claim that this in turn is equiva-

M.

TZTr(TZM)

which is clearly equivalent to

Q

(2)

M, N

in

Even if they do not intersect transversally, geo-

S.

metric intuition tells us that it should be possible to force them to do so

by an arbitrarily small translation of one manifold M We can make this feeling precise as follows.

claim that the S,

s

for which

Ms, N

justifying one's intuition.

mappings 50

under the translation of

M

the image of

F

:

M x S

-->

S

by

S

For

in some direction.

s s S

let

Ms

denote

defined by x H x + s.

We

intersect transversally will be dense in

To this end define a smooth family of smooth

(x, s) H x + s.

It is clear that

F

is a

submersion, so

is transverse to

F

is transverse to

fs

N

for a dense set of parameters

established by observing that

is transverse to

f

By the Basic Transversality Lemma

N.

N

-

Ms, N

Our claim is

s.

intersect transversally if and only if

which fact we leave as an exercise for the reader.

S

An Elementary Transversality Theorem

§3.

Our object in this section will be to show that given a smooth mapping f

IRn

:

ping to

g

f.

-+

and a smooth manifold

IRp,

: Rn -

which is transverse to

IRp

we can find a smooth map-

Q C IRp, Q,

and as "close" as we please

Of course the first problem is to say precisely what we mean by

smooth mappings being "close":

roughly speaking we shall take this to mean

that their values are "close", and that for each integer

vatives of order k

all smooth mappings

number

Rn ->

and an integer

bourhood in CO°(IRf, for which for all

their deri-

IRp,

IR.p)

k > 0

denote the set of

Let e(IRn, IRp)

and let

f

:

Mn - IRp

Given a (small) positive real number R,

1

are "close".

We make this precise as follows.

mapping.

k >.

a (large) positive real

e,

we associate to

be a given smooth

f

a fundamental neigh-

comprising all those smooth mappings

x E Rn with

Ixl

s R

g

: Rn, IRp

one has

51

Iljkf(x) - jkg(x)II with

set

XC

f

IRn

:

C

a fixed norm on the jet-space Jk(IRn, IRP).

II

II

'-

dense therein when given any smooth mapping

C-O (IRn, IRp) -,

and any fundamental neighbourhood V

IRp

smooth mapping

g

IRn ->

:

And we call a sub-

X with

in

IRp

g e V:

one can find a

f

of

intuitively, any mapping

can be approximated as closely as we please by mappings in

X.

'.`e are now in a position to state and prove an elementary transversality theorem.

The set of smooth mappings

(3.1)

submanifolds

Q1, ..., Qt of IRp is dense in

Proof

f

imate

Let f

:

IRn

->

IRp

be smooth.

n' ],p).

C-'O(IR

l7e have to show that we can approx-

as closely as we please by mappings transverse to

The idea is to construct a smooth family f,

transverse to given smooth

IRp

IRn --*

and with

F

transverse to

Transversality Lemma.

F

Q1, ..., Qt

lie shall make

F

:

IRn X'S - IRp

Q1, ..., Qt.

which contains

then one applies the Basic

transverse to

Q1, ..., Qt

by en-

suring that it is a submersion.

To motivate the construction recall that the

transversality of

to

transverse to

f

IRn x Q

:

1Rn -,

IRp

Q

is equivalent to the graph being

in the product IRn x

IRp.

If the graph is not already

transverse it seems reasonable that we might be able to force it to be so by

52

bodily translating it.

point.

The picture above illustrates the idea when

Therefore we take

(x, s) H f(x) + s.

and define

= IRp

S

F

:

Q

IRn x IRp , ]Rp by

Clearly, this is a submersion, so transverse to

Q1, ...) Qt.

By the Basic Transversality Lemma there is a dense set of

for which

is transverse to

fs

is that if

f0 = f: V

hood

then

0

s

All that remains to be shown

Q1, ..., Qt.

is close enough to

s

is a

is as close as we please to

fs

more formally, we have to check that given a fundamental neighbourof

we can find an

f

s / 0

for which

f

lies in

V.

And that

we can safely leave to the reader.

It is only fair to point out that this example of a transversality theorem is not particularly useful.

Its virtue lies rather in the fact that it is an

easily understood prototype of transversality theorems of greater complexity and application.

Our object here was simply to lay bare the underlying idea

behind the proofs of such theorems.

4.

Thom's Transversality Theorem

Our next transversality theorem is rather more useful than that of the preceding section.

Indeed it will suffice for all the applications we shall

require in this book.

Its statement does not possess quite the same immediate

intuitive appeal as our previous result:

however, in the next section we

shall discuss a simple application which should clarify the situation.

(4.1)

Let

Q1, ..., Qt

be smooth submanifolds of the jet space

The set of all smooth mappings f is transverse to

Q1, ..., Qt

:

IRn -a IRp

is dense in

J"( n, p).

for which j kf : Mn -> Jk(n, p)

Ca'(IRn,

IRp).

53

Note

It is essential that the reader appreciate the difference between

this transversality theorem and that of the preceding section. vious result we managed to make

f

transverse to a submanifold by using a

But in the present situation it is

constant deformation.

which we wish to make transverse to a submanifold. f

only

In the pre-

which we are allowed to deform.

not

jkf,

f,

On the other hand it is

A constant deformation will not

work here since it does not alter the derivatives of

What we do instead

f.

is to use a polynomial deformation.

Proof

Let

S = J (n, p).

Of course

:

IRn x S - Jk(n, p)

mersion, because for fixed Jk(n, p)

of

defined by x

is finite-dimensional, so can

Consider the smooth family

be identified with a Euclidean space. F

S

(x, s) _ jk(f + s)(x).

the mapping represents an affine isomorphism

with itself, hence transverse to all of

Q1, ..., Qt.

Basic Transversality Lemma there is a dense set of parameters the mapping

fs

:

IRn

->

Q1, ..., Qt.

we please to

by choosing

f

defined by x

Jk(n, p)

verse to all of

is a sub-

F

-+

s

for which

s

jk(f + s)(x)

And clearly we can make

f + s

By the

is trans-

as close as

to be sufficiently small.

First Order Singularity Sets

§5.

We shall use the Thom

This section is in the nature of an extended example.

Transversality Theorem to show that for a dense set of smooth mappings

f

:

IRn

-->

IRP

the singular set

smooth manifolds on each of which

Z f

f

can be partitioned into finitely many

has constant rank.

Recall that a singular point of a smooth mapping point

54

x E N

for which the rank of the differential

f

-> PP

T f

x

is a

falls below its

possible maximal value of

min (n, p).

A natural way of distinguishing one

singular point from another is by the actual value taken by the rank of the differential.

To this end we introduce the first order Thom singularity sets

Elf

=

{x E N

(Concerning the terminology :

:

T

later in this book we shall have something to

say about higher order Thom singularity sets.)

tition of

N

it.

xf has kernel rank

In this way we obtain a par-

into finitely'many sets on each of which

has constant rank.

g

One might reasonably hope that these sets will be submanifolds of N,

but

that (as the following examples will illustrate) is not necessarily the case.

Example 1

Take

f

IR2

:

-+

IR2

defined by the formula

(x, y)

-,

(x2, y2).

This we shall refer to as the "folded handkerchief" mapping for the following reason:

it is the composite of the two mappings

(x, Y)

~'

(x, y2)

:

(x, Y) _

(x2, Y)

which "fold" the plane along the x-axis, y-axis respectively.

One pictures

it something like this.

fold

fold

along x-axis

along y-axis

LL

x

The reader can easily check that the origin is the only on the axes being

Z1

points, and the rest being

ZO

ZL-point, other points

points.

(One need

hardly point out that this fits in with the fact that the origin is folded twice, other points on the axes just once, and the rest not at all.)

Here

55

Elf are all submanifolds.

the

Take the smooth mapping

Example 2

(x, y) H (x2 + y, y2). xy = 0,

given by

i.e.

E2

points.)

Take for instance fs

points are now given by xy = Thus as

moves from

s

However, it is well worth ob-

E1f

to be a submanifold by slightly

:

defined by

IR2 -+ 1R2

Another minor computation shows that the

(x2 + y, y2 + L.sx).

->

points are

But a small linear deformation will do

matrix, and is therefore of no use.

(x, y)

E1

Of course a constant deformation will not affect the Jacobian

f.

the trick.

given by

is the union of the two axes, so certainly not a

Elf

serving that in this example we can force deforming

IR2 ->7R2

:

A minor computation shows that the

(There are no

a submanifold.

f

s,

a hyperbola, so certainly a submanifold.

to a small value so the singularity set

0

E1

E1fs

changes from a pair of intersecting lines to a pair of disjoint curves. We shall now prove generally that the first-order singularity set

of a smooth mapping

f : N - P can be forced to be submanifolds, by the

device of slightly deforming

f.

For the sake of technical simplicity we

shall carry it through only in the case that

f

x E IRn

:

IRn

-+

IRp

Elf

N =IRn,

is our smooth mapping.

P = RP.

Suppose then

I'otice that given a point

the first order singularity set to which it belongs depends only on

the 1-jet of

f

of the jet space

at

x,

i.e.

J1(n, p)

Dxf.

So let us write

E1

for the subset

comprising all 1-jets which have kernel rank

i.

equal to

i

is a smooth submanifold of

1

J (n, p)

of codimension

i(p - n + i)

(5.1)

E

Proof

For this it will be convenient to identify a linear map with its

matrix relative to the standard bases.

56

We proceed in two steps.

Step

1

Let

matrix.

E

( D)

=

We claim that

E

p x n matrix with A

be a

has rank

k

D = CA 1B.

if and only if

this, observe that for an arbitrary matrix X

an invertible

the matrix

E

k x k To see

has the same

rank as

IIk X

0

A

B

A

B

Ip-k

C

D

XA + C

XB + D

and the claim follows on choosing X

Choose

Step 2

k

with

i + k

=

will be no restriction to suppose

such that

and let

n,

E0

XA + C

(CO

=

tible matrix. E0

matrix

U E

U 1\ E1

=

in it the

( D)

k x k

E0

f

is a submersion:

function of

D

is an invertible affine mapping.

j

1f

Proof

is invertible.

indeed if one fixes

is a smooth submanifold of codimension f-1(0)

=

of

is transverse to all the sets

El ,

--

Consider

D - CA 1B.

A. B, C

the resulting

It follows from (1.2) that (p - k)(n - k) = i(p - n + i):

U n E1,

There is a dense set of smooth mappings

smooth manifold of codimension

U

with the property that for any

f : U - J1(p - k, n - k) given by E

but by Step 1 the inverse image

inver-

0

matrix A

Notice that

(5.2)

k x k

a

It

a smooth submanifold of codimension i(p -n+i).

to be an open neighbourhood of

the smooth mapping

f-1(0)

AO

E1.

And it will suffice to produce an open neighbourhood

in J1(n, p) with

We take

with

DO)

0.

be a matrix in

EO

0

=

f

which ends the proof.

:

]Rn -+]Rp

and hence for which each

for which Ei f

is a

i(p - n + i).

The first statement is an immediate consequence of the Thom Transver-

sality Theorem (taking

k = 1).

And the second statement follows from the 57

fact that

Elf

is just the inverse image under

j1f

Z ,

of

so is a smooth

11

manifold of the same codimension by (1.2).

transverse to the

E i

will have codimension Elf

f

Consider a smooth mapping

Example 3

is void for

immersion

i.e.

that when

p 3 2n

:

in the special case when i(p - n + i)

i > 0,

>

is

fRn - ]Rp for which j 1f In this case

p > 2n.

i(n + i) > n,

unless

its germ at any point is immersive. any smooth mapping ]Rn -4IRp

Thus

i = 0.

which is the same thing as saying that

f

Elf

is an

Our theory says then

can be slightly deformed to

become an immersion, and indeed a linear deformation will suffice (as is clear For an explicit instance

from the proof of the Thom Transversality Theorem). take the cuspidal cubic curve

f

Here, for a small positive number

fs(t) = (t2, t3 - ts)

:

IR --]R2 s

given by

the curve

fs

:

is an immersion, obtained from

f(t) = IR ->IR2 f

(t2, t3).

given by

by a small linear

deformation.

image of fs

image of f Example 4

We shall consider, in some detail, the condition that

transverse to the

E1

for a function f

:

]Rn -> IR.

This means we are to

have

image of differential

tangent space J1(n, 1)

of 58

j 1f

at

a

to Ei at j 1f (a)

j1f

be

for all points

a e IRn.

We shall show that this is exactly the condition

that the so-called Hessian matrix a2f

H

l

(a)Jnxn

(ax. ax. i J

-

is non-singular at every critical point

where

a,

x1, ..., xn

are the stan-

dard co-ordinate functions in ]Rn.

To start with, let us simplify matters by identifying

with IRn

J1(n, 1)

by identifying a linear mapping IRn -IR with its matrix relative to the In particular

canonical bases.

Daf

is identified with

(

Of

Of

(a), ...,

Notice that the only Thus

respectively.

verse to it.

E

En-1

f is defined by the

and is just the critical set of

easily identified.

can be discounted. ate

IRn,

i.e.

that

1

j

j1f

at

f

at

a

*

*

is precisely

a

are H,

gen-

so its tangent space

is that the columns of

*

=

when

is the, subspace of IRn

a

is just the origin in IRn,

Thus the burden of H

= 0, ..., ax n

In that case the expressions in

so the image of the differential of

En

is automatically trans-

f

Thus we need only satisfy

f.

i = n.

j

0, n

of codimensions

conditions ax 1

n

The Jacobian matrix of

erated by its columns.

I

is an open set and

n

is a critical point, and

En-1, En

which arise are

El

a)

n

1

gener-

H

is non-singular, establishing our claim.

A critical point of a smooth function

f

:

IRn -+IR

matrix is non-singular is called non-degenerate. summed up by saying the condition for

j1f

at which the Hessian

Our discussion can then be

to be transverse to the

that every critical point should be non-degenerate.

E1

is

By way of illustration,

the reader may care to check that the function

f(x, Y)

=

x2 - 3xy2

has a degenerate critical point at the origin.

One pictures it like this.

59

0

In this example

j1f

cannot be transverse to the

E1.

According to the

theory of this section it should be possible to gain transversality to the E1

by a linear deformation of

is linear.

i.e.

f,

by taking

fs = f + Ls

where

Here is an explicit linear deformation which does just that

Ls -

the reader is urged to check this for himself.

fs(x, y)

The graph of

fs

=

x3 -

3x2

is rather difficult to draw.

happens is indicated in the pictures below. the case

s = 0,

fs > 0.

The left-hand picture refers to s > 0.

In both cases the

And in both figures the curves

have been drawn for small positive values of

60

However, some idea of what

the right-hand one to the case

shaded area is that where

- sx.

e.

fs = +e

III Unfoldings: the finite dimensional model

The kind of mathematics which we shall discuss later in this volume lies rather deep, at least in the sense that formal proofs of the main results rely However the underlying geometric ideas are

on hard theorems in analysis.

The main object of this chapter is to introduce these

rather straightforward.

ideas in a relatively simple situation where the mathematics is easy enough not it will also provide us with an opportunity of est-

to hinder understanding:

ablishing one or two little facts which we shall have occasion to use later.

Groups Acting on Sets

§1.

By an action of a group

(ii)

where

1

on a set

(g, x) + g.x,

usually written (i)

G

=

x

(gh).x

=

1.x

M

we mean a mapping

for which for all

iD

x E M.

: G x M -+ M,

and

g.(h.x)

denotes the identity of

G.

Given such an action we can define an equivalence relation agreeing that

x ti y when there exists an element

i.e.

x

.-

on

M

by

g E G for which y = g.x.

The equivalence classes are called the orbits under the action. the orbit through

g, h E G

Given X E M

is by definition the equivalence class which contains

x,

the set

G.x

=

{ g.x : g E Gi .

61

Before turning to examples let us observe one small geometric point.

with

M lying on the same orbit, so there exists

be points of

x1, x2 x2

=

is a transformation of serves orbits

-

M,

(i.e.

a bijection of

M

onto

g e G

a

defined by x

Observe that the mapping M -+ M

g.x1.

Let

-+

g.x

itself) which pre-

by which we mean that a point is always mapped to another

point lying on the same orbit

-

and which maps

What this

x2.

to

x1

amounts to is that one point on an orbit looks like any other; perty known as homogeneity of an orbit.

it is the pro-

The reader is urged to bear this

notion in mind as a guiding intuition. It is not the purpose of this chapter to pursue the general theory of groups acting on sets.

Rather, we wish to concentrate the reader's attention on a

class of geometric examples relevant to the mathematics of the next two chapters, and indeed providing genuine finite-dimensional analogues of the situations there studied.

§2.

To this end

Some Geometry of Jets

We start by recalling that in Chapter I we introduced a relation of equivalence on germs, so in particular on the set of all germs indeed two such germs

h, k

f, g

are equivalent when there exist invertible germs

for which

foh

=

k o g.

(1)

One can relativize this definition to d-jets as follows. to have equivalent d-jets when there exist invertible germs

jd(f o h)

In particular we can take

f, g

=

f, g h, k

are said for which

(2)

jd(k o g).

to be (germs at

cing an equivalence relation on the jet-space 62

(IRn, 0) -+(IRp, 0);

0

Jd(n, P).

of) d-jets, so induThe geometric

examples we have in mind arise from studying this equivalence relation not on

of all mappings

in -+

rather on the vector subspace

Hd(n, p)

each of whose components (relative to the standard

IRP

is a homogeneous polynomial of degree

co-ordinates on IRP) dard co-ordinates

but

Jd(n, p)

the whole jet-space

in

x1, ..., xn

d

in the stan-

The geometry has its genesis in

]Rn.

the following elementary, yet crucial, observation.

Two d-jets

(2.1)

f, g

are equivalent if and only if there

Hd(n, p)

in

exist invertible linear mappings

H. K for which

The condition is certainly sufficient.

Proof

f o H

=

K o g.

To establish necessity,

suppose there exist invertible germs

h, k for which (2) holds.

for the Taylor series of a germ

(IRn, 0)

k

with

:

H + terms of degree

>.

2

=

K + terms of degree

>

2

f, g

components of

cp

then we have

(]RP, 0):

=

Thus, bearing in mind that the

invertible linear mappings.

H, K

-->

Write

are homogeneous polynomials of degree

d,

(f

h)

=

(k

g)

= K o g + terms of degree > d

one has

f o H + terms of degree > d

so

f o H

=

jd(f o h)

=

We can re-phrase the above as follows.

jd(k o g)

Let

=

GL(s)

linear group of all invertible linear mappings ]Rs - IRs of composition. group

=

denote the general under the operation

The reader will readily check that we have an action of the

GL(n) x GL(p)

(H, K).f

K o g.

on the vector space

K o f o H 1.

Hd(n, p)

given by

and the condition for two d-jets

f, g

in 63

Hd(n, p)

to lie in the same orbit under this action is that there exist in-

f o H

= K o g.

classes described above.

=

f,

i.e.

GL(n) x GL(p)

are precisely the equivalence

The next step in our study is to indicate the We start with

connexions with geometry.

p = 1

The Case

In this case an element of degree

(H, K) . g

It follows immediately from (2.1) that the orbits in

under the action of

Hd(n, p)

for which

H, K

vertible linear mappings

d

in

x1, ..., xn

surface of degree

d

Hd(n, 1)

so (providing it is not zero) will define a hyper-

in real projective space

for instance when n = 2

we are dealing with

when n = 3 with a curve of degree And two elements of

is just a homogeneous polynomial of

d

P]Rn

d

of dimension

(n -1)

:

points on a projective line,

in the projective plane, and so on.

will lie in the same orbit if and only if the

Hd(n, 1)

corresponding hypersurfaces can be obtained from each other by an invertible linear change of co-ordinates,

are projectively equivalent.

i.e.

ficulty of listing the orbits increases sharply with

d:

The dif-

we shall allow our-

selves the luxury of describing in detail some of the simpler cases relevant to the mathematics of the next two chapters. d = 1.

One starts with

in n variables

is the vector space of linear forms

H1(n, 1)

x1, ..., xn and elementary linear algebra tells us that

there are just two orbits namely that containing the zero form, and that containing all the non-zero forms. The next case is forms in

n

d = 2.

variables

H2(n, 1)

x1, ..., xn

is the vector space of all quadratic

and elementary quadratic algebra tells

us that any such form can be brought into the shape x1 + ... + xs - xs+1 - .. - xr

64

by an appropriate change of co-ordinates.

The numbers

r, s

are called the rank, index of the form, and (by Sylvester's

Law of Inertia) are invariant under co-ordinate changes. tiply the form by

the rank remains invariant, though the index may change,

-1

so we work instead with the semi-index Thus quadratic forms in

x1, ..., x

s' = min(s, r - s)

x, y can be written

with the point

(a, b, c)

picture below.

The cone

in ]R3 b2 = ac

of the form.

are classified by rank and semi-index.

To make this even more explicit take the case when n = 2. in two variables

However, if we mul-

ax2 + 2bxy + cy2

A quadratic form so can be identified

We obtain four orbits as indicated in the comprises the forms of rank I (the Para-

bolic type) with the origin representing the zero form of rank bolic type).

(the IM-

0

The remainder of the space comprises the forms of rank 2:

indeed the inside of the cone corresponds to forms of semi-index

0

(the

elliptic type) and the outside of the cone to forms of semi-index I (the hyperbolic type).

A..

Let us continue with the case all cubic forms in

n

variables

d = 3.

H3(n, 1)

x1, ..., xn.

is the vector space of

This case differs from the

preceding ones in that a complete list of orbits is known only for The first non-trivial case is

n = 2,

n . 4.

and since this case will arise natur-

ally in the next chapter we shall take the opportunity to describe a once familiar bit of pure mathematics, the study of binary cubic forms

f(x, y)

=

ax3 + 3gx2y + 3yxy2 + Sy3.

65

It is easy to list the orbits using a little algebra. allow the variables forms.

Let us (temporarily)

x, y to be complex and restrict ourselves to non-zero

Recall that a non-zero complex homogeneous polynomial of degree

in two variables factorizes into

d

d linear factors (possibly with repetitions):

in particular

f(x, y)

=

(a1x + bly)(a2x + b2y)(a3x + b3y).

It follows that the zero set of equation

f(x, y)

=

0

f,

i.e.

the subset of

C2

defined by the

will comprise three lines through the origin.

We

shall distinguish four possible types of binary cubic form, corresponding to the following four types of triples.

(The reason for the terminology will

be explained shortly.)

elliptic

-

all distinct and real

hyperbolic

-

all distinct;

one real and two complex

parabolic

-

two distinct;

both real, one repeated twice

symbolic

-

one real line repeated thrice.

In fact these types are precisely the orbits we seek.

Certainly two bin-

ary cubic forms of the same type will lie in the same orbit, since given two triples of lines through the origin (of the same type) we can always find a non-singular real linear mapping of triple to those of the other.

C2

which maps the lines of the one

And conversely two binary cubic forms in the

same orbit are necessarily of the same type, since non-singular real linear mappings of

C2

preserve the above types of triples of lines.

We can ob-

tain normal forms for non-zero binary cubic forms by just choosing an example of each of the four types just described. given in the following table.

66

The standard choices are those

type

normal form

elliptic

x3 - xy2

hyperbolic

x3 +

xy2

parabolic

x2y

symbolic

x3

The reason for the terminology is as follows. f

Given a binary cubic form

one can associate with it a binary quadratic form

Hf

called the Hessian:

it is defined to be 2

H

f

=

2

a f

a f

ax2

axay

2

f

a2f

ayax

aye

1

37 a

and a little arithmetic will verify that it is given by the formula

82)x2 (ay -

+ (as - QY)xy +

((3S

- Y2)y2

.

It is now an easy matter to verify that a binary cubic form

f

is elliptic,

hyperbolic, parabolic or symbolic exactly according as the binary quadratic form

Hf

is;

and that is the reason these words are used to describe the

types of binary cubic form.

Although it is more difficult, one can obtain a visualization of binary cubic forms, just as we did for binary quadratic forms. cubic forms can be identified with IR4 ax3 + 3f3x2y + 3yxy2 + 8y3

The space of binary

by identifying

with the point

(a, (3, y, S):

non-zero forms into four types yields a partition of and by projecting this appropriately into

IR3

the partition of the

IR4 - [0; into four sets,

(we shall not go into the

67

details) one arrives at the following delightful picture, dubbed the umbilic bracelet.

It is obtained by rotating a deltoid (the curve traced by a fixed

point on a circle rolling inside another circle of three times its radius) about an axis with a twist of 2rr/3 for each full circle.

In this picture the

cusped edge of the bracelet corresponds to the symbolic forms, the rest of the

surface to the parabolic forms, the interior to the elliptic forms, and the exterior to the hyperbolic forms.

Let us pursue the possibility d = 3 consider is

n = 3,

i.e.

a little further.

ternary cubic forms in

x, y, z:

thought of as cubic curves in the real projective plane

The next case to these are best

The deriva-

PIR3.

tion of normal forms here is a lengthy exercise in the geometry of curves for details of which we refer the reader to texts on that subject; merely quote the results.

we shall

One starts with non-singular cubic curves.

non-singular cubic

zy2

- x3 + axz2

+ bz3

with 4a3 + 27b2 / 0

To such a curve one associates the so-called j-invariant

j = 4a3/4a3 + 27b2:

geometrically, this is the cross-ratio of the four lines through a point on the curve tangent to the curve elsewhere.

68

It can be proved that two curves

in this form lie in the same orbit if and only if the corresponding equal, and the the

a's

b's

have the same sign:

have the same sign.

cally irreducible:

or if

b = 0

j's

are

then if and only if

Of course, a non-singular curve is automati-

the remaining irreducible curves are those which are singu-

lar, and these have exactly one real singular point.

The tangents at the

singular point can be distinct (the nodal case) or coincide (the cuspidal case). And in the nodal case one can made the finer distinction between the crunodal

cubic (when the tangents are both real) and the acnodal cubic (when the tangents are complex conjugate).

In this way one obtains three singular irre-

ducible real cubic curves.

crunodaZ cubic

x3 + y2z + x2z

acnodaZ cubic

x3 + y2z - x2z

cuspidal cubic

x3 - y2z

We are left with a motley array of reducible cubic curves.

The cubic can

reduce to a conic and a line, which may or may not be tangent to the conic;

or it may reduce to three lines, giving rise to various possibilities.

69

conic

x(x2 ± y2 ± z2)

and Line

conic and tangent

y(xy - z2)

triangle

x(y2 ± z2)

three

x(x2 ± y2)

concurrent Lines

two lines

2

x y

one e repeated

triple Line

70

x3

Rather than attempting to enlarge still further on the various situations which can arise by choosing special values of

d, n

we shall widen our class

of examples by considering

The General Case

p a 2

One thinks of an element of Hd(n, 1)

sion



G

and inversion

G -> G are

smooth mappings.

The general linear group

Example I

GL(n)

is a Lie group.

To see this

it is convenient to identify it with the group of all non-singular real matrices; real

it is therefore an open subset of the vector space

n x n matrices, hence a smooth manifold.

matrices in

M(n)

smooth.

GL(n)

of all

Further, multiplication of

is a polynomial mapping, so certainly smooth, and its res-

triction to matrices in inversion in

M(n)

n x n

GL(n)

will likewise be smooth.

Finally, matrix

is a rational mapping (with nowhere zero denominator) so

Note incidentally that the tangent space to

GL(n)

at any point

can be identified with M(n).

Example 2

The reader is left the task of checking that the cartesian

product of two Lie groups is again a Lie group.

In particular the product

of two general linear groups will again be a Lie group.

73

By a smooth action of a Lie group action

iD

:

G x M -4

M for which

G

1)

on a smooth manifold M we mean an is smooth.

The reader will readily

check that the geometric examples given in the previous section are smooth actions of Lie groups on smooth manifolds.

We intend to restrict ourselves

entirely to the case when all the orbits are smooth submanifolds of M.

In

fact this turns out to be hardly any restriction at all, and it can be proved (though not too easily) that our geometric examples satisfy this condition.

Further information on this point can be found in Appendix B. under the natural action of

GL(n) x GL(p)

H1(n, p)

on

the partition by rank of linear maps ]Rn - ]Rp,

For instance,

the orbits provide

and we saw in Chapter II

that these are smooth manifolds.

In practice one needs to know how to compute the tangent space to an orbit at a point.

The procedure is based on the following proposition.

Let

(3.1)

smooth manifold

folds of

M.

:

G x M -> M.

M be a smooth action of a Lie group

For any point x e M the natural mapping i x : G - G.x of

I claim that

x

rank at the identity element

defined by

by y - h.y.

g - hg,

g - g.x

is a submersion.

has the same rank at every point in

x

fices to show that the rank of

G

on a

It is assumed that all the orbits are smooth submani-

the group onto the orbit given by

Step 1

G

I E G.

and let

0

at any point Let

0

h E G

G.

It suf-

coincides with its

denote the diffeomorphism of

denote the diffeomorphism of

M defined

The commuting diagram of smooth mappings on the left then gives

rise to the commuting diagram of differentials on the right.

The vertical

arrows in the diagram of differentials are linear isomorphisms, so T1i x The

74

x

have the same rank, as was claimed.

and

G.x

G

y G

Step 2

iD

X

Tx8

ThIx

y T G h

4,

G.x

In view of Step 1 it suffices to show that Ix

some point in

G,

Tx(G.x)

T18

8

8

x

T1G

y >Th.x(G.x)

is submersive at

and that follows immediately from Sard's Theorem, guaran-

teeing the existence of at least one regular value, necessarily in the image

.

of

tial TPx of the tangent space

is the image under the differen-

Tx(G.x)

Thus the required tangent space

T1G

to the group at the identity element.

And that is how one goes about computing Tx(G.x)

Let us return to our class of geometric examples namely the

Example

natural action of F

:

IRn

in practice.

GL(n) x GL(p)

on the vector space

Hd(n, p).

Let

-+ MP be an element of this vector space, with components

x1, ..., n We

each a form of degree

d

in

shall compute the tangent space at

F

to the orbit through

f1, ..., fp

n

variables

mapping of the group onto the orbit is given by

(H, K)

-+

F.

The natural

K o F o H 1,

we require the image of the differential of this mapping at the identity i.e.

pings

the vector sum of the images of the differentials at H-+ F o

11-1

and K -+ K o F.

first the mapping H -+ F o H 1. matrix inversion

H -+ H 1

I

and I,

of the map-

We consider these separately.

Take

Observe first that it is the composite of

with the mapping H -+ F o H:

since the differen-

tial of the former mapping (at any point) is invertible we see that the

75

images of the differentials at I of H - F o H 1, H -+ F o H will coincide so we need only consider the latter mapping. differential as follows.

ej

the curves

yij

close to

to

0.

ei,

F(x1,

through

M(n)

GL(n).

]Rn

-+ Mn be the linear mapping

I

..., xj

with respect to

0

all other

zero.

g's

Example 4

to the orbit through xi ax,

and

n

in

d

f

..., gp)

with just one

t

of

variables

=

E xi

p = 1.

Hd(n, 1)

will be the subspace of

the required tangent space is just the subspace of

with

1

-,c

i.,

Let

f

and

be a f

spanned by the

By Euler's Theorem xi

d

1

xi ax.

fj,

The tangent space at

x1, ..., xn.

so lies in the subspace spanned by the

,

ax.

aF

ex

and the image will be

Ja

f

xi

equal to an

gi

In fact it is simpler than that.

f.

of the curves

0

I,

It is worthwhile isolating the case when

form of degree

which is the

this is the restriction of a linear

mapping, so will coincide with its differential at

G = (g1,

t

yield a basis

which by the Chain Rule are the

Next, consider the mapping K -+ K o F:

spanned by all

with

It follows that the image of the differential at

txi, ..., xn),

+

I

of all linear mappings IRn -, ]Rn,

the derivatives at

i.e.

Consider now

0.

given by yij(t) = I + toij

will be spanned by the derivatives at

H -+ F o H

F o yij,

:

Evidently the derivatives of these curves at

tangent space to of

Ai j

denote the standard basis vectors

and the remaining basis vectors to

GL(n)

in

for the vector space

I

e1, ..., en

1 s i, j 5 n let

in IRn, and for which maps

Let

We compute the image of the

IL

.

thus

1

H (n,

1)

spanned by the

j < n.

J

Example 5

By way of explicit illustration take the case of binary cubic

forms,

d = 3,

i.e.

n = 2,

us that the tangent space at

76

p = 1. f

And take

f = x2y.

to the orbit through

f

The above tells

will be spanned by

xay .1

xax ,

x3,

x2 y,

yax , a

which are respectively the monomials

so a basis for the tangent space is given by

x2 y,

2x2y,

xy2,

2xy2, It

x3.

follows that the tangent space, hence the orbit, has dimension 3.

Example 6

For an example involving linear systems consider pencils of

binary quadratic forms, and in particular the pencil Example 3 the tangent space at

by xax , y ax , x ay , y ay

F

F = (x2, 0).

to the orbit through

basis for the tangent space is provided by

will be spanned

F

together with (x2, 0),

(0, x2 ); hence a (xy, 0),

(x2, 0),

By

(0, x2)

so

the tangent space, and hence the orbit, is of dimension 3. Let us return to the general situation of a smooth action of a Lie group

G

on a smooth manifold M, supposing that all the orbits are

smooth submanifolds of point

x E M

: G x M - M

We can define the codimension

M.

to be the codimension of the orbit

G.x

in

cod x M.

of any

As we have

seen, one of the broad problems is that of actually listing the orbits.

In

practice this tends to be a difficult, if not virtually impossible, question to answer.

Experience suggests that the difficulty of listing the orbits of

a given codimension may well increase with the value of the codimension: indeed, this is clear on an intuitive level, since the larger the codimension the more space there is in which the orbit can twist and turn, and hence the On this basis it should

larger the number of possible types of behaviour. be most feasible to list those orbits of codimension This is worth re-phrasing.

G when there exists a neighbourhood x' E U

lies in the same orbit as

tly small perturbations of

x e M

One calls

x:

U

of

x

0.

stable under the action of in

M

such that every point

intuitively, this means that sufficien-

x will not displace it from its orbit, hence the

terminology.

Note that the homogeneity property for orbits implies that if

one point x

on an orbit is stable then every point on that orbit is stable, 77

and hence that the orbit is open in stated in algebraic terms.

has codimension zero. that

x

We call

This geometric notion can be re-

M.

x e M

infinitesimally stable when it

It is now a trivial exercise in linear algebra to see

is stable if and only if it is infinitesimally stable, so that in the

present context the distinction between the two notions is a fine one.

How-

ever it is worth making, because in studying the singularities of smooth mappings one comes across analogous situations where both notions can be defined, but it is decidedly hard to show that they coincide.

Consider again the natural action on binary cubic forms.

Example 7

that there are five orbits, with representatives have respective codimensions

x3 +

Recall

0, x3, x2y, x3 + xy2;

these

so we have two stable orbits, namely

1+, 2, 1, 0

Xy 2.

Example 8

By way of contrast consider the natural action on ternary cubic

forms,

cubic curves.

i.e.

Here there are no stable orbits.

One could

prove this ad hoc by systematically computing the codimensions of the normal forms listed in §2.

There is however a more illuminating argument.

group in question is

GL(3) x GL(1).

If we forget about the second factor

here we obtain another action by the group

GL(3)

alone.

these two actions give rise to exactly the same orbits.

We claim that What one has to

show is that multiplication of a ternary cubic form by a scalar be achieved by a change of co-ordinates:

of co-ordinates Now the group

x t+ ,1/3x, GL(3)

yr x1/3y,

has dimension

any point must be of dimension

9,

5 9.

of ternary cubic form has dimension 10. dimension

78

>.

1.

The

4

0

can

and this is clear since the change

z -, A1/3z

achieves precisely that.

so the tangent space to any orbit at On the other hand the space

H3(3, 1)

It follows that any orbit has co-

Let us continue with the general situation of a smooth action t : G x M - M.

We wish to expand on the question of just how the action a point

x s M.

behaves close to

(I

The crude picture is that the orbits through points close to

x flow smoothly together past

and a very basic idea in studying this

x:

situation is to look at the way in which the orbits out a (small) cross-section of this flow.

(See Figure 1.)

Fig. 2

Consider a smooth sub-

This crude idea can be made precise as follows.

manifold

S

M,

of

with x E X.

which is transverse to the orbit

We shall certainly have

(See Figure 2.)

least possible dimension for the dimension of

S

identity, the slice

and in that case we call

is cod x,

S

M

given by m H g.m,

of

(3.1)

x

in

smooth manifold

:

G x M M,

following property:

-,

M

and let

smooth submanifolds of

x.

close to the

g E G

the basic pic-

will sweep out a small neighbourhood

S

M.

be a smooth action of a Lie group

The point

is a diffeomorphism onto

at

on a

x has a neighbourhood U with the

there is a smooth submanifold H S

G

It is assumed that all the orbits are

x e M.

identity element, and a slice H x S

with

when

i.e.

a slice at

S

the

The formal mathematical statement of this reads as follows.

M.

Let

dim(G.x):

+

will slide slightly along the orbits:

ture which we wish to convey is that U

dim S

.

will be when we have equality,

S

Under the diffeomorphism of

dim M

G.x.

x,

of

G

containing the

for which the restriction of

to

U.

79

G. x

For obvious reasons such a neighbourhood structure.

One thinks of

U

U

is said to have a product

as a small tube with the orbit

G.x

running

down the middle.

We start by isolating a lemma which will be used twice in the main

Step 1

Suppose that

body of the proof.

manifold Aa, manifold

and that of

Cc

a = b + c.

with

A with

Indeed by

bourhood of

mapping to

0

c = a - b

then

we claim that there exists a smooth sub-

x E C

which is transverse to

and for which

B

(1.2.2) there is a diffeomorphism of an open neighonto a relatively open neighbourhood of

in IRa

0

x s B:

is a smooth submanifold of a smooth

Bb

x,

0 x]Rc

and IRb x 0

corresponding to

maps to a smooth submanifold

in

A

If we put

B. C

x

A with the

of

desired properties.

As we have already pointed out the natural mapping

Step 2

(I

x : G - G.x

of the group onto the orbit is a submersion, so by (11.1.2) the inverse image Gx

x will be a smooth submanifold of

under this mapping of the point

of codimension the dimension of the orbit, with Gx

G

is a subgroup of

to

Gx

with

the orbit 80

dim G

G.x.

=

(Incidentally,

1 e Gx.

known as the isotropy subgroup at

there exists a smooth submanifold

H

of

dim Gx + dim H,

G

i.e.

with H

1

G

By Step

x.)

I

E H which is transverse

has the same dimension as

Again by Step I there exists a slice

S

at

x.

Consider

now the restriction

T

H x S - M

:

of the action

Observe that domain

and target have the same dimension, and that the differential at invertible.

It follows from the Inverse Function Theorem that if

small enough then

of x

'P

maps

H x S

(1, x)

H, S

is are

diffeomorphically onto a neighbourhood U

in M.

The reason for proving this result is to justify the intuitive idea that in

order to study the action near x the orbits out a slice

S.

it will suffice to study the way in which

From a purely practical point of view this pro-

vides a considerable simplification since the dimension of the slice well be very small indeed compared to that of the manifold

S

may

All this

M.

brings us to one of the central ideas of the subject, namely that of an "unfolding" of

roughly speaking a parametrized slice at

x.

Transversal Unfoldings

§4. Let

fold

x,

:

M.

G x M - M be a smooth action of a Lie group and let

x e M.

:

By an parameter unfolding of x we

M.

(7Rr, 0) - (M, x)

is transverse to the orbit

on a smooth mani-

As usual, it is tacitly understood that all the

orbits are smooth submanifolds of

mean a germ X

G

G.x

at

:

x

X

is said to be transverse when it

i.e.

T0X (a") + Tx(G.x)

=

TxM.

One pictures the situation something like this:

81

X .0

Now let

x have codimension

it follows immediately that

r = c

is a fairly straightforward matter.

X

the case when M foldings,

i.e.

is a transverse unfolding

we call

X

a minimal trans-

Let us restrict ourselves to

is a linear space, so it is natural to look for linear unthose of the form

X(u1, ..., u0)

where

when

r > c:

X

The construction of an explicit minimal transversal unfol-

versal unfolding.

ding

Given that

c.

b 1, ..., be

=

x + u1b1 + ... u b c

are fixed vectors in

M.

a

The condition for such an

X

to be transverse is that

IR{b1,

...,

bc} + Tx(G.x) = M

where the first term represents the subspace spanned by b1, ..., bc, the image of the differential of choose

b1, ..., b

0

X

at

0.

i.e*

Thus all we need to do is to

to be a basis for a supplement of the tangent space

Tx(G.x).

Example I

Consider the action of the product group

vector space

H3(2, 1)

GL(2) x GL(1)

of all binary cubic forms, discussed in §2.

compute transversal unfoldings for the non-zero normal forms. H3(2, 1)

has as basis the monomials

x3, x2y, xy2, y3.

on the

We shall

Recall that

We can compute the

tangent space to the orbit at each normal form using the results of §3. 82

Bases for these tangent spaces can be found in the table below.

It is a

happy accident that in each case basis vectors for the tangent space can be chosen from the list of basis monomials, and the list of remaining monomials provides a basis for a supplement. Transversal unfoldings are then provided by the formula given above.

Normal Form

Basis for tangent space

Transversal Unfolding

3.

x3 ± X Y

x3 + xy2

9x2 2y, x3'2, y3

x3

x 2y

, x2 y,

xy2

2

x y+

2 x3 + uxy + vy3

x3, x2y

x3

u3

Let us pursue this example further to see just how the transversal unfolding can yield rather explicit information concerning the action close to a point.

We shall describe the action close to

moves over the plane

IR2

so the transversal unfolding

cut all orbits through binary cubics close to values of type. 3ux 2

u, v

As the point

x3.

(u, v)

x3 + uxy2 + vy3

will

We shall determine the

x3.

for which the unfolding represents a binary cubic of a given

The associated Hessian quadratic form of the unfolding is 2 2 + 9vxy - uy

with discriminant a multiple of

is the zero quadratic form if and only if (u, v)

u = v = 0:

so the origin in the

plane is the only point which corresponds to a perfect cube.

now stay away from the origin.

=

0:

i.e.

when the discriminant

this is the equation of a cuspidal cubic curve in the plane.

(See the picture below.) More precisely:

Let us

The unfolding will represent a parabolic bi-

nary cubic when the Hessian is parabolic, v2 + 1+u3

The Hessian

27v2 + 1+u. 3

At all other points in the plane

inside the cuspidal cubic we have

v2 +

1+u3

v2 +


0

corresponding to hyperbolic binary cubics.

Thus we arrive

at the following picture, representing a cross-section of the flow of orbits close to

x3.

Some more interesting situations are provided by the transversal unfoldings of cubic curves.

Example 2 f

We shall study the transversal unfolding of the cubic curve

x(x2 + yz),

=

affine plane the conic is a parabola x = 0.)

f

and the chord is its axis

y = -x2,

A minor computation shows that the tangent space to the orbit at

is spanned by the monomials of degree so

3

in

y3, z3

with

x, y, z

f

deleted,

has codimension 2 and a transversal unfolding is provided by

F

=

x(x2 + yz) + uy3 + vz3.

As in the previous example we ask for the values of a given type.

Certainly, when u = 0, Suppose

and chord type.

u ,-E 0,

zero one finds a singular point at that the tangent lines there are Likewise when hand when

u = 0,

u X 0,

v / 0

v p( 0

v = 0:

v = 0

x = 0,

y = 0

u, v

f or which

F

has

we have the original conic

putting

(0, 0, 1),

2x ,

-r ,

8z

equal to

and further computation shows so one has a crunodal cubic.

one obtains a crunodal cubic.

On the other

are sufficiently small the curve is non-singular,

as a few lines of working will verify.

84

(In fact in the real

which is of the conic and chord type.

Thus we obtain the following picture,

representing a cross-section of the flow of orbits close to the conic and chord type.

non-singular erunodaZ

Another pleasant illustration is provided by the transversal

Example 3

f = xyz

unfolding of the cubic curve tangent space to the orbit at x, y, z

with

In this case the

is spanned by the monomials of degree 3 in

f

deleted, so

x3, y3, z3

of triangle type.

f

has codimension 3 and a transversal

unfolding is provided by

F

=

We ask first for the values of there exist

xyz + ux3 + vy3 + wz3.

u, v, w

aF OF

8z

or

=

yz + 3ux2

=

0

=

xz + 3vy2

=

0

=

xy + 3wz2

=

0.

v = 0,

or w = 0

Notice that if

u = 0,

is satisfied.

Let us see what happens when u

that if one of

x, y, z

z

is singular,

i.e.

not all zero for which

x, y, z

y/0,

F

for which

0.

then certainly this condition 0,

v

0,

w

0.

Observe

vanishes then all of them vanish, so we can suppose

But then our equations yield

85

=

(27uvw + 1)x2 y2 z2

0.

We are only considering what happens very close to

so we can suppose

x2y2z2

all / 0

=

which is impossible.

-

0

and small then

what happens in the plane

u = 0

in

(u, v, w) space.

are zero we recover the original triangle:

v, w

zero we have the conic and chord type; obtains a crunodal cubic. plane

v = 0,

are

Let us look in more detail at

is non-singular.

F

u, v, w

In other words if

u,

4 0,

to be very small, which means the expression within brackets is

v, w so

f,

Of course when both

when just one of

and when both

v, w

are

v, w

is

one

0

By symmetry one has similar situations in the

and the plane w = 0.

The nett result is the following pic-

ture of a cross-section of the flow or orbits past a cubic curve of triangle type.

crunodal conic and line

For an example involving linear systems let us look at pencils

Example 4

of binary quadratic forms. pencil (y2, 0),

F = (x2, 0).

tangent space at (0, x2)

(y2, 0):

by 86

The space of pencils is spanned by (0, xy),

(0, x2),

F

We shall study the transversal unfolding of the

(0, y2).

to the orbit through

F

(xy, 0),

In Example 6 of 3 we saw that the F

is spanned by

so a basis for a supplement is provided by thus

(x2, 0),

(x2, 0),

(0, y2),

(0, xy)

(xy, 3),

and

has codimension 3, and a transversal unfolding is given

(x2, 0) + u(0, y2) + v(0, xy) + w(y2, 0)

i.e.

(x2 + wy2,

uy2 + vxy).

The objective now is to determine the values of

u, V. w

cil has one of the seven possible types listed in 2. fied the space of binary quadratic forms with IR3 with

ax2 + 2bxy + cy2

the cone

b2 = ac,

for which this pen-

Recall that we identi-

by identifying

(a, b, c), that the singular forms then correspond to

and that the types of pencil are separated by the way in

which they lie relative to this singular cone.

The first thing is to deter-

mine when the pencil represents a line through the origin:

tion shows that the binary quadratics

x2 + wy2,

endent if and only if

i.e.

space.

v = 0,

In that case when w = 0

tangent to the cone; w < 0

u = 0,

when w > 0

a line outside the cone.

(pv)2

=

along the w-axis in

F,

a line

we have a line inside the cone, and when Now we ask what happens off the w-axis, when

i.e.

4,.(Aw + µw)

To this end we consider how the

Ax2 + pvxy + (Aw + µu)y2,

+ 4u)µ. - v2µ2

u2 + wv2

discriminant, and is a surface in IR3,

=

0

*

0.

=

This expression is itself a binary quadratic in The equation

which is singular

i.e.

4w.12

16(u2 + wv2 ).

(u, v, w)

A typical element of the pencil is

A(x2 + wy2) + p(uy2 + vxy)

when

are linearly dep-

we recover the original pencil

our pencil is a plane through the origin. plane cuts the cone.

uy2 + vxy

a simple computa-

A, u

with discriminant

represents the vanishing of this

called the Whitney umbrella.

Note

that it includes the whole w-axis, sometimes dubbed the handle of the umbrella.

On this surface, but off the handle, the equation : N

for a solution so the pencil meets the cone

represents a plane tangent to the cone. u2 + wv2 < 0

so the equation

*

*

has a unique ratio b2 = ac

in a line, so

Inside the umbrella one has

is satisfied by no real ratio

A : µ, which

means that the pencil represents a plane which does not intersect the cone.

87

plane not intersectin cone

plane tangent to cone

line outside cone line inside cone plane intersecting cone

And outside the umbrella

u2 + wv2 > 0,

two distinct real ratios

A : p

so the equation

*

is satisfied by

which means that the pencil represents a

plane intersecting the cone in a pair of lines through the origin.

Once

again then we have obtained a simple picture representing a cross-section of the flow of orbits.

Of course, as we have presented them, all these examples provide little more than amusing exercises in elementary geometry. than that;

But there is more to it

there are deep questions in singularity theory, lying beyond the

scope of this book, which one can only answer by going into the geometry of transversal unfoldings in considerable detail.

Our objective was simply to

lay bare the underlying ideas involved and to give the reader some feeling for the mechanics of the matter.

Later in this book we shall meet analogous

situations, lying just outside the present framework, where the idea of a transversal

unfolding can be used to study the possible ways in which a germ

of a smooth mapping can be deformed.

88

Versal Unfoldings

§5.

We shall conclude our discussion of finite-dimensional unfoldings by expanding somewhat on the sense in which a transversal unfolding describes an action this will provide us with a useful characterization of

close to a point;

stable points in terms of their unfoldings.

To this end we introduce a series

is a smooth action of a Lie group

G

on a smooth

of notions.

As usual,

manifold

and it is assumed that all the orbits are smooth submanifolds of

M,

We let

M.

iD

x e M.

Equivalence of Unfoldings Two r-parameter unfoldings

X1, X2

x

of

there exists an r-parameter unfolding tity element

I

I

are said to be equivalent when :

(IRr, 0)

-+

(G, 1)

of the iden-

in the group for which

X2(u)

=

I(u).X1(u).

Pictorially, the idea is that you can get from X1

to

X2

by sliding smoothly

down the orbits.

X1

e X2

0

Induced Unfoldings

X

Suppose that H

:

(IR5, 0)

folding of

.

x,

is an r-parameter unfolding of (IRr, 0)

is a germ.

said to be induced by

x,

Then Y = X0H H.

and that is an s-parameter un-

In this situation we refer to

H 89

as a change of parameter and write

Y = H X.

One pictures the situation

something like this.

X

A, HI

Y

M

Morphisms of Unfoldings

Let

X, Y

is a pair

be r, s-parameter unfoldings of with

(H, I)

with H : (]Rs, 0)

-->

I

A morphism from X

x.

Y

to

an r-parameter unfolding of the identity 1, and

(]Rr, 0) a change of parameter, for which X is equi-

valent to the induced unfolding H*Y

via

I.

When

r = s,

and

H

is inver-

tible we call the morphism an isomorphism.

Versal Unfoldings

An unfolding Y

of x is said to be versal when for any unfolding X

there exists a morphism from X

to

Y.

of

Intuitively, this means that

sufficiently large to allow all unfoldings of

x

to appear in it.

Y

x is

A versal

unfolding of minimal dimension is said to be universal.

The basic fact about unfoldings is the following result connecting the algebraic notion of transversality with the geometric notion of versality.

(5.1) M,

Let '

be a smooth action of a Lie group

all of whose orbits are smooth submanifolds of

icient condition for an unfolding X : 90

G

on a smooth manifold

M.

(]Rr, 0) -+ (M, x)

A necessary and suffto be versal is that

it should be transversal.

In which we establish necessity

Step 1

We have to show

sal.

X

of the condition.

is transversal,

TxM C T

0

that

i.e.

(1)

(]R, 0) - (M, x).

from Y

(H, I)

ver-

X (]Rr) + Tx(G.x).

Consider any 1-parameter unfolding Y : there exists a morphism

X

Suppose

to

X,

i.e.

X

As

is versal

Y(v) = I(v).X(H(v)).

Take differentials to get

T Y (IR) 0

C T0Y (IRr) + Tx(G.x).

(2)

(1) follows immediately from (2) since any tangent vector in gent vector to some curve through

Step 2

x,

We wish to show

X

versal.

is a tanY.

X

Suppose

Observe first that it will suffice

to show that some unfolding induced by X X

x

a 1-parameter unfolding

i.e.

In which we establish sufficiency of the condition.

transversal.

T M

is versal.

Let us therefore replace

by an induced unfolding J*X which is both transversal and minimal, and

continue to denote it by the same letter

linear mapping into Mr, T0X

taken by

(One can choose

X.

J

to be a

with domain of the correct dimension, whose image is

to a supplement for

Tx(G.x)

in

TxM.)

Such a germ

X

is

immersive, and the image of a sufficiently small representative will be a slice S

at

i.e.

x.

Recall now that

has a neighbourhood with a product structure,

there exists a smooth submanifold

ment

1

for which the germ

is invertible.

x.

x

'

at

( 1 ,

H x)

of

G

of the restriction of

Consider now an arbitrary unfolding Y

We define germs J, Z by J = IIH 0 T-1 0 Y,

SH, HS

through the identity ele-

are the germs at

(1, x)

Z=

of the projections of

:

1?

to H x S

(Es, 0) - (M, x)

11 S 0

H x S

-1

of

o Y where

onto

H, S

91

It is immediate that

respectively.

K

J(u).Z(u)

=

(sr, 0) -+ (7Rr, 0) by K = X 1 o Z we see that

:

Y

so

is equivalent

On the other hand if we define the change of parameter

Z.

to

Y(u)

induced from

and conclude that

X,

X

It follows that

(K, J)

is an unfolding

Z = K *X

is a morphism from Y

to

X.

is a versal unfolding.

An immediate consequence of this characterization is the following proposition justifying the use of the prefix in the term "universal".

Under the hypotheses of (5.1) any two universal unfoldings

(5.2)

x

X, X'

of

are isomorphic.

Suppose

Proof

X, X'

are r-parameter unfoldings.

is equivalent to some induced unfolding

And as

H*X.

X

As

is versal,

is universal

X'

subspace of that

TXM,

so the image of

is invertible, and hence

TH

T(X o H)

TH H

=

TX o TH

is an r-dimensional

is likewise r-dimensional: is invertible.

H*X

That means

is universal, hence is a minimal transversal unfolding by (5.1). that the image of the differential

X'

it follows

The result follows.

Finally, as promised at the beginning of this section, we shall make use of

To this end

these ideas to give a simple characterization of stable points. let us call an r-parameter unfolding

to the r-parameter constant unfolding

X

of

x

trivial when it is equivalent

(Mr, 0) -+ (M, x)

given by

u -+ x.

We then have

(5.3)

Under the hypotheses of (5.1) a necessary and sufficient condition for

an element

Necessity

92

x e M

to be stable is that every unfolding of

x

should be trivial.

Suppose x is stable, and that X is an r-parameter unfolding.

We have to show that X is equivalent to the r-parameter constant unfolding. Since

x

is stable the constant 0-parameter unfolding is transversal, hence

versal by (5.1).

Thus

X

is equivalent to an r-parameter unfolding induced

from it, which is necessarily the constant r-parameter unfolding.

Sufficiency

Conversely, suppose that every unfolding

valent to a constant unfolding.

In particular, take

unfolding.

Certainly then some representative of

orbit

so

G.x,

dimension

0,

X

X

X

X

of

x

is equi-

to be a transversal

is a mapping into the

can only be transversal to the orbit when

i.e.

x

G.x

has co-

is stable.

F7

The reader should be warned that the theory of this section is not particularly useful.

Its virtue lies rather in the fact that it is an easily

understood model for the ideas to be used in Chapter V.

93

IV Singular points of smooth functions

Some Basic Geometric Ideas

We come now to the meat of this book, the study of singular points of smooth mappings.

In accordance with our philosophy of treating the simplest situa-

tions first we shall restrict this chapter to the case of singular points of smooth functions

f : N -IR with N a smooth manifold.

And we shall sim-

plify life even further by studying germs of such functions under a rather finer notion of equivalence than that introduced in Chapter I:

at first sight

this may seem to be a complication rather than a simplification, but in fact it will enable us to finesse various algebraic difficulties. germs of functions

exists an invertible germ mutes.

right-equivalent

f1, f2 g

We call two

(or .R-equivalent)

when there

for which the following diagram of germs com-

Our broad intention

can now be formulated rather more accurately by saying that we wish to classify germs of functions under the relation of $-equivalence.

As such our pro-

gramme is far too ambitious, but we shall see that we can gain some distance by using a little common sense.

To avoid unnecessary symbolism we shall,

throughout this chapter, adhere to the convention that equivalence of two germs of functions is to mean 9Q-equivalence=

94

since we shall have no occasion to

refer to the more general notion there should be no confusion on this point.

And it will be useful to write

to mean that two germs

f2

f1

f1, f2

of

functions are equivalent.

The starting point is to try and set up our problem in such a way that it looks more like a problem which we know how to handle: we can proceed by analogy.

the hope then is that

That is the object of the present section.

setting up the basic geometric ideas we intend to be deliberately vague;

In the

reason for this is that any systematic theory covering the situation we wish All

to study would lie far and away above the intended level of this book.

we shall do is to argue heuristically to arrive at precise interpretations of vague geometric ideas:

we shall never use such arguments to prove propositions.

Our basic objects then are germs at

x we see that we can suppose

jects we wish to study is the set Of course

Sn

:

(N, x)

N = ]Rn,

En

(]R, y).

and x = 0.

of all germs

In fact

f

Taking a chart a Thus the set of ob-

(IRn, 0) - (]R, y).

:

&n has more algebraic structure than

the operation of multiplication on ]R

plication on

->

is a real vector space under the natural operations of addition

and scalar multiplication. that;

g

under which

8n,

Sn

induces an operation of multi-

becomes a real algebra.

More information

on the subject of real algebras can be found in Appendix C of this book.

Next, let Rn

denote the set of invertible germs

observe that Rn acts on

n

g :

(]Rn, 0) + (]Rn, 0), and

Now Rn

is a group under the operation of composition.

by composition,

(g, f) H f o g 1.

i.e.

we have an action R

Further, two germs in 9 n

n

x 9 n

-,

n

given by

will be equivalent if and

only if they lie in the same orbit under this action.

This observation, how-

ever trivial, is crucial in that it sets the scene in which we are to work.

However, pursuing the analogy with Chapter III is not quite that simple. What one would like to do is to introduce the "codimension" of a germ, and start the problem off by listing the orbits of fairly low codimension.

The 95

stumbling block here is that we have as yet only a group acting on a set,

whereas we would like to have a Lie group acting smoothly on a smooth manifold. It is an unfortunate fact that

and R n

& n

are not smooth manifolds in the That need not be, and

sense in which that word has been used in this book.

What we shall do is just to pretend

will not be, an insuperable difficulty.

that the action of Rn

on

6n

is a smooth action of a Lie group on a smooth

manifold, and proceed by analogy. This goes as follows.

the orbit through of

En,

f.

f

Choose a germ

in

We shall pretend that R.f

and look for a vector subspace of

gn

and write just R.f for

9n,

is a "smooth submanifold"

which has some reasonable claim

to the title of the "tangent space" to the orbit R.f

To do this we

f.

The natural mapping of the group onto

just mimic the theory of Chapter III.

the orbit is the mapping Rn - &n

at

g N f o g 1,

given by

and one would expect

the required "tangent space" to be the image of the "differential" at the identity 1 of this mapping.

g' 4 g'- g

A preliminary simplification is to observe that

f o g 1 is the composite of g -+ g 1

1

and

g F+ f o g:

one expects

to be a "diffeomorphism", so the required image should coincide with

that of the "differential" at the identity of

g - f o g.

Of course, one

would expect this "differential" to be a linear mapping between vector spaces, and the first thing to get straight is just what these vector spaces should be.

The domain should be the "tangent space" at the identity to the group Rn.

A

Observe that

Rn

little common sense will soon produce a candidate for this.

is a subset of the vector space &n

n

of all germs

moreover, we should really think of Rn

g :(IRn,o)

as an "open" subset of

(IRn, 0):

n

n

because

if we disturb an invertible germ very slightly we still expect to get an invertible germ.

to Rn,

On this basis we would certainly expect the "tangent space"

at any point, to be just &

n.

n 96

The target of the "differential"

should be the "tangent space" at we would expect this to be just linear mapping

do n - 8n.

for the "differential".

the components of

g

=

I + tg with

"differential" to

g

t

0

t

at

1:

The result of applying the sought for

0.

Just doing this naively

f(x1 + tg1, ..., xn + tgn). g1 ax

of the curve

0

the "differential" with res-

i.e.

t --+ f(y(t))

by the Chain Rule one gets the answer

at

given by

.Rn

should then be the "differential" at

pect to

n

Now we can realize this

g1, ..., gn.

of the curve through 1 in

close to

in n given by

R

an arbitrary "tangent vector" to

will be written

f

of

Thus our "differential" should be a

We start with an arbitrary germ

i.e.

through

0

is a vector space

The problem now is to find a sensible formula

germ as the "differential" at y(t)

9n

The line of attack to take here is suggested by

(ILn, 0) _ (IR', 0)

:

to n, and since

9n.

Example 3 in §3 of Chapter III. g

x

To sum up:

gn 8x .

+

we

n

1

80 n

expect the required "differential" to be the linear map

->

n

given

by the formula

(911 ..., gn) H

'f

'fn

+ ... + gn ax

91 ax

1

The required "tangent space" should be the image of this linear mapping, and The germs

this is neatly described in algebraic terms as follows. g1, ..., gn

all have zero target, so belong to the ideal

en

comprising all germs with zero target.

in

En

Let us denote by

generated by the partial derivatives

ax 1

the Jacobian ideal of

f.

the product ideal .An.Jf.

in the algebra

, , nn

The image of the linear mapping

Jf

,

*

the ideal

and call it above is just

In this way we come to the conclusion that a good

candidate for the "tangent space" to the orbit

f

However, that is not quite the end of the story.

is

nJf. We have described our

model in the language of germs, but could just as well have worked with

97

functions defined on a neighbourhood

U

germ case as the limiting case when U case we should replace

functions defined on

0

in IRn,

and thought of the In that

is infinitesimally small.

by the vector space

dn U.

of

G"O(U,

of all smooth

IR)

Rn by the group

And we should replace the group

of all diffeomorphisms U - U under the operation of composition.

Diff (U)

We would then have an action of Diff (U) on e (U, formula as before.

IR)

given by the same

And the same heuristic reasoning as before would bring

us to the conclusion that the required"tangent space" should be thought of as the image of a linear mapping given by the formula crucial difference.

In the group

origin, whilst in the group

Diff (U)

property, no matter how small model is oversimplified. gn

,Rn

U

*.

Note however one

our germs are forced to preserve the our diffeomorphisms need not have this All this indicates is that our

is.

We shall correct matters by allowing our germs

to have an arbitrary target, so to lie just in

.0

n

rather than

The only difference this will make to the final answer is

its ideal

that the image of

*

will be simply the Jacobian ideal

Jf.

On the basis of this heuristic reasoning we introduce the following formal Let

definition.

Jacobian ideal

Jf.

f e gn.

We define the tangent space to f to be the

And we define the codimension

codimension of the tangent space

cod f If

cod f

is finite we say that

compute the codimension.

=

f

8n,

of

to be the

i.e.

dim &n/Jf. is of finite codimension;

f

otherwise,

f

is of finite codimension, and if it is how to

There are some subtleties here which are worthy of

explanation, and require a detailed discussion of the algebra

98

f

In practice we shall need to know how to decide

is of infinite codimension. whether or not a given germ

in

Jf

cod f

&n:

this then

is the object of the next section.

The Algebra

§2.

gn

We have already observed that n is indeed a real algebra, in the technical sense of the word.

(See Appendix C.)

In this section we shall concern our-

selves with purely algebraic matters so as to provide a convenient reference for succeeding sections.

(2.1)

Observe first of all the following elementary fact.

A necessary and sufficient condition for a germ

vertible (as an element of the ring

Suppose

Necessity fg = 1:

then

inverse for

is that

f(O)

0.

is invertible, so we can find a germ

f

f(O)g(O)

=

1,

so

f(O)

f E n satisfies

Suppose

Sufficiency

n)

f e n to be in-

g e n with

0.

f(O) ,

0:

then

g = 1/f

is an

f.

of all germs in &n with target

This tells us that the ideal

has

0,

a rather special algebraic property.

(2.2)

Proof f E I

,/n is the unique maximal ideal in n Suppose

with

is an ideal with .,en C I C n,

I C g

f(O) t 0.

n f

follows that n is maximal.

so we can find an

is invertible by (2.1) and hence

It

I = 9n.

And the same argument establishes uniqueness.

lies in the fact that it

To some extent the importance of the ideal

n

allows a convenient algebraic description of certain ideals in gn which frequently come into consideration.

We have in mind the idea

2k+1

of all

99

f E dn

5 k

whose k-jet is zero (i.e.

vanish at

little fact

0).

-

all partial derivatives of

f

of order

As a preliminary we shall establish one extremely useful

sometimes called the Hadamard Lemma

-

which we shall also

have occasion to use in later chapters.

(2.3)

Let

U

be a convex neighbourhood of

smooth function defined on smooth functions

U x

f where

x1, ..., xn

which vanishes on

IRq

on U x IRq

f1, ..., fn

in IRn,

0

0

and let x IRq:

f

be a

there exist

with

xIf 1 + ... + xnfn

=

are the standard co-ordinate functions on IRn.

Proof f(x1, ..., xn, Y1, ..., Yq)

=

7t

J

0

(tx1, ..., txn) Y1, ..., yq)dt

xi axi (tx1' ...,

J0

xifi(x1,

txn, Y1, ..., Yq)dt

..., xn, Y1, ..., Yq)

i=1

if we take

fi(x1, ..., xn, Y1, ..., Y ) q

Now we can characterise the ideal

(2.4-)

3k =

J

0

ax

(tx1, ..., txn, Y1, ..., y )dt

i

q

7k+1'

(fin, and is generated by (the germs at

in x1, ..., xn of degree 5 k.

100

=

0

of) the monomials

First, we establish 2k

Proof

It is clear that ./l n C 2k, follow by induction from

that

f

that they lie in cular, that

21

thus

2k-1 =

It follows that

2k =.,Wn

mials of degree

k

is trivial.

k = I

3k C

n

,

which will

This we prove using the Hadamard

Indeed if

f E 2k

then

f(0) = 0,

so

and it is clear from the construction of the

f E J/ k-1.

fi

This argument shows, in parti-

is generated by (the germs at

.41n

in

so it suffices to show

q = 0.

x1 f1 + .., xnfn,

=

The case

.

n2k-1.

2k

Lemma, in the special case

n

=

is generated by (the germs at

x1, ..., xn.

of)

0 0

of)

the mono-

x1, ..., xn.

In particular (2.4-) shows that the ideals

This is worth remarking on because the ring every ideal is finitely-generated.

2k &n

are all finitely-generated. is not Noetherian,

not

i.e.

(We shall give an explicit example of

such an ideal shortly.)

At this juncture it is worth saying something about the exact connexion Of course a Taylor series is just a

between germs and their Taylor series. formal power series in several variables.

Given a germ

f

in n we shall

write its Taylor series as

f(0) + 11

so

f

ates

x1, ..., xn.

of algebras.

(2.5)

Ea

(0)xi + 2!

sJ

(O)xixj + ...

the algebra of formal real power series in

is in Vin,

given by f

a-LL xi

f,

n

indetermin-

In this way we obtain a natural mapping n -a

an

and one can check easily ennugh that it is a homomorphism

What is by no means so clear is the Bnrel Lemma.

The algebra homomorphism

-*

n

given by

f H f

is surjec-

tive.

101

The proof of the Borel Lemma is a slightly involved piece of analysis which

we have isolated in Appendix D of this book so as not to interrupt the flow of Notice that the kernel of the homomorphism

relevant ideas,

00n ilk eOo =k=1 n n

precisely

by (2.4.).

9n/-/4n

n in 9n

an is

It follows that =

'gn

We can get a more finite version of this, as follows.

phism n - -n

Cn ->

Under the epimor-

the maximal ideal n in 9n maps to the maximal ideal

comprising all formal power series with zero constant term, and

hence any power .Aln maps to the power J

en/fin

The nett result is that

J.

an/ n

The particular virtue of this relation is that it tells us that the quotient

Indeed .,lfn is generated by the in-

space on the left is finite-dimensional. determinates

x1, ..., xn

Ak

so .,lln is generated by the monomials of degree

k

in x1, ..., xn and the quotient space on the right can be identified with the real vector space of polynomials of degree

< k

in

x1, ..., xn,

which

is certainly of finite dimension.

In order to make further progress we shall require a rather pretty result from algebra, called the Nakayama Lemma.

(2.6)

Let

be a commutative ring with an identity element 1, and let

&

,,4rbe an ideal in

any

x E ,,'

9 with the property that

Let

b1, ..., bt

102

is invertible in

Further, let M be an C-module, and let

with A finitely generated.

Proof

I + x

a1, ..., at in

B.

If A C B + .A'.A

be generators for A.

and elements

AiJ

in

& for

be d-submodules

A. B

then A C B.

By hypothesis we can find

4 for which for

I

< i < t

we

can write

ai

Introduce a a

=

(a1,

t x t

with

& by A =

matrix over the ring

..., at),

M x ... x M,

bi + Ai1.a1 + ... + Ait.at

=

b t

=

(b1,

Then

I

is the identity

to show that

=

b

matrix over the ring

t x t

and take

can be re-written as

*

(I - A)a

where

:Lj

to be elements of the 8-module

..., bt)

factors.

( A.

It will suffice

d.

I - A is an invertible matrix, since then we can solve this

system of linear equations for the

which will show that

a1, ..., at

in terms of

a1, ..., at

lie in

B,

b1, ..., bt,

as was required.

To this end

recall from linear algebra that a square matrix (over a commutative ring with an identity) is invertible if and only if its determinant is an invertible in So it suffices to show that

the ring.

det(I - A)

is an invertible in d.

Observe that

det(I - A)

say, with

A e

=

1 _ (sum of products'

l of elements in_4tl

And by hypothesis

In practice the rings

8

I - A

is an invertible in

we have in mind are

dn,

n

9.

and the ideals

are n, In both of which satisfy the initial hypothesis of (2.6).

The Nakayama Lemma allows a very simple proof that n is not

Example 1

a Noetherian ring.

We take

00 A

= -,/,/ co

=

n .'n ,

B

M

to be the Sri module n And we take

the trivial ideal.

Clearly A C B + ,,/f .A.

If

n k=1 n were Noetherian then A would be finitely generated, and the Nakayama

Lemma would tell us that A C :B,

i.e.

that

A

is trivial.

However this

103

is false since there are standard examples in calculus of non-zero germs with

It follows that En

zero Taylor series.

Example 2 B

=

<x2

In + y3,

consider the ideals

g2

Clearly

y2 + x3>.

so need to show is the case.

M =

cannot be Noetherian.

B C A.

A

It is not immediately obvious to the eye that this

A C B.

Here the Nakayama Lemma provides an easy answer, because it

is generated by x3,

x2y,

xyy2,

I C M

We say that

M.

tient space

M/I,

i.e.

cod I

M/I

is finite-dimensional,

of

in

I

M

i.e.

the sub-

And in that case we

M.

to be the dimension of the quoThe point of the

the dimension of any supplement.

next proposition is that it gives us a useful algebraic criterion for be of finite codimension in

(2.7)

Let

g-submodule.

Sufficiency

I

to

M.

M be an &-module with a finite basis, and let A necessary and sufficient condition for

codimension in M

Let M

has finite codimension

I

admits a finite-dimensional supplement in

define the codimension

,,an.

,

be an &-submodule, so in particular a vector

subspace of the real vector space

in M when the quotient space

.4f2A

d will denote

In the following

dn, n and .,fl the corresponding maximal ideal A

be an 9-module, and let

I

And this is clear as

y3.

Now we can return to the main theme. either of

A = B,

We claim that indeed

tells us that it suffices to show A C B + -4f,2 A.

space

<x2, y2> and

=

is that there exists an integer k 3

Suppose there exi sts an integer k 3

1

I I

I C M

be an

to be of finite

with .Alk.M C I.

for which .,ffk.M C I.

Certainly then

dim M/I Now I claim that 101+

M/ lfkM

4

dim M/.,IIkM.

is finite-dimensional.

As

M

has a finite basis

we can suppose

M = d

s

for some integer

k

isomorphic with the product

s

k

x ... X &/

ds/// k.8s

Now

1.

>.

is naturally

with s factors, and since

(as we have already seen) each factor in this product is finite-dimensional the product is as well.

codimension in

Necessity

It follows that

dim M/I < co,

so

I

has finite

M.

Suppose

is of finite codimension in

I

Consider the

M.

descending sequence of d -submodules

I +,(l0. M D I

,,C11

.M2

... 2 I

Clearly, each strict inclusion in this sequence makes a contribution the codimension of

1

to

1

Since the codimension is finite the inclusions, from

I.

some point onwards, must all be equalities.

integer k >

>

In particular there exists an

for which

I

M=

I +,4f

k+1

which implies that .//k.M C I +

Since .,llk.M

it follows from the Nakayama Lemma that

tion ,llkM C I

ltk.M

C I,

is finitely-generated

as required.

As is clear from the final lines of the proof, the

Note one small point.

hypothesis .4 k.M C I

M

is equivalent to the apparently more complicated rela-

We chose the former relation for the statement

+.,llk+1. M.

of (2.7) because of its simplicity, but in practice it tends to be easier to verify the latter relation.

A further point is that one can extract a bit

more information from the proof of (2.7) than we have stated.

Keeping to the

same notation define codkI

=

dim

1+4-I + !k+1M

M

105

for k = 0, 1, 2, ...,

The quotient space which appears here is necessar-

.

+,Elk+1.M

ily of finite dimension, and its dimension is the codimension of

I

in I Under the hypotheses of (2.7) a necessary and sufficient condition

(2.8)

for

to be of finite codimension in

I C M

of the

codk I

is that all but finitely many

and in that case

vanish:

cod I

Suppose

Necessity

M

I

cod0 I + cod. I + ...

=

.

has finite codimension in M

so by (2.7) there

That implies codj I = 0 for

exists an integer k 3 I with .,Elk M C I.

all j3k. .,llk.M I

C

I+

codk I = 0

If

Sufficiency

.,Elk+1.M

then

I +.A M

=

+.J1k+1.M

I

so

which implies ,O/k M C I by the Nakayama Lemma,

is of finite codimension in

i.e.

M.

The second statement in (2.8) follows immediately from the proof of (2.7).

Next, let us expand a little on the kind of situation we have in mind. Write

n,p

with y E

for the real vector space of all germs Such a germ has components

]RP.

f

:

f1, ..., f

(]Rn, 0) - (lip, y)

relative to the

p

standard co-ordinates in ]Rn, think of

=

&n

each of which lies in

IRP

gn x ... X &n

(with

p

dn.

Thus we can

factors), and see that it is a

sP

M

natural example of an &n module shall be thinking of in the case

p = I

I

with a finite basis.

as the "tangent space" to a germ

we shall be thinking of

associated to a function-germ

f.

I = Jf,

In practice we f:

for instance

the Jacobian ideal

There is an important point to note here

in connexion with the practicalities of computing the dimension of

106

9n.,p/I.

Let us write

gn

-'

&n

n,p

-+

=

an.,p

n

x ... X 9n

(with

factors).

p

of the Borel Lemma gives rise to a surjective linear mapping

mapping

np

We claim the following.

I, say.

to

I

A necessary and sufficient condition for

(2.9)

The epimorphism

dimension in

is that

dn,p

I

to be of finite co-

should be of finite codimension in

I

9^

and in that case the two dimensions coincide.

Consider the composite of the surjective linear mappings

Proof -a

the second being the natural projection onto the

n,,p /f,

in,p

9n,p

I +.Sn,p so that by element-

By previous work the kernel is

quotient.

ary linear algebra

I +,Ilf°n.e n,p n Necessity

I +A

If

I

3

has finite codimension in

and hence

I

n,p

has finite codimension in

then so too does nip

by the displayed

isomorphism.

Sufficiency

If

I

finite codimension in

has finite codimension in n,p then by the displayed isomorphism.

J/n+1.&n,p.

has

n

It follows from

8n,P,

(2.7) that there exists an integer

I+

I +,.&n.,p

k >.

I

k with .,in.dn,P C I +

.dn. p

The Nakayama Lemma then tells us that we have

C

Si i,

and (2.7) that I has finite codimension in 9n,P Finally, when Ar

k.

n 9

n,p C - I

I

has finite codimension in

tells us that -4f"On'&n,p C -I

so

&n,p

the relation

I +.,e'O° . g n n,p

=

I,

and

the displayed isomorphism reads

an,P/I

,P/I

107

yielding the final statement of (2.9)

The practical consequence of (2.9) is that when computing the codimension of

in

I

we can, and shall, replace all our germs by their Taylor

np

series and handle them as formal power series, which are much easier to work Bear this point in mind when you come to do such computa-

with in practice.

Before resuming our session of algebra we shall digress to look at

tions.

the sheer practicalities of just how one computes the codimension of a germ

in 9n.

Of course the first thing one wants to know is whether

codimension:

f

has finite

f

the next proposition at least gives one a very simple way of

recognizing a germ of infinite codimension.

Suppose the germ

(2.10)

Then

has finite positive codimension.

is an isolated singular point of any representative of

the origin in ]Rn i.e.

in 9n

f

f,

there exists a neighbourhood of the origin in which the origin is the

only singular point of the representative.

Proof

Observe first that

(any representative of)

0 e ]Rn

Indeed if some

f.

be an invertible element of Since

zero codimension.

some integer xl, ..., xn

k > 1

must indeed be a singular point of

9n,

f

of (0)

2xi

by (2.1) so

(

0

and

Jf = 9n

has finite codimension we have

by (2.6).

then f

would

would have k

.,ll

of axi

C Jf

for

In particular, this means that the monomials

can be written as linear combinations of

8x

,

..., 2x

with

1

coefficients in

9n :

must vanish, so

x1,

in question must be

Example 3

at a singular point of ...,

all these partial derivatives

xn must vanish as well,

i.e.

the singular point

0.

Consider the germ

partial derivatives are 108

f

f(x, y, z)

=

y2 - z2x2 + x3.

Here the

of ax

=

-2 z 2x + 3x2

of .

ay

2y

=

of :

=

az

which vanish simultaneously precisely on the z-axis. is not an isolated singular point of of infinite codimension.

The set

-2zx2

Thus the origin in IR3

and (2.10) tells us that

f,

f = 0

is sketched below.

f

must be

Notice that

the z-axis is precisely the line of "double points" where the surface intersects itself.

Thus (2.10) provides us with a simple method of showing that a germ is of infinite codimension.

The strict converse of (2.10) is false, but there

seems no harm in mentioning that in the complex case the converse does hold,

though the proof requires considerable mathematical machinery. ple is provided by the germ

f(x, y)

=

(x2 + y2)2:

A good exam-

the reader will readily

verify that as a .real germ the origin is an isolated singular point, but that

as a complex germ it is not, since any point on the lines x - iy

=

0

is singular:

x + iy

=

0,

thus the germ is of infinite codimension.

Using

these remarks the reader should find it a relatively straightforward matter to decide whether a given germ has finite codimension. For the actual computation of the codimension we proceed as follows. Suppose that using (2.8).

f

The idea is to compute

is of finite codimension.

Let us abbreviate

codkJf

to

codkf:

cod f

we know that

1o9

cod f

=

so we must successively compute

cod0 f + cod1 f + ...

cod0 f,

cod1 f,

until we reach a zero

...

The practicalities

answer, and then add up the list of integers so obtained. of the matter are worth expanding upon.

a basis for a supplement of Jf

+_'#/k+1

in Jf

clearly, this can be

extracted from a list of monomials of degree

x1, ..., xn which do not lie in

if +

k = 1, 2, 3, ...;

and for each

k

k in

x1, ..., xn,

variables

The first step in this computa1.

cod0 f

n

In practice then one one has to decide, for

whether it lies in the ideal

A labour saving observation here is that if this condition holds

for some monomial i.e.

k+1.

in the

dim an/ n =

each monomial of degree k+1

Jf +J1

k

=

tion is always trivial, as takes successively

codk f we have to find

To compute

m

then it automatically holds for all its descendants,

m by multiplying it be some

the monomials which can be obtained from

other monomial.

The reader is warned that the decision as to whether a given

monomial lies in the relevant ideal may well involve quite a bit of work. A case which frequently appears in practice is in two variables

Here, the monomials

n = 2.

x, y can be conveniently displayed as the following array

y

x

x

2

y

xy

2

2

x3

x y

xY2

y3

One works from the top of the array downwards.

A simple way of recording

the results of the computations is as follows.

Suppose we have reached the

kth row of the array,

i.e.

the monomials of degree

monomials which appear in the relevant ideal

k. k+1:

Jf +.'ff

Underline those and then underline

all its descendants in the array, to avoid unnecessary work.

maining monomials select a supplement for Jf +,,ffk+1 in Jf 110

From the re-

+.,Cfk:

the

number of monomials selected is the integer

The computation finishes

codk f.

whenever one reaches a row of monomials which can all be underlined. dimension is the sum of these assuming that

cod0 = 1,

Here

Jf

<xs>.

=

of degree k in x, does.

xs

because we have to add

with

an integer . 1.

s

it is clear that the sole monomial

does not lie in the ideal Jf

namely xk,

It follows immediately that

Example 5

xs+1

=

f(x)

0 , k , s - 1

For

-

is a singular germ.

f

Consider the germ

Example 1+

plus one

codk f,

Here

Y

should have finite codimension.

f

x2

=

(See Figure 1 below.)

x, y.

x2, xy, y2

Of

2

Of

xy = -

4y3,

8Y

x

f ex

x3y, x2y2, xy3

,

,

but

in

4

Y

Jf +,J'4,

1

=

x, y lies in

does not;

y2 -

namely

x3, xy,

cod f

T+y8fay

_$1x 8xof

.

and hence that

5. l4

in the third

xy2

Of the monomials

Jf +.

The codimension is the total

We have, incidentally, shown that

C Jf,

x2, xy

Thus all the monomials of degree 4 lie

and the computation is finished.

=

as

y need be considered, and it lies in

number of monomials which have not be underlined (namely one, so

Jf +

thus we underline

in the fourth row and so on.

And of the monomials of degree 4 only +,115 as

Neither of

need be considered, and it does not lie in

of degree 3 only y3

if

= 2xy,

We write down the array of

the first two certainly lie in

and all their descendants in the array row, and

2x

vanish simultaneously (even in the complex case) only at the

of = llx2 + 4y3

Jf +

but

f(x, y) = x2y + y ,

which will appear in §4 as the so-called "parabolic umbilic".

monomials in

+.,/1k+1.

cod f = s.

We shall compute the codimension of the germ

origin, so

The co-

X. y, y2, y3) plus ..114

C Jf +

5

by the Nakayama Lemma.

III

2

y2

X 2

x y

x3 x3Y

X

x

x5

y3

xy2

Z

3

x2Y2 x2y3

x3y2

Y

Figure 1 y

x x

2

xY

x3

2

y3

,q,2

x 2y 2 2

x4

x5

y

xXy

y

x Y

x3Y

x3y2

x2Y3

'cam,

Y5

_

Figure 2

Example 6 indeed

The germ

cod f = 7.

we omit it.

f(x, y)

=

x3 + xy3

is of finite codimension,

The computation is very similar to that in Example 4 so

The array of monomials is illustrated in Figure 2.

Before going any further we should stop to verify that the codimension of a germ possesses the most basic property one would expect of it, namely that it is invariant under the relation of equivalence.

To this end we introduce

a new idea, which will be exploited later in this book. f

(7Rn, 0) - (Pp, 0)

A - A o f.

One checks easily that

to be induced by 112

we obtain a mapping

f.

f

f

: 9p -

Given a germ n

by the formula

is an algebra homomorphism, said

It is worth remarking that at the time of writing it

seems to be an open problem whether every algebra homomorphism 9p the form

f* for some germ

f

- (IRR, 0).

(IRn, 0)

:

-*

En has

To the reader we leave

the task of checking that the operation of taking induced homomorphisms has the following functorial properties,

(I)

(f o g)*

=

The germ at

(II)

the identity map &n

->

g* o f*. 0

of the identity map IRn - in induces

8n , and is indeed the only such germ.

(III) It follows from (I) and (II) that if then

is invertible and

h*

(2.11)

f

Let

(h* )-1

to be an iso-

Necessity of the condition is provided by III above.

For suffici-

morphism is that

ency suppose self.

onto it-

is an isomorphism, so maps the maximal ideal

f*

Indeed

f*

should be invertible.

f

will map the standard generators

f*

to the components

n.

A necessary and sufficient Sn - gn

condition for the induced algebra homomorphism

Proof

is invertible

(h-')*.

=

(Mn, 0) -, (IFtn, 0) be a germ.

:

h

f1, ..., fn

of

f,

for

x1, ..., xn

/n

which must likewise be generators for

Consequently we can find elements

A.

in

&n

for which

1

s i 5 n

we have

xi

Ai1f 1 + ... + llinfn.

If we differentiate both sides of this relation with respect to evaluate at

0,

6 i.j

and then

we obtain

sib where

xi,

of 1

A i1(o) axe (o) +

denotes the Kronecker delta.

of

... + xin(0) a- (o)

j

Thus

113

af. In

=

l

A( A(axl (0)) J

In

where the

denotes the identity n x n

Aij.

matrix, and A denotes the matrix of

It follows that the Jacobian matrix is invertible, and hence

(from the Inverse Function Theorem) that

is invertible.

f

Now we are in a position to establish the invariance of the codimension, in the following precise sense.

(2.12)

If the germs

Proof

Since

h

:

(]Rn, 0)

isomorphism

->

h*

f, g

f, g

cod f = cod g.

are equivalent then

in &n

are equivalent there exists an invertible germ for which

(]Rn, 0)

: 9n - 9n

of algebras.

will imply that the quotient spaces which the result follows.

g = f o h.

By (2.11)

We claim that

&n/Jf,

9n/Jg

h

induces an

h*(Jf) = Jg:

that

are isomorphic, from

To establish the claim note that by the Chain

Rule

n

D-

aah. (a j oh/ax

i

j=1

so

J9 C h*(Jf).

% h laxjl

The same reasoning shows

yielding the reverse inclusion

x.

j=1

h*(Jf)

Jf C

(h-1)#Jg

=

(h*)-1Jg

C Jg.

There is one final matter which we should clear up before passing to new things, namely the connexion between the codimensions of the ideals

Jf,,A'Jf.

First of all, (2.13)

The ideal .,AAJf

finite codimension.

114

has finite codimension if and only if

Jf

has

_1%'(f

4'Jf C Jf,

Proof

so certainly if

has finite codimension.

has finite codimension then

Conversely, suppose

again by (2.7), we see that

Let

(2.14)

f e n,

]RI

ax

subspace of

cod.!Jf =

then we have

n + cod Jf.

is the vector sum of the vector subspaces

if

+

u1 ax

+

...

observation follows on writing a.(x)

=

un

with the

ax

with at least one

ei(0)

for which

0,

1, ..., en

g

of i axi

=

as the components of a vector field

in in with

g(0)

can be "straightened",

on a neighbourhood of

f

we can find a diffeomorphism

i.e.

dn,

Think of

Recall now from (I.4.3) that the flow of

0.

in

To prove this it

0.

is convenient to confuse germs with their representatives.

en

and the

ui a &n;

(ui(x) - ui(O)) + ui(O).

Next we claim that it is not possible to find germs

g1,

Jf,J'Jf has

ax j , where the curly brackets indicate the vector n spanned by the first partial derivatives. Indeed an element

has the form

Jf

C .A'Jf so,

,

1

of

//k+1

and suppose that one (hence both) of

Observe first that and

But then

.AJf has finite codimension.

finite positive codimension:

Proof

has finite codimension, so

Jf

,,/,/k C Jf for some integer k > I by (2.7).

Jf

h

e

0

near 0

of a neighbour-

ah.

hood of

0

in ]Rn

with

for which

h(O) = 0

h1, ..., hn

are the components of

germ in

equivalent to

9n

Let us agree to write

f,

b = h(a).

h.

ei o h

=

axl ,

Observe now that

aa--

g = f o h

is a

so likewise of finite positive codimension.

Then -- (a)

ei(b) ax (b)

_

i

1

so

where

is identically zero on a neighbourhood of

0

in e, i.e.

=

g

0,

is

1

independent of the variable

x1:

by (2.10),the origin

point for

g,

so that every point near

point for

g,

and

g

0

on the

has infinite codimension.

0

is a singular

x1 - axis is a singular

This contradiction estab-

lishes our claim. 115

Taking

Now we can easily finish the proof. numbers we see that

ax

,

e1, , en

to be real

are linearly independent over the reals,

..., ax

n

1

and hence that the vector subspace

]R{ax

, , ax n

1

has dimension

n.

Further, it follows immediately from the above that the vector sum of and

]R{ax

is direct,

, , axn}

1

tl Jf

that the intersection is the trivial

i.e.

The result now follows.

subspace.

Determinacy of Germs

§3.

Having completed our session of algebra we return to the main theme, namely the classification of germs in

gn

of low codimension under R-equivalence.

The underlying philosophy is to reduce the problem to a finite-dimensional one.

And the idea to do this is an follows.

Take a germ

f e dn:

f

has

a Taylor series

f(0) +

x

1

i

1!

of (0) + axi

n

x.x. i

2!

2

a f (0) + ... axiaxj

i,j=1

i=1

whose initial segments (the so-called Taylor polynomials of better approximations (in some sense) to f

f.

f)

provide even

The hope is that a nice enough

will be equivalent to one of its Taylor polynomials, and then the problem

of classification reduces to one in a finite-dimensional vector space of polynomials.

We introduce therefore the following definition, integer.

A germ

f e Sn

k >.

0

be an

is said to be k-determined when any germ

with the same k-jet satisfies

116

Let

f - g:

g e En

in other words a knowledge of all

partial derivatives of order valence.

s k

completely determines the germ, up to equi-

Note that the property of k-determinacy will be invariant under

equivalence.

Example I

Any non-singular germ

(I.1.3) we know that

f

f s do

is 1-determined.

is equivalent to the germ

And the same applies to any germ

(x1,

Indeed by

..., xn)

having the same 1-jet as

g s 9n

x1.

-*

Hence

f.

the result.

The basic result of this section is the following sufficient condition for k-determinacy.

(3.1)

Let f s &n be such that .!/ k C

Proof

Let

Now

f, g

g s &n

have the same k-jet as

are points in the real vector space

straight line.

for t s IR.

f : then f is k-determined. f.

We have to prove

9n,

f - g.

so can be joined by a

More formally, we define F(x, t)

f1 = g.

_14//J

Thus

=

ft(x)

=

(1 - t)f(x) + tg(x)

(ft) is a 1-parameter family of germs with f0 = f,

We intend to show that any two germs in this family are equivalent,

which will prove the result. fice to show that given any

Since the real line is connected it will sufs

s IR

we have

ft N fs

for

t

close to

s.

We claim that for this it will suffice to show that there exists a germ at

(0, s) of a smooth mapping

H

(a)

H(x, s)

=

x

(b)

H(0, t)

=

0

(c)

F(H(x, t), t)

Mn x IR

:

=

- IRn for which

F(x, s).

117

To see this, write

ht(x)

It follows that for

t

Now (a) tells us that

H(x, t).

=

close to

the germ

s

the determinant of the Jacobian matrix at t.

hs = identity.

must be invertible, since

ht

depends continuously on

0

of

ht

Further, the condition (b) ensures that

ht

maps the origin in ]Rn

itself.

ft - fs

ft o ht

Finally, the condition (c) can be re-written as

for

t

close to

as was required.

s,

is automatically satisfied for

t = s,

=

to

f s,

so

Notice incidentally that (c)

Thus it will be suf-

in view of (a).

ficient to replace (c) by the condition that the left hand side does not depend on

t,

has a zero derivative with respect to

i.e.

Written out in full

t.

this is the condition. a

(CO )

i=1

tI (x' t) ax 1(H(x, t), t) +

at

(H(x, t), t)

=

0.

Our problem then is to construct a smooth mapping H which satisfies conditions

We claim that it will suffice to construct a germ

(a), (b), (c').

at (0, s) of a smooth mapping aF

d (

c

)

(e)

i axi ei(0, t)

For suppose such a mapping

dependent vector field on Mn.

]Rn x

]R

-,

]Rn

for which

aF - at

=

=

:

0.

a

exists.

We can think of

It has therefore a flow,

as a timei.e.

a mapping H

of the required type for which (f)

at (x, t)

_

e(H(x, t), t),

which can be supposed to satisfy the "initial condition" (a).

A minor com-

putation shows that (c') follows from (d) and (f) and that (b) follows from

(a), (e) and (f).

118

It remains to establish the existence of a mapping ties (d) and (e).

The argument is algebraic, and does not depend on the

particular value of

chosen:

s

we shall therefore suppose

since we wish to work simultaneously with functions of x1, ..., xn

and functions of (n + 1) variables

think of an element of t.

having the proper-

e

as an element of

9n

9

s = 0.

variables

n

x1, ..., xn, t n+1

n

at using the fact that

s en

As a preliminary, note that

.

OF

we shall

which does not depend on

And then condition (e) is simply the requirement that 1,

should lie in

Also,

_

aat (1 - t)f+tg

g - f

=

g-fE

.,//k+1 n

has zero k-jet, and (2.4).

Thus (d) will follow

from

k+1 C

aF

aF\ "' axn

'

ax 1

n

n Next we have

of \ ...s axn/

k / of llnk+1 C n C "en\ax1

aF

C

aF

s axn

+

n+

n

the first inclusion being trivial, the second being the hypothesis of the theorem, and the third following from the fact that

aF

of

to

-f

axi

k n+1

n

* now follows immediately from ** using the Nakayama Lemma, and the result is proved.

119

It is worth remarking that the condition (d) which appears in the above Recall that we took some germ

proof has a simple geometric interpretation. on the line joining

ft

along the line from

ft

f, g

and set out to show that if we moved slightly

we remained in the orbit through

ft:

thus we ex-

pected the direction of the line to be a "tangent vector" to the orbit through We think of the direction of the line being given by

ft.

(d) then says that at

lies in

.JTJf

the "tangent space" to the orbit

i.e.

,

The relation

at .

t

through

ft.

A word is also in order concerning the hypothesis of (3.1). Lemma tells us that the condition

Ak

is equivalent to the apparently

C.,#'Jf

Jf + .,

more complicated condition -.0"k C

The Nakayama

We chose the former con-

k+1

dition because it is rather simpler to remember though in practice it tends to be easier to establish the latter condition.

It is convenient to call mined for some

k >.

1.

f e &

n

finitely-determined when

f

is k-deter-

It follows then from (2.7) and (3.1) that if

of finite codimension then automatically practicalities of showing that a germ

f

f E gn

f

The

is finitely-determined.

is k-determined are very simiOne

lar to those involved in computing the codimension, discussed in §2. checks successively whether-,// k C -"//Jf

for a given k

+./1

is

k+1

for k = 1, 2, ...

:

and

this is equivalent to showing that each monomial of degree

k

+.,//k+1.

in

x1, ..., xn

lies in the ideal

f e 91

Consider a germ

Example 1

11Jf

for which the following conditions are

satisfied.

f(O)

=

0

of

akf (0)

=

0

ax

Clearly f E_/9/ 1 , f 120

:

...

(0)

axk 1+1.

=

0

:

ak+1f (0) k+1 ax

'

0.

It follows from the Hadamard Lemma that we

can write ax

=

aY,7f =

exk+1

+

=

k+1.

.!41

where

:

Then

g(O) ,' 0.

the expression in curly brackets is

(3.1) tells us that

if

=

<xk\

when

0

k

_

is (k + 1)-determined, so

f

An obvious change of co-ordinates then yields

c gE 0.

xk. The germ

Example 2

if

}

with

g e 91

so invertible by (2.1), and one deduces that

hence

f

for some

xk{(k + 1)g + x

x = 0,

f

xk+1g

=

f

<x2, y2>,

f(x, y)

=

x3 + y3

is 3-determined.

so .4'Jf = <x3, x2y, xy2, y3>

Here

and the claim

follows from (3.1) Remark

Another sufficient condition for

is that .,f k+1 C.,lr2Jf.

f E n to be k-determined

The proof of this is exactly the same as that of

one has simply to check that the last few lines of that proof go

(3.1):

through equally well with this hypothesis.

Of course, this is a slight im-

provement on (3.1) since ,,ayk+1 C2Jf is a weaker condition than ..l k C .,lvj The germ

Example 3 if -4/4

x3, y3

=

C 411f

=

.,I'Jf

=

=

+

y is

u-determined.

Here

<x4, x3y, xy3, y> and the condition

is not satisfied, so (3.1) does not allow us to conclude that

is L-determined. le if

so

x4

f(x, y)

On the other hand

<x5, xy, x3y2, x2y3, xy, y5> =

just mentioned allows us to conclude that

Notice that if

f

f e gn

_j3 f

so the improvement on (3.1) is indeed 4.-determined.

is k-determined then it is, in particular, equi-

valent to its kth Taylor polynomial.

Thus the study of germs in &n

finite codimension reduces to that of polynomial germs.

In this connexion

it is well to bear in mind that a polynomial germ of degree finitely-determined is not necessarily k-determined.

of

k which is

For instance the germ

121

f(x, y)

x5 + y5

=

is easily checked to be 6-determined, but can be shown

not to be 5-determined.

Classification of Germs of Codimension , 5

§1+.

We are now in a position to reap the rewards of our hard work in the previous

We

section, and achieve our object of classifiing germs of low codimension. start with germs of codimension

(1+.1)

0.

A necessary and sufficient condition for a germ

codimension

is that it be non-singular;

0

lent to the germ

(x1,

...,

Suppose

Necessity

identity element

8xf1

to have

and in that case it is equiva-

xn) - x1. has codimension

lies in

I

for which I = e1 for that at least one

f

f e 9n

Jf,

i.e.

+ ... + e 0,

aX (0)

so that

0,

Jf = 9n

and the in

e1, ..., en

we can find germs

If we evaluate at 0 we see

8x

and hence

f

is non-singular.

i

Suppose

Sufficiency

By (2.1) this means that Jf = &n,

so

f

f

is non-singular, so at least one ax (0)

If ax. 1

is an invertible element of

has codimension

122

and hence

0.

Finally, a non-singular germ in &n by (I.1.3).

1

6n,

0.

is equivalent to

(x1, ..., xn) -* x1

Remembering our discussion of

Next we consider germs of codimension 1.

2

functions in §5 of Chapter II we say that a germ

f E.Af

a singular

i.e.

,

germ, is non-degenerate when the Hessian matrix 2

_

a

l

f (0)J (axiaxj

Hf

The next proposition, classifying germs of codimension 1,

is non-singular.

is a classic result in the calculus of variations, called the Morse Lemma.

A necessary and sufficient condition for a germ

(4.2)

equivalent to a germ of the form

x2i

+ ... + xg - xs+1

" '

f

will be

x2

The main part of the proof consists of establishing that f e

Proof

non-degenerate if and only if ,A1 = Jf. the

n

and in that case

codimension 1 is that it be non-degenerate:

to be of

f E,,//2

all vanish at

ax

0.)

(Note:

is

1122,1

mil D Jf automatically since

To this end let Q denote the second polynomial

1

of

f - which is just a quadratic form in

x1, ..., xn.

Non-degeneracy of f

is the same thing as non-degeneracy (in the normal sense) of the quadratic form Q,

span the vector space

which in turn is equivalent to saying that the 1

of all linear polynomials in Since we have

1=

saying

ax

x1, ..., xn (by elementary linear algebra).

modulo .,11 2,

ax.

ax.

1

this last condition is equivalent to

1

2

Jf +.s1 , which in turn is equivalent to

vious application of the Nakayama Lemma.

Sufficiency

codim f

dim c/Jf

=

f

f

Jf

=

11,

and

f

Jf =,/,/by the above.

Then

1.

has codimension 1.

Jf C .,ll then dim &/Jf > 1,

And if

The rest of the proof is now easy.

non-degenerate, so

dim 8/.,11=

=

Suppose

Necessity

Thus

Suppose

.,ll = Jf, using an ob-

i.e.

Certainly then

codim f > 1,

dim &/,' =

1.

which is impossible.

is non-degenerate, by the above discussion. 123

Finally, we have to verify that saw above, non-degeneracy of

has the quoted normal form.

f

is equivalent to

f

It follows immediately from (3.1) that

..11 2 C ,E'Jf.

equivalent to its second Taylor polynomial algebra that

Q.

f

being a non-degenerate quadratic form, is equivalent (under

Q,

+ ... x2 - x2

s+1

s

1

is 2-determined, so

Recall now from quadratic

2 x1 + ... xs

a linear change of co-ordinates) to the normal form x

which implies

Jf

..//=

As we

- x2

-

n

where

is the index of

s

2 xs+1 -

That concludes

Q.

the proof.

The geometric meaning of the Morse we have a smooth function

f :IRn

degenerate critical point at

0.

-

correspond to sets

f-1(E)

with

IR

f(0) = 0

Suppose

admitting a non-

The Morse Lemma tells us that there is a

diffeomorphism of one neighbourhood of the fibres

Lemma is worth spelling out.

in 1Rn

I

onto another under which

x2 + ... + x2 - x2+1-..., - xn

=

E,

1

for small enough and for

c ? 0

c .

In particular,

the relevant part of

f 1(0)

corresponds to a quadric cone;

f-(E)

is a smooth manif old diffeo-

morphic to part of a non-singular quadric surface. case

n = 2,

Here is a picture for the

s = 1.

diffeomorphism

To make this even more convincing recall again our discussion of functions in §5 of Chapter II.

There we showed that any function

be approximated as closely as we please by one

124-

ft

:

f

:

1Rn -, IR

IRn - 1R

could

which admits

only non-degenerate critical points; from

f

be a linear deformation.

and indeed that

x3 - 3xY2

=

which admits a degenerate critical point at

as the approximating function, for erate critical point of

f

at

(+ f

points at the points

,

0

0).

could be obtained

One of our examples was the monkey saddle

f(x, Y)

ft(x, Y)

ft

=

0.

In this case we could take

x3 - 3xY2 - tx

t > 0

The degen-

as small as we please.

"splits" into two non-degenerate critical And the diagram of level curves

f_1(e)

3

is as below:

the reader will note that close to the critical points we have

precisely the situation described above.

To push the classification further for germs of codimension > 2 we shall require one more bit of machinery. 2.

Suppose that

Certainly then the Hessian matrix

the non-negative integer

c = n - r

Hf

f

n

has codimension

is singular, so has rank

is called the corank of

checks that it is invariant under equivalence.)

r < n:

(One easily

g.

The extra machinery which

we require is the following result, known as the Splitting Lemma, and proved originally under less restrictive hypotheses.

(4.3)

Let

2

f E,// n

be a finitely-determined germ of corank

c;

f

is

equivalent to a germ 125

2

xc) + Sc+1xc+1 +. . .+8x

g(x1s

where

g E _.*,1 and each

Proof

Si = + 1.

We shall adopt the following notation.

8n we write for which relation.

n

Given two germs

r when there exists an invertible germ

¢ o k, 0

have the same k-jet.

Observe that

We intend to show, by induction on

0, 0

in

h : (IRn 0) -> (IRn 0)

is an equivalence

k

that there exists a germ

k,

Ed'3, which is polynomial of degree 5 k, for which

gk

f(x1, ..., xl)

k

gk(x1,

Sc+1xc2

..03, xc) +

+1 + ... + Snxn

This will suffice to prove the result, since there exists a

k

for which f

is equivalent to its k-jet.

The induction starts when k = 2.

n

variables of rank r = n - c,

The 2-jet of

is a quadratic form in

f

so equivalent (under a linear change of co2

ordinates) to a quadratic form

Se+1x0+1 + ... + Snxn with each

by standard quadratic algebra:

and we can take

step we assume

g2 = 0.

Si

+1,

For the induction

Certainly then

*.

f(x1, ..., xn) k+1

gk(x1'

'

xc) + Sc+1xc+1 +

+Snxn + H(x1, ...) xn) **

where

H

is a homogeneous polynomial of degree (k + 1).

Now write, as we

evidently can do,

H(x1,

..., n

=

h(x1, ..., xc) + xc+1Hc+1(x1' ..., xn

+ ...

+ xnHn(x1, ..., x

where

h

)

is a homogeneous polynomial of degree (k + 1) and He+1' ...' Hn

are homogeneous polynomials of degree 126

n

k.

We use these latter polynomials

to define 0 : (IRn, 0) -+ (]Rn, 0) by taking its components to be 01(x1,

..., n) =

x1

00(x1,

..., xn)

xc

=

e+1)

1

0o+1(x1' " .'

n) =

on(x1, ..., xn)

=

2SH(x1, c+1

xc+1

xn

_

" ., n

1

2S Hn(x1' ..., xn). n

Note that the Jacobian matrix of

Substituting

invertible.

f(x1, ..., xn)

where

gk+1

=

is the identity n x n matrix, so

/1, ..., Sn

gk+1(x1V

k+1

"

for

x1, ..., xn in

**

0 is

we obtain

2

2

xc) + Sc+1xc+1 + .. + Snxn

That'completes the induction step, and establishes

gk + h.

the result.

It is worth remarking that the germs

f, g which appear in the statement

of the Splitting Lemma are necessarily of the same codimension. can safely be left as an exercise for the reader.

The proof

Our first application of

the Splitting Lemma is to establish a broad connexion between the corank and the codimension of a germ, which will be crucial to the remainder of the classification.

Roughly speaking, it is that as the corank increases so

does the codimension, only rather more quickly.

(4.4)

Let

codimension of

Precisely,

f e.,F2 be a germ of finite codimension, and corank f

is

>

1+

C (c

2

c:

the

1.

127

Proof

We have already observed in §3 that a germ of finite codimension is

automatically finitely-determined, so the Splitting Lemma applies, and tells us that

f

is equivalent to a germ g(x1, ..., xc) + xc+1

with

and (vide the remark above)

g e.,lf3:

Note that

g.

+ ... + xn

f

has the same codimension as

I = A Jg C !! o and that by (2.8)

cod I

=

cod0l + codf I + cod2 I + cod3 I + ...

But

cod0 I

I + 9 dim I

=

J

=

dim

=

Arc

I + ,,!/ codf I

dim

=

1

c dim

-/f c

-ff

I

2

cod2 I= dim

_ dim 3=

I + I c

C (c 2 1

ec

so

1 +c+cc2+

codI = c+codJg

using (2.14), from which the result follows.

Thus, for instance, we have the following rough estimates

c=1 c=2 c=3

,

,

codf>.2 codf >4 codf>>7

The consequence of (4.4) which we need to isolate is that if then corank

f -2 and s 5:

(4.5)

n

then (up to addition of

a non-degenerate quadratic form in further variables, and multiplication by ± 1)

f

is equivalent to one of the seven germs on the following list.

corank

codim

germ

2

x3

f old

3

x

cusp

4

x5

swallowtail

5

x6

butterfly

name

1

2

4

x3 - xy2

elliptic umbilic

4

x3 + y3

hyperbolic umbilic

5

4 x y + y

parabolic umbilic

2

The exact meaning of the statement will become clear as we proceed. Morse

Lemma give us a complete classification of germs of corank

The

We

0.

shall now classify all germs of corank 1 having finite codimension.

Let

(4.6)

2

f E

n

have corank 1, and finite codimension (k - 1):

f(x1, ..., xn) - +

1

± x2

+

then

2

+ xn

We know from §3 that a germ of finite codimension is finitely-

Proof

determined, so the Splitting Lemma applies and tells us that

f(x1, ..., xn) ^- g(x1) + x2 + ... + xn

where

gE

show that

I

also has codimension

g - + xk.

k.

Write

x = x1.

It suffices now to

Observe that there must exist a least integer

j

with 129

g E,%' , cod g

if not then J9C ll for all j,

g X.;/l1+1

cod ll

>

=

computation shows that

O(x)

xh(x)1'k:

=

k(x),

h(O) > 0, =

so

0

g - xk. g

xj-h(x))1/k

=

g(x)

(P'(0) # 0

so

0

with

xih(x)

(j - 1),

(IR, 0) -+ (IR, 0)

:

clearly,

and yields

g would have infinite codimen-

i.e.

has codimension

g

In that case take

g(x) _

O(x)

j,

The Hadamard Lemma tells us that

sion.

odd.

for all

j

so

so

h(O) J 0:

Suppose

j = k.

to be the germ given by is invertible.

Further,

The same reasoning applies when k

But when k

xk.

k

is even and

is even and

h(O) < 0

take

k

g - -x

to deduce

,

It is the germs obtained in the above result which are called the cuspoids. Of course, if we restrict ourselves to germs of codimension < 5 in the result,

then we obtain (up to addition of a non-degenerate quadratic form in further x3, xl+, x5, x6.

variables, and multiplication by ± 1) precisely the germs

And with this we have proved half of Thom's classification theorem (4.5). A singular point of a smooth function of sion

k

corank ,

is usually called an Ak singularity.

1

and finite codimen-

Such singularities have long

been in the repertoire of classical algebraic geometry, and the reader may find it of interest to recall how they arise naturally when studying double points of algebraic curves. We shall restrict our attention to real algebraic curves,

i.e.

of the plane IR2 defined by an equation f(x, y) = 0 where f

:

IR2 -1R

is a polynomial function:

we shall suppose that the curve

through the origin (so

f

has no constant term), and indeed that the origin

is a singular point of

f (so

f

f = 0

subsets

has no linear terms either).

passes

Recall that

the singular point at the origin is said to be of multiplicity m when m is the lowest degree of a monomial in have a double point, 130

when m = 3

f:

thus

m > 2.

When m = 2

a triple point, and so on.

we

The condition

on the origin, that it is a double point of the curve the singular point there has corank s 1.

+

xk+1

f = 0

the curve

k = 1, 2, 3, 4, 5

+ y2.

Ak

for some integer

xk+1 + y2 by (4.6).

by a non-zero scalar does not affect the curve normal form is

entails that

If we assume the singularity to be

of finite codimension then it will be of type so equivalent to the singularity

f = 0,

f = 0

k >>

1,

Multiplication of f so we can suppose the

And since we are not interested in the case when

is a point we consider only the case

xk+1 - y2.

The cases

give rise to the following pictures, which are tradition-

ally dubbed with the titles indicated.

A

1

: node

A2 : cusp

A 3 : tacnode

A4 : rcunphoid cusp

A 5

: oscnode

131

Example 1

Consider the real algebraic curve given by the equation

f = 0

where 2 2

f(x, y)

=

x4 + x y

- 2x2y - xy2 + y2.

Clearly, the curve has a singular point at the origin, indeed a double point. And computation verifies that we have a ramphoid cusp.

will have codimension 1+

f

at the origin, sn

The curve looks like this.

X

It is perhaps also of interest to see how

Ak

singularities turn up natu-

rally in studying the differential geometry of plane curves.

a smooth mapping with

:

IR -+ R2 whose image is a smooth submanifold of

mapping its domain diffeomorphically onto its image;

C

think of

as a parametrization of a plane curve.

standard scalar product on IR2, tion

I

Suppose we have

Let

.

IR2,

we wish to denote the

giving rise to the standard distance func-

(We work with the square to ensure smoothness at the origin.)

I2

Given a fixed point w

in the plane consider the smooth function fw

:

IR-+IR

defined by fw(t)

=

2'w - C(t)j

measuring, essentially, the distance from w parameter

t.

As

w

to the point on the curve with

varies we obtain a 2-parameter family F = (fw)

of

smooth functions, better thought of as a single smooth function, say

F

: IR2

of each

132

x IR -+ IR fw

given by F(w, t)

must have corank E I,

=

fw(t).

Clearly, the singular points

and those of finite codimension will be

classified by the Ak-series:

indeed the condition for an

singularity

Ak

will be that aF

at

2

ak+1F

0

akF

at_

=

atk+1

atk

Let us look at the geometric information contained in these relations. The first few derivatives, written out in full, are aF

2

(C - w)

=

C

=

3C.C + C.(S - w)

F

a

at2

a3F at3

where

denote the successive derivatives of

for the components of

Thus

C.

2)

_

Write

C1, C2

is the tangent vector to the

curve, and the normal direction is given by

the result of

i.e.

rotating the tangent vector through a right angle in an anticlockwise direction.

The condition at

tor

is perpendicular to

to the curve at

C(t):

=

then amounts to saying that the tangent vec-

0

S - w,

that

i.e.

the set of points

(w, t)

w

lies on the normal line in ]R2 x]R where

is therefore called the normal bundle space of the curve. a pair

(w, t)

we can write

w =

normal direction to the curve at

where

+ An

and

fi(t),

A

n = n(t)

at

= 0

Note that for such denotes the unit

is some scalar.

Now let us

2

add the further condition that just given in that for

0.

Substitute the expression for

w

at

a2

at

=

as

and equate the result to zero;

, 2

a few lines of

computation will then show that c

A

2 1

+

2, 2 3/2 1C2

133

which the reader will recognize as the radius procal of the curvature for which i.e.

at

=

0,

It follows that the points

K. a

of curvature,

the reci-

i.e.

(w, t)

in ]R2 X IR

2

=

0

are precisely those for which w =

1

+

at

with w a point on the evolute of

It is interesting to go one

stage further and ask what further geometric information we can glean by adding yet one more condition K = 0,

is equivalent to

a3F

= 0: we leave the reader to check that this at3 i.e. to the point C(t) being a vertex of the

Thus singularities of type

curve.

Ak with

bundle space for the curve, those with k . 2 those with

k 3 3

obtain

singularities with

Ak

to vertices.

k 3

I

correspond to the normal

to points on the evolute, and

It turns out that generically one does not in this situation, so that is the end

k > 4

of the story.

Example 2 If we write

Consider the parabola w = (u, v)

2F(w, t)

C

:1R - IR2

given by (t),= (t, t2).

then

=

t4 - tut - (2v - 1)t2 + u2 + v2

and comparison with Example 5 of §1 in Chapter II shows that defines a folded surfaces in IR2 xlER,

=

at

that the further condition

0 a

2

at

= 3

0

2 a defines the fold curve on the surface, and that the final requirement F = 0 7t3 defines the exceptional point in the fold curve where the two folds meet. Under the projection ]R2 x IR

n IR2 the fold curve maps to the evolute of

the parabola, which is a cuspidal cubic, the cusp arising from the vertex of the parabola at the origin.

134

It remains for us to deal with germs of corank 2 and codimension 5 5.

The

next result will complete Thom's classification theorem.

(1+.7)

Let

f E.//2 have corank 2 and codimension . 5:

n

then

f

is equi-

valent to one of the following germs

±

(x3 + x2)

±

x3

+

x3 ± ...

4

2

± (x1x2 + x2)

Proof step.

...

2

x2

+

± xn2

Again, it is the Splitting Lemma which provides us with the first It tells us that

f

is equivalent to a germ

g(x1, x2)

with

+

+

2

x3

± ... ± x2n

g e-//3 having the same codimension as

Taylor polynomial of

g

f,

is a binary cubic form in

namely . 5. x1, x2

The third

so can be

135

assumed x31

- by the discussion in the previous chapter

+ x2,

x21x2

-

to be

We have to decide which possibilities can arise.

this end, recall that a germ of corank 2 has codimension >. 4. (and hence

2 x1 - x1 x,

according as it is identically zero, symbolic, elliptic, hyper-

bolic or parabolic.

f

0,

has codimension 4 or 5.

g)

To

In particular

We shall consider the possi-

bilities separately.

g

has codimension 4

Since

Jg C,,ll2,

we have

g

To see this we argue as follows.

so 11Jg C .,113.

Put

I

codimension 4, and hence by (2.14) that

= ,%1J9.

In fact ,%'J, = .,113.

We are given that

has codimension 6.

I

Jg has

Also, from

(2.8) we know that

cod I But

I

cod0 I + cod1 I + cod2 I + ...

=

and hence I +.ll J = -,or j for

cod0 I

=

I

:

cod1 I

=

3,

3

2

:

.

so that

cod2 I

=

3

we deduce that

cod3 I which means that

43 C I

third Taylor polynomial.

0

as required.

= .AVJg,

It follows now from (3.1) that

=

g

is 3-determined so equivalent to its

By checking the five possible normal forms for

this one by one we see that the only ones of codimension 4 are the elliptic and hyperbolic types.

And these give us the first two germs on the list.

has codimension 5 .,(N1 x P1, z1)

(N1, x1)

1 ,(N1, x1)

H

h

y

12

h y

112

y

(N2, x2) --)(N2 x P2, 2)(N2, x2) where for

k = 1, 2

ik

=

germ at

xk

of the inclusion Nk - Nk x Pk

given by x H Sk

=

germ at

zk

of the projection Nk x Pk -

given by In other words,

H

(x, yk

Nk

(x, y) '- x.

is necessarily given by a formula 143

H(x, y)

with

O(x, y1)

=

(orX-equivalent)

y2.

(h(x), 9(x, y))

=

And we call our germs

with

N1

contact-equivalent

when there exists a contact-equivalence

H0(1, f1)

Thus one pictures

f1, f2

H

mapping onto

(1, f2) o h.

=

as mapping the graph of N2

via

for which

(h, H)

f1

onto the graph of

f2,

h.

P2

H

h

The reader will have no difficulty in checking that if A-equivalent then automatically they are X-equivalent. discussing._*=equivalence of germs

tion to suppose that

N = ]Rn,

(N, x)

x = 0

and

hand as we shall see two germs which are A-equivalent.

(P, y)

-+

P = ]Rp,

f1, f2

are

In particular, when it will be no restricy = 0.

On the other

equivalent are not necessarily

In the remainder of this section we shall justify our state-

ment that=equivalence is a fairly computable notion by deriving a more-orless algebraic criterion for two germs to be-7,-equivalent.

We do this in

The first step, which is the main one, is to look at a finer

two steps.

notion of equivalence called

8-Equivalence.

For the sake of simplicity let us restrict our attention to germs (]Rn,

0)

-+

(]RP, 0).

Two such germs

.equivalent via a-equivalence

144

(h, H)

f, g

are 6-equivalent when they are

having the special property that h

is the germ at f, g

h

0

of the identity mapping on IRn.

Note therefore that

are.-equivalent if and only if there exists an invertible germ

ORn, 0) -+ (IRn, 0)

:

f o h, g are g-equivalent.

for which

proposition provides us with a purely algebraic criterion for g-equivalent.

To simplify notation write

If =

The next to be

f, g

the ideal in

i.e. P

9n

generated by the components

f1, ..., fp

of

f, g in 90n,p

The following three conditions on germs

(2.1)

And similarly for

f.

Ig.

are equi -

valent.

(i)

(ii)

(iii)

f, g

are e,-equivalent.

The ideals

are equal.

If, 19

There exists an invertible entries in

for which

8n

p x p fi

matrix

=

(uij )

u..gj .

for

with

1 t.

the Jacobian module

(x2, 0),

when

We shall compute the,5.codimension of the germ

integers

and

p - I

=

codkf

there are (p - 1) such gen-

All the generators for .111.&1 "p

We conclude that the germ

Example 8

lie in

None of the generators for,// 1k.8 1,p

xiyj

is generated by the vectors

Clearly, if Tf +,1112+1.82

(0, yk).

i ,

I

and

j

1

>.

It remains to

2'

We claim that the first

two likewise lie in T f +4'2+1.92,02 : when k = I this is because (x, 0)

=

ay it is because

- (0, (xk, 0),

byb-1),

(Y, 0)

(yk, 0)

= ax -

are multiples of

appeared in our list of generators for Tf. lies in

158

Tf *2+1.82 2

only when

(0, axa-1),

k >, a,

and when k , 2

(x2, 0),

(y2, 0)

Finally, the vector

whilst the vector

which

(0, x

k (0, y )

k )

does

so only when k > b. therefore the I

The vectors which span

s k s b - 1.

Example 9

a > 3

f

:

(IR2, 0)

-

is an integer.

a).

2

+ y

(0, x

),

(IR2, 0)

Here if.

given by f(x, y) (2x, axa-1)

of = The ideal

with

will be

(x2 + y2, xa)

=

and

ay =

(2y, 0)

is generated by x2 + y2,

If

is generated by

so the submodule 2

f

)

a + b.

=

generate the Jacobian module

(0, x

(0, y

For our final illustration we shall compute the-T=codimension

of the germ

xa

k

and the

s k 5 a - 1,

1

It follows that the, -codimension of

2 + (a - 1) + (b - 1)

where

with

(0, xk)

the required supplements are

(x2 + y2, 0),

(x a, 0),

Following through a computation similar to that in

the previous example one finds that a supplement for the ,=tangent space in is provided by the (0, xi) so the,7e=codimension is

§3.

and the

(0, xl-1y)

2 + (a - 1) + (a - 1)

with

1

. i s a - 1,

2a.

=

Deformations Under Contact Equivalence

The next step in our programme is to set up the basic ideas for a theory of deformations of germs under,'-equivalence:

the relevance of this to the

problem of classifying stable germs will be discussed in §4.

Let us start with an r-parameter deformation of a germ

f

:

(IRm, 0)

-a (IRq, 0).

F

:

(IRr x IRm, 0) -+ (IRq, 0)

Pursuing the analogy with the Finite

Dimensional Model of Chapter III one expects a major role to be played by "transversal" deformations.

We need a formal interpretation for this in-

tuitive idea, so we argue heuristically.

Think of the deformation as a

159

"germ"

(IRr, 0) -,

(gmsq ,

f) given by u --> fu where fu(x)

F(u, x).

=

We wish this mapping to be "transverse" to theX-orbit through

f,

i.e.

we

want something like

image of the "differential" at 0 of this map

tangent space to X- orbit through f

+

tangent space

to

gm.,q

at f.

The only quantity here for which we do not yet have a concrete interpretation is the "differential" at ear mapping Iu=O

au

-,

IRr

m,q au

ordinates on IRr.

r

of the mapping

0

this should be the lin-

which sends the standard basic vectors for IRr to where we write

1u=0'

u - fu:

u1, ..., ur for the standard co-

As a matter of convenience we write

F.

= aui

(u=0

With this notation the image of our "differential" will be the real vector On this heuristic basis we introduce

dm,q.

of

subspace IR{F1, ..., FrJ

the following formal definition.

F

is aJr-transversal deformation of

f

when

IR{F1,

Notice therefore that

..., FrI + Tf = am

s q.

admits ax transversal deformation if and only if

f

it has finite.'-codimension

a,

say,

Assuming this to be the case one can

construct explicit X-transversal deformations by the same device used in the One looks for a deformation

Finite Dimensional Model.

F(u, x)

where the germs

F

=

f1, ..., fc

for which

f(x) + u1f1(x) + ... + ucfc(x)

are to be determined.

The coalition for this

to be -V-transversal is that IR{f1, ..., fc1 + Tf

This yields an entirely practical procedure. 160

F

=

gm,,q.

Simply choose f1, ..., fc

to

Tf,

be a supplement to

and then define

by

F

*.

In practice one is interested solely in X-transversal unfoldings of germs of rank

(Just why this is so will be made clear in the next section.)

0.

For such a germ

one can be a little more explicit about the form of a

f

transversal unfolding. product

e1, ..., eq for the standard basis vectors in 7Rq.

We write

is a germ of rank

f

f1,

find a basis

Tf

supplement of

then a moment's thought will convince the

0;

is a vector subspace of

Tf

reader that

&mpq with the

as a real vector space we can think of this as the direct sum of

8m :

IRq with E. Suppose

Make the usual identification of

"'s'

in

fr am

_4,q.

It will then be possible to

for a supplement of

Tf

is then provided by

e1, ..., eq,

in flm

a basis for a f1, ..., fr'

and we obtain a.5 transversal deformation

r F(u, w, x)

- J wi.e

_

+ f(x) +

i=1

i=1

where we insert the minus sign for a minor geometric reason which will be menHere are some examples of these computations,

tioned in the next section.

parallel to the computations of, '-codimension given in §2, where all the work

In each example the germ

was done.

The germ

Example I

has X-codimension J+: (1, 0),

provided by mation is

F

:

(IR4

f

(IR2, 0)

:

has rank

f

-+

(IR2, 0)

0.

given by f(x, y) = (x2, y2)

indeed, we saw that a supplement for Tf in 82,2 is (0, 1), x IR2, 0)

(y, 0),

(0, x).

(]R2, 0)

-+

with components given by

F1

=

x2 + u1y - w1

F2

=

y

2

Thus aX-transversal defor-

+ u2x - w2.

161

The germ

Example 2 f(x)

=

(0,

f

..., 0, xt+1),

indeed a supplement for for which

(IR, 0)

:

F

:

t

is provided by the deleted.

..., 0, xt) 0)

(0,

...,

xk, ...,

0)

Thus a,=transversal

with components given by

-+ (IRP, 0)

-w1 + u11x + ... + u1st-1x

t-1

+ u1'tx

t

1xt-1 + u2,txt

-w2 + u21x + ... +

=

c = pt + p - 1:

has_,*-codimension

1,

1,p

(0,

(]Re x IR,

F1

F2

in

f

with

0 s k . t

deformation is

T

with

defined by

-, (IR', 0)

,

2 t - 1

The germs

Example 3 f(x, y)

I

(1, 0)

F

:

(xy, xa + yb),

:

(IR2, 0)

with

indeed a supplement for

a + b:

with

=

f

c- i E a - 1,

and

(tea+b

a, b

(0, y

with

)

x IR2, 0) - (IR2, 0)

F1

=

xy - w1

F2

=

xa

The germ =

together with

with components

+

y

b-1

uixl +

+

f

(x2 + y2, xa),

:

(IR2, 0)

where

viyl - w2 i=1

-+

(IR2, 0)

a 3 3,

given by

has, -codimension

ment for Tf in 92,2 is provided by the (0, x') together with (1, 0) and (0, 1). 1 s i s a - 1,

deformation is the germ 162

(0, x1)

Thus a,=transversal deformation is the germ

(0, 1).

i=1

f(x, y)

5 i , b - 1,

1

a-1

Example 4

is provided by the

&2,2

i

and the

3, have X-codimension

integers

in

Tf

given by

(]R2, 0)

-+

F

:

(IR2ax IR2, 0)

-+

and

2a.

A supple-

(0, x1-1y) with

Thus a X-transversal

(IR2, 0) with components

F1

=

x2 + Y2 - w1

=

xa

a-1

a-1 F2

uixl

+

+

i=1

So much for examples.

viyi - w2

1=1

The next step is to pursue the analogy with the

Finite Dimensional Model further to see if we can characterize the algebraic notion of "transversality" by a geometric notion of "versality".

To this

end we introduce a series of notions for deformations of a germ

f

:

(1Rm, 0) -+ (IRq, 0).

Equivalence of Deformations Two r-parameter deformations

F1, F2

there exists an r-parameter unfolding mapping on ]Rm

f

of

are said to be.Y.equivalent when of the germ at

Im

0

of the identity

for which

Im(F1(.,llq)) In this situation we call

F2(

=

Im aX-equivalence of deformations.

(2.2) this relation implies that

F1, F2

Of course by

are-Z. .equivalent as germs:

however

it says more in that the change of co-ordinates at the source has to respect the product structure on ]Rr X ]Rm.

Induced Deformations Suppose

and that

F

(]Rr x Mm, 0) -+ (]Rq, 0)

:

H

deformation

(]Rs, 0) -+ (]Rr, 0)

:

G

:

is an r-parameter deformation of

is a germ.

(]Rs x IRm, 0) -+ (]Rq, 0) G(v, x)

=

of

f,

We obtain an s-parameter

f by putting

F(H(v), x)

.

163

One writes

situation

G H

H*F,

=

and calls

the deformation induced by

G

in this

H:

is a change of parameter.

Morphisms of Deformations

Let

F, G

be r, s-parameter deformations of

is a pair H

with

(H, I)

H G

under

I.

A morphism from

F

to

G

ai equivalence of r-parameter deformations, and

I

a change of parameter, for which

mation

f.

When

F

r = s

is,' equivalent to the induced deforand

H

is invertible we refer to the

morphism as an isomorphism.

Versal Deformations

A deformation

G

morphism from

F

f is 5- versal when for any deformation F there is a to G. When f has finite,=codimension c, say, a cof

parameter,'-versal deformation is said to be, -universal. The main result about deformations under contact equivalence is the following analogue of (111.5.1) which we dub the X-Versality Theorem.

(3.1)

Let F

of a germ f

for

F

:

:

(]Rr x

]Rm, 0) - (]Rq, 0) be an r-parameter deformation

(I1m, 0) -+ (]Rq, 0)

:

a necessary and sufficient condition

to bej -versal is that it should beX-transversal.

A word or two is in order concerning the analogy between this result and the Versality Theorem of Chapter III.

In that result the key was the exist-

ence of neighbourhoods having a product structure, which were produced by the Inverse Function Theorem.

But in the present situation the basic objects

lie in a vector space of germs:

one has no immediate analogue for the Inverse

Function Theorem, and is forced to adopt a different stratagem. 161+

The proof

that .7-transversal deformations are X-versal is by no means easy, and we shall not give it;

it uses an extension to the real case of a classical

theorem of Weierstrass in complex function theory.

A sketch of the result

can be found in the paper of J. Martinet quoted in Appendix E.

On the other

hand, the converse result is relatively trivial.

Proof of Necessity Suppose i.e.

is X-versal.

F

We have to show that

F

is -Y-transversal,

that

...,

]R(F1, Consider then a germ

g

in

,,q,

Fr} + Tf

=

m ,q

and the 1-parameter deformation

G

of

f

given by G(v, x)

G

By hypothesis tion say.

Thus

f(x) + vg(x).

is X -equivalent (as a deformation) to an induced deformawith

h

H(t, x)

=

H = h*F

=

:

(R, 0)

->

F(h(t), x)

having components

(]R", 0)

and

8h

H

-

8h

at (0)F1 + ... + at

establishing that H lies in ]R[F1, ..., FrJ. G, H

Tf,

(0)Fr

Starting from the fact that

are X-equivalent deformations a computation shows that

and hence that

G

lies in

h1, ..., hr

Fr} + Tf.

G - H

lies in

However G = g,

finishing the proof.

Just as in the Finite Dimensional Model we are now in a position to justify the use of the prefix in the term

"X-universal".

165

Let

3.2

f

:

F, G

be X -universal deformations of a germ

of finite '-codimension

(]Rm, 0) -+ (]Rq, 0)

,

then F, G are

c:

isomorphic deformations. Proof for which

G

is '-versal there exists a germ h : (]R', 0) -+ (]Rc, 0)

F

As

is .T'-equivalent to the induced deformation

there exists a c-parameter unfolding the germ at

:

(1R0

of the identity map on ]Rm

0

9-equivalent,

i.e.

We shall show that

F o 'P,

G

X ]Rm, 0) -+ (RC X ]Rm, 0)

for which

T =

are 9-equivalent, where

with entries in

for which F o

9m+c

= AG,

(h x 1) o -1.

q x q

one obtains for c

I

In

matrix

with the usual

Differentiating this relation with respect to

setting u = 0,

of

are

is an invertible germ, which we do as follows.

h

identifications.

G

h*F o

view of (2.1) this means that there exists an invertible

A = A(u, x)

That means

h F.

ui,

and

a relation of the form

s i s c

ahj au.

AOGi + T.

(0)Fj

j=1

where Now

F1,

A0

F. G

=

AO(x)

=

A(0, x),

and

lies in the ,5 -tangent space

Ti

are both X -transversal deformations of

..., Fc span a supplement for

Tf, as do

f

Tf

by (3.1) so

G1, ..., Gc.

In view of the

proof of (2.1+) the invertible matrix A0 induces a linear automorphism of the 8m-module

&m which leaves Tf invariant.

a supplement for ah. au. 1

(0),

i.e.

If.

It follows from

the Jacobian matrix of

h,

F, G

A0G1, ..., AOFc also span

that the matrix of coefficients

*

tion Theorem now enables us to deduce that It follows from the definition that

Thus

is invertible: h

the Inverse Func-

is invertible, as was required.

are -Y-isomorphic deformations.

In fact we can squeeze a little more information out of the proof than is actually stated in (3.2). 166

and let

C.

f

Let

(3.3)

:

-

m

(IR , 0)

(IRq, 0) be a germ of finite X-codimension

be a X -universal deformation of f.

F

parameter .X-versal deformation parameter constant deformation of versal deformations of

Proof

:

:

Thus

d 3 c any d-

is X-isomorphic to the (d - c)-

And hence any two d-parameter

F.

,

'-

-isomorphic.

Proceeding exactly as in (3.2) one comes to the conclusion that

(IRd, 0)

--1,

(IRc

0)

an invertible germ 7T

are

f

is ,Y-equivalent to an induced deformation

F'

h

f

of

F'

For

(3Rd, 0) F'

'

:

(3Re, 0)

a submersive germ.

F.

with

However by (1.1.3) there exists

0) - (3Rd, 0) for which

(3Ra

the projection given by

is -Y-isomorphic to

deformation of

h F,

ir'F,

i.e.

7r(u1,

h0o

with

i,., ud) _ (u1, ,

uc).

to the (d - c)-parameter constant

R

The rest is clear.

There is a small technicality which is worth mentioning at this point. "e have phrased the definition of X -isomorphism for deformations of a single germ

f.

One can of course phrase the definition for deformations of X-

equivalent germs and (3.3).

f, f'

and obtain results which correspond exactly to (3.2)

We shall leave this matter to the reader, and proceed rather to

the next step in our programme, which is to show how one can use the existence and uniqueness of

deformations to reduce the problem of classifying

stable germs under the relation of A-equivalence to that of classifying germs of finite

X-codimension under the relation of X-equivalence.

167

Classification of Stable Germs

§4.

The time has come to put together the bits and gain some distance into the

problem of producing explicit lists of stable germs in given dimensions. Our starting point is a very simple idea.

Let

(4.1)

G

:

(Itn, 0) - (Itp, 0) be a germ of rank r: then there

exists an invertible gem_ h

:

(]Rn, 0) -+ (]Rn, 0) for which F = G o h is

an r-parameter unfolding of a germ of rank 0. Proof

By making linear changes of co-ordinates at source and tar-

get we can suppose that the Jacobian matrix of

where g

:

Ir

is the identity

of

G.

an invertible germ h tion

Clearly, :

evaluated at

g

has rank

(]Rn, 0) - (Mn, 0)

so by

r,

for which

r

is

components

(1.1.3) there exists

g o h

And then F = G o h

(x1, ..., xn) - (x1, ..., xr).

0,

Consider the germ

matrix.

whose components are the first

(]Rn, 0) - (Itr, 0)

G1, ..., Gr

r x r

G,

is the projecis the required

germ.

F-I

The point of (4.1) as far as the present section is concerned is that since we are only classifying germs up to A-equivalence we can restrict our atten-

tion to r-parameter unfoldings

F : (]R" x ]Rm, O)

r: for such an unfolding write fF germ of rank F

-+

168

fF.

0

which

F

unfolds.

:

(]Rm, 0)

-*

-+

(]Rr x ]Rq, 0)

(]R q, 0)

of rank

for the unique

We wish now to study the assignation

The first step in this direction is provided by

Let F, F'

(4.2)

foldings of germs are

fF, fF,

]R m' 0)

(IRr x

.

fF, fF,

of rank

0:

x IRq, 0) be r-parameter un-

(7Rr if

F, F'

are A-equivalent then

,T-equivalent.

Since

Step I

F, F'

are A-equivalent there exist invertible germs

H, K for which the following diagram commutes.

( IRr x IRm,

0)

(Mr x ]Rq, 0)

F

T

T K

H

(Mr x 3P

0)

14

F

(1f?.r

x IRq, 0)

Notice first that in view of (1+.1) we can suppose that for some

0

(IRm, 0) -. (IRr, 0),

:

for some germ 0

:

g

:

and likewise that

(]Rm, 0) - (IRq, 0)

K(0, y) = (rb(y), y)

Write F(u, x) =

(IRq, 0) -+ (IRr, 0).

It should now be clear that

H(0, x) = (0(x), x)

(

f(u, x)).

,

is A-equivalent to the germ

fF,

given by g(x)

f(4 (x), x).

=

And it will suffice

to show that g, fF are X -equivalent. f(u, x) - fF(x)

Step 2

vanishes on

0 x

IRm,

so each of its

q

components does likewise, and the Hadamard Lemma allows us to write f(u, x )

=

fF(x) + M(u, x).u

entries are germs at

0

where

0(y)

=

0

write

B(x).g(x)

0(g(x))

=

where

A(y).y

whose entries are germs at _

is a

of functions on IRr x Fm,

And likewise we can write

q>(x)

M(u, x)

to the same matrix notation we can now write

matrix, for which

C(x)

a

q x r

g(x)

=

fF(x) + D(x).g(x)

with

C(0) = D(x)

a

0,

is an

_

0.

matrix

r x q

It follows that we can

an

B

matrix whose

M(0, 0)

and A(y)

of functions on IRq

with

q x r

r x q

matrix.

Keeping

g(x) = fF(x) + C(x).O(x)

with

and hence

q x q

matrix for which

D(0) = 0. 169

This last relation we can re-write as the identity

r x r

matrix.

fF(x)

Clearly,

and it follows from (2.1) that

g, fF

It follows that the assignation F

=

(I - D(x)).g(x)

I - D(x)

-+

fF

induces a mapping from A-

sive.

x ]Rm, 0)

-+

A sufficiently small representative of

f

the zero set

Vf

of codimension

= q.

f-1(0)

We wish to study

Vf

(]Rs

Suppose we have

which is submer-

(]Rq, 0)

will be a submersion, so

f will be a smooth submanifold of

of

We take

of the restriction to

:

lTf

:

is

are g-equivalent, hence -7,,"-equivalent.

this mapping in detail, and to do this we need a new idea. f

I

is an invertible matrix,

equivalence classes of germs to X-equivalence classes.

an s-parameter deformation

where

(Vf, 0)

of the projection

-,

v

(IRs, 0) :

]Rs

]Rs x ]Rm

to be the germ at

x ]Rm -> ]Rq.

0

We need

the following fact.

(4.3)

Let

f, g

:

(]Rs

x ]Rm, 0) -+ (]Rq, 0) be 9l-versal s-parameter

deformations of germs f0, g0

are X -equivalent then

g0

Proof

:

(]Rm, 0) -a (]Rq, 0)

1Tf, lTg

of rank 0:

if f0,

are A-equivalent.

It is convenient to make the preliminary observation that a

'-transversal deformation (and hence any X-versal deformation) of a germ of rank 0 is automatically submersive, so that the notation introduced above makes sense in the present context.

Observe also that

f, g

must be .Y-

isomorphic deformations, in view of (3.3).

Step I

f, g

Now consider first the case when f0 = g0.

are X-isomorphic deformations is expressed by the existence of a

commuting diagram of germs

170

The fact that

' (]RS x ]Rm, 0)

(IRS x ]R', 0)

Tr

IT

h

with 4), h

invertible for which (f o 4)#( ,,//q) induces a mapping from

relation ensures that

g' ( 1q)

=

Vg

onto

This last

. Vf,

yielding a

commuting diagram of germs

(Vg, 0)

'(Vg, 0)

Trg

iTf

h

(IRS, 0)

expressing the fact that

(]R5

0)

are A-equivalent.

frf, v

g

Consider next the general case when

Step 2

in

:

(]Rm, 0) -+ (]Rm, 0),

9m,

for which

deformation

g0(x)

g'(u, x)

=

follows from Step 1 that

fore to show that Irf, ?rg, will map

are just .-

By (2.1) that means that there exists an invertible germ

equivalent. h

f0, g0

Vg,

and an invertible

M(x).f(u, h(x)) Trg,

g0

of

is

are A-equivalent.

Trg,

M(x)

with entries

Evidently the s-parameter

M(x).f0(h(x)).

=

matrix

q x q

_

-versal as well.

And it suffices there-

that however is clear as

are A-equivalent:

It

1 x k

onto V.

11

The relevance of these ideas to the material of the present section is as follows.

Let

F

:

(]Rr

]If, 0)

-*

(]Rr x ]Rq, 0)

be an r-parameter un-

folding of a germ fF of rank 0, given by a formula F(u, x) = ( To

F we associate the germ

DF

:

(]Rr x

x

,

f(u, x)).

0) - (]Rq, 0) given by 171

-. w + f(u, x):

(u, w, x)

deformation of just graph

necting

f,

F, DF

F

(4.4)

fF.

thus

DF

is an (r + q)-parameter submersive

The geometric connexion between

F, DF

is that

VD

is

can be identified with

f.

F The basic theorem con-

is an A-stable germ if and only if

DF

is a - -versal deforma-

and

lTD

F

is

tion of fF. The proof of (4,4) does not use techniques lying outside the scope of this However a careful version would occupy more space than is available.

volume.

We shall therefore content ourselves with the statement of the result, and concentrate rather on showing the reader how one uses it to obtain explicit A sketch of the proof, sufficient

lists of stable germs in given dimensions.

for the competent reader, can be found in the paper of Jean Martinet mentioned in the Introduction to this book.

It is perhaps worthwhile spelling out the

fact that (4.4) together with the .'e'-Versality Theorem allows one to produce

explicit examples of stable germs.

Example 1

of the germ By the

We saw in Example I of §3 that a .'o-transversal deformation

is the germ given by

(x2, y2)

(x2 + u1y - w1, y2 + u2x - w2).

j -Versality Theorem this deformation is 5e-versal.

precisely the deformation associated to the unfolding y2 + u2x)

It is however

(u1, u2,

x2 + u1y,

so this germ must be stable.

We can now return to the main theme of this section by stating

(4.5)

Let F, G

:

(]Rr x ]Rm, 0)

->

unfoldings of germs fF, fG of rank 0: F, G are A-equivalent.

172

(]Rrx ]Rq, 0) be stable r-parameter

if fF, fG are X -equivalent then

(4.5) is an immediate consequence of (4.3) and (4.4).

In more homely

language the burden of this result is that if we restrict our attention to stable germs

then the assignation

F

F

tive mapping from A-equivalence classes of germs.

fF

--*

actually induces an injec-

f germs to

Our objective now is to determine the image of this mapping.

To this end we make the following observation. F

(]Rr x ]Rm, 0)

:

fF

germ

:

s r + q:

fF

equivalence classes

, (Mr x ]R.q, 0)

If

is a stable r-parameter unfolding of a

of rank 0 then fF has

(]Rm, 0) -+ (]Rq, 0)

.

'-codimension

indeed (4.4) tells us that the (r + q)-parameter deformation

is '-versal, hence X-transversal, so the '-codimension of

fF must be

,

,

the number of parameters,

i.e.

S = S(r, m, q)

With this in mind write

r + q.

.

set of A-equivalence classes of stable germs

=

(]Rr x ]Rm, 0) , K = K(r, in, q)

DF of

=

(]Rr x ]Rq, 0)

r.

rank

of

set of X-equivalence classes of germs

(]R; 0)

(3P, 0)

->

sion

r + q.



(]R, 0) of rank 0 and X-codimension

Certainly then, as in §3, we can construct an (r + q)-parameter

.- -transversal deformation of the form

deformation of

f0.

-w + f(u, x)

with f

This is precisely the deformation

the r-parameter unfolding

F

:

(]R r x ]Rm, 0)

->

DF

an r-parameter

associated to

(]Rr x ]Rq, 0)

given by 173

F(u, x)

(u, f(u, x)).

=

The -Y-Versality Theorem tells us that

5 -versal, and then (4.4) tells us that vation that

F

has rank

r

F

is stable.

DF

is

The trivial obser-

concludes the proof.

Thus the problem of classifying stable germs under the relation of A-equivalence reduces to the problem of classifying germs under the relation of J'-equivalence, up to a certain 5 -codimension. know, in principle, how to approach:

This latter problem we

it is just the analogue for map-germs

of the problem for function germs discussed at length in Chapter IV.

Indeed

the main function of Chapter IV is to provide the reader with a model on which to base his ideas for the problem now facing us.

At root we are trying to list

the matter are decidedly more complicated.

the germs

(]Rm, 0) _+ (]Rq, 0)

of type

However the mechanics of

Em

under the relation of ,'-equiva-

lence, at least up to a certain X-codimension.

But it may well happen that

it is simply too complicated to list all the possibilities which can occur, and in order to increase one's chances of obtaining complete lists it is necessary to restrict oneself to germs defined by finer invariants than the first order symbol

Thus the next aspect of the theory to which we

E1.

address our attention is the construction of such invariants.

§5.

Higher Order Singularity Sets

We have already seen that the symbol singular point. IRn -.

IR

of codimension E1

a1

is of type

En.)

In this section we shall

symbolism in a natural way to obtain finer in-

The underlying idea is maybe best understood by a detailed study

of an example.

174

is only a very crude invariant of a

(For instance any singular point of a smooth function

show how to extend the variants.

E1

Consider the smooth mapping ]R2 -, IR2 given (in complex numbers) by we consider the deformed map

Given a (small)

E > 0

z "

where the bar denotes complex conjugation.

z2

+ 2cz,

(x, y) H

is defined by

f

u

=

(u, v)

f

:

IR2

-+ IR2

z - z2.

given by

In real numbers

where

x2 - y2 + 2Ex

v

=

2xy - 2Ey

and has Jacobian matrix

(2x + 2E 2y

which has rank < 2 2

X

+ y

2

2

=

E

-2y 2x - 2E

when its determinant vanishes,

this then is its singular set.

:

i.e.

on the circle

The bifurcation set is soon

If we parametrize the singular set by putting

found as well.

x= E cos a

y=

E sine

then we obtain a parametrization of the bifurcation set in the form

u

=

E2 (cos 26 + 2cos e)

v

=

E2 (sin 2e

- 2 sine)

which is a standard representation of a tricuspidal hypocycloid, the curve traced by a fixed point on a circle rolling inside another circle of three times its radius.

175

In fact our circle set

Elf,

x2 + y2

=

E2

is precisely the first-order singularity

since clearly the Jacobian matrix cannot have rank

We are

0.

therefore unable to distinguish one point on the circle from another by just looking at the symbol

On the other hand there are three points on the

Z1.

circle (the complex cube roots of

E3

in fact) which very clearly need to be

distinguished from the others in that they map to the cusps on the hypocycloid.

A clue as to how we should distinguish these three points is obtained by further analysis.

Let us concentrate our attention on the way in which onto the hypocycloid.

maps the circle

f

fjz1f.

We look therefore at the restriction

compute the rank of the restriction at a point

(x, y)

on the circle.

Let us Recall

that the differential of the restriction is the restriction of the differential of

Now the tangent line to the circle

to the tangent line to the circle.

f

at the point vector.

(x, y)

is the line through the origin perpendicular to this

And a unit tangent vector will be

under the differential of

f

(x, y)

at

(-y/E,

The image of this

x/e).

will be obtained by applying the

Jacobian matrix to it, yielding the vector

Zx + 2E

-2y

-y/E

-2xy - Ey

Zy

2x - 2E

X/E

-y2 + x2 - EX

and it has rank

The differential of the restriction certainly has rank . 1;

zero only when this last vector vanishes, which happens precisely at the cube roots of

c3.

In other words our three points are distinguished precisely

by the fact that they are

E1

other points on the circle are

points for the restriction EO

:

176

]Rn -+ P.

whilst all

points for the restriction.

The next step in the theory becomes clear.

f

fIE1f,

Given a smooth mapping

we have the first-order singularity sets Zlf.

If these are

E1'3f = E3(flzlf).

submanifolds we can introduce second-order singularity sets

If these sets are submanifolds we can

And this process can be continued.

introduce third-order singularity sets E1'j'kf

=

Ek(ffEi'Jf).

And so on.

The sets obtained in this way are the higher order Thom singularity sets of f. As Thom observed when he introduced these sets there is an unsatisfactory element here in that the definitions only make sense as long as we continue to However, as we saw in Chapter II, there is a way out of

obtain submanifolds.

this difficulty submanifolds

Z

of the jet-space

for which the inverse images under

J1(n, p)

are precisely the required sets

j1f

and then for a generic mapping

Elf:

these sets will be smooth manifolds.

Thom proposed the problem of imitating

this procedure for the kth order singularity sets, manifolds

One defines

at least for the first-order singularity sets.

-

k

in the jet-space

E

J (n, p)

i.e.

of defining sub-

such that for a generic

k the pull-back under j f are smooth manifolds, and precisely 11,...,lk th f. The case k = 2 was solved by the k order singularity sets E

f

:

]Rn

-+

]Rp

H. Levine, but the general case waited till 1967 when it was solved by We cannot hope to give a full account of Boardman's solution in a

Boardman.

what we can do however is to describe the construction 1,...,ik is to be attached to the k-jet whereby one decides which symbol E

book of this nature:

of a given map-germ, since it is an

entirely practical one.

We start with an algebraic idea. in the algebra

s

let

f1, ..., fp

>.

1.

We define

ideal generated by all

A I

s x s

s

I

be a finitely generated ideal

be generators for

be a system of co-ordinates in

y1' , yn integer

dn,

Let

&n.

to be the ideal

I.

and let

Suppose one is given an I + I'

where

I'

is the

minors of the Jacobian matrix

177

aft

aft

ayI

ayn

aft ay1

The ideal

(5.1)

AsI

so obtained depends neither on the choice of gen-

erators, nor the choice of co-ordinates, tors for

I,

z1, ..., zn

and

i.e.

g1, ..., gq

if

are genera-

is a system of co-ordinates, then

cides with the ideal generated by

and the

I

s x s

E

I

coin-

minors of the Jacobian

matrix

ag1

ag1 8z1

azn 1

.....

\agg az1

ate/ azn

Proof

Clearly, it suffices to show that any

in psI.

Each

gi

s x s

lies

minor of

can be written as a linear combination of the

fk,

with

ag. coefficients in

fin.

Thus each

afk

2zl

j

bination of the az plus an element of

can be written as the same linear comI.

Using the multilinearity of the

J

determinant we see that any

by

178

I

and the

s x s

s x s

minor of

*=

lies in the ideal generated

minors of the Jacobian matrix

af

of I

..... az n

az1

az

.....

az1

Thus it will be enough to show that any

az azn

***

minor of

s x s

For this observe that the Chain Rule allows us to write each

lies in & I. as a linear

az J

combination of the

with coefficients in

,

The result now follows

&n

ayk

on appealing again to the multilinearity of the determinant.

In practice we shall work with the standard system of co-ordinates x1, ..., xn but it will be important for us to know that any system will do.

A I

Note that

when

I

=

s

C A nI C An-1I C

I

Let

Example I

L I

=

=

I

I

and the

<xk-1>.

And A

generated by

I

Let

Example 2 ated by

so

A1I

minors;

I,

=

and the

I

<xk> 1 x I

s

&1

C

...

with

I.

k > 1.

All

minors of the Jacobian matrix <xk>

for

=

=

<xy, x2 + y2>

A2I =

in

I

I x I

<x, y>.

clearly A2I

also that one has the inclusions of ideals

s > n,

is the ideal k-1 (10C

),

so

s > 2.

in

32.

is the ideal gener-

All

minors of the Jacobian matrix

is the ideal generated by

<x2, xy, y2>.

And

I

and the

A I = <xy, x2 + y2>

2 x 2

for

s

s>.3. 179

I C gn we shall adopt the notation

Given an ideal

s

0 I

and refer to the ideal

AII,

A2I,

dnI

An-s+1I as the successive Jacobian extensions of

In view of the sequence of inclusions above we have

I.

I Let us call

...,

=

I

=

...

AOI C All C A2I C

proper when

I = gn

(Note:

I A en.

C A nI

saying that if we take a finite set of generators for generator has constant term Jacobian extension of is proper.

I

Suppose

0.)

is the last ideal

0

1

I

then at least one

I

of successive critical Jacobian extension of

The critical

in the sequence

It has in turn a critical Jacobian extension A

In this way we obtain an ascending sequence

on.

*

is the same thing as

is proper.

I i

.

px1I,

A 20i1I,

and we say that

I.

And so

I.

2L

which

*

...

I

has

Boardman symbol

(i1,

Example 3

One checks easily that the Boardman symbol of the ideal

I

<xk>

=

(k - 1)

in

d2

in

g1

repeated

i2,

mentioned in Example I is Its.

(1,

1),

..., 1, 0, ...)

with

And the Boardman symbol of the ideal I = <xy, x2 (2, 0, 0, ...).

mentioned in Example 2 is

The Boardman symbol of a germ

+y

f

:

(]Rn, 0)

-+

(]RP,

0)

is defined to be

that of the ideal If generated by the components f1, ..., f p. (5.2)

invariant,

The Boardman symbol_ of a germ i.e.

(]Rn, 0)

-j

(]RR, 0)

is a contact

if two such germs are X -equivalent they have the same

Boardman symbol.

Proof 180

Suppose the germs are

f

with components

f1, ..., fp

and

2 >

fi, ..., f'.

with components

f'

Suppose

Step 1

f, f'

coincide, so the Boardman symbols of

If,

Step 2

Suppose

invertible germ

h

x1, ..., xn

Jacobian matrix of

yield a system of co-ordinates.

h

of

ft

If we

relative to the standard system

we get the same ideal as if we compute the relative to the system of co-ordinates

f1, ..., fp

Again it follows from (5.1) that

h1, ..., hn.

there exists an

i.e.

(]Rn, 0) - (IRn, 0) for which f o h, f' coincide.

:

compute the Jacobian matrix of of co-ordinates

If,

coincide, by (5.1).

are right-equivalent,

f, f'

his ..., hn

The components

By (2.1) the ideals

are g-equivalent.

f, f'

f, f'

have the same Boardman

symbol.

The required result is immediate from the two preceding steps.

Step 3

Now we can extend our definition.

is X -equivalent to a germ f0 symbol of

to be that of

f

:

f0.

Certainly any germ

(]Rn, 0) - (IRP, 0).

f :

(]Rr, x)-+(]R, y)

We define the Boardman

In view of (5.2) this definition is

unambiguous.

The first

(5.3)

f

k

integers in the Boardman symbol of a germ

(]Rn, x) - (IRp, y) depend only on the k-jet of

:

Clearly, we can suppose

Proof

x = 0,

f. Let

and let

(i1, i2, ...)

generated by the components

f1, ..., fp

the Boardman symbol of

It is evident, by induction on

i

Lsa k-1

I.

of

f,

be the ideal

y = 0.

k,

I

be

that an ideal

i

... A 1I

f1, ..., fp.

is generated by partial derivatives of order

. k

of

And whether this ideal is proper or not depends only on the

181

values of all these derivatives at

0:

thus

depends only on the k-jet

ik

f.

of

We return now to our objective of partitioning the jet-space

Given k

into kth order singularity sets. that a germ

f

(IRn, x) -+ (IRp, y)

:

Boardman symbol has the form

Jk(n, p)

i

('1,

integers i1, ..., ik we say ik,...,ik when its is of type E

.. ., ik;

...).

We define

E

11'"''lk

to

be the subset of the jet-space comprising those jets having a representative in view of (5.3) this definition is unambiguous.

germ of type

In the case

Example 4-

k = I

the above definition recovers the first-

order singularity sets studied in Chapter II. Consider a jet in

J1(n, p)

and target, and let Now

A3I

One sees this as follows.

having a representative germ with zero source

be the ideal generated by the components of the germ.

I

is generated by

I

and the minors of order

(n - s + 1)

of the

Jacobian matrix, and will be proper if and only if all the minors of order(n - s + 1)

are zero, S

it follows that rank precisely

A I s

-

i.e.

if and only if the Jacobian has kernel rank >.

will be critical if and only if the Jacobian has kernel which is the same thing as saying that the jet lies in

the first-order singularity set

Es,

as defined in Chapter II.

Before we turn to further examples we shall determine just when the kth k k in The answer order singularity set E J (n, p) is non-empty. is provided by

(5.4)

A necessary and sufficient condition for the set

i1,...,ik E

C Jk(n, p)

should be satisfied:

182

s:

to be non-empty is that the following conditions

n>i1i2>....>ik> 0

(i)

(ii)

i1

(iii)

if

Proof

n -p i1 = n - p

i1 = i2 _ ... = ik.

then

Note first that we need only concern ourselves with jets

having zero source and target.

Necessity

(i)

i1, i2, ..., 'k

and that

n > i1,

It is an immediate consequence of the definitions that

Take

we proceed as follows.

I

are all

to be the ideal in

components of a representative of some jet in Alt that

A1I lis generated by

...

To see that

> 0.

generated by the

8

Z

.

We can suppose

say, with

g1, ..., got

s

A A J... A 1I

Suppose

(n - s + 1)

is proper.

ii > ij+1

n - p. i

(iii)

immediately that

Sufficiency

i1 = n - p

If

i1 = i2 =

then

A 1I = Ap+1I = I,

and it follows

..

Suppose conditions (i), (ii) and (iii) are satisfied.

have to produce a germ say, which is of type

f Z

:

(]Rn, 0) -+ (]RR, 0)

with components

We

f1, ..., fp

We consider cases, leaving the computa-

tions as good exercises for the reader.

183

Case when

i1 = n - p

Case when

i1

In this case we choose

> n - p

f1 = x1, ..., fp = xp.

In this case the following choice will do.

(1 5 i 0.

for the germ of

21'j

f

By definition, we need to compute the

(IR2, 0) _, (IR2, 0)

which is X equivalent

An obvious choice for

f0

is given by

where

fuo(x, Y)

=

u(x + x0, Y + YO) - u(x0, YO)

vO(x, Y)

=

v(x + x0, Y + YO) - v(xO, YO)

is the ideal generated by

u0, v0

and

All

is the ideal generated

by

u0, v0

and the minors of order

2 The ideal

is generated by

I

of their Jacobian matrix

(3 - i)

au0

au-0

ax

ay

8v0

8v0

ax

ay

u0, v0

and the entries in the Jacobian, and au0

cannot be proper as two of its generators The ideal

A1I

is generated by

av0

ax

have constant term pL 0.

ay

,

and the determinant

u0, v0

D

of the above

Jacobian, and will be proper (hence critical) when the constant term 2

2

x0 + y0 - e

2

the germ of and the

in

D

vanishes:

has type E.

f

2 x 2

this then, as we saw before, is the set where The ideal

A1A1I is generated by

minors of their Jacobian:

u0, v0, D

a line or two of computation will

verify that the constant terms in these generators are

x0 + Y0 - E2;

And the ideal

A1A1I

vanish simultaneously,

2

Y0(2x0 + E);

2

Y0 - x0 + x0e.

will be critical exactly when the last three expressions i.e.

exactly at the three complex cube roots of

Thus the three exceptional points on the circle cisely by the fact that the germ of

f

21f

E3

are distinguished pre-

at these points has type

whereas at all other points on the circle it has type

E1'1'

E1,0

In Chapter II we proved that the first-order singularity sets

E1 C J1(n, p)

were smooth manifolds, and computed their codimensions.

We shall only state the much harder result of Boardman.

(5.5)

If the k

th

11$ order singularity set

E

...,lk

in

k

J (n, p)

is non-

empty then it is a smooth submanif old of codimension

185

(p - n + 11)µ(i1, ..., ik) - (11 - i2)µ(12, ..., ik) - ... - (ik-1 - ik)µ(ik) where

µ(i1, ..., ik)

denotes the number of sequences

(j

,

-

..., jk )

of

1

integers which satisfy the following conditions

j1 ' j2 ' ... % jk > 0

(1)

is 3 js

(ii)

Example 7

for all

In the case

dimension of

Z

i

in

J1

(3Rn ,

I

k = I

< s < k

and

we have

µ(i) = i

and hence the co-

which agrees with the

(p - n + i)i,

is

IRp)

j1 > 0.

formula we obtained in Chapter II.

Suppose

Example 8 µ(1, ...,

1)

=

k,

Clearly we have

E1'...$1

so the codimension of

Note that in the equidimensional case

(p - n + 1)k. just

k,

i1 = i2 = ... = ik = 1.

the number of repeated

In the case

Example 9

so the codimension of

El's

Jk(n, p)

p = n

will be

the answer is

1's.

k = 2 in

in

one has

J2(n, p)

µ(i, j)

=

i(j + 1) - J

-

1

2

'

is given by the formula

(p - n + i)i + 12{ (p - n + i) (2i - j + 1) - 2i + 2j] . k In view of (5.5) the singularity sets submanifolds of

Jk(n, p).

are called the Boardman

Z

The Thom Transversality Lemma, proved in §4 of

Chapter II, yields the following.

(5.6)

The set of all smooth mappings

f

:

IRn -+ IRp

for which

i1,...,ik transverse to all the Boardman submanifolds

C O° (IRn, RP) .

186

Z

is dense in

jkf

is

We shall call a smooth mapping f : ]Rn - ]Rp generic in the sense of k i1,...,zk Boardman when j f is transverse to all the Boardman submanifolds E , for every integer k 3 1.

For such a mapping the set

1 ,...,1 1

E

i ,...,i

kf

(jkf)-1(E

1

k)

=

i1, ..., ik

n will be a smooth submanifold of

having the same codimension as

]R

E

.

Boardman showed that

Let

(5.7)

f

k'1k+1f

]Rn - ]Rp

be generic in the sense of Boardman:

then

Ek+1(fiEkf)

=

In other words we have the following.

Any smooth mapping

f

:

]Rn -* ]Rp

can be forced to be generic in the sense of Boardman by an arbitrarily small i 1,..., i

perturbation:

moreover, for such a mapping the sets

k f

E

manifolds and coincide precisely with the Thom singularity sets.

are smooth

Note inci-

dentally one trivial consequence of (5.7), namely that

E11

Let

Example 10

f 2) f

:

2

E11

E11.41 2 13

f 2) ]R3

]R3 ->

be generic in the sense of Boardman. By Example 7 the codimension

We ask which Thom singularity sets can occur. of

J1(3, 3)

in

E1

clearly then

Elf

is

E1,1,1f

1, 2

hence

with codimension

ity set which can occur. dimensions

i2,

Elf

f 2 ...

Elf I

has codimension

2, 3

splits into

21'0 f and E1'1f with co-

respectively;

k,

so will not appear for

E1,1,0

f and

no further splitting can

take place since the kth order Thom singularity set sion

in X23:

is the only first-order Thom singular-

respectively, and E1'1f splits into

with codimensions

i2

E1' " '1f

has codimen-

k > 4.

187

A pleasant illustration is provided by the dovetail mapping

Example 11

f

: ]R3 -> ]R3

(x, y, z) H (u, v, w)

given by

u=x

v=y

w=z

where

-uz -vz2.

The reader will easily check that the possible Thom singularity sets

E1'" '1f are given by the equations below. aw as

E1f

0

aw az

E1,1f

=

0

Z= 0 E111,11f

:

1

Lw

a2w

and

=

E1,1,1f

aw 3

2 and

as

aZ

is the folded surface illustrated below:

E f

0

az2

=

0

and

=

0.

aZ

E1'1f is the fold curve, and

is the origin.

In succeeding sections it will be useful for us to know that the Boardman symbols

are invariant under unfolding in the following precise

E

sense.

(5.8)

Let

F

:

(]Rr

unfolding of the germ f Boardman symbol. 188

x ]Rn00) - (]Rr x ]RP, 0) be an r-parameter :

(]Rn, 0) - (]RR, 0): then f, F have the same

Proof

x1, ..., xn

Let

u1, ..., ur

those in ]R'.

u1, ..., ur,

f1, ..., fp

be the standard co-ordinates in ]Rn,

We take where

and

to be the germ whose components are

f'

f1, ..., fp

are the components of

f.

The proof proceeds in two steps.

We claim first that

Step I

Suppose

has Boardman symbol

f

i

A A

that the ideal

k,

on i

have the same Boardman symbol.

f, f'

(i1, i2, ...).

k-1 ...A 10

We shall show, by induction

is generated by

If,

and

u1, ....4 ur

i

A s A k-1 ... A O If.

Here we put

9n with a subset of

i0 = 0

where

n,

for convenience, and tacitly identify

n' = n + r.

It will follow immediately

A k

that

... A ° If,

is critical, so establishing the claim.

the assertion is trivial.

Suppose it holds for

When

Take

k.

J

k = 0

to be the

Alk Jacobian matrix of some fixed set of generators for to the standard co-ordinates

x1, ..., xn,

of the same set of generators augmented by

generated by

u1, ..., ur,

minors of order and the identity order

u1, ..., ur.

x1, ..., xn,

ordinates

(n' - s + 1) (n - s + 1 )

u1, ..., ur, (n - s + 1)

of P.

Now

of

010

If ,

k ... A 0 If'

and the

is the direct sum of

J'

J

coincides with the ideal generated by the minors

J'

of

J.

i.e.

k s Thus A A A k A

i

.. A 0 If,

J,

Step 2

We claim that

...

f', F

is generated by

1

it is generated by

i k ... AiOIf AA

valent.

...

s

A A

matrix, so the ideal generated by the minors of

the generators for of

relative to the co-

Consider the ideal k

If relative

to be the Jacobian matrix

u1, ..., ur

i

of order

J'

the generators for A

(n' - s + 1)

r x r

and

... 010

If, and the minors of order u1, ..., ur

and

are g-equivalent, and hence .gi'-equi-

It will then follow from (5.2) that

f', F have the same Board-man

symbol, which fact combined with Step 1 will clinch the result.

Since

F 189

unfolds

it has components

f

F1, ..., Fp

u1, ..., ur,

and for

I

s i 6 p

we have Fi(x1,

..., xn,

identically in

x1, ..., xn.

write each

=

by

F.

fi + Thus

u1, ..., ur.

=

fi(x1, ..., xn)

It follows from the Hadamard Lemma that we can where

Ci

u1, ..., ur,

generate the same ideal in

F1, ..., Fp f', F

i

0, ..., 0)

lies in the ideal in f1, ..., fp Sn

and

&n,

generated

u1, ..., ur,

so the corresponding germs

are P,-equivalent by (2.1).

The germ F

Example 12

(IRn, 0) -, (Mn, 0) with the components

:

F1, ..., Fn where

(1, isn-1)

xi

F.

n-1 F

_

xnn +1

n

x xl

+

i

n

i=I

is an unfolding of the germ f

(IR, 0) - (IR, 0) given by f(x) = xn+1

:

has type L1,...,1,0 with n repeated 1's using Example 3. Example 13

The germ F

:

(IRk, 0)

-

(IR , 0) with the components

F1, F2, F3, F4 where x1

F1

190

F2

=

x2

F3

=

x3x4

F4

=

x3 + x 4 + x1x3 + x2x4

2

so

is an unfolding of the germ f

f(x, y)

:

(IR2, 0)

(v, x2 + y2) so has type

=

-4

E2'0

(IR? 0)

given by

by Example 3.

Classifying Germs under JP-equivalence

§6.

In this section we shall consider the very simplest situations where it is possible to obtain explicit lists of germs under the relation of -Y-equivalence. As in Chapter IV the whole thing turns on the idea of "determinacy". a germ

g

f

(]Rn, 0) ,

:

(IRn, 0)

:

-+

(IRR, 0)

(IRR, 0) with jkf

We call

. -k-determined when any germ =

f.

jkg is X -equivalent to

By

analogy with (IV.3.1) one might reasonably expect the following result.

(6.1)

A sufficient condition for a germ

f

0) - (IRP, 0)

to be

-k-determined is that .,1V n+1. &n'p C Tf

more than

In fact the result is correct, and its proof turns out to be no a slightly complicated version of the proof of (IV.3.1): shall omit the proof. finitely

for that reason we f

By further analogy with Chapter IV we call

7,'-determined when it is

,5'-k-determined for some

k >.

follows immediately from (2.3) and (6.1) that a germ of finite

must be finitely -Y-determined.

1.

It

X-codimension

In fact that statement is the only appli-

cation we shall make of (6.1) in this book.

As a starting point, let us look again at the case of germs of functions, only this time under

-T'-equivalence rather than R-equivalence.

The first

step is the Splitting Lemma.

191

Let

(6.2)

f e (n be a germ of corank

c

and finite

f -codimension:

then f is X-equivalent to a germ g(x1P ..., xc)

with g

+

xc+1

-

...

-

xn

e,,o,13

The normal

The proof is exactly the same as that given in Chapter IV.

forms for germs of corank 0 are given by the Morse Lemma. of corank

then

one obtains almost exactly the same classification as before.

I

Let

(6.3) f

And for germs

f r =,//2n

have corank

I

and finite ,5'-codimension

is ,'Y'-equivalent to a germ of the form

xk+1

±

k >

1:

x2 + ... + xn 2

Again, the proof is exactly the same as that of the corresponding result in Chapter IV.

One can continue in this way, just as we did in Chapter IV,

and it turns out that up to a certain point the two classifications are more or less identical, but then they begin to diverge.

;Ye shall not pursue the

point further.

Let us now turn our attention to germs of smooth mappings, as opposed to germs of smooth functions.

we are now in a much more complicated situation,

and can only hope to obtain results in very special cases.

It is in just

this situation that the Boardman symbol of a germ proves to be useful, in that it allows us to distinguish special cases.

The first fact one ought

to be aware of is

(6.4)

Let

f

:

(IRn, 0 ) -- (]RP, 0)

its Boardman symbol must have the form integer

192

k >

1.

be a germ of finite -codimension: (i1,

..., ik, 01 0, ...)

for some

Proof

As we have remarked already

so X-equivalent to

g

:

(7Rn, 0)

given by a polynomial of degree Boardman symbol of

(IRR, 0)

-+

p

>>

E

1 .

2

is the case when

and

has the same form.

f

(]Rn, 0) - (IRP, 0)

The first order Boardman symbol is

n = 1.

Now suppose the germ has finite

We ask for the classification up to ,5

1

Let f

(6.5) .

Clearly, the

with Z0

or

In the former case the germ is non-singular and a normal form is pro-

vided by (I.1.4). E

case to study for germs

k 3 1.

..., uk, 0, ..., 0),

(u1,

since the Boardman symbol is a .5r-invariant, that of

Perhaps the simplest

is

each of whose components

for some integer

. k,

has the required form

g

must be finitely X -determined

f

(IR, 0)

:

Then

-codimension.

equivalence.

(]RP, 0) be a germ of type Z1,

-+

is necessarily of type

f

, '-codimension, and type

E(with

k

and in that case isX-equivalent to

k 3 1,

repetitions) for some integer

and finite

k+1).

(0, 0,

..., 0,

Proof

That

f

has the type indicated is an immediate consequence

Next,-we claim that

of (5.1+) and (6.1+).

f

E1'...'1'0

(with

k

Indeed by induction on

k

has type

repetitions) if and only if the following conditions are satisfied. (i)

ajf

ajf.

=

1 (0) axj

...,

0,

a

(0)

=

0

(j

k)

aJf. (ii)

some

0

(0)

axj

where

f1, ..., fp

denote the components of

one readily checks that ajfl tives

with

Al ...

1 , i s p,

AIIf 1

f.

is generated by

5 j 5 k.

If,

and the deriva-

The claim follows immediately.

ox" 193

In view of (IV.2.4) the conditions (i) and (ii) are equivalent to saying that If

=

k+1

<X

and then (2.1) tells us that this is the same as

g-equivalent to

So far so good. p f

..., xk+1).

(0, 0,

being

f

The result follows.

Let us press on further.

If we insist on both

n

and

being > 2, then the next simplest case to study is that of germs (IR2, 0)

:

E0

E1,

(]R2, 0).

-.

Here the possible first-order Boardman symbols are

In the first case the germ is non-singular, and a normal form

E2.

is provided by (1.1.3).

Let

(6.6)

f

:

For germs of type

(]R2, 0)

Then

'-codimension.

f

.

E1

we have the following result.

be a germ of type

(IR2, 0)

and finite

E1,

is necessarily of type E(with k

repetitions) for some integer k 3 1,

and in that case is .9i-equivalent to

k+1

(x, y

).

Proof f

As in the proof of the preceding proposition, the fact that

has the type indicated is an immediate consequence of (5.4) and (6.4).

As

f

has rank

I

we know from (4.1) that

parameter unfolding of a germ f0 21,...,1,0

(5.8) also has type

then X -equivalent to J '-equivalent to

:

(IR, 0) -

(with

(x, f0(y))

yk+1,

and hence

f

k

can be assumed to be a 1-

of rank 0, which by

(IR, 0)

repetitions).

And clearly

But (6.5) tells us that f

We have still to treat germs (IR2, 0)

is - -equivalent to

(]R2, 0)

f

f0(y)

(x)

is

is

yk+1

At this

of type E2.

point we are entering a decidedly more complex situation, and must proceed cautiously. E2,2

in order of increasing degeneracy.

plete result

194

The possible second-order Boardman symbols are

(due to J. Mather).

2210,

E2,1

For the first we can obtain a com-

(6.7)

Any germ

f

:

-

(]R2, 0)

(]R2, 0)

of type E2'0 and finite

,

'-

codimension is -Y-equivalent to one of the germs listed below.

Iab,

b

IIa,b

(xy, xa - y

IVa

(x2 + y2, xa)

Note

these germs.

b3a>2

(xy,xa+y)

b t a 3 2,

)

even

a 3 3

We have kept to Mather's notation

Ia b,

IIa b,

IVa

for

His list, which was for a more general situation, included in

addition certain germs denoted

IIIa ,

Proof

a

b'

Va'

The first thing to note is that the components of

no linear terms, so the 2-jet of

f

f

have

can be thought of as a pair of binary

quadratic forms

(a1x2 + 2b1xy + aIy2,

a x 2

2

+ 2b2xy + c2y2)

In Chapter III we saw that by applying linear changes of co-ordinates at source and target we can suppose that the 2-jet has one of the normal forms

written out in the table below. second-order Boardman symbol.

Beside each normal form we have written the The computations are very easy, and left to

the reader.

195

Boardman symbol

normal form

for pencil

E2,0

y2)

(xY, x2 + (xy,

x

E2,0

2

E2,0 (xy, 0) (x2 + y2,

E2,0

0)

(x2

0)

(0,

0)

E2,1

E2,2

Since the Boardman symbol is a Y -invariant we can discard the last two normal forms in the table.

xy

2-jet

The

xy

or

to

x2 + y2.

xy.

will have

f

We consider these cases separately.

The first component of

case

of corank

Certainly then the first component of

f

is a smooth germ

(]R2, 0) -i (7R, 0)

It follows from the Morse Lemma that this germ is R-equivalent

0.

Applying the same change of co-ordinates to (xy, c(x, y))

.5 -equivalent to a germ

is of finite

have supposed

f

and therefore

C(x, y)

where

C

f

we see that it is

has no linear terms.

, '-codimension, so finitely

can be supposed to be a polynomial.

We

-determined, It follows from

(2.1) that we shall not change the . 5'--gquivalence class by subtracting from a multiple of

xy

(xy, a(x) + Q(y))

in

92:

thus we can suppose our germ has the form

with a, (3 polynomials.

At this point a couple of remarks are in order. has order

a >>

2

(i.e.

xa

is the lowest power of

Suppose

196

becomes

+ xa.

And similarly, if R

0

a

0,

and

x which appears in a):

then certainly we can produce a change of co-ordinates a

5

x F-

and has order

X under which b > 2

we can

find a change of co-ordinates x. X

here that

under which

generate the same ideal in 91.

the same ideal in

&1,

becomes

(3

+ yb.

y, Y

and likewise

Note generate

Now we consider the possibilities regarding a, p.

are both zero

a, /3

y - Y

This case yields the germ

which is not

(xy, 0),

of finite .Z'codimension, so can be discarded.

Just one of

is zero

a, 0

apply in the case Q

Suppose

(A similar argument will

0.

a ,

x H X, y - y

A change of co-ordinates

0.)

brings the germ to the form

Xy, xy generate the

and since

+ xa),

(Xy,

same ideal this is X-equivalent by (2.1) to

(xy, xa).

But

(xy, xa)

is

also not of finite . -codimension, so this case can likewise be discarded.

A change of co-ordinates

are both non-zero

a, $

brings the germ to the form

(XY, + xa + y

point that we are at liberty to suppose either component by

-1.

Also, if

by a change of co-ordinates

xa

I

and since

a , b;

ab

xy, XY

generate the

(xy, + xa + yb).

Note at this

and of course we can multiply

is odd we can change the sign in front

a

x H -x,

is odd we can change the sign in front of a germ of type

)

X-equivalent by (2.1) to

same ideal this is

of

b

y H Y

x H X,

yb.

y H y:

and similarly if

b

Thus in every case we obtain

save when the signs are different, and

a, b

are both

s

even, and in that case we obtain a germ of type

The

x2 + y2

the

xy

where

S

f

is

T,-equivalent to a germ (x2 + y2, C(x, y))

is a polynomial with no terms of degree s 2.

change the .-equivalence class by subtracting from in f

92 :

b.

Following the same initial reasoning used in

Case

case we see that

IIa

in particular we can suppose

has the form

(x2 + y2,

C

a(x) + y/3(x))

C

By (2.1) we do not

a multiple of

has no terms with factor with

a, (3 polynomials.

x2 + y2

y2,

i.e.

Note that

197

the second component cannot be identically zero, as

5-codimension. order term is

Denote by

a > 3

pxq + gyxa-1,

we can suppose

p / 0,

the order of

(x2 + y2, 0)

a(x) + YP(x),

so the lowest

p, q ' 0.

say, with at least one of

q = 0.

has infinite

I claim

To this end consider a linear change of co-

ordinates

X= x cos 0+ y sin 6 y

Notice that ulo

x2 + y2

X2 + Y2,

=

=

-x sin e + y cos B

+ Y.

X2

Using the fact that

Y23 =(-1)jX2j

mod-

a straightforward computation yields

gyxa-1

pxa +

=

QYXa-1

PXa +

modulo

X2 + Y2

where

P

=

p cos ae - q sin ae

Q

=

p sin ae + q cos ae

and the claim follows an observing that we can choose P / 0,

Q = 0.

It follows that our germ is

which is the desired normal form of type

g

is such a way that

9l-equivalent to

(x2

+Y2 ,

xa),

IVa.

Of course one could go further, and take up the next case of germs of type E2,1

and finite ,Y-codimension.

At the time of writing no complete list is

available, but one could certainly work with increasing codimension and gradually generate a list.

However the reader can probaoly see for himself by

now that such computations will become increasingly complicated and uninteresting.

And that is as far as we shall pursue the problem of listing germs

under the relation of

The next step in our programme is to

spell out just how all this enables us to list some of the simplest types of stable germs under A-equivalence. 198

Some Examples of Classifying Stable Germs

§7.

We are now in a. position to write out explicit lists of stable germs (IRn, 0) -+

(IR',

Boardman symbol.

under certain restrictions on the dimensions and the

0)

Let us start with the case

situation is that of a non-singular germ.

n s p.

Such a germ is necessarily of type

automatically stable, and has normal form

E0,

x

x , ...

nj,

0

by

0)

We can reasonably expect the next simplest case to be stable germs

(1.1.4).

of type E.

Here we have a complete result, due to B. Morin.

Let n 5 p,

(7.1) of type

The simplest possible

Then

E1.

F

and let

F

(IRP, 0)

be a stable germ

is necessarily of type E(with k

titions) for some integer in that case

(IRn, 0) -

F :

with

k

q = p - n + 1.

And

(IRn, 0) - (IRP, 0)

with

5 k . n/q where

1

G

is A-equivalent to the germ

:

repe

components

1 G1

=

(1 s i 5 n - 1)

u. k

G

n+i

jk-1

Gp

i

(0

uik+jx3

=

q - 2)

=1

j

=

u(q-1)k+jx

+ x

k+1

j=1

where we write

Proof (6.4).

u1,

... J,

un-1'

The initial statement follows immediately from (5.4) and

The theory of §4 tells us that

parameter unfolding of a germ sion, and also of type f

is

for the standard co-ordinates on

x

-equivalent to

f

:

E1'...,1,0 (0,

F

(IR, 0)

must be A-equivalent to an (n-I)->

by (5.8).

..., 0, xk+1),

(Iaq, 0)

of finite 5

codimen-

In view of (6.5) we know that

of 1'-codimension

qk + q - I

by 199

The theory now tells us that

Example 7 in §2.

The deformation in question was computed in Example 2 of

xk+1).

..., 0,

is A-equivalent to the

, -versal deformation of the germ

stable germ associated to a (0,

F

and the germ written out above is clearly the associated stable germ.

§3,

Finally, the theory tells us that we need only consider those

the X-codimension

qk + q - 1

is

5 p,

for which

i.e.

k

for which

k < n/q. 1-1

It is probably worthwhile isolating the equidimensional case of this result, the case

i.e.

Let

(7.2)

F

F

:

1

: k E n,

(]Etn, 0)

:

-+

(]R1', 0)

(]Rn, 0)

-'

E1' " .'1'0

must be of type

with G

n = p.

(with

k

be a stable germ of type

E1.

repetitions) for some integer

k

and in that case is A-equivalent to the germ

(itn' 0) with components

(1 4i 4n-1) k -1 Gn

uxi +

=

xk+1

.

j=1

Of course, this is the germ which we have previously dubbed the "generalized" Whitney mapping, and which in the special case

n = 2, k = 2

yields the Whitney cusp mapping of the plane.

Let us now reverse the emphasis by taking up the case of stable germs n

(]R '0)

-+

(IR , 0)

with

n 3 p.

is that of a non-singular germ.

200

Again, the simplest possible situation Such a germ is necessarily of type

En-p

automatically stable, and has normal form

(x1,

..., x

by

)

(I.1.3).

We

p

can reasonably expect the next simplest case to be that of stable germs of type

n-p+1

The starting point here, as always, is that the theory of §1+

E

tells us that such a germ is A-equivalent to a (p - 1)-parameter unfolding of

a germ

f

(lip, 0)

:

degree of complication in

singular).

Let

1

f

:

Then

OmIf

' , axm

Notice that

c.

f

Indeed that is the case.

(]Rm, 0)

- OR, 0)

has type

be a germ of type

if and only if

Em'c

We start with the condition for

Proof The ideal

depends at root on its corank

It is therefore a good guess that the two invariants will

f.

be closely connected.

(7.3)

in = n - p + 1,

where

and the second order Boardman symbol,-both depend solely on

c,

the 2-jet of

f

n-p+1.

Our experience in Chapter IV tells us that the

and of finite X-codimension.

the corank

also of type

(7R, 0)

-+

is generated by

f

f

f

Em

i.e.

has corank

to be of type

c.

Em,c

and the partial derivatives

and is to be critical.

The ideal

the same list, together with the minors of order

AsAmIf

(m - s + 1)

is generated by of their

Jacobian of

...

ax1

a2f ...

of

axm

a2f axloxm

ax2

a2f

a2f ...

ax2

axin ax 1

201

The condition for the ideal

m

s

A A I

should have zero constant term, (m - s + 1)

should vanish at

row, since all the

to be proper is that all its generators

f

i.e.

In this matrix we can disregard the first

0.

vanish at

ax

that all the minors of order

so are left precisely with the Hessian

0,

J

matrix of

sdmIf

Thus the condition for the ideal

f.

the Hessian should have corank

and the condition for

> s,

critical is that the Hessian should have corank exactly we see that

f

has type

Em'c

if and only if

The simplest situation is when

to be proper is that

f

f

AsAmIf Taking

s.

has corank

has corank

0.

(IRn, 0)

(IRP, 0)

to be s = c

c.

This yields another

result of B. Morin. Let n > p,

(7.4) of type

En-p+1,0

:

and let

then

F

F

:

->

be a stable germ

is A-equivalent to a germ

(]Rn, 0)

(Pp, 0)

>

given by

(1 s i s p- I)

U.

Gp

Proof

=

2

+ xP +

...

+

xn-1

As was pointed out above,

(p - 1)-parameter unfolding of a germ f and finite k-codimension, where

2

+ xn

F :

is certainly A-equivalent to a

(IRm, 0) - (IR, 0)

m = n - p + 1.

By (7.3)

f

of type

Em'0

has corank

0,

so by the Morse Lemma is R-equivalent, hence X -equivalent, to a germ

+ x2

p

+

...

+

2 xn.

The theory of §4 now tells us that

F

is A-equivalent

to the stable germ associated to a p-parameter , -versal deformation of this germ.

It is a trivial computation to verify that the germ written out above

is the stable germ in question.

202

The next simplest case is when

has corank 1, giving rise to the foll-

f

owing result, also due to B. Morin.

Let

(7.5)

En-p+1,1

of type

k

(with

F

case

n

p,

and let

Then

F

F

:

(IRn, 0)

-+

(]RP, 0)

En-p+1,1,...,1,0

is necessarily of type

repetitions) for some integer k with is A-equivalent to a germ

G

:

(]Rn, 0)

be a stable germ

s k s q:

1

(]RR, 0)

-+

and in that given by

k 2

_ ±xp2

+ ... + xn-1 +

-

xn+2 + )

u in

J=1 Proof

The initial statement follows immediately from (5.1+) and

The rest of the proof follows exactly the same lines as that of

(6.4).

(7.4), save that this time by (6.3).

f

is X-equivalent to

+2

+ xP + ... + xn -1 +

The computation of the deformation is very straightforward, and

can safely be left to the reader.

Of course, one could push these techniques further, by systematically classifying germs of functions with increasing ,5-dimension.

However the

point of the last two results has already been made in that we have indicated the simplest complete results which can be obtained. in the equidimensional case

n = p

the last two results yield (7.2).

the next case to study is that of

E2

Let

Having dealt with germs of type

us pursue the equidimensional case further. E1,

Note incidentally that

germs.

In view of the discussion

of §6 it is clear that we can only expect a complete result in the case of E2'0

germs.

The result is due to J. Mather.

203

Let

(7.6)

F

:

(Mn, 0) - (]Rn, 0)

be a stable germ of type

then F is A-equivalent to one of the following germs G :

Types

a-l+j

E2'0:

(]R", 0) - (:u

U.

(1 5 i < a-1)

=

vj

(I s j

=

xy

=

xa ± yb + a-1 E ui x

0).



F

F

is A-equivalent to an (n -2)-

(7R2, 0)

By (6.7) this germ is .

And then

vixj-1y

of type

E2'0

and

-equivalent to one of the

will be A-equivalent to the stable

germ associated to a 9F-versal deformation of these germs.

The deformations

were computed in Examples 3, 4 of §3, and the corresponding stable germs are evidently the germs written out above.

With these results we have gone as far as is possible in a book of this nature, and have achieved our objective of indicating how one goes about classifying singular points of smooth mappings.

We shall however pursue

the matter a little further by looking briefly at certain mappings whose singular points are necessarily stable. 204

Singular Points of Stable Mappings

§8.

We have so far discussed only stable germs of smooth mappings

f

]Rn - ]RP.

:

But the underlying ideas can be introduced equally well for mappings themselves.

By analogy with germs there is a natural notion of "equivalence"

for smooth mappings; e

when there exist diffeomorphisms

,.lent

-

namely, we call two smooth mappings fV f2 g, h

:

]Rn -,]Rp

for which the following

diagram commutes

Now ]Rn,

So the pair

(g, h)

Diff(]Rn) x Diff(]RP) to be

(g, h).f

Diff(]R n)

is an element of the group

h

and

]RP.

is an element of the group

g

Diff(]RP)

of all diffeomorphisms of of all diffeomorphisms of

is an element of the product group

and this group acts on g-1.

h o f o

cf°(]Rn, ]RP)

Of course this action lies outside the frame-

work discussed in Chapter III because neither the group nor the set

C°O(]R', ]RP)

if we define

Diff(]Rn) x Diff(]RP)

are in any way finite-dimensional.

But for all

that there is no harm in proceeding by analogy, just as we did with germs. Recall that in Chapter III we introduced the equivalent notions of "stability" and "infinitesimal stability, the former being geometric in nature, and the latter algebraic.

When dealing with germs we interpreted stability in terms

of unfoldings, but for mappings it will be easier if we stick to the geometric idea.

We should therefore call

close"

g

:

f

: e-+ ]RP "stable" when all "sufficiently

]Rn - ]gP are equivalent to

f.

On this basis we introduce the

following formal definition.

205

A smooth mapping

f

IRn - ]Rp is stable when there exists a real number

:

e > 0

such that every smooth mapping

of

is equivalent to

f

g

is the e-neighbourhood

7Rn -+ ]Rp

:

Our first result gives us some idea of just how

f.

nice stable mappings are.

(8.1)

Any stable mapping

f

]Rn -r IR13

:

is generic in the sense of

Boardman. Proof

Let

k >>

I

be an integer.

transverse to all the Boardman submanifolds

We have to show that jkf is i1,...,ik To this end recall Z

from (5.6) that the set of all smooth mappings ]Rn - ]Rp is dense in

C O°(it , ]Rp):

thus, given any

with this property

we can find a

e > 0

]Rn - ]Rp in the e-neighbourhood of f for which jkg is transverse to i1, ...,ik And since f is stable we can choose a so small that all the E g

:

every map in the e-neighbourhood of to observe that if

f

ity condition, then f

is equivalent to

f

is equivalent to does as well

g,

and

g

It remains only

f.

satisfies a transversal-

which fact we leave as an exercise

-

7

for the reader.

(8.2)

Let

f

:

]Rn - ]Rp

be stable:

then the germ of

f

at any point

is stable.

If we now combine these results

We shall omit the proof of this result.

with the classifications obtained in the previous paragraphs we obtain some aesthetically very pleasing theorems describing completely all possible singu-

lar points of a stable mapping f

:

]Rn - IRp

for certain values of

Let us start by looking at the equidimensional case stable mapping

f

:

]Rn -, in.

first-order Boardman submanifolds

206

First of all, El

-

n = p.

n, p.

Consider a

f must be transverse to the

which by (5.1) in Chapter II have

codimension Elf

-

In particular, if

1 2.

n t 3

which have the same codimension

-

the first-order singularity sets

must be empty for

only possible singular points are those of type

f

(8.3)

In fact we can be more

E1.

k repetitions) has codimension

(with

thus the only singularity sets with

and the

Recall that in the equidimensional case the singularity set

explicit. E1'

i >> 2,

k 5 n.

Let

If we take

which can be non-void are those

Z

we obtain immediately from (7.2)

n = I

f :1R -+]R

k, by Example 8 of §5:

be a stable mapping:

its germ at any point is

equivalent to one of the following

E0

:

y1

=

E1,0

Taking the next case

x2

=

y

(regular)

x1

(simple minimum).

1

we recover a famous result of H. Whitney des-

n = 2

cribing the possible singularities of stable mappings from the plane to itself.

(8.4)

Let

f

]R2

:

-

be a stable mapping:

]R2

its germ at any point is

equivalent to one of the following

0 E

=

x2

y1

=

x1

Y2

_ =

x2

y1

=

x1

Y2

=

Y2

E1'0

E1,1,0

x1

y1

2

(regular)

(fold)

(cusp).

x2 + x1x2

207

In order to obtain geometric insight into these germs it is best to think of them respectively as composites

(

xI, x 2 )I--->

(x

(x Is

, ---) (x', x2 )

(xx2

F

)I

O

(x 1,

proj

x2, x2 )

x2, x2) x2, x

x2 )

;

(x 1, x2)

prod

+ x 1 x2 )

j

p Toj

(x

,

x

+ x 1 x2 )

2

2

where in each case the second mapping is (the restriction to the image of the first of) the projection

(x1, x2, x3)

ualize our maps as follows. x2

x1

A x2

xl

x2

x1

208

-*

(xV x3).

In this way we can vis-

Take for instance the situation of Example 1+ in §4 of Chapter II, namely the projection of a torus onto a plane.

Locally, such a mapping is from the

plane to itself, and generically such a mapping is stable shall have to ask our reader to accept points of the above three types.

-

-

a fact which we

so we should only see singular

Indeed this is the case.

Recall that the

set of critical values looks like this.

Here the curves which make up the picture are the images under the projection f

of the singularity sets

torus folding over.

E1'0 f:

at a point in such a set we simply see the

But the four exceptional points are the images of the

E1'1f: here we have the more complicated situation of two

singularity sets

folds coming together to form a "pleat".

We can of course gain one more result by taking the final case

(8.5)

Let

f

:

7R3

-

7R3

be a stable mapping:

n = 3.

its germ at any point is

equivalent to one of the following.

y1

x1

y2

=

x2

y3

=

x3

(regular)

209

E1,0

(fold)

x2

y2

2

y3

=

x3

E1,1,0

(cusp)

y3

1,1,1'0

x3 + x1x3

y1

=

x1

y2

=

x

y3

=

(dovetail)

2

X4

+ x 1 x 3 +

x2x3

Of course, when we go up to the next dimension n = 4 we no longer obtain only

As we have already pointed out

points.

E1

Elf

has oodimension

in the equidimensional case so when n = 4 we can'certainly have but avoid E2'2f

E1

points for

with codimensions

i 3 3.

4, 7, 10

three sets can be non-void. germs of type

Let

(8.6)

E2'0

f

:

Now

splits into

respectively:

points,

E2'0f, E2'1f,

thus only the first of these

Appealing to the classification of stable

given in §7 we obtain the following:

]R4 -* ]R4

be a stable mapping:

equivalent to one of the following.

{yi =

xi

EO

Y4 =

210

E2f

E2

i2

x4

its germ at any point is

(1 E i (3)

yi

f Y4 X.

yi

51,1,0

{

51,1,1, 0

Y4-

yi

{ Y4

xi

=

x4 + xx4 + x2x

f

(1 .

(p - n + 1)k >

3n - I

s

2p
3n - 1,

so

we ask which singular points of type

The condition

(p - n + 1)k 4 n,

E1,

3n - 4 < 2p < 4n.

Recall that the codimension of

(p - n + 1)k.

we

We should certainly

It is worthwhile analysing the situation a little further. have a pair

E1

But for

p < 2n.

2p > 3n - 4.

i.e.

(n, p)

has codimension

we do not wish our mapping to have singular points of type

i(p - n + i) > n for

n < p

admits only singular

If our mapping is to admit singular points of type

i(p - n + i).

i > 2

with

(n, p)

We can determine a useful range of such pairs

E1.

by a simple combinatoric exercise.

must have

Take for instance

p = 2n - 1

(8.7)

Let

f

:

]Rn -+ 1Rp

the metastable range:

then

f

be a stable mapping with the pair

(n, p)

only admits singular points of type

in

E1,0

and its germ at such a point is equivalent to the germ

(14isn-1) yn-1+i

-

(1 si.p - n)

xixn

2

yP

The case

n = 1

=

xn

of this result is simply (8.3).

something new, namely that a stable mapping lar points of type

E1,0

f

:

But when n = 2

7R2 - 7R3

we get

only admits singu-

and that the germ at such a point is equivalent to

the germ

whose image is the so-called Whitney umbrella depicted'below.

213

What about the case pairs

(n, p)

n 3 p?

Here again one could construct ranges of

for which a stable mapping

points of a very simple type.

]Rn

-+ RP assumes only singular

However the combinatorics are rather more com-

plex, and it is probably more illuminating to take specific pairs write out a complete list of possibilities.

(8.8)

Let

f

:

]R3 -+

One of the simplest cases is

be a stable mapping:

1R.2

(n, p) and

its germ at any point is

equivalent to one of the following.

y1

E1

'

32

x1

x2

=

x1

y1

E2'0 2

2

y2

=

+ x2 + x3

y1

_

x1

E2,1,0 Y2

Proof

In this situation Elf

be empty for

Now Elf

±

i > 3,

splits into

i.e.

x2 ± x3 + x1x3 2

has codimension

i(i - 1)

so must

we can only have singular points of type

E2'Of,

E2'1f,

E2'2f

with codimensions

E2.

2, 3, 5

respectively, so we need only consider the first two of these sets. singular point of type

E2'0

normal forms are provided by (7.4).

remaining possibilities are singular points of type repetitions):

this has codimension k + 2,

so we need only consider the cases (7.5).

214

k = 0, 1

E2,1,---,0

At a The only

(with

k

as an easy computation verifies,

with normal forms provided by

Appendix A

The theorem of Sard

The key idea here is borrowed from measure theory, and is that of a "null set"; this will provide our starting point.

in Mn when, given any real number

A set

duct of

n

V

e > 0,

union of cubes the sum of whose volumes is

V C ]Rn

is called a null set

is contained in a countable (A cube in ]Rn

< c.

is a pro-

open intervals, called its sides, and its volume is the product The reader will readily check that in using this defi-

of their lengths.)

nition one can assume that each cube has all its sides of equal length.

Note that when V

is compact the countable union of cubes can be assumed

finite.

The first fact we need is

(Al)

Any countable union V

of null sets

V.

in ]Rn

is again a null

set in ]an. Let

Proof

< e/2J:

of total volume

C J.,k

We can cover

e > 0.

of total volume

V.

by countably many cubes

Ci,k

hence we can cover V by the countably many cubes




are open sets in ]R5,

B

one has ]Rn

It

Let

Nn C ]Rs

be a

is called a null set in N when for every O(A 0 V)

respectively.

a null set in ]Rn:

of course A, B

Sard's Theorem is the following

proposition.

(A5)

Let

f : N - P

be a smooth mapping with

The set of critical values of

f

is a null set in

Nn, Pp

smooth manifolds.

P.

217

Proof

One proceeds by induction on

(It is a convention that IRO

that (A5) holds for

The case

n.

comprises a single point.)

and will deduce that it holds for

(n - 1),

N, P

One easily checks that

that its image under this mapping, namely

f

f(N).

is precisely

Proof

N x 0

n < p

Of course when n > p

is

given

is a null set in ]Rp, so

is a null set in ]Rp

f(N),

It remains only to observe that when

and one needs a more subtle argument: n > p,

n < p

Consider indeed the smooth mapping N x RP -n -+ P

(x, y) - f(x).

of

to be open

It is probably worthwhile pointing out that the case

relatively trivial.

(A3).

It fol-

n.

IRp.

Remark

by

is trivial.

We suppose therefore

lows immediately from the definitions that we can suppose

sets in ]Rn,

n = 0

by

the set of critical values

this reasoning breaks down

this shows that we may as well assume

so the critical points are precisely the singular points.

Write

(continued)

for the set of points

E

for the singular set of

f.

And write

Ek

x E N where all the partial derivatives of order

. k

Thus

are zero.

E 2 E 1 2 Z2 2.. We have to show that

f(E)

is a null set in IRp.

In view of (Al) it will

suffice to establish the following steps.

Step 1

:

f(E- E1)

Step 2

:

f(Ek

Step 3

:

f(Ek)

218

Ek+1)

is a null set in IR.p, for all k.

is a null set in IRp, for some k.

Note first that we can assume

Step 1

have

-

is a null set in ]Rp.

E = E1,

and there is nothing to prove.

p > 2

since when p = 1

we

By (A2) it suffices to show

that each point x image of

in

under the restriction flu

E - E1

in N for which the

has a neighbourhood U

E - E1

is a null set in ]Rp.

And

using (A3) it is clear that we can replace this restriction by any equivalent mapping, which we shall also denote

f.

f(x) = 0:

at

and since the rank of

f

In particular we can assume

x

is

>.

and

I

(by choosing linearly adapted local co-ordinates) that where

s

and each f

E ]R,

s

x = 0,

we can suppose

< p

f(s, x) = (

fs(x))

,

3n-1

is a smooth mapping from an open set in

to an open set in 3Rp-1.

V

We have to show that the set

of critical values of

Notice that the set of critical points of

]RP.

a countable union of compact sets:

of critical points under

With the notation of values of

fs,

f,

(Au-)

it follows that

is a closed set,

f

the image of the set

V,

observe that

Vs

is precisely the set of critical by the induction hypothesis for

V is a null set.

the theorem.

It follows from (A4) that

Step 2

By (A2) it suffices to show that each point

in N for which the image of

f

is a null set in ]RP.

f

is

/ 0,

axg- with

say

hence

is likewise a countable union of compact sets.

which is a null set in ]RP-1,

has a neighbourhood U

is a null set in

f

At x g

x in

U (1 (Ek

E1 - Ek+1

Ek+1)

-

under

(k + 1) th order partial derivative of

some

a kth order partial derivative of

f,

so there

J

certainly exists a neighbourhood set defined by

Ek C G :

g = 0

=

so its image under

U fl Ek

f

of

x whose intersection

is a smooth manifold of dimension

in particular x E G.

u n (Ek - Ek+1)

U

Clearly if

U

G

(n - 1).

with the Note that

is small enough

is contained in the set of critical points of fIG,

is a null set in ]Rp,

by the induction hypothesis of

the theorem.

Step 3

Let

a null set in ]Rp.

k be an integer

> n - 1.

We shall show that

By (A2) it is enough to show that

f(C r Ek)

f(Ek) is is a null 219

set in ]Rp tained in

ber

for any cube

in ]Rn,

C

for which If(x) - f(y)

divide

x, y in C

say, whose closure is con-

c

As a consequence of Taylor's Theorem there exists a real num-

N.

K > 0

for all

of side

above inequality 2K(o/r)k+1,

Introduce an arbitrary integer

C rl Ek.

R = rn

into

equal cubes

C1, ..., CR

r > 1,

of side

f(C n) Ek)

is contained in the union of

and sub-

By the

c/r.

is contained in a cube in e of side

f(C n Ek)

and hence

KIx _ ylk+1

s

rn

2

cubes in

of total volume

]Rp

rn.(2K)p(c/r)p(k+1)

=

Note that n - pk - p < 0 as k > P - 1,

so that this last expression -- 0

It follows immediately that f(C nzk) is a null set in ]RP.

r -oo.

as

r(n-pk-p).

constant x

In order to establish the consequence of Sard's Theorem used in Chapter II to prove the Basic Transversality Lemma we need

Let

(A6)

lement of

V

be a smooth manifold and V C P

is dense in

P = ]RP.

P.

Suppose (A6) were false, so we could find a cube

is a subset of

V.

V C U CI

e > 0,

(C1)

Vol (C)

Vol (C.1 ) 1=1

220

C.

a finite subcover

-

whose closure Since

find a countable family of cubes

and with Y vol(C.)
g

g

Here, given

we write x as an abbrevia-

..., an)

(a1,

= >2 fax ,

of all formal power series

xR

we set

f.g

=

h where

= > fa.g

hY

NN

V. W an

Given real algebras

mapping with O(x.y)

=

p(x).O(y)

The natural mapping &n

(C1)

algebra homomorphism for all

-+

do

: V- W is a linear

x, y in V.

given by f

is an algebra

homomorphism.

Here we keep to the notation of Chapter IV:

Proof

n that

226

n

denote the unique maximal ideals in

(f.g)

d, n .

in particular

We have to show

= f.g; it suffices to show that these are equal modulo an

n+1,

element in

f = fk,

g

Then, modulo

for all

gk modulo .,ll A// n+1

,

k >. 0.

k+

wi th

Choose

fk

k.

gk

By (IV.2.li-) we can write

polynomials of degree < k.

we have

(f.g) bearing in mind that the Taylor series of a polynomial function is precisely that polynomial. 1-1

227

Appendix D

The Borel lemma

Our starting point is the explicit construction of smooth functions having very special properties.

There exists a smooth function

(D1)

that

0(t)

=

t

for

5 1/2

ItJ

and

:

(P(t)

]R -+]R with the property

=

for

0

Certainly there exists a smooth function

Proof 0 .< 0(t)

4,

1

for all

for which

t,

0(t) = 0

(0

-1/t2

The required function 0

1.

>.

Iti

B

:

]R -+]R with

precisely when

when

t 5 0

when

t > 0

t 5 0, namely

.

is obtained by setting

t6(1-t) 2 Y

(t)

=

0(1-t

2)

+ O(t2-

)

Now we come to the Borel Lemma itself, as it was stated in Chapter IV.

The natural algebra homomorphism do _, do

(D2)

given by f -+

is

surjective.

x

=

(x1,

..., x ) n

integers 3 0,

of real numbers, and a sequence

we write

a

a1

=

(a1,

..., an)

of

an

x as an abbreviation for x1...xn.

have to show is that given an element

228

Given a sequence

We adopt the following notation.

Proof

What we

x in n there exists an

in Sn

f

element

for which

standard co-ordinates

?

is the Taylor series of

x1, ..., xn

i.e.

a, where we write a: = a I .... an: , Daf

fa = I Daf(0)

alai

=

axiI

To this end let

0

:

>Rn -+ IRn

the formula Cx1, ..., xn) trary sequence limit

Ek

=

f

... axnn

(

to be the smooth mapping defined by

(xJ), ...,

of real numbers with

(Ek) 0,

=

and

be a smooth function having the properties

IR -+ IR.

listed in (D1) and take

relative to the

for all choices of

+ ... + an

= a1

la l

f

O(xn)). 0 < Ek
Dafk

converges

k=0 It then follows from standard theorems in

a.

calculus that the sum-function

in IRn,

(Ek)

f

is smooth, and that

229

CO

_ X Dafk(0)

Daf(0)

a:fa

=

k=0 just as we required.

By the usual comparison test it will suffice to choose 00

(Ek)

in such a way that

ving

a)

5'

supn 'Dafkl

k=0

which converges for any choice of

an upper bound on D fk.

is dominated by a series (involClearly, we need to produce

a.

First of all, note that (by sheer differentiation)

we have

Dafk(X)

Ek'aI fR''(a, 9, k, X)

=

k where

as

n

(a, f3' k, x)

=H j=1

As the function 0

a.

CC

ax.J

k)

vanishes outside a compact set we have a well-defined

bound

B(a, g, k)

=

II'(a, 9, k, x)J

sup XE]Rn

so setting

B(a, k)

fR B(a, 6, k)

=

I/I=k we see that

sup . in and hence that the series

I Dafkl

5

Ek Ia I B(a, k)

sup IDaffi

is dominated by

k=0 All we need to do now is to choose the sequence latter series converges for all choices of

230

a:

(Ek)

(a, k).

k=0 in such a way that the

and that can be done by

choosing

ek

in such a way that

e.

>

2k

7,

B(a, k).

231

Appendix E

Guide to further reading

This guide is addressed largely to those who have read at least part of the present book and wish to pursue their interests, but who do not have the advantage of expert advice.

Singularity theory, in common with most areas of

mathematics, has an extensive technical literature, much of which is not readily available.

In preparing this guide I decided only to quote those

sources which I know to be available and which, in my view, a student can hope to gain from reading.

I have made no attempt to compile a comprehensive list

of references, on the grounds that such a list would add little to the value of the book.

In particular, virtually all the mathematics in the last two

chapters has its genesis in papers of J. Mather, listed for instance in [LS1].

(I)

Material Directly Related to this Book

On a general level I feel that anyone starting off in this subject should look at the classic survey paper [Al] by Arnol'd.

[LSI] still provides a very

good reference, and gives one a fair idea of the state of the subject at the turn of the decade.

As far as more systematic texts are concerned, [L] com-

plements the material of this volume in several respects, whilst [GG] provides a fairly formal treatment of much of the global theory. Here are some more detailed suggestions. topology has several good expositions. excellent little volume [M1].

The beginnings of differential

First and foremost I recommend the

The reader who wishes to graduate to the

abstract idea of a smooth manifold will find good accounts in [GP] and [BJ].

These two volumes also contain rather more material on the subject of transversality, as does the more advanced [GG]. 232

The classification of function

germs has now been taken far beyond the beginning indicated in Chapter IV,

though good accounts of this fascinating area of the subject have yet to appear.

Certainly the next step in this direction would be the first few

pages of [A2], whilst some idea of how the subject develops can be found in [A3].

For a lucid exposition of the connexions with other areas of mathema-

tics try [A4].

The reader who wishes to fill in the gaps in Chapter V should

refer to [M], though it is not all easy going.

For this one certainly needs

to be familiar with the Preparation Theorem.

Here I recommend Wall's

"Introduction to the Preparation Theorem" in [LS1] followed by the appropriate chapter of [BL].

Further information on the classification of stable germs

can only be obtained by dipping into the research literature on the subject.

(II)

Material Not Directly Related to this Book

Beyond the differentiable theory of smooth mappings, where the changes of coordinates are diffeomorphisms, lies the important topological theory where the changes of co-ordinates are just homeomorphisms.

Some ideas concerning the

local theory can be found in the latter chapters of [L], whilst an exposition of part of the global theory can be found in [GWPL].

leans heavily on the idea of a "stratification":

The topological theory

unfortunately, there is as

yet no real introduction to the theory of stratifications, but some idea of what it is all about can be gleaned from the references just quoted.

Another

aspect of the topological theory is the subject matter of the classic [M2].

(III)

Applications

In Chapter IV we indicated how singularities of smooth mappings arose naturally in both differential geometry and algebraic geometry.

And it is prin-

cipally to these areas that one looks for applications of the theory within

233

mathematics.

As far as differential geometry is concerned I suggest looking

at [P], whilst the reader who wishes to see further ahead will find solid reading in [W].

The applications of singularity theory to algebraic geometry

are still at a rudimentary level, and no account of this is likely to appear

for some timne. The applications of singularity theory to the physical sciences have centred around the catastrophe theory of Thom.

Those with a serious interest in such

matters will want to look at Thom's own book [T].

Quick introductions to the

ideas can be found in [Z] and [S) while [SP] provides a comprehensive account of the state of the subject at the time of writing.

For the mathematics of

the matter, the account in [BL] extends the material in Chapter IV of the

present volume, whilst more appears in [ TZ] .

I recommend [LS2] for further

ideas relating singularity theory to problems in the physical sciences.

References [Al]

Arnol'd, V. I. Surveys

Singularities of Smooth Mappings.

(1968) 1 - 43.

Russian Math.

Translated from Uspehi Math. Nauk 23 (1968)

3 - 44. [A2]

Arnol'd, V. I. Points, larities.

[A3]

Normal Forms for Functions near Degenerate Critical

the Weyl Groups of

Ak, Dk

and

F.k,

Functional Anal. Appl. 6, 254-272,

Arnol'd, V. I.

and Lagrangian singu1972.

Normal Forms of Functions in Neighbourhoods of

Degenerate Critical Points.

Russian Math. Surveys

29,

10-50,

1974.

[Al

Arnol'd, V. I.

Critical Points of Smooth Functions.

ernat. Congr. Math.

234

Vancouver

1974,

19-39.

Proc. Int-

[BL]

BrBcker, T. & Lander, L.

Differential Germs and Catastrophes.

London Mathematical Society Lecture Notes 17. Press, Cambridge. [BJ]

1975

BrBcker, T. & JAnich, K.

Einftlhrung in die Differentialtopologie.

Springer, Berlin & New York. [GG]

New York, [GP]

1970

Golubitsky, M. & Guillemin, V. larities.

Cambridge University

Stable Mappings and their Singu-

Graduate Texts in Mathematics 14.

Springer, Berlin &

1973.

Guillemin, V. & Pollack, A.

Differential Topology.

Prentice Hall,

1974.

[GWPL]

Gibson, C. G.,

E. J. N.

Wirthmialler, K.,

du Plessis, A. A.,

Looijenga,

Topological Stability of Smooth Mappings.

Lecture Notes in Mathematics

552.

Springer,

Berlin & New York,

1977. [L]

Lu, Y. C. Theory.

[LS1]

Singularity Theory and An Introduction to Catastrophe Springer, Berlin & New York,

I.

and New York, [LS2]

Wall, C. T. C. Symposium

II.

Lecture Notes in Mathematics 192.

Martinet, J.

Springer, Berlin

1971.

Proceedings of Liverpool Singularities

(Ed.)

Lecture Notes in Mathematics 209.

Berlin & New York, [M]

Proceedings of Liverpool Singularities

Wall, C. T. C. (Ed.) Symposium

1976.

Springer,

1971.

Deploiements Versels des Applications Differen-

tiables et Classifications des Applications Stable.

Notes In Mathematics

535

Springer,

Lecture

Berlin & New York,

1975.

235

[M1]

Topology

Milnor, J.

From the Differentiable Viewpoint.

-

The University Press of Virginia. [M2]

Milnor, J.

Princeton University Press.

61

Stewart, I. N. 68,

[SP]

447-454,

[T]

London

Wall, C. T. C. Manifolds.

Lecture Notes Zeeman, E. C. 65-83, [TZ]

1978.

Benjamin-Addison Wesley,

from Geometry and Topology III

New York.

1975

Proceedings of

pp. 707-774

Springer

no. 597.

Catastrophe Theory.

Scientific American

234,

1976.

Catastrophes of Codimension s 5.

236

(trans-

Geometric Properties of Generic Differentiable

Trotman, D. J. A. & Zeeman, E. C.

525

New Scientist

Catastrophe Theory and its Applica-

Symp. at IMPA, Rio de Janeiro, July 1976.

[Z]

543-564.

Structural Stability and Morphogenesis

lated by D. H. Fowler) [W]

(1971)

1975.

Pitman.

Thom, R.

Vol. 5,

The Seven Elementary Catastrophes.

Stewart, I. N. & Poston, T. tions.

1968

The Normal Singularities of a Submanifold.

Porteous, Z. R.

Journal of Differential Geometry. [S]

Annals

Singular Points of Complex Hypersurfaces.

of Maths. Studies [P]

Charlottesville, 1965.

Springer.

Berlin & New York.

Classification of Elementary Lecture Notes in Mathematics. 1976

pp. 263-327.

Index

Action of a group

61

Critical value

Algebra

226

Cubic curves

Algebra product

226

Cuspoids

homomorphism

226

222

Algebraic set

Bifurcation set

40 68-70 130

Curve

Deformation

25

of germ

140

induced

163

44

Binary cubic form

linear

65, 66

Boardman symbol

180,

submanifold

181

186 101, 228

Borel Lemma g-equivalence

144

Chain Rule

9, 21, 23

Change of basis

72

of co-ordinates

72

of parameter

90

Chart

13

Codimension of function germ

98

56

transversal

160

universal

164

Dense

52

Descendant

110

116, 191

Determinacy finite

Determinacy,

120, 191

Diffeomorphism

8, 12

Differential

8, 20

Double point

130

Equivalence of deformations

163

152

of germs

of orbit

77

of maps

16, 205

of submanifold

15

of unfoldings

89, 142

of map germ

Co-ordinates,

local

13

Corank of function germ Critical point Critical point,

139

125 29, 40

degenerate

59

Fold curve

46

Folded handkerchief mapping

55

Flow line

26

237

Fundamental neighbourhood

General linear g roup

Generic Germ,

51

Jets,

equivalence of

62

of germ

37

63 187

-r- e quivalence

143 ,

144

equivalence of

35

immersive

35

Lie group

73

invertible

34

Linear systems

71

rank of

35

representative of

34

Local Existence and Uniqueness Theorem

28

singular

35

source of

34

submersive

35

target of

34

Graph

28

flow

Manifold

12

Metastable range

212

Morphism of deformation

164

Multiplicity

130

Nakayama Lemma

102

Normal bundle space

133

Null set

215

21

Hadamard Lemma

100

Hessian

59, 67

Homogeneity Homomorphism,

62

induced

Infinitesimally stable point Inverse function theorem Isomorphism of ideals

Isotropy subgroup

Jacobian extension ideal

module

112

78

Orbit

61

Parametrization

12

Pencil

71

Product structure

80

Quadratic form

64

10

149 80

180

97 151

Quadratic form,

index of

65

Jet extension

37

rank of

65

space

37

semi-index of

65

Jets 238

37

R -equivalence

94

Transversality Lemma, Basic

49 47

Rational mapping

223

Transversality Theorem

Regular point

224

Transversality Theorem,

value

40

Sard's Theorem

1+9, 217

elementary

51

of Thom

53

Transverse

39 38

Semialgebraic set

222

Transverse intersections

Semi-direct product

150

Triple point

Singular point

44

Singular point,

non-degenerate

123

of type Ak

130

Singularity set

130

44, 54

Singularity set,

first order

higher order Slice

55 177

79

Smooth mapping

Umbilic bracelet

68

Unfolding

81

Unfolding

of germ

141

induced

89

minimal

81

morphism of

90

transverse

81

universal

90

versal

90

8, 12

Splitting Lemma

125

Vector field

Stable germ

142

Vector field,

map

point

206 77

Submanifold

15

Submersion

40

24

equivalence of

30

time dependent

32

time independent

33

Whitney cusp mapping Whitney umbrella

Tangent bundle

87, 213

24

bundle space

22

mapping

22

space

16

Tarski-Seidenberg Theorem

44

224

239