Research Notes in Mathematics
CGGibson
Singular points of smooth mappings
Pitman LONDON· SAN FRANCISCO· MELBOURNE
25
CGGibson 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, CG Singular points of smooth mappings - (Research notes in mathematics; 25). I. Mappings (Mathematics) 2. Singularities (Mathematics) I. Title II. 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 Acknowledgements Chapter I
7 Smooth Manifolds and Mappings
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
Chapter II
Transversality
§2.
Chapter III
8
The Notion of Transversali ty
38
The Basic Transversality Lemma
48
An Elementary Transversality Theorem
51
Thom's Transversality Theorem
53
First Order Singularity Sets
54
Unfoldings
The Finite Dimensional Model
Groups Acting on Sets
61
Some Geometry of Jets
62
Smooth Actions of Lie Groups on Smooth Manifolds
73
§4.
Transversal Unfoldings
81
§5.
Versal Unfoldings
89
Chapter IV
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
122
Stable Singularities of Smooth Mappings
Chapter V
§1.
The Basic Ideas
139
§2.
Contact Equivalence
143
§3.
Deformations underJF-Equivalence
159
§4.
Classification of Stable Germs
168
§5.
Hi gher Order Singularity Sets
174
§6.
Classifying Germs underJr-Equivalence
191
§7.
Some Examples of Classifying Stable Germs
199
§8.
Singular Points of Stable Mappings
205
Appendix A
The Theorem of Sard
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 with ever,y point q = f(p)
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. 1
Fig. 2
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.
sents the simplest situation one can imagine, with the surface over at
p:
such points
p
are called told
point~.
Figure 1 repreX folding
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
A still more compli-
X is a torus,
i.e.
a dough-
nut - shaped surface.
Here one has again curves of fold points, on which lie
Fig. J four cusp points:
but in addition we have two simple crossing points on the
apparent outline, where the curves cross over properly. any point
Thus, if we take
q on the apparent outline and look at the nature of the outline
very close to
q,
we can distinguish just three possibilities.
II
III
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 slightest 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.
change
position slightly
To make the point even clearer, here are three further types of behaviour
q
which are all inessential, because they could be eliminated by the slightest change in position of simple crossing points
X to yield situations exhibiting only fold, cusp and 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, Yare replaced by smooth manifolds, which are natural generalizations to higher dimensions of the familiar notions of curve and surface. the projection of the surface onto the plane is replaced by an mapping f :
X~
Y.
arbitra~
And smooth
For such mappings we introduce the general notion of
singularity and begin to list the simplest singularities which can occur, idealizing each type by a model.
We shall, for reasons of simplicity, con-
cern ourselves principally with the
~
behaviour of the mapping,
behaviour ver,y 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. Scien-
There is nothing particularly new in the notion of a singularity.
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 Rene Thom pointed
out the analogy with more finite-dimensional situations and indicated the general lines along which a
theo~
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 Amol'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
theo~
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
theo~
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 n
Chapter I as subsets of]R
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 respecti vely.
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.
Singulari ty
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. of the list of singularities of codimension case of mappings is taken up:
~
The result is the derivation
5.
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 whioh 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 pict"J.res, 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
1.
A Preliminary Review of
Calculus
S~me
[In this section it will be understood that ~p. ~q
Let
U. V. Ware open sets in ~n.
respectively.] f: U
~ V
be a mapping with components
We call
f 1 ••••• f p •
f
smooth when all the partial derivatives
exist and are continuous in
U.
fice to observe that if
.•. , f
then U
onto
f
For the purposes of this book it will sufp
will be smooth.
are all given by polynomials in And we call
V when it is a bijection. and both
Now suppose
f : U ~ V is smooth.
linear mapping
a diffeomorphism of
f
are smooth.
For every point
a E U there is a
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
~a) ax n
8
Take, for instance, the case when F: JRn ~ ]RP;
linear map
then
f
f : U ~ V is the restriction of a is certainly smooth, and explicit compu-
tation of the Jacobian will verify that A particular case is provided when mapping;
in this case
D f
a
=
U C V and
D f = 1 at every point a
at every point
F
f
a E U.
U ~ V is the inclusion a
E
U,
with 1 the iden-
ti ty mapping on ]Rn. One of the basic facts we shall need to know about the differential is the Chain Rule, which we express as folloVls.
If we have a commuting triangle
of smooth maps
And a point
a E U with
f(a)
= b,
then the corresponding triangle of dif-
ferentials commutes.
This provides us with a simple proof of the following fact. ( 1 .1)
If
differential Proof
f : U ~ V is a diffeomorphism, then at each point D f : ]Rn ~]RP a
is invertible, so that necessarily
a E U the n = p.
Indeed we shall have the commuting triangle of smooth mappings
9
Now let
a E U,
and put
b = f(a).
By the Chain Rule we have a commuting
diagram of differentials
which entails that
Df
o
is invertible, with inverse
a
The reader should be reminded that the direct converse of (1.1) does not hold.
Nevertheless, there is a partial converse, the Inverse Function
Theorem, which provides one of the really basic theorems of the calculus. If
(1.2)
f : U ~ V is smooth, and
a E U is such that
is invertible, then there exist open neighbourhoods respectively such that
f
maps
U'
U', V'
diffeomorphicallyonto
: lRn ~ lRn
D f
a
of
a, f(a)
V'.
The next two propositions are important consequences of the Inverse Function The0rem and provide the keys to several useful results. Let rank
p
f : U ~ :rn.P be a smooth mapping with
(so necessarily
some neighbourhood of
0
n
~
p):
in :rn.n
and
DOf
of
h
of
then there exists a diffeomorphism onto another such that
=
10
f(O) = 0
h(O) = 0
and
Proof
The Jacobian matrix of
the leading F
p x p
... , x where
h
will have rank
n
0
)
in R n
is invertible.
satisfies
h(O) = O.
R n ~ RP
ff
••• , x )
(so necessarily
some neighbourhood of
0
n ~ p):
n x n
Computation will verify that
0
o
f(O) = 0
onto another such that
... , x n '
f
will have rank
submatrix which is invertible.
=
0
is the required
and
Dof
of
k
of
then there exists a diffeomorphism
RP
in
from a neighbourhood of
Theorem
h
=
The Jacobian matrix of
the leading
To see that
onto another,
to be the projection defined by
= Proof
has a restriction
in R n
f : U ~ R P be a smooth mapping with
Let n
F
0
=
n
of
Computation verifies that
and observe that
=
F
f.
By the Inverse Function Theorem
diffeomorphism take
rank
Define a smooth mapping
=
which is a diffeomorphism of a neighbourhood of
and automatically
we can suppose it is
onto another by
are the components of
( 1.4)
n:
submatrix which is invertible.
from a neighbourhood of
DOF
f
in R P
k(O) = 0
and
0, ••• , 0). n:
we can suppose it is
Define a smooth mapping
onto another by
.•. , x ) + (0, ... , 0, x l' ... , x ). n n+ p
DOF
is invertible.
By the Inverse Function
F has a restriction which is a diffeomorphism of a neighbourhood in R P
onto another.
Denote the inverse of this restriction by
k:
clearly it is the required diffeomorphism as 11
D That does not conclude our review of calculus, for we shall also require standard results concerning the existence of solutions for ordinary differHowever we shall postpone that discussion till §4 when
ential equations.
we shall have available the language necessary to state results in a succinct geometric form.
~2.
Smooth Manifolds
It is an easy matter to extend the basic concept of "smoothness" to mappings between arbitrary subsets when for any point JR j
(depending on
u n X. f
k
JR.
We call
f
: X --+ Y smooth
X there exists an open neighbourhood
X E
x)
j
X, Y of JR ,
k
F:U--+JR
and a smooth mapping
And, by analogy wi th
is bijective, and both
~1,
we call
f
U
with
x
of
f =
in
F on
X --+ Y a diffeomorphism when
f, f- 1 are smooth:
in that case
X, Yare said
to be diffeomorphic. However, we do not wish to consider completely arbitrary subsets of Euclidean spaces:
instead we shall isolate a particularly useful class of subsets.
First we need some definitions.
An n-dimensional parametrization of a set
X f JRj is a smooth mapping
V --+ JRj
which
(V) = X and the mapping
Suppose now that of
x
which
in
k
N C JR
and that
N is a set of the form
x E U.
cf>
:
X E
U
n
V --+ cf>(V) N:
V an open set in JRn
a relatively open neighbourhood
N with
n
for
is a diffeomorphism.
U C JRk an open set for
Now we are ready to introduce the main idea.
N C JRk a smooth manifold of dimension 12
with
when every point
We call x E N has a
U n N admitting an n-dimensional parametriza-
relatively open neighbourhood
some authors term the inverse chart at
cf;
-1
un N
x, and call its components local co-ordinates at
adhere to the convention of writing Nn n; ,ani sometimes we shall write Any open set
Example
(and in particular
JRk
N ~ JRk
Here, given
parametrization of
N.
N one takes
E
N C JRk
JRn ~ JRk
of dimension n.
Given
with image
n
k.
U = JRk ,
is a k-dimensional parametrization of
'space) is a smooth manifold of dimension and then any linear mapping
x
a
We shall
N C JRk is a smooth manifold of dimension
Any vector subspace
Example 2
V
N has dimension
dim N for the dimension of
is one).
and. then the inclusion
to indicate that
x.
~
U n N.
(as a vector
x E N we take
U = JRk
N will be an n-dimensional
U n N.
In both these examples the whole set
N admits a parametrization.
The
next example is more subtle in that the set in question is compact, so cannot possibly admit a parametrization; Let
Example 3 (xO' x 1'
... , xn )
Sn C with
manifold of dimension south pole of
Sn
N
JRn + 1 be the n-sphere, 2
o
X
n.
however it is still a smooth manifold.
+ xn
+
2
=
1.
i.e.
the set of all points
I claim that
Sn
is a smooth
For convenience we take the north pole and the
to be the points
= (1, 0, ••• , 0)
S
= (-1, 0, ••• , 0)
and we take cf;N cf;S
Sn _
{N}
~
JRn
Sn _ {S}
~
JRn
to be respectively stereographic projection from the north, south pole, i.e.
13
the mappings defined by
= (1
xi
- Xo
,
••• , 1
xi = (1 + x '
o
cover
it will suffice to show that both
these mappings are diffeomorphisms.
They are certainly smooth, being the
Since the domains of
~N' ~S
restrictions of the smooth mappings defined by the same formulae with domains the open sets defined by tion will verify that
Xo
~N' ~S
I
1,
xo
1-1
respectively.
A minor computa-
are bijections with inverses defined by
where y
2 + Yn •
=
And clearly the inverses are smooth mappings as well. n = 2
one pictures
~N' ~S
In the case when
like this.
The reader will no doubt note, with increasing apprehension, that even for such a simple example as
14
Sn
a fair amount of work is involved in verifying
that it is indeed a smooth manifold.
However in Chapter II we shall estab-
lish a proposition which provides a very simple way of showing that quite complicated sets are smooth manifolds.
For the time being we shall stick to
our rather short list, but observe that one can increase the stock of examples using (2.1)
N
respecti vely:
then
S
lRk
be smooth manifolds of dimensions
M x N C lR j +k
m, n
is a smooth manifold of dimension
The proof of (2.1) is sufficiently obvious to justify its omission. S1 x S1 C lR4
way of illustration the torus
(m + n). By
is a smooth manifold of dimen-
sion 2. Let
~
C lRk
smooth manifold
1m C lRk
M in N.
of
A smooth submanifold of N is a
be a smooth manifold. with
One calls
MeN.
n - m the codimension
The next result tells us that, locally, a smooth submanifold
of a smooth manifold looks like a subspace of a vector space.
~f1 be a smooth submanifold of a smooth manifold ;f
Let
(2.2) any point
x E M has a relatively open neighbourhood in
metrization
~,
v
be parametrizations of dimensions x
open neighbourhoods of ~(O)
= x,
v(O)
subset of lRm m. k
= x.
in Now
into lRn
(Note that therefore
with
M, N respectively.
=
v-1
o
f(O)
=
0
f
¢
m, n
of relatively
Clearly, one can suppose
is a smooth mapping of an open
~
whose differential at
0
has rank
By (1.4) there exists a diffeomorphism
m " n. )
of some open neLghbourhood of
and then
N having a para-
¢
Let
Proof
C lRk:
0
in lRn
onto another for which
has the desired properties.
o 15
In differential topology (the study of smooth manifolds and smooth mappings between them) one regards two smooth manifolds when they are diffeomorphic. f 1 : N1 --t P l'
N1 , N2
as being the "same"
Likewise we regard two smooth mappings
f 2 : N2 --t P 2
as being the "same", formally one calls them
equivalent, when there exist diffeomorphisms
h, k
for which the following
diagram commutes.
§3.
The Differential of a Smooth Mapping
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. cp : U --t Ek in
N
Let
~ C Ek be a smooth manifold, let x
E
N and let
be a parametrization of a relatively open neighbourhood of
with cp(u)
= x,
say.
We define the tangent space
to be the image of the differential
n
Dcp:E u
k --tEo
T N x
parallel to the affine subspace of Ek
x
N close to
16
x.
x
to
One pictures
the vector subspace of Ek which best approximates
at
x N
as through
Of course, for the above definition to make sense we must show that it is independent of the choice of parametrization.
So let
I/J : V ~:m.k
another parametrization of a relatively open neighbourhood of
I/J(v) u
=
x,
say.
Then X
= I/J -1
onto an open neighbourhood
gram
0
cP
V1 of
x
in
be N with
will map an open neighbourhood V,
U1
of
and we shall have a commuting dia-
of smooth mappings :m. k
/\
U
X .,V 1 (diffeomorphism) 1
giving rise to a commuting diagram of differentials
from which it is immediate that
D cP, D I/J u v
have the same image, as was
required.
17
Let gent space
N C lRk be a smooth manifold, and let
TxN
is a vector subspace of lRk
x EN:
of the same dimension
We keep to the notation of the above discussion, is a parametrization of a relatively open neighbourhood of
cp(u)
=
open set
x.
then the tan-
i. e. x
in
n
cp: N,
as
N.
U --+ lRk
with
Now cp -1 : cp(u)
--+ U
is smooth, so (by definition) there is an
VI C lRn with cp-1
=
on
verse image under cp
of
Vi
n cp( u)
VI n cp(U).
Taking
U'
to be the in-
we obtain a commuting diagram of smooth
mappings
/w~ )lRn
U' (inclusion) hence a commuting diagram of differentials
It is immediate that
TxN,
the image of
D cp,
has the required dimension
u
n.
D The reader will readily check for himself that the tangent space at a point to an open set in lRk
is precisely lRk
point to a vector subspace
also, that the tangent space at a
V of lRk is precisely V.
is to show that the tangent space at a point is precisely the subspace perpendicular to
x x.
A harder exercise
to the n-sphere
Sn C lRn+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 taneent space is that it allows one to intr0duce the differential at a point of a smooth mapping defined on a smooth manifold, rather than just on an open set. observation. f : M -+ N that
f
ll.P ,
~1f1
Let
:rn.P ,
c
The definition will require a preliminary
I'f c
:rn. q be two smooth manifolds, let
be a smooth mapping, let
x
E
and put
M,
is smooth requires that there is an open neighbourhood
and a smooth mapping
F at
the differential of
-+:rn. q with
F: Vi
x
=F
f
on
x,
if;(v) = y,
suppose that by
g=if;
-1
c
ofo
x, y
in
Of course
M, N respectively with
in
By taking U to be sufficiently small we can
say. W,
M.
x
T M into the tangent x V -+:rn. q be parametriza-
Y
=
n
of
Vie
D F necessarily maps the tangent space x To prove this let : U -+:rn.P , if; T N.
tions of relatively open neighbourhoods of (u)
W
YT
is a linear mapping
claim that space
The fact
y = f(x).
and that
f
0
£.
if; (V) ,
is a well-defined smooth mapping.
so
g : U -+ V given We now have a com-
muting diagram of smooth mappings
U--------.;0.:> V
yielding a commuting diagram of differentials
DF
:rn.p _ _ _-=x_ _ _~:> :rn. q
Dupi
Dug
I
Dv
p
:rn.m _ _ _ _ _ _ _ _:> :rn.n It is immediate that
D F maps the image of x
D
into the image of
D if;, v
19
the tangent space
i.e.
T M
x
The restriction of the linear mapping D F : JRP x T f x
: TM x
Tf
as the best linear approximation to
x
-4
-4
T N and call the differential of
f
y
x = 0
is drawn for the case
T N,
into the tangent space
f
at
and
y
x:
Y
F at
to
as was claimed. T M we write x
x.
One pictures
the illustration below
=0
T:r/ ..
-
f
T
It follows from the very construction of the choice of
that it does not depend on f x Notice also that in the particular case when M, N are
F.
open sets in JRP , JR q
the differential
as previously defined.
T f
x
is the the differential
Dx f
It is an easy matter to derive the basic properties
of this more general notion of differential from those mentioned in §1.
Vie
shall mention two of these, leaving the proofs to the reader. Take, for instance, the situation when smooth manifold X E
T f x
11
N,
and
the tangent space '1'
x
11
-4
-4
N is the inclusion mapping.
At any point
and the differential T M is a subspace of x One pictures the situation something T N is the inclusion. x
like this.
M
20
f : M
M is a smooth submanifold of a
Likewise, we have the Chain Rule, which Vfe express as follows.
Suppose
A, B, C are smooth manifolds, and we have a commuting triangle of smooth mappings
and a point
a
E
A with
f(a)
= b,
g(b)
=c
then the corresponding triangle
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.
Vie define the
~ll
graph f
(3.2)
Let
=
of a mapping f : {(x, y) EN x P
P
y
to be the set
=f(x)}.
f : N -+ P be a smooth mapping between smooth manifolds:
graph f
is a smooth submanifold of N x P.
graph f
at any point
Proof
N -+
F : N
-+
(x, f(X»)
graph f
!~so
then
the taneent space to
is precisely graph
Txf.
given by F(x) = (x, f(X»)
is a smooth bijec-
tion, indeed a diffeomorphism since its inverse F- 1 is the restriction to graph f
of the smooth projection N x P
-+
N.
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
o
T f.
graph
i.e.
and the
x
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
(x, y)
lvI,
N be smooth manifolds.
to the product
The tangent space at a point
M x N is the product
T 11 x T N.
Y
x
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
Tf.
N C JRk be a smooth manifold.
Formally, we proceed as follows. lfle
TN
Let bundle space
ly open set in F
:
U x
! (x,
N,
v) E N x JRk : vET xNl •
N.
defined by
~
P
(u, x)
~ (f(U), Duf(X)) is a 2n-
be a smooth mapping with
1,'!e define the tangent mapping
22
2n.
be an n-dimensional parametriz ation of a rela ti ve-
dimensional parametrization of a relatively open set in
f: N
the tangent
The pror:osition follows from the observation that
JRn ~ JRk x JRk
Now let
n:
JR2k is a smooth manifold of dimension
f : U ~ JRk
N to
i.e.
N C JRk be a smooth manifold of dimension TN C
Let
Proof
=
Let
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
o
(3.5)
Let
Tf : TN
~
f: N ~ P
TP
be a smooth mapping:
is likewise smooth.
Proof
(xO' vO) E TN and an
Choose a point
open set F : U
the tangent mapping
U C ]l.s with
~:rn.
smooth. )
t
wi th
F = f
on
U for vlhich there exists a smooth mapping U II N.
(This is possible since
Observe that the tangent mapping of
this mapping is smooth.
(Xo' vo):
Xo E
also we have
definition of
Tf.
Notice that if
But TF
U x]l.s
= Tf
on
It follows that
11
C
F
:rn. s x:rn. s
is open and contains going back to the
M~
D
is smooth.
is a smooth submanifold of a smooth manifold
the tangent mapping of the inclusion
is
is the map
(U x :rn. s ) II TN, Tf
f
N will be the inclusion
N then TM
~
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
TA
y:~
~TC
23
otherwise expressed, T(g
0
f)
Tg
=
Notice also that given a smooth manifold 7TN : TN
tangent bundle of
N.
N there is a smooth projection
(x, v)
N given by the formula
-+
Tf.
0
x:
-+
this mapping is the
It follows that given any smooth mapping
between smooth manifolds
Vie
f : N
-+
P
have a commuting diagram of smooth mappings. Tf
N -------~ .. P
§4.
Vector Fields and Flows
~
Let
be a smooth manifold with tangent bundle
smooth vector field on mapping
e
N -+ TN
x
to each point
E
7T
N we mean a smooth section of such that
7T
0
e
=
e(x)
N a tangent vector
7T,
i.e.
In that case
TN
is
N x]Rn
e
assigns
x.
at
N C lRn
For the purpose of illustration take the special case when an open set.
a smooth
In other words
iN.
By a
TN -+ N.
and
7T : N x lRn
-+
is
N is pro-
jection on the first factor, so that a smooth vector field is a smooth mapping
e
N
-+
smooth mapping.
N x lRn
of the form
Suppose
e1,
x
••• , en
-+
(x,
e(x))
are the components of
vector field aoes no more than associate with each point
... , 24
e :
with
x
E
e:
N -+
lRn
a
then our
N 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 Here are some examples, which the reader is advised
to check for himself:
e
indeed a good exercise is to invent simple formulae for
and sketch the corresponding vector fields.
taken
In all these examples we have
2
N=JR.
~\V: 7]-
1\"
y
1>'
x
x
;'
61 = x
6
2
=Y
91 = Y
2 9 2 = -p x
Y
Y
x
2 2 61 = x-y : 6 2 = 2xy
!) ~ ~(f
x
61 = x : 9 2 = -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.
~
N we mean a smooth mapping
an open interval of positive length.
o
Let
be a smooth manif: I
~
N with
I
For convenience we shall suppose that
And we say that the curve starts at
Xo
= f(O).
25
N
-
I
o
t
f
Recall now that we have the following commuting diagram of smooth mappings. Tf
TI--------.,.~ TN
1
'I
1 f
I----------;.~
Here, of course, TI field
e i
vector at at
f(t).
at
f(t)
on t,
I
I x
N
Suppose now that we have a smooth vector
]t.
We wish to make precise what we mean by saying that f'
on N.
e.
a "flow-line" for field
=
'N
To this end we introduce a "canonical" smooth vector
by writing and
is
i ( t)
Tf(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
e.
i.e.
the tangent vector
In other words we want the follow-
ing diagram of smooth mappings to commute Tf
i
TI--------~~
TN
1
1<
I _ _ _ _.::.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
The composite mapping Tf
f: I 0
i
is
~
N
usually written
f':
with that notation the condition for
f
to be a flow-
line is that f'(t)
=
e(f(t)).
---------------------*
To make this as concrete as possible it might help to go back to the case when
N is an open set in ~n,
functions
e1 ,
••• ,
functions
en
••• , f
on
and the vector field is given by
N.
A smooth curve
as its components.
n
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 systern of
n
simultaneous equations
Thus the problem of constructing a flow-line ving simultaneously the above
n
f
for
f 1, ••• , f n •
(x 1 ' ••• , xn)
in
is that the solution should satisfy the "initial conditions"
q(o) Example 1 82
is that of sol-
differential equations for
And the condition for the flow-line to start at the point N
e
=-p
2
x
=
X
1 : •••
: f'
n
(0)
Take the smooth vector field on ~2 with
p
I-
0
a real numb er.
=
x. n
defined by
e1 = y,
(We sketched this one ab ove. )
To
find flow-lines we have to solve the simultaneous differential equations
f'1
=
f' 2
=
2 -p f 1"
These yield the harmonic equation
27
it 2f f1 + p 1
=
0
having solutions
wi th
c, a
E
the ellipses
22222 Notice that p f 1 + f 2 = P c so the flow-lines are 22222 p x + y = pc. (Compare with the sketch given above.)
JR..
The most basic fact which it is necessary to know about flow-lines is what we shall term the Local Existence and Uniqueness Theorem.
(4.1)
Let
e
be a smooth vector field on a smooth manifold
N,
and
~et
The following assertions hold. (i)
e
There exists a flow-line for
e
(ii) Any two flow-lines for hood of
0
in JR.n •
which starts at
which start at
xo
(Existence)
xO.
agree on some neighbour-
(Uniqueness)
The reader will readily check that it suffices to prove the result when N is an open subset of JR.n •
And in that case the result is just the Local
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 single flow-line through a point:
we need to know ttat we can simultaneously This idea
parametrize all the flovl-lines through points near to a given one.
field on a smooth manifold local flow for
e
at
xo
N,
xo
the property that for any point
xo
in N,
be a smooth vector
be a point in
we mean a smooth mapping
U is an open neighbourhood of
28
and let
e
Let
is captured by the following formal definition.
and
I
x E U the mapping
N.
F : Ux I
By -+
a smooth
N (where
is an open interval) with f
x
I
-+
N defined by
f x (t)
=
F(x, t)
is a flow-line for
e
which starts at
x.
One pictures
the situation something like this, with I
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 we assume the reader to be familiar with from calculus. Let
(4.2) N:
Xo E
e
be a smooth vector field on a smooth manifold
there exists a smooth local flow for
e
at
N,
and let
xO.
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.
e(x)
= O.
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
x 1, x 2
are points in smooth manifolds
e1, e2 •
defined smooth vector fields ly equivalent at Ul' U2
of
x 1, x2
x 1, x2
N1 , N2
on which are
e1 , e2
We shall say that
are smooth-
when there exist relatively open neighbourhoods
in Nl' N2
and a diffeomorphism if> : U1
--+
U2
for which
the following diagram commutes, and
<,
r
Here of course we are abusing notation by allowing their restrictions to
U1 , U2
something like this with if>
respectively.
constant when there exists a vector x E N.
v
I:
e 0
also to denote
One pictures the situation
mapping the flow-lines for
Now let us call a smooth vector field
all
e1 , e2
e1
to those for
on a smooth manifold
for which
e(x)
N
= (x, v)
Of course, such a vector field has no critical points.
for We
contend the following. Let Xo E
e
be a smooth vector field on a smooth manifold
N be a non-critical point.
N smoothly eguivalent to
30
e
at
N and let
There exists a constant vector field on xO.
e2 •
The picture for this result is the same as that above save that one of the vector fields is constant.
Proof
Clearly, it is no restriction to suppose that
b ourh 00 d
0f
F : Ux I
Xo
=
0
~n JRn • ~
~ JRn with
open interval containing critical point.
diffeomorphism
By (4.2) there exists a smooth local flow
U an open neighbourhood of O.
I claim that
stant vector field on JRn
N is an open neigh-
v = ';(0):
Let .;
then
I v
and
-I
0
as
is smoothly equivalent at
given by
0
0
I
an
is a non-
to the con-
For this we have to construct a
v.
of some neighbourhood of
0
in JRn
onto another which
satisfies
(2)
Dl(v)
To begin with, note that any vector form
x = 1T(X) + t(x)v
=
(1)
where t(x)
x
0
= ';(
JRn can be written uniquely in the
is a real number, and
1T denotes ortho-
gonal projection on the orthogonal complement of the line spanned by Observe that both that
1T (U) ~
u,
1T, t
are linear mappings.
t(U) C I.
We can suppose
v.
U chosen so
Define the smooth mapping
by
-------------------* Intuitively, to get to the point
you start at the point 1T(X)
slide up the flow-line through that point for time
t(x).
and
The pictorial idea
31
is as follows.
certainly (1) is satisfied because one has For (2) we proceed as follows. mind that
rr, t
= ~ rr(x),
x E U implies that
aF
also that ~o, 0)
.
p~ng
Thus
-mn. on .w.
x
for all
*,
Taking differentials in
t(x))
with obvious meanings attached to
scalar.
=
x E U.
and bearing in
are linear, one obtains
Dx?£
for all
F(x, 0)
0
rr +
~ rr(x),
aF aF au' at'
~x, 0)
t(x))
0
t
The fact that
F(x, 0)
is the identity map on En.
Note
represents multiplication of the fixed vector
DO?£ (x)
=
rr(x) + t(x)v
=
x
so
DO?£
=x
v
by a
is the identity map-
w
It follows from the Inverse Function Theorem that
will map
its domain U diffeomorphically onto its image, provided U is small enough.
D
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 with
J
defined by et(x) thinks of
32
e
=
e(x,
t)
Formally, a time-dependent
N is a smooth mapping
an open interval, such that for each
t E J
e:
(et)tEJ
Nx J
the mapping
is a smooth vector field on N.
as a 1-parameter family
i.e.
et
~
TN,
: N ~ TN
Thus one
of smooth vector fields on N.
is as follows.
....
certainly (1) is satisfied because one has For (2) we proceed as follows. mind that
rr, t
x E U implies that
aF
~o,
scalar. •
p~ng
Thus
on "lOn. .In.
its domain
x
E
U.
and bearing in
are linear, one obtains t(x))
with obvious meanings attached to
also that
for all
*,
Taking differentials in
Dl = ~ rr(x),
for all
F(x, 0) = x
0)
0
rr +
-lli rr(x),
aF aF au' at.
~~(x, 0)
t(x))
0
t
The fact that
F(x, 0) = x
is the identity map on Bn.
Note
represents multiplication of the fixed vector
DO~ (x)
=
rr(x) + t(x)v
=
x
so
DO~
v
by a
is the identity map-
w
It follows from the Inverse Function Theorem that
will map
U diffeomorphically onto its image, provided U is small enough.
o
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 with
J
defined by et(x) thinks of
32
e
=
e(x, t)
as a 1-parameter
Formally, a time-dependent
N is a smooth mapping
an open interval, such that for each
t E J
~
(~)
"'t tEJ
e :
Nx J
et
the mapping
is a smooth vector field on fam~ly
i.e.
N.
of smooth vector
TN,
--+
: N --+ TN
Thus one f~elds ~
on
N.
To maintain the contrast one can then refer to an ordinary vector field as time-independent. For a time-dependent vector field
e:
Nx J
~
TN
one can mimic the
definitions already introduced for time-independent vector fields.
e
flow-line for I
starting at
Xo
an open interval containing 0, f'(t)
And a smooth local flow for (where
e
f
: I
x
~
and f(O)
=
= xo'
e
Let
and let
Xo E N:
N with
for which
is a smooth mapping F : U x I
at
Xo
in N,
and
with the property that for any point
0)
~
= F(x, t)
I
~
N
is an x E U the
is a flow-line for
The analogue of (4.2) is
x.
(4.3)
: I
e(f(t), t).
defined by the rule
N
starting at
§5.
N is a smooth mapping f
U is a relatively open neighbourhood of
open interval containing map
e
E
Thus a
be a time-dependent vector field on a smooth manifold
e
then there exists a smooth local flow for
at
N,
xO.
Germs of Smooth Mappings
The subject matter of this book is concerned principally with the behaviour N~ P
of a smooth mapping smooth manifolds. point
x E N,
close to a point in its domain, where
We can make this precise as follows.
valence relation we write
f1
Suppose given a
and consider the set of all smooth mappings
ain U is a neighbourhood of
~
f2
~.
x
in N.
N, Pare
U ~ P whose dom-
On this set we introduce an equi-
Given two such mappings
f1 : U1 ~ P,
when there exists a neighbourhood
U of
f 2 : U2 x
in
~
P
N
33
depending
on f1
and
f2
for which the restrictions
f1lu,
f21u
coincide
The equivalence classes under this relation are called smooth germs of mappings N~ P
at
x, and elements of the equivalence class are called representatives
of the germ.
Notice that if
then f 1(x) = f 2 (x) , y, f
say, at
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 (p, y)
for the germ, and to call
source, target of the germ. notation
i~l
which they are derived. (p, y) ~ (R, z)
g
V ~ R with
x
=P
we use the
of the identity mapping
much the same way as one handles the mappings from (N, x) ~ (p, y)
f
and
one just chooses representatives
g
0
f
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 has
respectively the
we can "compose" them to obtain a germ
feU) 5; V,
of their composite
for the germ at
For instance given germs
: (N, x) ~ (R, z):
go f
x, y
In the particular case when N
1N: (N, x) ~ (N, x)
One can handle germs
g
are representatives of the same germ
fog
of f.
= 1p '
g
0
f
=
Likewise, one can pursue the A germ
g: (p, y) ~ (N, x)
1N: and in that case
Further, to a germ f : (N, x) ~ (p, y)
tial, denoted as one would expect by Txf : TxN the differential at
x
f : (N, x) ~ (p, y)
of any representative:
does not depend on the choice of representative.
~
for which one
g 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
invertible if and only if its differential is invertible. f
~
: (N, x)
rank equals
(p, y)
The rank of a germ
is defined to be that of its differential:
dim N the germ is immersive, and when it equals
when the
dim P
it is
Thus a germ will be invertible if and only if i t is both immer-
submersi ve.
sive and submersive.
A germ which is neither immersive nor submersive is
called singular. Example 1
Consider the plane curve
Its image is the cuspidal cubic at any point
x 3 = y2
f:JR.~JR
pictured below.
t lOis immersive, whilst that at
the only value of
t
2
t = 0
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.
f
o
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
35
Example 2
Consider the germs
(JR 2 , 0)
~ (JR2, 0)
(x 2 + y 2 , xy)
given by the formulae 2
(x ,
2
y ).
These germs are equivalent because in the above diagram we can take
=
hex, 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
f
(N, x)
~
(p, y)
is equivalent to
It is for this reason that much of this 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. Example 3
Proposition (1.3) tells us that any submersive germ
(JRn , 0) ~ (JRP , 0) is equivalent to the germ at (x 1, ••• , xn ) ~ immersive germ
(x 1' ••• , x p ).
(JRn , 0)
natural inclusion
(X 1'
~ ••• ,
0
of the projection
And Proposi tion (1.4) tells us that any
(JRP , 0) is equivalent to the germ at
x) n
~
of the
(x 1, ••• , 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
target) and the jets of the germ correspond to the initial finite segments of the Taylor series.
Let us make this more precise.
By the jet-space
fen, p) we mean the real vector space of all mappings f : ]Rn -+]RP each of whose components is a polynomial of degree .. k in ]Rn be called k-jets. that
a
E
]Rn.
with zero cons tant term: Suppose now that
f
in the standard co-ordinates
the elements of
: ]Rn -+]RP
If in the Taylor series of
~(n, p) will
is a smooth mapping, and
f(x) - f(a)
ressed in terms of the standard co-ordinates on ]Rn, ]Rp)
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
at
: lRn -+ ~(n, p)
jkf
called the k-jet extension of
f
f:
a.
In thi s way we
given by
a -+
this mapping will playa crucial role in
the final sections of the next chapter.
37
II llransversality
§1.
The Notion of Transversality
The notion of objects intersecting transversally (or in general position) has become quite fundamental to singularity
look at is two subspaces of a vector space transversally when their vector sum is
transverse
The simplest situation to
theo~.
V:
we say that they intersect
V.
not transverse
transverse
The notion is easily extended to smooth submanifolds of a smooth manifold. We say that two smooth submanifolds sect transversally at
x E N1
intersect transversally in when they do so at
eve~
n N2
TxN:
point in
N1, N2
of a smooth manifold N inter-
when the tangent spaces and N1 , N2 N1
n
N2 •
TxN1, TxN2
intersect transversally in
Probably the best way to under-
stand the idea is to look at a series of pictures.
) transverse
38
not transverse
N
transverse
not transverse
transverse
transverse
The idea which we really want to exploit is this. smooth mapping, and let
Q be a smooth submanifold of p.
N x Q are smooth submanifolds of N x P.
graph f
and
verse to
Q and (write
in N x P.
Let
f ~ Q)
f: N ~ P
be a
Recall that both We say f
is trans-
f, N x Q intersect transversally
when graph
One pictures it thus. P
gra;ph
--~~~~~~--------~~---
f
Nx Q
N
(1.1)
Let
manifold of with Y
~ ~ pp
f P.
= f(x)
be a smooth mapping, and let
An equivalent condition for E
x
x
The condition for graph
N x P is that for all points points
(x, f(x))
f ~ Q is that for all
x E N
Q we have T f(T N) + T Q
Proof
Q be a smooth sub-
with
z
y
=
T P • ---------* y
f, N x Q to intersect transversally in
= (x,
f(x) E Q,
y)
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~.3.3) this may be re-written as graph T f + T N x T Q = T N x T P x x Y x Y which is clearly equivalent to *.
D 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. Cas~-1
There are some special cases well worthy of separate mention. Take the case when
is submersive.
f
is a submersion
In that case
p-dimensional vector space
Txf (TxN)
TyP,
i.e.
its germ at any point
is a p-dimensional subspace of the We conclude that a
hence equal to it.
submersion f : N ~ P must be transverse to every submanifold Case 2
A further particular situation is prompted by the observation that holds (for some
..
if
codim Q > dim N then transversality of
to the image avoiding
Case 3
f(N)
x)
then certainly
Q,
being disjoint from
~
codim Q
if
f
Q ~ P.
dim N.
f: N ~ P
to
Consequently Q
is equivalent
or (expressed more vividly) to
Q.
One last special case is when
diagram above) • lar value of
f.
P
A point in A point
x
value is a critical Eoint of
have rank < n.
to which
f
N for which
E
f,
the condition for this is that
Q comprises a single point (see the
and Tf x
f(x)
is transverse is called a f(x)
is a critical value.
= P =:m.
the graph of
f
c
Clearly, i.e.
the cri tical points are pre-
cisely the points where the derivative vanishes; precisely the real numbers
fails to be a regular
should fail to be surjective,
For instance, when N
~-
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
f 1, f2
one has a commuting diagram of smooth mappings
40
are equivalent,
i.e.
with
h, k
diffeomorphisms;
submanifolds of
Pi' P2
respectively corresponding under k.
of the invariance of transversality is that only if
f2
Qi' Q2 are smooth
and suppose further that
is transverse to
fi
The statement
Qi
is transverse to
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. (1.2)
f : ~ ~ pp
Let
manifold of
P with
f
rh
~1
= f(x)
y
in
=
then M
Q:
Q,
having the same codimension as in N with
Q
f-1 Q is a smooth submanifold of N Further, for any point
or is empty.
one has
TM x
= Txf- 1
dimension
=
r
suppose that
p - q.
N, P
:m.q
x
Put
Y =
:m.n , :m.P
and that
Let f
to
1f
in
:m.n
denote the l)rojection of
Q tells us that
onto another with
1f 0
f
eve~
0
1f 0
h
point x E M
x = 0,
x, y
we can
y = 0:
Q is the intersection of
f
:m.P
k
The
on
has rank
then (1.1.3) tells us that there is a diffeomorphism 0
Q).
By taking charts at
f(x).
are open sets in
o.
transversalitv of
of
y
x
M is a smooth manifold of co-
indeed (1.2.2) allows us to assume further that with
(T
For the first proposition it suffices to show that
has a neighbourhood whose intersection with
P
Qq be a smooth sub-
be a smooth mapping. and let
r
at
0,
and
of a neighbourhood
the projection
The inverse image of
0
under this mapping
41
· an open sub se t 1S that
0 fOx
"..,n-r, £ correspond'1ng under
M is a smooth manifold of codimension
Step 2
The tangent space
h
M:
to
that shows
r. The commuting dia-
TxM is computed as follows.
gram of smooth mappings on the left gives rise to the commuting diagram of differentials on the right. f
N inc
I
;oP
1
inc
flM
M
T f x
TN x incj T M x
;oQ
from which it is clear that
T MeT f-\T Q). x - x y
;>
T P y
j inc ;> T Q y
T (fIM) x
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 tricting Ty Q
n
T f- 1 (T Q) x y
Txf: note that it has the same kernel as
~ Ty Q given by resTxf,
and image
The desired equality now follows on using the fact that
Txf(T xN).
the dimension of the domain is the sum of the dimensions of kernel and image,
D
together with the definition of transversality. Note the special case of this result when that case the hypothesis is that is that
f - 1 (y)
Example
y
Q is a single point y.
is a regular point of
is a smooth manifold of dimension
Let
A
=
(a .. )
Consider the smooth mapping
n - p,
be an invertible symmetric
1J
f
:
:rnn
x
value of
and
f,
is the linear mapping i.e.
a smooth manifold of dimension 42
v ~ 2(Ax. v),
the conclusion
or empty, and that
n x n
~:rn. which is given by
where • denotes the standard scalar product on :rn.n • at the point
f·,
f(x)
= Ax.x,
so 1 is a regular --'
or empty.
matrix.
Here the differential
the central quadric ) ' a .. x.x. (n - 1),
In
1J 1 J
=
1,
is
And the tangent space
at
x
to this quadric is the kernel of the differential,
plane perpendicular to the vector Ax. we see that
S
n-1
at a point expect.
x
of dimension
(n - 1),
x,
as we would
n:: 3 we see that ellipsoids, hyperbolic cylinders, ellip-
tic cylinders, and hyperboloids are all smooth manifolds of dimension Consider a smooth mapping
Example 2
f
: m2
-+
f- 1 (O) : (1.2) tells us that
smooth submanifold of m2
of dimension 1,
F : m
3
- {OJ
-+
:ill.
given by
i.e.
f- 1 (O)
so likewise
is then a
a smooth curve in the plane.
"J(x, y, z)
::
This too has the property that the differential has rank F- 1(O),
2.
m whose differential has
rank 1 at every point in
Consider now
a
moreover the tangent space to this sphere
is precisely the hyperplane perpendicular to
Taking
the hyper-
Taking A to be the identity matrix
is a smooth submanifold of mn
fact we already knew from Chapter I;
i.e.
2 f(,/x 2 + y ,
z).
at every point in
is a smooth submanifold of m3 of dimension 2:
F- 1(0)
it is just the surface of revolution obtained by rotating the curve fey, z) :: 0, tion let
y > 0
a, b
about the z-axis in m3 •
be real numbers with
Clearly the differential of is the circle radius F- 1(0)
b
f
By way of specific illustra-
a > b > 0,
and let
f
has raru( 1 at every point in
centred at
be given by
f- 1(O),
which
The surface of revolution
(a, 0).
is a torus, and a minor computation will verify that it is given by
the equation ::
Example 3
Suppose
are smooth submanifolds of a smooth manifold
CC intersecting transversally in or a smooth submanifold of
222 4a (x + y ).
C:
the intersection A n B will be empty,
C of dimension
a + b - c.
This follows on
43
A ~ C, B ~ C are transverse to
observing that the inclusions
B, A respec-
tively.
A singular point (or singularity) of a smooth mapping f : N ~ P is a x E N where the germ is singular.
point
(Of course a singular point is a
critical point, though the converse does not necessarily hold.) set
E f
of
f
is the set of all its singular points:
The singular
the image of
Ef
is
sometimes called the bifurcation set. Example 4 f
: JR2
The Vlliitney cusp mapping of the plane is the smooth mapping
~ :rn.2
•
g~ven
by
( x, y )
~
( u, v )
h were
u = x,
v -_ y3 - xy.
The
singular set is the set of points where the Jacobian matrix has rank < 2, the parabola parabola under
2
x = 3y •
f
i.e.
And the bifurcation set is the image of this
the cuspidal cubic having the equation
4u3 - 27v2 = 0.
v f u
The Whitney cusp mapping is best understood on a geometric level as the composite of the mapping
u = x,
v =
(u, v, w)
y3 _ xy,
~
ded surface
(u, v):
s
g :
w=y
:rn.2 ~:rn.3 given by (x, y) ~ (u, v, w) with
and the projection
see the diagram below.
defined by
of dimension 2 by (1.2).
v - w3 + uw
=
0,
1T
::rn.3 ~
given by
JR2
The image of
g
is the fol-
a smooth submanifold of
:rn.3
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.
3, "CJR _
surf ace plane
oJ
a 2-dimensional smooth subrnanifold of
P C:R3 through the origin:
tion onto of P
. l.e.
P,
Consider a smooth
Vie take
7T
:rn.3 , and a
to denote orthogonal projec-
and ask for the singular set of the restriction
7T/S.
Think
as "horizontal", and the line perpendicular to it as "vertical".
each point
XES
For
the commuting diagram of smooth mappings on the left gives
rise to a commuting diagram of differentials on the right.
The condition for should map
x
to be a singular point of
T S onto a proper subspace of x
Tx S should be "vertical"; be a curve in
P,
7T/S i.e.
is therefore that
that the tangent plane
thus, in general, the singular set of 7Tls
S projecting to a curve in
P,
7T
will
the bifurcation set.
45
s
o
Example 5
Let us return to the model situation presented by the last
example.
The folded surface was given by the equation
and was projected onto the plane kernel of the differential of ( oF
of
OP\
ou' ov' ow], Le.
to
w
= O.
2 = 3t,
u
2
(w,1,u-3w):
v
- w3 + uw
= 0,
is the
S
i.e. the plane perpendicular to the vector
F,
Thus the singular set of
the equations
=v
The tangent plane to
and the tangent plane will be
"vertical" when this vector lies in the plane v = -2w.
F
= -2t 3,
w
w
= 0,
when
u
= 3w2 ,
1T/S
is the "fold curve" parametrized by
= t;
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. differs qualitatively from the rest;
it is the one point where two folds on
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. Example 6 the plane
Consider the projection y
= z.
Recall that
1T
of the torus
S
of Example 2 onto
S was given by an equation ( 1)
The tangent plane to
S is the kernel of the differential of
G,
i.e.
the
aG aG) (aG ax' ay' Tz : and the 1Tls is that this vector
plane perpendicular to the vector point to be a singular point of plane
y
= z,
i.e.
8G 8y
that
=
condition for a should lie in the
8G
Tz' which computation verifies to be the
c ondi ti on that
o.
=
The singular set of
1Tls
is given therefore by (1), (2),
intersection of two surfaces of deGrees of degree 12. set;
i.e.
(2)
it is the
4, 3 respectively, hence a space curve
And its projection onto the plane
y = z
is the bifurcation
it is what Vie "see" if we imagine the torus made of glass and viewed
from a distant point on the line perpendicular to the plane
So much for transversality per see
y = z.
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 ~ P.
f : N~ P
which fails to be
Common sense tells one that
Q 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.
§2. Let
The Basic Transversality Lemma F: N x S
P
~
be a smooth mapping.
family of smooth maps
fs : N
the elements
Suppose
s E S.
'lle ask whether
fs
r/I
Q
~
P
This we think of as a smooth f
where
s
F ~ Q with
for all parameters
(x)
=
F(x, s)
parametrized by
Q a smooth submanifold of
s?
P.
That the answer can well be
in the negative is shown by Example 1 F : N fs
Take
S
x
~
maps JR2
the 2-sphere
N
2
= JR, S = JR,
defined by the formula
P
onto the horizontal plane S2.
F
and consider the smooth family
( (x, y) , s) z = s
is a diffeomorphism, so
~
in :rn.3 • F ~ Q.
And take
z = s
is tangent to
Q to be
On the other hand
only provided we avoid the two exceptional parameters the plane
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 "eood" values, for which
fs ~ Q.
Transversality Lemma: smooth manifold
those
That this holds generally is the content of the Basic for technical reasons we prefer to replace the single
Q by several smooth manifolds
Q1' •• " Qt: the reason will
become clear in the final section of this chapter. Sard's
i.e.
s
The crucial tool here is
Theorem, one of the truly fundamental results in differential topology,
which we state in the followine form.
(2.1)
Let
fi
: Ni
~
be a countable family of smooth mappings.
P
set of common regular values of the
f.
~
is dense in
The
P.
The proof of SardIs 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 Let
F: N x S
P be a smooth family of smooth mappings trans-
~
verse to smooth submanifolds of parameters Proof
By
s (1.2)
for which Mi
sider the restriction
=
Q1' ••• , Qt fs
then there is a dense set
is transverse to all of
F- 1(Qi)
1TIMi
of P:
Q1' ••• , Qt.
is a smooth submanifold of
of the projection
1T
:
Nx S ~
N x S. S.
Con-
We shall
show that
Vlhich will prove the result, by SardIs Theorem. 'lie can, and shall, drop the index
i
Now to the proof of
since it plays no further role.
*. The
reader is urged to keep the following picture in mind.
s
- - - -N x {a}
N
To start with, note that the con(li tion for that for all
z = (x, s)
in
N x S Vii th
F to be transverse to
w = F(z)
TzF(T xN x Ts S) + TwQ
=
in
Q is
Q we have
TwP. - - - - - - ( 1 )
49
Consider now the condition for dition that for the same
x
f
Q:
to be transverse to
s
this is the con-
as in (1)
T P. - - - - - - - ( 2 ) w
Furthermore, the condition for the same
x
N x {sJ to be transverse to
as in (1)
Tx N x Ts S - - - - - - - ( 3 )
=
where
T M z
=
T F- 1(T Q) z
(1)
If we assume (3) and apply
by (1.2).
w
both sides then
tells us that
(2)
Thus the condition for N x {sJ
f
s
lent to the condition that
M.
'.'le
x T
z
as in
1T(T zM)
(2)
(1)
=
{sl:
indeed this last con-
we have T S s
---------(4)
(3) .
which is clearly equivalent to
D
By way of explicit illustration consider two smooth manifolds some Euclidean space
S.
(3).
implies
claim that this in turn is equiva-
1T/M is transverse to
dition says that for the same
to
Q is precisely the condition
to be transverse to
to be transverse to
TzF
And conversely it is a
holds.
minor exercise in linear algebra to see directly that
for
M is that for
M, N in
Even if they do not intersect transversally, geo-
metric intuition tells us that it should be possible to force them to do so by an arbitrarily small translation of one manifold We can make this feeling precise as follows. the image of claim that the S,
s
for which
F : Mx S
-+
S
s E S let
defined by
x
H
M
s
X
denote + s.
\'le
1,ls' N intersect transversally Vlill be dense in
justifying one's intuition.
mappings
50
M under the translation of
For
M in some direction.
S by
To this end define a smooth family of smooth (x, s)
1-+
X
+ s.
It is clear that
F is a
submersion, so f
s
F
is transverse to
is transverse to
s
is transverse to
By the Basic Transversality Lemma
N for a dense set of parameters
established by observing that f
N.
N -
s.
Our claim is
Ms ,N intersect transversally if and only if
which fact we leave as an exercise for the reader.
N
translate M
§3.
An Elementary Transversality Theorem
Our object in this section will be to show that given a smooth mapping f:lRn~lRP ~ ,and a smooth manifold
ping to
g : lRn
f.
-t:rn.P
Q
p . S; lR, we can flnd a smooth map'"
which is transverse to
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
lRn
all smooth mappings
number
-t
JRP ,
~(~n, lRP )
Let
and let
f
: lRn
Given a (small) positive real number R,
bourhood in
and an integer CCO (lRn, lRP )
for which for all
their deri-
are "close".
We make this precise as follows.
mapping.
k?; 1
x
E
k?; 0
-t
e,
we associate to
denote the set of
lRP be a given smooth a (large) positive real
f
a fundamental neigh-
comprising all those smooth mappings
ll.n with
g : JRn
-t
~p
Ixl .. R one has
51
II II
with
a fixed norm on the jet-space
set f
Jk(:m.n,:rn.P ).
And we call a sub-
dense therein when given any smooth mapping
~ :rn.P
: :rn.n
and any fundamental neighbourhood g : :rn.n ~ :rn.P
smooth mapping
in
X with
g
V:
E
of
V
f
one can find a
intuitively, any mapping
X.
can be approximated as closely as we please by mappings in
",'e are now in a position to state and prove an elementary transversali ty theorem. :rn.n ~:rn.P
The set of smooth mappings
... ,
submanifolds
Proof imate
Let f
: :rn.n
f
is dense in
~ :rn.P be smooth.
';'Ie have to show that we can approx-
as closely as Vie please by mappinGs transverse to
The idea is to construct a smooth family f,
transverse to given smooth
and with
F
transverse to
Transversality Lemma.
Ql' ... , Qt:
'ife shall make
suring that it is a submersion. transversali ty of
f
F
: :rn.n ~:rn.p
F
:
:m.n x
'S
Ql ' ••. , Qt.
~:rn.p
which contains
then one applies the Basic
transverse to
Q1' ••• , Qt
by en-
To motivate the construction recall that the to
Q is equivalent to the graph being
transverse to :rn.n x Q in the product :rn.n x :rn.p •
If the graph is not already
transverse it seems reasonable that we might be able to force it to be so by
-------+-----------
52
IFf
x 0
bodily translating it. point.
The picture above illustrates the idea when
Therefore we take
(x, s)
f(x) + s.
t-+
f
is that if
F : :rn.n x :rn.P ~:rn.P
and define
by
Clearly, this is a submersion, so transverse to
By the Basic Transversality Lemma there is a dense set of
Ql' ••• , Qt· for which
S =:rn.P
Q is a
is transverse to
s
All that remains to be shown
••• , Qt·
is close enough to
s
then
f
s
is as close as we please to
s
more formally, we have to check that given a fundamental neighbourhood
V
of
f
we can find an
s
I
0
for which
f
s
lies in
V.
And that
o
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 is transverse to
n "'1' ••• ,
n"'t
f
: :rn.n ~:rn.P
for which
jkf
~(n, p).
::rn.n ~ ~(n, p)
is dense .~n COO (-rnn, -mP). ...1£\..1£\.
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 constant deformation.
f
transverse to a submanifold by using a
But in the present situation it is
which we wish to make transverse to a subll1anifold. only f
In the pre-
which we are allowed to deform.
jkf,
not
f,
On the other hand it is
A constant deformation will not
work here since it does not alter the derivatives of
f.
Vfuat we do instead
is to use a polynomial deformation. Proof
Let
S
= J«n,
p).
Of course
be identified with a Euclidean space. F
:
x S ~
:rn.n
;CCn, p)
mersion, because for fixed of
J
k
(n, p)
defined by x
S is finite-dimensional, so can Consider the smooth family
(x, s)
~ l(f
the mapping represents an affine isomorphism By the
with itself, hence transverse to all of
Basic Transversality Lemma there is a dense set of parameters the mapping
f
s
::rn.n ~ ;CCn, p) defined by x ~ /Cf
verse to all of
Q1' ••• , Qt.
we please to
by choosing
§5.
F is a sub-
+ s)(x).
f
And clearly we can make s
s
for which
+ s)(x)
f + s
is trans-
as close as
o
to be sufficiently small.
First Order Singularity Sets VIe shall use the Thorn
This section is in the nature of an extended exaoplo.
Transversality Theorem to show that for a dense set of smooth mappings f
:
:rn.n ~ :rn.P the singular set
smooth manifolds on each of which
Ef
f
can be parti tioned 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
falls below its
is a
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 Thorn singularity sets =
[x
N
E
Txf
iJ.
has kernel rank
(Concerning the terminology: later in this book we shall have something to say about higher order Thorn singularity sets.) tition of
In this way we obtain a par-
N into finitely many sets on each of which
g
has constant rank.
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
Take
f
: ]R2
-+
]R2
defined by the formula
(x, y)
-+
(x2 , y 2 ).
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, y)
H
H
(x 2 , y)
which "fold" the plane along the x-axis, y-axis respectively.
One pictures
it something like this. y
fold along x-axis
fold along
~
~------~----~~
..
y-axis
x
The reader can easily check that the origin is the only on the axes being
E1
points, and the rest being
EO
E~ -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
the
Zi f
are all submanifolds. Take the smooth mapping
Example 2 2
(x, y) ....
2
= 0,
xy
a. submanifold.
i.e.
Z1
points are
is the union of the two axes, so certainly not a
Zi f
(There are no
points.)
Z2
serving that in this example we can force deforming
given by
A minor computation shows that the
(x + y, y ).
given by
: JR2 ~ JR2
f
However, i t is well worth obto be a. submanifold by slightly
Zi f
Of course a constant deformation will not affect the Jacobian
f.
matrix, and is therefore of no use.
But a small linear deformation will do
the trick.
JR2 ~ JR2
(x, y)
Take for instance 2 2 (x + y, y + 4sx).
~
points are now given by Thus as
f
s
°
moves from
=
xy
s
defined by
Another minor computation shows that the s,
Z1
a hyperbola, so certainly a submanifold.
to a small value so the singularity set
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
device of slightly deforming f.
can be forced to be submanifolds,
X E
f JRn
: JRn ~ :rn.P
n
Suppose then
N=JR,
is our smooth mapping.
Ibtice that given a point
the first order singularity set to which it belongs depends only on
the i-jet of
f
of the jet space equal to
by the
For the sake of technical simplicity we
shall carry it through only in the case that
zi f
at J
x, 1
i.e.
(n, p)
Dxf.
So let us write
Zi
for the subset
comprising all i-jets which have kernel rank
i. is a smooth submanifold of
J 1Cn, p)
(5.1)
Zi
Proof
For this it will be convenient to identify a linear map with its
matrix relative to the standard bases.
of codimension
T,'Te proceed in two steps.
i(p - n +i}
Step 1 matrix.
Let
E
=
( AC
BD)
We claim that
be a
p x n
E has rank
matrix with
k
A an invertible
if and only if
D
=
CA- 1B.
k x k To see
X the matrix E has the same
this, observe that for an arbitrary matrix rank as
=
and the claim follows on choosing Step 2
Choose
k
with
X
=
i + k
tible matrix. in
We take matrix
n,
and let
(0Co
J1(n, p)
with
U (\ Zi
EO
=
(~ ~)
E =
in it the f : U
k x k
EO
matrix
~ J1(p - k, n - k)
is a submersion:
. by gJ.ven
function of
D is an invertible affine mapping.
indeed if one fixes
is a smooth submanifold of codimension
but by Step 1 the inverse image
f- 1(0)
=
is transverse to all the sets
zi ,
i(p -n+i).
E ~ D - CA- 1B. ~ A, B, C the resulting
It follows from (1.2) that (p - k)(n - k)
i U (\ Z,
There is a dense set of smooth mappings
smooth manifold of codimension Proof
It
Consider
A is invertible.
f
j 1f
Zi.
be a matrix in
EO
with the property that for any
Notice that
(5.2)
O.
a smooth submanifold of codimension
U to be an open neighbourhood of
the smooth mapping
f- 1(0)
XA + C =
BO) with AO a k x k inverDO And it will suffice to produce an open neighbourhood U of
will be no restriction to suppose
EO
such that
= i(p
- n + i):
which ends the proof.
f : JRn ~ JRP
0
for which
and hence for which each
Zi f
is a
i(p - n + i).
The first statement is an immediate consequence of the Thorn Transver-
sality Theorem (taking
k
= 1).
And the second statement follows from the
57
fact that
Zi f
is just the inverse image under
so is a smooth
o
manifold of the same codimension by (1.2).
ExamEle 3
Consider a smooth mapping
Zi
transverse to the
will have codimension Zif
in the special case when
immersion
i.e.
that when
p;!: 2n
;!:
i(p - n + i)
i > 0,
is void for
: JRn ~ JRP
f
p ;!: 2n.
i(n + i) > n,
any smooth mapping JRn ~ JRP
is Zi f
In this case
unless
i
which is the same thing as saying that
its germ at any point is immersive.
j1f
for which
= O. f
Thus
is an
Our theory says then
can be slightly deformed to
become an immersion, and indeed a linear deformation will suffice (as is clear from the proof of the Thom Transversality Theorem). take the cuspidal cubic curve
f
Here, for a small positive number f
s
(t)
=
(t 2 , t 3 - ts)
:JR~JR
s
For an explicit instance
2
the curve
fs
is an immersion, obtained from
JR ~ JR2 f
given by
by a small linear
deformation.
image of f
Example 4
image Of fs
We shall consider, in some detail, the condition that
transverse to the
Zi
for a function
f
: JRn
~ JR.
tangent space +
58
be
This means we are to
have image of differential
j1f
1
= J (n, 1)
for all points
a
E
:ntn •
l're
shall show that this is exactly the condition
that the so-called Hessian matrix
H
if = (ax, ax ,
(a) ) rum
J
~
is non-singular at every critical point
are the stan-
where
a,
dard co-ordinate functions in :ntn • 1
J (n, 1)
To start with, let us simplify matters by identifying
by identifying a linear mapping :ntn ~:nt with its matrix relative to the canonical bases.
In particular
Notice that the only
Zi
respectively.
Z
verse to it.
znf
"
wh~ch ar~se
n-1
Thus
f.
i = n.
En
A
't'~ca1
cr~
that 't
po~n
"
codimens~ons
af
aX 1
=0,
••• ,
j1f
at
at
a
af
*
is just the origin in :ntn ,
are
is precisely H,
is the, subspace of :ntn
a
= o
aXn
* when a
In that case the expressions in j1f
0, n
is automatically trans-
conditions
The Jacobian matrix of
erated by its columns.
i.e.
n
j1f
of
Thus we need only satisfy
so the image of the differential of
ate :ntn ,
Zn-1 ,u",n
are
is defined by the
is a critical point, and
can be discounted.
is identified with (-:-:-1(a), ••• , :n (a) ).
is an open set and
and is just the critical set of
easily identified.
Da f
gen-
so its tangent space
* is that the columns of H gener-
Thus the burden of
H is non-singular, establishing our claim. 0f
a smoo th f unc t'~on
=
f:"TO n ~ ~
j1f
'1T>
at wh~ch the Hess~an ~ ~
Our discussion can then be
matrix is non-singular is called non-degenerate. summed up by saying the condition for
=
to be transverse to the
that every critical point should be non-degenerate.
Zi
is
By way of illustration,
the reader may care to check that the function f(x, y)
=
has a degenerate critical point at the origin.
One pictures it like this.
59
In this example
j1f
cannot be transverse to the
Ei.
According to the
theory of this section it should be possible to gain transversality to the by a linear deformation of is linear.
by taking
f,
fs
=f
+ 1s
where
Here is an explicit linear deformation which does just that
the reader is urged to check this for himself.
= The graph of fs
x
3
is rather difficult to draw.
happens is indicated in the pictures below. the case
s = 0,
However, some idea of what The left-hand picture refers to
the right-hand one to the case
shaded area is that where
fs > O.
s > O.
In both cases the
And in both figures the curves
have been drawn for small positive values of
60
2
- 3xy - sx.
(.
fs
=
+(
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 on hard theorems in analysis. rather
However the underlying geometric ideas are
The main object of this chapter is to introduce these
str~ightforward.
ideas in a relatively simple situation where the mathematics is easy enough not to hinder understanding:
it will also provide us with an opportunity of est-
ablishing one or two little facts which we shall have occasion to use later.
§1.
Groups Acting on Sets
By an action of a group
G on a set
usually written
~
(i)
(ii) where
(g, x)
=
1.x
(gh).x
g.x,
M we mean a mapping t
for which for all
x E M,
: Gx
and
=
~
y
G.
when there exists an element
i.e.
x
~
on
g E G for which
The equivalence classes are called the orbits under the action. the orbit through
g, h E G
g.(h.x)
Given such an action we can define an equivalence relation x
M,
x
denotes the identity of
agreeing that
M~
M by y
Given
= g.x. x
E
is by definition the equivalance class which contains
x,
the set G.x
=
{g.x
g E
M
GJ.
61
Before turning to examples let us observe one small geometric point. x 1, x2 be points of with
M lying on the same orbit, so there exists Observe that the mapping
=
is a transformation of serves orbits
M,
(i.e.
defined by
M onto
a bijection of
a
Let g
E
G
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
to
amounts to is that one point on an orbit looks like perty known as homogeneity of an orbit.
a~
What this other;
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 equi-
valence 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 f
0
h
=
k
0
g • - - - - - - - - ( 1)
One can relativize this definition to d-jets as follows. to have equivalent d-jets when there exist invertible germs =
In particular we can take
f, g
jd(k
f, g h, k
are said for which
g). - - - - - - - ( 2 )
0
to be (germs at
cing an equivalence relation on the jet-space
62
(]tn, 0) ~(]tP, 0);
d
0
J (n, p).
of) d-jets, so induThe geometric
examples we have in mind arise from studying this equivalence relation not on d
J (n, p)
the whole jet-space
~n~
of all mappings
co-ordinates on ~p)
mP
but
rather on the vector subspace
d H (n,
p)
each of whose components (relative to the standard
is a homogeneous polynomial of degree JRn •
in
dard co-ordinates
d
in the stan-
The geometry has its genesis in
the following elementary, yet crucial, observation. (2.1)
Two d-,jets
f, g in Hd(n, p)
H, K for which f
exist invertible linear mappings Proof
are equivalent if and only if there
The condition is certainly sufficient.
H
0
=
K 0 g.
To establish necessity,
suppose there exist invertible germs
h, k for which (2) holds.
for the Taylor series of a germ
(~n, 0)
rf> :
~
=
H + terms of degree
~
2
k
=
K +
terms of degree
~
2
then we have
f, g are homogeneous polynomials of degree
d,
(f ~ h)
=
f
(k ~ g)
=
K 0 g + terms of degree > d
0
~
rp
Thus, bearing in mind that the
with H, K invertible linear mappings. components of
~ (~p, 0):
Write
one has
H + terms of degree > d
so = K
=
We can re-phrase the above as follows.
Let
GL(s)
linear group of all invertible linear mappings JRs ~
GL(n) x GL(p)
(H, K).f = K
D
g.
denote the general
mS under the operation
The reader will readily check that we have an action of the
of composition. group
0
0
f
0
on the vector space
H-1 :
d
H
(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-
vertible linear mappings f
0
H
=
d
H (n, p)
K
0
H, K for which
classes described above.
GL(n) x GL(p)
We start with
d H (n, 1)
In this case an element of in x 1' ••• , xn
surface of degree
d
for instance when n when n
are precisely the equivalence
=1
p
d
i.e.
The next step in our study is to indicate the
connexions with geometry.
degree
= f,
It follows immediately from (2.1) that the orbits in
g.
under the action of
The Case
(H, K) • g
=3
so (providing it is not zero) will define a hyper-
in real projective space
=2
we are dealing with
with a curve of degree
And two elements of
is just a homogeneous polynomial of
d H (n, 1)
d
pJRn
of dimension
(n -1) :
d points on a projective line,
in the projective plane, and so on.
will lie in the same orbit if and only if the
corresponding hypersurfaces can be obtained from each other by an invertible linear change of co-ordinates,
i.e.
are projectively equivalent.
ficulty of listing the orbits increases sharply with selves the
luxu~
d:
The dif-
we shall allow our-
of describing in detail some of the simpler cases relevant
to the mathematics of the next two chapters. One starts with
= 1.
d
1
H (n, 1)
in n variables x 1' ••• , xn and
is the vector space of linear forms
elementa~
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
variables
= 2.
2
H (n, 1)
x 1' ••• , xn and
is the vector space of all quadratic elementa~
quadratic algebra tells
us that any such form can be brought into the shape 222 2 x 1 + ••• + Xs - xs+1 - ••• - xr by an appropriate change of co-ordinates.
The numbers
r, s
~
are called the rank,
of the form, and (by Sylvester's
Law of Inertia) are invariant under co-ordinate changes. tiply the form by
-1
However, if we mul-
the rank remains invariant, though the index may change,
so we work instead with the semi-index
= mines,
s'
r - s)
of the form.
Thus quadratic forms in x 1' ••• , xn are classified by rank and semi-index. To make this even more explicit take the case when n = 2. in two variables with the point picture below.
x, y
(a b
2
2
can be written ax + 2bxy + cy
c)
'n ",3. .Il\.
The cone
so can be identified
We obtain four orbits as indicated in the
l.
"
A quadratic form
b2 = ac
comprises the forms of rank
bolic type) with the origin representing the zero form of rank bolic type).
(the para0
(the~-
The remainder of the space comprises the forms of rank 2:
indeed the inside of the cone corresponds to forms of semi-index 0 elliptic type) and the outside of the cone to forms of semi-index bolic type).
(the (the hyper-
a a
Let us continue with the case all cubic forms in
n variables
d
= 3.
H3(n, 1)
x 1' ••• , 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,
~d
n
~
4.
since this case will arise natur-
ally in the next chapter we shall take the opportunity to describe a onoe familiar bit of pure mathematics, the study of binary cubic forms f(x, y)
= 65
It is easy to list the orbits using a little algebra. allow the variables x, y forms.
Let us (temporarily)
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
It follows that the zero set of f, equation f(x, y)
=
i.e.
the subset of
c2
defined by the
0 will comprise three lines through the origin.
shall distinguish four possible types of the following four types of triples.
bina~
We
cubic form, corresponding to
(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. a~
Certainly two bin-
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 C2 which maps the lines of the one triple to those of the other.
And conversely two
bina~
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.
tain normal forms for non-zero
bina~
cubic forms by just choosing an example
of each of the four types just described. given in the following table.
66
We can ob-
The standard choices are those
type
normal form
- x:y2
elliptic
x3
hyperbolic
~ + x:y2
parabolic
2 xy
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
Hf
=
if
if
ai
axay
a2f ayax
if
1
3b
ai
and a little arithmetic will verify that it is given by the formula
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.
The space of binary
cubic forms can be identified with JR4 by identifying
a~
+
3~x2y
+ 3yxy2 + Sy3
with the point
(a,~, y, S): the partition of the
non-zero forms into four types yields a partition of JR4 - {oj into four sets, and by projecting this appropriately into JR3
{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 a little further. consider is
n = 3,
i.e.
terna~
cubic forms in
The next
x, y, z:
thought of as cubic curves in the real projective plane
cas~
to
these are best The deriva-
p]t3.
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.
0
we shall
One starts with non-singular cubic curves.
non-singuw.r cubic
zy 2 - x 3 + axz
'With
2
+ bz 3
4a 3 + 2?b 2 I 0
To such a curve one associates the so-called j-invariant
j
= 4a3/4a3
+ 27b 2:
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.
e>« •
-<
2
2
2
2
3
+ y z + x z
3
+ Y z - x z
crunodal cubic
x
acnodal cubic
x
cuspidal cubic
x
3
- y
2
z
We are left with a motley array of reducible cubic curves.
The cubic can
reduce to a conic and a line, which mayor may not be tangent to the conic; or it may reduce to three lines, giving rise to various possibilities.
69
conic a:n.d "Line
conic a:n.d tangent
2 Y (xy - z )
triangle
three coneurrent 7,ines
70
two lines one repeated
x 2y
triple line
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
~
2
One thinks of an element of Hd(n, p) d
of dimension ~ p, namely the subspace spanned by its components:
H (n, 1)
for that reason elements of d
of dimension ~ p.
H (n, 1) sion
~
as determining a vector subspace of
2,
Hd (n, p)
can be called linear systems in
The simplest linear systems are those of dimen-
3 which have long been known as pencils, nets, respectively, and
whose study has whiled away the leisure hours of many a geometer in the past. We shall, reluctantly, confine ourselves to the simplest example of all, namely pencils of binary quadratic forms
(i.e.
d
= 2,
n
= 2,
P
= 2)
which
we shall have occasion to recall in Chapter V. Let us spell out one or two points. will be a pair (~, q2)
A pencil of binary quadratic forms
of quadratic forms in two variables x, y
be identified with the subspace of H2 (2, 1)
and is to
spanned by these forms:
it will
therefore be a plane through the origin, a line through the origin, or just the origin itself.
Recall that
identifying the quadratic form
H2 (2, 1)
can be identified with m3 by
ax 2 + 2bxy + cy2
with the point
(a, b, c),
and that the various types of binary quadratic form are separated exactly by the cone
b
2
= ac.
It is convenient to list the possible pencils by the way
in which they lie relative to this cone.
We claim that there are exactly
seven possible pencils, as described in the following table.
It should be
clear to the reader first that any pencil of binary quadratic forms must be one of the seven types under the action,
d~ribed,
i.e.
and second that these types are invariant
if we apply an element of the group to a pencil of
a given type the resulting pencil will be of the same type. 71
normal form 2 (xy, x + y 2 ) 2 (xy, x - y 2 )
geometric description plane intersecting the cone plane lying outside the cone
(xy, x 2 ) 2 2 (x + y , 0)
plane tangent to the cone line inside the cone
line tangent to the cone
(xy, 0) 2 (x , 0)
the origin
(0,
line outside the cone
0)
What remains to be shown is that a pencil of a given type can be reduced to the normal form shown.
The group acting here is
To simpli-
GL(2) x GL(2).
fy proofs we refer to elements of the first factor as changes of co-ordinates, I t is
and elements of the second factor as changes of basis in the pencil. easier to work backwards through the list. represents a line through the origin;
Suppose our pencil
(q1' q2)
by a change of basis we can suppose
into one of q2 = 0, and then a change of co-ordinates enables us to put 222 the forms x + y, xy, x, according as it is ellipti~hyperbolic or parabolic, so yielding the three possible normal forms shown for the pencil. Suppose next that the pencil
(q1' q2)
represents a plane through the origin,
tangent to the cone, so by a change of basis we can suppose
2
cannot involve Y ,
q1 = xy,
as was required.
and a change of basis allows us to suppose
Finally, suppose the pencil represents a plane
through the origin not tangent to the cone. suppose
q1
By a change of basis we can
hyperbolic, and then by a change of co-ordinates that
subtracting off an appropriate multiple of this from the form 72
singular, and
the plane being tangent to
then by a change of co-ordinates that the cone
q2
ax2 + f3i
with a
-J
0, f3
I
0
q2
we reduce
q1
= xy: q2
to
else the plane would be tangent to
the cone.
Obvious changes of co-ordinates and basis now bring the pencil to 2
(xy, x +
the form
the + sign corresponding to the case when the plane
intersects the cone, and the - sign to the case when it fails to intersect the cone.
§3.
And that completes the derivation of the list.
Smooth Actions of Lie Groups on Smooth Manifolds
We wish now to specialize the idea of an action if? of a group
G on a set M
to the case when all three objects are "smooth", in a sense to be made precise. First, we consider the group.
A Lie grOUp is a group
mnaifold, and for which multiplication
G which is a smooth
Gx G
~
G and inversion
GL(n)
is a Lie group.
G~
G are
smooth mappings. Example 1
The general linear 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.
x
n
of all
Further, mUltiplication of
is a polynomial mapping, so certainly smooth, and its res-
triction to matrices in GL(n) inversion in GL(n)
M(n)
n
will likmvise 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
G on a smooth manifold M we mean an
action if? : G x M -+ M for which if?
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)
1
on H (n, p)
the partition by rank of linear maps JRn -+ JRP,
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 if? : G x M -+ M be a smooth action of a Lie grouE
(3.1)
smooth manifold M. folds of the
~ouE
SteE 1
G on a
I t is assumed that all the orbits are smooth submani-
M the natural maEping if? : G -+ G.x of x onto the orbit given bI g -+ g.x is a submersion.
M.
For any Eoint
I claim that if? x
x
E
has the same rank at every point in
fices to show that the rank of if?x rank at the identity element G defined by g-+ hg, by y-+ h.y.
1 E G.
and let e
at any point Let
e
G.
h E G 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 Thif? x
74-
It suf-
have the same rank, as was claimed.
T1if?x and
~x
G
j
8
~
G
step 2
x
>G.x
T 1~x
T1G
T,. j
1·
>G.x
Th~x
ThG
In view of step 1 it suffices to show that
some point in
GJ
> Tx (G.x)
~
x
1Tx· ) Th .x (G.x)
is submersive at
and that follows immediately from SardIs Theorem J guaran-
teeing the existence of at least one regular value J necessarily in the image
Thus the required tangent space tial
T1~x
of the tangent space
Tx(G.X)
is the image under the differen-
T1G to the group at the identity element.
And that is how one goes about computing Tx (G.x) Example 3
in practice.
Let us return to our class of geometric examples namely the
natural action of F : ]in -+ ]iP
Let
on the vector space
GL(n) x GL(p)
be an element of this vector space J with components
each a form of degree
d
in
n
variables
F to the orbit through F.
shall compute the tangent space at
mapping of the group onto the orbit is given by
(H, K)
We
x 1J ••• , xn •
-+
K
0
The natural
F
0
H-1 ,
and
we require the image of the differential of this mapping at the identity I J i.e. pings
the vector sum of the images of the differentials at -1
H-+FoH
and K -+ K
first the mapping H -+ F
0
-1
H •
0
F.
I
of the map-
We consider these separately.
Take
Observe first that it is the composite of
matrix inversion H -+ H- 1 with the mapping H -+ F
0
H:
since the differen-
tial of the former mapping (at any point) is invertible we see that the
75
images of the differentials at
of H ~ F
I
so we need only consider the latter mapping. differential as follows. in ]{n,
and for to
which maps
e.
the curves
Yij
J
o.
close to
F
0
Yij'
We compute the image of the
••• , en denote the standard basis vectors
:> n
A.. : ]{n ~ ]{n be the linear mapping
ei ,
and the remaining basis vectors to
GL(n)
~J
through
I
M(n)
GL(n).
i.e.
G = (g1' ••• ,~)
all other
zero.
d
in
to the orbit through and
0
with respect to
0
F:
f
=
of
I,
n
with just one
f
variables
gi
ax.
equal to an f j ,
• • • J
= 1.
Let
f
and
be a
The tangent space at
x • n
will be the subspace of
Hd(n, 1)
f
spanned by the
By Euler's Theorem Xi
of : ax.
thus
~
d
spanned by the
H (n, 1)
1:> i, j :> n.
By way of explicit illustration take the case of
forms,
d =
= 2,
p
us that the tangent space at
f
76
~
and the image will be
Example 5 i.e.
of
X.--J
the required tangent space is just the subspace of with
of
so lies in the subspace spanned by the
dE Xi OXi
of X.--~ oX j
t
this is the restriction of a linear
In fact it is simpler than that.
f.
J 1
which is the
of the curves
0
It is worthwhile isolating the case when p
form of degree
t
yield a basis
which by the Chain Rule are the
spanned by all
Example 4
with
It follows that the image of the differential at
the derivatives at
g's
I
of all linear mappings ]{n ~ ]{n,
H will be spanned by the derivatives at
0
Consider now
given by y •. ( t) = I + tA .. ~J ~J
mapping, so will coincide with its differential at
x.
o.
Evidently the derivatives of these curves at
Next, consider the mapping K ~ K
~
H will coincide
0
Let
+
x. oOf
H~ F
j
in
tangent space to of H ~ F
H- 1,
:> i,
for the vector space
I
0
3, n
= 1.
And take
f
= x 2y.
to the orbit through
f
bina~
cubic
The above tells will be spanned by
af
~,
2
x y,
af
:rax '
2
which are respectively the monomials
2
2xy ,
2
x 2y,
so a basis for the tangent space is given by
2x y,
It
x:y ,
follows that the tangent space, hence the orbit, has dimension 3. Example 6
For an example involving linear systems consider pencils of 2
F = (x , 0).
binary quadratic forms, and in particular the pencil Example 3 the tangent space at by x
aF
F
to the orbit through 2
(x ,
together with
ax'
basis for the tangent space is provided by
F will be spanned
0), (0,
(x2 , 0),
By
2
x);
(xy, 0),
hence a (0,
i)
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 x M~ M
G on a smooth manifold M, supposing that al1 the orbits are
smooth submanifolds of point
~
M.
We can define the codimension
x E 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 Experience suggests that the difficulty of listing the orbits of
to answer.
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 larger the number of possible types of behaviour. be most feasible to list those orbits of codimension This is worth re-phrasing.
One calls
G when there exists a neighbourhood Xl
E
U lies in the same orbit as
tly small perturbations of terminology. one point
x
x
x:
U of
On this basis it should O.
x E M stable under the action of x
in
M such that every point
intuitively, this means that sufficien-
will not displace it from its orbit, hence the
Note that the homogeneity property for orbits implies that if 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
M.
This geometric notion can be re-
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. Example 7
Consider again the natural action on binary cubic forms.
that there are five orbits, with representatives have respective codimensions x3
.±.
3 2 3 0, x , x y, x
~
Recall
2
these
xy ;
4, 2, 1, 0 so we have two stable orbits, namely
x/.
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-+
GL(3)
A1/3x ,
y
A1/3y ,
t-+
has dimension
any point must be of dimension
~
9,
9.
of ternary cubic form has dimension 10. dimension
78
~
1.
The
AI 0
can
and this is clear since the change z
H
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 ~
We wish to expand on the question of just how the action a point
x E M.
~
: G x M~ M.
behaves close to
The crude picture is that the orbits through points close to
x flow smoothly together past x:
and a very basic idea in studying this
situation is to look at the way in which the orbits cut a (small) cross-section of this flow.
(See Figure 1.)
C.x x
Fig. 2
This crude idea can be made precise as follows. manifold
S of M,
(See Figure 2.)
with
x E X,
the dimension of
which is transverse to the orbit
We shall certainly have
least possible dimension for
identity, the slice
M given by mH g.m,
(3.1)
x
in M.
with
i.e.
when
E
G close to the
S will slide slightly along the orbits:
the basic pic-
g
S will sweep out a small neighbourhood
The formal mathematical statement of this reads as follows.
and let
smooth submanifolds of following property:
the
S a slice at x.
Let ~ : G x M ~ M be a smooth action of a Lie group
smooth manifold M,
G.x.
dim S + dim(G.x):
and in that case we call
ture which we wish to convey is that
U of
dim M ~
S will be when we have equality,
S is cod x,
Under the diffeomorphism of
Consider a smooth sub-
M.
x
E
M.
It is assumed that all the orbits are
The point
x has a neighbourhood U with the
there is a smooth submanifold H of
identity element, and a slice
S at
G on a
x,
G containing the
for which the restriction of
~
to
H x S is a diffeomorphism onto U.
79
I
- - •x ____ 1__ ,, '
For obvious reasons such a neighbourhood structure.
One thinks of
G.x
U is said to have a product
U as a small tube with the orbit
G.x
running
down the middle. step
We start by isolating a lemma which will be used twice in the main Suppose that
body of the proof. manifold Aa ,
CC of
manifold a
=b
o
in N a
0 mapping to
c = a - b
then
x E B:
A with
Indeed by
c.
+
bourhood of wi th
and that
Bb
is a smooth submanifold of a smooth
we claim that there exists a smooth sub-
x E C which is transverse to
B and for which
(1.2.2) there is a diffeomorphism of an open neighonto a relatively open neighbourhood of
x,
0 x NC
and Nb x 0
corresponding to
maps to a smooth submanifold
x
in
A
I f we put
B. C of
A with the
desired properties. Step 2
As we have already pointed out the natural mapping
~x
: G ~ G.x
of the group onto the orbit is a submersion, so by (11.1.2) the inverse image Gx
under this mapping of the point
x will be a smooth submanifold of
of codimension the dimension of the orbit, with Gx
is a subgroup of
to
Gx
with
the orbit
80
dim G
G.x.
(Incidentally,
1 E Gx '
G known as ths isotropy subgroup at
there exists a smooth submanifold
=
H of
dim Gx + dim H,
G with i.e.
G
By Step
x.)
1 E H which is transverse
H has the same dimension as
Again by step 1 there exists a slice
S
at
x.
Consider
now the restriction
: Hx S
~
~
M of the action
~.
Observe that domain
and target have the same dimension, and that the differential at invertible.
x
maps
~
is
H, S are
It follows from the Inverse Function Theorem that if
small enough then of
(1, x)
H x S diffeomorphically onto a neighbourhood U
D
in M.
The reason for proving this result is to justify the intuitive idea that in order to study the action near the orbits cut a slice
S.
x
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
M.
S may
All this
brings us to one of the central ideas of the subject, namely that of an "unfolding" of
§4. Let foli
x,
roughly speaking a parametrized slice at
x.
Transversal Unfoldings Gx M
~
M,
~
and let
M be a smooth action of a Lie group x E M.
As usual, it is tacitly understood that all the
orbits are smooth submanifolds of mean a germ
X:
(:nl,
0)
is transverse to the orbit
G on a smooth mani-
~ G.x
M.
By an r-parameter unfolding of x
we
(M, x) : X is said to be transverse when it at
x
i.e.
=
Tx M.
One pictures the situation something like this:
81
M
Now let
x
have codimension
it follows immediately that versal unfolding.
r
c. ~
c:
Given that when
r
=c
X is a transverse unfolding we call
X a minimal trans-
The construction of an explicit minimal transversal unfol-
ding X is a fairly straightforward matter.
Let us restrict ourselves to
the case when M is a linear space, so it is natural to look for linear unfoldings,
i.e.
those of the form ••• , uc )
where
b 1, ••• , b c
=
+ ••• u b
c c
are fixed vectors in M.
The condition for such an
X
to be transverse is that
where the first term represents the subspace spanned by b 1 , ••• , b c ' the image of the differential of choose
b 1 , ••• , b c
X at
o.
i.e.
Thus all we need to do is to
to be a basis for a supplement of the tangent space
T (G.x).
x
Example 1
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. has as basis the monomials
x3 , x2y, xy2 , y 3 •
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
y}
.:!:.
~
Basis for tangent space
2
x3 , x 2y,
2 xy
y},
2
2, y3
x3 +
~
2
2 xy+
U;
2 x y,
y},
y}
~
Transversal Unfolding
-
~
y}
x 2y
2 +~
+ Vy3
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 ]R2
u, v
As the point
y}.
We shall determine the
for which the unfolding represents a binary cubic of a given
The associated Hessian quadratic form of the unfolding is type. 2 2 2 3ux + 9v~ - u Y with discriminant a multiple of is the zero quadratic form if and only if (u, v)
(u, v)
so the transversal unfolding x 3 + uxy2 + v~ will
cut all orbits through binary cubics close to values of
y}.
u
=v
= 0:
The Hessian
so the origin in the
plane is the only point which corresponds to a perfect cube.
now stay away from the origin.
The unfolding will represent a parabolic bi-
nary cubic when the Hessian is parabolic, v2 + 4u3
=
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
At all other points in the plane
inside the cuspidal cubic we have
l
v2 + 4u3
+ 4u3 < 0
I o. -
correspon-
ding to elliptic binary cubics, and outside the cuspidal cubic we have
83
v 2 + 4u3 > 0 corresponding to hyperbolic binar,y cubics.
Thus we arrive
at the £ollowing picture, representing a cross-section o£ the £low o£ orbits close to
~.
v hyperbolic eUiptic
u hyperbolic
Some more interesting situations are provided by the transversal un£oldings o£ cubic curves. Example 2
=
f
We shall study the transversal un£olding of the cubic curve
x(x2 + yz),
which is of the conic and chord type.
affine plane the conic is a parabola y = _x2 , x
=0.) f
and the chord is its axis
A minor computation shows that the tangent space to the orbit at
is spanned by the monomials of degree so
3
f
in x, y, z with y3, z3 deleted,
has codimension 2 and a transversal unfolding is provided by
As in the previous example we ask for the values of a given type.
Certainly, when u u ,l 0,
Suppose
and chord type.
= 0, v
that the tangent lines there are Likewise when u = 0, hand when u
I
0,
v
v
I
I
°
= 0:
v
=0
x
= 0,
u, v £or which F has
we have the original conic
putting
(0, 0, 1),
zero one £inds a singular point at
of
ax ,
of of Ty' az
equal to
and £urther computation shows
y = 0 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
Thus we obtain the following picture,
representing a cross-section
the
o~
orbits close to the" conic and
~low o~
chord type.
non-singular crunodal
Another pleasant illustration is provided by the transversal
Example 3
the cubic curve
~olding o~
tangent space to the orbit at x, y, z
~olding
3
~
with
= xyz
~
o~
triangle type.
In this case the
is spanned by the monomials of degree 3 in
f
3 deleted, so f
X-, Y , z
has codimension 3 and a transversal
is provided by F
=
We ask first for the values of there exist
xyz + ux3 + v,y3 + wz 3 • u, v, w for whioh
F
i.e.
x, y, z not all zero for which
aF ax
aF ay
2
=
yz +3ux
=
xz + 3v,y
aF
az =
2
=
0
=
O.
is satisfied.
Let us see what happens when u ~ 0,
that if
~
of
x, y, z
x ~ 0,
y
0,
z
O.
or w = 0
0
u = 0,
I
or v = 0,
xy + 3wz
2
=
Notioe that if
I
is singular,
then certainly this condition v ~ 0,
w ~ O.
Obsene
vanishes then all of them vanish, so we can suppose But then our equations yield
85
We are only considering what happens very close to
f,
so we can suppose
v, w to be very small, which means the expression within brackets is 222 x Yz
so
I
all
0
=
0
-
which is impossible.
and small then
what happens in the plane
In other words if
F is non-singular. u
=0
in
obtains a crunodal cubic. plane
v = 0,
and the plane
0,
u, v, ware
Let us look in more detail at
(u, v, w) space.
v, ware zero we recover the original triangle: zero we have the conic and chord type.
I
u,
Of course when both
when just one of
and when both
v, ware
I
v, w is 0
one
By symmetry one has similar situations in the w
= O.
The nett result is the following pio-
ture of a cross-section of the flow or orbits past a cubic curve of triangle type.
D
cY'unodal conic and line
Example 4
For an example involving linear systems let us look at pencils
of binary quadratio forms.
=
We shall study the transversal uni'olding of the
2
2
pencil F (x , 0). The space of penoils is spanned by (x, 0), (xy, 0), 2 2 2 (y ,0), (0, x), (0, xy), (0, y). In Example 6 of ~3 we saw that the tangent space at (0, x 2 ) 2
(y ,0): by
86
F
to the orbit through
F is spanned by
so a basis for a supplement is provided by thus
(x2 , 0),
(0, y2),
(0, xy)
(xy, J), and
F has codimension 3, and a transversal unfolding is given
(x2 , 0) + u(O, y2) + v(O, xy) + w(y2, 0)
i.e.
(x 2 + wy2,
uy2 + vxy).
The objective now is to determine the values of
u, v, w for which this pen-
cil has one of the seven possible types listed in
~2.
Recall that we identi-
fied the space of binary quadratic forms with :rn.3 by identifying 2 2 ax + 2bxy + cy with (a, b, c), that the singular forms then correspond to the cone
ac,
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: 2
2
tion shows that the binary quadratics
x
endent if and only if
i.e.
v = 0,
tangent to the cone;
when w > 0
along the w-axis in
=
4A(AW
To this end we consider how the
+ pw)
i.e.
Ax2 + pvxy +
(AW
+ pu)/ j
O. - - - - - - - - - - - *
This expression is itself a binary quadratic in A, p 2
2
which is singular
i.e.
= 16(u + wv ).
a line
A typical element of the pencil is
A(X2 + wy2) + p(uy2 + vxy)
(pv)2
F,
Now we ask what happens off the w-axis, when
a line outside the cone.
plane cuts the cone.
(u, v, w)
we have a line inside the cone, and when
our pencil is a plane through the origin.
when
2
uy + vxy are linearly dep-
In that case when w = 0 we recover the original pencil
space.
w< 0
u = 0,
+ wy ,
a simple computa-
The equation u 2 + wv 2
discriminant, and is a surface in :rn.3,
= o
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 A : p
for a solution so the pencil meets the cone
represents a plane tangent to the cone. u
2
+ wv
2
< 0
so the equation
*
* b
has a unique ratio 2
= ac
in a line, so
Inside the umbrella one has
is satisfied by no real ratio
A: P, which
means that the pencil represents a plane which does not intersect the cone.
87
not in tersectin
p~ane
~ine
outside cone
~ine
inside cone
w 2
And outside the umbrella
u
two distinct real ratios
A:
+ wv ~
2
>
0,
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
§5.
Versal Unfoldings
We shall conclude our discussion of finite-dimensional unfoldings by expanding somewhat on the sense in which a transversal unfolding describes an action close to a point;
this will provide us with a useful characterization of
stable points in terms of their unfoldings.
To this end we introduce a series
of notions.
As usual,
manifold
and it is assumed that all the orbits are smooth submanifolds of
M.
M,
We let
iP
is a smooth action of a Lie group
G on a smooth
x E M.
Equivalence of Unfoldings Two r-parameter unfoldings
X1, X2
of
x
there exists an r-parameter unfolding
I
are said to be equivalent when : (JR r, 0)
~
(G, 1)
of the iden-
in the group for which
tHy element
Pictorially, the idea is that you can get from X1 to
X2 by sliding
smooth~
down the orbits.
o
Induced Unfoldings Suppose that
X is an r-parameter unfolding of Then
folding of
x,
said to be induced by
H.
x,
Y = XOH
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
M
MOrphisms of Unfoldings Let X, Y be r, s-parameter unfoldings of is a pair
(H, I)
with
with H: (JRs , 0)
-+
I
x.
A morphism from
X to
Y
an r-parameter unfolding of the identity 1, and
(lRr , 0)
a change of parameter, for which
valent to the induced unfolding H"'y
via
I.
When
r
= s,
and
X is equiH is inver-
tible we call the morphism an isomorphism.
Versal Unfoldings Anunfolding Y of x
is said to be versal when for any unfolding X of
there exists a morphism from X to
Y.
Intuitively, this means that
sufficiently large to allow all unfoldings of
x
to appear in it.
x
Y 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~)__~L~e~t~ ~
M,
be a smooth action of a Lie group
all of whose orbits are smooth submanifolds of M.
icient condition for an unfolding X : (lRr , 0)
90
G on a smooth manifold
-+
(M, x)
A necessary and suffto be versal is that
it should be transversal. step
In which we establish necessity of the condition.
sal.
We have to show X is transversal,
i.e.
Consider any 1-parameter unfolding Y: (JR, 0) there exists a morphism
(H, I)
from
Y to
Suppose X ver-
that
-+
X,
(M, x).
i.e.
As
X is versal
= I(v).X(H(v».
Y(v)
Take differentials to get
(1) follows immediately from (2) since any tangent vector in T M is a tanx gent vector to some curve through Step 2
x,
i.e.
In which we establish sufficiency of the condition.
transversal.
We wish to show
Y.
a 1-parameter unfolding
X versal.
Suppose X
Observe first that it will suffice
to show that some unfolding induced by X is versal.
Let us therefore replace
X by an induced unfolding J*X which is both transversal and minimal, and continue to denote it by the same letter linear mapping into JRr , taken by TOX
X.
(One can choose
J
to be a
with domain of the correct dimension, whose image is
to a supplement for
in TxM.)
Tx(G.x)
Such a germ X is
immersive, and the image of a sufficiently small representative will be a slice S
x.
at
i.e.
Recall now that
there exists a smooth submanifold H of
ment
for which the gem
is invertible. x. IT H,
x has a neighbourhood with a product structure,
at
(1, x)
of the restriction of if?
Consider now an arbitrary unfolding Y : (JRs , 0)
We define germs TIS
IjJ
G through the identity ele-
J, Z
are the germs at
by J = TIft (1, x)
0
IjJ
-1
0
Y,
Z
= IT S
0
IjJ
to H x S
-+
-1 0
(M, x) of
Y where
of the projections of H x S onto H, S
91
respectively.
z.
to K
.
It is immediate that
=
Y(u)
so
J(u).Z(u)
Y is equivalent
On the other hand if we define the change of parameter
(]Rr, 0) ~ (]Rr, 0) by K = X- 1
induced from
and conclude that
X,
It follows that
we see that
Z
o
(K, J)
Z
* = KX
is an unfolding
is a morphism from
Y to
X.
X is a versal unfolding.
o
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 x
X, X' of
are isomorphic.
X, X'
Suppose
are r-parameter unfoldings.
is equivalent to some induced unfolding H*X.
As
And as
X'
=
that the image of the differential
T(X
0
subspace of
TH
is likewise r-dimensional:
that
TH
T~,
so the image of
is invertible, and hence
TX
0
TH
H is invertible.
H*X
is universal
That means
is universal, hence is a minimal transversal unfolding by (5.1). H)
X'
X is versal,
is an r-dimensional it follows
The result follows.
o
Finally, as promised at the beginning of this section, we shall make use of these ideas to give a simple characterization of stable points. let us call an r-parameter unfolding ' t o th e r-parame t er cons t ant unf 0 ld ~ng
To this end
X of x
trivial when it is equivalent
=.
given by
(",r. 0) ~ ~ (M, x)
u
~
x.
We then have Under the hypotheses of (5.1) a necessaEJ 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 O-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
valent to a constant unfolding.
eve~
unfolding
In particular, take
unfolding.
Certainly then some representative of
orbit
so
G.x,
dimension
0,
i.e.
x
x
is equi-
X to be a transversal
X is a mapping into the
X can only be transversal to the orbit when
G.x
has co-
D
is stable.
The reader should be warned that the cularly useful.
X of
theo~
of this section is not parti-
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
§1.
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 -+ lR 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
f i , f2
exists an invertible germ mutes.
right-equivalent
We call two
(or ~-equivalent)
when there
g 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 unnecessar,y symbolism we shall,
throughout this chapter. adhere to the convention that equivalence of two germs of functions is to mean
94
~-equivalence:
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
f1
~
f2
to mean that two germs
f 1, f2
of
functions are equivalent. The starting point is to tr,y 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 theor,y covering the situation we wish to study would lie far and away above the intended level of this book.
All
we shall do is to argue heuristically to arrive at precise interpretations of vague geometric ideas:
we shall never use such arguments to prove
Our basic objects then are germs at
x we see that we can suppose
g: (N, x) n
N=lR,
jects we wish to study is the set Bn Of course
Bn
(lR, y).
Thus the set of ob-
and x = O.
of all germs
f
Taking a chart a
: (lRn, 0)
-+
(lR, y).
is a real vector space under the natural operations of addition
and scalar multiplication. that;
-+
propositi~.
In fact
Bn has more algebraic structure than
the operation of multiplication on lR induces an operation of multi-
plication on en'
under which
B n
becomes a real algebra.
More information
on the subject of real algebras can be found in Appendix C of this book. Next, let )£n
denote the set of invertible germs
observe that )£ n
1-+
f
0
-+
(lRn, 0), and
is a group under the operation of composition.
acts on Bn by composition, (g, f)
g: (lRn, 0)
-1
g
•
i.e.
we have an action )£ x B -+B n n n
Further, two germs in B
n
Now )£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. and :Jt
It is an unfortunate fact that
n
are not smooth manifolds in the
sense in which that word has been used in this book. will not be, an insuperable difficulty. that the action of :Jtn
on
Gn
That need not be, and
What we shall do is just to pretend
is a smooth action of a Lie group on a smooth
manifold, and proceed by analogy. This goes as follows. the orbit through
f.
Choose a germ
f
8n ,
in
We shall pretend that :JtS
and look for a vector subspace of
8 n
and write just :Jt.f for is a "smooth submanifold"
which has some reasonable claim
to the title of the "tangent space" to the orbit fr..f just mimic the theor,y of Chapter III. the orbit is the mapping :Jt n
~
G
n
at
To do this we
f.
The natural mapping of the group onto gH fog -1 ,
given by
and one would expect
the required "tangent space" to be the image of the "differential" at the identity 1 g~
~f
this mapping.
f o g
gr+ g -1
-1
A preliminary simplification is to observe that
is the composite of
gH g
-1
gH f
and
0
g:
one expects
to be a "diffeomorphism", so the required image should coincide with
that of the "differential II at the identity of
g
-+
fog.
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 IItangent space" at the identity to the group fr. n •
A
Observe that
:ltn
little common sense will soon produce a candidate for this. is a subset of the vector space 8 0 n,n
of all germs
moreover, we should really think of :Jt n
g: (lRn, 0) ~ (lRn , 0):
as an "open" subset of
because
i f we disturb an invertible germ very slightly we still expect to get an in-
vertible germ. to :R.n'
96
On this basis we would certainly expect the "tangent space"
at any point, to be just 8°n,n·
The target of the "differential"
should be the "tangent space" at
x
we would expect this to be just
Gn •
linear mapping
3 in §3 of Chapter
the components of
en
is a vector space
Thus our "differential" should be a
through
f
in
8n
pect to
t
at
o
an arbitrary "tangent vector" to 1t
g will be written
1 + tg with
"differential" to
We start with an arbitrary germ
III.
germ as the "differential" at
=
and since
The line of attack to take here is suggested by
i.e.
y(t)
Gn ,
The problem now is to find a sensible formula
for the "differential". Example
to
g
t
0
close to
of the curve through 1 in
of
1:
1t
given by
n
The result of applying the sought for
should then be the "differential" at given by
at
Now we can realize this
g1' ••• , gn.
O.
n
t ~ f(y(t))
the "differential" with res-
i.e.
f(x 1 + tg 1, ••• , xn + t~).
Just doing this naively
af To sum up: we ~ax· 1 n 80 ~ 8 given expect the required "differential" to be the linear map n,n n
by the Chain Rule one gets the answer
g1
af
of the curve
0
a:x-
+ •••
by the formula af
•
···+~ax· n
The required "tangent space" shoula. be the image of this linear mapping, and The germs
this is neatly described in algebraic terms as follows. g1' ••• ,
Gn in
~
all have zero target, so belong to the ideal
comprising all germs with zero target.
f.
the produot ideal Afn.Jfo
in the algebra
Let us denote by
generated by the partial derivatives
the Jacobian ideal of
~
af aX 1 ,
af
••• J
axn
The image of the linear mapping
Jf
the ideal
and call it
•
above is just
In this way we oome 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
Afn.JrWe have described our
model in the language of germs, but could just as well have worked with
97
n
0 in lR ,
functions defined on a neighbourhood U of
and thought of the
germ case as the limiting case when U is infinitesimally small. case we should replace
&n
functions defined on U. Diff
(U)
by the vector space
lR)
of all smooth
And we should replace the group
of all diffeomorphisms
U~ U
We would then have an action of Diff formula as before.
~(U,
(U)
In that
'R
n
by the group
under the operation of composition. on
CO(U, lR)
given by the same
And the same heuristic reasoning as before would bring
us to the conclusi(1n 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 'Rn
origin, whilst in the group
Diff (U)
U is.
property, no matter how small model is oversimplified.
*.
Note however one
our germs are forced to preserve the our diff'eomorphisms need not have this All this indicates is that our
We shall correct matters by allowing our germs
to have an arbitrary target, so to lie just in
en
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 defini tion.
Let
Jacobian ideal
Jr.
f E en'
We define the tangent space to f to be the
And we define the codimension cod f
codimension of the tangent space
If
cod f
is finite we
s~
that
is of infinite codimension.
f
in en'
to be the
i.e.
is of finite codimension;
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
whether or not a given germ f compute the codimension.
Jf
of
en:
this then
is the object of the next section.
§2.
The Algebra
Gn
We have already observed that sense of the word.
Gn
is indeed a real algebra, in the technical
(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. A
necessa~
Observe first or all the following elementary fact.
and sufficient condition for a germ
vertible (as an element of the ring Necessity fg = 1:
Suppose then r(o)g(o)
Sufficiency inverse for
Suppose
f
r E & to be inn
is that reo) ~ O.
Gn )
is invertible, so we can find a germ =
1,
f E Gn
so
g
E
& with n
f(O) ~ O.
satisfies
f(O) ~ 0:
then
g = 1/f
is an
o
f.
This tells us that the ideal
of all germs in G with target n
0,
has
a rather special algebraic property • ..A'n is the unique maximal ideal in Proof f E I
Suppose with
is an ideal with Ln C I C G,
I C G - n
f(O) ~ O.
-
f
follows that .-R' is maximal. n
Gn •
n
so we can find an
is invertible by (2.1) and hence
I = &. n
It
And the same argument establishes uniqueness.
o To some extent the importance of the ideal
~
n
lies in the fact that it
allows a convenient algebraic descriptinn of certain ideals in Gn which frequently come into consideration.
VIe
have in mind the idea
J k+ 1
of all
99
f E 8 n whose k-jet is zero (i.e.
all partial derivatives of
' k vanish at
0).
little fact
sometimes called the Hadamard Lemma
f
of order
As a preliminary we shall establish one extremely useful -
which we shall also
have occasion to use in later chapters. Let
U
be a convex neighbourhood of
in ~n,
0
smooth function defined on U x ~q which vanishes on smooth functions
f l' ••• , f n
~ x 1 ' ••• , xn
and let
0 x ~q:
f
be a
there exist
on U x ~q with
are the standard co-ordinate functions on ~n.
Proof
s: f = 2:
=
i=1
Xi
aO:.
(tx 1 , ••• , txn , Yl' ••• ,
~
y~dt
Xi f i (x 1 ' ••• , xn ' y l' •••• y q)
i=1 i f we take
D Now we can characterise the ideal
(2.4)
dk
= 1~.
in x 1' ••• , xn
100
d k + 1•
and is generated by (the germs at
of degree
'k.
0
of) the monomials
First, we establish
Proof
It is clear
that.ff~
C :ilk'
follow by induction from
that
f
~:ilk_1.
q = O.
is trivial.
k =
k
so it suffices to show ~
:ilk
Lemma, in the special case
The case
:ilk ~.ff n '
which will
This we prove using the Hadamard
Indeed if
f E:il k
= 0,
then f(O)
so
and it is clear from the construction of the
=
This argument shows, in parti:il 1
=
.ffn
It follows that
:ilk
=.ff~ is generated by (the germs at
mials of degree
k
in
cular, that
In particular
(2.4)
is generated by (the germs at
of)
0
x 1' ••• , x n •
of)
0
the mono-
o
x 1 ' ••• , x n •
shows that the ideals
This is worth remarking on because the ring every ideal is finitely-generated.
:ilk 8
are all finitely-generated. is not Noetherian,
n
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
we shall
8n
write its Taylor series as
= f(O) + so
f
ates
... , xn.
given by f of algebras.
H
•
W
oaf Xi
(O)x. + 21,. l.
if
~ -""'-"'-- (O)x.x. + ••• ~
ax. ax. l.
l.
J
the algebra of formal real power series in
is in en' x 1'
-f. )
f,
In this way we obtain a natural mapping
J
n
indetermin-
8
n
->
Sn
and one can check easily ennugh that it is a homomorphism
What is by nn means so clear is the Bnrel Lemma. A
The algebra homomorphism
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 relevant ideas, precisely 1 00 n
Notice that the kernel of the homomorphism
=
n
k=1
1'k n
by (2.4).
It follows that
We can get a more finite version of this, as follows.
Under the epimor-
phism 8 n -+ 8n the maximal ideal .L'n in 8 n maps to the maximal ideal J in en comprising all formal power series with zero constant term, and n
k
Ak
hence any power 1 n maps to the power l ' n •
The nett result is that
The particular virtue of this relation is that it tells us that the quotient Indeed ~ is generated by the in-
space on the left is finite-dimensional. detem.inates
"k x 1 ' ••• , xn so 1 n is generated by the monomials of degree
k
in x 1 ' ••• , xn and the quotient space on the right can be identified with the real vector space of polynomials of degree
< k
in x 1 '
.",~,
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
8 be a commutative ring with an identity element 1, and let
l ' be an ideal in
8 wi th the property that
+ x
is invertible in
Further, let M be an 8-module, and let with A finitely generated. Proof
Let
A, B be 8-submodules
If A £. B + 1'. A then A £. B.
a 1 , ••• , at be generators for A. and elements
102
8 for
'A ••
l.J
in
.J{
By hypothesis we can find
for which for
1
~
i
~
t
we
can write
* Introduce a a
=
txt
matrix over the ring
(a 1 , ••• , at)'
M x ••• x M,
with
=
b t
G by A
(b 1 , ••• , b t )
factors.
Then
where
I
is the identity t x t
to show that
(Aij )
and take
to be elements of the 8-module
*
can be re-written as
=
(I - A)a
=
b
matrix over the ring
8.
It will suffice
I - A is an invertible matrix, since then we can solve this
system of linear equations for the which will show that
a 1 , ••• , at lie in
B,
in terms of b 1 , ••• , b t , To this end
as was required.
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 8.
Observe that det(I - A) say, with
A E ~.
=
1 _
(sum of prod~cts \ of elements ~n.A1
And by hypothesis
=
1 - A
1 - A is an invertible in 8.
o In practice the rings are ~,1. n
n
and the ideals
both of which satisfy the initial hypothesis of (2.6).
a Noetherian ring.
~oon
Bn
The Nakayama Lemma allows a very simple proof that
Example 1
A =
8 we have in mind are 8n ,
=
ri ~kn'
k=1
We take
M to be the G -module n
B the trivial ideal.
8. n
8
n
is not
And we take
Clearly A S; B +
~n.A.
If
Gn were Noetherian then A would be finitely generated, and the Nakayama Lemma would tell us that A S; B,
i.e.
that
A is trivial.
However this 103
is false since there are standard examples in calculus of non-zero germs with zero Taylor series. Example 2 B
= <x2
It follows that 8 n
In M
3 +y,
y
2
=82
is the case.
consider the ideals
'5'-..
Clearly B S; A.
+x/.
so need to show A S; B.
cannot be Noetherian. A
Here the Nakayama Lemma provides an easy answer, because it And this is clear as .12A
3 y.
Now we can return to the main theme.
In the following
8 will denote
8n ,8n and.1 the corresponding maximal ideal .1n '
either of
be an &-module, and let
in M when the quotient space
M.
Mil
We say that
I
has finite codimension
is finite-dimensional,
cod I
of
I
i.e.
the sub-
And in that case we
admits a finite-dimensional supplement in M.
define the codimension
Let M
.Jln •
I C M be an 8-submodule, so in particular a vector
subspace of the real vector space
tient space
A = B,
It is not immediately obvious to the eye that this
3 xy, 22 is generated by x, x y,
I
y2) and
We claim that indeed
tells us that it suffices to show AS; B + Af2 0A.
space
=
in M to be the dimension of the quo-
Mil, i.e. the dimension of any supplement.
The point of the
next proposition is that it gives us a useful algebraic criterion for
I
to
be of finite codimension in M.
(2.7)
Let
8-submodule.
M be an &-module with a finite basis, and let A necessary and sufficient condition for
I
I eM
be an
to be of finite
codimension in M is that there exists an integer k ~ 1 with 1k.M C Sufficiency
Suppose there exists an integer k ~
1
for which 1k oM
C
I. 1.
Certainly then dim Mil Now I claim that 101;.
MIAf~ is finite-dimensional.
As
M has a finite basis
we can suppose
M= 8 s
for some integer
isomorphic with the product 811
k
s ~ 1.
x ••• x 8/.A'
k
with
s
factors, and since
(as we have already seen) each factor in this product is finite-dimensional the product is as well.
It follows that
dim Mil <
00,
so
I
has finite
codimension in M. Necessity
Suppose
I
is of finite codimension in M.
Consider the
descending sequenoe of 8 -submodules
Clearly, eaoh striot inclusion in this sequenoe makes a oontribution the oodimension of
~
1
1
to
Sinoe the codimension is finite the inolusions, from
I.
In partioular there exists an
some point onwards, must all be equalities. integer k
~
for whioh k I +.-11 .M
whioh implies that 1k.M C
= Since 1k.M
I + .-II.C)-.M). k
it follows from the Nakayama Lemma that 1.M
is finitely-generated
S I, as required.
o Note one small point. hypothesis 1k.M C I
As is olear from the final lines of the proof, the is equivalent to the apparently more complicated rela-
k 1M +..At.u + ••
We chose the former relati on for the statement
of (2.7) beoause of its simplioity, but in praotioe it tends to be easier to verify the latter relation.
A further point is that one oan extraot a bit
more information from the proof of (2.7) than we have stated.
Keeping to the
same notation define =
105
for
The quotient space which appears here is necessar-
k = 0, 1, 2, ••• , •
ily of finite dimension, and its dimension is the codimension of in
+Af~M.
I
(2.8) for
(2.7)
Under the hypotheses of I
co~
I
vanish:
Suppose
Necessity
exists an integer j
~
M is that all but finitely many
and in that case cod I
all
a necessary and sufficient condition
M to be of finite codimension in
~
of the
I
=
codO I + cod 1 I + ••••
has finite codimension in
k ~ 1
with Afk. M ~ I.
co~
=0
M
so by
That implies
(2.7)
cod. I J
there
=0
for
k.
Sufficiency
I
I +Afk+1.M
If
I
then
k
=
I +1.M
so
is of finite codimension in M. The second statement in
(2.8)
follows immediately from the proof of
(2.7).
D Next, let us expand a little on the kind of situation we have in mind. Write
Sn,p
with y E
for the real vector space of all germs
:m.p •
Such a germ has components
standard co-ordinates in
:m.n , :m.P
=
think of
in the case
p
=1
p
as the "tangent space" to a germ
we shall be thinking of
associated to a function-germ
f.
(:m.n ,
0) ~
y)
relative to the Sn.
Thus we can
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
(:m.P ,
factors), and see that it is a
M with a finite basis.
n
I
f 1 , ••• , fp
each of which lies in (with
natural example of an S -module shall be thinking of
f:
S n,p /1.
Let us write 8n ~
8n
8n ,p = 8n
x
x
8n
(with
p
factors).
The epimorphism
of the Borel Lemma gives rise to a surjective linear mapping
.I,
to
say.
We claim the following.
A necessaEY and sufficient condition for
dimension in 8n,p
I
is that
I
to be of finite co-
should be of finite codimension in
8n,p :
and in that case the two dimensions coincide.
Proof
Consider the composite of the surjective linear mappings
8n,p ~ &n,p ~ &n,p If, quotient.
the second being the natural projection onto the
By previous work the kernel is
I +~.8n,p
so that by element-
ary linear algebra
I + //00 e: ;At
Necessity
If
I
has finite c odimension in &n,p
and hence
I +/.8
n n,p ,
n· n,p
has f ini te c odimension in
then so too does
8n,p by the displayed
isomorphism. Sufficiency
1 has finite codimension in
If
finite codimension in 8 n,p ,
by the
displ~ed
then 1+/.& has n n,p It follows from isomorphism. k
(2.7) that there exists an integer k ~ 1 with 1n.8n,p C I +f.n.Gn ,p C k k+1 The Nakayama Lemma then tells us that we have 1 .8 C I, I + 1 n .8n,p • n n,pand (2.7) that I has finite codimension in [; n,p • Finally, when
1~ .&n,p
C
I
I
has finite codimension in
tells us that 100. G C I n n,p
so
the relation
I, and
the displayed isomorphism reads
107
o
yielding the final statement of (2.9)
The practical consequence of (2.9) is that when computing the codimension of
I
&n,p we can, and shall, replace all our germs by their Taylor
in
series and handle them as formal power series, which are much easier to work with in practice. tions.
Bear this point in mind when you come to do such computa-
Before resuming our session of algebra we shall digress to look at
the sheer practicalities of just how one computes the codimension of a germ f Of course the first thing one wants to know is whether f codimension:
has finite
the next proposition at least gives one a very simple way of
recognizing a germ of infinite codimension.
Suppose the germ f
(2.10)
&
n
Then
has finite positive codimension.
the origin in En is an isolated singular point of any representative of i.e.
f,
there exists a neighbourhood of the origin in which the origin is the
only singular point of the representative. Observe first that (any representative of)
0 E En
Indeed if some
f.
be an invertible element of zero codimension.
Since
some integer k
by
k
k
x 1 ' ••• , xn
~
8n , f
lo J r = en'
by (2.1) so
in question must be
O.
k
-
r for
In particular, this means that the monomials
f
0f
af af , ... , ax-a-x 1
wlo.th
n
all these partial derivatives
must vanish as well,
Consider the germ
partial derivatives are
f
n
at a singular point of
... , xn
and
Xi would have
has finite codimension we have 1k C J
(2.6).
must vanish, so
108
ar aaf (0) f. 0 then -a-would x.
can be wrlo. tt en as 1·lonear comb·lona t·loons
coefficients in
Example 3
must indeed be a singular point of
i.e.
the singular point
D rex, y, z)
=
y
222 -
z x
3
+ x •
Here the
of ax
=
of
2 2 -2z x + 3:x:
ay
of
2y
=
which vanish simultaneously precisely on the z-axis. is not an isolated singular point of
f,
of infinite codimension.
=0
The set
f
-2zx
2
Thus the ori gin in
and (2.10) tells us that is sketched below.
f
:m.3
must be
Notice that
the z-axis is precisely the line of "dmble 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. pIe is provided by the germ
f(x, y)
(x
=
222
+ Y) :
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 x - iy
=
0
is singular:
a~
point on the lines
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
is of finite codimension. Let us abbreviate
co~Jf
The idea is to compute to
co~f:
cod f
we know that
109
=
cod f
codO f + cod 1 f + •••
so we must successively compute
cod O f,
cod 1 f,
until we reach a zero
answer, and then add up the list of integers so obtained. of the matter are worth expanding upon.
To compute
codk f
a basis for a supplement of
k
= 1,
each monomial of degree J f +vn//k+1 •
in the
2, 3, ••• ;
.
The
= dim SLl/1n
=
1. k
variables
st ep·~n
f~rst
and for each
x 1 ' ••• , x n '
n
.
th~s
computa-
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.
codO f
k in
k
J f +vn//k+1•
x 1 ' ••• , xn which do not lie in
takes successively
we have to find
clearly, this can be
extracted from a list of monomials of degree
tion is always trivial, as
The practicalities
m then it automatically holds for all its descendants,
the monomials which can be obtained
other monomial.
~rom
m by multiplying it be some
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 n = 2. in two variables
x, y
Here, the monomials
can be conveniently displayed as the following array x
x
2
y
2 Y
xy
2
xy
xy
2
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.
J f +1k+\
Underline those and then underline
all its descendants in the array, to avoid unnecessary work. maining monomials select a supplement for 110
J f +1k+1 in
From the re-
number of monomials selected is the integer
c~
f.
The computation finishes
whenever one reaches a row of monomials which can all be underlined. co~
dimension is the sum of these codO = 1,
assuming that
Example 4 Here
Jf
s
For
does.
in x,
because we have to add
= x s +1 with
f(x)
0
an integer
s
1.
;I
~
It follows immediately that
Example 5
-
k ~ s - 1 i t is clear that the sole monomial namely xk , does not lie in the ideal J f +1k+1.• but
= (xs). k
plus one
is a singular germ.
Consider the germ
of degree x
f
f,
The co-
cod f = s. 2 4 f(x, y) =xy+y,
We shall compute the codimension of the germ
which will appear in §4 as the so-called "parabolic umbilic".
of ax
Here
= 2xy,
of =1x2 + 4y3 vanish simultaneously (even in the complex case) only at the oy origin, so
f
should have finite codimension.
monomials in x, y. Jf
x
2
2
+1 •
Of
x, xy, y
of 3 oy - 4.y,
=
(See Figure 1 below.)
2
2
1 x:y = '2
x4 ,
3y,
X
X
of ax '
2 3 Y2 , x:y
of degree 3 only y3
Neither of
x~
y
lies in
the first two certainly lie in J f + 1 but
y2
and all their descendants in the array row, and
We write down the array of
d oes no; t -
+
as 2
thus we und er1·~ne
namely x 3 ,
X§,
in the fourth row and so on.
x:y
2
x , x:y
in the third
Of the monomials
+~.
need be considered, and it does not lie in J f
And of the monomials of degree 4 only y4 need be considered, and it lies in
5 as y4
J f +1
in J f +14,
1
of
1
= Ij:y oy - 'S'x
of ax.
Thus all the monomials of degree 4 lie
and the computation is finished.
The codimension is the total
number of monomials which have not be underlined (namely one, so
cod f
=
5.
x, y, y2,
We have, incidentally, shown that 14
and hence that 14 C J f ,
S
~) plus Jf
+
5
by the Nakayama Lemma.
111
x x
2
x5
xy
2
2
2 2
x 3y x4y
y
El 2 xy
x3 x4
Y
y3
-d
~
4
;
d:..
x2y3
x 3/
L
Figure 1
x x
2
x5
y
xy
2 2
A X4y
2
xy2
2 xy
x3 x4
y
y3
i y3
x 3/
y4
~
~
d:..
y5
Figure 2
ExamEle 6
The germ f(x, y)
indeed
= 7.
cod f
we omit it.
=
x 3 + xy3
is of finite c0dimension,
The computation is very similar to that in Rxample 4 so
The array of monomials is illustrated in Figure 2.
Before going any further we should stop to verify that the conimension of a germ possesses the most basic property one wouln 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
:
A. -+
(:Ilt, 0) A.
0
f.
-+
(:m.P, 0)
One checks easily that
to be induced by f. 112
we obtain a mapping f
...
f
Given a germ
...
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 5 p the form
f* for some germ
~ (:rn.P, 0).
: (:rn.n , 0)
f
5n
has
To the reader we leave
the task of checking that the operation of taking induced homomorphisms has the following functorial properties,
= g'"
g) *
(I)
(f
(II)
The germ at
the identity map
Gn
~
0
0
*•
f
of the identity map :rn.n ~:rn.n
0
Gn ' and is indeed the only such germ.
(III) It foll ows from (I) and (II) that if then
h '"
(2.11)
is invertible and
Let
(h*)-1
.;;:;.b..:;e-::::a--"1ge::.;rm::.:::,!.!...-....::.:A:....:;:n;:.e;:.c;:,;es::.,:s::.:a::.:ry:..L. . . :;an=dc.. .:::.suf=..:f:...:i::..:c:..::i:..::e~n:..::.t f
to be an iso-
Necessity of the condition is provided by III above.
For suffici-
morphism is that
f
: Gn
~
should be invertible.
is an isomorphi sm, so maps the maximal ideal vA'n onto i t-
ency suppose self.
'"
Gn
condition for the induced algebra homomorphism
Proof
h is invertible
(h-1)*.
=
: (:rn.n , 0) ~ (:nf, 0)
f
induces
Indeed
f
to the components
•
will map the standard generators f 1 , ••• , fn
of
f,
x 1 ' ••• , xn
for vA'n
which must likewise be generators for
Consequently we can find elements
A••
J.J
in
5.
n
for which
we have
= A.J. 1f 1 + ••• + A.J.nf n • If we differentiate both sides of this relation with respect to evaluate at
0,
x., J
and then
we obtain 8 •. J.J
where
8.. J.J
denotes the Kronecker delta.
Thus
113
of.
In = A( 0/ (0)
)
J
where
I
n
denotes the identity n x n matrix, and A denotes the matrix of
A...
the
It follows that the Jacobian matrix is invertible, and hence
~J
(from the Inverse Function Theorem) that
f
is invertible.
o
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 : (JRn , 0) isomorphism
-4
f, g in Gn
are equivalent then
cod f = cod g.
f, g are equivalent there exists an invertible germ
(JRn , 0)
h* : Gn
for which Gn
-4
of algebras.
h.
0
By (2.11)
We claim that
G~Jf'
will imply that the quotient spaces which the result follows.
g = f
8 /J
n
h
induces an
h * (Jf ) = J g: that
are isomorphic, from
g
To establish the claim note that by the Chain
Rule
n
k
ax.
=
~
so
Jg
c
h*(Jf )·
I (::. j=1
J
o
h
ht.ax. J
n
=
~
h*(Jf )
J
j=1
The same reasoning shows
yielding the reverse inclusion
oh.
L h*(::J --..J.. ax. Jf
C
~
(h-1)*J
g =
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,AfJ f •
First of all, (2.13)
The ideal AfJ f
finite codimension.
114
has finite codimension if and only if
Jf
has
Proof
.A'Jf
f
Jf ,
so certainly if 1Jf
has finite codimension.
/
f
Conversely, suppose
(2.14)
Let
f
E
Bn'
Proof
Observe first that
Jf
1Jf and
af lR{-a-"
•• ,
Gh
subspace of
has finite codimension, so
But then .A'k+1 f
cod1Jf
=
n + cod Jr-
is the vector sum of the vector subspaces
af ax J,
where the curly brackets indicate the vector
n
af u 1 aX 1 +
observation follows on wri ting
+ u :f nux 11.. 1
(x)
=
wi th the
ei(O) ~ 0,
Indeed an element u.
••• ,
en
for which ~
--'
e.
1
=
:f "Xi
as the components of a vector field
e(0)
in lRn with
~ O.
e1 ,
0
i.e.
in lRn with
O.
h(O)
=0
are the components of
germ in
equivalent to f,
Let us agree to write
b = h(a).
for which e i h.
0
in 80 ,
To prove this it Think of
on a neighbourhood of
we can find a diffeomorphism h
h 1, ••• , hn 8n
e
en
••• ,
e
Recall now from (1.4.3) that the flow of
can be "straightened", hood of
and the
(u.(x) - u.(O») + u.(O). 1 1 1
is convenient to confuse germs with their representatives.
e1 ,
8 ; n
E
1
n
Next we claim that it is not possible to find germs with at least one
so,
Jf,.A'Jf has
spanned by the first partial derivatives.
has the form
1Jf
o
and suppose that one (hence both) of then we have
x1
Jf
has finite codimension.
finite positive codimension:
Jf
Jf
for some integer k ~ 1 by (2.7).
Jf
again by (2.7), we see that 1Jf
of
has finite codimension then
h
=
Observe now that
0
near 0
of a neighbour-
ah. ax~' where g = f
0
h
is a
so likewise of finite positive codimension. Then ~ x 1 (a)
= L ei (b)
::. (b) 1
,.....
.
= 0,
g is so ~ is identically zero on a neighbourhood of O ....~n lRn ~ e "x 1 independent of the variable x 1 : by (2.10),the origin 0 is a singular
point for
g,
so that every point near
point for
g,
and
0
on the
g has infinite codimension.
x 1 - axis is a singular This contradictinn estab-
lishes our claim. 115
to be real
Now we can easily finish the proof. numbers we see that
of
of ox
-0-' .•. , x1
are linearly independent over the reals,
n
and hence th a t th e vec t or sub space
of ' JR{ ox 1
Of} has d'1menS1on . ••• , oX
n.
n
~Jf
Further, it follows immediately from the above that the vector sum of
of , ••• , ox Of} JR{ ox
and
subspace.
§3.
is direct,
i.e.
that the intersection is the trivial
n
1
The result now follows.
o
Determinacy of Germs
Having completed our session of algebra we return to the main theme, namely the classification of germs in
en
of low codimension under R-equivalence.
The underlying philosophy is to reduce the problem to a finite-dimensi0nal one.
And the idea to do this is an fol101'1S.
Take a germ
f E en:
f
has
a Taylor series 1 f(O) + -1' •
f
if
of-(0 ) + -2' 1 ~ ~-=-"- (0) + ... x.1 -" /. x.x. uX. • '--J 1 J oX.ux. 1 J •1= 1 1 • . 1 1,J=
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 en
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.
~
k
completely determines the germ, up to equi-
Note that the property of k-determinacy will be invariant under
equivalence.
Example 1
Any non-singular germ
(1.1.3) we know that
f
f E Gn
is 1-determined.
is equivalent to the germ
And the same applies to any germ
g E G
Indeed by
(x 1' ••• , x n ) ~
having the same 1-jet as
n
x 1" Hence
f.
the result. The basic result of this section is the following sufficient condition for k-determinacy.
0.1 )
Let
f E Gn be suoh that 1k C 1Jf :
Proof
Let
g E G
Now
n
have the same k-jet as
f1
t E JR"
= g.
is k-determined.
We have to prove
Gn ,
f
g.
~
so can be joined by a
More formally, we define F(x, t)
for
f.
f, g are points in the real vector space
straight line.
then f
Thus
(f t )
=
ft(x)
=
(1 - t)f(x) + tg(x)
is a 1-parameter family of germs with
We intend to show that
= f,
two germs in this family are equivalent,
Since the real line is connected it will suf-
which will prove the result. fice to show that given any
a~
fO
s E:rn. we have
ft
~
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 : lRn x lR (a)
H(x, s)
=
(b)
H(O, t)
= 0
(c)
F(H(X, t), t)
~:rn.n for which
x
= 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
s
the germ
the determinant of the Jacobian matrix at t.
ht
= identity.
hs
must be invertible, since
0
of
ht
Further, the condition (b) .ensures that
ht
maps the origin in JRn
itself. f t ~ fa
depends continuously on
Finally, the condition (c) can be re-written as for t
close to
s,
as was required.
is automatically satisfied for
t
= s,
ft
0
=
ht
to
f s'
so
Notice incidentally that (c)
in view of (a).
Thus it will be suf-
ficient to replace (c) by the condition that the left hand side does not depend on t,
i.e.
has a zero derivative with respect to
t.
Written out in full
this is the condition.
~ oH. (x, t) ax7 of ( H(x, t), t ) ~
0:
i=1
+
aF ( at H(x,
(0,
at
(a), (b), (c'). s)
)
=
o.
1
Our problem then is to construct a smooth mapping ditions
t), t
H which satisfies con-
We claim that it will suffice to construct a germ
e:
of a smooth mapping (d)
of )' e.1 ax. 1
(e)
e.(O, 1
=
~
t)
=
For suppose such a mapping dependent vector field on
:m.n •
JRn
x
JR
-+
:m.n for which
of - at •
O.
e
exists.
We can think of
It has therefore a flow,
e i.e.
as a timea mapping H
of the required type for which
~~
(f)
(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 fram
(a), (e) and
118
(f).
It remains to establish the existenoe of a mapping e ties (d) and (e).
having the proper-
The argument is algebraio, and does not depend on the
partioular value of
s
ohosen:
we shall therefore suppose
s
= O.
Also,
since we wish to work simultaneously with funotions of n variables x 1 ' ••• , xn and funotions of (n + 1) variables
x 1 ' ••• , xn ' t
we shall
think of an element of &n as an element of Gn+ 1 whioh does not depend on t. And then oondition (e) is simply the requirement that 1 , ••• , en
e
should lie in " ' . n
~! using the faot that
As a preliminary, note that
:t
=
g - f
(1 - t)f + tg
=
(2.4).
has zero k-jet, and
Thus (d) will follow
from
f
A'f'n
aF ... , axaF) .
n
Next we have
..EL) ax n
.k+1 + A'f'n+1.A'n
**
the first inclusion being trivial, the second being the hypothesis of the theorem, and the third following from the faot that
aF
=
L
ax: - ax. ].
*
].
now follows immediately from
proved.
**
ta(g-f)
aXi
E
using the Nakayama Lemma, and the result is
o 119
It is worth remarking that the condition Cd) which appears in the above proof has a simple geometric interpretation. ft
Recall that we took some germ
on the line joining f, g and set out to show that if we moved slightly
along the line from
ft
we remained in the orbit through
ft :
thus we ex-
pected the directi on of the line to be a "tangent vector" to the orbit through We think of the direction of the line being given by
aF
at
(d) then says that through
lies in
:~ •
The re la ti on
the "tangent space" to the orbit
i.e.
ft.
A word is also in order concerning the hypothesis of (3.1). Lemma tells us that the condition .A'k :;..-#'Jf
s;. 1Jf
more complicated condition.A'k
The Nakayama
is equivalent to the apparently
+ 1k+1.
We chose the former con-
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
8
n
fini tely-determined when
f
is k-deter-
It follows then from (2.7) and (3.1) that if
of finite codimension then automatically f practicalities of showing that a germ
f E 8n
f
is finitely-determined.
The
ve~
is k-determined are
lar to those involved in computing the codimension, discussed in §2. . k checks successl.vely whether..-#' :; 1Jf for a given in
k
k+1
for
simiOne
k = 1, 2,
and
this is equivalent to showing that each monomial of degree
x 1 ' ••• , xn
lies in the ideal.A'Jf Consider a germ
Example 1
+..-#'
is
+1k+1.
f E 8 1 for which the following conditions are
satisfied.
f( 0) " f Cl ear.l.Y
120
= o
= ,vk1 '
E.$
f.J r= . $//k1+1.
k
k
il (0) k
ax
=
0
I
O.
It follows from the Hadamard Lemma that we
can write
~~ = x
= 0,
hence
f
=
xk+1g for some
xk{(k + 1) g + x
*}:
g E 8 1 with
g(O)
f '" exk+1
f
=
1k+\
wh ere
(3.1) tells us that 0•
J F
c
Then
the expression in curly brackets is
so invertible by (2.1), and one deduces that 1 J
F O.
f
Jf
= <xk)
F 0 when = 1k
is (k + 1)-determined, so
An obvious change of co-ordinates then yields
k f '" .:!:. x •
Example 2
The germ
= ~2, i),
Jf
f(x, y)
so 1Jf
= <~,
=
x 3 + y3
is 3-determined.
x2y, xy2, ; )
=
1 3,
Here
and the claim
follows from (3.1) Remark
Another sufficient condition for
f E 8n to be k-determined The proof of this is exactly the same as that of
is that 1k+1 S;1 2 Jf' (3.1):
one has simply to check that the last few lines of that proof go
through equally well with this hypothesis. provement on (3.1) since 1k+1 S;12Jf
Of course, this is a slight im-
is a weaker condition than
,A'k S; 1Jf • Example 3
=
Jf
x 3 , y3
£ 1Jf
,A'4
The germ f(x, y) 1Jf
=
x4 + y4 is 4-determined.
Here
~4, x3y, xy3, y4) and the condition
is not satisfied, so (3.1) does not allow us to conclude that
is 4-determined. ,A'2J f
so
=
On the other hand
i,
= <x5 , x4y, x 3
x2y3, xy4, ; )
=1
just mentioned allows us to conclude that Notice that if
f
f
E
f
3 so the improvement on (3.1) is indeed 4-determined.
&n is k-determined then it is, in particular, equi-
valent to its kth Taylor polynomial.
Thus the study of germs in &
finite codimension reduces to that of polynomial germs.
n
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)
is easily checked to be 6-determined, but can be shown
=
not to be 5-determined.
§4.
~
Classification of Germs of Codimension
5
We are now in a position to reap the rewards of our hard work in the previous section, and achieve our object of classifiing germs of low codimension. start with germs of codimension
0.
A necessary and sufficient condition for a germ f codimension
° is that it be non-singular;
Necessity
Suppose
identity element 8n
for which
1
that at least one
1
f
By (2.1) this means that
f
i.e.
+ ••• + en
af (0) I ax.J.
Suppose
Su:fficiency
has codimension
lies in J f , f
= e1 ax1
0,
so that
0,
Jf
n and hence
to have
= 8n
and the
el ' ... , en
we can find germs
af ax •
E ~
and in that case it is eguiva-
in
If we evaluate at f
we see ° is non-singular.
is non-singular, so at least one
aaf x.
We
::. (0)
I o.
J.
is an invertible element of
8n ,
and hence
J.
f
has codimension
0.
Finally, a non-singular germ in 8 n by (I .1.3).
122
,,
is equivalent to
(x 1 ' ••• , xn ) ~ x 1
D
Next we consider germs of codimension 1.
2
§5
functions in
Remembering our discussion of
of Chapter II we say that a germ f EAfn'
i.e.
a singular
germ, is non-degenerate when the Hessian matrix
= (ax.ifax. (0) ) ~
J
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 f E~2
to be of
n
codimension
is that it be non-degenerate:
equivalent to a germ of the form
Proof
and in that case
x 21 + ••• + x 2 _ x 2 s
s+1
f
will be 2
-
- xn
The main part of the proof consists of establishing that f E Af2 is n
=
..-#'.2 J f
non-degenerate if and only if
Af
the
To this end let Q denote the second polynomial
;f all vanish at x1
0.)
Jfo
of f - which is just a quadratic form in
(Note:
x1'
000,
xn •
automatically since
Non-degeneracy of f
is the same thing as non-degeneracy (in the normal sense.) of the quadratic fom.
Q,
which in turn is equivalent to saying that the ~ span the vector space
ax.
~
of all linear polynomials in x 1 ' Since we have
-f.9.vX.
= :f
vX. 1
~
'00,
xn (by
vious application of the Nakayama Lemma. Suppose
Sufficiency codim f
=
dim 8/Jf
Jf C 1 Jf
= Af,
f
and
f
J f = 1by the above.
Then
1.
has codimension 1.
then dim 8/Jf > 1,
..-#'= J f , using an ob-
The rest of the proof is now easy.
non-degenerate, so dim 8/ Af =
=
Suppose
Necessity And if
f
linear algebra).
modulo Af2, this last condition is equivalent to
1= J f +Af2, which in turn is equivalent to
saying
Thus
elementa~
i.e.
Certainly then
codim f > 1,
dim 8/Af =
1.
which is impossible.
is non-degenerate, by the above discussion. 123
Finally, we have to verify that saw above, non-degeneracy of f
f
has the quoted normal form. ~=
is equivalent to
It follows immediately from (3.1) that
Q.
equivalent to its second Taylor polynomial
Q,
algebra that
which implies
Jf
f
As we
is 2-determined, so
Recall now from quadratic
being a non-degenerate quadratic form, is equivalent (under
a linear change of co-ordinates) to the normal form 2 -xn where
is the index of
s
2
2
x1 +
2
xs+1 - •••
Xs -
Q.
That concludes
o
the proof.
The geometric meaning of the Morse we have a smooth function
f : lRn
degenerate critioal point at
O.
-+
f
-1(.r) ..
for small enough and for
J (F
correspond to sets £.
lR with f(O) = 0 admitting a non-
The Morse Lemma tells us that there is a
diffeomorphism of one neighbourhood of the fibres
In partioular,
0 the relevant part of
in lRn
I 2
n = 2,
s
onto another under which 2
2
2
x 1 + ••• + Xs - x s + 1- ••• , - xn
f- 1 (O)
=
t,
corresponds to a quadric cone;
f1 )man~fold ' ~ 1 is ( a smooth
morphio to part of a non-singular quadric surface. case
Suppose
Lemma is worth spelling out.
diffeo-
Here is a picture for the
= 1. diffeomozrphism
To make this even more oonvincing reoall again our discussion of functions
§5
There we showed that any funotion n be approximated as olosely as we please by one f t : lR in
124
of Chapter II.
f --+
: lRn
-+
lR
could
lR whioh admits
only non-degenerate critical points; from
f
be a linear deformation.
and indeed that
could be obtained
ft
One of our examples was the monkey saddle
which admits a degenerate critical point at
O.
In this case we could take
=
as the approximating function, for erate critical point of
at
(.±.';
points at the points is as below:
f
0
J' 0).
t > 0 II
as small as we please.
The degen-
spli ts II into two non-degenerate critical And the diagram of level curves
f;1(!)
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 require one more bit of machinery. ~
2.
Suppose that
Certainly then the Hessian matrix Hf
the non-negative integer
c = n - r
~
2 we shall
f E12 has codimension n
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. Let
f E12 be a finitely-determined germ of c orank n
c·,
f
is
equivalent to a germ 125
·.. , xo ) ~ g
E
1~ and each 8i
+ 8
2
x c+1 0+1 + ••• +
= + 1.
We shall adopt the following notation. Sn we write ¢ k ¢ for which relation. gk E1~,
¢
0
k, ¢
h : (JR~ 0)
when there exists an invertible gem have the same k-jet.
Observe that
We intend to show, by induction on k, whioh is polynomial of degree:!; k,
¢, ¢ in
Given two germs
-+
(JRZ; 0)
k is an equivalence
that there exists a germ
for whioh ••• +
This will suffioe to prove the result, sinoe there exists a
k
>I<
for which f
is equivalent to its k-jet. The induotion starts when k n
= 2.
The 2-jet of
2
2
0+1 x c+1 + ••• + 8nxn with each
ordinates) to a quadratic form
8
by standard quadratic algebra:
and we can take
>I<
is a quadratic form in
so equivalent (under a linear ohange of co-
variables of rank r = n - 0,
step we assume
f
g2 = O.
8.
~
=
_+1,
For the induction
Certainly the n
** where
H
is a homogeneous polynomial of degree (k +
1).
Now write, as we
evidently can do,
+ xnn H (x 1 , ••• , x ) n where
h is a homogeneous polynomial of degree (k + 1) and Hc +1 ' ••• , Hn
are homogeneous polynomials of degree 126
k.
We use these latter polynomials
(:m.n , 0) -. (:m.n,O) by taking its components to be
to define ¢
... , xn )
=
xc 1
= xc+1 - ~ c+1
Note that the Jacobian matrix of ¢ invertible.
~+1
Substituting ¢1' ••• , ¢n for
=
c+
1(x 1 , ••• , xn)
is the identity n x n
matrix, so ¢ is
x 1, ••• , xn in
•.. , xn) where
H
**
we obtain
••• +
That 'completes the induction step, and establishes
~ + h.
0
~re~.
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) C
Let
odimensi on of
Precisely,
be a germ of fir-ite codimension. and corank f
is
~
c(c + 1) + 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
... with
and (vide the remark above) Note that
g.
I
=1 cJ g
cod I
=
f
+ x
-
2
n
has the same codimension as
C ",,3 and that by (2.8) c
-
COdal + cod 1 I + cod 2 I + cod3 I + •••
But coda I
=
cod 1 I
=
di
I + 8
c m I +1c I +~ dim I +~
8 di mc 1c
=
=
1di m~ c 1c
=
c
=
1c dim - 1 3c
=
c~c + 1) 2
=
2
?
cod2 I
=
I +1-c dim I +13 c
so 1 + c + c(c + 1) 2
~
=
cod I
c + cod J g
using (2.14), from which the result follows. Thu~for
D
instance, we have the following rough estimates c
=1
cod f
II
2
c
=2 =3
cod f
II
4
cod f
II
7
c
The consequence of (4.4) which we need to isolate is that if then corank f 1 and corank 2.
128
~
2.
cod f
~
5
It will therefore suffice to look at germs of corank
We are now in a position to state the remainder of the
classification up to codimension 5:
the result is due to R. Thom.
Let f e.L2 have codimension ~ 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
germ
2
x3
fold
3
x4
cusp
4
x5
swallowtail
5
x
1
2
name
codim
6
butterfly
4
~ -xy2
elliptic umbilic
4
x 3 + y3
hyperbolic umbilic
5
x 2y + y4
parabolic umbilic
The exact meaning of the statement will become clear as we proceed. Morse
Lemma give us a complete classification of germs of corank
The
O.
We
shall now classify all germs of corank 1 having finite codimension.
(4.6)
Let
f e.L~ have corank 1, and finite codimension (k - 1):
then
We know from §3 that a germ of tini te codimension is fini telydetermined, so the Splitting Lemma applies and tells us that
where
g
show that
E1;
also has codimension k. k
g ~ ~ x.
Write
x = x 1•
It suffices now to
Observe that there must exist a least integer
j
with 129
J .oj+1 gF.A'r 1
cod g ~ sion.
cod 1~
• •
=
j
¢(x) g(x)
j,
i.e.
=
xh(x) 11k:
=¢ k (x), =
so
(:m.,
clearly, ¢' (0) k
g ~ x.
g(x)
0)
-+
I
0
=
xjh(x)
(j - 1),
(:m.,
0)
so ¢
so
with
g
~
0:
Suppose k
to be the germ given by is invertible.
But when k is even and to deduce
I
h(O)
= k.
j
Further,
The same reasoning applies when k
and yields x(-h(X))1!k
g would have infinite codimen-
g has codimension
In that case take ¢ :
h(O) > 0, ¢(x)
for all
so
j,
g-
The Hadamard Lemma tells us that
computation shows that odd.
C 1 1J" for all
if not then J
is even and
h(O) < 0
take
k
-x •
o
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 variables, and mUltiplication by + 1) precisely the germs
~,x4,
x 5 , x 6•
And with this we have proved half of Thom's classification theorem (4.5). A singular point of a smooth function of
corank
sion k is usually called an ~ 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, of the plane
:m.2
defined by an equation f(x, y)
=0
i.e.
where
subsets
f f
=0
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
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 f: have a double point, 130
thus
m > 2.
When m
when m = 3 a triple point, and so on.
=2
we
The condi ti on
f = 0,
on the origin, that it is a double point of the curve the singular point there has corank
~
1.
entails that
If we assume the singularity to be
of finite codimension then it will be of type so equivalent to the singularity .:!:. xk+1 .:!:. /
~
for some integer k
by (4.6).
~
1,
Multiplication of f
°
so we can suppose the by a non-zero scalar does not affect the curve f = k+1 2 normal form is x .:!:.y. And since we are not interested in the case when 2 k+1 The cases is a point we consider only the case x the curve f = - y •
°
k = 1, 2, 3, 4, 5 give rise to the following pictures, which are traditionally dubbed with the titles indicated.
node
cusp
tacnode
ramphoid cusp
oscnode
131
Example
Consider the real algebraic curve given by the equation
f = 0
where f(x, y)
= x
422 2 2 2 + x y - 2x Y - x:y + y .
Clearly, the curve has a singular point at the origin, indeed a double voint. And computation verifies that we have a ramphoid cusp.
f
will have codimensinn 4 at the (lrigin,
Sl1
The curve looks like this.
r
----------~:--~~----------~
~
It is perhaps also of interest to see how geomet~
rally in studying the differential a smooth mapping with
~
~ : JR
-+ JR2
of plane curves.
as a parametrization of a plane curve.
~
standard scalar product on JR2, 1 12.
Suppose we have
whose image is a smooth submanifold of JR2,
mapping its domain diffeomorphically onto its image:
think of
tion
singularities turn up natu-
Let
we wish to denote the
giving rise to the standard distance func-
(We work with the square to ensure smoothness at the origin.)
Given a fixed point w in the plane consider the smooth function f w : JR-+ JR defined by f
w(t)
= ~Iw
- ~(t)1
measuring, essentially, the distance from w to the point on the curve with parameter
t.
As
w varies we obtain a 2-parameter family F
= (fw)
of
smooth functions, better thought of as a single smooth function, say 2
F : JR x JR -+ JR of each f 132
w
given by F(w, t)
must have corank
=
f
w(t).
Clearly, the singular points
,1, and those of finite codimension will be
classified by the Ak-series:
indeed the condition for an
~
singularity
will be that
aF at =
o
= o
o
-I o.
Let us look at the geometric information contained in these relations. The first few derivatives, written out in full, are
(.( + (.«( - w)
=
where
(, ( ,
denote the successive derivatives of
for the components of
(.
. . .
Thus
(.
Write
(1' (2
is the tangent vector to the
(= «(1' (2)
(-(2' (1) i.e. the result of
curve, and the normal direction is given by
rotating the tangent vector through a right angle in an anticlockwise direction. tor
The condition (
:;
=
is perpendicular to
to the curve at
( (t):
0
then amounts to saying that the tangent vec-
( - w,
i.e.
that
the set of points
(w, t)
w lies on the normal line in ]t2 x]t where
is therefore called the normal bundle space of the curve. a pair
(w, t)
we can write
w = ( + An where
normal direction to the curve at add the further condition that just given in that for
and
(t),
iF at 2
O.
=
n = n( t)
:~
=0
Note that for such denotes the unit
A is some scalar.
Now let us
Substitute the expression for
w
iF --2-' and equate the result to zero; a few lines of at
computation will then show that
A =
{(~
+
(1(2 -
(~13/2 (1(2
133
which the reader will recognize as the radius procal of the curvature of
for which i.e.
=
at
It follows that the points
K.
02F
=
at 2
0,
of curvature,
o
i.e.
the reci-
(w, t) in :m.2 x:m.
are precisely those for which w = ~ + in, K
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 add-
3
"ye t one more condit"loon long is equivalent to curve.
K
= 0,
a F = 0: --:3
at i.e.
we leave the reader to check that this
Thus singularities of type
-\: with k
bundle space for the curve, those with k those with obtain
~
k
~
~(t)
to the point
3 to vertices.
~
2
~ 1
being a vertex of the correspond to the no:noal
to points on the evolute, and
It turns out that generically one does not
singularities with k
~
4 in this situation, so that is the end
of the story. Example 2 If
Consider the parabola
we write w = (u, v)
~
then
2F(w, t)
+ v
2
and comparison with Example 5 of §1 in Chapter II shows that 2
defines a folded surfaces in :m. x:m.,
:i =
that the further condition
0
2 2a F
at defines the fold ourve on the surface, and that the final requirement
=
0
a3F :::J = 0 ot
defines the exceptional point in the fold ourve where the two folds meet. Under the projeotion :m.2 x:m.
~ JR2 the fold curve maps to the evolute
of
the parabola, whioh 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.
The
next result will complete Thom's classification theorem. f E12 have corank 2 and codimension n valent to one of the following germs
(4.7)
Let
2 3 2 ::!:. (x 1 - x 1x2 ) ::!:. x3 ±
± (x~
+
~)
2 4 ::!:. (x 1x 2 + x 2 )
Proof step.
with
2
± x3 +
+
5: then f
is egui-
xn2
2 ::!:. xn
+
2 x3 +
~
...
2 ::!:. xn
Again, it is the Splitting Lemma which provides us with the first It tells us that
f
is equivalent to a germ
g E1~ having the same codimension as
Taylor polynomial of
g is a
bina~
f,
namely ~ 5.
The third
cubic form in x 1' x 2 so can be
135
assumed
x;
-
by the discussion in the previous chapter
+~, x~x2
-
x~ - X1X~.
We have to decide whioh possibilities can arise.
this end, reoall that a germ of oorank 2 has oodimension (and hence
0,
according as it is identioally zero, symbolio, elliptio, hyper-
bolio or parabolio.
f
to be
has codimension 4 or 5.
g)
~
4.
To
In particular
We shall oonsider the possi-
bilities separately.
4
g has codimension
g e13 we have
Since
J
so 1Jg f 1 3 •
g
To see this we argue as follows.
Put
oodimension 4, and hence by (2.14) that
(2.8)
I
= 1J.g
In fact 1 J g
=1
We are given that
I has oodimension 6.
J
g
3• has
Also, from
we know that
=
cod I
=
ood O I
ood O I + ood 1 I + ood 2 I + ••••
1
we deduce that ood3 I which means that 1 3 f
I
=1 Jg,
It follows now from (3.1) that third Taylor polynomial.
= o
as required. g is 3-determined so equivalent to its
By oheoking the five possible normal forms for
this one by one we see that the only ones of oodimension 4 are the elliptio and hyperbolio types.
g
has oodimension 5 Again, sinoe
we put
136
And these give us the first two germs on the list.
I
g
=1Jg.
And again
e Jt 3 we have This time
J
g
has oodimension 5 so by (2.14) the
codimension of
I
is
cod I
7.
=
Further, we know that
codO I + cod 1 I + cod2 I + cod3 I + ••••
As before
=
cod 1 I
2
so we must have &
cod4 I
= O.
By checking the five normal forms for the third Taylor polynomial one by one we see that the only one for which 2
x 1x 2 •
And the fact that
cod4 I
=
cod3 I
=
1
is the parabolic type
0 tells us that 14 C 1J -
g
so
g is
4-determined by (3.1), hence equivalent to its fourth Taylor polynomial. This has the form
where
~
is a homogeneous polynomial of degree 4 inC x 1' x 2 •
In more detail,
we can write this as
where
¢(x1 ' x 2 )
is a quadratic form in x 1 ' x2 "
Notice that if we change
co-ordinates, and take the fourth Taylor polynomial of the result, we shall obtain an equivalent germ, since 4-determinacy is invariant under equivalence. In particular, if we replace
x1
take the fourth Taylor polynomial of the result, we obtain the equivalent germ
Notice that necessarily
a
I
0
(for otherwise the germ has infinite
137
codimension).
Finally, if we replace
x 1 by px 1 , x2 by
qx 2 ,
and choose
p, q suitably we see that our germ is equivalent to +
(the sign depending on that of
a).
~2
And that gives us the last germ on the
list.
Of course, by combining the results
D (4.6)
and
(4.7)
we have proved
and hence obtained the complete classification of germs of codimension
138
(4.5) ~
5.
v §1.
Stable singularities of smooth mappings
The Basic Ideas
We come now to the general problem of discussing singular points of smooth mappings.
Although the basic geometric ideas are similar to those for func-
tions, the fact is that we are in a much more complex situation and cannot hope to achieve anything like the results of the previous chapter.
However,
we shall see that it is possible to discuss some decidedly interesting situations if one is prepared to accept one deep result without proof. The basic ideas are soon set up.
Recall that two germs
equivalent when there exist invertible germs
f 1, f2
are
g, h for which the following
diagram commutes.
Sometimes it will be convenient to say in this situation that A-equivalent:
f 1 , f2
are
the reason for this is that later we shall wish to maintain a
clear distinction between this notion of equivalence and another to be introduced in §2.
Our problem is to classify germs under A-equivalence.
We can
at least make a start by setting up things in the right way, as we did for functions. charts at
The basic objects are germs x, y
we can suppose
N
= :m.n ,
f : (N, x) ~ x = 0 and P
(p, y)
= :m.P ,
and taking y = O.
139
of all germs (:m.n , 0)
Thus we wish to study the set
o
that
As in Chapter IV we take
Write
~ (:m.n , 0), An,p
xin,p
A
n,p
note
has the structure of a real vector space induced from that on :m.p •
Sn,p
(:m.n , 0)
~ (:m.P , 0);
=
~n
~
in,p
~n
to be the set of invertible germs
which is a group under the operation of composition.
x
~p
We then have an action
for the product group.
. by th e f ormuI a glven
and the problem of classifying germs in
( g, h) •f
SO
n,p
=
h
0
f
0
g-1
and
under this relation of equi-
valence is precisely the problem of classifying the orbits under this action. One might hope to proceed just as we did in the previous chapter and attempt to classify germs of low "codimension".
But, as we have already said, we are
now in a much more complex situation, and it turns out that such a programme What we. shall do, to make at least some progress, is
is far too ambitious.
to restrict ourselves to the germs of lowest possible "codimension",
i.e.
Even this limited objec-
the "stable" germs in the language of Chapter III.
tive will prove too much for us, but we shall manage to discuss some very special cases which will suffice to yield complete lists when the dimensions n, p
are fairly small.
Our immediate problem is to say just what we mean by a "stable" germ. What we shall do is to proceed by analogy with Chapter III and introduce this The heuristics of the matter proceed as
notion via that of an "unfolding". follows.
One would like to think of an r-parameter "unfolding" of
f
0) ~ (:m. q
. (:m.m
0'"
0)
as a "germ" (:m.r , 0) ~
In a natural way that determines (u, x)
~
given by
fu(x), (u, x)
~ (u, feu, x»).
m,q
(:m.r x :m.m, 0)
and this in turn determines
formal definitions.
,fo)'
~ (:m. q , 0)
say
u
~
fu.
given by
F
On the basis we introduce the following
J..IJ. r-parameter deformation of
is a germ f : (:m.r x :m.m, 0) 140
f:
(SO
~ (:m.\ 0)
with
fO: (:m.m, 0) ~OR\ 0)
f(O, x)
=
fo(x)
and the
corresponding r-parameter unfolding is the germ F : (]{ x Em, 0)
-l>
(Er x :nl.\ 0)
=
given by F(u, x)
(u, f(u, x)
).
There is precious little difference between the two notions just introduced, and the choice of which one to work with is largely a matter of convenience. In this section we prefer to work with unfoldings, but later in this chapter it will prove more convenient to work with deformations. Example 1
Let
f
(E2, 0)
: (E, 0)
be a germ.
Think of this as
a (small part of a) curve through the origin in the plane. we have the distance squared function, f
is the germ F : (E2
x
E, 0)
If12.
(E, 0)
-l>
=
F(u, x)
Associated to f
A 2-parameter deformation of given by
If(x) _ u1 2 •
As we saw in Chapter IV one can gain interesting geometric information about the curve by studying the singularities of the deformation Write
Example 2 x
for that in E.
components
u 1 , ••• , u n _ 1
for the standard co-ordinates in E n- 1 ,
The germ F : (En - 1 x:nl., 0)
-l>
em.n- 1 x E, 0) with
F1 , ••• , Fn where =
u.~
is an (n - 1)-parameter unfolding of f(x)
F.
(1 " i " n - 1)
f
(E, 0)
-l>
(E, 0)
given by
= xn+1.
Let us now head directly for our initial objective of introducing "stable" germs.
For this we need the analogy, in the present situation, of just one
of the notions for unfoldings introduced in Chapter III, namely that of "equivalence".
Based on the Finite Dimensional Model we make the following 141
definition.
Two r-parameter unfoldings
of a germ f: (JRm,O) .... (JR q, 0)
F 1 , F2
are said to be A-equivalent when there exist r-parameter unfoldings the germs at
0
of the identity maps on JRm, JR q
I ,I
m
q
of
respectively for which the
following diagram commutes. (JRr x JRm, 0) I
m
1
F1
I\
F
2 >(JRrx]Rq, 0)
(JRrxJRm,O)
In this situation one refers to the pair between
Note that
F1 , F2 •
) (JR r x JR\ 0)
1
=
q
are necessarily invertible germs.
1m' Iq
Further, we shall call an r-parameter unfolding trivial)
=
=
as an A-equivalence
(1m, 1 )
F
of
f
A-trivial
(or just
when it is A-equivalent to the constant r-parameter unfolding
(mrx mm, 0)
~ (mr
=
~
finally we call
f
X
=
.",q, 0)
A-stable
0f
f
. by glven
(or just stable)
( u, x ) ....
(u,f ( x )) •
when every unfolding of
And f
is
trivial. We now have a
sen~ible
definition of a stable germ;
but we have as yet no
idea how to produce explicit examples of such germs, and even less of how to go about classifying them.
The key to these questions lies in a very simple
and beautiful idea whose discussion must of necessity be postponed till The next two sections are devoted to the necessary preliminaries.
142
§4.
§2.
Contact Equivalence
The object of this section is to introduce and study another natural equivalence relation on germs
(N, x)
~
(p, y)
with
N, P
smooth manifolds.
This new relation, calledJr.:equivalence, is coarser than A-equivalence and has the advantage that it is fairly computable:
in §4 we shall see that there
are intimate connexions between the two relations directly relevant to the problem of classifying stable germs.
We turn our attention to the graphs of representatives, which will be smooth submanifolds of
N1 x P 1 ,
N2 x P2
through the
What we wish to do is to formalize the idea of these graphs having the same "contact" with
N1 , N2
lence) to be a pair
(h, H)
z1' z2
(or a~equiva-
To this end we define a contact equivalence
respectively.
at
of invertible germs for which one has a commu-
ting diagram
where for
k = 1, 2 ik
=
germ at
~
of the inclusion given by
x
1-+
Nk
~
Nk x Pk
(x, Yk)
nk = germ at zk of the projection Nk x Pk given by In other words,
(x, y)
1-+
~
Nk
X.
H is necessarily given by a formula
143
H(x, y)
(h(x), e(x, y))
=
And we call our germs
f 1 , f2
contact-equivalent
(or3V-equivalent)
when there exists a contact-equivalence
Thus one pictures
H as mapping the graph of
with
N1
mapping onto
N2
via
f1
(h, H)
for which
onto the graph of
f2'
h.
H
------h
The reader will have no difficulty in checking that if A-equivalent then automatically they are3.r-equivalent. discussing.J.r-equivalence of germs tion to suppose that
n
N = JR ,
~
(N, x) and
x = 0
hand as we shall see two germs which
P
f 1 , f2
are
In particular, when
(p, y)
it will be no restric-
= JRP ,
y =
are~equivalent
On the other
O.
are not necessarily
In the remainder of this section we shall justify our state-
A-equivalent.
ment that~equivalence is a fairly computable notion by deriving a more-orless algebraic criterion for two germs to two steps.
be~equivalent.
Vie do this in
The first step, which is the main one, is to look at a finer
notion of equivalence called ~-Equivalence.
For the sake of simplicity let us restrict our attention to germs Two such germs J.r-equivalent via
144
a~equivalence
(h, H)
f, g
are
~-equivalent
when they are
having the special property that h
is the germ at f, g h:
0
of the identity mapping on N n •
are~equivalent
oan , 0)
Note therefore that
if and only if there exists an invertible germ
~ (Nn , 0)
for which
f
0
proposition provides us with a purely algebraic criterion for e;-equi valent. 8n
If = f*(~)
To simplify notation write
generated by the components
The next
h, g are e;-equivalent.
f 1 , ••• , fp
of
Le.
to be
the ideal in
And similarly for
f.
The following three conditions on germs
f, g
f, g
I g•
are equi-
in
valent. (i) (ii)
(iii)
f, g
are e;-equivalent.
The ideals
If' Ig
are egual.
There exists an invertible entries in
for which f.
8n
l.
with
(u .. )
p x p matrix
l.J
' u .. g. = ) "-' l. J J
for
1
~
i .. p.
j
It will be convenient to split the proof into a number of fairly easy steps. Step 1
Suppose
In which we show that (i) implies (ii).
e;-equivalent.
I t will suffice to establish
If
£.
I g,
f, g
since then
We must therefore show that each component
follows by symmetry.
are Ig
f.l.
£.
If
can be
written in the form f.
l.
= ) a .. g. --' l.J J j
with the
a..
l.J
in
8n •
Since
f, g are e;-equivalent there exists a
~-equivalence (1, H) for which H(x, g(x») H(x, y) x,
=
(x,
e(x,
y»)
where
and hence its components
e
satisfies
=
(x, f(X)).
e(x,o)
satisfy
=
Write 0
identically in
e.(x,o) = l.
0
identically
145
in x.
By the Hadamard Lemma we can write
= )'
G. (x, y) ~
~
y.G .. (x, y) J
~J
j
:rn.P ,
the standard co-ordinates in
Y1' ••• , y p
with
and the
in
G••
~J
But now
= )'
=
f.(x) ~
g.(x)e .. (x, g(x)) J
~
~J
j
a .. (x)
which has the desired form (1) if we put
In which we isolate a lemma from linear algebra tn be used in the
steE 2 next step.
We claim that given real
real
matrix
I
=
~J
p x p
is the identity
C for which p x p
U
matrix. )
p x p
matrices
A, B there exists a
= C(I - AB) + B is invertible. To this end let
:rn.P
a, b :
-+
(Here
:rn.P be the
linear mappings whose matrices, relative to the standard bases, are respectively A, B. for
r + 1
b(e 1 ),
Choose a basis i
.;;
... , b(e r )
ded by vectors
p,
.;;
where
,
e'r+1' ••• , e p
r + 1 .;; i
.;;
u(e. ) 1
for
c(e.) = 0 ~
u
1
then
:rn.P ,
so can be exten-
Now let i .;; r,
c(1 - ab) + b
=
~J
{ b(ei)
f
A.. e! ~J
~
~
and
c :
:rn.P
c(e.) ~
-+
~
is invertible, and the matrix
C of
c
e1~
has the property
i .;; r
for
1
for
r + 1 .;; r .;; p
.;;
:rn.P be
=
j=r+1
e!
146
.;;
B:
b(e. ) = 0 ~
one has
A..
=
:rn.p •
to a basis for
+
u
for which
denotes the rank of
r
Observe that
p.
that for some scalars
so
:rn.P
for
are linearly independent vectors in
the linear mapping defined by for
... , ep
e 1,
is the required matrix.
Step 3
In which we show that (ii) implies (iii).
Assume that the ideals
If' Ig are equal, so we can certainly write
= )___
g.
~
a •. f.J ~J
j
f~
)b .. g. --' ~J J
=
...
j
where the coefficients the real
p
x
a .. , b.. ~J
p matrices with entries
Step 2 we can find a real invertible.
lie in
~J
p
p
x
8n •
For each
a .. (x) ,
b .. (x)
~J
x close to
respecti vely.
~J
= CO(I
matrix Co with Uo
It follows that for
0
By
- AOBO) + BO
the matrix Let
of
x let Ax' Bx be
u .. (x) ~J
be the entries
It is now a matter of computation to check that
Ux •
f~
...
= )' u .. g. --' J ~J
j
In which we complete the proof of (2.1) by showing that (iii) implies
Step 4
(i) •
Suppose there exists an invertible
properties stated in (iii).
(u .. ) with the
p x p matrix
e :
Let us define a germ
~J
(lRn x lRP , 0) --. (lRP , 0)
by taking its components to be
e. (x, ~
y)
= )'
~
(1~i~p).
y.u .. (x) J
~J
j
Certainly then
e(x, 0) = 0
H : (lRn x lRP , 0)
identically in x.
--. (lRn x lRP , 0)
by the formula
One checks easily that the Jacobian matrix of is an invertible germ, and struction
eJx,
H(X, g(x»)
=
g(x»)
(x, f(x»)
=
(1, H) fi (x)
Now define a germ
H at
H(x, y)
e(x,
entailing that
g(x»)
(x, e(x, y»).
0 is invertible, so H
is a e-equivalence. so
=
= f(x)
Moreover, by conand
f, g are e-equivalent.
o 147
Example 1 ~-equivalent,
and
f(x)
The germs
(x 2 , 0)
=
and
g(x)
are
=
since the ideals generated by the components clearly coincide,
hence~equivalent.
On the other hand
f, g
are not A-equivalent.
Suppose indeed that we could find invertible germs k : (:rn. 2 , 0)
--->
(:rn.2 , ) 0 • for which
tatives, we see that
k
f
would map
0
=
h
k
Imf - [OJ
0
to
h
(:rn., 0)
--->
(:rn., 0) and
Working with represen-
g.
Img - [OJ:
that however
is impossible since the former set is connected, whilst the latter is disconnected. Example 2
Consider a germ
(:rn., 0)
f
(:rn., 0)
--->
for which the follow-
ing conditions are satisfied. £teO) =
ox
0
f -_ x k +1g
It follo,ns " that
g: (JR , 0) ---> (-m =, 0)
for some
(The argument appeared twice in Chapter IV. ) algebra
x
&1
~> = ~k+1>
so
Now
·th
Wl
(0) r -L . 0 g
g is invertible in the
and we conclude that
f
is
~-equivalent
to
k+1
Let
Example 3 with components
f, g : (:rn.n , 0)
fi' gi
--->
respectively.
same image we can write each
f.1
(:rn.P , 0)
If these linear mappings have the
as a linear combination of
with scalar coefficients, and similarly each f 1 , ••• , fp
with scalar coefficients:
coincide, so are
f, g
~-equivalent
wri te each
f.
1
and each
are
~-equivalent.
g.
as a linear combination of 1
1
as a linear combination of
Conversely, we claim that if
as a linear combination of
f., g.
1
certainly then the ideals
then they have the same image.
However as the
148
be germs of linear mappings,
f, g
Indeed in this case we can ••• , g
p
with coefficients in with coefficients
are linear, the coefficients must be scalars
so
f, g
give rise to the same image.
Back to %-Equi valence Although we managed to produce a purely algebraic criterion for two germs to be
~-equivalent,
we have to be content with somewhat less
As a preliminary, call two ideals in algebra isomorphism of
&n
Two ideals in
when there exists an isomorphism of h : (lRn , 0)
-+
&n isomorphic when there exists an
to itself which maps the one ideal onto the other.
We need a more special notion.
germ
for~equivalence.
(JRn, 0),
&n
&n are induced isomorphic
to itself, of the form
h* for some
which maps the one ideal onto the other.
Note that by (IV.2.11) the germ
h
is necessarily invertible.
We contend
that A necessary and sufficient condition for two germs
are
to beJV--eguivalent is that the ideals
f, g
induced isomorphic. Suppose
Necessity germ h
(lRn, 0)
-+
f, g
are~equivalent,
(JRn , 0)
for which
follows from (2.1) that the ideals . morph~sm
h*
will map Suppose
Sufficiency isomorphism h * which maps
of
(2.1) the germs
f
If' Ig
&n'
onto
If
onto
0
I g•
with
f
0
I foh ' Ig Ig,
so there exists an invertible
h, g are ~-equivalent. are equal.
The induced iso-
so these ideals are induced isomorphic.
induced isomorphic, so there exists an induced h : (lRn, 0)
-+
(JRn , 0)
That means the ideals
h, g are
It
~-equivalent,
an invertible germ,
I foh ' I g
and hence
are equal, so by
f, g areX-equivalent.
o Example 4 and
g(x, y)
We have already observed that the germs
=
(x2 + y 2 , x:y )
f (x, y )
=
are A-equivalent hence%-equivalent.
2 y2) (x, On the 149
other hand these germs are not ~-equivalent since (for instance) the compo2 2 nent xy does not lie in the ideal generated by x , y •
Thus far we have vindicated our claim putable notion.
that~equivalence
is a fairly com-
And being a coarser relation than A-equivalence one has
rather more hope of being able to classifY germs
under~equivalence.
In §4
we shall see that for stable germs the two notions coincide, and this is the reaSon why we are going to pursue the problem of classifying germs under JYequivalence. The first step in this programme is to mimic the approach adopted in Chapter IV to deal with germs of functions,
i.e.
we shall set up the problem as
one of classifying the orbits under a group action, and then proceed by analogy with the Finite Dimensional Model. This goes as follows. (h, H)
lences H :
(:rn.n
x
with h
lRP , 0)
-+
Observe first that the set (lRn , 0)
(:rn.n
given by composition of the lences
(1, H)
x
-+
:rn.P , H.
(lRD., 0)
~,p
of
all~equiva
and
forms a group, the operation being
0)
Note that the set
is a subgroup of % n,p
~n,p
of all
~-equiva-
also that we can identify the group
discussed in Chapter III with the subgroup of %n,p comprising all JYWith this identification it is clear that any eleequivalences (h, h x 1).
1?
n
ment
(h, H)
in % n,p
can be written uniquely as the product of the element
in ~ n,p where HI = (h x 1)-1 0 H : one expresses this by saying that JY is the semin,p o direct product of 1C n and ~n,p The group ~,p acts on 8n,p by agree(h, h x 1)
in 1?n
and the element
(1, HI)
.
ing that
(1, (h, H).f)
=H
0
(1, f)
and it should be clear that two germs (lRn , 0) 150
-+
0
h- 1 (lRP , 0)
lie in the same
are~equivalent.
orbit under this action if and only if they lem of classifying germs
under~equivalence
Thus the prob-
is that of classifying the orbits
under the above action. Now we proceed as in Chapter IV. 80 n,p
We pretend that the action of %
on
n,P
is that of a Lie group on a smooth manifold, and look for a candidate
for the "tangent space" to the orbit %of
through the germ
We expect this to be the image of the "differential" at the identity of the natural mapping of the group onto the orbi t through f, which we can think of as a mapping
~
n
~
x
~
n,p
/!, o • n,p
Since the domain is
a product we expect this image to be the vector sum of the images of the "differentials" at the identities of the component mappings
~
and
~
n
We consider these separately.
The Mapping
~n
This is given by the formula Write
Chapter IV.
h
... , f
~
foh
-1
•
One reasons exactly as in
for the components of
p
f,
and.
af ax.
for
J.
afp
the germ with components
••• ,
standard co-ordinates in mn.
We expect the required "differential" at the
--- where
aXi
x 1 ' ••• , xn
denote the
identity to be the linear mapping + ••• +
with each
gi a germ (mn , 0) ~ (m, 0).
~
af
ax
*
n
For the reasons explained in
Chapter IV we do not wish to restrict these germs to have zero target, so allow them to be arbitrary germs in ping given by
*
/!'n'
will be the submodule
Thus the image of the linear map/!,n{:! ' ... , 1
aa; }
module
this submodule is called the Jacobian module of
written
Of course in the case
p
=1
of the
n
f,
we recover the Jacobian ideal 151
introduced in Chapter IV.
The Mapping To simplify life identify a ~-equivalence with
e.
(1, H)
where
= (x,
H(x, y)
e(x, Y~
Then our mapping is given by composition on the right with (1, f)
so is the restriction of a linear mapping, and should be its own "different1a.l" at any point.
e(x,
y)
We need therefore the image of this mapping taken over germs
which vanish identically in x when y
for the components of
Each component
8.
identically in x when y
= 0,
f 1,
... , YP
... , f p
on mP •
Thus
generated by the components of f:
comprises all germs in
If
likewise vanishes
on'
is a linear combination of i.e.
an element of the ideal
If
reversing the steps one sees moreover
can be so obtained.
°n,p
y)
of the) standard co-ordinates
0
ei(x, f(x) )
with coefficients in
that every element of
(X,
so by the Hadamard Lemma must lie in the
ideal generated by the (germs at
Y1'
8i
= o.
with components in
Thus the image of our mapping i.e.
the 0n-submodule
On this heuristic basis we arrive at the following formal definitions. Gi ven a germ f
: (mn , 0) ~ (mP , 0)
we define the %-tangent space to
f
And we define the ~codimension to be the 8 n -submodule J f + I f ·8 n,p • of f to be the codimension of this vector subspace in 8 • In view of n,p (IV.2.7) we have the following criterion for a germ to be of finite ~codimension. A necessary and sufficient condition for a germ
to be of k 1.8
n
152
finite~codimension
CT f • n,p-
f : (mn, o)~ (JRP, 0)
is that there exists an integer k
~ 1
with
This is fairly straightforward to apply in practice. of
as
f
Then
••• , f
I fo 8
n,p
and think of
p
f.
..£!.. ax.
1 = ( af ax. ' ••• , ~ ax. ) •.
~
in any position, whilst ~
vectors.
0).
••• J
n
thus
~
k
Now 1
x 1 , ••• ,~,
(p
as
n,p
and hence 1 n .8 n ,p
f. , ~
times).
••• , 0),
with
Jf
is generated by the
Tf
is generated by the list of all these
is generated by all monomials k
... ,
(0,
will be generated by all p-tuples
the
~
8
Write the components
m of degree
k
(0,
is generated by all p-tuples
And to check the condition
in
... , m,
one has only to check
that each such p-tuple can be written as a linear combination (with coefficients in
8n )
of the generators for
k = 1, 2, 3, •••
k
for which the condition is satisfied, or see that
it cannot be satisfied for any
o af ax
Doihg this successively for
there is at least a sporting chance that in a given example
one will either find a
Example 5
Tf •
k.
We shall show that the germ
f(x, y)
=
2 2 (x , y ), the germ at
of the "folded handkerchief" mapping, has finite %-codimension.
= (2x,
0),
af ay
(0, x 2 ),
(/, 0),
= (0,
(0, 2y). 2
y ).
Also
(x , 0),
is generated by
Clearly then the %-tangent space
generated by
(x,
0),
2
(y , 0),
(0,
is the
y)
and
We start by trying to verify the condition of (2.3) in the case The ideal
..Ai
Here 2
is generated by the monomials
1~.82 ,2 is generated by (x, 0),
(y, 0),
x, y (0, x),
so the (0, y).
(0,
2
x ).
k = 1.
82-submodule The question
now is whether each of these four vectors can be written as linear combinations of the four generators we obtained for is trivial.
But clearly we cannot express
Tf • (y,
For
(x,
0), (0, y) this
0), (0, x) as linear com-
binations of the four generators, so the condition fails. We continue therefore by trying to verify the condition of (2.3) in the case
k
= 2.
The ideal
A'~ is generated by the monomials x 2 ,
x:y, y
2
so 153
2 the 82-submodule 1 2 .8 2 , 2 is generated by
(0, xy), (0,
y2).
2 2 2 (x, 0), (xy, 0), (y , 0), (0, x ),
Again, the question is whether each of these vectors can
be written as a linear combination of the four generators for this is the case, so
1~.82,2 .s; Tf , and f
Tf •
Clearly,
has finite%-codimension.
Before going any further we should check that the%-codimension of a germ is actually a3F-invariant,
(2.4)
i.e.
(lRn , 0)
If two germs
that -t
(lRP,O)
are 3F-equivalent then they have
the same %-codimension.
= 8n,p
For the purposes of the proof we shall identify 8 the product
8 n x ••• x 8n
(p times) by identifying a germ
the p-tuple
(f l '
of its components relative to the standard co-
... , f p )
ordinates on lRn , lRP • Step 1 in
Let
8n •
We take
given by
f~
We claim that
u
= (u .. )
U maps
Tg C U(Tf ) ,
and hence
U{Tf ) CT. g
~
8
p x p
matrix with entries
to be the isomorphism of real vector spaces f
as a column vector.
isomorphically onto
Tf
Tg •
g
Put
= u.f.
It will suffice to
for then similar reasoning establishes
Tf .s; U- 1 (Tg)
Clearly then the problem reduces to that of estab-
lishing the two inclusions.
Cramer's Rule.
be an invertible
~J
where we think of
show that
observe that
with
8
We proceed in steps.
U: 8
u.f,
f
in
wi th
I g .8 .s; I f0 8
I g .8.s; U{T f )
=
U(I f .8)
and
.s; U(Tf ),
J g .s; U(Tf ).
For the first,
the equality following fran
And for the second observe that
J g .s; U(J f ) + I f .8
= U(Jf ) +U(If .8)
=
U(Tf ),
the inclusion followingfrom
the rule for differentiating a product of two functions. Let
Step 2 ~
: 8
154
~
8
¢ : (lRn, 0)
-t
(lRn , 0)
be an invertible germ, and let
be the isomorphism of real vector spaces given by f
-t
f
0
¢.
Put
g = f
0
¢.
~
We claim that
maps
Tf
isomorphically onto C ~ (Tf) g -
in step 1 it suffices to establish an inclusion to establishing the two inclusions
s;.
\.8
T
~ (Tf)
and
Tg.
which reduces
J g C q,(Tf ).
first inclusion follows immediately from the fact that
As
~(If.8).
I.8 C g
The
-
And
the second inclusion follows from the fact that, by the Chain Rule, we have, for
1:1: i
n,
:I:
n
..EL =~ a¢j ax.1 ax.1 :.-I
j=1
Let f, h : (:mn , 0) --. (:mP , 0)
Step 3
there exists an invertible germ
¢ for which g
And by (2.1) there exists an invertible in
8n
for which
morphism
U
0
h
=
f, h
p x p
have the
=f
0
matrix
Then
¢, h are e-equi valent. u
.. ) = (u1J
wi th entries
It follows from the previous steps that the autot
of the real vector space
~
so the quotient spaces i.e.
u.g.
be%-equivalent germs.
8
maps
Tf
isomorphically onto Th,
81Tf' 81Tn are isomorphic and have the same dimension,
D
same~codimension.
We come now to the question of actually computing theJr-codimension of a germ. Before describing a fairly systematic method of doing this it may be worthwhile looking at the familiar case
p = 1
of germs of functions to see how this difConsider then a germ
fers from the situation studied in Chapter IV. f
:
(:mn , 0) --. (:m, 0).
case arises when
f E Jf ,
TheJr-tangent space to
2
J f + If.
A special Jf ,
~-codimension
as defined in
For instance, this applies to the Ak-singularities 2
k+1
.:t. Xi .:!:. ••• .:!:. x n _ 1 .:!:. xn bolic umbilic
is
since then thejf:tangent space reduces to just
and the jf:codimension will coincide with the Chapter IV.
f
x 3 + y3,
to the elliptic umbilic
x
and to the parabolic umbilic
3
2
- xy,
to the hyper-
x 2y + y4.
But in
general the relation between the two codimensions is not well understood at
155
the time of writing. ~codimension
It is known that a germ of a function is of finite ~-codimension,
if and only if it is of finite
though there seems
to be no easy proof of this fact. As in the case of germs of functions, the method we use to compute codimensions of germs of mappings is based upon (IV.2.8).
Consider a germ
One writes cod f for the %-codimension of just
cod f,
when it is quite clear that we are dealing
f,
or
with~equivalence.
The proposition just mentioned tells us that
where di
T
f
k +1.8
n
n,p
..k:1
m
•
Tf +1::+ .8
n
What we do is to compute the integers as follows.
n,p
cOdOf, cod 1f, •••
Suppose we wish to compute
successively, as
As we have already pointed
co~f.
out we have an explicit finite list of generators for Af~.8n,p p-tuples
(0,
... ,
m, 0, ••• , 0)
with
m a monomial of degree
namely the k
in
The first thing one has to do is to check which of these generators lies in Tf +Afk+1. 8 this usually amounts to doing a small comn n,p putation for each generator in turn. There is a practical point to note !Ere k+1 namely that we need only consider the generators for Tf modulo Af n .8 n ,p , which means in practice that for each such generator we can put equal to zero all terms of degree
~
k + 1.
The next thing to do is to select from the
generators which do not lie in Tf +Af~+1.8n,p Tf +Af~.8nJP
a basis for a supplement in Of course,
the number of basis elements is
the process comes to an end when one finds an integer k k
the generators forAf n.8n,p the list of integers cOdOf,
156
for which all
Finally, one adds up
lie in
cod 1f, •••
~
to get
cod f.
Note that if none
of the components Tf
.s 1 n •8n ,p
f1' ••• , fp
and hence
cOdOf
of
f
=
p:
involves linear terms then this remark will apply to all the
examples below. Example 6
We shall compute the %-codimension of the germ
2
2
f : (JR , 0) --. (JR ,0) of the "folded handkerchief" mapping given by 2 2 f(x, y) (x , y). As we saw in Example 5 the~tangent space Tf is the
=
82,2
submodule of
generated by
To compute for 1 2 .8 2 ,2
(y, 0),
+~.82,2
whilst
supplement for
Tf
Tf
+~.82 ,2
Now
0), (0,
.A2. 8 2 ,2
(x, 0),
of these
(y, 0),
(y,
y),
cOdf
=
Example 7
=
f(x)
2 + 2
=
(0, y)
(0, x) do not. in
(0 , ... , 0 ,xt+1)
mOth
Indeed these vectors form a so
Tf +12 .8 2 ,2
cod 1f
(JR, 0) --. (JRP , 0)
f t
=
We need go
2.
(0, •.. , x t+1 , ••• , 0 ) :
1~
Tf •
is generated by
generated by the vectors
(0,
erators lies in the submodule
defined by
is generated
it follows that the ~tangent space
( 0, ••• , x t+1 , ... , 0 )
not appear in the last component, and The ideal
.s
Here the Jacobian mod-
an integer.
~
(0, ••• , 0, x ), and the ideal
is generated by all vectors
codkf.
(x, 0),
obviously lie in
t
is generated by
Tf
Here
4.
Consider the germ
by the vectors
x).
is generated by
no further now because, as we saw in Example 5, we have 1 22 .8 2 2 , Thus
2
(0,
we have first to determine which generators
Tf +1 2 .8 22 , •
(0, y):
2
) 0,
2
lie in
(0, x),
cod 1f
(x,
(0, 0, xj,
t
••• , x ).
where the power does Let us compute k
so the submodule1 1 .8 1,p
k
••• , x , ••• , 0).
is
We ask which of these genThere are three cases to con-
sider. The Case
k < t
Tf +11k+1 .8 1 ,p:
k
None of the generators for 1 1 .8 1 .. ,p there are
p
lies in
such vectors, and they form a supplement, so
157
=
c~f
p
for
k < t.
=
k The Case k t None of the generators for..ff 1.8 1 ,p lie in k+1 t Tf +..ff 1 .8 1,p save the last one (0, ••• , x): there are (p - 1) such gen-
The Case
=0
for
f
~
3.
the submodule
(0,
I f .8 2 ,2
verify that the
2 (x , 0),
with
i + j
Xiyj)
k
(x , 0),
=
where
(x, byb-1)
0),
(xy,
(xa + yb,
a, b are
generate x
a
+ y
b
so
0), (0, xy)
a
(0, x),
(0, xy),
of
of
ax' (0,
oy
together with
b y ).
We wish to com-
The ideal..ff~ is generated by the monomials
Xiyj
so the sUbmodu1e..ff~.82 , 2 is generated by the vectors
these vectors lie in consider
(xy, x a + yb)
space is generated by
2 (y, 0),
c~f for k ~ 1.
0), (0,
of the germ
is generated by xy,
If
is generated by
pute
(xiyj,
=
(y, axa-1) and of oy
The ideal
~tangent
k,
the~codimension
pt + p - 1.
The reader will find it a straightforward exercise to
(xy, 0),
=
=
Jf.
xa + yb).
~-codimension
has
given by f(x, y)
of ax
Here
the Jacobian module
and
f
We shall compute
: (:m.2 , 0) ~ (JR2, 0)
integers
= t.
k
k > t.
We conclude that the germ Example 8
p - 1 when
k · . T . Hk+1 All the generators for..ff 1.81,p 11e 1n f +Jff 1 .8 1 ,p'
k > t
c~f
so
=
co~f
erators, forming a supplement, so
=
with i + j Tf ,
k.
so certainly in
(yk , 0),
k
Clearly, if
k+1 Tf +..ff2 .8 2 , 2. k
(0, x),
i ~
(0, y).
and
j
~
It remains to
VIe claim that the first
k+1 Tf +..ff2 .8 2 ,2: when k = 1 this is because of (x, 0) = Q.f.- ( 0, byb-1) , (y, 0) = a;c(0, axa-1) , and when k ~ 2 oy 2 k k 2 it is because (x , 0), (y , 0) are multiples of (x , 0), (y , 0) which
two likewise lie in
appeared in our list of generators for Tf • lies in
158
only when k
~
a,
Finally, .~ he vector whilst the vector
k
( 0, x )
(0, yk) does
so only when
k
therefore the
1)
(a -
+
Example 9
a
with
1 ~ k ~ a - 1,
It follows that
=
(b - 1)
a +
x
f
(0, x
+
of f
2
2
3 is an integer.
~
Here
2
y), (0,
I f .&2,2 a
x ).
given by f(x, y)
af
ax
(
=
a-i) 2x, ax
The ideal
will be
the~codimension
= and
af ay
=
(2y, 0) 2 is generated by x + Y ,
If
(x2 + y2,
is generated by
0),
(xa ,
0),
Following through a computation similar to that in the~tangent
( 0, x i-i) y
and the so the~codimension is
with
2
the previous example one finds that a supplement for
§3.
(0, yk)
b.
: (]l , 0) -.. (]l , 0)
so the submodule 2
and the
the~codimension
generate the Jacobian module a
the required supplements are
For our final illustration we shall compute
of the germ
where
The vectors which span
(0, xk)
~k~b-1.
2+
b.
~
2 + (a - 1) + (a - 1)
=
space in
with
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
: (]lm, 0) ~ (]l\ 0).
F : (]lr x ]lm, 0) ~ (]lq, 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 fonnal interpretation for this in-
tuitive idea, so we argue heuristically.
Think of the defonnation as a
159
"germ" (Jl{, 0) -+
(8
m,q'
f)
given by u-+ fu where
fu(x)
We wish this mapping to be "transverse" to the%-orbit through
= f,
F(u, x). i.e.
we
want something like image of the "differential" at 0 of this map
tangent space to %- orbit through f
+
tangent space to (;m,q at f.
=
The only quantity here for which we do not yet have a concrete interpretation
o
is the "differential" at • ear mappl.ng
of au
1
u -+ f : this should be the linu which sends the standard basic vectors for lRr to
lRr
of the mapping
-+ 8m,q u:0' ••• , aF -0' where we write
I
aur L
ordina tes on lRr.
u 1 ' ••• , u r
for the standard co-
As a matter of convenience we write
F.l.
I
= l!.. au.l. u=O
•
With this notation the image of our "differential" will be the real vector
xl
of
subspace lR{ Fl ' ••• , F
On this heuristic basis we introduce
the following formal definition.
F is
a~transversal
deformation of
f
when
... , Notice therefore that
f
admits a3.r-transversal deformation if and only if
it has finite%-codimension
c,
say,
Assuming this to be the case one can
construct explicit%-transversal deformations by the same device used in the Finite Dimensional Model.
One looks for a deformation
F for which
* where the germs F to be
f 1 , ••• , fc
~transversal
are to be determined.
is that
This yields an entirely practical procedure. 160
The condition for this
Simply choose f l ' ••• , f c
to
be a supplement to
Tf ,
and then define
unfoldings of germs
(Just why this is so will be made clear in the next section.)
O.
For such a germ
f
one can be a little more explicit about the form of a
with the m,q q • as a real vector space we can think of this as the direct sum of 8m •
~transversal
product
unfolding.
q :IRq with .Lm•
Suppose
*.
in~transversal
In practice one is interested solely of rank
by
F
f
Make the usual identification of
e 1 , ••• , e q for the standard basis vectors in lR q •
We wri te
is a germ of rank
reader that
...
find a basis supplement of and we obtain
O·• then a moment's thought will convince the
is a vector subspace of .L~.
Tf
Tf
)
f
8
It will then be possible to a basis for a
for a supplement of Tf
r
in 8~ is then provided by e 1 , ••• , e q ,
a~transversal
deformation
= -f
F(u, w, x)
f 1 , ••• , f r ,
.....J
r w•• e. + f(x) + ~ u .• f.(x) l.
l.
--'
i=1
l.
l.
i=1
where we insert the minus sign for a minor geometric reason which will be menHere are some examples
tioned in the next section.
o~
these computations,
parallel to the computations of~codimension given in §2, where all the work was done. Example 1
In each example the germ r The germ
has~codimension
provided by mation is
4:
( 1, 0),
r
: (lR2 , 0)
has rank
-+
(lR2 , 0)
O. given by rex, y)
indeed, we saw that a supplement for (0, 1),
F : (lR4 x :m.2 , 0)
(y J 0),
(0, x).
(lR 2 , 0)
-+
2
y 2)
J
Tf in 82,2 is
Thus a %-transversal defor-
with components given by
+ u 1y - w1
=
x
=
Y + u 2x
2
= (2 x
- w20
161
Example 2 f(x)
The germ
f
: (]R, 0)
(0, ••• , 0, x t+1 ) , wi th t
=
indeed a supplement for
°
for which
~ k ~
deformation is
t
f(x, y)
&1,P
hasX-codimension
(]RP, 0)
-+
F2
=
F
=
with components given by
X
f
2
with
E i , a - 1,
Thus -+
(0,
in
,
yi)
1 ~ i , b - 1,
with
= xy - w1
F2
= x
( x 2 + y 2 ,xa) ,
1 , i , a - 1,
in
y
b
b-1
a-1
+
I
uixi
+
where
a
together with
Viyi
(1,
0)
W2
-
.
i=1 -+
~
I
2 (]R ,0)
given by
3, hasjV-codimension 2a. and
&2 , 2 is provided by the
deformation is the germ
162
2 : (]R , 0)
together with
xi)
deformation is the germ
i=1 f
(0,
with components
F1
The germ
given by
&2 2 is provided by the
a~transversal
±
t+1
integers ~ 3, haveY.r-codimension
a, b
(]R2, 0)
a
2
Tf
and the
(1, 0) and (0, 1).
Tf
X
: (]R , 0) -+ (]R , 0)
with
ment for
X
U
+ x
indeed a supplement for
=
Thus a~transversal
X
a + b:
f(x, y)
k
... + u 1, t_1 t-1 + 1,t t -w 2 + u 21 x + ... + u 2 t_1 t-1 + u 2 t t , ,
=
(xy, xa ± yb),
Example 4
c = pt + p - 1:
(0, ••• , x , ••• ,0)
is provided by the
-w 1 + u 11 x +
F1
F : (]Ra+b x ]R2, 0)
1,
~
defined by
t (0, ••• , 0, x) deleted.
with
The germs
=
in
F : (]Rc x]R, 0)
p
Example 3
Tf
(]RP, 0)
-+
and
2 F : ( ]R2a x:m., 0)
(0, -+
1).
(]R2 , ) 0
( 0, xi-1) y
A supplewith
Thus aX-transversal with components
F1
= x
2
+ y2
F2
a = x
+)'
- w1
a-1 ~
a-1
+)'
u.i ~
;...-.
i=1
So much for examples.
v./ - w2 ~
.
i=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 "transversali ty" by a geometric notion of "versali ty".
To this
end we introduce a series of notions for deformations of a germ
Equivalence of Deformations Two r-parameter deformations
F1 , F2
of
there exists an r-parameter unfolding
f
1m
are said to
be~equivalent
of the germ at
0
when
of the identity
mapping on ]Rm for which
In this situation we call
I
m
a~equivalence
(2.2) this relation implies that
F1 , F2
of deformations.
are~equivalent
Of course by
as germs:
however
it says more in that the change of co-ordinates at the source has to respect the produc t s t ruc t ure on ]Rr x .",m. Ja
Induced Deformations Suppose and that
F : (]Rr x ]Rm, 0) ~ (]R\ 0) H : (]Rs, 0) ~ (]Rr, 0)
deformation
is an r-parameter deformation of
is a germ.
G : (]Rs x :Rm, 0) ~ (:R\ 0)
G(v, x)
=
of
l~Te
obtain an s-parameter
f
by putting
f,
F(H(v), x) 163
One writes
G-
= H'" F,
and calls
the deformation induced by H:
G-
in this
situation H is a change of parameter.
MOrphisms of Deformations Let F, G- be r, s-parameter deformations of f. is a pair
(H, I)
with
I
I.
F to
G
a~equivalence of r-parameter deformations, and
H a change of parameter, for which mation H'" G under
A morphi sm from
When r
F
=s
is~equivalent
to the induced defor-
and H is invertible we refer to the
morphism as an isomorphism.
Versal Deformations A deformation
G of
f
morphism from
F to
G.
parameter~versal
is~versal
When f
when for any deformation
F there is a
has fini te ~codimension c, say,
a c-
deformation is said to bejF-universal.
The main result about deformations under contaot equivalence is the following analogue of (II1.5.1) which we dub the~Versality Theorem.
~ F : (:n{ x lRm, 0) of a germ f ~
: (lRm, 0)
-+
-+
(lRq , 0)
(lR.\ 0) :
be an r-parameter deformation
a neoessary and sufficient condition
F to bejF-versal is that it should bejF-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.
164
The proof
that
~transversa1
~versal
deformations are
shall not give it;
is by no means easy, and we
it uses an extension to the real case of a classical
theorem of Weierstrass in complex function
A sketch of the result
theo~.
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.
F
is
We have to show that
~versal.
F is
~transversal,
that
... ,
]({ Fl ' Consider then a germ
g
=
8
m,q
and the 1-parameter deformation
in
G
of
f
given by
G(v,
say.
H
=
Thus
= f(x) + vg(x).
G is ~equivalent (as a deformation) to an induced deforma-
By hypothesis tion
x)
h*F
with
H(t, x)
h :
OR,
0) ~ (]{r, 0)
= F(h(t), x)
having cqmponents
h 1 , ._ •• , hr
and
.
H = establishing that H lies in ]{{F 1 ,
... , Fl· r
Starting from the fact that
G, Hare 3Y-equivalent deformations a computation shows that Tf ,
and hence that
G lies
in
]{{F1' ••• ,
Frl + Tf •
finishing the proof.
G- H
However
lies in
G. = g,
D
Just as in the Finite Dimensional Model we are now in a position to justify the use of the prefix in the term
"%-uni versal" •
165
3.2 f
F, G be 3r-universal deformations of a germ
Let
: (JRm, 0)
-+
(JR\ 0)
of finite 3r-codimension
c:
then
F, G are
Y.r-isomorphic deformations. As
h : (:R c , 0)
is 3r-versal there exists a germ
F
for which G is 3r-equivalent to the induced deformation there exists a c-parameter unfolding II? : ORc x :Rm, 0) the germ at
-+
i.e.
We shall show that
F
0
q"
G are t;-equivalent, where
IjJ
x)
with entries in 8m+ c
identifications. setting u = 0,
--'
8h j
au:-
Now
Ao(x)
F, G are both
=
~
i
and
~transversal
=
(h
x 1)
0
q x q matrix with the usual
q which leaves 8m
Tf
ui'
and
...
lies in the Y.r-tangent space
~
proof of (2.4) the invertible matrix AO
In
.
T.
Tf ,
II?
a relation of the form
deformations of
span a supplement for
8 m-module
of
AOGi + T.~
=
J
A(O, x),
c
~
(O)F.
~
j=1
=
for which F 0 q, = AG,
one obtains for c
AO
(:IRc x JRm, 0)
Differentiating this relation with respect to
')
where
That means
h F.
h is an invertible germ, which we do as follows.
view of (2.1) this means that there exists an invertible
= A(u,
(:R c, 0)
-+
of the identity map on JRm for which
0
t;-equivalent,
A
...
as do
f
by
(3.1) so In view of the
induces a linear automorphism of the
invariant.
Thus AoG1 , ••• , AOFc
also span
It follows from ... that the matrix of coefficients a supplement for Tf • ah. ~8 (0), i.e. the Jacobian matrix of h, is invertible: the Inverse Funcu. ~ tion Theorem now enables us to deduce that h is invertible, as was required. It follows from the definition that
F, G are 3r-isomorphic deformations.
o In fact we can squeeze a little more information out of the proof than is actually stated in (3.2). 166
Let c,
and let
f F
: (:rnm, 0) be a
~
(:rn.\
~universal
parameter X-versal deformation
deformation of
F'
Proof
f
~
c
any d-
are ,J("-isomorphic.
(3.2)
one comes to the conclusion that h*F,
with
a submersive germ. However by (I.1.3) there exists d : (JR~ 0) ~ (JR , 0) for which hO¢ = 1T with
: (JRd, 0) ~ (JRc, 0)
F'
d
0) ~ (JRc , 0)
an invertible germ ¢
Thus
For
And hence any two d-parameter %-
Proceeding exactly as in
h : (JRd ,
f.
is ~isomorphic to the (d - c)-
is X-equivalent to an induced deformation
F'
1T
f
of
F.
parameter constant deformation of versal deformations of
be a germ of fini te %-codimension
0)
the projection given by 1T(U 1, .t., ud ) = (u 1 ' ••• , uc)·
is %-isomorphic t"
defomation of
F.
1T
F,
i.e.
to the (d - c)-parameter constant
o
The rest is clear.
There is a small technicality which is worth mentioning at this point. We have phrased the definition of X-isomorphism for deformations of a single germ
f.
One can of course phrase the definition for
equivalent germs and (3.3).
f, f'
~eformations
of %-
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 %-versal 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
§4.
Classification of Stable Germs
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 ver,y simple idea. ~ G : (JRn , 0)
(liP, 0)
-+
exists an invertible germ _ h : (JRn , 0)
-+
be a germ of rank (JRn , 0)
an r-parameter unfolding of a germ of rank
r:
for which
F
then there
=G
h
0
is
O.
By making linear changes of co-ordinates at source and target we can suppose that the Jacobian matrix of
where
I
r
is the identity
g : (JRn , 0)
-+
(JRr , 0)
r x r
g has rank
an invertible germ h : (JRn , 0) (x1 , ••• , xn )
-+
evaluated at
-+
(lin, 0)
(x1' ••• , x r ).
0,
is
Consider the germ
whose components are the first
Clearly,
tion
matrix.
G,
r,
so by
r
components
(1.1.3) there exists
for which
And then F = G
g
0
h is the projec-
0
h
is the required
o
germ.
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 attention to r-parameter unfoldings r:
for such an unfolding write
germ of rank F
-+
168
f F•
0 which
F : (JRr x JRm, 0) fF : (JRm, 0)
F unfolds.
-+
-+
(JR r x JR q , 0)
(JR q, 0)
of
rank
for the unique
We wish now to study the assignation
The first step in this direction is provided by
~ F, F' :
(4.2)
foldings of germs f F, fF'
are
(ltr x JR m, 0)
fF' fF'
of rank
0:
-+
(:Rr x JR\ 0) be r-parameter un-
if
F, F'
are A-equivalent then
,%"-equivalent.
Step 1
Since F, F'
are A-equivalent there exist invertible germs
H, K for which the following diagram commutes.
Notice first that in view of (4.1) we can suppose that for some ¢ : (JRm, 0)
-+
(lRr , 0),
and likewise that K(O, Write
It should now be clear that fF' g : (lRm, 0)
-+
to show that
(lRq, 0)
H(O, x)
F(u, x)
= (¢(x),
x)
y) = (¢(y), y)
= (u,
feu, x»).
is A-equivalent to the germ
given by g(x)
= f(¢(X), x).
And
it will suffice
g, fF are jr-equivalent.
Step 2
feu, x) - fF(x)
vanishes on 0 x lRm,
so each of its
q
components does likewise, and the Hadamard Lemma allows us to write feu, x)
=
fF(x)
+
entries are germs at And
M(u, x).u where M{u, x) e
of functions on lR r x lRm,
likewise we can write
whose entries are germs at wri te ¢(x)
= ¢( g(x») =
¢(y)
=
g(x)
=
q x r
0 of' functions on JRq.
matrix whose
and M(e, 0)
with
=
O.
It follows that we can
B{x) .g(x) wi th B an r x q matrix.
matrix, for which C(O) = 0,
fF(X) + D(x).g(x)
q x r
A(y).y where A{y) is an r x q matrix
to the same matrix notation we can now write c(x) a
is a
D(x)
a
Keeping
g(x) = f F(x) + C(x).¢ (x) with and hence
q x q matrix for which
D(e)
= o. 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 equivalence classes of germs to
F
(I - D(x)).g(x)
=
I - D(x)
sive.
o
~
induces a mapping from A-
fF
~-equivalence
classes.
~
A sufficiently small representative of
f
Vf
of codimension
We wish to study Suppose we have which is submer-
will be a submersion, so
f- 1 (0)
of f
will be a smooth submanifold of E. s x E.m
We take
1Tf :
(Vf , 0)
=
q.
is
are ~-equivalent, hence ~-equivalent.
m) f : ( lR s xlR,O
the zero set
I
is an invertible matrix,
this mapping in detail, and to do this we need a new idea. an s-parameter deformation
where
of the restriction to
Vf
of the projection
~ 1T
(lR s , 0) :
to be the germ at 0
E. s x lRm ~ E. q •
We need
the following fact.
Let
0) f, g • (lRs x lRm, 0) ~ ~ (",q, =
deformations of germs go
are
~-equivalent
Proof ~-transversal
be
f O' go : (lRm, 0) ~ (lR\ 0) then 1Tf,1T g
-versaI s-parame t er
Q:Y JZ
of rank
0:
if
f O'
are A-equivalent.
It is convenient to make the preliminary observation that a deformation (and hence any ~-versal deformation) of a germ
of rank 0 is automatically submersive, so that the notation introduced above makes sense in the present context. isomorphic deformations, in view of Step 1
Observe also that
f, g
must be %-
(3.3).
Now consider first the case when
fo = gO.
The fact that
f, g are jV-isomorphic deformations is expressed by the existence of a commuting diagram of germs
170
m ( :m.s x :m. , 0)
~
·1
1·
h
(:m. s , 0)
with
m ) (:m.s x :m. , 0)
) (:m. s , 0)
This last
invertible for which
~,h
relation ensures that
~
induces a mapping from
Vg
onto
Vf '
yielding a
commuting diagram of germs ~
(v , 0)
)(V , 0) g
.: 1 (:m. S, 0)
expressing the fact that Step 2
are A-equivalent.
7Tf' 7T g
and an invertible
=
for which
deformation
g' (u, x)
=
follows from Step 1 that
q x q matrix
are just y,r-
7T g'
7T g'
M(x)
with entries
Evidently the s-parameter
M(X).fO(h(X)).
M(X).f( u, hex))
go
of
It
is y,r-versal as well.
are A-equivalent.
And it suffices there-
that however is clear as
are A-equivalent:
fore to show that will map
) (:m. S, 0)
By (2.1) that means that there exists an invertible germ
h : (:m.m, 0) ~ (:m.m, 0), m
h
Consider next the general case when fO' go
equivalent.
in 8 ,
1
7Tf
x k
o
v, g
The relevance of these ideas to the material of the present section is as follows.
Let
F : (:m. r x
folding of a germ To
fF
JRll,
of rank
F we associate the germ
0)
~ (:m. r x :m.\ 0)
0,
given by a formula
DF : (:m. r x
:nil
x
JIf,
be an r-parameter unF(u, x)
= (u,
0) ~ (:m.\ 0)
feu, x)).
given by 171
(u, w, x) ~
-w +
f(u, x):
deformation of f F• just graph f,
thus
DF is an (r + q)-parameter submersive
The geometric connexion between F, DF is that
and 7rD
can be identified with f.
VD is F The basic the orem con-
F
necting F, DF is F is an A-stable germ if and only if
~-versal
DF is a
deforma-
tion of f F • The proof of volume.
(4.4) does not use techniques lying outside the scope of this
However a careful version would occupy more space than is available.
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 lists of stable germs in given dimensions.
A sketch of the proof, sufficient
for the competent reader, can be found in the paper of Jean Martinet mentioned in the Introduction to this book. fact that
(4.4) together with the
It is perhaps worthwhile spelling out the ~-Versality
Theorem allows one to produce
explicit examples of stable germs.
Example 1 of the germ By the
We saw in Example 1 of §3 that a ~-transversal deformation 2 2 (x , y ) is the germ given by (x 2 + u 1y - w1, Y2 + u2x - w2 ) •
~-Versality
Theorem this deformation is
~-versal.
precisely the deformation associated to the unfolding 2
Y + u2x)
It is however
(u 1 ' u2 '
so this germ must be stable.
We can now return to the main theme of this section by stating
(4.5)
~ F, G : (lR r
unfoldings of germs f F , fG F, G are A-equivalent.
172
x
lRm, 0) ~ (lR r x lR\ 0)
of rank
0:
if f F , fG
be stable r-parameter are
~-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
F then the assignation F
tive mapping from A-equivalence classes of germs.
fF
~
actually induces an injec-
f germs to
F : (JR r x JRm, 0) ~ (JR r x JR\ 0) fF : (JRm, 0) ~ (JR q , 0)
, r + q: fF ,
classes
Our objective now is to determine the image of this mapping.
To this end we make the following observation.
germ
~equivalence
indeed
If
is a stable r-parameter unfolding of a
of rank 0 then fF
(4.4) tells us that the (r
has %-codimension
+ q)-parameter deformation
is %-versal, hence %-transversal, so the J,f-codimension of fF the number of parameters, S
=
i.e.
=
S(r, m, q)
,
r + q.
must be
With this in mind write
set of A-equivalence classes of stable germs
(JRr x JRm, 0) ~ K = K(r, m, q)
DF of
(lRr x lR\ 0)
of
rank
r.
= set of J,f-equivalence classes of germs
(lRm, 0)
~
sion
r + q.
,
(lRq, 0)
of rank 0 and J,f-codimen-
We can now state the main result.
(4.6)
The mapping
S
~
K induced by the assignation
F
~
fF
is a
bijection. Proof
Only surjectivity remains to be established.
therefore a germ f 0 : (lRm, 0) ,
r + q.
~
Certainly then, as in
(lRq, 0)
deformation of fOe
of rank 0 and J,f-codimension
§3, we can construct an (r
3V-transversal deformation of the form
-w + feu, x)
F : (lR r x JRm, 0)
~
+ q)-parameter
with f
This is precisely the deformation
the r-parameter unf olding
We consider
DF
(lRr x lR \
an r-parameter
associated to 0)
given by 173
=
F(u, x)
( u, f(u,
X») •
The
~Versality
%-versal, and then (4.4) tell s us that vation that
F has rank
r
Theorem tells us that
F is stable.
DF is
The trivial obser-
concludes the proof.
0
Thus the problem of classifying stable germs under the relation of A-equivalence reduces to the problem of classifying germs under the relation of %-equivalence, up to a certain %-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. the matter are decidedly more complicated. the germs
(JRm, 0)
-+
(JR\ 0)
of type
However the mechanics of
At root we are trying to list
~m
under the relation of %-equiva-
lence, at least up to a certain %-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
~i.
Thus the next aspect of the theory to which we
address our attention is the construction of such invariants.
§5.
Higher Order SingUlarity Sets
We have already seen that the symbol singular point. JRn
-+
JR
~ 1 is of type
of codimension ~
i
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.
Ei
Consider the smooth mapping JR2 -+ ]R2 given (in complex numbers) by Given a (small) 2 z + 2~ Z, z
~ >
we consider the deformed map
: JR2 -+ JR2
f
where the bar denotes complex conjugati on.
~
f
0
(x, y) ~
is defined by
u
= x
2
(u, v) - y
2
+
z-+ z 2 •
given by
In real numbers
where
=
v
2~x
2xy -
2~y
and has Jacobian matrix
(2X + 2~ \.
-2y ) 2x -
2y
2~
which has rank < 2 when its determinant vanishes, x
222 + Y
=
on the circle
The bifurcation set is soon
this then is its singular set.
~
i.e.
If we parametrize the singular set by putting
found as well.
x
:::
!
cos
e
y
=
~
sin e
then we obtain a parametrization of the bifurcation set in the form u
=
~
2
(cos 2e + 2cos e)
v
=
l(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 x 1
set Z f,
2
+ y
2
=
£
2
is precisely the first-order singularity
since clearly the Jacobian matrix cannot have rank
0.
We are
therefore unable to distinguish one point on the circle from another by just looking at the symbol
Zi.
On the other hand there are three points on the
circle (the complex cube roots of
£3
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 f onto the hypocycloid.
maps the circle
We look therefore at the restriction fiz 1f.
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
f
to the tangent line to the circle.
at the point
(x, y)
Now the tangent line to the circle
is the line through the origin perpendicular to this
And a unit tangent vector will be
vector.
under the differential of
f
at
(x, y)
(-y/£,
x/£).
The image of this
will be obtained by applying the
Jacobian matrix to it, yielding the vector 2X + 2£
-2y )(-y/£\
( 2y
2x - 2£
= ~(
x/£)
£
-2xy-
2
-y + x
The differential of the restriction certainly has rank
2
~
1;
and it has rank
zero only when this last vector vanishes, which happens precisely at the cube roots of
£
3•
In other words our three points are distinguished precisely
by the fact that they are
Z1
other points on the circle are
points for the restriction fiz1f, ZO
points for the restriction.
The next step in the theory becomes clear. f
: JRn
176
-+
whilst all
Given a smooth mapping
JRP we have the first-order singularity sets Zif •
If these are
submanifolds we can introduce second-order singularity sets And this process can be continued.
Ei,jf
= Ej(f/Eif).
If these sets are submanifolds we can . j k
introduce third-order singularity sets El., , f
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 unsatisfactor,y element here in that the definitions only make sense as long as we continue to obtain submanifolds. this difficulty submanifolds j1f
-
Ei
However, as we saw in Chapter II, there is a way out of
of the jet-space
J1(n, p)
are preoisely the required sets
Eif:
these sets will be smooth manifolds.
for which the inverse images under and then for a generic mapping
Thom proposed the problem of imitating
this prooedure for the kth order singularity sets,
i 1 ,···,ik
manifolds
E
f : ]tn -+]tP
One defines
at least for the first-order singularity sets.
k
in the jet-space
the pull-baok under
J (n, p)
jkf
i.e.
of defining sub-
such that for a generic
are smooth manifolds, and preoisely
the kth order singularity sets
The case
k = 2 was solved by
H. Levine, but the general case waited till 1967 when it was solved by We oannot hope to give a full account of Boardman's solution in a
Boardman.
book of this nature:
what we can do however is to describe the construotion
whereby one decides which symbol
i 1,···,ik
E
of a given map-germ, since it is an
entirely practical one. Let
We start with an algebraic idea. in the algebra
••• , f
is to be attached to the k-jet
P
I
be a finitely generated ideal
be generators for
Y1' ••• , Yn be a system of co-ordinates in 8n • integer
s
~
1.
We define
ideal generated by all
s x s
~
s
I
to be the ideal
I, and let
Suppose one is given an I + I'
where
I'
is the
minors of the Jacobian matrix
177
of i
ofi oY1
* of
-2.
oYi
The ideal 6 s I
so obtained depends neither on the choice of gen-
erators, nor the choice of co-ordinates, tors for
I,
and
zi' ••• , z
--
n
i.e.
if
gi' ••• , gq
are genera-
is a system of co-ordinates, then
cides with the ideal generated by
and the
I
s x s
6s I
coin-
minors of the Jacobian
matrix
**
og
-=:!l. az n
Clearly, it suffices to show that any in 6 I. s
x
s
minor of
can be wri tten as a linear combination of the
Each
coefficients in Sn.
Thus each
af k bination of the
s
az.
agi oz.
**
lies with
can be written as the same linear com-
J
plus an element of
I.
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
*** of
~
OZ
Thus it will be enough to show that any
n
s x s
***
minor of
lies in
~sI.
For this observe that the Chain Rule allows us to write each __ 0_ as a linear OZj
a
combination of the
with coefficients in
oYk '
The result now follows
on appealing again to the multilinearity of the determinant.
D
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.
Note that
~
s
I
=
Example 1
Let
generated by =
I
k-1 <x >.
Example 2 ated by
I
And
and the
s > n,
=
I
and the
Let I,
when
~
k
I
s
~
~
1.
minors of the Jacobian matrix k
I
=
=
<xy, x
<x > for 2
s
~
(kxk- 1),
so
2.
2 + y >
is the ideal gener-
x 1 minors of the Jacobian matrix
is the ideal generated by
~2I
minors;
k
with
<x >
1 x
s
also that one has the inclusions of ideals
=
2
2
<x , xy, y >.
And
I
and the
2 x 2 for
3. 179
Given an ideal
I C 8 -
we shall adopt the notation
n
= /: :, n-s+1 I
... ,
and refer to the ideal
I.
/:::,nI
as the successive Jacobian extensions of
In view of the sequence of inclusions above we have
* Let us call
I
proper when
I ~ 8n •
(Note:
= 8n
I
saying that if we take a finite set of generators for generator has constant term
I 0.)
Suppose
I
is the same thing as I
then at least one
is proper.
The critical
i
Jacobian extension of
I
is the last ideal
/: :, 11
i
is proper.
i
In this way we obtain an ascending sequence
of successive critical Jacobian extension of
i
/:::, 2/:::, 11
and we say that
And so
...
, I
has
Boardman symbol
(i 1 ,
Example 3
One checks easily that the Boardman symbol of the ideal
I
=
<xk>
(k - 1)
i 2,
I,
/: :, 11 ,
which
i
It has in turn a critical Jacobian extension /:::, 2/:::, 11 • i
on.
*
in the sequence
••• ).
in 8 1 mentioned in Example 1 is
repeated
1's.
with 2 2 And the Boardman symbol of the ideal I = <xy, x + y >
in 8 2 mentioned in Example 2 is The Boardman symbol of a genn that of the ideal
If
(2, 0, 0, ••• ). f
: (lRn , 0)
i.e.
--+
generated by the components
The Boardman symbol of a genn invariant,
(1, 1, ••• , 1, 0, ••• )
if two such genns are
(lRn, 0)
(:ffiP,
0)
is defined to be
f1' ••• , f p •
-+
~-equivalent
(lRP , 0) is a contact they have the same
Boardman symbol. Proof 180
Suppose the germs are
f
with components
f 1, ••• , fp
and
... , f'
with components
f'
p.
Suppose I f,
f, f'
are e-equivalent.
coincide, so the Boardman symbols of f, f' Suppose
invertible germ The components
f, fl
h : (JRn , 0) h 1 , ••• , hn
-+
of
x 1 ' ••• , xn
Jacobian matrix of
coincide, by (5.1).
are right-equivalent, (JRn , 0)
for which
f
i.e.
there exists an
h, fl
0
coincide.
h yield a system of co-ordinates.
compute the Jacobian matrix of of co-ordinates
By (2.1) the ideals If'
••• , fl
P
If we
relative to the standard system
we get the same ideal as if we compute the
••• , f
p
relative to the system of co-ordinates
Again it follows from (5.1) that
f, fl
have the same Boardman
symbol. Step 3
The required result is immediate from the two preceding steps.
o
Now we can extend our definition. is X-equivalent to a germ symbol of
f
Certainly any germ f: (JRn, x) -+ (JRP, y)
fo : (JRn , 0)
to be that of f o •
-+
(JRP , 0).
We define the Boardman
In view of (5.2) this definition is
unambiguous.
The first f
:
(JRn , x)
-+
k
(JRP , y)
integers in the Boardman symbol of a germ depend only on the k-jet of f.
Clearly, we can suppose
x = 0,
y = O.
Let
of f,
and let
(i 1 , i 2 , ••• )
generated by the components
f 1 , ••• , fp
the Boardman symbol of
It is evident, by induction on
i
I.
k,
I
be the ideal be
that an ideal
i
t:. s t:. k-1 ••• t:. 1I
is generated by partial deri vati ves of order
'" k
of
And whether this ideal is proper or not depends only on the
181
values of all these derivatives at of
0:
thus
ik
depends only on the k-jet
o
f.
We return now to our objective of partitioning the jet-space into kth order singularity sets. that a germ
f
: (JR n , x)
-+
Given
(JRP, y)
k
J (n, p)
k
integers i 1 , ••• , i k we say ik,···,i k when its is of ~ E We define
Boardman symbol has the form
i 1 ,···,i k E 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 Example 4
In the case
k = 1 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
~sI
I
One sees this as follows.
having a representative germ with zero source
be the ideal generated by the components of the germ.
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, ~sI
it follows that rank precisely
s
-
i.e.
if and only if the Jacobian has kernel rank
~
s:
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
i 1 ,···,i k E
order singularity set
in
J
k
(n, p)
is non-empty.
The answer
is provided by A necessary and sufficient condition for the set i 1 ,··· ,i k
E
C
j
J-(n,
should be satisfied:
182
p)
to be non-empty is that the following conditions
(i)
n
(ii)
i1
(iii)
if
~
i1
~
i2
~
...
ik
~
~
0
n - p
~
i1
=
n - p
then
= i2 =
i1
... =
i k•
Note first that we need only concern ourselves with jets having zero source and target. Necessity
(i)
It is an immediate consequence of the definitions that To see that
we proceed as follows.
Take
I
to be the
idea~
~1'···'~k
components of a representative of some jet in Z ,it ,i 1I that u u is generated by g l ' ••• , ga
~
•
We can suppose
say, with
.{
t
is proper.
Suppose
(n - s + 1)
ij+1
genera ted by the
8~
in
ij
.
...
This implies that all the minors of order
of the Jacobian matrix of
g1' ••• , ga,
vanish, and therefore
J
in particular all the minors of order
(n - s + 1)
vanish, so that the ideal
g1' ••• , ga _ j 1 hence s ~ i,.
Taking
J
s
=
i j+1
we obtain
the kernel rank of the jet, so certainly (iii) immediately that
is proper, and i j+1
~
i .• J
As we observed above in Example 4 the first index i1
(ii)
Suffi ciency
of the Jacobian matrix of
If
i1
=
i1 i2
= =
n - p
=
then
~
n - p. i t. 11
= t.P+1I =
I,
and it follows
i k•
Suppose conditions (i), (ii) and (iii) are satisfied.
We
with components
have to produce a germ say, which is of type
is
i 1 ,···,i k Z •
We consider cases, leaving the computa-
tions as good exercises for the reader.
Case when
n - p
Case when
i1 > n - p f.
~
=
In this case we choose
In this case the following choice will do.
x.
~
n-i 1 f
~
=
>"
=
n-i 1+1
f.
= xp •
••• , fp
n-i 2 x. J
--J
+
)'3 x~ J
+
-'
n-i 1+1
n-i 2+1
0
(n - i1 + 2 '" i
'" p)
o One immediate consequence of this result is that the partition of i 1 ,···,ik by the non-empty E is actually finite. Example 5 are
2,2
E
The only non-empty singularity sets in the jet-space
E2 ,1,
,
E2 ,0,
E1 ,1,
beginning of this section, -t
(u, v)
at any point
with
U
(x o ' YO)
i.e. =
X
2
the mapping 2 - Y + 2E"x,
in the plane.
Boardman symbol of any germ to the germ of
Here, 184
1-+
I
(n, p)
J2(2, 2)
EO'O.
f
f
(u O' v O)
at
-t
JR2
Ei,j
defined by and
E" > O.
for the genn of
f
By definition, we need to compute the
fo : (JR2, 0)
(x o ' YO)·
: JR2
v = 2xy - 2E"Y
We shall compute the second-order Boardman symbol
(x, y)
k
Let us return to the example we studied in detail at the
Example 6
(x, y)
E1 ,0 and
J
-t
2 (JR , 0)
which is 3V-equivalent
An obvious choice for
f
0
is given by
where
is the ideal generated by
u o ' vo
and
~iI is the ideal generated
by u O' vo
and the minors of order
(3 - i)
of their Jacobian matrix
~2I is generated by u O' vo and the entries in the Jacobian, and
The ideal
cannot be proper as two of its generators
auO ax '
avO
10.
have constant term
ay
~1I is generated by u O' vo and the determinant D of the above
The ideal
Jacobian, and will be proper (hence critical) when the constant term
Xo2
2
+ YO - e
2
the germ of and the
in f
2 x 2
D vanishes:
has type
E1.
this then, as we saw before, is the set where The ideal
minors of their Jacobian:
~ 1~ 1I is generated by u O' v O'
D
a line or two of computatinn will
verify that the constant terms in these generators are 222 x o + YO - ! ; And the ideal
A1~1I will be critical exactly when the last three expressions
vanish simultaneously,
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
E1f
!
3•
are distinguished pre-
at these points has type
whereas at all other points on the circle it has type
E1,1,
E1,O.
In Chapter II we proved that the first-order singularity sets i1 1 E C J (n, p) were smooth ~anifolds, and computed their codimensions. We shall only state the much harder result of Boardman.
(5.5)
If the kth order singularity set
is non-
empty then it is a smooth submanifold of codimension
185
where
~(i1' ••• , i k )
denotes the number of sequences
(j1' ••• , jk)
of
integers which satisfy the following conditions
Example 7
In the case
dimension of
Ei
in
k
=1
~(i)
we have
J1(~n, ~p) is
=i
(p - n + i)i,
and hence the cowhich agrees with the
formula we obtained in Chapter II. Example 8
Suppose
~(1, ••• , 1)
=
(p - n + 1)k. just
k,
k,
= i2
i1
= •••
=i k
= 1.
Clearly we have
E1 , ••• ,1
so the codimension of
Note that in the equidimensional case
the number of repeated
Example 9
In the case
so the codimension of
Ei,j
(p - n + i)i +
k
Jk(n, p).
Jk(n, p)
p = n
will be
the answer is
1's.
=2
in
~
one has
J2(n, p)
(i, j)
=
i(j + 1) _ j(j-1) 2
is given by the formula
ir (p - n + i) (2i -
In view of (5.5) the singularity sets submanifolds of
in
j + 1) - 2i + 2j].
i 1 ,···,ik
E
are called the Boardman
The Thom Transversality Lemma, proved in §4 of
Chapter II, yields the following. The set of all smooth mappings transverse to all the Boardman submanifolds
186
f
: ~n
--+ ~p for which i 1 ,···,i k
E
is dense in
jkf
is
: lRn
We shall call a smooth mapping f Boardman when
-+
lRP
generic in the sense of
is transverse to all the Boardman submanifolds
for every integer k
~
1.
i Z
1,···, ~
,
For such a mapping the set
will be a smooth submanifold of lRn having the same codimension as Boardman showed that
=
lRn -+ lRP be generic in the sense of Boardman: ik+1 i 1 ,···,i k Z (f Z f).
I
In other words we have the following.
Any smooth mapping f
then
: lRn
-+
lRP
can be forced to be generic in the sense of Boardman by an arbitrarily small perturbation:
i 1 ,···,i k
moreover, for such a mapping the sets
Z
manifolds and coincide precisely with the Thom singularity sets. dentally one trivial consequence of
Example 10
Let
f
: lR3
:m.3
-+
(5.7),
f are smooth Note inci-
namely that
be generic in the sense of Boardman. By Example 7 the codimension 2 has codimension i
We ask which Thom singularity sets can occur. is
.2
J. ,
i
hence
Zf
clearly then Z1f with codimension 1 is the only first-order Thom singularity set which can occur. dimensions
1, 2
Z1f
splits into
respectively, and Z1,1 f
Z1,1,1 f with codimensions
2, 3
Z1 ,0 f and Z1,1 f with cosplits into
respectively;
no further splitting can
take place since the kth order Thom singularity set sion k,
so will not appear for
k
~
Z1 ,1, 0 f and
Z1, ••• ,1 f
has codimen-
4.
187
A pleasant illustration is provided by the dovetail mapping
ExamEle 11 f
: lR3
-+
lR3
given by
(x, y, z)
=x
u
v
t-+
(u, v, w)
=y
w
= z4
where - uz - vz
2
.
The reader will easily check that the possible Thom singularity sets E1, ••• , 1 fare given by the equations below. E1f
ow
=
oz
E:!!.
E1, 1f
oz = ow
E1 ,1,1, f
E1f
0
and
0
oz =
0
and
iw
0/ = o2w
oz2 =
is the folded surface illustrated below:
",1,1,1 f
lJ
.
~s
th e
0
0
and
l: 1,1 f
o3w
az 3
= o.
is the fold curve, and
•.
or~gJ.n.
In succeeding sections it will be useful for us to know that the Boardman i 1' •• • i k are invariant under unfolding in the following precise symbols l: sense.
unfolding of the germ Boardman symbol. 188
f
Proof
Let
... ,
u 1,
u
those in JR r •
r
... , ur'
u 1,
x 1 ' ••• , xn
... , f p
f 1,
be the standard co-ordinates in JRn , and
We take where
f 1,
f'
to be the germ whose components are
... , f p
are the components of f.
The proof proceeds in two steps. We claim first that
Step 1 Suppose on
f
(i 1 , i 2 , s ~k-1 iO I:;,. I:;,. ••• I:;,. If I
'-' u
that the ideal iO k-1 ••• UA I f. Here we put
8
with a subset of
ASA
n
i
I:;,. k
that
8 I n
i
I:;,. 0 1
have the same Boardman symbol.
... ) .
has Boardman s~bol
k, i
f, f'
We shall show, by induction
is generated by
n l = n + r.
It will follow immediately
is critical, so establishing the claim.
fl
Suppose it holds for
k.
Jacobian matrix of some fixed set of generators for JI
I:;,.
of the same set of generators augmented by
u 1" ••• , u r
and the identity order
-
the generators for
u 1 ' ••• , -
k = 0
to be the
i !::,.
0I
f
relative
to be the Jacobian matrix
Consider the
x 1 ' ••• , x n '
(n l
J
Take
and
minors of order
When
ik
to the standard co-ordinates
generated by
and
for convenience, and tacitly identify
where
the assertion is trivial.
ordinates
u 1 ' ••• , u r
s + 1)
r x r
Now
JI
ik
and the
is the direct sum of
J
matrix, so the ideal generated by the minors of JI
coincides with the ideal generated by the minors i iO I:;,. I f, is generated by of order (n - s + 1) of J. Thus I:;,.sl:;,. k ik i1 , ur ' the generators for I:;,. ••• I:;,. If' and the minors of order u 1' (n l
s + 1)
of
of JI.
I:;,.
...
i.e.
Step 2 valent.
it is generated by
We claim that
and
f', Fare e-equivalent, and hence JV-equi-
It will then follow from (5.2) that
f', F have the same Boardman
symbol, which fact combined with Step 1 will clinch the result.
Since
F 189
unfolds
f
it has components
and for
u 1 ' ••• , ur '
we have
0, ••• , 0)
=
fi + Si
f.(x 1 , ••• , xn ) ~
It follows from the Hadamard Lemma that we can
identically in x 1' ••• , xn • write each Fi
=
where
Si
lies in the ideal in Sn'
generated
Thus ••• , F
p
generate the same ideal in
so the corresponding germs
f', F are ~-equivalent by (2.1).
Example 12 F1 ,
(:m.n , 0) -. (:m.n , 0) with the component s
The germ F
... , Fn
where
{::
=
x.~
=
n+1 x n
(1 :;; i
Example 13
n - 1)
i x.x ~ n
)'
+
,.-oJ
i=1
E1 , ••• ,1,0
with
f n
:
(:m., 0) -. (:m., 0)
repeated
l' s
(:m.4 , 0) -.
The germ F
F1
= x1
F2
= x2
F3
= x 3x4
F4
=
given by f(x)
= xn+1
using Example 3.
(:m.4, 0)
F1 , F2 , Fy F4 where
190
:!;
n-1
is an unfolding of the germ has type
D
2 2 x3 + x4 + x 1x 3 + x 2x4
with the components
so
is an unfolding of the germ f(x, y)
§6.
=
(~, x
2
2 + y)
f
: (JR2 , 0)
OR,2
~
Z2,0
so has type
0)
given by
by Example 3.
Classifying Germs under %-equivalence
In this section we shall consider the
ve~
simplest situations where it is ~equivalence.
possible to obtain explicit lists of germs under the relation of
As in Chapter IV the whole thing turns on the idea of "detenninacy". a germ f
: (JRn , 0)
g : (JRn , 0)
~
(JRP, 0)
~ (JRP, 0) with
We call
%-k-determined when any germ
If = 19
is %-equivalent to
f.
By
analogy with (IV.3.1) one might reasonably expect the following result. (6.1)
A sufficient condition for a germ
jY-k-determined is that l ' ~+1 • Sn,p
f
f
(JRn , 0) ~ (JRP, 0)
Tf •
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
more than
for that reason we
By further analogy with Chapter IV we call f
%-determined when it is
%-k-determined for some
k
follows immediately from (2.3) and (6.1) that a germ of finite must be finitely
to be
%-determined.
~
1.
It
~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
%-equivalence rather than
~-equivalence.
The first
step is the Splitting Lemma. 191
(6.2) then
Let f
2
f E../f n
be a germ of corank
c
and finite 5.r-codimension:
is 3.f-equivalent to a germ
with The proof is exactly the same as that given in Chapter IV. forms for germs of corank 0 are given by the Morse "Lemma.
The normal And for germs
of corank 1 one obtains almost exactly the same classification as before.
(6.3) then
Let f
f E../f 2 n
have corank 1 and finite 3.f-codimension
k ~ 1:
is 3.f-eguivalent to a germ of the form
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.
We shall not pursue the
point further. Let us now turn our attention to germs of smooth mappings, as opposed to germs of smooth functions.
Vve are now in a much more complicated situation,
and can only hope to obtain results in
ve~
special cases.
I t 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
: (JRn , 0)
-+
(JRP , 0)
its Boardman symbol must have the form integer
192
k
~
1.
be a germ of finite 5.r-codimension: (i 1 , ••• , i k , 0, 0, ••• )
for some
Proof
As we have remarked already f g : (:m.n , 0)
so Jr-equi valent to
~
given by a polynomial of degree Boardman symbol of
(:m.P , 0)
--+
k,
must be finitely Jr-determined each of whose components
for some integer
g has the required form
k
~
is
Clearly, the
1.
(u 1 , ••• , ~, 0, ••• , 0),
since the Boardman symbol is a Jr-invariant, that of
f
and
has the same form.
o Perhaps the simplest
case to study for germs
= 1.
(:m.n , 0)
--+
(:m.P , 0)
with
EO
The first order Boardman symbol is
or
p ~ 2
is the case when
E1.
In the former case the germ is non-singular and a normal form is pro-
vided by 1
E.
(1.1.4).
n
Now suppose the germ has finite
We ask for the classification up to JY-equivalence.
(6.5)
Let
f
:
Jr-codimension.
(:m., 0)
Then
f
(:m.P , 0)
--+
(0, 0,
be a germ of tyPe
is necessarily of type k ~ 1,
repetitions) for some integer
1
and finite
E ,
E1, ••• ,1,0
(with
k
and in that case is Jr-equivalent to
..., °, xk+1) •
Proof of
Jr-codimension, and type
That
(5.4)
and
(6.4).
f
has the type indicated is an immediate consequence
Next,-we claim that
f
has type
E1, ••• ,1,O
(with
k
repetitions) if and only if the following conditions are satisfied. (i)
a jf . ~(O) ax J
=
... ,
0,
ajf ~ (0)
ax j
=
0
(j
~
(j
=k
ajf.
(li)
some
-+J (0) ax
where
f 1 , ••• , fp
I
0
denote the components of
...
one readily checks that t:. 1 ajf. tives ~ with 1 ~ i ~ p, bx j
f.
~
j
~
k.
+ 1)
Indeed by induction on k
is generated by
1
k)
If'
and the deriva-
The claim follows immediately. 193
In view of (IV.2.4) the conditions (i) and (ii) are equivalent to saying that k+1 If <x >, and then (2.1) tells us that this is the same as f being
=
~-equivalent
( 0, 0, ••• , xk+1) •
to
So far so good. p
being
~
Let us press on further.
If we insist on both
nand
2, then the next simplest case to study is that of germs
(JR.2, 0)
f
o
The result follows.
(JR.2, 0).
-+
Here the possible first-order Boardman symbols are
In the first case the germ is non-singular, and a normal form is provided by (I.1.3).
Let
(6.6)
f
Z1 we have the following result.
For germs of type
2
: (JR. , 0)
~%.;...-_c::.:o::.:d=i:.::m~e;.:;n=s=i.::.on~._...::T:..::.h~e=n f
1 Z ,
be a germ of type
-+
is nece s sari ly of type
repetitions) for some integer k ~ 1,
Z 1 , ... , 1 , 0
and finite (with
k
and in that case is 3F-equivalent to
( x, Yk+1) • 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
we know from (4.1) that
f
can be assumed to be a 1-
fO : (JR., 0)
-+
(E, 0)
parameter unfolding of a germ
(S.8)
also has type
then 3F-equivalent to
Z1, ••• ,1,0
(with
and hence
We have still to treat germs
repetitions).
0,
f
(JR.2 , 0)
is ,%-equivalent to -+
(JR.2, 0)
which by
And clearly f
But (6.5) tells us that
(x, fO(y)).
~-equivalent to yk+1,
k
of rank
of type
fO(y)
is is
(x, y k+1 ) • 2
Z •
D
At this
point we are entering a decidedly more complex situation, and must proceed cautiously.
Z2,2
in order of increasing degeneracy.
plete result
194
The possible second-order Boardman symbols are
(due to J. Mather).
Z2,O,
Z2,1,
For the first we can obtain a com-
(6.7)
AnY germ
f
: (lR. 2 , 0) ~
(lR.2 , 0)
of tyPe
E2 ,0
and finite
%-
codimension is JY-equivalent to one of the germs listed below.
a,b
(xy, xa + yb )
b ) a ) 2
a,b
(xy, x a - yb)
b ) a
2 (x2 + y , xa )
a ) 3
I II
IV
a
Note these germs.
We have kept to Mather's notation
a,
b'
II
a,
b'
2,
a
even
IV a
for
His list, which was for a more general situation, included in
addition certain germs denoted Proof
I
~
III a, b'
va •
The first thing to note is that the components of f
no linear terms, so the 2-jet of f
have
can be thought of as a pair of binary
quadratic foms
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 nomal form we have written the The computations are very easy, and left to
the reader.
195
normal form for pencil
Boardman symbol
2
2 ±.y)
E2 ,0
x2 )
Z2,0
(xy, x
(xy,
Z2,0
(xy, 0) (x
2
2
+ y ,
Z2,0
0)
2 (x , 0)
Z2,1
0)
Z2,2
(0,
Since the Boardman symbol is a jf-invariant we can discard the last two norCertainly then the first component of
mal forms in the table. 2-jet
The
xy
xy
or
xy.
2
case
2
f
2 (JR, 0)
is a smooth germ
It follows from the Morse Lemma that this germ is
Applying the same change of co-ordinates to
~-equivalent to a germ (xy, ~(x, y)) where ~ have supposed
f
and therefore
~(x,
will have
We consider these cases separately.
+ y •
The first component of
O.
of corank to
x
f
is of finite y)
~-codimension,
f
-+
(JR,O)
~-equivalent
we see that it is
has no linear terms.
We
so finitely jf-determined,
can be supposed to be a polynomial.
It follows from
(2.1) that we shall not change the ~-~uivalence class by subtracting from a multiple of
xy
(Xy, a (x) + [3 (y) )
in
thus we can suppose our germ has the form
82 :
wi th
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 x
then certainly we can produce a change of co-ordinates becomes 196
a
±. x •
~
And similarly, if [3
I
0
a
I
0,
and
which appears in a): x
X under which
1-+
and has order
b
~
2
we can
find a change of co-ordinates
H>
Y under which
x, X generate the same ideal in 8 1,
here that
the same. ideal in 8 1 • a J (3
Y
becomes
~
and likewise
~
b y •
y, Y generate a,~.
Now we consider the possibilities regarding
are both zero
(xy, 0),
This case yields the germ
Note
which is not
of finite JV-codimension, so can be discarded. a, (3
Just one of
is zero
apply in the case ~
I o.
Suppose a
I 0.)
(A similar argument will
A change of co-oroinates
brings the germ to the form
+ x a ),
(Xy,
x
and since
X,
H>
Xy, xy
Y
Y
H>
generate the
(xy, x a ).
same ideal this is 3r-equivalent by (2.1) to
also not of finite 3r-codimension, so this case can likewise be discarded.
(II
A change of co-oroinates x
are both non-zero
(3
(XY,
brings the germ to the form same ideal this is
-+
x a + yb)
xa
Also, if
a
by a change of co-ordinates
x
-1.
a
b;
~
-x,
is odd we can change the sign in front of
Y •
a germ of type
I
a,b
and of course we can multiply
b
Y
2
y:
H>
and similarly if b
Thus in ever.y case we obtain
save when the signs are different, and
even, and in that case we obtain a germ of type 2
Note at this
is odd we can change the sign in front H>
of
Y
H>
Y
xy,XY generate the
(xy J ~ xa ~ yb).
~-equivalent by (2.1) to
point that we are at liberty to suppose either component by
and since
-
X,
H>
a, b
are both
II a, b.
The
x
Case
Following the same initial reasoning used in
the
xy case we see that f
is 3Y-equivalent to a germ (x 2 + y2, ,(x, y))
where
+Y
,
is a polynomial with no terms of degree
~
2.
change the JV-equivalence class by subtracting from
,
in 8 2 : in particular we can suppose f
has the form
(x 2 + y2,
,
a(x) + YJ6(x))
By (2.1) we do not a multiple of
has no terms with factor with a,
~ polynomials.
x 2
y,
2
+ y
2
i.e.
Note that
197
the second component cannot be identically zero, as %-codimension.
Denote by
pxq + qyxa-1
. ord er t erm ~s
we can suppose
p
I
2
2
+ y ,0)
a (x) + yf3 (x),
the order of
say, with at least one of
= 0.
q
0,
a ~ 3
(x
p, q
I
has infini~ so the lowest 0.
I claim
To this end consider a linear change of co-
ordinates
{: Notice that
x
x2 +~,
ulo
2
+ y
2
=
e + sin e +
= x cos
y sin
-x
y cos
=
X2 + Y2 •
e e
Using the fact that
mod-
a straightforward computation yields px
a
a-1
modulo
+ qyx
where =
{:
p cos ae - q sin ae p sin ae + q cos ae
and the claim follows an observing that we can choose p
I
Q = 0.
0,
It follows that our germ is
which is the desired normal form of type
e
is such a way that
%-equivalent to
(x 2 + y2, x a),
IVa.
o
Of course one could go further, and take up the next case of germs of type 1;2,1
and finite .5f"-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 .5f"-equi valence.
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
§7.
Some Examples of Classifying Stable Germs
We are now in a.position to write out explicit lists of stable germs
(JRn , 0)
--+
(JRP, 0)
Boardman symbol.
under certain restrictions on the dimensions and the Let us start with the case
situation is that of a non-singular germ.
zO,
n
of type
(x 1 ' ••• , xn ' 0, ••• , 0)
by
We can reasonably expect the next simplest case to be stable germs Z 1.
(7.1) of type
The simplest possible
Such a germ is necessarily of type
automatically stable, and has normal form
(1.1.4).
p.
~
Here we have a complete result, due to B. Morin. n ~ p,
Let
Th en
Z 1.
F
. · necessarlly lS of t ype
titions) for some integer in that case
F : (JRn , 0)
and let
k
with
1
~
~
k
(JRP, 0)
--+
~1,
~
be a stable germ
••• ,1,0
k
q = P - n + 1.
n/q where
G : (JRn , 0)
F is A-equivalent to the germ
(wi th
--+
(JRP,O)
repeAnd with
components G.
1
= u.1 k
G n+i
)'
=
--'
uik+jx
j
(1
~
i
~
n - 1)
(0
~
i
~
q - 2)
j=1 k-1
G p
=
k+1 )' u x j +x (q-1)k+j ~
j=1 where we write
The initial statement follows immediately from (5.4.) and
Proof (6.4) •
The theory of §4 tells us that
parameter unfolding of a germ . slon, and al so f
for the standard co-ordinates on
x
0
f t ype
is ~equivalent to
f
: (JR, 0)
~1, ••• ,1,0
u
F
must be A-equivalent to an (n - 1)--+
by (5.8).
(0, ••• , 0, x k+1) ,
of
(JR\ 0)
of finite %-codimen-
In view of (6.5) we know that ~-codimension
qk + q - 1 by
199
Example 7 in §2.
The theor,y now tells us that
stable germ associated to a
(0, ••• , 0, xk+1) •
~versal
F
is A-equivalent to the
deformation of the germ
The deformation in question was computed in Example 2 of
§3, and the germ written out above is clearly the associated stable germ. Finally, the theor,y tells us that we need only consider those ~codimension
the
qk + q -
is
lSi
p,
i.e.
for which
k
k liii
for which
n/q.
o
It is probably worthwhile isolating the equidimensional case of this result,
i.e.
the case
(7.2)
n = p.
Let
F : (lRn, 0) ,,1, ••• ,1,0
F mus t b e 0 f t ype with
lSi
k
G : (lRn , 0)
lSi
n, -+
(lRn , 0)
-+
u
('th k
~
be a stable germ of tYpe
E 1.
repetitions ) for some integer
k
and in that case is A-equivalent to the germ (JRn, 0)
with components
G.
= u.l.
Gn
=
l.
(1
k-1
)' u.x j --J
J
liii
i
!Ii n -
1)
+ xk+1
j=1
Of course, this is the germ which we have previously dubbed the "generali zed II Whitney mapping, and whi ch 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 (lRn,O)
-+
(lRP , 0)
with
n ~ p.
is that of a non-singular germ. 200
Again, the simplest possible situation n-p Such a germ is necessarily of type E ,
automatically stable, and has normal form
••• , x ) p
(1.1.3) •
by
We
can reasonably expect the next simplest case to be that of stable germs of type
Zn-p+1.
The starting point here, as always, is that the
theo~
of §4
tells us that such a germ is ~-equivalent to a (p - i)-parameter unfolding of a germ
f
m (:m,
:
0)
( :rn., ) 0
->
degree of complication in the corank
c,
the 2-jet of
where
m = n - p + 1,
f
depends at root on its corank
c.
Notice that
and the second order Boardman symbol, -both depend solely on f.
It is therefore a good guess that the two invariants will Indeed that is the case.
be closely connected.
Let .)
f
:
,;;.s;;;J.n;;.g\l...,;;;I""a:;,;r;;..<..;.:.---,T:;,;;h""e=n f
(:rn.m, ,
nas
0) t
(:rn.,
->
;lPe
.,!ll,C
'-'
0)
be a germ of type
if and only if
We start with the condition for
Proof The ideal
".,n-p +1,
Our experience in Chapter IV tells us that the
~codimension.
and of finite
also of type
~mIf
is generated by f
f
f
Zm
(i.e.
has corank
to be of type
c.
Zm,c.
and the partial derivatives
..Qf.... af ax ' ••• , ax
is generated by The ideal and is to be critical. m the same list, together with the minors of order (m - s + 1) of their 1
Jacobian
aX 1
af aXm
a2 f aX 1
a2 f aX 1aXm
--2
201
The condition for the ideal
to be proper is that all its generators
should have zero constant term, (m - s + 1)
should vanish at
row, since all the matrix of
f.
af
i.e.
In this matrix we can disregard the first
0.
vanish at
aX j
that all the minors of order
so are left precisely with the Hessian
0,
Thus the condition for the ideal
the Hessian should have corank
;J! s,
to be proper is that
and the condition for
critical is that the Hessian should have corank exactly we see that
f
mc E'
has type
if and only if
The simplest situation is when
f
f
to be
s.
Taking
has corank
has corank
0.
(:rn.n , 0)
(:rn.P , 0)
s
c
o
c.
This yields another
resul t of B. Morin.
(7.4)
Let
of type
n;J! p,
En - p +1 ,0:
and let
then
F
F :
--+
is A-equivalent to a germ
G-
be a stable germ : (J[ln, 0) --+ (:rn.P, 0)
given by
{::
=
u.
(1
~
2
= -+ xp
+
.:!:.
and finite Jr-codimension, where so by the Morse Lemma is +x
-
2
+ •••
p -
2 +x.
-
n
f
F
:!(
P - 1)
is certainly A-equivalent to a
: (:rn.m,O)
m = n - p + 1.
~-equivalent,
i
x n2 _ 1 + x 2 n
As was pointed out above, (p - 1)-parameter unfolding of a germ
:\0
(:rn., 0)
--+
By
(7.3)
f
of type
Em,O
has corank
0,
hence 3V-equivalent, to a germ
rhe theory of §4 now tells us that
F
is A-equivalent
to the stable germ associated to a p-parameter Jr-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 f
has corank 1, giving rise to the foll-
owing result, also due to B. Morin.
Let
(7.5)
",n-p+1,1.
of tyPe (with
lJ
k
case
n ~ p,
F
and let
Then
is A-equivalent to a germ
{:: Proof
k with
G : (]Rn, 0)
~
~
k
~
q:
(]RP, 0)
and in that given by
u.
~
2
2
k+2 .:t. X n _ 1 -+ x n +
= -+ xp-+
k
I
j u.x J n
j=1
The initial statement follows immediately from (5.4) and
(6.4) •
The rest of the proof follows exactly the same lines as that of
(7.4), save that this time by
be a stable germ
F is necessarily of type En- p +1,1, ••• ,1,0
repetitions) for some integer
=
~ (]RP, 0)
F : (]Rn, 0)
(6.3).
f
is jV-equivalent to
The computation of the deformation is very straightforward, and
D
can safely be left to the reader.
Of course, one could push these techniques further, by systematically classifying germs of functions with increasing
~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).
us pursue the equidimensional case further.
E1,
Note incidentally that
the next case to study is that of
E2
Let
Having dealt with germs of type 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 F : (JRn , 0) -. (JRn , 0)
(7.6)
be a stable germ of type
then F is A-equivalent to one of the following germs
G.
l.
Types I
a,b and Hab ,
u.
(1 .. i .. a-1)
=
v.
(1 .. j .. b - 1)
=
x:y
=
x
l.
J
a
.:!:. y
b
(1 .. i .. a - 1) =
Vj
=
x
2
(1 .. j " a - 1) + y
2
By the theory of §4 the germ F is A-equivalent to an (n - 2)-
parameter unfolding of a germ f finite
G: (JR~ 0) -. (JR~ 0).
=
G.l.
Proof
Z2,0
3F-codimension.
By
IVa •
: (JR2, 0) -. (JR2, 0)
of type
and
(6.7) this germ is 3F-equivalent to one of the And then
F will be A-equivalent to the stable
germ associated to a 5F-versal deformation of these germs. were computed in Examples
Z2,0
The deformations
3, 4 of §3, and the corresponding stable germs are
evidently the germs written out above.
o
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
§8.
Singular Points of Stable Mappings
We have so far discussed only stable germs of smooth mappings
f
: JRn -+
:m.P.
But the xnderlying ideas can be introduced equally well for mappings themselves.
By analogy with germs there is a natural notion of "equivalence"
for smooth mappings;
e·
.. Lent
namely, we call two smooth mappings
when there exist diffeomorphisms
g, h
f l' f 2 : JRn -+ JRP
for which the following
diagram commutes
Now JRn ,
g
and
lRP •
is an element of the group h
is an element of the group
So the pair
(g, h)
Diff(JRn ) x Diff(JRP )
(g, h).f
Diff(JR n)
to be
Diff(JR~
of all diffeomorphisms of of all diffeomorphisms of
is an element of the product group
and this group acts on -1
hofog.
cf'(JRn , JRP)
Of course this action lies outside the frame-
work discussed in Chapter III because neither the group nor the set
COO (JRn , JRP)
if we define
Diff(JRn ) x Diff(JRP )
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 latter algebraic.
stability'~
the former being geometric in nature, and the
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
f
: JRn -+ JRP "stable" when all "sufficient:l,y
close"
g : JRn -+:m.P are equivalent to
f.
On this basis we introduce the
following formal definition.
205
: lRn -+]RP is stable when there exists a real number
A smooth mapping f e > 0
such that every smooth mapping
of
is equivalent to
f
f.
g : lRn -+ lRP is the e-neighbourhood
Our first result gives us some idea of just how
nice stable mappings are. Any stable mapping
lRn -+ lRP is generic in the sense of
f
Boardman. Proof
Let
k
~
1 be an integer.
transverse to all the Boardman submanifolds
We have to show that i 1 ,···,i k
Z
•
jkr is
To this end recall
from (5.6) that the set of all smooth mappings lRn -+ lRP with this property is dense in
COO (lRn , lRP ):
thus, given any
g : lRn -+ lRP in the e-neighbourhood of f And since
all the
f
is equivalent to
is equivalent to
ity condition, then f
for which
jkg is transverse to
is stable we can choose
every map in the e-neighbourhood of f to observe that if f
e > 0 we can find a
does as well
g,
-
and
f.
so small that It remains only
g satisfies a transversal-
which fact we leave as an exercise
o
for the reader.
(8.2)
lRn -+ lRP be stable:
then the germ of f
at any point
is stable. We shall omit the proof of this result.
If we now combine these results
with the classifications obtained in the previous paragraphs we obtain some aesthetically very pleasing theorems describing completely all possible singular points of a stable mapping f
: lRn -+ lRP for certain values of
Let us start by looking at the equidimensional case stable mapping f
: lRn -+ lRn.
first-order Boardman submanifolds
206
First of all, zi
-
f
n = p.
n, p.
Consider a
must be transverse to the
which by (5.1) in Chapter II have
.2
codimension
1.
In particular, if
•
n ( 3 the first-order singularity sets
~if _ which have the same codimension - must be empty for i ~ 2, and the Z1 •
only possible singular points are those of type explicit.
Recall that in the equidimensional case the singularity set
~i, ••• ,i,Of (with
k repetitions) has codimension k, by Example 8 of §5.
thus the only singularity sets with k ( n.
(8.3)
In fact we can be more
Let
n =
If we take
f
Z 1, ••• ,1,0 f
which can be non-void are those
we obtain immediately from (7.2)
: ]R ~]R be a stable mapping:
its germ at any point is
equivalent to one of the following
zO Z 1,0 :
Taking the next case
Yi
=
Xi
(regular)
Yi
=
2 Xi
(simple minimum).
n
=2
we recover a famous result of H. Whitney des-
cribing the possible singularities of stable mappings from the plane to itself.
(8.4)
Let
f
: ]R2 ~]R2
be a stable mapping:
its germ at any point is
equivalent to one of the following
zO
r' r' Y2
zi,O
Y2
"',1,0{Y1 Y2
= Xi
(regular)
= x2
=
Xi 2 = x2
(fold)
= Xi =
~
(cusp) • + x i X2
207
In order to obtain geometric insight into these germs it is best to think of them respectively as composites (x1 , x 2 )1
) (x 1 , x 2 ' x 2 )
I
(x 1' x 2 ) I
2 ) (x1 , x 2 ' x 2 )
J
(x1, x 2 )1
> (x 1 ' x 2 '
~
proj proj
+ x 1x 2 ) I
) (x 1 ' x 2 ) 2 ) (x 1' x 2 ) proj
) (x 1'
x~
+
x1~)
where in each case the second mapping is (the restriction to the image of the first of) the projection
(x1 , x 2 ' x 3) ~
ualize our maps as follows. x2
x1
x2
x1
x2
x1
208
(x1 , x 3).
In this way we can vis-
Take for instance the situation of Example 4 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.
PC Here the curves which make up the picture are the images under the projection f
of the singularity sets
But the four exceptional points are the images of the
torus folding over. singularity sets
E1,0 f: at a point in such a set we simply see the
E 1,1 f:
here we have the more complicated situation of two
folds coming together to form a "pleat". We can of course gain one more result by taking the final case
(8.5)
Let
f
: ]R3
-+
]R3 be a stable mapping:
n
= 3.
its germ at any point is
equivalent to one of the following.
EO
Y1
=
x1
Y2
=
x2
Y3
=
x3
(regular)
209
E1,0
E1 ,1,0
E1,1,1,0
Y1
=
X1
Y2
=
x2
Y3
=
2 x3
Y1
=
x1
Y2
=
x2
Y3
=~
Y1
=
x1
Y2
=
x2
Y3
=
4 2 x3 + x 1x 3 + x 2x3
(fold)
(cusp) + x 1x 3
(dovetail)
Of course, when we go up to the next dimension n only E1 points.
As we have already pointed out
in the equidimensional case so when n but avoid Z2,2f
Ei
i ~ 3.
points for
three sets can be non-void. E2 ,0
(8.6)
given in
JR.4
Now Z2f
-+
Eif
EO
r' Y4
210
=
x4
E2
points,
thus only the first of these
Appealing to the classification of stable
§7
we obtain the following: its germ at any point is
eguivalent to one of the following.
x.l-
has codimension i 2
splits into E2 ,Of, E2 ,1 r ,
JR.4 be a stable mapping:
=
we no longer obtain
we can 'certainly have
4, 7, 10 respectively:
with codimensions
germs of type
=4
=4
(1
~
i ~ 3)
{:: ::; {:: : :t =
1:;1,1,1,0
(1 .. i .. 3)
(1 .. i .. 3) + x, x4
x.l.
(1 .. i .. 3)
1:; 1,1, 1,1,0
Y1
=
x1
1:;2,0
Y2
=
x2
12?2
Y3
=
Y4
=
x3x4 2 2 x3 + x4 + x 1x3 + x2x4
Y3
=
x3x4
Y4
=
x3 - x4 + x 1x3 + x2x4 •
2
2
The reader is invited to continue the listing process for stable mappings That is as far as he will get by just dipping into the results of the previous sections. germs of type 1:;2,1,
stable mappings JR.7 ... JR.7
and these we have not listed:
can have
however, the enterprising 211
reader will find that he can obtain a complete list in this case too, using a little common sense and the techniques already expounded. Let us diverge from the equidimensional case to see what further gain can be derived from the results of the previous three sections. the case n
p.
~
We ask if there are pairs of integers
having the property that a stable mapping JRn 1 points of type Z.
We know that
admits only singular
Zi
has codimension
i(p - n + i) ~ n in the case
i = 1,
i.e.
Z1
i(p - n + i) > n for
then be looking at pairs
i ~ 2,
(n, p)
i.e.
2p > 3n - 4.
in the range
Z1, ••• ,1,0
(n, p)
can arise.
repetitions) is k - 1
we have
of having Z1,0 E1, ••• ,1,0
i.e.
2p
~
in this range:
Suppose we
we ask which singular points of type
so we are not excluding the possibility
We shall certainly avoid singular points of type
(with two or more repetitions)
if
for k ~ 2,
(p - n + 1)k > n
imposing only a very slight further restriction.
ural range to consider then is given by 3n - 1 times called the metastable range.
~
2p < 4n,
is due to H. Whitney.
A nat-
which is some-
Combining these arguments with the
resul ts of §7 we deduce the following result, which in the case
212
(with k
The condition p < 2n ensures that for k - 1
(p - n + 1)k ~ n,
3n - 1,
so
3n - 4 < 2p < 4n.
Recall that the codimension of z1, ••• ,1,0
(p - n + 1)k.
points.
Zi,
We should certainly
It is worthwhile analysing the situation a little fUrther. have a pair
we
But for
p < 2n.
i ~ 2 we do not wish our mapping to have singular points of type need
(n, p)
If our mapping is to admit singular points of type
i(p - n + i).
instan~
with n ~ p
(n, p)
We can determine a useful range of such pairs
by a simple combinatoric exercise.
must have
-..:m.P
Take for
p
=
2n - 1
(8.7)
Let
f
: JRn
-4
JRP
_t_h"'-e_m_e.;..t....;a.:.:s'-t;.;;.a..;;;b.;;;l..;;;e~ran=g ...e;...:'--_t;..;.h....;e..;..;.n
be a stable mapping with the pair f
(n, p)
in
only admits singular points of type
and its germ at such a point is equivalent to the germ Yi
n
=1
x.
J.
Yn-1+i
=
=
2 x n
YP The case
=
X.x J. n
:10
n - 1)
(1 , i
:10
P - n)
of this result is simply (8.3).
something new, namely that a stable mapping f lar points of type
(1 , i
Z1,O
: JR2
But when n -4
JR3
=2
we get
only admits singu-
and that the germ at such a point is equivalent to
the germ
whose image is the so-called Vlliitney umbrella depicted'below.
213
What about the case pairs
(n, p)
n
~
p?
Here again one could construct ranges of
for which a stable mapping
points of a ver,y 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
(n, p) and
One of the simplest cases is
: ]R3 -+]R2 be a stable mapping:
its germ at any point is
equivalent to one of the following.
{
,E1
Y1
=
x1
Y2
=
x2
Y1
=
x1
Y2
=
2 2 .:!:. x 2 .:!:. x3
Y1
=
x1
Y2
=
.:!:.~ +
{ {
In this situation ,Eif
~
be empty for Now ,E2f
2
i
~
3,
splits into
i.e.
~
+ x 1x3 •
has codimension i(i - 1)
we can only have singular points of type
,E2,Of,
,E2,1 f ,
,E2,2f
with codimensions
normal forms are provided by
remaining possibilities are singular points of type repetitions):
this has codimension k + 2,
so we need only consider the cases k
(7.5). 214
= 0,
(7.4).
,E2,1, ••• ,O
,E2.
2, 3, 5
respectively, so we need only consider the first two of these sets. singular point of type ,E2,O
so must
At a The only
(with
k
as an easy computation verifies, 1 with normal forms provided by
o
AppendixA
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 lRn
when, given any real number
A set I:
> 0,
union of cubes the sum of whose volumes is duct of
n
open intervals, called its
V C lRn
is contained in a countable
V
<
is called a null set
(A cube in lRn is a pro-
1:.
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
(A 1)
Any countable union V of null sets
set in
J
in lRn is again a null
:mn •
Proof
Let
I:
of total volume
We can cover V. J
> O.
< €/2 j :
of total volume
c.J, k
V.
by countably many cubes
V by the countably many cubes
hence we can cover
< )
~
t:/2 j '"
c.J, k
o
€.
J
It seems worthwhile spelling out a simple consequence of (A1) which we shall use several times in the proof of SardIs Theorem.
Let
(A2) point
x
E
set in lRP
f : X -t Y be a mapping with
X has a nei ghbourhood then
f(X)
Ux
X
in lRn
So lRn,
Y
for which
So
lRP •
feU x n X)
If each is a null
is a null set in lRP.
215
Proof
From the cover
subcover, so f(X) set in ~p,
(UX)XEX for
X one can extract a countable
is a countable union of null sets in ~p,
hence a null
by (A1).
The next fact that we need to know is that the property of being a null set is invariant under smooth mappings, in the following precise sense.
Let f : N~ P be a smooth mapping with
(A3) if
V C N is a null set in ~n
then likewise
N, P
fey)
£
open sets in ~n: P is a null set in
~n.
By (A2) it will suffice to show that fCc n V)
Proof in ~n
is a null set
C in ~n whose closure is contained in N.
for each cube
It
follows from the Mean Value Theorem that there exists a real number K > 0 such that for any x, y
in
C
~ Klx - yl •
If(x) - f(y)1 Choose
E:
> O.
Since
countably many cubes
Ci
C n V is a null set in mn
£
c,
each
C.
J.
it can be covered by
having all its sides of equal length
cJ..'
and of total volume ) ' cr; < E: • It follows that each ~ J. J. tained in a cube having all its sides of equal length that fCC n V)
f(Ci)
and hence
is covered by a countable family of cubes of total volume It follows that
fCC n V)
is a null set in mn.
The next proposition is a very special case of Fubini's Theorem. it we need a little notation. define
216
is con-
Suppose
V
£
n
~
=m x
n-1
~
•
For
o
To state s
E
m
Let V f JRn be countable union of compact sets such that for all
(A4) 5
E JR
the set
step
V C 5
:nf- 1
•
~s
a nu11 se t ·~n lRn - 1 :
V is compact,
Suppose first that
interval
I
for which V C I x JRn-1.
By hypothesis
Vs
sEI
and let C
•
S,J
<>0.
in JRn- 1 of
indeed it follows from the compactness of V that there
is an open interval
Is'
srune cubes cover all the I
Nowlet
there exists a closed
can be covered by finitely many cubes
total volume < <:
open intervals
50
then V is a null set
with mid-point Vt
for
t
s,
having the property that the
in I s •
From this cover of
we can extract a finite subcover
s
oertainly suppose that this suboover is minimal,
I
by
I
, ••• , Is one oan s1 p i.e. that if we delete any
one of these intervals the remaining intervals will fail to cover
I,
and
hence that the sum of their lengths is not more than twice the length L of I. Thus we have finitely many cubes
I
It follows that
volume is Step 2
si
x C s. .
covering V,
V is a null set in JRn • It
Now suppose V is a countable union of compact sets.
follows immediately from Step 1 and (A1) that
So much for technical preliminaries.
V is a null set in JRn •
smooth manifold.
A set
¢ : AnN -+
Let
~ C JRs
be a
V C N is called a null set in N when for every
B one has
are open sets in lRs, lRn
o
The next step is to extend the basic
notion of a null set to subsets of smooth manifolds.
chart
and their total
~,J
¢(A n V)
respecti vely.
a null set in JRn:
of course A, B
Sard's Theorem is the following
proposi tion. (A5)
Let
f: N-+ P be a smooth mapping with
The set of oritical values of
f
~,pP
smooth manifolds.
is a null set in P. 217
One proceeds by induction on
n.
The case
n = 0 is trivial.
(It is a convention that lR O comprises a single point.) that (A5) holds for
(n - 1),
We suppose therefore
and will deduce that it holds for
lows immediately from the definitions that we can suppose
N, P
n.
It fol-
to be open
sets in lRn , lRP •
Remark
It is probably worthwhile pointing out that the case
relatively trivial. by
(x, y)
Consider indeed the smooth mapping
f(x).
--+
One easily checks that
of
f
It remains only to observe that when is precisely feN).
n .. P, Proof
P
given
is a null set in lRP by
n < p
Of course when n .. p
and one needs a more subtle argument:
--+
N x 0 is a null set in lRP , so
that its image under this mapping, namely feN),
(A3) •
N x lRP- n
n < p is
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. (continued)
Write
for the set of points are zero.
E for the singular set of f.
x E N where all the partial derivatives of order
~
Thus
We have to show that
f(E)
is a null set in lRP •
In view of (A1) it will
suffice to establish the following steps. f(E- E1 )
Step 2
f(E k - Ek + 1 )
Step 3
f(E k )
Step have
218
is a null set in lRP •
Step 1
is a null set in lRP I
is a null set in lRP I
for some
Note first that we can assume E
= El '
for all
and there is nothing to prove.
p
~
2
k.
k. since when P
=1
we
By (A2) it suffices to show
k
that each point x
in E - E has a neighbourhood U in N for which the 1
E - E1
image of
under the restriction flu
is a null set in JRP.
And
using (A3) it is clear that we can replace this restriction by any equivalent mapping, which we shall also denote f(x)
= 0:
and since the rank of
f
f. at
In particular we can assume x
~ 1
is
and
(by choosing linearly adapted local co-ordinates) that where
s
E
< p we can suppose
r(s, x)
= (s,
fs(x))
is a smooth mapping fran an open set in JRn-1
and each f s
JR,
x = 0,
to an open set in JRP- 1 • We have to show that the set V of critical values of ~p.
Notice that the set of critical points of f
a countable union of compact sets: of critical points under
f,
(A4)
With the notation of
it follows that
s
is a null set in
is a closed set,
V, the image of the set
observe that
V
s
is precisely the set of critical by the induction hypothesis for
the theorem.
It follows from (A4) that
Step 2
By (A2) it suffices to show that each point x
V is a null set. in E1 - Ek+1
has a neighbourhood U in N for which the image of U n (E k - Ek +1) f
is a null set in JRP.
f
is
I
0,
At
say ~ with
x
some
under
(k + 1)th order partial derivative of
g a kth order partial derivative of f,
hence
is likewise a countable union of compact sets.
which is a null set in JRp- 1 ,
values of f ,
f
so there
J
certainly exists a neighbourhood set defined by
Ek
£
g
=0
=
un Ek
so its image under f
x whose intersection G with the
is a smooth manifold of dimension
G: in particular x
U n (Ek - Ek +1 )
U of
E
G.
(n - 1).
Note that
Clearly if U is small enough
is contained in the set of critical points of fiG,
is a null set in JRP ,
by the induction hypothesis of
the theorem. step 3
Let
a null set in JRP.
k be an integer > ~ - 1.
We shall show that f(E k ) is
By (A2) it is enough to show that
f(C
n Ek )
is a null 219
set in ~p
tained in N. ber
> 0
K
say, whose closure is con-
R
= rn
Introduce an arbitrar,y integer r
equal cubes
and hence
C1 '
••• ,
CR
of side
is contained in a cube in ~p
above inequality fCC n i: k )
~p
c
for which
C into
2K(c/r)k+1,
of side
As a consequence of Taylor's Theorem there exists a real num-
x, y in C n Ek •
for all divide
C in ~n,
for any cube
fCC n Ek )
~
and sub-
1,
c/r.
By the
of side
is contained in the union of
rn
2 cubes in
of total volume constant x r(n-pk-p).
= Note that as
r
n - pk - p < 0
~ 00.
as
k >~ - 1, p
so that this last expression~ 0 fCC n Ek )
I t follows immediately that
is a null set in ~p.
D In order to establish the consequence of SardIs Theorem used in Chapter II to prove the Basic Transversality Lemma we need (A6)
Let P be a smooth manifold and V C P a null set:
lement of
V is dense in p. It follows immediately from the definitions that we can assume
~
Suppose (A6) were false, so we could find a cube C is a subset of
V C U C. -
1
~
(C i )
vol (C)
220
e > 0,
It
Since
find a countable family of cubes
and with ) ' vol (C.) < e.
from the open cover
C whose closure
One derives a contradiction as follows.
V.
a null set one can, given wi th
then the com-
1
(C i )
As C is compact we can extract
a finite subcover C1 ' ••• , Ct ' It
V is
)'
.:..-.
vol (C.) < e 1
say;
then
and
e being arbitrary one deduces that
vol (C)
=
0,
which is the desired
contradiction.
o
Note that we assume in the above proof that if
t C C .U C. -
t
cally vol reader.
(A7) N., P 1
(C) ~ ~ vol (C.): we leave this fact as 1=1 1
1=1 1
an
then automati-
exercise for the
Finally, we deduce Let
fi : Ni
-+
P be a countable family of smooth mappings with the
smooth manifolds:
the set of oommon regular values of the
f.
1
is
-
dense in p. Proof
The set
C. 1
of oritioal values of f.
by SardIs Theorem, so the union
1
C
= u C.1
is a null set in P,
is a null set by
of oommon regular values is the oomplement of
C,
(A1).
The set
so dense in P by (A6).
o
221
AppendixB
Let
~
fold
: Gx M M.
-t
Semialgebraic group actions
M be a smooth action of a Lie group
G on a smooth mani-
The objeotive of this appendix is to show that if the aotion is
"semialgebraio", in a sense to be made precise below, then automatically all the orbits will be smooth submanifolds of M.
The virtue of this fact is
that most of the examples which arise in this area of mathematics turn out to be "semialgebraic", so that automatically the theory of Chapter III applies to them.
At the time of writing the only reference for the theory of semi-
algebraic sets and mappings is the set of research notes by S. Lojasiewicz entitled "Ensembles Semi-Analytiques": easily available.
and these, unfortunately, are not
Until a more accessible account of the theory appears
these notes must remain the sole reference for the two basic results we need, namely (B2) and (B3) below. The ideas involved have their genesis in real algebraic geometry.
Recall
that a set A s;. JRn is algebraic when it can be obtained by finitely many applications of the operation of intersection, starting from sets of the form (x
E
lRn : f(x)
= 01
wi th f
a polynomial function on lRn.
a little linear algebra shows that any linear subspace of lRn
For instance, is algebraic.
However, there are areas of mathematics where it is profitable to introduce a wider class of sets, closed under as many set theoretic and topological operations as is possible.
One such class is obtained by calling A C ll.n semi-
algebraic when it can be obtained by finitely many applications of the operations of intersection, union and set difference starting from sets of the form (x
222
E
lRn : f(x) >
01
wi th
f
a polynomial function on lRn.
The reader
will readily check that an algebraic set is automatically semialgebraic. good example is provided by the general linear group set of the linear space
M(s)
of all real
obvious way with a Euclidean space: subset of M(s), plement
GL(s)
s x s
GL(s).
-+ lRP
in lRn
the singular matrices form an algebraic
given by the vanishing of the determinant, so that the comis semialgebraic.
with
x lRP •
A C lRn
is semialgebraic when graph f
arises from considering rational mappings which each component n
A mapping
is semialgebraic
Linear projections provide simple examples of semialgebraic
mappings, since their graphs are linear spaces.
on lR ,
This is a sub-
matrices, identified in an
The idea can be extended to mappings in an obvious way. f : A
A
with ¢.l.
f.l.
=
¢./¢. l. l.
where
nowhere zero on A.
f:
A wider class of examples A -+ lRP ,
¢l..' ¢i
i.e.
those for
are polynomial functions
It seems worthwhile spelling out the
following proposition.
(B1)
f
If
A
f
lRn
is semialgebraic, and
f
A -+ lRP
is rational, then
is semialgebraic.
Proof
With the above notation define pOlynomial functions 8 1, ••• , 8p
= wh ere
x = ( x 1 ' ••• , xn )
The vanishing of
¢.(x) - Yi¢.(x) l. l.
is l.. n lRn,
defines an algebraic subset
H.l.
The
proposition follows on noting that graph f is necessarily semialgebraic.
=
o 223
One of the basic facts about semialgebraic mappings is the Tarski-Seidenberg theorem. (B2)
Let X C JRn be semialgebraic. and let
algebraic with
A C lRn : then the image
f
: A
-+
JRP
be semi-
is semialgebraic in lR P •
f(X)
It follows from the Tarski-Seidenberg theorem that the domain of any semialgebraic mapping is necessarily semialgebraic, being the image of the graph under a linear projection.
And by the same argument, if a product A x B is
seroialgebraic then so too are the factors ested primarily in group actions in the sense just described.
~
We are going to be inter-
A, B.
: Gx M
-+
M which are semialgebraic
It follows from the preceding remarks that
then automatically G, M have to be semialgebraic. We need one more fact. A point
x
E
Let A C lRn.
First, a preliminary definition.
A is regular (of dimension
d)
when x
has a neighbourhood U
in lRn for which UnA is a smooth submanifold of lRn (of dimension d). (B3)
Let
A C JRn be a non-void semialgebraic set:
then A has at
least one regular point. Now we can put the bits together to obtain the main result.
(B4)
Let
~
; Gx M
smooth manifold M.
-+
M be a smooth action of a Lie group
G on a
And suppose that the action is semialgebraic.
Then
all the orbits are smooth submanifolds of M. Proof ~
Let
Xo E M.
of the semialgebraic set
The orbit Gx
{xol,
G.x O through Xo
is the image under
hence semialgebraic by the Tarski-
Seidenberg theorem.
By (B3) the orbit has at least one regular point, of
dimension d,
But the homogeneity property for orbits implies that
say.
then every point on the orbit is regular, of dimension d, 224
i.e.
the orbit
is a smooth submanifold of M of that dimension.
o
By way of explicit illustration consider the class of geometric actions studied in some detail in Chapter III. The natural action of
(B5)
(g, h).f
=
-1
hofog
GL(n)
We have d
GL(p) ~ H (n, p)
x
given by
is semialgebraic, and hence all the orbits are smooth
manifolds. Proof
GL(n) ,
product,
GL(p)
and
~(n, p) are all semialgebraic, so their
the domain of the action, is likewise semialgebraic.
i.e.
of (B1) it will now suf'fice to show that the action is rational. will be convenient to think of our linear mappings
g, h
In view
For this it
as square matrices
We have to show that each coefficient of each component of
h
0
f
0
g-1
is a rational function of the entries in
the coefficients of components of is precisely hi1F1 + ••• + hipFp' of
F = fog -1
f.
where F 1 , ••• , Fp
0
f
0
g-1
denote the components
F is a rational function of the entries in g,
and the coefficients of components of
Certainly the entries
f.
are rational functions of the entries
~j
And the claim follows since
F1 ,
denominator.
ith component of h
it will therefore be sufficient to show that each coeffi-
cient of each component of
g-1
Now the
g, hand
tivelyfromthe components
f 1 , ••• , fp
of
g~l.J.x1 + ••• + g~l.nxn for the variable xi·
f
of
g,
... , Fp
*
~j
of
with nowhere-zero are obtained respec-
by substituting
D
225
Real algebras
AppendixC
By a real algebra is meant a real vector space mapping V x V
v, written (x, y)
--+
--+
V together with a bilinear
x.y and called the algebra product.
In this book only two examples of algebras are of interest.
Example 1 f
The real vector space
: (lRn , 0)
--+
(m,
8n
of all germs
y) of smooth functions.
This is endowed with the
algebra product induced from that on the reals •
Example 2
f
.
The real vector space
=
L:
=
(x, ••• , x)
fa xa
in n
8n
of all formal power series
real indeterminates
Here, given
=
(a l ' ••• , an) we write x a as an abbreviaa 1 a2 an The algebra product is defined as tion for the expression x 1 x 2 ••• xn
x
and
n
1
a
follows.
h
we set
= I. hyXY
and
=
hy
I.
. .
f.g
=
..
h where
fa'~'
y=a+{3
Given real algebras mapping with cJ>(xoY)
V, W an
=
cJ>(x).cJ>(y)
algebra homomorphism cJ> for all
.
x, y
V --+ W is a linear
in V.
..
f
is an algebra
Here we keep to the notation of Chapter IV:
in particular
The natural mapping 8n
--+
8n
given by f
--+
homomorphism.
.. b .A't'n' ...... n b
that
226
denote the unique maximal ideals in 8,
(f :g)
= f.g;
..8
n •
We have to show
it suffices to show that these are equal modulo an
• k+1
element in~ n
'
for all
• k+1
9c Then, modulo
A
k ~ O.
modulo 1 n k+1
1 n
with
Choose f k,
k• ~
By (IV.2.4) we can write polynomials of degree
~
k.
,we have
bearing in mind that the Taylor series of a polynomial function is precisely that polynomial.
o
227
The Borel lemma
AppendixD
Our starting point is the explicit construction of smooth functions having very special properties.
(D1)
=
that ¢(t)
t
for
It I ,,1/2
and ¢(t)
=
Itl
0 for
certainly there exists a smooth function
Proof
o"
:m. -+:m. wi th the property
There exists a smooth function ¢
e(t) ,,1
for all
t,
for which
e(t)
The required function ¢
_ -
=0
e(t)
{O
e-1/ t2
~ 1.
e : :m. -+:m.
precisely when
when
t" 0
when
t > 0 •
with
t" 0, namely
is obtained by setting
¢(t)
=
te(1-t 2 ) e(1_t 2 ) + e(t2 _
t)
o
Now we come to the Borel Lemma itself, as it was stated in Chapter IV. The natural algebra homomorphism Gn -.. 8n
(D2)
given by f
-+
r
is
surjective. We adopt the following notation.
x
=
~
0,
of real numbers, and a sequence a = (a 1 , a a1 an we write x as an abbreviation for x 1 ••• xn •
of
(x 1, ••• , x n )
integers
have to show is that given an element 228
Given a sequence
r
=
I
l'axa
in
8n
What we
there exists an
element f
in Gn
standard co-ordinates
is the Taylor series of
= a 1~
••• an!'
listed in (D1) and take the formula
limit
~ :
~(x1' ••• , xn) (~k)
trary sequence 0,
-+
Ia I = a 1 +
f
relative to the
for all choices of
••• + an
and
=
Iff
To this end let ¢ : JR
= ~;Daf(O)
fa
x 1 ' ••• , xn i.e.
where we write a 1
a,
f
for which
JR be a smooth function having the properties
JJf -+
=
JRn
to be the smooth mapping defined by
(¢(x 1 ), ••• ,
¢(xn )).
We consider an arbi-
0 < ~k < 1 for which
of real numbers with
and the corresponding function
f
: JRn
-+
JR defined by
k-+oo
f(x)
=
Note first that finitely many terms
f
is well-defined, since for a given value of fk(x)
x
only
are non-zero.
near the origin, and hence that
=
{O ~ if
k
alf • a
if
Iial k
= lal
We claim that it is possible to choose the sequence that, on some neighbourhood of uniformly for any choice of
o
a.
calculus that the sum-function f
•
.".,n
(~k)
in such a way
L 00
the series
Daf k converges k=O It then follows from standard theorems in ~n
.It\.
,
is smooth, and that 229
rff(O)
f
=
=
Dafk(O)
a!fa
k=O just as we required.
By the usual comparison test it will suffice to choose 00
ving a)
~ sup iDafki is dominated by a series (invol"-' xeJRn k=O whioh oonverges for a~ choice of a. Clearly, we need to produce First of all, note that (by sheer differentiation)
an upper bound on we have
= where 1jI (a,
As the function ¢
(3, k, x)
n
a. {3j
j=1
ax.J
IT a J!.~
=
.. (~k')
J
vanishes outside a compact set we have a well-defined
bound B(a, (3, k)
=
sup xEJRn
i1jl(a, (3, k, x)i
so setting B(a, k)
=
L
f{3 B(a, (3, k)
i{3i=k we see that
and henoe that the series
is dominated by
All we need to do now is to choose the sequence latter series converges for all choices of a: 230
(e k )
!
ekB(a, k).
k=O in such a w~ that the
and that oan be done by
choosing
~k
in such a way that
D
231
AppendixE
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 [1S1). (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 [A1] by Arnolld.
[1S1] 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, [1) 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 232
[GG1.
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 tr,y [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 [1S1] 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 theor,y can be found in the latter chapters of [L), whilst an exposition of part of the global theor,y can be found in [GWPL). leans heavily on the idea of a "stratification":
The topOlogical theor,y
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 geometr,y and algebraic geometr,y.
And it is prin-
cipally to these areas that one looks for applications of the theor,y within
233
mathematics.
As far as differential
geomet~
is c0ncerned I suggest I00king
at [PJ, whilst the reader who wishes to see further ahead will find solid reading in
[WJ.
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 tijlle. The applications of singularity around the catastrophe
theo~
theo~
of Thom.
to the physical sciences have centred
Those with a serious interest in such
matters will want to look at Thom's own book [TJ.
Quick introductions to the
ideas can be found in [ZJ and [SJ while [SPJ provides a comprehensive account of the state of the subject at the time of writing.
For the mathematics of
the matter, the account in [BLJ extends the material in Chapter IV of the present volume, whilst more appears in [TZJ. ideas relating singularity
theo~
I recommend [LS2J for further
to problems in the
p~sical
sciences.
References [A1]
Amol'd, V. I.
Singularities of Smooth Mappings.
Surveys
- 43.
(1968)
Russian Math.
Translated from Uspehi Math. Nauk 23 (1968)
3 - 44. [A2]
Amol'd, V. I. Points, larities.
[A3J
Normal Forms for Functions near Degenerate Critical
the Weyl Groups of
~,
Dk and
Ek,
Functional Anal. Appl. 6, 254-272,
Amol'd, V. I.
and Lagrangian singu1972.
Normal Fonns of Functions in Neighbourhoods of
Degenerate Critical Points.
Russian Math. Surveys
29,
10-50,
1974. [A4]
Amol'd, V. I.
Critical Points of Smooth Functions.
ernat. Congr. Math. 234
Vancouver
1974,
19-39.
Pree. Int-
[BL]
BrBcker, T. & Lander, L.
Differential Germs and Catastrophes.
London Mathematical Society Lecture Notes 17. Press, Cambridge. [BJ]
1975
BrBcker, T. & J~ch, K.
EinfUhrung 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.
WirthmUller, 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,
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 I.
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 ]
Milnor, J.
Topology
-
From the Differentiable Viewpoint.
The University Press of Virginia.
[M2]
Milnor, J.
Singular Points of Complex Hypersurfaces.
of Maths. Studies
[p]
61
Princeton University Press.
Stewart, I. N. 68,
['SP]
447-454,
[T)
Wall, C. T. C.
Lecture Notes
[TZ]
Zeeman, E. C.
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
236
(trans-
Geometric Properties of Generic Differentiable
Trotman, D. J. A. & Zeeman, E. C.
525
New Scientist
1978.
Symp. at IMPA, Rio de Janeiro, July 1976.
65-83,
543-564.
Structural Stability and Morphogenesis
Manifolds.
[Z]
(1971)
Catastrophe Theory and its Applica-
London
lated by D. H. Fowler)
[WJ
Vol. 5,
1975.
Pitman.
Thom, R.
1968
The Seven Elementary Catastrophes.
Stewart, I. N. & Poston, T. tions.
Annals
The Normal Singularities of a Submanifold.
Porteous, I. R.
Journal of Differential Geometry.
[S]
Charlottesville, 1965.
Springer.
~
5.
Berlin & New York.
Classification of Elementary Lecture Notes in Mathematics. 1976
pp. 263-327.
Index
61
Action of a group Algebra
226
Cubic curves
Algebra product
226 226
Cuspoids
homomorphism
222
Algebraic set
40
Cri tical value
68-70 130 25
Curve Deformation
140 163 56 160 164
of germ induced
Bifurcation set
44
linear
65, 66
Binary cubic form
transversal universal
Boardman symbol submanifold
180, 181 186
Dense
Borel Lemma
101, 228
Descendant
144
~-equi valence
9, 21, 23
Chain Rule
72 72 90
Change of basis of co-ordinates of parameter Chart
52 110 116, 191
Determinacy Determinacy,
finite
120, 191
Diffeomorphism
8, 12
Differential
8, 20
Double point
130
13
Codimension of function germ of map germ of orbit
Cri tical point,
of maps of unfoldings
163 139 16, 205 89, 142
13
local
Corank of function germ Critical point
Equivalence of deformations of germs
77 15
of submamfold Co-ordinates,
98 152
125 29, 40
degenerate
59
Fold curve
46
Folded handkerchief mapping
55
Flow line
26
237
Fundamental neighbourhood
51
General linear group
63
Generic Germ,
187
equivalence of
62
of germ
37
%- equivalence
143, 144
Jets,
equivalence of immersive
35 35
Lie group
73
invertible
34
Linear systems
71
rank of
35
representative of
34
Local Existence and Uniqueness Theorem
28
singular source of
35
submersive target of
35
Graph
28
flow
34 34 21
Manifold
12
Metastable range
212
Morphism of deformation Hadamard Lemma
100
Multiplicity
130
Nakayama Lemma
102
Normal bundle space
133
Null set
215
59, 67
Hessian
62
Homogeneity Homomorphism,
induced
112
Infinitesimally stable point
78
Inverse function theorem
10
Isomorphism of ideals Isotropy subgroup
Jacobian extension ideal module
149
80
Orbit
61
Parametrization
12
Pencil
71
Product structure
80
Quadratic form
64
180
97 151
Quadratic form,
index of
65
Jet extension space
37
rank of
65
37
semi-index of
65
Jets
37
238
:R - equi valence
94
Transversality Lemma, Basic
49 47
Rational mapping
223
Transversality Theorem
Regular point value
224 40
Transversality Theorem, elementary
51 53
of Thom
49, 217
Sard I s Theorem
Transverse
39 38
Semialgebraic set
222
Transverse intersections
Semi-direct product
150
Triple point
411-
Singular point Singular point,
123 130
non7'iegenerate of type
130
~
Singularity set
44, 54
Singularity set, first order higher order
55 177
Umbilic bracelet
68
Unfolding
81
Unfolding
of germ induced
141 89
minimal
81
morphism of transverse
90 81
90 90
universal
79
Slice Smooth mapping
versal
8, 12
Spli tting Lemma
125
Vector field
Stable germ map
142 206 77
Vector field,
point Submanifold
15
Submersion
40
24 equivalence of time dependent
30 32
time independent
33
Whitney cusp mapping Whitney umbrella
Tangent bundle bundle space mapping space Tarski-Seidenberg Theorem
41187, 213
24 22 22 16 224 239
