Research Notes in Mathematics
C G Gibson
Singular points of smooth mappings
Pitman LONDON SAN FRANCISCO - MELBOURNE
25
C G Gibson University of Liverpool
Singular points of smooth mappings
Pitman LONDON SAN FRANCISCO MELBOURNE
PITMAN PUBLISHING LIMITED 39 Parker Street, London WC2B 5PB FEARON PITMAN PUBLISHERS INC. 6 Davis Drive, Belmont, California 94002, USA Associated Companies Copp Clark Pitman, Toronto Pitman Publishing New Zealand Ltd, Wellington Pitman Publishing Pty Ltd, Melbourne First published 1979
AMS Subject Classification: 57D45
British Library Cataloguing in Publication Data Gibson, C G Singular points of smooth mappings - (Research notes in mathematics; 25). 1. Mappings (Mathematics) 2. Singularities (Mathematics) 1. Title H. Series 515 QA360 ISBN 0 273 08410 0 © C G Gibson 1979
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TO
DORLE
Contents
Introduction
I
Acknowledgements
7
Chapter I
Chapter II
Smooth Manifolds and Mappings
-
§1.
A Preliminary Review of Some Calculus
§2.
Smooth Manifolds
12
§3-
The Differential of a Smooth Mapping
16
§4.
Vector Fields and Flows
24
§5.
Germs of Smooth Mappings
33
Transversality
-
§1
The Notion of Transversality
38
§2.
The Basic Transversality Lemma
48
An Elementary Transversality Theorem
51
§4.
Thom's Transversality Theorem
53
§5.
First Order Singularity Sets
54
3,
Chapter III
8
-
Unfoldings
The Finite Dimensional Model
§1.
Groups Acting on Sets
61
§2.
Some Geometry of Jets
62
§3.
Smooth Actions of Lie Groups on Smooth
Manifolds
73
Chapter IV
Chapter V
§4.
Transversal Unfoldings
81
§5.
Versal Unfoldings
89
-
Singular Points of Smooth Functions
§1.
Some Basic Geometric Ideas
94
§2.
The Algebra
99
§3.
Determinacy of Germs
§4.
Classification of Germs of Codimension
-
n
116
5 5
122
Stable Singularities of Smooth Mappings
§1.
The Basic Ideas
139
§2.
Contact Equivalence
143
§3.
Deformations under.'-Equivalence
159
§4.
Classification of Stable Germs
168
§5.
Higher Order Singularity Sets
174
§6.
Classifying Germs under,11-Equivalence
191
§7.
Some Examples of Classifying Stable Germs
199
§8.
Singular Points of Stable Mappings
205
Appendix A
-
The Theorem of Sara
215
Appendix B
-
Semialgebraic Group Actions
222
Appendix C
-
Real Algebras
226
Appendix D
-
The Borel Lemma
228
Appendix E
-
Guide to Further Reading
232
Index
237
Introduction
Suppose you take a smooth curved surface
X
made of some transparent material
and imagine it projected downwards onto a plane surface of light from above. on the surface
Think of this as a map
X projected down to a point
Y by shining a beam
f : X - Y
q = f(p)
with every point in the plane
Y.
p
On
the plane you will see the apparent outline of the surface, as it would appear from below.
Here are two simple examples.
Fig. 2
Fig. 1
It is not hard to see which points on X
give rise to the apparent outline
they are precisely the points where the tangent plane to the surface is vertical, the so-called singularities of the mapping
f : X - Y.
Figure 1 repre-
sents the simplest situation one can imagine, with the surface X folding over at
p:
such points
p
are called fold points.
Figure 2 represents a
more complex situation, a curve of fold points on which lies an exceptional point
p
where two folds meet, a so-called cusp point.
cated situation is provided by Figure 3 where
X
A still more compli-
is a torus,
i.e.
a dough-
1
nut - shaped surface.
Here one has again curves of fold points, on which lie
Fig. 3 four cusp points:
but in addition we have two simple crossing points on the
apparent outline, where the curves cross over properly. any point
q
very close to
Thus, if we take
on the apparent outline and look at the nature of the outline we can distinguish just three possibilities.
q,
I
II
III
q
q
q
Common sense, and a certain amount of experimentation, will soon convince one that these are the only essential types of behaviour which can arise, in the sense that any other type of behaviour could be eliminated by the slight-
est change of position of X in space.
For instance our torus might be so
positioned that the apparent outline was as in Figure 4, with the outline touching itself at some point:
2
but clearly we could just nudge
X
slightly
to get back to the previous situation where only possibilities I, II, III can occur.
To make the point even clearer, here are three further types of behaviour
which are all inessential, because they could be eliminated by the slightest change in position of simple crossing points
to yield situations exhibiting only fold, cusp and
X -
as follows.
The broad objective of this book is to introduce the reader to the less technical aspects of a mathematical theory of singularities which seeks to make precise the kind of heuristic reasoning just described.
The basic
3
objects
X, Y are replaced by smooth manifolds, which are natural generaliza-
tions to higher dimensions of the familiar notions of curve and surface.
And
the projection of the surface onto the plane is replaced by an arbitrary smooth mapping
f : X-+ Y.
For such mappings we introduce the general notion of
singularity and begin to list the simplest singularities which can occur,
We shall, for reasons of simplicity, con-
idealizing each type by a model.
cern ourselves principally with the local behaviour of the mapping, behaviour very close to a single point in the domain:
i.e.
its
thus in the situation
discussed above we interest ourselves solely in the fold and cusp points, and neglect simple crossing points which arise from considering what happens close to two distinct points in the domain.
There is nothing particularly new in the notion of a singularity.
Scien-
tists and geometers have recognized them, and appreciated their significance, for a long time now.
But no-one seems to have systematically set about
studying the singularities of smooth mappings till the pioneering work of Hassler Whitney in the mid 1950's.
Around the same time Rend Thom pointed
out the analogy with more finite-dimensional situations and indicated the general lines along which a theory might proceed.
So it was in the 1960's
that a number of mathematicians, principally John Mather, laid the foundations of a general theory.
That was the position in 1967 when Vladimir Arnol'd put
together the bits and wrote his now classic survey paper, a model of lucid descriptive writing.
It was a time of great promise.
Singularity theory
itself threw up a number of provocative problems, and the range of possible applications (both within and without mathematics) added to the excitement. Without question, the intervening years have justified that promise, and singularity theory can hold its own as a flourishing area of mathematics.
I feel that the time has come to provide prospective students with readable introductions to the subject.
4
It is my personal conviction that the way to
get into any area of mathematics is to concentrate on understanding the simplest situations first, so as to build up some intuitive feeling for what is going on, and to leave the deeper matters till later in life.
Singular Points
of Smooth Mappings is the result of following this guiding philosophy.
I have
taken a small number of intuitively appealing ideas and used them to pursue the problem of listing singularity types, one of the goals of the local theory.
It is the kind of book which I would expect a postgraduate student in mathematics to read with little difficulty, and I rather hope that others will find it within their compass as well.
A guide to further reading has been inclu-
ded to help the reader pursue those matters which interest him most. A few words are in order concerning the structure of the book.
In accor-
dance with the philosophy outlined above smooth manifolds are introduced in
Chapter I as subsets of in enjoying certain properties: way everyone should meet them.
I think this is the
Anyone who wants to get to grips with singu-
larity theory should be familiar with the basic ideas of transversality and of unfolding, so these topics provide the subject matter for Chapters II and III respectively.
Here again I have kept to the simplest situations which can
arise, imposing restrictions whenever I felt it was possible to suppress undue technicality:
in particular, unfoldings have been introduced in a finite-
dimensional situation where they are much easier to understand.
Singularity
theory proper is taken up in Chapter IV with the study of functions;
this
enables one to make some distance fairly easily, without getting involved in the subtleties associated to general mappings.
The result is the derivation
of the list of singularities of codimension : 5. case of mappings is taken up:
In Chapter V the general
it is inevitable that one must quote more and
prove less, but I have tried to expose the less technical aspects and give a fairly coherent account of just how one uses the theory to obtain explicit lists of the simplest singularities which turn up.
Of course one can pursue 5
the listing process much further than is indicated in this book, but one soon comes up against much deeper matters which lie beyond the scope of an introductory account.
I have not attempted systematically to attribute results to their authors, mainly on the ground that such a practice is out of place in a book at this level.
In any case, the material of the first four chapters is now pretty
well an established part of the subject.
I should say however that the open-
ing sections on differential topology follow closely the exposition given by John Milnor in his excellent little book "Topology, from the Differentiable The material in the final chapter, basically Mather's classi-
Viewpoint".
fication of stable germs by their local algebras, is not as well-known as it should be.
Here I decided to follow the elegant account of Jean Martinet
(see the Guide to Further Reading) in which the unfolding idea plays the central role.
The key result in this development, namely the characteriza-
tion of versal unfoldings, turns on a real version of the Weierstrass Preparation Theorem which I do not discuss;
I felt it was more important at this
level to place proper emphasis on the underlying geometric ideas, and to leave an exposition of the Preparation Theorem to a volume with more ambitious aims. I decided also to say nothing about the applications of singularity theory, mainly because I feel each area of application is probably worthy of a volume in itself.
For instance, Thom's catastrophe theory is already the subject
matter of several volumes.
Also, the applications within mathematics itself
all seem to be at too early a stage to merit writing-up. that my guide to further reading will prove to be useful.
Liverpool February 1978
6
Here again I hope
Acknowledgements
To Les Lander who drew the pictures,
and helped me find enthusiasm at a time when it was all but lost.
To Peter Giblin who undertook the considerable task of correcting the manuscript,
and whose suggestions have contributed much to the final form of this book.
And to Ann Garfield who did the typing and produced an excellent job in difficult circumstances.
7
I
Smooth manifolds and mappings
A Preliminary Review of Some Calculus
1.
[In this section it will be understood that
]RP, 7Rq Let
U, V, W
are open sets in
Rn
respectively.] f : U --> V be a mapping with components
f1, ..., f
We call
.
f
P
smooth when all the partial derivatives
aaf a
(a = a 1 + ... + an )
a
ax 11...8xnn
exist and are continuous in fice to observe that if
For the purposes of this book it will suf-
U.
f1, ..., f
are all given by polynomials in
P
x1, ..., xn U
onto
then
will be smooth.
f
And we call
V when it is a bijection, and both
Now suppose
linear mapping
f : U -+ V
Daf
:
is smooth.
7Rn - ]RP
f, f-1
f
a diffeomorphism of are smooth.
For every point
there is a
a E U
called the differential of
f
at
a:
it
is precisely the linear mapping whose matrix (relative to the standard bases) is the so-called Jacobian
(af
1
1
of
aa) 1
8
...
of axP-
n
Take, for instance, the case when linear map
F : Mn -+
IRP;
then
f
f : U -+ V
is certainly smooth, and explicit compu-
tation of the Jacobian will verify that
D f a
A particular case is provided when U C V mapping;
in this case
Daf =
I
is the restriction of a
=
and
F
at every point
f : U -+ V
a e U,
at every point
a E U.
is the inclusion with
1
the iden-
tity mapping on Mn. One of the basic facts we shall need to know about the differential is the Chain Rule, which we express as follows.
If we have a commuting triangle
of smooth maps
V
/hNc W
U
And a point
with
a e U
f(a) = b,
then the corresponding triangle of dif-
ferentials commutes.
D h a
This provides us with a simple proof of the following fact.
(1.1)
If
differential
Proof
f : U -+ V
D f a
:
is a diffeomorphism, then at each point
]Rn -+ IRp
is invertible, so that necessarily
a r= U
the
n = p.
Indeed we shall have the commuting triangle of smooth mappings
9
1U
Now let
a e U,
and put
b = f(a).
By the Chain Rule we have a commuting
diagram of differentials
which entails that
is invertible, with inverse
D f a
Dbf-1.
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 f
(1.2)
: 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'
D
a
of
f
:
En
]Rn
a, f(a)
diffeomorphically onto V.
The next two propositions are important consequences of the Inverse Function Theorem and provide the keys to several useful results.
Let
(1.3)
rank
p
f : U - IRp
(so necessarily
some neighbourhood of
0
be a smooth mapping with n : p):
in IRn
and
D0f
of
h
of
then there exists a diffeomorphism onto another such that
f0h-1(x1, ..., xn)
10
f(O) = 0
=
(x1, ..., xp).
h(0) = 0
and
Proof
The Jacobian matrix of
the leading F
p x p
where
will have rank
0
..., x ) n
f1, ..., f
in IRn
Define a smooth mapping
onto another by
(f1(x),
=
we can suppose it is
n:
submatrix which is invertible.
from a neighbourhood of
F(x1,
f
..., fp(x), xp+1' ..., xn/
are the components of
f.
Computation verifies that
p
D0F h
By the Inverse Function Theorem
is invertible.
which is a diffeomorphism of a neighbourhood of
and automatically
diffeomorphism take Tr(x1,
..., x ) n
Let n
rank
f
Proof
=
0
--
(x1, ..., xp).
n , p):
then there exists a diffeomorphism
in
onto another such that
IRP
The Jacobian matrix of
f
will have rank
submatrix which is invertible.
p
)
=
0
in IRP
D0F
n:
k
of
we can suppose it is
Define a smooth mapping
onto another by
f(x1, ..., x ) n
Computation will verify that F
1-1
and
k(0) = 0
(x1, ..., xn, 0, ..., 0).
from a neighbourhood of
0
ir0F0 h 1(x1, ..., xn)
=
F(x1, ..., x
of
and observe that
kOf(x1, ..., xn)
the leading n x n
Theorem
is the required
h
be a smooth mapping with f (O) = 0 and D0f of
IRP
(so necessarily
some neighbourhood of
F
..., xp)
..., xn)
: U -+
onto another,
IRn - ItP to be the projection defined by
:
(x1,
=
f0h- 1(x1,
(1.1+)
it
has a restriction
in IRn
0
To see that
h(0) = 0.
satisfies
F
+
(0,
..., 0, xn+1' ..., x
is invertible.
).
p
By the Inverse Function
has a restriction which is a diffeomorphism of a neighbourhood
in IRP
onto another.
Denote the inverse of this restriction by
k:
clearly it is the required diffeomorphism as 11
f(x1, ...) x
n
)
=
F(x1, ..., xn, 0, ..., 0).
That does not conclude our review of calculus, for we shall also require standard results concerning the existence of solutions for ordinary differ-
However we shall postpone that discussion till A 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
X - Y
between arbitrary subsets
X, Y
when for any point
there exists an open neighbourhood
IRj
(depending on x)
U (1 X.
f
x E X
of IRj,
and a smooth mapping
And, by analogy with §1, we call
is bijective, and both
I
f, f
We call
IRk.
F
f
:U
--I
: X - Y
are smooth:
f
IRk
:
U
smooth
x
of
with f = F
in
on
a diffeomorphism when
in that case
X, Y
are said
to be diffeomorphic.
However, we do not wish to consider completely arbitrary subsets of Euclidinstead we shall isolate a particularly useful class of subsets.
ean spaces:
An n-dimensional parametrization of a set
First we need some definitions. X C IR'
is a smooth mapping
:
which (V) = X
and the mapping
Suppose now that
N C IRk and that
of
x
which
in
N
x E U.
is a set of the form
V -+ IRj
:
V - c(V)
x e N:
u n N
V
an open set in
n
IRn
for
is a diffeomorphism.
a relatively open neighbourhood with
U C IRk an open set for
Now we are ready to introduce the main idea.
N C ]Rk a smooth manifold of dimension
12
with
when every point
We call x e N
has a
relatively open neighbourhood
tion
(p
:
chart at
admitting an n-dimensional parametriza-
U fl N
some authors term the inverse -1
V -+ IRk:
U n N -+ V a
:
x, and call its components local co-ordinates at
adhere to the convention of writing Nn n; and sometimes we shall write
Example I
Any open set
(and in particular
dim N
and then the inclusion N --'IRk
has dimension
N
for the dimension of
Here, given
N.
N C
one takes
of dimension
k
space) is a smooth manifold of dimension and then any linear mapping
x m N
Given
n.
with image
IRn -+ IRk
n
X E N
k.
U = IRk
is a k-dimensional parametrization of
Any vector subspace
Example 2
to indicate that
N C IRk is a smooth manifold of dimension
is one).
IRk
We shall
x.
u fl N.
(as a vector we take
U = IRk
N will be an n-dimensional
parametrization of U fl N. N
In both these examples the whole set
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 (x0, x1,
..., x
n
)
Sn C IRn+1 with
manifold of dimension south pole of
Sn
N
however it is still a smooth manifold.
be the n-sphere,
x02 + ... + xn2
=
i.e.
the set of all points
I claim that
1.
Sn
is a smooth
For convenience we take the north pole and the
n.
to be the points
=
(1, 0,
...) 0)
(-1, 0, ..., 0)
S
=
Sn - {N}
-4
Sn -
-3 iRn
and we take 'PN
S
{Sj
]n
to be respectively stereographic projection from the north, south pole, i.e.
13
the mappings defined by xn
x1
N(x0, ... $ xn)
=
OS(x0, ..., xn)
=
l1 - x0 .1
..., 1 - x0
xn
x1
Since the domains of N' OS
cover
(
l1 + x0 '
l
...,
1 + x01
it will suffice to show that both
Sn
these mappings are diffeomorphisms.
1
They are certainly smooth, being the
restrictions of the smooth mappings defined by the same formulae with domains the open sets defined by tion will verify that
x0 ? 1,
A minor computa-
(IYI2 - 1, 2y1) ..., 2yn/
_ I
S1(y1,
respectively.
-1
are bijections with inverses defined by
ON, OS
ON1(y1, ..., Yn)
x0
+1IYI2 (1
..., Yn)
IYI2, 2y1, ..., 2Yn/
=
1 + 1IYI2
where
y
2
=
(Y1' ...) Yn)
IYI2
= YI
And clearly the inverses are smooth mappings as well. n = 2
one pictures
4N'
h
2
+ ... + yri
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)
Let
respectively:
N C IRk
M C IRS,
be smooth manifolds of dimensions
then M x N C IRS+k
m, n
is a smooth manifold of dimension
The proof of (2.1) is sufficiently obvious to justify its omission. way of illustration the torus
S1 X S1
C
IR4
(m + n).
By
is a smooth manifold of dimen-
sion 2.
Let
Nn C IRk
smooth manifold of
M
in
be a smooth manifold. Mm C IRk
M C N.
with
N
A smooth submanifold of n - m
One calls
is a
the codimension
The next result tells us that, locally, a smooth submanifold
N.
of a smooth manifold looks like a subspace of a vector space.
Let
(2.2)
any point
x e M
metrization Proof
Mm be a smooth submanifold of a smooth manifold
:
Let
has a relatively open neighbourhood in
V -> IRk
p, v
x
M.
k
in
Now
v(0) = X.
subset of IRm
x
into IRn
with
(Note that therefore
M, N
respectively.
f(O) = 0
Clearly, one can suppose
whose differential at
0
has rank
By (1.i+) there exists a diffeomorphism
m . n.)
in IRn
k0f(x1, ..., x ) m
=
(x1,
v0k-1
of relatively
is a smooth mapping of an open
o p
0
=
m, n
= (V) 0 M.
v-1
f =
of some open neighbourhood of
and then
having a para-
and O(IRmx 0)
be parametrizations of dimensions
open neighbourhoods of
p(0) = x,
for which
N
Nn C IRk:
onto another for which
..., xm, 0, ..., 0)
has the desired properties. 15
In differential topology (the study of smooth manifolds and smooth mappings between them) one regards two smooth manifolds when they are diffeomorphic.
f1
:
N1 -+ P1,
N1, N2
as being the "same"
Likewise we regard two smooth mappings
as being the "same", formally one calls them
f2 : N2 -+ P2
equivalent, when there exist diffeomorphisms
h, k
for which the following
diagram commutes.
f
P
N1
k
h
f2
N2
> P2
The Differential of a Smooth Mapping
§3.
Our objective in this section is to introduce the concept of "differential" for smooth mappings with domain a smooth manifold, rather than just an open The following definition provides a necessary
set in some Euclidean space. preliminary. :
in
U -+7Rk
Let
be a smooth manifold, let
Nn C IRk
x E N
and let
be a parametrization of a relatively open neighbourhood of
N with 4(u) = x,
say.
We define the tangent space
to be the image of the differential
Du(P
:
IRn ->IRk.
TxN
One pictures
the vector subspace of IRk
parallel to the affine subspace of IRk
x which best approximates
N
16
close to
x.
at
x
x
N
to
TxN through
as
Of course, for the above definition to make sense we must show that it is independent of the choice of parametrization.
So let
J
:
V -*IRk
another parametrization of a relatively open neighbourhood of
0(v) = x, u
say.
Then X =
onto an open neighbourhood
gram
x
in
be
N with
0_1
o 0 V1
of
will map an open neighbourhood V,
U1
of
and we shall have a commuting dia-
of smooth mappings
giving rise to a commuting diagram of differentials
from which it is immediate that
Duc, Dvo
have the same image, as was
required.
17
Let
(3.1)
gent space
Proof
N C IR.k be a smooth manifold, and let
TXN
is a vector subspace of IRk
x E N
of the same dimension
We keep to the notation of the above discussion,
= x.
in
n
as
N.
k
i.e.
is a parametrization of a relatively open neighbourhood of x (u)
then the tan-
:
U ->]R N.
with
Now-1
open set W C IRn with verse image under
(u) -> U -1 =
of
it
is smooth, so (by definition) there is an
on W n 4(U).
Taking
U'
to be the in-
we obtain a commuting diagram of smooth
mappings
U'
(inclusion)
hence a commuting diagram of differentials
It is immediate that
TxN,
the image of
has the required dimension
Duo,
n.
The reader will readily check for himself that the tangent space at a point to an open set in IRk
is precisely IRk:
point to a vector subspace
V
of IRk
also, that the tangent space at a
is precisely
x
is to show that the tangent space at a point is precisely the subspace perpendicular to
x.
V.
A harder exercise
to the n-sphere
Sn C IR
n+1
However for such examples as
this (where the smooth manifold is defined by equations) we shall establish 18
much simpler ways of computing tangent spaces in Chapter II. The importance of the tangent space is that it allows one to introduce the differential at a point of a smooth mapping defined on a smooth manifold, rather than just on an open set.
observation.
f
:
that IRp,
M --> N f
Let Mm C ]RP,
The definition will require a preliminary
Nn C
IRq
be two smooth manifolds, let
be a smooth mapping, let x E N, and put y = f(x).
The fact
is smooth requires that there is an open neighbourhood W of
and a smooth mapping
the differential of claim that
space TyN.
DxF
F
F : W -*]Rq
x
at
with
is a linear mapping
suppose that
D F
To prove this let 0
:
U - IRp,
0(v) = y,
By taking U
say.
q(U) C W,
and that
0
TxM
in
Of course
]Rp -+ 1Rq.
We
into the tangent
V - 7Rq be parametriza-
:
x, y
:
x
necessarily maps the tangent space
tions of relatively open neighbourhoods of
0(u) = x,
on w n M.
f = F
x
in
M, N
respectively with
to be sufficiently small we can
f o 4(U) C O(V),
so
g
:
U --r V
given
0-1
by
g =
o f o 0
is a well-defined smooth mapping.
We now have a com-
muting diagram of smooth mappings
F ]R.q
W
0 I
g
U
1
V
yielding a commuting diagram of differentials D F IRp
x
Ftq
T
D 0 v
D u'p
Dug IRm
It is immediate that
D F
x
maps the image of
n IR
Duo
into the image of
Dv', 19
the tangent space
i.e.
TXM
into the tangent space
The restriction of the linear mapping DXF T f
X
:
T f
TXM
TyN
->
IRp -> 11
and call the differential of
as the best linear approximation to
x
:
is drawn for the case
x = 0
f
at
and y = 0
in lip,
TyN,
f
at
to
as was claimed. TXM
x.
we write One pictures
the illustration below
x:
in IRq.
TXf f
It follows from the very construction of the choice of
T f
x
that it does not depend on
Notice also that in the particular case when
F.
open sets in IRp,
the differential
IRq
as previously defined.
Txf
M, N
is the the differential
are Dxf
It is an easy matter to derive the basic properties
of this more general notion of differential from those mentioned in 1.
We
shall mention two of these, leaving the proofs to the reader.
Take, for instance, the situation when M smooth manifold
x E M TXf
N,
and
the tangent space :
TXM
-+ TXN
f : M - N T M X
is a smooth submanifold of a
is the inclusion mapping.
is a subspace of
is the inclusion.
T N, X
At any point
and the differential
One pictures the situation something
like this.
TXN
20
Likewise, we have the Chain Rule, which we express as follows. A, BY C
Suppose
are smooth manifolds, and we have a commuting triangle of smooth map-
pings
and a point
a E A
with
f(a) = b,
then the corresponding triangle
g(b) = c
of differentials commutes
And as in 1 it follows from these two basic properties that at any point the differential of a diffeomorphism is an isomorphism of vector spaces, so that in particular domain and target have the same dimension.
Our next
result provides us with further examples of smooth manifolds.
First a defi-
nition.
We define the
rash of a mapping
graph f= {(x, y) E N (3.2)
Let
f : N -+ P
is a smooth submanifold of
graph f
at any point
F
:
X P
:
f(x)}.
(x, f(x))
N - graph f
is precisely graph
given by
F(x)
of the smooth projection N x P
-+
then
Also the tangent space to
N x F.
=
tion, indeed a diffeomorphism since its inverse graph f
y
to be the set
be a smooth mapping between smooth manifolds:
graph f
Proof
f : N - P
N.
Txf.
( x, f(x)) F-1
is a smooth bijec-
is the restriction to
It follows immediately that 21
graph f
is a smooth manifold.
tangent space to
graph f
Now
T F
x
is the mapping
will be its image,
(1, T f)
x
graph
i.e.
and the
Txf.
At this point is is convenient to augment (2.1) by the following proposition, determining the tangent space to a product;
the proof is left to the
reader.
Let
(3.3) (x, y)
M, N be smooth manifolds.
to the product
M x N
The tangent space at a point TxM x TyN.
is the product
So far we have talked only about the differential f
at a point
x.
of a smooth mapping
T f
x
We shall conclude this section by mentioning an elegant
globalisation of this notion which is essential to any serious study of differential topology.
The idea is to glue together all the linear mappings
to obtain a single mapping
Formally, we proceed as follows.
Tf.
N C 7Rk be a smooth manifold.
TN
Let
(3.1+)
Let
ly open set in F
:
I (x, v) E N x I R
N,
:
N.
the tangent
2n.
The proposition follows from the observation that -+
(f(u), Duf(x)) is a 2n-
dimensional parametrization of a relatively open set in
f : N --> P
be a smooth mapping with
Ve define the tangent mapping
22
n:
be an n-dimensional parametrization of a relative-
U x IRn - ]Rk x IRk defined by (u, x)
Now let
to
V E TXN1 .
N C 1Rk be a smooth manifold of dimension
f : U -+ IRk
N
i.e.
TN C IR2k is a smooth manifold of dimension
bundle space
Proof
=
Let
9Je define the tangent bundle space to
be the set of all possible tangent vectors to
Txf
Tf
:
TN - TP
N, P
TN.
smooth manifolds.
by the formula
Tf(x, v)
Let
(3.5)
Tf
:
f : N -a P
be a smooth mapping:
Let N C IRs, P C IEtt
open set
U C
TF
]Rs
with
x0 E U
Choose a point
:
-->
IRt x IRt
this mapping is smooth. (x0, v0):
also we have
definition of
M
(x, v)
U X as C
TF = Tf
on
It follows that
Tf.
Notice that if
given by But
and an
(This is possible since f is
Observe that the tangent mapping of
U x IRs
(x0, v0) e TN
for which there exists a smooth mapping
: U -+ IRt with F = f on U (1 N.
smooth.)
the tangent mapping
is likewise smooth.
TN -+ TP
Proof
F
(f(x), Txf(v)).
=
IRs
F
--
is the map
(F(x), DxF(v)) and that
X IRs
is open and contains
(U x IRs) (1 TN,
Tf
going back to the
is smooth.
is a smooth submanifold of a smooth manifold
the tangent mapping of the inclusion
M -+ N
will be the inclusion
N then TM -i TN.
One of the advantages of the tangent mapping is that it allows a simple and elegant formulation of the Chain Rule, namely that a commuting diagram of smooth mappings
gives rise to a commuting diagram of tangent mappings
TB
TA"
TC
23
otherwise expressed,
T(gof)
TgoTf.
=
Notice also that given a smooth manifold N :
TN -+ N
given by the formula
(x, v)
-
there is a smooth projection this mapping is the
x:
1TN
tangent bundle of
N.
It follows that given any smooth mapping
f : N -+ P
between smooth manifolds we have a commuting diagram of smooth mappings.
Tf
f
Vector Fields and Flows
§4.
Let
Nn be a smooth manifold with tangent bundle
v
smooth vector field on N we mean a smooth section of mapping
e
:
N - TN
to each point
x e N
such that
it o e
a tangent vector
=
at
By a
a smooth
i.e.
v,
In other words
1N.
e(x)
TN - N.
:
assigns
x.
For the purpose of illustration take the special case when N C an open set.
In that case
TN
is
N x 7Rn
and
it
:
IRn
is
is pro-
N x IRn -+ N
jection on the first factor, so that a smooth vector field is a smooth map-
ping e
:
N -+ N x IRn of the form x -+ ( x, 9(x)) with
smooth mapping.
Suppose
Bl, ..., 0n
are the components of
vector field does no more than associate with each point (01(x), ..., 6n(x))
24
in IRn.
0
x E N
N -+ IRn
:
B:
a
then our
the vector
Thus, for instance, in the case when
n = 2
we can sketch a vector field by the simple device of drawing at each point
x e N an arrow starting at that point, and in the direction given by the vector
indeed a good exercise is to invent simple formulae for
to check for himself: 6
Here are some examples, which the reader is advised
(61(x), 02(x)).
and sketch the corresponding vector fields.
taken
In all these examples we have
N = 7R2
ay
aY
61 =x: 62=y
61 = y : 02 = -p 2x
61 = x2-y 2 : 0 2 = 2xy
61=x: 02=-y
These pictures have a certain didactic value.
They suggest that if you
start at a point and follow the arrows you will move along a curve, a "flow line" if you like. fold.
We shall make this precise.
By a smooth curve in
is the mid-point of
I.
Nn
N we mean a smooth mapping
an open interval of positive length. 0
Let
be a smooth manif
:
I - N
with
I
For convenience we shall suppose that
And we say that the curve starts at
x0 = f(0).
25
f
I 0
Recall now that we have the following commuting diagram of smooth mappings.
Tf
TI
)TN "N
ITI
f
I Here, of course, TI = I x 1R. field
6
on
i
vector at at
f(t).
at
f(t)
on t,
I
Suppose now that we have a smooth vector
We wish to make precise what we mean by saying that f
N.
a "flow-line" for e. field
N
To this end we introduce a "canonical" smooth vector
by writing and
is
Tf(i(t))
i(t) =
(t, 1).
Thus
i(t)
is a unit tangent
is the corresponding tangent vector to the curve
We want this to be precisely e(f(t)), given by our vector field
i.e.
the tangent vector
In other words we want the follow-
e.
ing diagram of smooth mappings to commute
Tf
-> TN
6
f
-N
Thus we are led to define a flow-line for for which the above diagram commutes.
26
e
to be a smooth curve
f
The composite mapping Tf o i
:
I -> N is
usually written
f':
with that notation the condition for
to be a flow-
f
line is that
f' (t)
=
a(f(t)).
To make this as concrete as possible it might help to go back to the case
when N
and the vector field is given by n
is an open set in ]Rn,
functions
81, ..., Bn
on
A smooth curve
functions
f1, ..., fn
as its components.
N.
smooth
in N will have
f
And the condition
*
n
for
smooth f
to
be a flow-line of the vector field is, written out in full, the following system of
n
simultaneous equations
..., fn(t))
f1(t)
=
01(f1(t),
f' (t)
=
0n(f1(t), ..., fn(t)).
n
Thus the problem of constructing a flow-line n
ving simultaneously the above
f
for
is that of sol-
differential equations for f1, ..., fn.
And the condition for the flow-line to start at the point N
e
(x1, ..., xn)
in
is that the solution should satisfy the "initial conditions"
=
fl.(0)
Example 1
...
x1
fn(O)
Take the smooth vector field on IR2
02 = -p2x with
p I 0 a real number.
=
xz
defined by
01 = y,
(We sketched this one above.)
To
find flow-lines we have to solve the simultaneous differential equations
= 1
f2
f2
=
-p2f 1.
These yield the harmonic equation
27
f
p2f1
+
0
=
having solutions
=
f1(t)
with
p2 x2 + y2
=
f2(t)
2 2
Notice that
c, a E ]R.
the ellipses
c sin(pt + a)
p f
p2c2.
+ f2
=
2
=
p c
plc cos(pt + a)
2
so the flow-lines are
,
(Compare with the sketch given above.)
The most basic fact which it is necessary to know about flow-lines is what we shall term the Local Existence and Uniqueness Theorem.
Let
(4.1)
x0 E N.
(i)
be a smooth vector field on a smooth manifold
e
and let
The following assertions hold.
There exists a flow-line for
(ii) Any two flow-lines for hood of
N,
0
in 3Rn.
e
which starts at
which start at
e
(Existence)
x0.
agree on some neighbour-
x0
(Uniqueness)
The reader will readily check that it suffices to prove the result when N
And in that case the result is just the Local
is an open subset of IRn.
Existence and Uniqueness Theorem for ordinary differential equations, which we assume him to be familiar with from calculus.
In fact we shall need to know rather more than the mere existence of a we need to know that we can simultaneously
single flow-line through a point:
parametrize all the flow-lines through points near to a given one. is captured by the following formal definition. field on a smooth manifold
local flow for U
e
at
x0
N.
x0
is an open neighbourhood of
x0
in
x E U
N,
6
be a smooth vector
be a point in
we mean a smooth mapping
the property that for any point
28
and let
Let
and
I
the mapping
This idea
N.
F : U x I
By a smooth -+
N
(where
is an open interval) with fx
I -> N
defined by
fx(t) = F(x, t)
is a flow-line for
which starts at
x.
One pictures
the situation something like this, with
N F U
the vertical lines in the left-hand picture mapping to the flow-lines in the right-hand picture.
The fact which we need to know is the following result,
which again vie assume the reader to be familiar with from calculus.
(4.2)
Let
xO E N:
there exists a smooth local flow for
be a smooth vector field on a smooth manifold e
at
N,
and let
x0.
The next thing we need to know is that we can "straighten-out" a local flow.
However, for this we need an additional proviso, embodied in the folA critical point for a smooth vector field
lowing definition. smooth manifold
N
is a point
x e N for which
at such a point the flow-line is constant, (Indeed a constant curve through
x
i.e.
fi(x) = 0.
e
on a
Notice that
its image is a point.
is certainly a flow-line, and the Local
Existence and Uniqueness Theorem tells us that it is the only one.)
a critical point
another critical point
a non-critical point
29
We formalise the idea of "straightening out" a vector field as follows. Suppose that
U1, U2
of
N1, N2
in
are smooth-
when there exist relatively open neighbourhoods
x1, x2
x1, x2
on which are
We shall say that 1, e2
defined smooth vector fields i, 2. ly equivalent at
NV N2
are points in smooth manifolds
x1, x2
and a diffeomorphism
:
U1 - U2
for which
the following diagram commutes, and iP(x1) = x2.
TU1
TU2
62
e1
'U2
UI
Here of course we are abusing notation by allowing
their restrictions to UV U2
respectively.
Now let us call a smooth vector field v
constant when there exists a vector x E N.
One pictures the situation
mapping the flow-lines for
something like this with
all
e1, e2 also to denote
e 0
eI
to those for
on a smooth manifold N for which
e(x) =
(x, v)
Of course, such a vector field has no critical points.
for We
contend the following.
(4.3)
Let
e
be a smooth vector field on a smooth manifold N
x0 E N be a non-critical point.
N 30
smoothly equivalent to
e
at
and let
There exists a constant vector field on x0.
e2.
The picture for this result is the same as that above save that one of the vector fields is constant.
Clearly, it is no restriction to suppose that
Proof
bourhood of F
:
U x I
x0 = ->
IRn
0
open interval containing
an open neighbourhood of 0.
stant vector field on IRn (I
v = e(O):
Let
I claim that
critical point.
diffeomorphism
U
is an open neigh-
By (4.2) there exists a smooth local flow
in ]Rn.
with
N
then
v
as
0
is smoothly equivalent at
6
given by
and
in IRn,
I
I
an
is a non-
0
to the con-
0
For this we have to construct a
v.
of some neighbourhood of
0
in IRn
onto another which
satisfies
(1)
(0)
(2)
DX (v)
=
To begin with, note that any vector x e form
x = 7T(x) + t(x)v
where t(x)
0
_
n can be written uniquely in the
is a real number, and
denotes ortho-
v
gonal projection on the orthogonal complement of the line spanned by v. Observe that both
vr,
t
are linear mappings.
that Yi(U) C U, t(U) C I.
We can suppose
Define the smooth mapping
fi(x)
Intuitively, to get to the point
=
:
U
chosen so
U - IRn by
F(7r(x), t(x)).
fi(x)
you start at the point
slide up the flow-line through that point for time
t(x).
7r(x)
and
The pictorial idea
31
is as follows.
Certainly (1) is satisfied because one has For (2) we proceed as follows.
v, t
mind that
Taking differentials in *,
fi(x), t(x)) o
=
with obvious meanings attached to x E U
Thus
and bearing in
aF au ,
t(x)) o t
OF at
The fact that
F(x, 0) = x
represents multiplication of the fixed vector
D0c (x)
=
U
Note
is the identity map on ]Rn.
ir(x) + t(x)v
=
x
so
D0i
v
by a
is the identity map-
It follows from the Inverse Function Theorem that )
ping on ]Rn.
its domain
OF
IT
implies that aU x, 0)
also that at 0, 0) scalar.
x E U.
are linear, one obtains
Dx
for all
F(x, 0) = x for all
will map
is small enough.
diffeomorphically onto its image, provided U
And that completes the proof.
There is one last point which we should clear up, namely that much of what we have said can be extended to an apparently more general situation. In practice one often comes across "time-dependent" vector fields, one-parameter families of smooth vector fields.
vector field on a smooth manifold N with
J
defined by et(x) thinks of
32
e
=
e(x, t)
Formally, a time-dependent
is a smooth mapping
an open interval, such that for each
t E J
6
:
N x J - TN,
the mapping t : N - TN
is a smooth vector field on
as a 1-parameter family
i.e.
N.
Thus one
of smooth vector fields on
N.
To maintain the contrast one can then refer to an ordinary vector field as time-independent.
For a time-dependent vector field
one can mimic the
: N x J -> TN
e
definitions already introduced for time-independent vector fields. flow-line for I
x0 e N
starting at
e
an open interval containing
0,
f'(t)
And a .smooth local flow for (where
U
at
e
is a smooth mapping f
f(0) = x0,
and
=
x0
Let
and let
x0 E N:
x0
in
U x I -+ N
:
and
N,
with the property that for any point
0)
=
is an
I
x E U
F(x, t) is a flow-line for
be a time-dependent vector field on a smooth manifold
e
the
The analogue of (4.2) is
x.
(4.3)
for which
is a smooth mapping F
map x : I - N defined by the rule fx(t) starting at
I - N with
:
*(t), t).
is a relatively open neighbourhood of
open interval containing
Thus a
then there exists a smooth local flow for
6
at
N,
x0.
Germs of Smooth Mappings
§5.
The subject matter of this book is concerned principally with the behaviour of a smooth mapping smooth manifolds. x E N,
point ain
U
N -+ P
and consider the set of all smooth mappings
is a neighbourhood of
we write
f1 - f2
x
in
N.
are
Suppose given a
We can make this precise as follows.
valence relation - .
N, P
close to a point in its domain, where
whose dom-
U --r P
On this set we introduce an equi-
Given two such mappings
fI
when there exists a neighbourhood
f2
: U1 -> P, U
of
x
in
:
U2 -+ P N
33
depending
on
f1
for which the restrictions
f2
and
f2IU
f11U,
coincide
The equivalence classes under this relation are called smooth germs of mappings x, and elements of the equivalence class are called representatives
at
N -> P
Notice that if
of the germ.
f1(x) = f2(x),
then
say, at
y, f
x:
(N, x)
:
->
f1, f2
so all representatives of the germ take the same value
in view of this fact it is usual to adopt the notation
1N
:
x, y respectively the
for the germ, and to call
(P, y)
In the particular case when N = P
source, target of the germ.
notation
are representatives of the same germ
(N, x) - (N, x)
for the germ at
x
we use the
of the identity mapping
N - N. One can handle germs in much the same way as one handles the mappings from which they are derived. g
(P, y) -> (R, z)
:
gof
:
For instance given germs
f(U) C V,
of their composite
g o f
U - P,
which is evidently possible, and then the germ
U - R
:
at
x
can be defined to be the composite
It is a trivial matter to check that this definition does not depend
germ.
on the particular choices of representatives. analogy with mappings to introduce "inverses".
is invertible when there exists a germ
of
and
we can "compose" them to obtain a germ
(N, x) - (R, z): one just chooses representatives f
g : V - R with
has
f : (N, x) --# (P, y)
f o g f.
=
1P,
g o f
=
Further, to a germ
:
x
f
:
(P, y) --> (N, x)
(N, x) --> (P, y)
tial, denoted as one would expect by Txf : TxN the differential at
A germ
and in that case
1N
f
g :
Likewise, one can pursue the
of any representative:
does not depend on the choice of representative.
-'
g
(N, x) -> (P, y)
for which one
is called the inverse
we associate a differenTyP,
and defined to be
once again the definition And we shall leave the
reader to write out in full for himself the obvious version of the Chain Rule for germs.
It is perhaps worth pointing out that the Inverse Mapping Theorem
admits a neat statement in the language of germs, namely that a germ is 34
invertible if and only if its differential is invertible. f
:
(N, x)
(P, y)
-+
dim N
rank equals submersive.
The rank of a germ
is defined to be that of its differential:
the germ is immersive, and when it equals
when the
dim P
it is
Thus a germ will be invertible if and only if it is both immer-
sive and submersive.
A germ which is neither immersive nor submersive is
called singular.
Consider the plane curve
Example 1
Its image is the cuspidal cubic at any point
:
t
defined by
i. -+IR2
pictured below.
is immersive, whilst that at
t j1 0
the only value of
x3 = y2
f
t = 0
t -,
(t2, t3).
Clearly, the germ is singular.
Thus
for which we get a singular germ corresponds to the one
rather exceptional point on the curve, namely the cusp point.
0
There are various equivalence relations under which it is sensible to study smooth germs.
A good starting point from which to develop is the following.
By analogy with the definition of equivalence given for mappings in §2 we call two germs
f1, f2
equivalent when there exist invertible germs
h, k
for
which the following diagram commutes
f1
(Ni, x1) T
h
k f2
(N2, x2)
I
(P2, Y2)
35
Example 2
Consider the germs
(p2,
0)
-
(]R2,
0)
given by the formulae
(x2
f1(x, Y)
=
+ y2, xY)
f2(x) Y)
=
(x2, y2)
These germs are equivalent because in the above diagram we can take
h(x, y)
(x + y, x - y)
=
k(x, y)
(2x + 2y, x - y)
=
which are invertible because the formulae represent vector space isomorphisms, hence diffeomorphisms.
It is maybe worthwhile spelling out one simple consequence of the above definition, namely that any germ some germ
(Mn,
0) -
(IR',
f
:
(N, x) - (P, y)
is equivalent to
It is for this reason that much of this
0).
book is restricted to studying germs of this apparently special type.
Our
broad objective is to introduce the reader to the basic ideas relevant to the problem of classifying germs under the above equivalence relation.
Proposition (1.3) tells us that any submersive germ
Example 3
(Ian, 0)
-+
(]RT, 0)
..., xn )
(x1,
immersive germ
is equivalent to the germ at (x1,
->
..., xP ).
(]Rn, 0)
natural inclusion
->
of the projection
And Proposition (1.4) tells us that any
(]RR, 0)
(x1, ..., xn)
0
-+
is equivalent to the germ at (x1,
of the
..., xn, 0, ..., 0).
This example takes care of the non-singular germs. ones?
0
What of the singular
Here one is up against serious mathematical difficulties, and it will
take us to the final chapter of this book even to outline how one sets about resolving these difficulties.
Finally, we wish to introduce "jets" of smooth mappings, to be thought of as finite approximations to germs of smooth mappings, in the following rough sense:
given a germ we can write down a corresponding Taylor series (of some
representative with respect to appropriate local co-ordinates at source and 36
target) and the jets of the germ correspond to the initial finite segments of the Taylor series. J (n, p)
Let us make this more precise.
we mean the real vector space of all mappings
of whose components is a polynomial of degree 5 k xi, ..., xn
in IRn
be called k-jets. that
By the jet-space
a E IRn.
with zero constant term: Suppose now that
f
:
IRn
If in the Taylor series of
f
:
IRn
in the standard co-ordinates
the elements of --,
each
-, IRp
IRp
will
J (n, p)
is a smooth mapping, and
f(x) - f(a)
ressed in terms of the standard co-ordinates on Mn, IRP)
at the origin
(exp-
we delete all terms
of degree > k the result can be thought of as a k-jet, which one writes jkf(a)
and calls the k-jet of (the germ of)
arrive at a smooth mapping
jkf
called the k-jet extension of
:
f:
IRn
-,
f
at
dk(n, p)
a.
In this way we
given by
a -
jkf(a)
this mapping will play a crucial role in
the final sections of the next chapter.
37
II Transversality
§1.
The Notion of Transversality
The notion of objects intersecting transversally (or in general position) has The simplest situation to
become quite fundamental to singularity theory. look at is two subspaces of a vector space transversally when their vector sum is
transverse
V:
we say that they intersect
V.
not transverse
transverse
The notion is easily extended to smooth submanifolds of a smooth manifold. N1, N2
We say that two smooth submanifolds sect transversally at
x e N1 (1 N2
intersect transversally in
TxN:
when they do so at every point in
of a smooth manifold N
when the tangent spaces and
N1, N2
N1 n N2.
TxN1, T N x 2
intersect transversally in
38
not transverse
N
Probably the best way to under-
stand the idea is to look at a series of pictures.
transverse
inter-
transverse
not transverse
transverse
The idea which we really want to exploit is this. smooth mapping, and let graph
and
f
verse to in
Q
N x P.
N x Q
be a smooth submanifold of
Q
are smooth submanifolds of
and (write
Let
when graph
f (j1 Q)
P.
One pictures it thus. P
be a
Recall that both
We say
N x P.
f, N x Q
f : N -i P
f
is trans-
intersect transversally
graph f
N x Q
N
(1.1)
Let
f
manifold of P. with
y = f(x) E Q
be a smooth mapping, and let
Nn - Pp
An equivalent condition for
N x P points
The condition for graph is that for all points (x, f(x))
be a smooth sub-
is that for all
xEN
we have Txf(TxN) + TyQ
Proof
f (I1 Q
Q
with
to intersect transversally in
f, N x Q
z = (x, y)
f(x) E Q.
TyP .
=
in the intersection,
i,e.
all
we have
Tz(graph f) + Tz(N x Q)
=
Tz(N x P).
In view of (I.3.2) and (1.3.3) this may be re-written as
graph Txf +TxN x TyQ
=
TxN x Ty
which is clearly equivalent to . 39
Some authors use the relation tainly in practice
*
as a definition of
*
is probably easier to work with.
f ( Q, and cerHowever we prefer
the definition given above on the ground that it has more immediate geometric content.
Case 1
There are some special cases well worthy of separate mention.
Take the case when
is submersive.
In that case
p-dimensional vector space submersion
Case 2
f : N -> P
holds (for some
*
if
codim Q > dim N
to the image avoiding
Case 3
Txf (TIN)
is a p-dimensional subspace of the
hence equal to it.
TyP,
its germ at any point
i.e.
We conclude that a
must be transverse to every submanifold
Q C P.
A further particular situation is prompted by the observation that
if
f
is a submersion
f
f(N)
x)
then certainly
then transversality of
being disjoint from
Consequently
codim Q , dim N. f
:
N--+ P
to
Q
is equivalent
or (expressed more vividly) to
Q,
Q.
One last special case is when
diagram above). lar value of
f.
P
A point in A point
f,
the condition for this is that
have rank < n.
to which
x e N
value is a critical point of
Q
comprises a single point (see the is transverse is called a re u-
f
for which and
Txf
f(x)
is a critical value.
should fail to be surjective,
For instance, when N = P =1R
the graph of
f
c
Clearly, i.e.
the critical points are pre-
cisely the points where the derivative vanishes; precisely the real numbers
fails to be a regular
f(x)
for which the line
and the critical values are
y = c
fails to intersect
transversally.
It is maybe worthwhile spelling out the fact that the notion of transversality is invariant under equivalence of smooth mappings, in the following precise sense.
Suppose the smooth mappings
one has a commuting diagram of smooth mappings 40
f1, f2
are equivalent,
i.e.
h
1
k
I
f2
N
P2
2
h, k
with
diffeomorphisms;
respectively corresponding under
P1, P2
submanifolds of
of the invariance of transversality is that only if
is transverse to
f2
are smooth
Q1, Q2
and suppose further that
The statement
k.
is transverse to
f1
Q1
if and
The proof is a straightforward deduction
Q2.
from the definitions, and is left as an exercise for the reader.
Bear this
point in mind when reading the proof of the next proposition, which provides us with a painless procedure for extracting further examples of smooth manifolds.
Let
(1.2)
P
manifold of
be a smooth mapping, and let
f : Nn -a PP
with
having the same codimension as N
in
y = f(x)
with
in
M = f-1Q
then
Q:
f tj
Q
one has
T
N
Further, for any point
x
M = Txf-1 (T Q).
x
For the first proposition it suffices to show that every point x s M
Step I
M
has a neighbourhood whose intersection with dimension
Put
r = p - q.
suppose that
y = f(x).
are open sets in IRn,
N, P
is a smooth manifold of co-
with IRq x
Let
0.
IRp
and that
transversalitv of
f
to
Q
tells us that
7T o f
(x1,
0
in IRn ..., xn)
onto another with -*
(x1,
..., xr).
i
o f o h
on
IRp
has rank
then (I.1.3) tells us that there is a diffeomorphism of
x = 0,
y = 0:
is the intersection of
Q
denote the projection of
7
x, y we can
By taking charts at
indeed (I.2.2) allows us to assume further that P
be a smooth sub-
is a smooth submanifold of
or is empty.
Q,
Qq
k
0 x
r at
IRr.
0,
The and
of a neighbourhood
the projection
The inverse image of
0
under this mapping 41
0 x ,n r,
is an open subset of
M
that
is a smooth manifold of codimension
Step 2
The tangent space
f
h
corresponding under
TxM
to
that shows
M:
r.
The commuting dia-
is computed as follows.
gram of smooth mappings on the left gives rise to the commuting diagram of differentials on the right.
N
inc T M
TQ
f M
T f x
T N
3, P
X
inc
inc
y
inc
T (f IM)
x
TxM
TXM C Txf-1(TyQ).
from which it is clear that
T P
>T
y
Q
To show that these vector
spaces are equal it will suffice to show that their dimensions are equal.
For this one considers the linear mapping
Txf-1 (TyQ)
-->
given by res-
T Q
y tricting T Q
Txf :
Txf(TxN).
(1
y
Txf,
note that it has the same kernel as
and image
The desired equality now follows on using the fact that
the dimension of the domain is the sum of the dimensions of kernel and image, together with the definition of transversality.
Note the special case of this result when
T
xM
f -'(y)
is a smooth manifold of dimension
Let
A = (aid)
Consider the smooth mapping .
In
the conclusion
f;
or empty, and that
n - p,
be an invertible symmetric f
:
IRn -> IR
x
value of
and
f,
is the linear mapping f-1(1),
i.e.
a smooth manifold of dimension
v - 2(Ax.v),
the central quadric (n - 1),
n x n
which is given by
denotes the standard scalar product on IRn.
at the point
42
y.
is the kernel of Txf.
Example I
where
is a single point
y is a regular point of
that case the hypothesis is that is that
Q
or empty.
matrix.
f(x) = Ax.x,
Here the differential so
I
is a regular
a..x.x.
1J 1 J
=
1,
is
And the tangent space
at
x
to this quadric is the kernel of the differential,
Taking A
plane perpendicular to the vector Ax. we see that
is a smooth submanifold of IRn
Sn-1
fact we already knew from Chapter I;
Taking
expect.
the hyper-
to be the identity matrix of dimension
(n - 1),
x,
as we would
we see that ellipsoids, hyperbolic cylinders, ellip-
n = 3
tic cylinders, and hyperboloids are all smooth manifolds of dimension
Consider a smooth mapping
Example 2 rank
I
at every point in
f-1(0)
:
IR.3
f
- {0J -->
IR2
:
---, IR
i.e.
f-1(0)
is then a
a smooth curve in the plane.
given by F(x, y, z)
IR
2.
whose differential has
(1.2) tells us that
:
of dimension 1,
smooth submanifold of IR2
Consider now F
a
moreover the tangent space to this sphere
is precisely the hyperplane perpendicular to
x
at a point
i.e.
=
f(Ix2
+
y2,
z).
This too has the property that the differential has rank 1 at every point in so likewise
F 1(0),
is a smooth submanifold of
F-1(0)
IR3
of dimension 2:
it is just the surface of revolution obtained by rotating the curve f(y, z) = 0, tion let
y > 0
a, b
about the z-axis in
f
Clearly the differential of is the circle radius F 1(0)
a > b > 0,
be real numbers with
f(y, z)
b
By way of specific illustra-
IR3.
and let
f
be given by
(y - a) 2 + z2 - b2
=
has rank I at every point in
centred at
f-1(0),
which
The surface of revolution
(a, 0).
is a torus, and a minor computation will verify that it is given by
the equation
(x2 + y2 + z2 + a2 - b2)2
Example 3 Cc
Suppose
Aa, Bb
C
4a2(x2 + y2).
are smooth submanifolds of a smooth manifold
intersecting transversally in
or a smooth submanifold of
=
C:
the intersection A n B
of dimension
a + b - c.
will be empty,
This follows on 43
observing that the inclusions
A - C, B -4 C
are transverse to
B, A
respec-
tively.
A singular point (or singularity) of a smooth mapping point
x E N where the germ is singular.
f : N - P
(Of course a singular point is a
critical point, though the converse does not necessarily hold.)
E f of
set
is the set of all its singular points:
f
is a
The singular
the image of
Z f is
sometimes called the bifurcation set.
Example 4
f
:
The Whitney cusp mapping of the plane is the smooth mapping
given by (x, y)
]R2 -+ IR2
->
(u, v)
u = x, v =
where
y3
- xy.
The
singular set is the set of points where the Jacobian matrix has rank < 2, i.e.
the parabola
parabola under 4u3 - 27v2
=
x = 3y2.
f ,
And the bifurcation set is the image of this
the cuspidal cubic having the equation
i.e.
0.
f
The Whitney cusp mapping is best understood on a geometric level as the composite of the mapping
g
:
IR2 - IR3 given by
u = x, v = y3 - xy, w = y and the projection (u, v, w) ded surface
(u, v): 5
see the diagram below.
defined by
of dimension 2 by (1.2).
44
v - w3 + uw
=
0,
v
(x, y) :
->
(u, v, w) with given by
IR3 -> IEt2
The image of
g
is the fol-
a smooth submanifold of
IR3
It is maybe worth saying a little more about the projection of a surface onto a plane, as it helps to strengthen one's intuition. surface plane
S C ]R3,
P C R3
tion onto of
P
i.e.
Consider a smooth
a 2-dimensional smooth submanifold of ]R3,
through the origin:
we take
to denote orthogonal projec-
v
and ask for the singular set of the restriction
P,
and a
7rIS.
Think
For
as "horizontal", and the line perpendicular to it as "vertical".
each point
x E S
the commuting diagram of smooth mappings on the left gives
rise to a commuting diagram of differentials on the right.
T S
IT
The condition for x should map T S
x
to be a singular point of
onto a proper subspace of
T S
x
should be "vertical";
be a curve in
S
IT
x
P,
HIS i.e.
is therefore that
that the tangent plane
thus, in general, the singular set of 1rIS
projecting to a curve in
P,
IT
will
the bifurcation set.
4+5
Example 5
Let us return to the model situation presented by the last The folded surface was given by the equation
example.
and was projected onto the plane kernel of the differential of alF
lau' 8v'
i.e.
w),
to
w = 0.
The tangent plane to
(w, 1, u - 3w2):
Thus the singular set of
the equations
u = 3t2,
S
is the
i.e. the plane perpendicular to the vector
F,
and the tangent plane will be
"vertical" when this vector lies in the plane v = -2w.
F = v - w3 + uw = 0,
zrJS
v = -2t3, w = t;
w = 0,
when
i.e.
u = 3w2,
is the "fold curve" parametrized by the geometrically inclined reader
will recognise this as a twisted cubic, projecting onto the cuspidal cubic. Note that one point on the "fold curve", namely that projecting to the cusp. it is the one point where two folds on
differs qualitatively from the rest;
the surface meet, whereas all the others are just points where the surface In some sense therefore we have two types of singular point.
folds over.
Consider the projection
Example 6
the plane y = z.
G
=
Recall that
S
of the torus
S
of Example 2 onto
was given by an equation
(x2 + y2 + z2 + a2 - b2)2
The tangent plane to 46
S
it
-
4a2(x2 + y2)
is the kernel of the differential of
=
(1)
0.
G,
i.e.
the
plane perpendicular to the vector point to be a singular point of plane
y = z,
i.e.
that
ay
raG aG aGll tax, ay, az/:
iris
=
and the condition for a
is that this vector should lie in the
az ,
which computation verifies to be the
condition that
+ y2 + z2 + a2 - b2) + 2a2y
(z - Y X X
The singular set of
iris
=
is given therefore by (1), (2),
0.
i.e.
(2)
it is the
intersection of two surfaces of degrees 4, 3 respectively, hence a space curve of degree 12. set;
And its projection onto the plane
y = z
is the bifurcation
it is what we "see" if we imagine the torus made of glass and viewed
from a distant point on the line perpendicular to the plane y = z.
So much for transversality per se.
The remainder of this chapter is
devoted to the development of a very basic intuition associated with transversality.
Suppose we have a smooth mapping
transverse to a submanifold forced to be transverse to
Q C P. Q
f : N --).P
which fails to be
Common sense tells one that
by arbitrarily small perturbations.
f
can be
The for-
mal development of this intuition leads to a host of results, reaching high degrees of complexity and subtlety, called transversality theorems.
In recent years transversality theorems have assumed a role of increasing importance in differential topology, indeed Singularity Theory could hardly exist without them.
Essentially they all say the same thing, namely that by
an arbitrarily small perturbation a given smooth mapping (or some closely
47
related smooth mapping) can be made transverse to a smooth manifold.
The
object of the next section is to establish a very basic lemma which provides the key to most transversality theorems.
The Basic Transversality Lemma
§2.
F : N x S
Let
P
-4
be a smooth mapping.
family of smooth maps
fs
the elements
Suppose
s e S.
We ask whether
fs (t1 Q
: N -> P
This we think of as a smooth
where with
F (I1 Q
for all parameters
F(x, s)
=
fs(x)
parametrized by
a smooth submanifold of
Q
s?
P.
That the answer can well be
in the negative is shown by
Example 1 F
fs
:
Take
N x S maps
P
-*
IR2
the 2-sphere
N = ]R 2,
S = ]R,
P = IR3
defined by the formula
onto the horizontal plane F
S2.
and consider the smooth family
((x, y), s) in IR3.
z = s
is a diffeomorphism, so
->
F
Q.
And take
the plane
is tangent to
z = s
to be
Q
On the other hand
only provided vie avoid the two exceptional parameters
fs (h Q
Thus
(x, y, s).
s = +1
when
Q.
However, it is clear in this example that any value of the parameter can be approximated as closely as we please by "good" values, for which
Transversality Lemma: Q
for technical reasons we prefer to replace the single
by several smooth manifolds
Q1, ..., Qt :
become clear in the final section of this chapter. Sard's
the reason will
The crucial tool here is
Theorem, one of the truly fundamental results in differential topology,
which we state in the following form.
48
those
That this holds generally is the content of the Basic
fs (f Q.
smooth manifold
i.e.
s
(2.1)
Let
fi
:
Ni
->
P
be a countable family of smooth mappings.
set of common regular values of the
fi
is dense in
The
P.
The proof of Sard's Theorem is quite lengthy, so we have isolated it as Appendix A to this book in order not to break the flow of the text.
The
Basic Transversality Lemma is
(2.2)
Let
F : N x S
P
->
verse to smooth submanifolds of parameters
Proof
B
for which
s
1.2
M.
sider the restriction
= in Ni
be a smooth family of smooth mappings transof
Q1, ..., Qt fs
then there is a dense set
P:
is transverse to all of
F(Q1)
Q1,
..., Qt.
is a smooth submanifold of
of the projection
in
: N x S
->
N x S. S.
Con-
l'Te shall
show that
vjMi
if and only if fs
{sJ
which will prove the result, by Sard's Theorem. ',4'e can, and shall, drop the index
i
41
Qi
Now to the proof of
since it plays no further role.
*.
The
reader is urged to keep the following picture in mind.
N To start with, note that the condition for that for all
z
=
(x, s)
in
N x S
with
TzF(TxN x TSS) + T
w
F
to be transverse to
w = F(z)
Q
=
TWP.
in
Q
Q
is
we have
(1)
49
Consider now the condition for dition that for the same
x
to be transverse to
fs
Q:
as in (1) =
TZF(TXN x 0) + TwQ
TwP.
(2)
Furthermore, the condition for N x (s} to be transverse to x
the same
M = T F-1(TWQ)
T
M
is that for
as in (1)
TXN x 0 + TZM
where
this is the con-
=
TXN x TsS
(3)
If we assume (3) and apply TF to
by (1.2).
z
z
both sides then
(1)
tells us that
And conversely it is a
holds.
(2)
minor exercise in linear algebra to see directly that Thus the condition for
for N x {s}
fs
to be transverse to
to be transverse to
lent to the condition that
7vIM
x
dition says that for the same
implies
is precisely the condition
is transverse to as in
(s}:
indeed this last con-
we have
(1)
TSS
(4)
(3).
By way of explicit illustration consider two smooth manifolds some Euclidean space
(3).
We claim that this in turn is equiva-
M.
TZTr(TZM)
which is clearly equivalent to
Q
(2)
M, N
in
Even if they do not intersect transversally, geo-
S.
metric intuition tells us that it should be possible to force them to do so
by an arbitrarily small translation of one manifold M We can make this feeling precise as follows.
claim that the S,
s
for which
Ms, N
justifying one's intuition.
mappings 50
under the translation of
M
the image of
F
:
M x S
-->
S
by
S
For
in some direction.
s s S
let
Ms
denote
defined by x H x + s.
We
intersect transversally will be dense in
To this end define a smooth family of smooth
(x, s) H x + s.
It is clear that
F
is a
submersion, so
is transverse to
F
is transverse to
fs
N
for a dense set of parameters
established by observing that
is transverse to
f
By the Basic Transversality Lemma
N.
N
-
Ms, N
Our claim is
s.
intersect transversally if and only if
which fact we leave as an exercise for the reader.
S
An Elementary Transversality Theorem
§3.
Our object in this section will be to show that given a smooth mapping f
IRn
:
ping to
g
f.
-+
and a smooth manifold
IRp,
: Rn -
which is transverse to
IRp
we can find a smooth map-
Q C IRp, Q,
and as "close" as we please
Of course the first problem is to say precisely what we mean by
smooth mappings being "close":
roughly speaking we shall take this to mean
that their values are "close", and that for each integer
vatives of order k
all smooth mappings
number
Rn ->
and an integer
bourhood in CO°(IRf, for which for all
their deri-
IRp,
IR.p)
k > 0
denote the set of
Let e(IRn, IRp)
and let
f
:
Mn - IRp
Given a (small) positive real number R,
1
are "close".
We make this precise as follows.
mapping.
k >.
a (large) positive real
e,
we associate to
be a given smooth
f
a fundamental neigh-
comprising all those smooth mappings
x E Rn with
Ixl
s R
g
: Rn, IRp
one has
51
Iljkf(x) - jkg(x)II with
set
XC
f
IRn
:
C
a fixed norm on the jet-space Jk(IRn, IRP).
II
II
'-
dense therein when given any smooth mapping
C-O (IRn, IRp) -,
and any fundamental neighbourhood V
IRp
smooth mapping
g
IRn ->
:
And we call a sub-
X with
in
IRp
g e V:
one can find a
f
of
intuitively, any mapping
can be approximated as closely as we please by mappings in
X.
'.`e are now in a position to state and prove an elementary transversality theorem.
The set of smooth mappings
(3.1)
submanifolds
Q1, ..., Qt of IRp is dense in
Proof
f
imate
Let f
:
IRn
->
IRp
be smooth.
n' ],p).
C-'O(IR
l7e have to show that we can approx-
as closely as we please by mappings transverse to
The idea is to construct a smooth family f,
transverse to given smooth
IRp
IRn --*
and with
F
transverse to
Transversality Lemma.
F
Q1, ..., Qt
lie shall make
F
:
IRn X'S - IRp
Q1, ..., Qt.
which contains
then one applies the Basic
transverse to
Q1, ..., Qt
by en-
suring that it is a submersion.
To motivate the construction recall that the
transversality of
to
transverse to
f
IRn x Q
:
1Rn -,
IRp
Q
is equivalent to the graph being
in the product IRn x
IRp.
If the graph is not already
transverse it seems reasonable that we might be able to force it to be so by
52
bodily translating it.
point.
The picture above illustrates the idea when
Therefore we take
(x, s) H f(x) + s.
and define
= IRp
S
F
:
Q
IRn x IRp , ]Rp by
Clearly, this is a submersion, so transverse to
Q1, ...) Qt.
By the Basic Transversality Lemma there is a dense set of
for which
is transverse to
fs
is that if
f0 = f: V
hood
then
0
s
All that remains to be shown
Q1, ..., Qt.
is close enough to
s
is a
is as close as we please to
fs
more formally, we have to check that given a fundamental neighbourof
we can find an
f
s / 0
for which
f
lies in
V.
And that
we can safely leave to the reader.
It is only fair to point out that this example of a transversality theorem is not particularly useful.
Its virtue lies rather in the fact that it is an
easily understood prototype of transversality theorems of greater complexity and application.
Our object here was simply to lay bare the underlying idea
behind the proofs of such theorems.
4.
Thom's Transversality Theorem
Our next transversality theorem is rather more useful than that of the preceding section.
Indeed it will suffice for all the applications we shall
require in this book.
Its statement does not possess quite the same immediate
intuitive appeal as our previous result:
however, in the next section we
shall discuss a simple application which should clarify the situation.
(4.1)
Let
Q1, ..., Qt
be smooth submanifolds of the jet space
The set of all smooth mappings f is transverse to
Q1, ..., Qt
:
IRn -a IRp
is dense in
J"( n, p).
for which j kf : Mn -> Jk(n, p)
Ca'(IRn,
IRp).
53
Note
It is essential that the reader appreciate the difference between
this transversality theorem and that of the preceding section. vious result we managed to make
f
transverse to a submanifold by using a
But in the present situation it is
constant deformation.
which we wish to make transverse to a submanifold. f
only
In the pre-
which we are allowed to deform.
not
jkf,
f,
On the other hand it is
A constant deformation will not
work here since it does not alter the derivatives of
What we do instead
f.
is to use a polynomial deformation.
Proof
Let
S = J (n, p).
Of course
:
IRn x S - Jk(n, p)
mersion, because for fixed Jk(n, p)
of
defined by x
is finite-dimensional, so can
Consider the smooth family
be identified with a Euclidean space. F
S
(x, s) _ jk(f + s)(x).
the mapping represents an affine isomorphism
with itself, hence transverse to all of
Q1, ..., Qt.
Basic Transversality Lemma there is a dense set of parameters the mapping
fs
:
IRn
->
Q1, ..., Qt.
we please to
by choosing
f
defined by x
Jk(n, p)
verse to all of
is a sub-
F
-+
s
for which
s
jk(f + s)(x)
And clearly we can make
f + s
By the
is trans-
as close as
to be sufficiently small.
First Order Singularity Sets
§5.
We shall use the Thom
This section is in the nature of an extended example.
Transversality Theorem to show that for a dense set of smooth mappings
f
:
IRn
-->
IRP
the singular set
smooth manifolds on each of which
Z f
f
can be partitioned into finitely many
has constant rank.
Recall that a singular point of a smooth mapping point
54
x E N
for which the rank of the differential
f
-> PP
T f
x
is a
falls below its
possible maximal value of
min (n, p).
A natural way of distinguishing one
singular point from another is by the actual value taken by the rank of the differential.
To this end we introduce the first order Thom singularity sets
Elf
=
{x E N
(Concerning the terminology :
:
T
later in this book we shall have something to
say about higher order Thom singularity sets.)
tition of
N
it.
xf has kernel rank
In this way we obtain a par-
into finitely'many sets on each of which
has constant rank.
g
One might reasonably hope that these sets will be submanifolds of N,
but
that (as the following examples will illustrate) is not necessarily the case.
Example 1
Take
f
IR2
:
-+
IR2
defined by the formula
(x, y)
-,
(x2, y2).
This we shall refer to as the "folded handkerchief" mapping for the following reason:
it is the composite of the two mappings
(x, Y)
~'
(x, y2)
:
(x, Y) _
(x2, Y)
which "fold" the plane along the x-axis, y-axis respectively.
One pictures
it something like this.
fold
fold
along x-axis
along y-axis
LL
x
The reader can easily check that the origin is the only on the axes being
Z1
points, and the rest being
ZO
ZL-point, other points
points.
(One need
hardly point out that this fits in with the fact that the origin is folded twice, other points on the axes just once, and the rest not at all.)
Here
55
Elf are all submanifolds.
the
Take the smooth mapping
Example 2
(x, y) H (x2 + y, y2). xy = 0,
given by
i.e.
E2
points.)
Take for instance fs
points are now given by xy = Thus as
moves from
s
However, it is well worth ob-
E1f
to be a submanifold by slightly
:
defined by
IR2 -+ 1R2
Another minor computation shows that the
(x2 + y, y2 + L.sx).
->
points are
But a small linear deformation will do
matrix, and is therefore of no use.
(x, y)
E1
Of course a constant deformation will not affect the Jacobian
f.
the trick.
given by
is the union of the two axes, so certainly not a
Elf
serving that in this example we can force deforming
IR2 ->7R2
:
A minor computation shows that the
(There are no
a submanifold.
f
s,
a hyperbola, so certainly a submanifold.
to a small value so the singularity set
0
E1
E1fs
changes from a pair of intersecting lines to a pair of disjoint curves. We shall now prove generally that the first-order singularity set
of a smooth mapping
f : N - P can be forced to be submanifolds, by the
device of slightly deforming
f.
For the sake of technical simplicity we
shall carry it through only in the case that
f
x E IRn
:
IRn
-+
IRp
Elf
N =IRn,
is our smooth mapping.
P = RP.
Suppose then
I'otice that given a point
the first order singularity set to which it belongs depends only on
the 1-jet of
f
of the jet space
at
x,
i.e.
J1(n, p)
Dxf.
So let us write
E1
for the subset
comprising all 1-jets which have kernel rank
i.
equal to
i
is a smooth submanifold of
1
J (n, p)
of codimension
i(p - n + i)
(5.1)
E
Proof
For this it will be convenient to identify a linear map with its
matrix relative to the standard bases.
56
We proceed in two steps.
Step
1
Let
matrix.
E
( D)
=
We claim that
E
p x n matrix with A
be a
has rank
k
D = CA 1B.
if and only if
this, observe that for an arbitrary matrix X
an invertible
the matrix
E
k x k To see
has the same
rank as
IIk X
0
A
B
A
B
Ip-k
C
D
XA + C
XB + D
and the claim follows on choosing X
Choose
Step 2
k
with
i + k
=
will be no restriction to suppose
such that
and let
n,
E0
XA + C
(CO
=
tible matrix. E0
matrix
U E
U 1\ E1
=
in it the
( D)
k x k
E0
f
is a submersion:
function of
D
is an invertible affine mapping.
j
1f
Proof
is invertible.
indeed if one fixes
is a smooth submanifold of codimension f-1(0)
=
of
is transverse to all the sets
El ,
--
Consider
D - CA 1B.
A. B, C
the resulting
It follows from (1.2) that (p - k)(n - k) = i(p - n + i):
U n E1,
There is a dense set of smooth mappings
smooth manifold of codimension
U
with the property that for any
f : U - J1(p - k, n - k) given by E
but by Step 1 the inverse image
inver-
0
matrix A
Notice that
(5.2)
k x k
a
It
a smooth submanifold of codimension i(p -n+i).
to be an open neighbourhood of
the smooth mapping
f-1(0)
AO
E1.
And it will suffice to produce an open neighbourhood
in J1(n, p) with
We take
with
DO)
0.
be a matrix in
EO
0
=
f
which ends the proof.
:
]Rn -+]Rp
and hence for which each
for which Ei f
is a
i(p - n + i).
The first statement is an immediate consequence of the Thom Transver-
sality Theorem (taking
k = 1).
And the second statement follows from the 57
fact that
Elf
is just the inverse image under
j1f
Z ,
of
so is a smooth
11
manifold of the same codimension by (1.2).
transverse to the
E i
will have codimension Elf
f
Consider a smooth mapping
Example 3
is void for
immersion
i.e.
that when
p 3 2n
:
in the special case when i(p - n + i)
i > 0,
>
is
fRn - ]Rp for which j 1f In this case
p > 2n.
i(n + i) > n,
unless
its germ at any point is immersive. any smooth mapping ]Rn -4IRp
Thus
i = 0.
which is the same thing as saying that
f
Elf
is an
Our theory says then
can be slightly deformed to
become an immersion, and indeed a linear deformation will suffice (as is clear For an explicit instance
from the proof of the Thom Transversality Theorem). take the cuspidal cubic curve
f
Here, for a small positive number
fs(t) = (t2, t3 - ts)
:
IR --]R2 s
given by
the curve
fs
:
is an immersion, obtained from
f(t) = IR ->IR2 f
(t2, t3).
given by
by a small linear
deformation.
image of fs
image of f Example 4
We shall consider, in some detail, the condition that
transverse to the
E1
for a function f
:
]Rn -> IR.
This means we are to
have
image of differential
tangent space J1(n, 1)
of 58
j 1f
at
a
to Ei at j 1f (a)
j1f
be
for all points
a e IRn.
We shall show that this is exactly the condition
that the so-called Hessian matrix a2f
H
l
(a)Jnxn
(ax. ax. i J
-
is non-singular at every critical point
where
a,
x1, ..., xn
are the stan-
dard co-ordinate functions in ]Rn.
To start with, let us simplify matters by identifying
with IRn
J1(n, 1)
by identifying a linear mapping IRn -IR with its matrix relative to the In particular
canonical bases.
Daf
is identified with
(
Of
Of
(a), ...,
Notice that the only Thus
respectively.
verse to it.
E
En-1
f is defined by the
and is just the critical set of
easily identified.
can be discounted. ate
IRn,
i.e.
that
1
j
j1f
at
f
at
a
*
*
is precisely
a
are H,
gen-
so its tangent space
is that the columns of
*
=
when
is the, subspace of IRn
a
is just the origin in IRn,
Thus the burden of H
= 0, ..., ax n
In that case the expressions in
so the image of the differential of
En
is automatically trans-
f
Thus we need only satisfy
f.
i = n.
j
0, n
of codimensions
conditions ax 1
n
The Jacobian matrix of
erated by its columns.
I
is an open set and
n
is a critical point, and
En-1, En
which arise are
El
a)
n
1
gener-
H
is non-singular, establishing our claim.
A critical point of a smooth function
f
:
IRn -+IR
matrix is non-singular is called non-degenerate. summed up by saying the condition for
j1f
at which the Hessian
Our discussion can then be
to be transverse to the
that every critical point should be non-degenerate.
E1
is
By way of illustration,
the reader may care to check that the function
f(x, Y)
=
x2 - 3xy2
has a degenerate critical point at the origin.
One pictures it like this.
59
0
In this example
j1f
cannot be transverse to the
E1.
According to the
theory of this section it should be possible to gain transversality to the E1
by a linear deformation of
is linear.
i.e.
f,
by taking
fs = f + Ls
where
Here is an explicit linear deformation which does just that
Ls -
the reader is urged to check this for himself.
fs(x, y)
The graph of
fs
=
x3 -
3x2
is rather difficult to draw.
happens is indicated in the pictures below. the case
s = 0,
fs > 0.
The left-hand picture refers to s > 0.
In both cases the
And in both figures the curves
have been drawn for small positive values of
60
However, some idea of what
the right-hand one to the case
shaded area is that where
- sx.
e.
fs = +e
III Unfoldings: the finite dimensional model
The kind of mathematics which we shall discuss later in this volume lies rather deep, at least in the sense that formal proofs of the main results rely However the underlying geometric ideas are
on hard theorems in analysis.
The main object of this chapter is to introduce these
rather straightforward.
ideas in a relatively simple situation where the mathematics is easy enough not it will also provide us with an opportunity of est-
to hinder understanding:
ablishing one or two little facts which we shall have occasion to use later.
Groups Acting on Sets
§1.
By an action of a group
(ii)
where
1
on a set
(g, x) + g.x,
usually written (i)
G
=
x
(gh).x
=
1.x
M
we mean a mapping
for which for all
iD
x E M.
: G x M -+ M,
and
g.(h.x)
denotes the identity of
G.
Given such an action we can define an equivalence relation agreeing that
x ti y when there exists an element
i.e.
x
.-
on
M
by
g E G for which y = g.x.
The equivalence classes are called the orbits under the action. the orbit through
g, h E G
Given X E M
is by definition the equivalence class which contains
x,
the set
G.x
=
{ g.x : g E Gi .
61
Before turning to examples let us observe one small geometric point.
with
M lying on the same orbit, so there exists
be points of
x1, x2 x2
=
is a transformation of serves orbits
-
M,
(i.e.
a bijection of
M
onto
g e G
a
defined by x
Observe that the mapping M -+ M
g.x1.
Let
-+
g.x
itself) which pre-
by which we mean that a point is always mapped to another
point lying on the same orbit
-
and which maps
What this
x2.
to
x1
amounts to is that one point on an orbit looks like any other; perty known as homogeneity of an orbit.
it is the pro-
The reader is urged to bear this
notion in mind as a guiding intuition. It is not the purpose of this chapter to pursue the general theory of groups acting on sets.
Rather, we wish to concentrate the reader's attention on a
class of geometric examples relevant to the mathematics of the next two chapters, and indeed providing genuine finite-dimensional analogues of the situations there studied.
§2.
To this end
Some Geometry of Jets
We start by recalling that in Chapter I we introduced a relation of equivalence on germs, so in particular on the set of all germs indeed two such germs
h, k
f, g
are equivalent when there exist invertible germs
for which
foh
=
k o g.
(1)
One can relativize this definition to d-jets as follows. to have equivalent d-jets when there exist invertible germs
jd(f o h)
In particular we can take
f, g
=
f, g h, k
are said for which
(2)
jd(k o g).
to be (germs at
cing an equivalence relation on the jet-space 62
(IRn, 0) -+(IRp, 0);
0
Jd(n, P).
of) d-jets, so induThe geometric
examples we have in mind arise from studying this equivalence relation not on
of all mappings
in -+
rather on the vector subspace
Hd(n, p)
each of whose components (relative to the standard
IRP
is a homogeneous polynomial of degree
co-ordinates on IRP) dard co-ordinates
but
Jd(n, p)
the whole jet-space
in
x1, ..., xn
d
in the stan-
The geometry has its genesis in
]Rn.
the following elementary, yet crucial, observation.
Two d-jets
(2.1)
f, g
are equivalent if and only if there
Hd(n, p)
in
exist invertible linear mappings
H. K for which
The condition is certainly sufficient.
Proof
f o H
=
K o g.
To establish necessity,
suppose there exist invertible germs
h, k for which (2) holds.
for the Taylor series of a germ
(IRn, 0)
k
with
:
H + terms of degree
>.
2
=
K + terms of degree
>
2
f, g
components of
cp
then we have
(]RP, 0):
=
Thus, bearing in mind that the
invertible linear mappings.
H, K
-->
Write
are homogeneous polynomials of degree
d,
(f
h)
=
(k
g)
= K o g + terms of degree > d
one has
f o H + terms of degree > d
so
f o H
=
jd(f o h)
=
We can re-phrase the above as follows.
jd(k o g)
Let
=
GL(s)
linear group of all invertible linear mappings ]Rs - IRs of composition. group
=
denote the general under the operation
The reader will readily check that we have an action of the
GL(n) x GL(p)
(H, K).f
K o g.
on the vector space
K o f o H 1.
Hd(n, p)
given by
and the condition for two d-jets
f, g
in 63
Hd(n, p)
to lie in the same orbit under this action is that there exist in-
f o H
= K o g.
classes described above.
=
f,
i.e.
GL(n) x GL(p)
are precisely the equivalence
The next step in our study is to indicate the We start with
connexions with geometry.
p = 1
The Case
In this case an element of degree
(H, K) . g
It follows immediately from (2.1) that the orbits in
under the action of
Hd(n, p)
for which
H, K
vertible linear mappings
d
in
x1, ..., xn
surface of degree
d
Hd(n, 1)
so (providing it is not zero) will define a hyper-
in real projective space
for instance when n = 2
we are dealing with
when n = 3 with a curve of degree And two elements of
is just a homogeneous polynomial of
d
P]Rn
d
of dimension
(n -1)
:
points on a projective line,
in the projective plane, and so on.
will lie in the same orbit if and only if the
Hd(n, 1)
corresponding hypersurfaces can be obtained from each other by an invertible linear change of co-ordinates,
are projectively equivalent.
i.e.
ficulty of listing the orbits increases sharply with
d:
The dif-
we shall allow our-
selves the luxury of describing in detail some of the simpler cases relevant to the mathematics of the next two chapters. d = 1.
One starts with
in n variables
is the vector space of linear forms
H1(n, 1)
x1, ..., xn and elementary linear algebra tells us that
there are just two orbits namely that containing the zero form, and that containing all the non-zero forms. The next case is forms in
n
d = 2.
variables
H2(n, 1)
x1, ..., xn
is the vector space of all quadratic
and elementary quadratic algebra tells
us that any such form can be brought into the shape x1 + ... + xs - xs+1 - .. - xr
64
by an appropriate change of co-ordinates.
The numbers
r, s
are called the rank, index of the form, and (by Sylvester's
Law of Inertia) are invariant under co-ordinate changes. tiply the form by
the rank remains invariant, though the index may change,
-1
so we work instead with the semi-index Thus quadratic forms in
x1, ..., x
s' = min(s, r - s)
x, y can be written
with the point
(a, b, c)
picture below.
The cone
in ]R3 b2 = ac
of the form.
are classified by rank and semi-index.
To make this even more explicit take the case when n = 2. in two variables
However, if we mul-
ax2 + 2bxy + cy2
A quadratic form so can be identified
We obtain four orbits as indicated in the comprises the forms of rank I (the Para-
bolic type) with the origin representing the zero form of rank bolic type).
(the IM-
0
The remainder of the space comprises the forms of rank 2:
indeed the inside of the cone corresponds to forms of semi-index
0
(the
elliptic type) and the outside of the cone to forms of semi-index I (the hyperbolic type).
A..
Let us continue with the case all cubic forms in
n
variables
d = 3.
H3(n, 1)
x1, ..., xn.
is the vector space of
This case differs from the
preceding ones in that a complete list of orbits is known only for The first non-trivial case is
n = 2,
n . 4.
and since this case will arise natur-
ally in the next chapter we shall take the opportunity to describe a once familiar bit of pure mathematics, the study of binary cubic forms
f(x, y)
=
ax3 + 3gx2y + 3yxy2 + Sy3.
65
It is easy to list the orbits using a little algebra. allow the variables forms.
Let us (temporarily)
x, y to be complex and restrict ourselves to non-zero
Recall that a non-zero complex homogeneous polynomial of degree
in two variables factorizes into
d
d linear factors (possibly with repetitions):
in particular
f(x, y)
=
(a1x + bly)(a2x + b2y)(a3x + b3y).
It follows that the zero set of equation
f(x, y)
=
0
f,
i.e.
the subset of
C2
defined by the
will comprise three lines through the origin.
We
shall distinguish four possible types of binary cubic form, corresponding to the following four types of triples.
(The reason for the terminology will
be explained shortly.)
elliptic
-
all distinct and real
hyperbolic
-
all distinct;
one real and two complex
parabolic
-
two distinct;
both real, one repeated twice
symbolic
-
one real line repeated thrice.
In fact these types are precisely the orbits we seek.
Certainly two bin-
ary cubic forms of the same type will lie in the same orbit, since given two triples of lines through the origin (of the same type) we can always find a non-singular real linear mapping of triple to those of the other.
C2
which maps the lines of the one
And conversely two binary cubic forms in the
same orbit are necessarily of the same type, since non-singular real linear mappings of
C2
preserve the above types of triples of lines.
We can ob-
tain normal forms for non-zero binary cubic forms by just choosing an example of each of the four types just described. given in the following table.
66
The standard choices are those
type
normal form
elliptic
x3 - xy2
hyperbolic
x3 +
xy2
parabolic
x2y
symbolic
x3
The reason for the terminology is as follows. f
Given a binary cubic form
one can associate with it a binary quadratic form
Hf
called the Hessian:
it is defined to be 2
H
f
=
2
a f
a f
ax2
axay
2
f
a2f
ayax
aye
1
37 a
and a little arithmetic will verify that it is given by the formula
82)x2 (ay -
+ (as - QY)xy +
((3S
- Y2)y2
.
It is now an easy matter to verify that a binary cubic form
f
is elliptic,
hyperbolic, parabolic or symbolic exactly according as the binary quadratic form
Hf
is;
and that is the reason these words are used to describe the
types of binary cubic form.
Although it is more difficult, one can obtain a visualization of binary cubic forms, just as we did for binary quadratic forms. cubic forms can be identified with IR4 ax3 + 3f3x2y + 3yxy2 + 8y3
The space of binary
by identifying
with the point
(a, (3, y, S):
non-zero forms into four types yields a partition of and by projecting this appropriately into
IR3
the partition of the
IR4 - [0; into four sets,
(we shall not go into the
67
details) one arrives at the following delightful picture, dubbed the umbilic bracelet.
It is obtained by rotating a deltoid (the curve traced by a fixed
point on a circle rolling inside another circle of three times its radius) about an axis with a twist of 2rr/3 for each full circle.
In this picture the
cusped edge of the bracelet corresponds to the symbolic forms, the rest of the
surface to the parabolic forms, the interior to the elliptic forms, and the exterior to the hyperbolic forms.
Let us pursue the possibility d = 3 consider is
n = 3,
i.e.
a little further.
ternary cubic forms in
x, y, z:
thought of as cubic curves in the real projective plane
The next case to these are best
The deriva-
PIR3.
tion of normal forms here is a lengthy exercise in the geometry of curves for details of which we refer the reader to texts on that subject; merely quote the results.
we shall
One starts with non-singular cubic curves.
non-singular cubic
zy2
- x3 + axz2
+ bz3
with 4a3 + 27b2 / 0
To such a curve one associates the so-called j-invariant
j = 4a3/4a3 + 27b2:
geometrically, this is the cross-ratio of the four lines through a point on the curve tangent to the curve elsewhere.
68
It can be proved that two curves
in this form lie in the same orbit if and only if the corresponding equal, and the the
a's
b's
have the same sign:
have the same sign.
cally irreducible:
or if
b = 0
j's
are
then if and only if
Of course, a non-singular curve is automati-
the remaining irreducible curves are those which are singu-
lar, and these have exactly one real singular point.
The tangents at the
singular point can be distinct (the nodal case) or coincide (the cuspidal case). And in the nodal case one can made the finer distinction between the crunodal
cubic (when the tangents are both real) and the acnodal cubic (when the tangents are complex conjugate).
In this way one obtains three singular irre-
ducible real cubic curves.
crunodaZ cubic
x3 + y2z + x2z
acnodaZ cubic
x3 + y2z - x2z
cuspidal cubic
x3 - y2z
We are left with a motley array of reducible cubic curves.
The cubic can
reduce to a conic and a line, which may or may not be tangent to the conic;
or it may reduce to three lines, giving rise to various possibilities.
69
conic
x(x2 ± y2 ± z2)
and Line
conic and tangent
y(xy - z2)
triangle
x(y2 ± z2)
three
x(x2 ± y2)
concurrent Lines
two lines
2
x y
one e repeated
triple Line
70
x3
Rather than attempting to enlarge still further on the various situations which can arise by choosing special values of
d, n
we shall widen our class
of examples by considering
The General Case
p a 2
One thinks of an element of Hd(n, 1)
sion
<.
as determining a vector subspace of
of dimension s p, namely the subspace spanned by its components:
for that reason elements of Hd(n, 1)
Hd(n, p)
Hd(n, p)
of dimension s p. 2, 3
can be called linear systems in
The simplest linear systems are those of dimen-
which have long been known as
cils, nets, respectively, and
enp
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 (q1, q2)
A pencil of binary quadratic forms
of quadratic forms in two variables
be identified with the subspace of
H2 (2, 1)
x, y 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 ]R3
ax2 + 2bxy + cy2
with the point
by
(a, b, c),
and that the various types of binary quadratic form are separated exactly by the cone
b2 = 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 described, and second that these types are invariant under the action,
i.e.
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
geometric description
normal form
plane intersecting the cone
(xy, x2 + y2)
plane lying outside the cone
(xy, x2 -
plane tangent to the cone
(xy,
line inside the cone
y2)
x2)
(x2 + y2,
0)
line outside the cone
(xy, 0)
line tangent to the cone
(x2, 0)
the origin
(0,
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
GL(2) x GL(2).
To simpli-
fy proofs we refer to elements of the first factor as changes of co-ordinates, 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
x2 + y2,
xy,
x2,
(qj, q2)
by a change of basis we can suppose
q2 = 0, and then a change of co-ordinates enables us to put the forms
It is
q1
into one of
according as it is elliptic,hyperbolic or para-
bolic, so yielding the three possible normal forms shown for the pencil. Suppose next that the pencil
represents a plane through the origin,
(q1, q2)
tangent to the cone, so by a change of basis we can suppose then by a change of co-ordinates that cannot involve
the cone
q,
q, = xy,
as was required.
y2,
q2 = x2:
q,
the plane being tangent to
Finally, suppose the pencil represents a plane 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
and a change of basis allows us to suppose
through the origin not tangent to the cone. suppose
q2
ax2 + /9y2
with
a
0,
Q / 0
q2
we reduce
q, = xy: q2
to
else the plane would be tangent to
Obvious changes of co-ordinates and basis now bring the pencil to
the cone.
the form
(xy, x2 + y2),
the + sign corresponding to the case when the plane
intersects the cone, and the - sign to the case when it fails to intersect And that completes the derivation of the list.
the cone.
Smooth Actions of Lie Groups on Smooth Manifolds
§3.
We wish now to specialize the idea of an action
1)
of a group
G
M
on a set
to the case when all three objects are "smooth", in a sense to be made precise.
First, we consider the group.
G which is a smooth
A Lie group is a group
mnnifold, and for which multiplication
G x G
->
G
and inversion
G -> G are
smooth mappings.
The general linear group
Example I
GL(n)
is a Lie group.
To see this
it is convenient to identify it with the group of all non-singular real matrices; real
it is therefore an open subset of the vector space
n x n matrices, hence a smooth manifold.
matrices in
M(n)
smooth.
GL(n)
of all
Further, multiplication of
is a polynomial mapping, so certainly smooth, and its res-
triction to matrices in inversion in
M(n)
n x n
GL(n)
will likewise be smooth.
Finally, matrix
is a rational mapping (with nowhere zero denominator) so
Note incidentally that the tangent space to
GL(n)
at any point
can be identified with M(n).
Example 2
The reader is left the task of checking that the cartesian
product of two Lie groups is again a Lie group.
In particular the product
of two general linear groups will again be a Lie group.
73
By a smooth action of a Lie group action
iD
:
G x M -4
M for which
G
1)
on a smooth manifold M we mean an is smooth.
The reader will readily
check that the geometric examples given in the previous section are smooth actions of Lie groups on smooth manifolds.
We intend to restrict ourselves
entirely to the case when all the orbits are smooth submanifolds of M.
In
fact this turns out to be hardly any restriction at all, and it can be proved (though not too easily) that our geometric examples satisfy this condition.
Further information on this point can be found in Appendix B. under the natural action of
GL(n) x GL(p)
H1(n, p)
on
the partition by rank of linear maps ]Rn - ]Rp,
For instance,
the orbits provide
and we saw in Chapter II
that these are smooth manifolds.
In practice one needs to know how to compute the tangent space to an orbit at a point.
The procedure is based on the following proposition.
Let
(3.1)
smooth manifold
folds of
M.
:
G x M -> M.
M be a smooth action of a Lie group
For any point x e M the natural mapping i x : G - G.x of
I claim that
x
rank at the identity element
defined by
by y - h.y.
g - hg,
g - g.x
is a submersion.
has the same rank at every point in
x
fices to show that the rank of
G
on a
It is assumed that all the orbits are smooth submani-
the group onto the orbit given by
Step 1
G
I E G.
and let
0
at any point Let
0
h E G
G.
It suf-
coincides with its
denote the diffeomorphism of
denote the diffeomorphism of
M defined
The commuting diagram of smooth mappings on the left then gives
rise to the commuting diagram of differentials on the right.
The vertical
arrows in the diagram of differentials are linear isomorphisms, so T1i x The
74
x
have the same rank, as was claimed.
and
G.x
G
y G
Step 2
iD
X
Tx8
ThIx
y T G h
4,
G.x
In view of Step 1 it suffices to show that Ix
some point in
G,
Tx(G.x)
T18
8
8
x
T1G
y >Th.x(G.x)
is submersive at
and that follows immediately from Sard's Theorem, guaran-
teeing the existence of at least one regular value, necessarily in the image
.
of
tial TPx of the tangent space
is the image under the differen-
Tx(G.x)
Thus the required tangent space
T1G
to the group at the identity element.
And that is how one goes about computing Tx(G.x)
Let us return to our class of geometric examples namely the
Example
natural action of F
:
IRn
in practice.
GL(n) x GL(p)
on the vector space
Hd(n, p).
Let
-+ MP be an element of this vector space, with components
x1, ..., n We
each a form of degree
d
in
shall compute the tangent space at
F
to the orbit through
f1, ..., fp
n
variables
mapping of the group onto the orbit is given by
(H, K)
-+
F.
The natural
K o F o H 1,
we require the image of the differential of this mapping at the identity i.e.
pings
the vector sum of the images of the differentials at H-+ F o
11-1
and K -+ K o F.
first the mapping H -+ F o H 1. matrix inversion
H -+ H 1
I
and I,
of the map-
We consider these separately.
Take
Observe first that it is the composite of
with the mapping H -+ F o H:
since the differen-
tial of the former mapping (at any point) is invertible we see that the
75
images of the differentials at I of H - F o H 1, H -+ F o H will coincide so we need only consider the latter mapping. differential as follows.
ej
the curves
yij
close to
to
0.
ei,
F(x1,
through
M(n)
GL(n).
]Rn
-+ Mn be the linear mapping
I
..., xj
with respect to
0
all other
zero.
g's
Example 4
to the orbit through xi ax,
and
n
in
d
f
..., gp)
with just one
t
of
variables
=
E xi
p = 1.
Hd(n, 1)
will be the subspace of
the required tangent space is just the subspace of
with
1
-,c
i.,
Let
f
and
be a f
spanned by the
By Euler's Theorem xi
d
1
xi ax.
fj,
The tangent space at
x1, ..., xn.
so lies in the subspace spanned by the
,
ax.
aF
ex
and the image will be
Ja
f
xi
equal to an
gi
In fact it is simpler than that.
f.
of the curves
0
I,
It is worthwhile isolating the case when
form of degree
which is the
this is the restriction of a linear
mapping, so will coincide with its differential at
G = (g1,
t
yield a basis
which by the Chain Rule are the
Next, consider the mapping K -+ K o F:
spanned by all
with
It follows that the image of the differential at
txi, ..., xn),
+
I
of all linear mappings IRn -, ]Rn,
the derivatives at
i.e.
Consider now
0.
given by yij(t) = I + toij
will be spanned by the derivatives at
H -+ F o H
F o yij,
:
Evidently the derivatives of these curves at
tangent space to of
Ai j
denote the standard basis vectors
and the remaining basis vectors to
GL(n)
in
for the vector space
I
e1, ..., en
1 s i, j 5 n let
in IRn, and for which maps
Let
We compute the image of the
IL
.
thus
1
H (n,
1)
spanned by the
j < n.
J
Example 5
By way of explicit illustration take the case of binary cubic
forms,
d = 3,
i.e.
n = 2,
us that the tangent space at
76
p = 1. f
And take
f = x2y.
to the orbit through
f
The above tells
will be spanned by
xay .1
xax ,
x3,
x2 y,
yax , a
which are respectively the monomials
so a basis for the tangent space is given by
x2 y,
2x2y,
xy2,
2xy2, It
x3.
follows that the tangent space, hence the orbit, has dimension 3.
Example 6
For an example involving linear systems consider pencils of
binary quadratic forms, and in particular the pencil Example 3 the tangent space at
by xax , y ax , x ay , y ay
F
F = (x2, 0).
to the orbit through
basis for the tangent space is provided by
will be spanned
F
together with (x2, 0),
(0, x2 ); hence a (xy, 0),
(x2, 0),
By
(0, x2)
so
the tangent space, and hence the orbit, is of dimension 3. Let us return to the general situation of a smooth action of a Lie group
G
on a smooth manifold M, supposing that all the orbits are
smooth submanifolds of point
x E M
: G x M - M
We can define the codimension
M.
to be the codimension of the orbit
G.x
in
cod x M.
of any
As we have
seen, one of the broad problems is that of actually listing the orbits.
In
practice this tends to be a difficult, if not virtually impossible, question to answer.
Experience suggests that the difficulty of listing the orbits of
a given codimension may well increase with the value of the codimension: indeed, this is clear on an intuitive level, since the larger the codimension the more space there is in which the orbit can twist and turn, and hence the On this basis it should
larger the number of possible types of behaviour. be most feasible to list those orbits of codimension This is worth re-phrasing.
G when there exists a neighbourhood x' E U
lies in the same orbit as
tly small perturbations of
x e M
One calls
x:
U
of
x
0.
stable under the action of in
M
such that every point
intuitively, this means that sufficien-
x will not displace it from its orbit, hence the
terminology.
Note that the homogeneity property for orbits implies that if
one point x
on an orbit is stable then every point on that orbit is stable, 77
and hence that the orbit is open in stated in algebraic terms.
has codimension zero. that
x
We call
This geometric notion can be re-
M.
x e M
infinitesimally stable when it
It is now a trivial exercise in linear algebra to see
is stable if and only if it is infinitesimally stable, so that in the
present context the distinction between the two notions is a fine one.
How-
ever it is worth making, because in studying the singularities of smooth mappings one comes across analogous situations where both notions can be defined, but it is decidedly hard to show that they coincide.
Consider again the natural action on binary cubic forms.
Example 7
that there are five orbits, with representatives have respective codimensions
x3 +
Recall
0, x3, x2y, x3 + xy2;
these
so we have two stable orbits, namely
1+, 2, 1, 0
Xy 2.
Example 8
By way of contrast consider the natural action on ternary cubic
forms,
cubic curves.
i.e.
Here there are no stable orbits.
One could
prove this ad hoc by systematically computing the codimensions of the normal forms listed in §2.
There is however a more illuminating argument.
group in question is
GL(3) x GL(1).
If we forget about the second factor
here we obtain another action by the group
GL(3)
alone.
these two actions give rise to exactly the same orbits.
We claim that What one has to
show is that multiplication of a ternary cubic form by a scalar be achieved by a change of co-ordinates:
of co-ordinates Now the group
x t+ ,1/3x, GL(3)
yr x1/3y,
has dimension
any point must be of dimension
9,
5 9.
of ternary cubic form has dimension 10. dimension
78
>.
1.
The
4
0
can
and this is clear since the change
z -, A1/3z
achieves precisely that.
so the tangent space to any orbit at On the other hand the space
H3(3, 1)
It follows that any orbit has co-
Let us continue with the general situation of a smooth action t : G x M - M.
We wish to expand on the question of just how the action a point
x s M.
behaves close to
(I
The crude picture is that the orbits through points close to
x flow smoothly together past
and a very basic idea in studying this
x:
situation is to look at the way in which the orbits out a (small) cross-section of this flow.
(See Figure 1.)
Fig. 2
Consider a smooth sub-
This crude idea can be made precise as follows.
manifold
S
M,
of
with x E X.
which is transverse to the orbit
We shall certainly have
(See Figure 2.)
least possible dimension for the dimension of
S
identity, the slice
and in that case we call
is cod x,
S
M
given by m H g.m,
of
(3.1)
x
in
smooth manifold
:
G x M M,
following property:
-,
M
and let
smooth submanifolds of
x.
close to the
g E G
the basic pic-
will sweep out a small neighbourhood
S
M.
be a smooth action of a Lie group
The point
is a diffeomorphism onto
at
on a
x has a neighbourhood U with the
there is a smooth submanifold H S
G
It is assumed that all the orbits are
x e M.
identity element, and a slice H x S
with
when
i.e.
a slice at
S
the
The formal mathematical statement of this reads as follows.
M.
Let
dim(G.x):
+
will slide slightly along the orbits:
ture which we wish to convey is that U
dim S
.
will be when we have equality,
S
Under the diffeomorphism of
dim M
G.x.
x,
of
G
containing the
for which the restriction of
to
U.
79
G. x
For obvious reasons such a neighbourhood structure.
One thinks of
U
U
is said to have a product
as a small tube with the orbit
G.x
running
down the middle.
We start by isolating a lemma which will be used twice in the main
Step 1
Suppose that
body of the proof.
manifold Aa, manifold
and that of
Cc
a = b + c.
with
A with
Indeed by
bourhood of
mapping to
0
c = a - b
then
we claim that there exists a smooth sub-
x E C
which is transverse to
and for which
B
(1.2.2) there is a diffeomorphism of an open neighonto a relatively open neighbourhood of
in IRa
0
x s B:
is a smooth submanifold of a smooth
Bb
x,
0 x]Rc
and IRb x 0
corresponding to
maps to a smooth submanifold
in
A
If we put
B. C
x
A with the
of
desired properties.
As we have already pointed out the natural mapping
Step 2
(I
x : G - G.x
of the group onto the orbit is a submersion, so by (11.1.2) the inverse image Gx
x will be a smooth submanifold of
under this mapping of the point
of codimension the dimension of the orbit, with Gx
G
is a subgroup of
to
Gx
with
the orbit 80
dim G
G.x.
=
(Incidentally,
1 e Gx.
known as the isotropy subgroup at
there exists a smooth submanifold
H
of
dim Gx + dim H,
G
i.e.
with H
1
G
By Step
x.)
I
E H which is transverse
has the same dimension as
Again by Step I there exists a slice
S
at
x.
Consider
now the restriction
T
H x S - M
:
of the action
Observe that domain
and target have the same dimension, and that the differential at invertible.
It follows from the Inverse Function Theorem that if
small enough then
of x
'P
maps
H x S
(1, x)
H, S
is are
diffeomorphically onto a neighbourhood U
in M.
The reason for proving this result is to justify the intuitive idea that in
order to study the action near x the orbits out a slice
S.
it will suffice to study the way in which
From a purely practical point of view this pro-
vides a considerable simplification since the dimension of the slice well be very small indeed compared to that of the manifold
S
may
All this
M.
brings us to one of the central ideas of the subject, namely that of an "unfolding" of
roughly speaking a parametrized slice at
x.
Transversal Unfoldings
§4. Let
fold
x,
:
M.
G x M - M be a smooth action of a Lie group and let
x e M.
:
By an parameter unfolding of x we
M.
(7Rr, 0) - (M, x)
is transverse to the orbit
on a smooth mani-
As usual, it is tacitly understood that all the
orbits are smooth submanifolds of
mean a germ X
G
G.x
at
:
x
X
is said to be transverse when it
i.e.
T0X (a") + Tx(G.x)
=
TxM.
One pictures the situation something like this:
81
X .0
Now let
x have codimension
it follows immediately that
r = c
is a fairly straightforward matter.
X
the case when M foldings,
i.e.
is a transverse unfolding
we call
X
a minimal trans-
Let us restrict ourselves to
is a linear space, so it is natural to look for linear unthose of the form
X(u1, ..., u0)
where
when
r > c:
X
The construction of an explicit minimal transversal unfol-
versal unfolding.
ding
Given that
c.
b 1, ..., be
=
x + u1b1 + ... u b c
are fixed vectors in
M.
a
The condition for such an
X
to be transverse is that
IR{b1,
...,
bc} + Tx(G.x) = M
where the first term represents the subspace spanned by b1, ..., bc, the image of the differential of choose
b1, ..., b
0
X
at
0.
i.e*
Thus all we need to do is to
to be a basis for a supplement of the tangent space
Tx(G.x).
Example I
Consider the action of the product group
vector space
H3(2, 1)
GL(2) x GL(1)
of all binary cubic forms, discussed in §2.
compute transversal unfoldings for the non-zero normal forms. H3(2, 1)
has as basis the monomials
x3, x2y, xy2, y3.
on the
We shall
Recall that
We can compute the
tangent space to the orbit at each normal form using the results of §3. 82
Bases for these tangent spaces can be found in the table below.
It is a
happy accident that in each case basis vectors for the tangent space can be chosen from the list of basis monomials, and the list of remaining monomials provides a basis for a supplement. Transversal unfoldings are then provided by the formula given above.
Normal Form
Basis for tangent space
Transversal Unfolding
3.
x3 ± X Y
x3 + xy2
9x2 2y, x3'2, y3
x3
x 2y
, x2 y,
xy2
2
x y+
2 x3 + uxy + vy3
x3, x2y
x3
u3
Let us pursue this example further to see just how the transversal unfolding can yield rather explicit information concerning the action close to a point.
We shall describe the action close to
moves over the plane
IR2
so the transversal unfolding
cut all orbits through binary cubics close to values of type. 3ux 2
u, v
As the point
x3.
(u, v)
x3 + uxy2 + vy3
will
We shall determine the
x3.
for which the unfolding represents a binary cubic of a given
The associated Hessian quadratic form of the unfolding is 2 2 + 9vxy - uy
with discriminant a multiple of
is the zero quadratic form if and only if (u, v)
u = v = 0:
so the origin in the
plane is the only point which corresponds to a perfect cube.
now stay away from the origin.
=
0:
i.e.
when the discriminant
this is the equation of a cuspidal cubic curve in the plane.
(See the picture below.) More precisely:
Let us
The unfolding will represent a parabolic bi-
nary cubic when the Hessian is parabolic, v2 + 1+u3
The Hessian
27v2 + 1+u. 3
At all other points in the plane
inside the cuspidal cubic we have
v2 +
1+u3
v2 + <
0
1-u3
0.
correspon-
ding to elliptic binary cubits, and outside the cuspidal cubic we have 83
v2 + 4u3
>
0
corresponding to hyperbolic binary cubics.
Thus we arrive
at the following picture, representing a cross-section of the flow of orbits close to
x3.
Some more interesting situations are provided by the transversal unfoldings of cubic curves.
Example 2 f
We shall study the transversal unfolding of the cubic curve
x(x2 + yz),
=
affine plane the conic is a parabola x = 0.)
f
and the chord is its axis
y = -x2,
A minor computation shows that the tangent space to the orbit at
is spanned by the monomials of degree so
3
in
y3, z3
with
x, y, z
f
deleted,
has codimension 2 and a transversal unfolding is provided by
F
=
x(x2 + yz) + uy3 + vz3.
As in the previous example we ask for the values of a given type.
Certainly, when u = 0, Suppose
and chord type.
u ,-E 0,
zero one finds a singular point at that the tangent lines there are Likewise when hand when
u = 0,
u X 0,
v / 0
v p( 0
v = 0:
v = 0
x = 0,
y = 0
u, v
f or which
F
has
we have the original conic
putting
(0, 0, 1),
2x ,
-r ,
8z
equal to
and further computation shows so one has a crunodal cubic.
one obtains a crunodal cubic.
On the other
are sufficiently small the curve is non-singular,
as a few lines of working will verify.
84
(In fact in the real
which is of the conic and chord type.
Thus we obtain the following picture,
representing a cross-section of the flow of orbits close to the conic and chord type.
non-singular erunodaZ
Another pleasant illustration is provided by the transversal
Example 3
f = xyz
unfolding of the cubic curve tangent space to the orbit at x, y, z
with
In this case the
is spanned by the monomials of degree 3 in
f
deleted, so
x3, y3, z3
of triangle type.
f
has codimension 3 and a transversal
unfolding is provided by
F
=
We ask first for the values of there exist
xyz + ux3 + vy3 + wz3.
u, v, w
aF OF
8z
or
=
yz + 3ux2
=
0
=
xz + 3vy2
=
0
=
xy + 3wz2
=
0.
v = 0,
or w = 0
Notice that if
u = 0,
is satisfied.
Let us see what happens when u
that if one of
x, y, z
z
is singular,
i.e.
not all zero for which
x, y, z
y/0,
F
for which
0.
then certainly this condition 0,
v
0,
w
0.
Observe
vanishes then all of them vanish, so we can suppose
But then our equations yield
85
=
(27uvw + 1)x2 y2 z2
0.
We are only considering what happens very close to
so we can suppose
x2y2z2
all / 0
=
which is impossible.
-
0
and small then
what happens in the plane
u = 0
in
(u, v, w) space.
are zero we recover the original triangle:
v, w
zero we have the conic and chord type; obtains a crunodal cubic. plane
v = 0,
are
Let us look in more detail at
is non-singular.
F
u, v, w
In other words if
u,
4 0,
to be very small, which means the expression within brackets is
v, w so
f,
Of course when both
when just one of
and when both
v, w
are
v, w
is
one
0
By symmetry one has similar situations in the
and the plane w = 0.
The nett result is the following pic-
ture of a cross-section of the flow or orbits past a cubic curve of triangle type.
crunodal conic and line
For an example involving linear systems let us look at pencils
Example 4
of binary quadratic forms. pencil (y2, 0),
F = (x2, 0).
tangent space at (0, x2)
(y2, 0):
by 86
The space of pencils is spanned by (0, xy),
(0, x2),
F
We shall study the transversal unfolding of the
(0, y2).
to the orbit through
F
(xy, 0),
In Example 6 of 3 we saw that the F
is spanned by
so a basis for a supplement is provided by thus
(x2, 0),
(x2, 0),
(0, y2),
(0, xy)
(xy, 3),
and
has codimension 3, and a transversal unfolding is given
(x2, 0) + u(0, y2) + v(0, xy) + w(y2, 0)
i.e.
(x2 + wy2,
uy2 + vxy).
The objective now is to determine the values of
u, V. w
cil has one of the seven possible types listed in 2. fied the space of binary quadratic forms with IR3 with
ax2 + 2bxy + cy2
the cone
b2 = ac,
for which this pen-
Recall that we identi-
by identifying
(a, b, c), that the singular forms then correspond to
and that the types of pencil are separated by the way in
which they lie relative to this singular cone.
The first thing is to deter-
mine when the pencil represents a line through the origin:
tion shows that the binary quadratics
x2 + wy2,
endent if and only if
i.e.
space.
v = 0,
In that case when w = 0
tangent to the cone; w < 0
u = 0,
when w > 0
a line outside the cone.
(pv)2
=
along the w-axis in
F,
a line
we have a line inside the cone, and when Now we ask what happens off the w-axis, when
i.e.
4,.(Aw + µw)
To this end we consider how the
Ax2 + pvxy + (Aw + µu)y2,
+ 4u)µ. - v2µ2
u2 + wv2
discriminant, and is a surface in IR3,
=
0
*
0.
=
This expression is itself a binary quadratic in The equation
which is singular
i.e.
4w.12
16(u2 + wv2 ).
(u, v, w)
A typical element of the pencil is
A(x2 + wy2) + p(uy2 + vxy)
when
are linearly dep-
we recover the original pencil
our pencil is a plane through the origin. plane cuts the cone.
uy2 + vxy
a simple computa-
A, u
with discriminant
represents the vanishing of this
called the Whitney umbrella.
Note
that it includes the whole w-axis, sometimes dubbed the handle of the umbrella.
On this surface, but off the handle, the equation : N
for a solution so the pencil meets the cone
represents a plane tangent to the cone. u2 + wv2 < 0
so the equation
*
*
has a unique ratio b2 = ac
in a line, so
Inside the umbrella one has
is satisfied by no real ratio
A : µ, which
means that the pencil represents a plane which does not intersect the cone.
87
plane not intersectin cone
plane tangent to cone
line outside cone line inside cone plane intersecting cone
And outside the umbrella
u2 + wv2 > 0,
two distinct real ratios
A : p
so the equation
*
is satisfied by
which means that the pencil represents a
plane intersecting the cone in a pair of lines through the origin.
Once
again then we have obtained a simple picture representing a cross-section of the flow of orbits.
Of course, as we have presented them, all these examples provide little more than amusing exercises in elementary geometry. than that;
But there is more to it
there are deep questions in singularity theory, lying beyond the
scope of this book, which one can only answer by going into the geometry of transversal unfoldings in considerable detail.
Our objective was simply to
lay bare the underlying ideas involved and to give the reader some feeling for the mechanics of the matter.
Later in this book we shall meet analogous
situations, lying just outside the present framework, where the idea of a transversal
unfolding can be used to study the possible ways in which a germ
of a smooth mapping can be deformed.
88
Versal Unfoldings
§5.
We shall conclude our discussion of finite-dimensional unfoldings by expanding somewhat on the sense in which a transversal unfolding describes an action this will provide us with a useful characterization of
close to a point;
stable points in terms of their unfoldings.
To this end we introduce a series
is a smooth action of a Lie group
G
on a smooth
of notions.
As usual,
manifold
and it is assumed that all the orbits are smooth submanifolds of
M,
We let
M.
iD
x e M.
Equivalence of Unfoldings Two r-parameter unfoldings
X1, X2
x
of
there exists an r-parameter unfolding tity element
I
I
are said to be equivalent when :
(IRr, 0)
-+
(G, 1)
of the iden-
in the group for which
X2(u)
=
I(u).X1(u).
Pictorially, the idea is that you can get from X1
to
X2
by sliding smoothly
down the orbits.
X1
e X2
0
Induced Unfoldings
X
Suppose that H
:
(IR5, 0)
folding of
.
x,
is an r-parameter unfolding of (IRr, 0)
is a germ.
said to be induced by
x,
Then Y = X0H H.
and that is an s-parameter un-
In this situation we refer to
H 89
as a change of parameter and write
Y = H X.
One pictures the situation
something like this.
X
A, HI
Y
M
Morphisms of Unfoldings
Let
X, Y
is a pair
be r, s-parameter unfoldings of with
(H, I)
with H : (]Rs, 0)
-->
I
A morphism from X
x.
Y
to
an r-parameter unfolding of the identity 1, and
(]Rr, 0) a change of parameter, for which X is equi-
valent to the induced unfolding H*Y
via
I.
When
r = s,
and
H
is inver-
tible we call the morphism an isomorphism.
Versal Unfoldings
An unfolding Y
of x is said to be versal when for any unfolding X
there exists a morphism from X
to
Y.
of
Intuitively, this means that
sufficiently large to allow all unfoldings of
x
to appear in it.
Y
x is
A versal
unfolding of minimal dimension is said to be universal.
The basic fact about unfoldings is the following result connecting the algebraic notion of transversality with the geometric notion of versality.
(5.1) M,
Let '
be a smooth action of a Lie group
all of whose orbits are smooth submanifolds of
icient condition for an unfolding X : 90
G
on a smooth manifold
M.
(]Rr, 0) -+ (M, x)
A necessary and suffto be versal is that
it should be transversal.
In which we establish necessity
Step 1
We have to show
sal.
X
of the condition.
is transversal,
TxM C T
0
that
i.e.
(1)
(]R, 0) - (M, x).
from Y
(H, I)
ver-
X (]Rr) + Tx(G.x).
Consider any 1-parameter unfolding Y : there exists a morphism
X
Suppose
to
X,
i.e.
X
As
is versal
Y(v) = I(v).X(H(v)).
Take differentials to get
T Y (IR) 0
C T0Y (IRr) + Tx(G.x).
(2)
(1) follows immediately from (2) since any tangent vector in gent vector to some curve through
Step 2
x,
We wish to show
X
versal.
is a tanY.
X
Suppose
Observe first that it will suffice
to show that some unfolding induced by X X
x
a 1-parameter unfolding
i.e.
In which we establish sufficiency of the condition.
transversal.
T M
is versal.
Let us therefore replace
by an induced unfolding J*X which is both transversal and minimal, and
continue to denote it by the same letter
linear mapping into Mr, T0X
taken by
(One can choose
X.
J
to be a
with domain of the correct dimension, whose image is
to a supplement for
Tx(G.x)
in
TxM.)
Such a germ
X
is
immersive, and the image of a sufficiently small representative will be a slice S
at
i.e.
x.
Recall now that
has a neighbourhood with a product structure,
there exists a smooth submanifold
ment
1
for which the germ
is invertible.
x.
x
'
at
( 1 ,
H x)
of
G
of the restriction of
Consider now an arbitrary unfolding Y
We define germs J, Z by J = IIH 0 T-1 0 Y,
SH, HS
through the identity ele-
are the germs at
(1, x)
Z=
of the projections of
:
1?
to H x S
(Es, 0) - (M, x)
11 S 0
H x S
-1
of
o Y where
onto
H, S
91
It is immediate that
respectively.
K
J(u).Z(u)
=
(sr, 0) -+ (7Rr, 0) by K = X 1 o Z we see that
:
Y
so
is equivalent
On the other hand if we define the change of parameter
Z.
to
Y(u)
induced from
and conclude that
X,
X
It follows that
(K, J)
is an unfolding
Z = K *X
is a morphism from Y
to
X.
is a versal unfolding.
An immediate consequence of this characterization is the following proposition justifying the use of the prefix in the term "universal".
Under the hypotheses of (5.1) any two universal unfoldings
(5.2)
x
X, X'
of
are isomorphic.
Suppose
Proof
X, X'
are r-parameter unfoldings.
is equivalent to some induced unfolding
And as
H*X.
X
As
is versal,
is universal
X'
subspace of that
TXM,
so the image of
is invertible, and hence
TH
T(X o H)
TH H
=
TX o TH
is an r-dimensional
is likewise r-dimensional: is invertible.
H*X
That means
is universal, hence is a minimal transversal unfolding by (5.1). that the image of the differential
X'
it follows
The result follows.
Finally, as promised at the beginning of this section, we shall make use of
To this end
these ideas to give a simple characterization of stable points. let us call an r-parameter unfolding
to the r-parameter constant unfolding
X
of
x
trivial when it is equivalent
(Mr, 0) -+ (M, x)
given by
u -+ x.
We then have
(5.3)
Under the hypotheses of (5.1) a necessary and sufficient condition for
an element
Necessity
92
x e M
to be stable is that every unfolding of
x
should be trivial.
Suppose x is stable, and that X is an r-parameter unfolding.
We have to show that X is equivalent to the r-parameter constant unfolding. Since
x
is stable the constant 0-parameter unfolding is transversal, hence
versal by (5.1).
Thus
X
is equivalent to an r-parameter unfolding induced
from it, which is necessarily the constant r-parameter unfolding.
Sufficiency
Conversely, suppose that every unfolding
valent to a constant unfolding.
In particular, take
unfolding.
Certainly then some representative of
orbit
so
G.x,
dimension
0,
X
X
X
X
of
x
is equi-
to be a transversal
is a mapping into the
can only be transversal to the orbit when
i.e.
x
G.x
has co-
is stable.
F7
The reader should be warned that the theory of this section is not particularly useful.
Its virtue lies rather in the fact that it is an easily
understood model for the ideas to be used in Chapter V.
93
IV Singular points of smooth functions
Some Basic Geometric Ideas
We come now to the meat of this book, the study of singular points of smooth mappings.
In accordance with our philosophy of treating the simplest situa-
tions first we shall restrict this chapter to the case of singular points of smooth functions
f : N -IR with N a smooth manifold.
And we shall sim-
plify life even further by studying germs of such functions under a rather finer notion of equivalence than that introduced in Chapter I:
at first sight
this may seem to be a complication rather than a simplification, but in fact it will enable us to finesse various algebraic difficulties. germs of functions
exists an invertible germ mutes.
right-equivalent
f1, f2 g
We call two
(or .R-equivalent)
when there
for which the following diagram of germs com-
Our broad intention
can now be formulated rather more accurately by saying that we wish to classify germs of functions under the relation of $-equivalence.
As such our pro-
gramme is far too ambitious, but we shall see that we can gain some distance by using a little common sense.
To avoid unnecessary symbolism we shall,
throughout this chapter, adhere to the convention that equivalence of two germs of functions is to mean 9Q-equivalence=
94
since we shall have no occasion to
refer to the more general notion there should be no confusion on this point.
And it will be useful to write
to mean that two germs
f2
f1
f1, f2
of
functions are equivalent.
The starting point is to try and set up our problem in such a way that it looks more like a problem which we know how to handle: we can proceed by analogy.
the hope then is that
That is the object of the present section.
setting up the basic geometric ideas we intend to be deliberately vague;
In the
reason for this is that any systematic theory covering the situation we wish All
to study would lie far and away above the intended level of this book.
we shall do is to argue heuristically to arrive at precise interpretations of vague geometric ideas:
we shall never use such arguments to prove propositions.
Our basic objects then are germs at
x we see that we can suppose
jects we wish to study is the set Of course
Sn
:
(N, x)
N = ]Rn,
En
(]R, y).
and x = 0.
of all germs
In fact
f
Taking a chart a Thus the set of ob-
(IRn, 0) - (]R, y).
:
&n has more algebraic structure than
the operation of multiplication on ]R
plication on
->
is a real vector space under the natural operations of addition
and scalar multiplication. that;
g
under which
8n,
Sn
induces an operation of multi-
becomes a real algebra.
More information
on the subject of real algebras can be found in Appendix C of this book.
Next, let Rn
denote the set of invertible germs
observe that Rn acts on
n
g :
(]Rn, 0) + (]Rn, 0), and
Now Rn
is a group under the operation of composition.
by composition,
(g, f) H f o g 1.
i.e.
we have an action R
Further, two germs in 9 n
n
x 9 n
-,
n
given by
will be equivalent if and
only if they lie in the same orbit under this action.
This observation, how-
ever trivial, is crucial in that it sets the scene in which we are to work.
However, pursuing the analogy with Chapter III is not quite that simple. What one would like to do is to introduce the "codimension" of a germ, and start the problem off by listing the orbits of fairly low codimension.
The 95
stumbling block here is that we have as yet only a group acting on a set,
whereas we would like to have a Lie group acting smoothly on a smooth manifold. It is an unfortunate fact that
and R n
& n
are not smooth manifolds in the That need not be, and
sense in which that word has been used in this book.
What we shall do is just to pretend
will not be, an insuperable difficulty.
that the action of Rn
on
6n
is a smooth action of a Lie group on a smooth
manifold, and proceed by analogy. This goes as follows.
the orbit through of
En,
f.
f
Choose a germ
in
We shall pretend that R.f
and look for a vector subspace of
gn
and write just R.f for
9n,
is a "smooth submanifold"
which has some reasonable claim
to the title of the "tangent space" to the orbit R.f
To do this we
f.
The natural mapping of the group onto
just mimic the theory of Chapter III.
the orbit is the mapping Rn - &n
at
g N f o g 1,
given by
and one would expect
the required "tangent space" to be the image of the "differential" at the identity 1 of this mapping.
g' 4 g'- g
A preliminary simplification is to observe that
f o g 1 is the composite of g -+ g 1
1
and
g F+ f o g:
one expects
to be a "diffeomorphism", so the required image should coincide with
that of the "differential" at the identity of
g - f o g.
Of course, one
would expect this "differential" to be a linear mapping between vector spaces, and the first thing to get straight is just what these vector spaces should be.
The domain should be the "tangent space" at the identity to the group Rn.
A
Observe that
Rn
little common sense will soon produce a candidate for this.
is a subset of the vector space &n
n
of all germs
moreover, we should really think of Rn
g :(IRn,o)
as an "open" subset of
(IRn, 0):
n
n
because
if we disturb an invertible germ very slightly we still expect to get an invertible germ.
to Rn,
On this basis we would certainly expect the "tangent space"
at any point, to be just &
n.
n 96
The target of the "differential"
should be the "tangent space" at we would expect this to be just linear mapping
do n - 8n.
for the "differential".
the components of
g
=
I + tg with
"differential" to
g
t
0
t
at
1:
The result of applying the sought for
0.
Just doing this naively
f(x1 + tg1, ..., xn + tgn). g1 ax
of the curve
0
the "differential" with res-
i.e.
t --+ f(y(t))
by the Chain Rule one gets the answer
at
given by
.Rn
should then be the "differential" at
pect to
n
Now we can realize this
g1, ..., gn.
of the curve through 1 in
close to
in n given by
R
an arbitrary "tangent vector" to
will be written
f
of
Thus our "differential" should be a
We start with an arbitrary germ
i.e.
through
0
is a vector space
The problem now is to find a sensible formula
germ as the "differential" at y(t)
9n
The line of attack to take here is suggested by
(ILn, 0) _ (IR', 0)
:
to n, and since
9n.
Example 3 in §3 of Chapter III. g
x
To sum up:
gn 8x .
+
we
n
1
80 n
expect the required "differential" to be the linear map
->
n
given
by the formula
(911 ..., gn) H
'f
'fn
+ ... + gn ax
91 ax
1
The required "tangent space" should be the image of this linear mapping, and The germs
this is neatly described in algebraic terms as follows. g1, ..., gn
all have zero target, so belong to the ideal
en
comprising all germs with zero target.
in
En
Let us denote by
generated by the partial derivatives
ax 1
the Jacobian ideal of
f.
the product ideal .An.Jf.
in the algebra
, , nn
The image of the linear mapping
Jf
,
*
the ideal
and call it above is just
In this way we come to the conclusion that a good
candidate for the "tangent space" to the orbit
f
However, that is not quite the end of the story.
is
nJf. We have described our
model in the language of germs, but could just as well have worked with
97
functions defined on a neighbourhood
U
germ case as the limiting case when U case we should replace
functions defined on
0
in IRn,
and thought of the In that
is infinitesimally small.
by the vector space
dn U.
of
G"O(U,
of all smooth
IR)
Rn by the group
And we should replace the group
of all diffeomorphisms U - U under the operation of composition.
Diff (U)
We would then have an action of Diff (U) on e (U, formula as before.
IR)
given by the same
And the same heuristic reasoning as before would bring
us to the conclusion that the required"tangent space" should be thought of as the image of a linear mapping given by the formula crucial difference.
In the group
origin, whilst in the group
Diff (U)
property, no matter how small model is oversimplified. gn
,Rn
U
*.
Note however one
our germs are forced to preserve the our diffeomorphisms need not have this All this indicates is that our
is.
We shall correct matters by allowing our germs
to have an arbitrary target, so to lie just in
.0
n
rather than
The only difference this will make to the final answer is
its ideal
that the image of
*
will be simply the Jacobian ideal
Jf.
On the basis of this heuristic reasoning we introduce the following formal Let
definition.
Jacobian ideal
Jf.
f e gn.
We define the tangent space to f to be the
And we define the codimension
codimension of the tangent space
cod f If
cod f
is finite we say that
compute the codimension.
=
f
8n,
of
to be the
i.e.
dim &n/Jf. is of finite codimension;
f
otherwise,
f
is of finite codimension, and if it is how to
There are some subtleties here which are worthy of
explanation, and require a detailed discussion of the algebra
98
f
In practice we shall need to know how to decide
is of infinite codimension. whether or not a given germ
in
Jf
cod f
&n:
this then
is the object of the next section.
The Algebra
§2.
gn
We have already observed that n is indeed a real algebra, in the technical sense of the word.
(See Appendix C.)
In this section we shall concern our-
selves with purely algebraic matters so as to provide a convenient reference for succeeding sections.
(2.1)
Observe first of all the following elementary fact.
A necessary and sufficient condition for a germ
vertible (as an element of the ring
Suppose
Necessity fg = 1:
then
inverse for
is that
f(O)
0.
is invertible, so we can find a germ
f
f(O)g(O)
=
1,
so
f(O)
f E n satisfies
Suppose
Sufficiency
n)
f e n to be in-
g e n with
0.
f(O) ,
0:
then
g = 1/f
is an
f.
of all germs in &n with target
This tells us that the ideal
has
0,
a rather special algebraic property.
(2.2)
Proof f E I
,/n is the unique maximal ideal in n Suppose
with
is an ideal with .,en C I C n,
I C g
f(O) t 0.
n f
follows that n is maximal.
so we can find an
is invertible by (2.1) and hence
It
I = 9n.
And the same argument establishes uniqueness.
lies in the fact that it
To some extent the importance of the ideal
n
allows a convenient algebraic description of certain ideals in gn which frequently come into consideration.
We have in mind the idea
2k+1
of all
99
f E dn
5 k
whose k-jet is zero (i.e.
vanish at
little fact
0).
-
all partial derivatives of
f
of order
As a preliminary we shall establish one extremely useful
sometimes called the Hadamard Lemma
-
which we shall also
have occasion to use in later chapters.
(2.3)
Let
U
be a convex neighbourhood of
smooth function defined on smooth functions
U x
f where
x1, ..., xn
which vanishes on
IRq
on U x IRq
f1, ..., fn
in IRn,
0
0
and let x IRq:
f
be a
there exist
with
xIf 1 + ... + xnfn
=
are the standard co-ordinate functions on IRn.
Proof f(x1, ..., xn, Y1, ..., Yq)
=
7t
J
0
(tx1, ..., txn) Y1, ..., yq)dt
xi axi (tx1' ...,
J0
xifi(x1,
txn, Y1, ..., Yq)dt
..., xn, Y1, ..., Yq)
i=1
if we take
fi(x1, ..., xn, Y1, ..., Y ) q
Now we can characterise the ideal
(2.4-)
3k =
J
0
ax
(tx1, ..., txn, Y1, ..., y )dt
i
q
7k+1'
(fin, and is generated by (the germs at
in x1, ..., xn of degree 5 k.
100
=
0
of) the monomials
First, we establish 2k
Proof
It is clear that ./l n C 2k, follow by induction from
that
f
that they lie in cular, that
21
thus
2k-1 =
It follows that
2k =.,Wn
mials of degree
k
is trivial.
k = I
3k C
n
,
which will
This we prove using the Hadamard
Indeed if
f E 2k
then
f(0) = 0,
so
and it is clear from the construction of the
f E J/ k-1.
fi
This argument shows, in parti-
is generated by (the germs at
.41n
in
so it suffices to show
q = 0.
x1 f1 + .., xnfn,
=
The case
.
n2k-1.
2k
Lemma, in the special case
n
=
is generated by (the germs at
x1, ..., xn.
of)
0 0
of)
the mono-
x1, ..., xn.
In particular (2.4-) shows that the ideals
This is worth remarking on because the ring every ideal is finitely-generated.
2k &n
are all finitely-generated. is not Noetherian,
not
i.e.
(We shall give an explicit example of
such an ideal shortly.)
At this juncture it is worth saying something about the exact connexion Of course a Taylor series is just a
between germs and their Taylor series. formal power series in several variables.
Given a germ
f
in n we shall
write its Taylor series as
f(0) + 11
so
f
ates
x1, ..., xn.
of algebras.
(2.5)
Ea
(0)xi + 2!
sJ
(O)xixj + ...
the algebra of formal real power series in
is in Vin,
given by f
a-LL xi
f,
n
indetermin-
In this way we obtain a natural mapping n -a
an
and one can check easily ennugh that it is a homomorphism
What is by no means so clear is the Bnrel Lemma.
The algebra homomorphism
-*
n
given by
f H f
is surjec-
tive.
101
The proof of the Borel Lemma is a slightly involved piece of analysis which
we have isolated in Appendix D of this book so as not to interrupt the flow of Notice that the kernel of the homomorphism
relevant ideas,
00n ilk eOo =k=1 n n
precisely
by (2.4.).
9n/-/4n
n in 9n
an is
It follows that =
'gn
We can get a more finite version of this, as follows.
phism n - -n
Cn ->
Under the epimor-
the maximal ideal n in 9n maps to the maximal ideal
comprising all formal power series with zero constant term, and
hence any power .Aln maps to the power J
en/fin
The nett result is that
J.
an/ n
The particular virtue of this relation is that it tells us that the quotient
Indeed .,lfn is generated by the in-
space on the left is finite-dimensional. determinates
x1, ..., xn
Ak
so .,lln is generated by the monomials of degree
k
in x1, ..., xn and the quotient space on the right can be identified with the real vector space of polynomials of degree
< k
in
x1, ..., xn,
which
is certainly of finite dimension.
In order to make further progress we shall require a rather pretty result from algebra, called the Nakayama Lemma.
(2.6)
Let
be a commutative ring with an identity element 1, and let
&
,,4rbe an ideal in
any
x E ,,'
9 with the property that
Let
b1, ..., bt
102
is invertible in
Further, let M be an C-module, and let
with A finitely generated.
Proof
I + x
a1, ..., at in
B.
If A C B + .A'.A
be generators for A.
and elements
AiJ
in
& for
be d-submodules
A. B
then A C B.
By hypothesis we can find
4 for which for
I
< i < t
we
can write
ai
Introduce a a
=
(a1,
t x t
with
& by A =
matrix over the ring
..., at),
M x ... x M,
bi + Ai1.a1 + ... + Ait.at
=
b t
=
(b1,
Then
I
is the identity
to show that
=
b
matrix over the ring
t x t
and take
can be re-written as
*
(I - A)a
where
:Lj
to be elements of the 8-module
..., bt)
factors.
( A.
It will suffice
d.
I - A is an invertible matrix, since then we can solve this
system of linear equations for the
which will show that
a1, ..., at
in terms of
a1, ..., at
lie in
B,
b1, ..., bt,
as was required.
To this end
recall from linear algebra that a square matrix (over a commutative ring with an identity) is invertible if and only if its determinant is an invertible in So it suffices to show that
the ring.
det(I - A)
is an invertible in d.
Observe that
det(I - A)
say, with
A e
=
1 _ (sum of products'
l of elements in_4tl
And by hypothesis
In practice the rings
8
I - A
is an invertible in
we have in mind are
dn,
n
9.
and the ideals
are n, In both of which satisfy the initial hypothesis of (2.6).
The Nakayama Lemma allows a very simple proof that n is not
Example 1
a Noetherian ring.
We take
00 A
= -,/,/ co
=
n .'n ,
B
M
to be the Sri module n And we take
the trivial ideal.
Clearly A C B + ,,/f .A.
If
n k=1 n were Noetherian then A would be finitely generated, and the Nakayama
Lemma would tell us that A C :B,
i.e.
that
A
is trivial.
However this
103
is false since there are standard examples in calculus of non-zero germs with
It follows that En
zero Taylor series.
Example 2 B
=
<x2
In + y3,
consider the ideals
g2
Clearly
y2 + x3>.
so need to show is the case.
M =
cannot be Noetherian.
B C A.
A
It is not immediately obvious to the eye that this
A C B.
Here the Nakayama Lemma provides an easy answer, because it
is generated by x3,
x2y,
xyy2,
I C M
We say that
M.
tient space
M/I,
i.e.
cod I
M/I
is finite-dimensional,
of
in
I
M
i.e.
the sub-
And in that case we
M.
to be the dimension of the quoThe point of the
the dimension of any supplement.
next proposition is that it gives us a useful algebraic criterion for be of finite codimension in
(2.7)
Let
g-submodule.
Sufficiency
I
to
M.
M be an &-module with a finite basis, and let A necessary and sufficient condition for
codimension in M
Let M
has finite codimension
I
admits a finite-dimensional supplement in
define the codimension
,,an.
,
be an &-submodule, so in particular a vector
subspace of the real vector space
in M when the quotient space
.4f2A
d will denote
In the following
dn, n and .,fl the corresponding maximal ideal A
be an 9-module, and let
I
And this is clear as
y3.
Now we can return to the main theme. either of
A = B,
We claim that indeed
tells us that it suffices to show A C B + -4f,2 A.
space
<x2, y2> and
=
is that there exists an integer k 3
Suppose there exi sts an integer k 3
1
I I
I C M
be an
to be of finite
with .Alk.M C I.
for which .,ffk.M C I.
Certainly then
dim M/I Now I claim that 101+
M/ lfkM
4
dim M/.,IIkM.
is finite-dimensional.
As
M
has a finite basis
we can suppose
M = d
s
for some integer
k
isomorphic with the product
s
k
x ... X &/
ds/// k.8s
Now
1.
>.
is naturally
with s factors, and since
(as we have already seen) each factor in this product is finite-dimensional the product is as well.
codimension in
Necessity
It follows that
dim M/I < co,
so
I
has finite
M.
Suppose
is of finite codimension in
I
Consider the
M.
descending sequence of d -submodules
I +,(l0. M D I
,,C11
.M2
... 2 I
Clearly, each strict inclusion in this sequence makes a contribution the codimension of
1
to
1
Since the codimension is finite the inclusions, from
I.
some point onwards, must all be equalities.
integer k >
>
In particular there exists an
for which
I
M=
I +,4f
k+1
which implies that .//k.M C I +
Since .,llk.M
it follows from the Nakayama Lemma that
tion ,llkM C I
ltk.M
C I,
is finitely-generated
as required.
As is clear from the final lines of the proof, the
Note one small point.
hypothesis .4 k.M C I
M
is equivalent to the apparently more complicated rela-
We chose the former relation for the statement
+.,llk+1. M.
of (2.7) because of its simplicity, but in practice it tends to be easier to verify the latter relation.
A further point is that one can extract a bit
more information from the proof of (2.7) than we have stated.
Keeping to the
same notation define codkI
=
dim
1+4-I + !k+1M
M
105
for k = 0, 1, 2, ...,
The quotient space which appears here is necessar-
.
+,Elk+1.M
ily of finite dimension, and its dimension is the codimension of
I
in I Under the hypotheses of (2.7) a necessary and sufficient condition
(2.8)
for
to be of finite codimension in
I C M
of the
codk I
is that all but finitely many
and in that case
vanish:
cod I
Suppose
Necessity
M
I
cod0 I + cod. I + ...
=
.
has finite codimension in M
so by (2.7) there
That implies codj I = 0 for
exists an integer k 3 I with .,Elk M C I.
all j3k. .,llk.M I
C
I+
codk I = 0
If
Sufficiency
.,Elk+1.M
then
I +.A M
=
+.J1k+1.M
I
so
which implies ,O/k M C I by the Nakayama Lemma,
is of finite codimension in
i.e.
M.
The second statement in (2.8) follows immediately from the proof of (2.7).
Next, let us expand a little on the kind of situation we have in mind. Write
n,p
with y E
for the real vector space of all germs Such a germ has components
]RP.
f
:
f1, ..., f
(]Rn, 0) - (lip, y)
relative to the
p
standard co-ordinates in ]Rn, think of
=
&n
each of which lies in
IRP
gn x ... X &n
(with
p
dn.
Thus we can
factors), and see that it is a
sP
M
natural example of an &n module shall be thinking of in the case
p = I
I
with a finite basis.
as the "tangent space" to a germ
we shall be thinking of
associated to a function-germ
f.
I = Jf,
In practice we f:
for instance
the Jacobian ideal
There is an important point to note here
in connexion with the practicalities of computing the dimension of
106
9n.,p/I.
Let us write
gn
-'
&n
n,p
-+
=
an.,p
n
x ... X 9n
(with
factors).
p
of the Borel Lemma gives rise to a surjective linear mapping
mapping
np
We claim the following.
I, say.
to
I
A necessary and sufficient condition for
(2.9)
The epimorphism
dimension in
is that
dn,p
I
to be of finite co-
should be of finite codimension in
I
9^
and in that case the two dimensions coincide.
Consider the composite of the surjective linear mappings
Proof -a
the second being the natural projection onto the
n,,p /f,
in,p
9n,p
I +.Sn,p so that by element-
By previous work the kernel is
quotient.
ary linear algebra
I +,Ilf°n.e n,p n Necessity
I +A
If
I
3
has finite codimension in
and hence
I
n,p
has finite codimension in
then so too does nip
by the displayed
isomorphism.
Sufficiency
If
I
finite codimension in
has finite codimension in n,p then by the displayed isomorphism.
J/n+1.&n,p.
has
n
It follows from
8n,P,
(2.7) that there exists an integer
I+
I +,.&n.,p
k >.
I
k with .,in.dn,P C I +
.dn. p
The Nakayama Lemma then tells us that we have
C
Si i,
and (2.7) that I has finite codimension in 9n,P Finally, when Ar
k.
n 9
n,p C - I
I
has finite codimension in
tells us that -4f"On'&n,p C -I
so
&n,p
the relation
I +.,e'O° . g n n,p
=
I,
and
the displayed isomorphism reads
an,P/I
,P/I
107
yielding the final statement of (2.9)
The practical consequence of (2.9) is that when computing the codimension of
in
I
we can, and shall, replace all our germs by their Taylor
np
series and handle them as formal power series, which are much easier to work Bear this point in mind when you come to do such computa-
with in practice.
Before resuming our session of algebra we shall digress to look at
tions.
the sheer practicalities of just how one computes the codimension of a germ
in 9n.
Of course the first thing one wants to know is whether
codimension:
f
has finite
f
the next proposition at least gives one a very simple way of
recognizing a germ of infinite codimension.
Suppose the germ
(2.10)
Then
has finite positive codimension.
is an isolated singular point of any representative of
the origin in ]Rn i.e.
in 9n
f
f,
there exists a neighbourhood of the origin in which the origin is the
only singular point of the representative.
Proof
Observe first that
(any representative of)
0 e ]Rn
Indeed if some
f.
be an invertible element of Since
zero codimension.
some integer xl, ..., xn
k > 1
must indeed be a singular point of
9n,
f
of (0)
2xi
by (2.1) so
(
0
and
Jf = 9n
has finite codimension we have
by (2.6).
then f
would
would have k
.,ll
of axi
C Jf
for
In particular, this means that the monomials
can be written as linear combinations of
8x
,
..., 2x
with
1
coefficients in
9n :
must vanish, so
x1,
in question must be
Example 3
at a singular point of ...,
all these partial derivatives
xn must vanish as well,
i.e.
the singular point
0.
Consider the germ
partial derivatives are 108
f
f(x, y, z)
=
y2 - z2x2 + x3.
Here the
of ax
=
-2 z 2x + 3x2
of .
ay
2y
=
of :
=
az
which vanish simultaneously precisely on the z-axis. is not an isolated singular point of of infinite codimension.
The set
-2zx2
Thus the origin in IR3
and (2.10) tells us that
f,
f = 0
is sketched below.
f
must be
Notice that
the z-axis is precisely the line of "double points" where the surface intersects itself.
Thus (2.10) provides us with a simple method of showing that a germ is of infinite codimension.
The strict converse of (2.10) is false, but there
seems no harm in mentioning that in the complex case the converse does hold,
though the proof requires considerable mathematical machinery. ple is provided by the germ
f(x, y)
=
(x2 + y2)2:
A good exam-
the reader will readily
verify that as a .real germ the origin is an isolated singular point, but that
as a complex germ it is not, since any point on the lines x - iy
=
0
is singular:
x + iy
=
0,
thus the germ is of infinite codimension.
Using
these remarks the reader should find it a relatively straightforward matter to decide whether a given germ has finite codimension. For the actual computation of the codimension we proceed as follows. Suppose that using (2.8).
f
The idea is to compute
is of finite codimension.
Let us abbreviate
codkJf
to
codkf:
cod f
we know that
1o9
cod f
=
so we must successively compute
cod0 f + cod1 f + ...
cod0 f,
cod1 f,
until we reach a zero
...
The practicalities
answer, and then add up the list of integers so obtained. of the matter are worth expanding upon.
a basis for a supplement of Jf
+_'#/k+1
in Jf
clearly, this can be
extracted from a list of monomials of degree
x1, ..., xn which do not lie in
if +
k = 1, 2, 3, ...;
and for each
k
k in
x1, ..., xn,
variables
The first step in this computa1.
cod0 f
n
In practice then one one has to decide, for
whether it lies in the ideal
A labour saving observation here is that if this condition holds
for some monomial i.e.
k+1.
in the
dim an/ n =
each monomial of degree k+1
Jf +J1
k
=
tion is always trivial, as takes successively
codk f we have to find
To compute
m
then it automatically holds for all its descendants,
m by multiplying it be some
the monomials which can be obtained from
other monomial.
The reader is warned that the decision as to whether a given
monomial lies in the relevant ideal may well involve quite a bit of work. A case which frequently appears in practice is in two variables
Here, the monomials
n = 2.
x, y can be conveniently displayed as the following array
y
x
x
2
y
xy
2
2
x3
x y
xY2
y3
One works from the top of the array downwards.
A simple way of recording
the results of the computations is as follows.
Suppose we have reached the
kth row of the array,
i.e.
the monomials of degree
monomials which appear in the relevant ideal
k. k+1:
Jf +.'ff
Underline those and then underline
all its descendants in the array, to avoid unnecessary work.
maining monomials select a supplement for Jf +,,ffk+1 in Jf 110
From the re-
+.,Cfk:
the
number of monomials selected is the integer
The computation finishes
codk f.
whenever one reaches a row of monomials which can all be underlined. dimension is the sum of these assuming that
cod0 = 1,
Here
Jf
<xs>.
=
of degree k in x, does.
xs
because we have to add
with
an integer . 1.
s
it is clear that the sole monomial
does not lie in the ideal Jf
namely xk,
It follows immediately that
Example 5
xs+1
=
f(x)
0 , k , s - 1
For
-
is a singular germ.
f
Consider the germ
Example 1+
plus one
codk f,
Here
Y
should have finite codimension.
f
x2
=
(See Figure 1 below.)
x, y.
x2, xy, y2
Of
2
Of
xy = -
4y3,
8Y
x
f ex
x3y, x2y2, xy3
,
,
but
in
4
Y
Jf +,J'4,
1
=
x, y lies in
does not;
y2 -
namely
x3, xy,
cod f
T+y8fay
_$1x 8xof
.
and hence that
5. l4
in the third
xy2
Of the monomials
Jf +.
The codimension is the total
We have, incidentally, shown that
C Jf,
x2, xy
Thus all the monomials of degree 4 lie
and the computation is finished.
=
as
y need be considered, and it lies in
number of monomials which have not be underlined (namely one, so
Jf +
thus we underline
in the fourth row and so on.
And of the monomials of degree 4 only +,115 as
Neither of
need be considered, and it does not lie in
of degree 3 only y3
if
= 2xy,
We write down the array of
the first two certainly lie in
and all their descendants in the array row, and
2x
vanish simultaneously (even in the complex case) only at the
of = llx2 + 4y3
Jf +
but
f(x, y) = x2y + y ,
which will appear in §4 as the so-called "parabolic umbilic".
monomials in
+.,/1k+1.
cod f = s.
We shall compute the codimension of the germ
origin, so
The co-
X. y, y2, y3) plus ..114
C Jf +
5
by the Nakayama Lemma.
III
2
y2
X 2
x y
x3 x3Y
X
x
x5
y3
xy2
Z
3
x2Y2 x2y3
x3y2
Y
Figure 1 y
x x
2
xY
x3
2
y3
,q,2
x 2y 2 2
x4
x5
y
xXy
y
x Y
x3Y
x3y2
x2Y3
'cam,
Y5
_
Figure 2
Example 6 indeed
The germ
cod f = 7.
we omit it.
f(x, y)
=
x3 + xy3
is of finite codimension,
The computation is very similar to that in Example 4 so
The array of monomials is illustrated in Figure 2.
Before going any further we should stop to verify that the codimension of a germ possesses the most basic property one would expect of it, namely that it is invariant under the relation of equivalence.
To this end we introduce
a new idea, which will be exploited later in this book. f
(7Rn, 0) - (Pp, 0)
A - A o f.
One checks easily that
to be induced by 112
we obtain a mapping
f.
f
f
: 9p -
Given a germ n
by the formula
is an algebra homomorphism, said
It is worth remarking that at the time of writing it
seems to be an open problem whether every algebra homomorphism 9p the form
f* for some germ
f
- (IRR, 0).
(IRn, 0)
:
-*
En has
To the reader we leave
the task of checking that the operation of taking induced homomorphisms has the following functorial properties,
(I)
(f o g)*
=
The germ at
(II)
the identity map &n
->
g* o f*. 0
of the identity map IRn - in induces
8n , and is indeed the only such germ.
(III) It follows from (I) and (II) that if then
is invertible and
h*
(2.11)
f
Let
(h* )-1
to be an iso-
Necessity of the condition is provided by III above.
For suffici-
morphism is that
ency suppose self.
onto it-
is an isomorphism, so maps the maximal ideal
f*
Indeed
f*
should be invertible.
f
will map the standard generators
f*
to the components
n.
A necessary and sufficient Sn - gn
condition for the induced algebra homomorphism
Proof
is invertible
(h-')*.
=
(Mn, 0) -, (IFtn, 0) be a germ.
:
h
f1, ..., fn
of
f,
for
x1, ..., xn
/n
which must likewise be generators for
Consequently we can find elements
A.
in
&n
for which
1
s i 5 n
we have
xi
Ai1f 1 + ... + llinfn.
If we differentiate both sides of this relation with respect to evaluate at
0,
6 i.j
and then
we obtain
sib where
xi,
of 1
A i1(o) axe (o) +
denotes the Kronecker delta.
of
... + xin(0) a- (o)
j
Thus
113
af. In
=
l
A( A(axl (0)) J
In
where the
denotes the identity n x n
Aij.
matrix, and A denotes the matrix of
It follows that the Jacobian matrix is invertible, and hence
(from the Inverse Function Theorem) that
is invertible.
f
Now we are in a position to establish the invariance of the codimension, in the following precise sense.
(2.12)
If the germs
Proof
Since
h
:
(]Rn, 0)
isomorphism
->
h*
f, g
f, g
cod f = cod g.
are equivalent then
in &n
are equivalent there exists an invertible germ for which
(]Rn, 0)
: 9n - 9n
of algebras.
will imply that the quotient spaces which the result follows.
g = f o h.
By (2.11)
We claim that
&n/Jf,
9n/Jg
h
induces an
h*(Jf) = Jg:
that
are isomorphic, from
To establish the claim note that by the Chain
Rule
n
D-
aah. (a j oh/ax
i
j=1
so
J9 C h*(Jf).
% h laxjl
The same reasoning shows
yielding the reverse inclusion
x.
j=1
h*(Jf)
Jf C
(h-1)#Jg
=
(h*)-1Jg
C Jg.
There is one final matter which we should clear up before passing to new things, namely the connexion between the codimensions of the ideals
Jf,,A'Jf.
First of all, (2.13)
The ideal .,AAJf
finite codimension.
114
has finite codimension if and only if
Jf
has
_1%'(f
4'Jf C Jf,
Proof
so certainly if
has finite codimension.
has finite codimension then
Conversely, suppose
again by (2.7), we see that
Let
(2.14)
f e n,
]RI
ax
subspace of
cod.!Jf =
then we have
n + cod Jf.
is the vector sum of the vector subspaces
if
+
u1 ax
+
...
observation follows on writing a.(x)
=
un
with the
ax
with at least one
ei(0)
for which
0,
1, ..., en
g
of i axi
=
as the components of a vector field
in in with
g(0)
can be "straightened",
on a neighbourhood of
f
we can find a diffeomorphism
i.e.
dn,
Think of
Recall now from (I.4.3) that the flow of
0.
in
To prove this it
0.
is convenient to confuse germs with their representatives.
en
and the
ui a &n;
(ui(x) - ui(O)) + ui(O).
Next we claim that it is not possible to find germs
g1,
Jf,J'Jf has
ax j , where the curly brackets indicate the vector n spanned by the first partial derivatives. Indeed an element
has the form
Jf
C .A'Jf so,
,
1
of
//k+1
and suppose that one (hence both) of
Observe first that and
But then
.AJf has finite codimension.
finite positive codimension:
Proof
has finite codimension, so
Jf
,,/,/k C Jf for some integer k > I by (2.7).
Jf
h
e
0
near 0
of a neighbour-
ah.
hood of
0
in ]Rn
with
for which
h(O) = 0
h1, ..., hn
are the components of
germ in
equivalent to
9n
Let us agree to write
f,
b = h(a).
h.
ei o h
=
axl ,
Observe now that
aa--
g = f o h
is a
so likewise of finite positive codimension.
Then -- (a)
ei(b) ax (b)
_
i
1
so
where
is identically zero on a neighbourhood of
0
in e, i.e.
=
g
0,
is
1
independent of the variable
x1:
by (2.10),the origin
point for
g,
so that every point near
point for
g,
and
g
0
on the
has infinite codimension.
0
is a singular
x1 - axis is a singular
This contradiction estab-
lishes our claim. 115
Taking
Now we can easily finish the proof. numbers we see that
ax
,
e1, , en
to be real
are linearly independent over the reals,
..., ax
n
1
and hence that the vector subspace
]R{ax
, , ax n
1
has dimension
n.
Further, it follows immediately from the above that the vector sum of and
]R{ax
is direct,
, , axn}
1
tl Jf
that the intersection is the trivial
i.e.
The result now follows.
subspace.
Determinacy of Germs
§3.
Having completed our session of algebra we return to the main theme, namely the classification of germs in
gn
of low codimension under R-equivalence.
The underlying philosophy is to reduce the problem to a finite-dimensional one.
And the idea to do this is an follows.
Take a germ
f e dn:
f
has
a Taylor series
f(0) +
x
1
i
1!
of (0) + axi
n
x.x. i
2!
2
a f (0) + ... axiaxj
i,j=1
i=1
whose initial segments (the so-called Taylor polynomials of better approximations (in some sense) to f
f.
f)
provide even
The hope is that a nice enough
will be equivalent to one of its Taylor polynomials, and then the problem
of classification reduces to one in a finite-dimensional vector space of polynomials.
We introduce therefore the following definition, integer.
A germ
f e Sn
k >.
0
be an
is said to be k-determined when any germ
with the same k-jet satisfies
116
Let
f - g:
g e En
in other words a knowledge of all
partial derivatives of order valence.
s k
completely determines the germ, up to equi-
Note that the property of k-determinacy will be invariant under
equivalence.
Example I
Any non-singular germ
(I.1.3) we know that
f
f s do
is 1-determined.
is equivalent to the germ
And the same applies to any germ
(x1,
Indeed by
..., xn)
having the same 1-jet as
g s 9n
x1.
-*
Hence
f.
the result.
The basic result of this section is the following sufficient condition for k-determinacy.
(3.1)
Let f s &n be such that .!/ k C
Proof
Let
Now
f, g
g s &n
have the same k-jet as
are points in the real vector space
straight line.
for t s IR.
f : then f is k-determined. f.
We have to prove
9n,
f - g.
so can be joined by a
More formally, we define F(x, t)
f1 = g.
_14//J
Thus
=
ft(x)
=
(1 - t)f(x) + tg(x)
(ft) is a 1-parameter family of germs with f0 = f,
We intend to show that any two germs in this family are equivalent,
which will prove the result. fice to show that given any
Since the real line is connected it will sufs
s IR
we have
ft N fs
for
t
close to
s.
We claim that for this it will suffice to show that there exists a germ at
(0, s) of a smooth mapping
H
(a)
H(x, s)
=
x
(b)
H(0, t)
=
0
(c)
F(H(x, t), t)
Mn x IR
:
=
- IRn for which
F(x, s).
117
To see this, write
ht(x)
It follows that for
t
Now (a) tells us that
H(x, t).
=
close to
the germ
s
the determinant of the Jacobian matrix at t.
hs = identity.
must be invertible, since
ht
depends continuously on
0
of
ht
Further, the condition (b) ensures that
ht
maps the origin in ]Rn
itself.
ft - fs
ft o ht
Finally, the condition (c) can be re-written as
for
t
close to
as was required.
s,
is automatically satisfied for
t = s,
=
to
f s,
so
Notice incidentally that (c)
Thus it will be suf-
in view of (a).
ficient to replace (c) by the condition that the left hand side does not depend on
t,
has a zero derivative with respect to
i.e.
Written out in full
t.
this is the condition. a
(CO )
i=1
tI (x' t) ax 1(H(x, t), t) +
at
(H(x, t), t)
=
0.
Our problem then is to construct a smooth mapping H which satisfies conditions
We claim that it will suffice to construct a germ
(a), (b), (c').
at (0, s) of a smooth mapping aF
d (
c
)
(e)
i axi ei(0, t)
For suppose such a mapping
dependent vector field on Mn.
]Rn x
]R
-,
]Rn
for which
aF - at
=
=
:
0.
a
exists.
We can think of
It has therefore a flow,
as a timei.e.
a mapping H
of the required type for which (f)
at (x, t)
_
e(H(x, t), t),
which can be supposed to satisfy the "initial condition" (a).
A minor com-
putation shows that (c') follows from (d) and (f) and that (b) follows from
(a), (e) and (f).
118
It remains to establish the existence of a mapping ties (d) and (e).
The argument is algebraic, and does not depend on the
particular value of
chosen:
s
we shall therefore suppose
since we wish to work simultaneously with functions of x1, ..., xn
and functions of (n + 1) variables
think of an element of t.
having the proper-
e
as an element of
9n
9
s = 0.
variables
n
x1, ..., xn, t n+1
n
at using the fact that
s en
As a preliminary, note that
.
OF
we shall
which does not depend on
And then condition (e) is simply the requirement that 1,
should lie in
Also,
_
aat (1 - t)f+tg
g - f
=
g-fE
.,//k+1 n
has zero k-jet, and (2.4).
Thus (d) will follow
from
k+1 C
aF
aF\ "' axn
'
ax 1
n
n Next we have
of \ ...s axn/
k / of llnk+1 C n C "en\ax1
aF
C
aF
s axn
+
n+
n
the first inclusion being trivial, the second being the hypothesis of the theorem, and the third following from the fact that
aF
of
to
-f
axi
k n+1
n
* now follows immediately from ** using the Nakayama Lemma, and the result is proved.
119
It is worth remarking that the condition (d) which appears in the above Recall that we took some germ
proof has a simple geometric interpretation. on the line joining
ft
along the line from
ft
f, g
and set out to show that if we moved slightly
we remained in the orbit through
ft:
thus we ex-
pected the direction of the line to be a "tangent vector" to the orbit through We think of the direction of the line being given by
ft.
(d) then says that at
lies in
.JTJf
the "tangent space" to the orbit
i.e.
,
The relation
at .
t
through
ft.
A word is also in order concerning the hypothesis of (3.1). Lemma tells us that the condition
Ak
is equivalent to the apparently
C.,#'Jf
Jf + .,
more complicated condition -.0"k C
The Nakayama
We chose the former con-
k+1
dition because it is rather simpler to remember though in practice it tends to be easier to establish the latter condition.
It is convenient to call mined for some
k >.
1.
f e &
n
finitely-determined when
f
is k-deter-
It follows then from (2.7) and (3.1) that if
of finite codimension then automatically practicalities of showing that a germ
f
f E gn
f
The
is finitely-determined.
is k-determined are very simiOne
lar to those involved in computing the codimension, discussed in §2. checks successively whether-,// k C -"//Jf
for a given k
+./1
is
k+1
for k = 1, 2, ...
:
and
this is equivalent to showing that each monomial of degree
k
+.,//k+1.
in
x1, ..., xn
lies in the ideal
f e 91
Consider a germ
Example 1
11Jf
for which the following conditions are
satisfied.
f(O)
=
0
of
akf (0)
=
0
ax
Clearly f E_/9/ 1 , f 120
:
...
(0)
axk 1+1.
=
0
:
ak+1f (0) k+1 ax
'
0.
It follows from the Hadamard Lemma that we
can write ax
=
aY,7f =
exk+1
+
=
k+1.
.!41
where
:
Then
g(O) ,' 0.
the expression in curly brackets is
(3.1) tells us that
if
=
<xk\
when
0
k
_
is (k + 1)-determined, so
f
An obvious change of co-ordinates then yields
c gE 0.
xk. The germ
Example 2
if
}
with
g e 91
so invertible by (2.1), and one deduces that
hence
f
for some
xk{(k + 1)g + x
x = 0,
f
xk+1g
=
f
<x2, y2>,
f(x, y)
=
x3 + y3
is 3-determined.
so .4'Jf = <x3, x2y, xy2, y3>
Here
and the claim
follows from (3.1) Remark
Another sufficient condition for
is that .,f k+1 C.,lr2Jf.
f E n to be k-determined
The proof of this is exactly the same as that of
one has simply to check that the last few lines of that proof go
(3.1):
through equally well with this hypothesis.
Of course, this is a slight im-
provement on (3.1) since ,,ayk+1 C2Jf is a weaker condition than ..l k C .,lvj The germ
Example 3 if -4/4
x3, y3
=
C 411f
=
.,I'Jf
=
=
+
y is
u-determined.
Here
<x4, x3y, xy3, y> and the condition
is not satisfied, so (3.1) does not allow us to conclude that
is L-determined. le if
so
x4
f(x, y)
On the other hand
<x5, xy, x3y2, x2y3, xy, y5> =
just mentioned allows us to conclude that
Notice that if
f
f e gn
_j3 f
so the improvement on (3.1) is indeed 4.-determined.
is k-determined then it is, in particular, equi-
valent to its kth Taylor polynomial.
Thus the study of germs in &n
finite codimension reduces to that of polynomial germs.
In this connexion
it is well to bear in mind that a polynomial germ of degree finitely-determined is not necessarily k-determined.
of
k which is
For instance the germ
121
f(x, y)
x5 + y5
=
is easily checked to be 6-determined, but can be shown
not to be 5-determined.
Classification of Germs of Codimension , 5
§1+.
We are now in a position to reap the rewards of our hard work in the previous
We
section, and achieve our object of classifiing germs of low codimension. start with germs of codimension
(1+.1)
0.
A necessary and sufficient condition for a germ
codimension
is that it be non-singular;
0
lent to the germ
(x1,
...,
Suppose
Necessity
identity element
8xf1
to have
and in that case it is equiva-
xn) - x1. has codimension
lies in
I
for which I = e1 for that at least one
f
f e 9n
Jf,
i.e.
+ ... + e 0,
aX (0)
so that
0,
Jf = 9n
and the in
e1, ..., en
we can find germs
If we evaluate at 0 we see
8x
and hence
f
is non-singular.
i
Suppose
Sufficiency
By (2.1) this means that Jf = &n,
so
f
f
is non-singular, so at least one ax (0)
If ax. 1
is an invertible element of
has codimension
122
and hence
0.
Finally, a non-singular germ in &n by (I.1.3).
1
6n,
0.
is equivalent to
(x1, ..., xn) -* x1
Remembering our discussion of
Next we consider germs of codimension 1.
2
functions in §5 of Chapter II we say that a germ
f E.Af
a singular
i.e.
,
germ, is non-degenerate when the Hessian matrix 2
_
a
l
f (0)J (axiaxj
Hf
The next proposition, classifying germs of codimension 1,
is non-singular.
is a classic result in the calculus of variations, called the Morse Lemma.
A necessary and sufficient condition for a germ
(4.2)
equivalent to a germ of the form
x2i
+ ... + xg - xs+1
" '
f
will be
x2
The main part of the proof consists of establishing that f e
Proof
non-degenerate if and only if ,A1 = Jf. the
n
and in that case
codimension 1 is that it be non-degenerate:
to be of
f E,,//2
all vanish at
ax
0.)
(Note:
is
1122,1
mil D Jf automatically since
To this end let Q denote the second polynomial
1
of
f - which is just a quadratic form in
x1, ..., xn.
Non-degeneracy of f
is the same thing as non-degeneracy (in the normal sense) of the quadratic form Q,
span the vector space
which in turn is equivalent to saying that the 1
of all linear polynomials in Since we have
1=
saying
ax
x1, ..., xn (by elementary linear algebra).
modulo .,11 2,
ax.
ax.
1
this last condition is equivalent to
1
2
Jf +.s1 , which in turn is equivalent to
vious application of the Nakayama Lemma.
Sufficiency
codim f
dim c/Jf
=
f
f
Jf
=
11,
and
f
Jf =,/,/by the above.
Then
1.
has codimension 1.
Jf C .,ll then dim &/Jf > 1,
And if
The rest of the proof is now easy.
non-degenerate, so
dim 8/.,11=
=
Suppose
Necessity
Thus
Suppose
.,ll = Jf, using an ob-
i.e.
Certainly then
codim f > 1,
dim &/,' =
1.
which is impossible.
is non-degenerate, by the above discussion. 123
Finally, we have to verify that saw above, non-degeneracy of
has the quoted normal form.
f
is equivalent to
f
It follows immediately from (3.1) that
..11 2 C ,E'Jf.
equivalent to its second Taylor polynomial algebra that
Q.
f
being a non-degenerate quadratic form, is equivalent (under
Q,
+ ... x2 - x2
s+1
s
1
is 2-determined, so
Recall now from quadratic
2 x1 + ... xs
a linear change of co-ordinates) to the normal form x
which implies
Jf
..//=
As we
- x2
-
n
where
is the index of
s
2 xs+1 -
That concludes
Q.
the proof.
The geometric meaning of the Morse we have a smooth function
f :IRn
degenerate critical point at
0.
-
correspond to sets
f-1(E)
with
IR
f(0) = 0
Suppose
admitting a non-
The Morse Lemma tells us that there is a
diffeomorphism of one neighbourhood of the fibres
Lemma is worth spelling out.
in 1Rn
I
onto another under which
x2 + ... + x2 - x2+1-..., - xn
=
E,
1
for small enough and for
c ? 0
c .
In particular,
the relevant part of
f 1(0)
corresponds to a quadric cone;
f-(E)
is a smooth manif old diffeo-
morphic to part of a non-singular quadric surface. case
n = 2,
Here is a picture for the
s = 1.
diffeomorphism
To make this even more convincing recall again our discussion of functions in §5 of Chapter II.
There we showed that any function
be approximated as closely as we please by one
124-
ft
:
f
:
1Rn -, IR
IRn - 1R
could
which admits
only non-degenerate critical points; from
f
be a linear deformation.
and indeed that
x3 - 3xY2
=
which admits a degenerate critical point at
as the approximating function, for erate critical point of
f
at
(+ f
points at the points
,
0
0).
could be obtained
One of our examples was the monkey saddle
f(x, Y)
ft(x, Y)
ft
=
0.
In this case we could take
x3 - 3xY2 - tx
t > 0
The degen-
as small as we please.
"splits" into two non-degenerate critical And the diagram of level curves
f_1(e)
3
is as below:
the reader will note that close to the critical points we have
precisely the situation described above.
To push the classification further for germs of codimension > 2 we shall require one more bit of machinery. 2.
Suppose that
Certainly then the Hessian matrix
the non-negative integer
c = n - r
Hf
f
n
has codimension
is singular, so has rank
is called the corank of
checks that it is invariant under equivalence.)
r < n:
(One easily
g.
The extra machinery which
we require is the following result, known as the Splitting Lemma, and proved originally under less restrictive hypotheses.
(4.3)
Let
2
f E,// n
be a finitely-determined germ of corank
c;
f
is
equivalent to a germ 125
2
xc) + Sc+1xc+1 +. . .+8x
g(x1s
where
g E _.*,1 and each
Proof
Si = + 1.
We shall adopt the following notation.
8n we write for which relation.
n
Given two germs
r when there exists an invertible germ
¢ o k, 0
have the same k-jet.
Observe that
We intend to show, by induction on
0, 0
in
h : (IRn 0) -> (IRn 0)
is an equivalence
k
that there exists a germ
k,
Ed'3, which is polynomial of degree 5 k, for which
gk
f(x1, ..., xl)
k
gk(x1,
Sc+1xc2
..03, xc) +
+1 + ... + Snxn
This will suffice to prove the result, since there exists a
k
for which f
is equivalent to its k-jet.
The induction starts when k = 2.
n
variables of rank r = n - c,
The 2-jet of
is a quadratic form in
f
so equivalent (under a linear change of co2
ordinates) to a quadratic form
Se+1x0+1 + ... + Snxn with each
by standard quadratic algebra:
and we can take
step we assume
g2 = 0.
Si
+1,
For the induction
Certainly then
*.
f(x1, ..., xn) k+1
gk(x1'
'
xc) + Sc+1xc+1 +
+Snxn + H(x1, ...) xn) **
where
H
is a homogeneous polynomial of degree (k + 1).
Now write, as we
evidently can do,
H(x1,
..., n
=
h(x1, ..., xc) + xc+1Hc+1(x1' ..., xn
+ ...
+ xnHn(x1, ..., x
where
h
)
is a homogeneous polynomial of degree (k + 1) and He+1' ...' Hn
are homogeneous polynomials of degree 126
n
k.
We use these latter polynomials
to define 0 : (IRn, 0) -+ (]Rn, 0) by taking its components to be 01(x1,
..., n) =
x1
00(x1,
..., xn)
xc
=
e+1)
1
0o+1(x1' " .'
n) =
on(x1, ..., xn)
=
2SH(x1, c+1
xc+1
xn
_
" ., n
1
2S Hn(x1' ..., xn). n
Note that the Jacobian matrix of
Substituting
invertible.
f(x1, ..., xn)
where
gk+1
=
is the identity n x n matrix, so
/1, ..., Sn
gk+1(x1V
k+1
"
for
x1, ..., xn in
**
0 is
we obtain
2
2
xc) + Sc+1xc+1 + .. + Snxn
That'completes the induction step, and establishes
gk + h.
the result.
It is worth remarking that the germs
f, g which appear in the statement
of the Splitting Lemma are necessarily of the same codimension. can safely be left as an exercise for the reader.
The proof
Our first application of
the Splitting Lemma is to establish a broad connexion between the corank and the codimension of a germ, which will be crucial to the remainder of the classification.
Roughly speaking, it is that as the corank increases so
does the codimension, only rather more quickly.
(4.4)
Let
codimension of
Precisely,
f e.,F2 be a germ of finite codimension, and corank f
is
>
1+
C (c
2
c:
the
1.
127
Proof
We have already observed in §3 that a germ of finite codimension is
automatically finitely-determined, so the Splitting Lemma applies, and tells us that
f
is equivalent to a germ g(x1, ..., xc) + xc+1
with
and (vide the remark above)
g e.,lf3:
Note that
g.
+ ... + xn
f
has the same codimension as
I = A Jg C !! o and that by (2.8)
cod I
=
cod0l + codf I + cod2 I + cod3 I + ...
But
cod0 I
I + 9 dim I
=
J
=
dim
=
Arc
I + ,,!/ codf I
dim
=
1
c dim
-/f c
-ff
I
2
cod2 I= dim
_ dim 3=
I + I c
C (c 2 1
ec
so
1 +c+cc2+
codI = c+codJg
using (2.14), from which the result follows.
Thus, for instance, we have the following rough estimates
c=1 c=2 c=3
,
,
codf>.2 codf >4 codf>>7
The consequence of (4.4) which we need to isolate is that if then corank
f <. 2.
1 and corank 2.
128
cod f s 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-./K2 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
codim
germ
2
x3
f old
3
x
cusp
4
x5
swallowtail
5
x6
butterfly
name
1
2
4
x3 - xy2
elliptic umbilic
4
x3 + y3
hyperbolic umbilic
5
4 x y + y
parabolic umbilic
2
The exact meaning of the statement will become clear as we proceed. Morse
Lemma give us a complete classification of germs of corank
The
We
0.
shall now classify all germs of corank 1 having finite codimension.
Let
(4.6)
2
f E
n
have corank 1, and finite codimension (k - 1):
f(x1, ..., xn) - +
1
± x2
+
then
2
+ xn
We know from §3 that a germ of finite codimension is finitely-
Proof
determined, so the Splitting Lemma applies and tells us that
f(x1, ..., xn) ^- g(x1) + x2 + ... + xn
where
gE
show that
I
also has codimension
g - + xk.
k.
Write
x = x1.
It suffices now to
Observe that there must exist a least integer
j
with 129
g E,%' , cod g
if not then J9C ll for all j,
g X.;/l1+1
cod ll
>
=
computation shows that
O(x)
xh(x)1'k:
=
k(x),
h(O) > 0, =
so
0
g - xk. g
xj-h(x))1/k
=
g(x)
(P'(0) # 0
so
0
with
xih(x)
(j - 1),
(IR, 0) -+ (IR, 0)
:
clearly,
and yields
g would have infinite codimen-
i.e.
has codimension
g
In that case take
g(x) _
O(x)
j,
The Hadamard Lemma tells us that
sion.
odd.
for all
j
so
so
h(O) J 0:
Suppose
j = k.
to be the germ given by is invertible.
Further,
The same reasoning applies when k
But when k
xk.
k
is even and
is even and
h(O) < 0
take
k
g - -x
to deduce
,
It is the germs obtained in the above result which are called the cuspoids. Of course, if we restrict ourselves to germs of codimension < 5 in the result,
then we obtain (up to addition of a non-degenerate quadratic form in further x3, xl+, x5, x6.
variables, and multiplication by ± 1) precisely the germs
And with this we have proved half of Thom's classification theorem (4.5). A singular point of a smooth function of sion
k
corank ,
is usually called an Ak singularity.
1
and finite codimen-
Such singularities have long
been in the repertoire of classical algebraic geometry, and the reader may find it of interest to recall how they arise naturally when studying double points of algebraic curves. We shall restrict our attention to real algebraic curves,
i.e.
of the plane IR2 defined by an equation f(x, y) = 0 where f
:
IR2 -1R
is a polynomial function:
we shall suppose that the curve
through the origin (so
f
has no constant term), and indeed that the origin
is a singular point of
f (so
f
f = 0
subsets
has no linear terms either).
passes
Recall that
the singular point at the origin is said to be of multiplicity m when m is the lowest degree of a monomial in have a double point, 130
when m = 3
f:
thus
m > 2.
When m = 2
a triple point, and so on.
we
The condition
on the origin, that it is a double point of the curve the singular point there has corank s 1.
+
xk+1
f = 0
the curve
k = 1, 2, 3, 4, 5
+ y2.
Ak
for some integer
xk+1 + y2 by (4.6).
by a non-zero scalar does not affect the curve normal form is
entails that
If we assume the singularity to be
of finite codimension then it will be of type so equivalent to the singularity
f = 0,
f = 0
k >>
1,
Multiplication of f so we can suppose the
And since we are not interested in the case when
is a point we consider only the case
xk+1 - y2.
The cases
give rise to the following pictures, which are tradition-
ally dubbed with the titles indicated.
A
1
: node
A2 : cusp
A 3 : tacnode
A4 : rcunphoid cusp
A 5
: oscnode
131
Example 1
Consider the real algebraic curve given by the equation
f = 0
where 2 2
f(x, y)
=
x4 + x y
- 2x2y - xy2 + y2.
Clearly, the curve has a singular point at the origin, indeed a double point. And computation verifies that we have a ramphoid cusp.
will have codimension 1+
f
at the origin, sn
The curve looks like this.
X
It is perhaps also of interest to see how
Ak
singularities turn up natu-
rally in studying the differential geometry of plane curves.
a smooth mapping with
:
IR -+ R2 whose image is a smooth submanifold of
mapping its domain diffeomorphically onto its image;
C
think of
as a parametrization of a plane curve.
standard scalar product on IR2, tion
I
Suppose we have
Let
.
IR2,
we wish to denote the
giving rise to the standard distance func-
(We work with the square to ensure smoothness at the origin.)
I2
Given a fixed point w
in the plane consider the smooth function fw
:
IR-+IR
defined by fw(t)
=
2'w - C(t)j
measuring, essentially, the distance from w parameter
t.
As
w
to the point on the curve with
varies we obtain a 2-parameter family F = (fw)
of
smooth functions, better thought of as a single smooth function, say
F
: IR2
of each
132
x IR -+ IR fw
given by F(w, t)
must have corank E I,
=
fw(t).
Clearly, the singular points
and those of finite codimension will be
classified by the Ak-series:
indeed the condition for an
singularity
Ak
will be that aF
at
2
ak+1F
0
akF
at_
=
atk+1
atk
Let us look at the geometric information contained in these relations. The first few derivatives, written out in full, are aF
2
(C - w)
=
C
=
3C.C + C.(S - w)
F
a
at2
a3F at3
where
denote the successive derivatives of
for the components of
Thus
C.
2)
_
Write
C1, C2
is the tangent vector to the
curve, and the normal direction is given by
the result of
i.e.
rotating the tangent vector through a right angle in an anticlockwise direction.
The condition at
tor
is perpendicular to
to the curve at
C(t):
=
then amounts to saying that the tangent vec-
0
S - w,
that
i.e.
the set of points
(w, t)
w
lies on the normal line in ]R2 x]R where
is therefore called the normal bundle space of the curve. a pair
(w, t)
we can write
w =
normal direction to the curve at
where
+ An
and
fi(t),
A
n = n(t)
at
= 0
Note that for such denotes the unit
is some scalar.
Now let us
2
add the further condition that just given in that for
0.
Substitute the expression for
w
at
a2
at
=
as
and equate the result to zero;
, 2
a few lines of
computation will then show that c
A
2 1
+
2, 2 3/2 1C2
133
which the reader will recognize as the radius procal of the curvature for which i.e.
at
=
0,
It follows that the points
K. a
of curvature,
the reci-
i.e.
(w, t)
in ]R2 X IR
2
=
0
are precisely those for which w =
1
+
at
with w a point on the evolute of
It is interesting to go one
stage further and ask what further geometric information we can glean by adding yet one more condition K = 0,
is equivalent to
a3F
= 0: we leave the reader to check that this at3 i.e. to the point C(t) being a vertex of the
Thus singularities of type
curve.
Ak with
bundle space for the curve, those with k . 2 those with
k 3 3
obtain
singularities with
Ak
to vertices.
k 3
I
correspond to the normal
to points on the evolute, and
It turns out that generically one does not in this situation, so that is the end
k > 4
of the story.
Example 2 If we write
Consider the parabola w = (u, v)
2F(w, t)
C
:1R - IR2
given by (t),= (t, t2).
then
=
t4 - tut - (2v - 1)t2 + u2 + v2
and comparison with Example 5 of §1 in Chapter II shows that defines a folded surfaces in IR2 xlER,
=
at
that the further condition
0 a
2
at
= 3
0
2 a defines the fold curve on the surface, and that the final requirement F = 0 7t3 defines the exceptional point in the fold curve where the two folds meet. Under the projection ]R2 x IR
n IR2 the fold curve maps to the evolute of
the parabola, which is a cuspidal cubic, the cusp arising from the vertex of the parabola at the origin.
134
It remains for us to deal with germs of corank 2 and codimension 5 5.
The
next result will complete Thom's classification theorem.
(1+.7)
Let
f E.//2 have corank 2 and codimension . 5:
n
then
f
is equi-
valent to one of the following germs
±
(x3 + x2)
±
x3
+
x3 ± ...
4
2
± (x1x2 + x2)
Proof step.
...
2
x2
+
± xn2
Again, it is the Splitting Lemma which provides us with the first It tells us that
f
is equivalent to a germ
g(x1, x2)
with
+
+
2
x3
± ... ± x2n
g e-//3 having the same codimension as
Taylor polynomial of
g
f,
is a binary cubic form in
namely . 5. x1, x2
The third
so can be
135
assumed x31
- by the discussion in the previous chapter
+ x2,
x21x2
-
to be
We have to decide which possibilities can arise.
this end, recall that a germ of corank 2 has codimension >. 4. (and hence
2 x1 - x1 x,
according as it is identically zero, symbolic, elliptic, hyper-
bolic or parabolic.
f
0,
has codimension 4 or 5.
g)
To
In particular
We shall consider the possi-
bilities separately.
g
has codimension 4
Since
Jg C,,ll2,
we have
g
To see this we argue as follows.
so 11Jg C .,113.
Put
I
codimension 4, and hence by (2.14) that
= ,%1J9.
In fact ,%'J, = .,113.
We are given that
has codimension 6.
I
Jg has
Also, from
(2.8) we know that
cod I But
I
cod0 I + cod1 I + cod2 I + ...
=
and hence I +.ll J = -,or j for
cod0 I
=
I
:
cod1 I
=
3,
3
2
:
.
so that
cod2 I
=
3
we deduce that
cod3 I which means that
43 C I
third Taylor polynomial.
0
as required.
= .AVJg,
It follows now from (3.1) that
=
g
is 3-determined so equivalent to its
By checking the five possible normal forms for
this one by one we see that the only ones of codimension 4 are the elliptic and hyperbolic types.
And these give us the first two germs on the list.
has codimension 5 .,
I
= ,%VJg.
This time
Jg
And again
has codimension 5 so by (2.14) the
codimension of
is
I
cod I
As
Further, we know that
7.
cod0 I + codI I + cod2 I + cod3 I + ...
.
before cod0 0
=
codI I
:
1
2
=
cod2 I
:
=
3
so we must have
cod3 I
=
& cod4 I
1
0.
=
By checking the five normal forms for the third Taylor polynomial one by one we see that the only one for which x2 x2.
And the fact that
cod4, I
=
_
cod3 I 0
is the parabolic type
I
tells us that --44 C .,11Jg
so
is
g
4.-determined by (3.1), hence equivalent to its fourth Taylor polynomial.
This has the form x1x2 + I (x1,
where
x2)
is a homogeneous polynomial of degree 4 in'
x1, x2.
In more detail,
we can write this as
x1x2
+ t
where
c(x1, x2)
2+
bx2x1 + x20(x1, x2)}
is a quadratic form in x1, 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
by
x1 -
2x2
,
x2
by
x2 - O(X1) x2), and
take the fourth Taylor polynomial of the result, we obtain the equivalent germ
xIx2
Notice that necessarily
a 9 0
+ ax2
(for otherwise the germ has infinite
137
codimension).
p, q
Finally, if we replace
x1
by px1, x2
by
qx2,
and choose
suitably we see that our germ is equivalent to
(the sign depending on that of
a).
And that gives us the last germ on the
list.
Of course, by combining the results (4.6) and (4.7) we have proved (4.5) and hence obtained the complete classification of germs of codimension . 5.
138
V Stable singularities of smooth mappings
§1.
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. Recall that two germs
The basic ideas are soon set up.
equivalent when there exist invertible germs
f1, f2
are
for which the following
g, h
diagram commutes.
f 1
(N1, x1)
g
(P1, Y1)
I
I
f
h
1
2
(P2, Y2)
(N2, x2)
Sometimes it will be convenient to say in this situation that A-equivalent.
f1, 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 = IRn,
f :
(N, x) -
x = 0
and
(P, y)
P = IRp,
and taking y = 0.
139
Thus we wish to study the set that
dn,p
(IRn, 0) - (IR", 0),
A
n,p
(IRn, 0)
-'
note
(IRp, 0);
has the structure of a real vector space induced from that on IRp. Rn
As in Chapter IV we take
Write
of all germs
dn,p
=
An,p
Rn
which is a group under the operation of composition.
n,p
Rp
x
0n,p
x EO
to be the set of invertible germs
for the product group.
given by the formula
and the problem of classifying germs in
(g, h).f
9n,p
We then have an action =
h o f o 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 is far too ambitious.
What we, shall do, to make at least some progress, is
to restrict ourselves to the germs of lowest possible "codimension", the "stable" germs in the language of Chapter III.
i.e.
Even this limited objec-
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.
f0
One would like to think of an r-parameter "unfolding" of
(IEtm, 0)
-+
(IRq, 0) as a "germ" (IRr, 0) -'
In a natural way that determines f
:
(IRr x IRm, 0)
(9m
,q
$1
f0), say u -' fu.
-' (IR.q, 0) given by
(u, x) -' fu(x), and this in turn determines F : (]Rr x IRm, 0) -' (IRr x]Rq, 0) On the basis we introduce the following given by (u, x) - (u, f(u, x)). formal definitions. is a germ f 14-0
:
An r-parameter deformation of f0 : (IRm, 0) -+ (OR% 0)
(IRr x IRm, 0)
--'
(]Rq, 0)
with f(0, x)
=
f0(x) and the
corresponding r-parameter unfolding is the germ
F
: (]Rr x
e, 0) , (]Rr x ]R`1, 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
:
(IR, 0)
(IR2, 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
:
(IR2 x
Associated to
f
A 2-parameter deformation of
If I2.
IR, 0) - (]R, 0) given by F(u, X)
=
If(x)
- u12.
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 IR.
components
u1, ..., un-1
The germ
F1, ..., Fn
for the standard co-ordinates in IRn-1
x IR, 0)
(IRn-1
F
:
=
ui
F.
-+
(Rn-1
x
IR, 0)
with
where
Fi(u,x)
Fn(u, x)
=
in-1)
(1 xn+1 +
uixl i=1
is an (n - 1)-parameter unfolding of f
:
(IR, 0) -+ (IR, 0)
given by
f(x) _ xn+1 Let us now head directly for our initial objective of introducing 'btable" 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 F1, F2 of a germ f :
(]Rm, 0) -+ (]Rq, 0)
are said to be A-equivalent when there exist r-parameter unfoldings the germs at
0
of the identity maps on IRm,
Im, Iq of
respectively for which the
IRq
following diagram commutes.
(]Rrx JRm,
F
(IRrx ]Rq, 0)
0)
Im
I F
I
q
I
(IRr x e, 0) 2 >(]Rr x Iftq, 0) In this situation one refers to the pair between
Note that
F1, F2.
Im, Iq
I = (Im, Iq)
are necessarily invertible germs.
Further, we shall call an r-parameter unfolding trivial)
as an A-equivalence
F
f
of
A-trivial
(or just
when it is A-equivalent to the constant r-parameter unfolding
(IRrx IRm, 0) - (IRr x IRq, 0) of f given by (u, x) finally we call
f
A-stable
(or just stable)
-->
(u, f(x)).
when every unfolding of
And f
is
trivial.
We now have a sensible 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 §1+.
The next two sections are devoted to the necessary preliminaries.
14
Contact Equivalence
§2.
The object of this section is to introduce and study another natural equivalence relation on germs
(N, x) -
(P, y)
N, P
with
smooth manifolds.
This new relation, calledX-equivalence, is coarser than A-equivalence and has the advantage that it is fairly computable:
in §1+ we shall see that there
are intimate connexions between the two relations directly relevant to the problem of classifying stable germs.
Suppose then that we have germs f2
:
(N1, x1) -
f1
tatives, which will be smooth submanifolds of z1 = (x1, y1),
N1 x P1,
N2 x P2
N1, N2
at
z1, z2
(or aXequiva-
To this end we define a contact equivalence
lence) to be a pair
through the
What we wish to do is to formalize
z2 = (x2, y2).
the idea of these graphs having the same "contact" with respectively.
and
We turn our attention to the graphs of represen-
(N2, x2) - (P2' y2).
points
(P1, y1)
of invertible germs for which one has a commu-
(h, H)
ting diagram >(N1 x P1, z1)
(N1, x1)
1 ,(N1, x1)
H
h
y
12
h y
112
y
(N2, x2) --)(N2 x P2, 2)(N2, x2) where for
k = 1, 2
ik
=
germ at
xk
of the inclusion Nk - Nk x Pk
given by x H Sk
=
germ at
zk
of the projection Nk x Pk -
given by In other words,
H
(x, yk
Nk
(x, y) '- x.
is necessarily given by a formula 143
H(x, y)
with
O(x, y1)
=
(orX-equivalent)
y2.
(h(x), 9(x, y))
=
And we call our germs
with
N1
contact-equivalent
when there exists a contact-equivalence
H0(1, f1)
Thus one pictures
f1, f2
H
mapping onto
(1, f2) o h.
=
as mapping the graph of N2
via
for which
(h, H)
f1
onto the graph of
f2,
h.
P2
H
h
The reader will have no difficulty in checking that if A-equivalent then automatically they are X-equivalent. discussing._*=equivalence of germs
tion to suppose that
N = ]Rn,
(N, x)
x = 0
and
hand as we shall see two germs which are A-equivalent.
(P, y)
-+
P = ]Rp,
f1, f2
are
In particular, when it will be no restricy = 0.
On the other
equivalent are not necessarily
In the remainder of this section we shall justify our state-
ment that=equivalence is a fairly computable notion by deriving a more-orless algebraic criterion for two germs to be-7,-equivalent.
We do this in
The first step, which is the main one, is to look at a finer
two steps.
notion of equivalence called
8-Equivalence.
For the sake of simplicity let us restrict our attention to germs (]Rn,
0)
-+
(]RP, 0).
Two such germs
.equivalent via a-equivalence
144
(h, H)
f, g
are 6-equivalent when they are
having the special property that h
is the germ at f, g
h
0
of the identity mapping on IRn.
Note therefore that
are.-equivalent if and only if there exists an invertible germ
ORn, 0) -+ (IRn, 0)
:
f o h, g are g-equivalent.
for which
proposition provides us with a purely algebraic criterion for g-equivalent.
To simplify notation write
If =
The next to be
f, g
the ideal in
i.e. P
9n
generated by the components
f1, ..., fp
of
f, g in 90n,p
The following three conditions on germs
(2.1)
And similarly for
f.
Ig.
are equi -
valent.
(i)
(ii)
(iii)
f, g
are e,-equivalent.
The ideals
are equal.
If, 19
There exists an invertible entries in
for which
8n
p x p fi
matrix
=
(uij )
u..gj .
for
with
1
J
It will be convenient to split the proof into a number of fairly easy steps.
In which we show that (i) implies (ii).
Step 1
g-equivalent.
It will suffice to establish
follows by symmetry.
f, g
Suppose
If C Ig,
are
Ig C If
since then
We must therefore show that each component
fi
can be
written in the form
f.1 aid
with the
g-equivalence H(x, y) x,
=
in
(1, H)
f, g
Since
9n
where
and hence its components
0
(1)
are g-equivalent there exists a
H(x, g(x))
for which
(x, O(x, y))
- ai g.
_
satisfies
B1, ..., Bp
=
Write
(x, f(x)). 0(x, 0)
satisfy
=
0
identically in
6i(x, 0) = 0
identically
145
in
By the Hadamard Lemma we can write
x.
6.(x, Y)
Yj61J(x, Y)
_ J
with
the standard co-ordinates in IRP,
y1, ..., yp
and the
0ij
in
9n.
But now
fi(x)
6.(x, g(x))
=
S' gj(x)eij(x, g(x))
=
J
which has the desired form (1) if we put
Step 2
I
0ij(x, g(x)).
=
In which we isolate a lemma from linear algebra to be used in the We claim that given real
next step.
matrix
p x p
real
aij(x)
is the identity
C
matrices
U = C(I - AB) + B
for which
p x p
p x p
To this end let
matrix.)
A, B
there exists a
is invertible.
(Here
a, b
be the
IRP - IRP
:
linear mappings whose matrices, relative to the standard bases, are respectively
Choose a basis
A, B.
e1, ..., e
for which
for IRP
b(ei) = 0
p
for
r + I
5
i
p,
.
b(e1), ..., b(er)
where
denotes the rank of
r
the linear mapping defined by r + 1
:
i
.
so can be exten-
are linearly independent vectors in IRP,
ded by vectors ex,+1, ..., ep to a basis for IRP.
for
then
B:
p.
that for some scalars
Observe that A..
for
c(ei) = 0 u
=
1
Now let i s r,
c(1 - ab) + b
c :
and
IRP - IRP be
c(ei) =
has the property
one has
for
je!
+
1
si
r
j=r+1
for r+ 1 s r s p so
146
u
is invertible, and the matrix
C
of
c
ei
is the required matrix.
Assume that the ideals
In which we show that (ii) implies (iii).
Step 3
are equal, so we can certainly write
If, 19
a
gi
f
13 d
J
J
where the coefficients the real
p x p
lie in
ai3, bi3
matrices with entries
Step 2 we can find a real
For each
a ..(x), 1j
b l..(x)
matrix
p x p
It follows that for
AXBx) + Bx
fin.
C0
with
x close to
0
x let
Ax, Bx
respectively.
be By
U0 = CO(I - A0B0) + BO
the matrix
is invertible as well.
ui3(x)
Let
be the entries
It is now a matter of computation to check that
fi
Y' u..gj
=
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).
matrix
p x p
Let us define a germ
0
:
(ui3)
with the
(1Rn x ]RP, 0) - (1Rp, 0)
by taking its components to be
6.(x, Y)
y.ui3(x)
_
(1 5 i 5 p).
3
Certainly then H
:
0(x, 0)
=
0
identically in
x.
by the formula
(]Rn x ]Rp, 0) , (]Rn x 1Rp, 0)
One checks easily that the Jacobian matrix of is an invertible germ, and struction H(x, g(x))
81(x, g(x)) =
(x, f(x))
=
(1, H)
fi(x)
Now define a germ
H
at
H(x, y) = (x, 0(x, y)). 0
is a 9-equivalence. so
0(x, g(x))
entailing that
f, g
= f(x)
is invertible, so
H
Moreover, by conand
are g-equivalent.
11 147
Example I
The germs
f(x)
(x2, 0)
=
and
g(x)
(x2, x3)
=
are
g-equivalent, since the ideals generated by the components clearly coincide, and hence.'-equivalent.
On the other hand
are not A-equivalent.
f, g
Suppose indeed that we could find invertible germs
k
:
(IR2, 0) _ (IR2, 0). for which f o h k would map
tatives, we see that
h
k o g.
=
-* (IR, 0)
and
"forking with representhat however
Im g - {0}:
to
Im f - {0}
(IR, 0)
:
is impossible since the former set is connected, whilst the latter is disconnected.
Consider a germ f
Example 2
:
(IR, 0)
for which the follow-
(IR, 0)
-+
ing conditions are satisfied. k+1 0
It follows that f =
xk+1g
for some
g
(IR, 0) - (IR, 0) with g(O) / 0.
:
(The argument appeared twice in Chapter IV.) algebra
so
<xk+1>
0.
axk+1 (°)
Now
g
and we conclude that
is invertible in the f
is g-equivalent to
k+1
x
Let
Example 3 with components
f, g
fi, gi
(IRn, 0) -+ (PP, 0) be germs of linear mappings,
:
respectively.
same image we can write each
fi
If these linear mappings have the
as a linear combination of
with scalar coefficients, and similarly each f1, ..., fp
with scalar coefficients: f, g
coincide, so
are 9-equivalent.
gi
8n,
in
11+8
fi
and each gn.
fi, gi
f, g
Indeed in this case we can
g1, ..., gp
as a linear combination of
However as the
If, 19
Conversely, we claim that if
as a linear combination of gi
as a linear combination of
certainly then the ideals
are g-equivalent then they have the same image. write each
g1, ..., gp
with coefficients in
f1, ..., fp
with coefficients
are linear, the coefficients must be scalars
so
f, g
give rise to the same image.
Back to X-Equivalence Although we managed to produce a purely algebraic criterion for two germs to be g-equivalent, we have to be content with somewhat less for ge'-equivalence.
As a preliminary, call two ideals in algebra isomorphism of
an
do
to itself which maps the one ideal onto the other.
We need a more special notion.
Two ideals in
when there exists an isomorphism of germ
h
:
isomorphic when there exists an
9n
are induced isomorphic
to itself, of the form
&n
h*
for some
which maps the one ideal onto the other.
(IRn, 0) -+ (IR", 0),
Note that by (IV.2.11) the germ
h
is necessarily invertible.
We contend
that
A necessary and sufficient condition for two germs
(2.2)
f, g
:
(]Rn, 0) -+ (IRn, 0)
to be-equivalent is that the ideals If, 19 are
induced isomorphic. Necessity
germ
h
:
Suppose
are X-equivalent, so there exists an invertible
f, g
(Mn, 0) -+ (IRn, 0) for which
follows from (2.1) that the ideals
morphism
will map
h*
which maps
h*
Sn, Ig.
f o h, g
with
Ifoh, 19 Ig,
are equal.
=
The induced iso-
induced isomorphic, so there exists an induced h
:
(Mn, 0) -+ (IRn, 0)
That means the ideals
an invertible germ,
Ifoh, 19
are g-equivalent, and hence
(x2 + y2, xy)
It
so these ideals are induced isomorphic.
We have already observed that the germs
Example 4 g(x, y)
of
onto
If,
(2.1) the germs
and
onto
Suppose If 19
Sufficiency isomorphism
If,
f o h, g are g-equivalent.
are equal, so by
f, g are -equivalent.
f(x, y)
=
are A-equivalent hence X -equivalent.
(x2, y2)
On the 149
other hand these germs are not 9-equivalent since (for instance) the compo-
nent xy does not lie in the ideal generated by x2, y2.
Thus far we have vindicated our claim that.'-equivalence is a fairly computable notion.
And being a coarser relation than A-equivalence one has
rather more hope of being able to classify germs under_5i'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 Xequivalence.
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.
Observe first that the set X
This goes as follows.
lences (h, H) H
:
with h
(]Rn x IRp, 0)
-+
:
(]Rn, 0) -+ (]Rn-, 0)
(1, H)
H.
and
forms a group, the operation being
(IRn x ]R>, 0)
given by composition of the lences
of all,'-equiva-
n,p
Note that the set
is a subgroup of 9l n,p
:
g
n,p
of all g-equiva-
also that we can identify the group
Rn discussed in Chapter III with the subgroup of n,p comprising all equivalences
(h, h x 1).
ment
in 9ln,p
(h, H)
(h, h x 1) H'
=
in
can be written uniquely as the product of the element
and the element
R n
(h x 1)-1 o H
direct product of
With this identification it is clear that any ele-
SEn
(1, H')
in
d
n,p
where
one expresses this by saying that 9l n,p and
gn,p
The group
nip
is the semiby agree-
acts on dnsp
ing that (1, (h, H).f) = H o (1, f) o h-1
and it should be clear that two germs
150
(IRn, 0) - (IR&, 0)
lie in the same
orbit under this action if and only if they are.5 equivalent.
Thus the prob-
lem of classifying germs under,'-equivalence is that of classifying the orbits under the above action.
Now we proceed as in Chapter IV.
We pretend that the action of mil n,p
on
is that of a Lie group on a smooth manifold, and look for a candidate n,p
for the "tangent space" to the orbit Zf through the germ
f
:
(]Rn, 0) - (]RP, 0).
We expect this to be the image of the "differential"
at the identity of the natural mapping of the group onto the orbit through
which we can think of as a mapping
n
x g
- 9n , p
n,p
f,
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 gn,p
The Mapping Rn
-'
and
8n,p
0
This is given by the formula
Write
Chapter IV.
_+
We consider these separately.
8n,p
_+
Rn
h
for the components of
f1, ..., fp
the germ with components
of
of 1
ax1
,
standard co-ordinates in ]Rn.
One reasons exactly as in
f o h 1.
-,
..., ai
where
f,
for
and
ax1 denote the
x1, ..., xn
We expect the required "differential" at the
identity to be the linear mapping
(g1,
gn)
If-
g1 ax
of + ... + gn axn
*
1
with each
gi a germ (]Rn, 0) - (]R, 0).
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 module
written
g
n,p Jf.
*
:
en.
will be the submodule
Thus the image of the linear mapen {ax1
,
of } ..., axn
of the
this submodule is called the Jacobian module of Of course in the case
p = I
f,
d n and is
we recover the Jacobian ideal 151
introduced in Chapter IV.
9n0
The Mapping
gn,p
-,
,p
To simplify life identify a g-equivalence
with
where
H(x, y) = ( x, 0(x, y))
Then our mapping is given by composition on the right with (1, f)
0.
so is the restriction of a linear at any point.
mapping, and should be its own "differential"
We need therefore the image of this mapping taken over germs
for the components of
0.
Each component
identically in x when y = 0,
on IRp.
f1, ..., fp
with coefficients in
that every element of comprises all germs in
I.
likewise vanishes
of the) standard co-ordinates
0
y1, ..., yp
01(x, f(x))
generated by the components of
Oi(x, y)
01, ..., 0p
so by the Hadamard Lemma must lie in the
ideal generated by the (germs at Thus
Write
x when y = 0.
which vanish identically in
0(x, y)
(1, H)
f:
6n,
is a linear combination of i.e.
an element of the ideal
If
reversing the steps one sees moreover Thus the image of our mapping
can be so obtained.
8n,P with components in
If,
i.e.
submodule
the &
If.8n,p* On this heuristic basis we arrive at the following formal definitions. Given a germ
f
:
(IRn, 0)
to be the &n submodule of
f
Tf
-->
=
(IRP, 0)
we define the X-tangent space to
if + If.9n,p.
And we define the
to be the codimension of this vector subspace in
g
n,p'
f
'-codimension
,
In view of
(IV.2.7) we have the following criterion for a germ to be of finite X codimension.
(2.3)
A necessary and sufficient condition for a germ
f
:
(IR,
to be of finite-Y-codimension is that there exists an integer k >.
-//n . En,p C Tf 152
0)
1
with
This is fairly straightforward to apply in practice. f
of
Then the
f1, ..., fp
as
If.9
aft
ax
( la x.1
=
1
an,p
as
&n x ... X en
will be generated by all p-tuples
n,p
in any position, whilst
fi
8f
and think of
0),
with
Jf
is generated by the
Tf
is generated by the list of all these
of thus
..., ax 1 k
,
x1, ..., xn,
and hence ,llri gn,p
m
k
of degree
is generated by all p-tuples
in
(0,
...J,
m,
k
And to check the condition
..., 0).
times).
(p
-$ fi, ...,
(0,
Now _ll n is generated by all monomials
vectors.
Write the components
.dln.8no C Tf,
one has only to check
that each such p-tuple can be written as a linear combination (with coefficients in
of the generators for
9n)
k = 1, 2, 3, ...
Doing this successively for
there is at least a sporting chance that in a given example
one will either find a
k
for which the condition is satisfied, or see that
it cannot be satisfied for any
Example 5
Tf.
k.
We shall show that the germ
f(x, y)
=
(x2, y2), the germ at
of the "folded handkerchief" mapping, has finite. -codimension.
0
= (2x, 0),
ax
(y2, 0),
ay
(0, x2),
a2-submodule of
=
(0, y2).
2$2
Also
(0, 2y).
If.92.2
is generated by
Clearly then the X-tangent space
generated by
(x, 0),
(y2, 0),
(0, y)
(x2, 0), Tf
and
We start by trying to verify the condition of (2.3) in the case The ideal 2 s
2
.,lli
is generated by the monomials
is generated by
(x, 0),
(y, 0),
x, y (0, x),
so the (0, y).
Here
is the (0, x2).
k = 1.
92-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.
For
(x, 0), (0, y)
(y, 0), (0, x)
this
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 Yl 2
is generated by the monomials
x2, xy, y2
so
153
the
(x2, 0), (xy, 0), (y2, 0), (0, x2),
is generated by
2
Again, the question is whether each of these vectors can
(0, xy), (0, y2).
be written as a linear combination of the four generators for this is the case, so .,12.92,2 C If,
and
f
Tf.
Clearly,
has finite.'- codimension.
Before going any further we should check that the X-codimension of a germ
is actually aX-invariant,
(2.4)
i.e.
that
(IRn, 0) - (IRI; 0) are .5'-equivalent then they have
If two germs
the same. -codimension. For the purposes of the proof we shall identify
Proof the product
gn x ... X don
the p-tuple
(f1,
..., f
(p times) by identifying a germ
f
d =
in
8np g
with
with
of its components relative to the standard co-
)
p
ordinates on ]Rn,
Let
Step 1 9n.
in
given by
u = (u.j )
We take
U
U
f
matrix with entries
as a column vector.
isomorphically onto
Tf
maps
p x p
to be the isomorphism of real vector spaces
-+ d
Tg.
T9 C U(Tf),
and hence
U(Tf) C Tg.
Put
g = u.f.
It will suffice to
for then similar reasoning establishes
show that
Tf C U-1(Tg)
Clearly then the problem reduces to that of estabIg.uo C U(Tf)
lishing the two inclusions.
observe that Ig.d C If,& Cramer's Rule.
be an invertible
where we think of
f -p u.f,
We claim that
We proceed in steps.
IR12.
=
and
U(If.&) C U(Tf),
Jg C U(Tf).
For the first,
the equality following from
And for the second observe that
Jg C U(if) + If.&
=
U(Jf) +U(If.&)
=
U(Tf),
the inclusion following from
the rule for differentiating a product of two functions. Step 2 :
154
& -+ e
Let 0
:
(]Rn, 0) - (IR.n, 0) be an invertible germ, and let
be the isomorphism of real vector spaces given by f - f o 0.
g = f o 0.
Put
We claim that
45
maps
isomorphically onto
Tf
in Step 1 it suffices to establish an inclusion to establishing the two inclusions
Tg C 45(Tf)
and
Ig.9 C 9 (Tf)
first inclusion follows immediately from the fact that
As
Tg.
which reduces
J9 C i (Tf).
The
Ig.& C 'I(If.&).
And
the second inclusion follows from the fact that, by the Chain Rule, we have,
1 si sn,
for
a
axi
-7
a
j ( af
axi
laxj
0
0)
.
j=1
Step 3
Let f, ,h
:
(IRn, 0) - (IRP, 0) bey-equivalent germs.
there exists an invertible germ 0
for which
And by (2.1) there exists an invertible in
9n
for which
morphism
U o
h = u.g.
i.e.
f, h
matrix
are g-equivalent.
u = (uij)
with entries
It follows from the previous steps that the autoa
of the real vector space
so the quotient spaces
p x p
g = f o 0, h
Then
S/Tf, &/Tn
&
maps
Tf
isomorphically onto Th,
are isomorphic and have the same dimension,
have the same X-codimension.
We come now to the question of actually computing the y -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 dif-
fers from the situation studied in Chapter IV.
f
:
(IRa, 0) -+(]R, 0).
case arises when
f E Jf,
Consider then a germ
They'-tangent space to f is Jf + If.
A special
since then they-tangent space reduces to just
Jf,
and the X-codimension will coincide with the R-codimension as defined in For instance, this applies to the Ak singularities
Chapter IV.
1+
k+1
- n-1 +- xn
x2 + ... + x2
bolic umbilic
x3 + y3,
to the elliptic umbilic
x3 - xy2,
and to the parabolic umbilic
x2y +
to the hyperBut in
general the relation between the two codimensions is not well understood at
155
the time of writing.
It is known that a germ of a function is of finite
-codimension if and only if it is of finite R-codimension, 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).
f
:
(]RR, 0) - (]RR, 0).
just
cod f,
Consider a germ
f,
One writes cod f for the X-codimension of
or
when it is quite clear that we are dealing withx-equivalence.
The proposition just mentioned tells us that
cod f
=
cod0f + cod If + cod2f + ...
where
Tf + lln 9n
dimTf
codkf
n
What we do is to compute the integers as follows.
,p
+ ,P
cod0f, cod1f, ...
Suppose we wish to compute
codkf.
successively, as
As we have already pointed namely the
out we have an explicit finite list of generators for .,llnk.gn,p
p-tuples
(0,
x1, ..., xn.
with m
..., in, 0, ..., 0)
a monomial of degree
k
in
The first thing one has to do is to check which of these gen-
erators lies in Tf
+,/n+1.&n'p
:
this usually amounts to doing a small com-
putation for each generator in turn.
There is a practical point to note here
namely that we need only consider the generators for
n+1.8n,p,
Tf
which means in practice that for each such generator we can put equal to zero all terms of degree
The next thing to do is to select from the
>. k + 1.
generators which do not lie in
Tf +-n.cn
Tf +-I/n+1.9n,P
a basis for a supplement in
the number of basis elements is the number
codkf.
Of course,
sp
the process comes to an end when one finds an integer k >
the generators
for.,lln.En,P
the list of integers 156
cod0f,
lie in
Tf #,/1n+1.gnaP.
cod If, ...
to get
1
for which all
Finally, one adds up
cod f.
Note that if none
of the components
f1, ..., fp
n.8nP and hence
Tf C
f
of
involves linear terms then
cod0f = p:
this remark will apply to all the
examples below.
Example 6
f
:
We shall compute the 5 .codimension of the germ
0) - (IR2, 0) of the "folded handkerchief" mapping given by
(IR2
f(x, y)
(x2, y2).
_
As we saw in Example 5 the
generated by
submodule of
822,2
cod0f = 2.
To compute
for 112.2,2
lie in
(y, 0), T
(0, x),
supplement for
(y2, 0),
(0, y),
(0, x2).
Now 2.82,2
Tf +,, 2.92,2 of these
(y, 0),
(0, y)
do not.
(0, x)
in
Tf +,-//2-'g2.2
(x, 0),
is generated by
codf
2+2
=
_
ule
Jf
(0, ..., 0, xt+1)
is generated by
by the vectors Tf
(0,
with (0,
Indeed these vectors form a so
Tf +,"(/2-62.42
f
cod If =
(IR, 0) - (IRP, 0)
:
t ,
an integer.
1
..., 0, xt),
..., xt+1P
...,
is generated by all vectors
We need go
2.
2 C Tf.
(0,
...,
(0, ..., xk,
erators lies in the submodule Tf
Here the Jacobian mod-
and the ideal
xt+1S ..., 0)
(0, 0, ..., xt).
The ideal ,,/f k is generated by xi ,
generated by the vectors
defined by
If
is generated
it follows that the -tangent space
0):
not appear in the last component, and
codkf.
(x, 0),
4.
Consider the germ
Example 7 f(x)
=
Here
obviously lie in
no further now because, as we saw in Example 5, we have,,l12.92
Thus
is the
cod1f we have first to determine which generators
(0, y):
whilst
2.82 s 2
(x, 0),
Tf
tangent space
,P
Let us compute
so the submodule .,111.8is lop
..., 0).
+,,111+1.81
where the power does
We ask which of these gen-
There are three cases to con-
sider.
The Case Tf
k < t
*d11+1.81P :
None of the generators for.,llk 8 1' 1,p
there are
p
lies in
such vectors, and they form a supplement, so
,
157
codkf
k=t
The Case Tf +4/1
for k < t.
p
=
+1.81,p
save the last one
(0, ..., xt):
erators, forming a supplement, so
k>t
The Case
codkf =
so
0
f
:
(IR2, 0)
-+
3.
the submodule
f
Here 2x =
+-*/I+1.81
Tf
jp,
is generated by
If.g2,2
(y2, 0),
pute
codkf
for k 3 1.
with
i + j
=
(0, xa),
(X' +
The ideal-A12
with
these vectors lie in
Tf,
i + j
=
y
(0, xy)
0),
,
k.
(0, xk),
,
sy
together with
We wish to com-
(0, yb).
is generated by the monomials
so certainly in
(yk, 0),
ax
(0, xy),
so the submodule ,112.82 2
(0, xiy )
(xk, 0),
xa + y so
The reader will find it a straightforward exercise to
(xy, 0),
k,
b
(xy, 0),
generate
(x, by -1)
is generated by xy,
If
a, b are
where
(xy, xa + yb)
and ay =
(y, axa-1)
The ideal
Jf.
=
f(x, y)
verify that the X-tangent space is generated by
consider
lie in
has Y-codimension pt + p - 1.
given by
(IR2, 0)
(0, xa + yb).
(x1y3, 0),
k = t.
for k > t.
the Jacobian module
(x2, 0),
when
We shall compute the,5.codimension of the germ
integers
and
p - I
=
codkf
there are (p - 1) such gen-
All the generators for .111.&1 "p
We conclude that the germ
Example 8
lie in
None of the generators for,// 1k.8 1,p
xiyj
is generated by the vectors
Clearly, if Tf +,1112+1.82
(0, yk).
i ,
I
and
j
1
>.
It remains to
2'
We claim that the first
two likewise lie in T f +4'2+1.92,02 : when k = I this is because (x, 0)
=
ay it is because
- (0, (xk, 0),
byb-1),
(Y, 0)
(yk, 0)
= ax -
are multiples of
appeared in our list of generators for Tf. lies in
158
Tf *2+1.82 2
only when
(0, axa-1),
k >, a,
and when k , 2
(x2, 0),
(y2, 0)
Finally, the vector
whilst the vector
which
(0, x
k (0, y )
k )
does
so only when k > b. therefore the I
The vectors which span
s k s b - 1.
Example 9
a > 3
f
:
(IR2, 0)
-
is an integer.
a).
2
+ y
(0, x
),
(IR2, 0)
Here if.
given by f(x, y) (2x, axa-1)
of = The ideal
with
will be
(x2 + y2, xa)
=
and
ay =
(2y, 0)
is generated by x2 + y2,
If
is generated by
so the submodule 2
f
)
a + b.
=
generate the Jacobian module
(0, x
(0, y
For our final illustration we shall compute the-T=codimension
of the germ
xa
k
and the
s k 5 a - 1,
1
It follows that the, -codimension of
2 + (a - 1) + (b - 1)
where
with
(0, xk)
the required supplements are
(x2 + y2, 0),
(x a, 0),
Following through a computation similar to that in
the previous example one finds that a supplement for the ,=tangent space in is provided by the (0, xi) so the,7e=codimension is
§3.
and the
(0, xl-1y)
2 + (a - 1) + (a - 1)
with
1
. i s a - 1,
2a.
=
Deformations Under Contact Equivalence
The next step in our programme is to set up the basic ideas for a theory of deformations of germs under,'-equivalence:
the relevance of this to the
problem of classifying stable germs will be discussed in §4.
Let us start with an r-parameter deformation of a germ
f
:
(IRm, 0)
-a (IRq, 0).
F
:
(IRr x IRm, 0) -+ (IRq, 0)
Pursuing the analogy with the Finite
Dimensional Model of Chapter III one expects a major role to be played by "transversal" deformations.
We need a formal interpretation for this in-
tuitive idea, so we argue heuristically.
Think of the deformation as a
159
"germ"
(IRr, 0) -,
(gmsq ,
f) given by u --> fu where fu(x)
F(u, x).
=
We wish this mapping to be "transverse" to theX-orbit through
f,
i.e.
we
want something like
image of the "differential" at 0 of this map
tangent space to X- orbit through f
+
tangent space
to
gm.,q
at f.
The only quantity here for which we do not yet have a concrete interpretation is the "differential" at ear mapping Iu=O
au
-,
IRr
m,q au
ordinates on IRr.
r
of the mapping
0
this should be the lin-
which sends the standard basic vectors for IRr to where we write
1u=0'
u - fu:
u1, ..., ur for the standard co-
As a matter of convenience we write
F.
= aui
(u=0
With this notation the image of our "differential" will be the real vector On this heuristic basis we introduce
dm,q.
of
subspace IR{F1, ..., FrJ
the following formal definition.
F
is aJr-transversal deformation of
f
when
IR{F1,
Notice therefore that
..., FrI + Tf = am
s q.
admits ax transversal deformation if and only if
f
it has finite.'-codimension
a,
say,
Assuming this to be the case one can
construct explicit X-transversal deformations by the same device used in the One looks for a deformation
Finite Dimensional Model.
F(u, x)
where the germs
F
=
f1, ..., fc
for which
f(x) + u1f1(x) + ... + ucfc(x)
are to be determined.
The coalition for this
to be -V-transversal is that IR{f1, ..., fc1 + Tf
This yields an entirely practical procedure. 160
F
=
gm,,q.
Simply choose f1, ..., fc
to
Tf,
be a supplement to
and then define
by
F
*.
In practice one is interested solely in X-transversal unfoldings of germs of rank
(Just why this is so will be made clear in the next section.)
0.
For such a germ
one can be a little more explicit about the form of a
f
transversal unfolding. product
e1, ..., eq for the standard basis vectors in 7Rq.
We write
is a germ of rank
f
f1,
find a basis
Tf
supplement of
then a moment's thought will convince the
0;
is a vector subspace of
Tf
reader that
&mpq with the
as a real vector space we can think of this as the direct sum of
8m :
IRq with E. Suppose
Make the usual identification of
"'s'
in
fr am
_4,q.
It will then be possible to
for a supplement of
Tf
is then provided by
e1, ..., eq,
in flm
a basis for a f1, ..., fr'
and we obtain a.5 transversal deformation
r F(u, w, x)
- J wi.e
_
+ f(x) +
i=1
i=1
where we insert the minus sign for a minor geometric reason which will be menHere are some examples of these computations,
tioned in the next section.
parallel to the computations of, '-codimension given in §2, where all the work
In each example the germ
was done.
The germ
Example I
has X-codimension J+: (1, 0),
provided by mation is
F
:
(IR4
f
(IR2, 0)
:
has rank
f
-+
(IR2, 0)
0.
given by f(x, y) = (x2, y2)
indeed, we saw that a supplement for Tf in 82,2 is (0, 1), x IR2, 0)
(y, 0),
(0, x).
(]R2, 0)
-+
with components given by
F1
=
x2 + u1y - w1
F2
=
y
2
Thus aX-transversal defor-
+ u2x - w2.
161
The germ
Example 2 f(x)
=
(0,
f
..., 0, xt+1),
indeed a supplement for for which
(IR, 0)
:
F
:
t
is provided by the deleted.
..., 0, xt) 0)
(0,
...,
xk, ...,
0)
Thus a,=transversal
with components given by
-+ (IRP, 0)
-w1 + u11x + ... + u1st-1x
t-1
+ u1'tx
t
1xt-1 + u2,txt
-w2 + u21x + ... +
=
c = pt + p - 1:
has_,*-codimension
1,
1,p
(0,
(]Re x IR,
F1
F2
in
f
with
0 s k . t
deformation is
T
with
defined by
-, (IR', 0)
,
2 t - 1
The germs
Example 3 f(x, y)
I
(1, 0)
F
:
(xy, xa + yb),
:
(IR2, 0)
with
indeed a supplement for
a + b:
with
=
f
c- i E a - 1,
and
(tea+b
a, b
(0, y
with
)
x IR2, 0) - (IR2, 0)
F1
=
xy - w1
F2
=
xa
The germ =
together with
with components
+
y
b-1
uixl +
+
f
(x2 + y2, xa),
:
(IR2, 0)
where
viyl - w2 i=1
-+
(IR2, 0)
a 3 3,
given by
has, -codimension
ment for Tf in 92,2 is provided by the (0, x') together with (1, 0) and (0, 1). 1 s i s a - 1,
deformation is the germ 162
(0, x1)
Thus a,=transversal deformation is the germ
(0, 1).
i=1
f(x, y)
5 i , b - 1,
1
a-1
Example 4
is provided by the
&2,2
i
and the
3, have X-codimension
integers
in
Tf
given by
(]R2, 0)
-+
F
:
(IR2ax IR2, 0)
-+
and
2a.
A supple-
(0, x1-1y) with
Thus a X-transversal
(IR2, 0) with components
F1
=
x2 + Y2 - w1
=
xa
a-1
a-1 F2
uixl
+
+
i=1
So much for examples.
viyi - w2
1=1
The next step is to pursue the analogy with the
Finite Dimensional Model further to see if we can characterize the algebraic notion of "transversality" by a geometric notion of "versality".
To this
end we introduce a series of notions for deformations of a germ
f
:
(1Rm, 0) -+ (IRq, 0).
Equivalence of Deformations Two r-parameter deformations
F1, F2
there exists an r-parameter unfolding mapping on ]Rm
f
of
are said to be.Y.equivalent when of the germ at
Im
0
of the identity
for which
Im(F1(.,llq)) In this situation we call
F2(
=
Im aX-equivalence of deformations.
(2.2) this relation implies that
F1, F2
Of course by
are-Z. .equivalent as germs:
however
it says more in that the change of co-ordinates at the source has to respect the product structure on ]Rr X ]Rm.
Induced Deformations Suppose
and that
F
(]Rr x Mm, 0) -+ (]Rq, 0)
:
H
deformation
(]Rs, 0) -+ (]Rr, 0)
:
G
:
is an r-parameter deformation of
is a germ.
(]Rs x IRm, 0) -+ (]Rq, 0) G(v, x)
=
of
f,
We obtain an s-parameter
f by putting
F(H(v), x)
.
163
One writes
situation
G H
H*F,
=
and calls
the deformation induced by
G
in this
H:
is a change of parameter.
Morphisms of Deformations
Let
F, G
be r, s-parameter deformations of
is a pair H
with
(H, I)
H G
under
I.
A morphism from
F
to
G
ai equivalence of r-parameter deformations, and
I
a change of parameter, for which
mation
f.
When
F
r = s
is,' equivalent to the induced deforand
H
is invertible we refer to the
morphism as an isomorphism.
Versal Deformations
A deformation
G
morphism from
F
f is 5- versal when for any deformation F there is a to G. When f has finite,=codimension c, say, a cof
parameter,'-versal deformation is said to be, -universal. The main result about deformations under contact equivalence is the following analogue of (111.5.1) which we dub the X-Versality Theorem.
(3.1)
Let F
of a germ f
for
F
:
:
(]Rr x
]Rm, 0) - (]Rq, 0) be an r-parameter deformation
(I1m, 0) -+ (]Rq, 0)
:
a necessary and sufficient condition
to bej -versal is that it should beX-transversal.
A word or two is in order concerning the analogy between this result and the Versality Theorem of Chapter III.
In that result the key was the exist-
ence of neighbourhoods having a product structure, which were produced by the Inverse Function Theorem.
But in the present situation the basic objects
lie in a vector space of germs:
one has no immediate analogue for the Inverse
Function Theorem, and is forced to adopt a different stratagem. 161+
The proof
that .7-transversal deformations are X-versal is by no means easy, and we shall not give it;
it uses an extension to the real case of a classical
theorem of Weierstrass in complex function theory.
A sketch of the result
can be found in the paper of J. Martinet quoted in Appendix E.
On the other
hand, the converse result is relatively trivial.
Proof of Necessity Suppose i.e.
is X-versal.
F
We have to show that
F
is -Y-transversal,
that
...,
]R(F1, Consider then a germ
g
in
,,q,
Fr} + Tf
=
m ,q
and the 1-parameter deformation
G
of
f
given by G(v, x)
G
By hypothesis tion say.
Thus
f(x) + vg(x).
is X -equivalent (as a deformation) to an induced deformawith
h
H(t, x)
=
H = h*F
=
:
(R, 0)
->
F(h(t), x)
having components
(]R", 0)
and
8h
H
-
8h
at (0)F1 + ... + at
establishing that H lies in ]R[F1, ..., FrJ. G, H
Tf,
(0)Fr
Starting from the fact that
are X-equivalent deformations a computation shows that
and hence that
G
lies in
h1, ..., hr
Fr} + Tf.
G - H
lies in
However G = g,
finishing the proof.
Just as in the Finite Dimensional Model we are now in a position to justify the use of the prefix in the term
"X-universal".
165
Let
3.2
f
:
F, G
be X -universal deformations of a germ
of finite '-codimension
(]Rm, 0) -+ (]Rq, 0)
,
then F, G are
c:
isomorphic deformations. Proof for which
G
is '-versal there exists a germ h : (]R', 0) -+ (]Rc, 0)
F
As
is .T'-equivalent to the induced deformation
there exists a c-parameter unfolding the germ at
:
(1R0
of the identity map on ]Rm
0
9-equivalent,
i.e.
We shall show that
F o 'P,
G
X ]Rm, 0) -+ (RC X ]Rm, 0)
for which
T =
are 9-equivalent, where
with entries in
for which F o
9m+c
= AG,
(h x 1) o -1.
q x q
one obtains for c
I
In
matrix
with the usual
Differentiating this relation with respect to
setting u = 0,
of
are
is an invertible germ, which we do as follows.
h
identifications.
G
h*F o
view of (2.1) this means that there exists an invertible
A = A(u, x)
That means
h F.
ui,
and
a relation of the form
s i s c
ahj au.
AOGi + T.
(0)Fj
j=1
where Now
F1,
A0
F. G
=
AO(x)
=
A(0, x),
and
lies in the ,5 -tangent space
Ti
are both X -transversal deformations of
..., Fc span a supplement for
Tf, as do
f
Tf
by (3.1) so
G1, ..., Gc.
In view of the
proof of (2.1+) the invertible matrix A0 induces a linear automorphism of the 8m-module
&m which leaves Tf invariant.
a supplement for ah. au. 1
(0),
i.e.
If.
It follows from
the Jacobian matrix of
h,
F, G
A0G1, ..., AOFc also span
that the matrix of coefficients
*
tion Theorem now enables us to deduce that It follows from the definition that
Thus
is invertible: h
the Inverse Func-
is invertible, as was required.
are -Y-isomorphic deformations.
In fact we can squeeze a little more information out of the proof than is actually stated in (3.2). 166
and let
C.
f
Let
(3.3)
:
-
m
(IR , 0)
(IRq, 0) be a germ of finite X-codimension
be a X -universal deformation of f.
F
parameter .X-versal deformation parameter constant deformation of versal deformations of
Proof
:
:
Thus
d 3 c any d-
is X-isomorphic to the (d - c)-
And hence any two d-parameter
F.
,
'-
-isomorphic.
Proceeding exactly as in (3.2) one comes to the conclusion that
(IRd, 0)
--1,
(IRc
0)
an invertible germ 7T
are
f
is ,Y-equivalent to an induced deformation
F'
h
f
of
F'
For
(3Rd, 0) F'
'
:
(3Re, 0)
a submersive germ.
F.
with
However by (1.1.3) there exists
0) - (3Rd, 0) for which
(3Ra
the projection given by
is -Y-isomorphic to
deformation of
h F,
ir'F,
i.e.
7r(u1,
h0o
with
i,., ud) _ (u1, ,
uc).
to the (d - c)-parameter constant
R
The rest is clear.
There is a small technicality which is worth mentioning at this point. "e have phrased the definition of X -isomorphism for deformations of a single germ
f.
One can of course phrase the definition for deformations of X-
equivalent germs and (3.3).
f, f'
and obtain results which correspond exactly to (3.2)
We shall leave this matter to the reader, and proceed rather to
the next step in our programme, which is to show how one can use the existence and uniqueness of
deformations to reduce the problem of classifying
stable germs under the relation of A-equivalence to that of classifying germs of finite
X-codimension under the relation of X-equivalence.
167
Classification of Stable Germs
§4.
The time has come to put together the bits and gain some distance into the
problem of producing explicit lists of stable germs in given dimensions. Our starting point is a very simple idea.
Let
(4.1)
G
:
(Itn, 0) - (Itp, 0) be a germ of rank r: then there
exists an invertible gem_ h
:
(]Rn, 0) -+ (]Rn, 0) for which F = G o h is
an r-parameter unfolding of a germ of rank 0. Proof
By making linear changes of co-ordinates at source and tar-
get we can suppose that the Jacobian matrix of
where g
:
Ir
is the identity
of
G.
an invertible germ h tion
Clearly, :
evaluated at
g
has rank
(]Rn, 0) - (Mn, 0)
so by
r,
for which
r
is
components
(1.1.3) there exists
g o h
And then F = G o h
(x1, ..., xn) - (x1, ..., xr).
0,
Consider the germ
matrix.
whose components are the first
(]Rn, 0) - (Itr, 0)
G1, ..., Gr
r x r
G,
is the projecis the required
germ.
F-I
The point of (4.1) as far as the present section is concerned is that since we are only classifying germs up to A-equivalence we can restrict our atten-
tion to r-parameter unfoldings
F : (]R" x ]Rm, O)
r: for such an unfolding write fF germ of rank F
-+
168
fF.
0
which
F
unfolds.
:
(]Rm, 0)
-*
-+
(]Rr x ]Rq, 0)
(]R q, 0)
of rank
for the unique
We wish now to study the assignation
The first step in this direction is provided by
Let F, F'
(4.2)
foldings of germs are
fF, fF,
]R m' 0)
(IRr x
.
fF, fF,
of rank
0:
x IRq, 0) be r-parameter un-
(7Rr if
F, F'
are A-equivalent then
,T-equivalent.
Since
Step I
F, F'
are A-equivalent there exist invertible germs
H, K for which the following diagram commutes.
( IRr x IRm,
0)
(Mr x ]Rq, 0)
F
T
T K
H
(Mr x 3P
0)
14
F
(1f?.r
x IRq, 0)
Notice first that in view of (1+.1) we can suppose that for some
0
(IRm, 0) -. (IRr, 0),
:
for some germ 0
:
g
:
and likewise that
(]Rm, 0) - (IRq, 0)
K(0, y) = (rb(y), y)
Write F(u, x) =
(IRq, 0) -+ (IRr, 0).
It should now be clear that
H(0, x) = (0(x), x)
(
f(u, x)).
,
is A-equivalent to the germ
fF,
given by g(x)
f(4 (x), x).
=
And it will suffice
to show that g, fF are X -equivalent. f(u, x) - fF(x)
Step 2
vanishes on
0 x
IRm,
so each of its
q
components does likewise, and the Hadamard Lemma allows us to write f(u, x )
=
fF(x) + M(u, x).u
entries are germs at
0
where
0(y)
=
0
write
B(x).g(x)
0(g(x))
=
where
A(y).y
whose entries are germs at _
is a
of functions on IRr x Fm,
And likewise we can write
q>(x)
M(u, x)
to the same matrix notation we can now write
matrix, for which
C(x)
a
q x r
g(x)
=
fF(x) + D(x).g(x)
with
C(0) = D(x)
a
0,
is an
_
0.
matrix
r x q
It follows that we can
an
B
matrix whose
M(0, 0)
and A(y)
of functions on IRq
with
q x r
r x q
matrix.
Keeping
g(x) = fF(x) + C(x).O(x)
with
and hence
q x q
matrix for which
D(0) = 0. 169
This last relation we can re-write as the identity
r x r
matrix.
fF(x)
Clearly,
and it follows from (2.1) that
g, fF
It follows that the assignation F
=
(I - D(x)).g(x)
I - D(x)
-+
fF
induces a mapping from A-
sive.
x ]Rm, 0)
-+
A sufficiently small representative of
f
the zero set
Vf
of codimension
= q.
f-1(0)
We wish to study
Vf
(]Rs
Suppose we have
which is submer-
(]Rq, 0)
will be a submersion, so
f will be a smooth submanifold of
of
We take
of the restriction to
:
lTf
:
is
are g-equivalent, hence -7,,"-equivalent.
this mapping in detail, and to do this we need a new idea. f
I
is an invertible matrix,
equivalence classes of germs to X-equivalence classes.
an s-parameter deformation
where
(Vf, 0)
of the projection
-,
v
(IRs, 0) :
]Rs
]Rs x ]Rm
to be the germ at
x ]Rm -> ]Rq.
0
We need
the following fact.
(4.3)
Let
f, g
:
(]Rs
x ]Rm, 0) -+ (]Rq, 0) be 9l-versal s-parameter
deformations of germs f0, g0
are X -equivalent then
g0
Proof
:
(]Rm, 0) -a (]Rq, 0)
1Tf, lTg
of rank 0:
if f0,
are A-equivalent.
It is convenient to make the preliminary observation that a
'-transversal deformation (and hence any X-versal deformation) of a germ of rank 0 is automatically submersive, so that the notation introduced above makes sense in the present context.
Observe also that
f, g
must be .Y-
isomorphic deformations, in view of (3.3).
Step I
f, g
Now consider first the case when f0 = g0.
are X-isomorphic deformations is expressed by the existence of a
commuting diagram of germs
170
The fact that
' (]RS x ]Rm, 0)
(IRS x ]R', 0)
Tr
IT
h
with 4), h
invertible for which (f o 4)#( ,,//q) induces a mapping from
relation ensures that
g' ( 1q)
=
Vg
onto
This last
. Vf,
yielding a
commuting diagram of germs
(Vg, 0)
'(Vg, 0)
Trg
iTf
h
(IRS, 0)
expressing the fact that
(]R5
0)
are A-equivalent.
frf, v
g
Consider next the general case when
Step 2
in
:
(]Rm, 0) -+ (]Rm, 0),
9m,
for which
deformation
g0(x)
g'(u, x)
=
follows from Step 1 that
fore to show that Irf, ?rg, will map
are just .-
By (2.1) that means that there exists an invertible germ
equivalent. h
f0, g0
Vg,
and an invertible
M(x).f(u, h(x)) Trg,
g0
of
is
are A-equivalent.
Trg,
M(x)
with entries
Evidently the s-parameter
M(x).f0(h(x)).
=
matrix
q x q
_
-versal as well.
And it suffices there-
that however is clear as
are A-equivalent:
It
1 x k
onto V.
11
The relevance of these ideas to the material of the present section is as follows.
Let
F
:
(]Rr
]If, 0)
-*
(]Rr x ]Rq, 0)
be an r-parameter un-
folding of a germ fF of rank 0, given by a formula F(u, x) = ( To
F we associate the germ
DF
:
(]Rr x
x
,
f(u, x)).
0) - (]Rq, 0) given by 171
-. w + f(u, x):
(u, w, x)
deformation of just graph
necting
f,
F, DF
F
(4.4)
fF.
thus
DF
is an (r + q)-parameter submersive
The geometric connexion between
F, DF
is that
VD
is
can be identified with
f.
F The basic theorem con-
is an A-stable germ if and only if
DF
is a - -versal deforma-
and
lTD
F
is
tion of fF. The proof of (4,4) does not use techniques lying outside the scope of this However a careful version would occupy more space than is available.
volume.
We shall therefore content ourselves with the statement of the result, and concentrate rather on showing the reader how one uses it to obtain explicit A sketch of the proof, sufficient
lists of stable germs in given dimensions.
for the competent reader, can be found in the paper of Jean Martinet mentioned in the Introduction to this book.
It is perhaps worthwhile spelling out the
fact that (4.4) together with the .'e'-Versality Theorem allows one to produce
explicit examples of stable germs.
Example 1
of the germ By the
We saw in Example I of §3 that a .'o-transversal deformation
is the germ given by
(x2, y2)
(x2 + u1y - w1, y2 + u2x - w2).
j -Versality Theorem this deformation is 5e-versal.
precisely the deformation associated to the unfolding y2 + u2x)
It is however
(u1, u2,
x2 + u1y,
so this germ must be stable.
We can now return to the main theme of this section by stating
(4.5)
Let F, G
:
(]Rr x ]Rm, 0)
->
unfoldings of germs fF, fG of rank 0: F, G are A-equivalent.
172
(]Rrx ]Rq, 0) be stable r-parameter
if fF, fG are X -equivalent then
(4.5) is an immediate consequence of (4.3) and (4.4).
In more homely
language the burden of this result is that if we restrict our attention to stable germs
then the assignation
F
F
tive mapping from A-equivalence classes of germs.
fF
--*
actually induces an injec-
f germs to
Our objective now is to determine the image of this mapping.
To this end we make the following observation. F
(]Rr x ]Rm, 0)
:
fF
germ
:
s r + q:
fF
equivalence classes
, (Mr x ]R.q, 0)
If
is a stable r-parameter unfolding of a
of rank 0 then fF has
(]Rm, 0) -+ (]Rq, 0)
.
'-codimension
indeed (4.4) tells us that the (r + q)-parameter deformation
is '-versal, hence X-transversal, so the '-codimension of
fF must be
,
,
the number of parameters,
i.e.
S = S(r, m, q)
With this in mind write
r + q.
.
set of A-equivalence classes of stable germs
=
(]Rr x ]Rm, 0) , K = K(r, in, q)
DF of
=
(]Rr x ]Rq, 0)
r.
rank
of
set of X-equivalence classes of germs
(]R; 0)
(3P, 0)
->
sion
r + q.
<,
of rank 0 and
,
-c odimen-
We can now state the main result.
(4.6)
The mapping
S -, K induced by the assignation
F - fF
is a
bijection.
Proof
therefore a germ f0 5
r + q.
We consider
Only surjectivity remains to be established. :
(Jm, 0)
->
(]R, 0) of rank 0 and X-codimension
Certainly then, as in §3, we can construct an (r + q)-parameter
.- -transversal deformation of the form
deformation of
f0.
-w + f(u, x)
with f
This is precisely the deformation
the r-parameter unfolding
F
:
(]R r x ]Rm, 0)
->
DF
an r-parameter
associated to
(]Rr x ]Rq, 0)
given by 173
F(u, x)
(u, f(u, x)).
=
The -Y-Versality Theorem tells us that
5 -versal, and then (4.4) tells us that vation that
F
has rank
r
F
is stable.
DF
is
The trivial obser-
concludes the proof.
Thus the problem of classifying stable germs under the relation of A-equivalence reduces to the problem of classifying germs under the relation of J'-equivalence, up to a certain 5 -codimension. know, in principle, how to approach:
This latter problem we
it is just the analogue for map-germs
of the problem for function germs discussed at length in Chapter IV.
Indeed
the main function of Chapter IV is to provide the reader with a model on which to base his ideas for the problem now facing us.
At root we are trying to list
the matter are decidedly more complicated.
the germs
(]Rm, 0) _+ (]Rq, 0)
of type
However the mechanics of
Em
under the relation of ,'-equiva-
lence, at least up to a certain X-codimension.
But it may well happen that
it is simply too complicated to list all the possibilities which can occur, and in order to increase one's chances of obtaining complete lists it is necessary to restrict oneself to germs defined by finer invariants than the first order symbol
Thus the next aspect of the theory to which we
E1.
address our attention is the construction of such invariants.
§5.
Higher Order Singularity Sets
We have already seen that the symbol singular point. IRn -.
IR
of codimension E1
a1
is of type
En.)
In this section we shall
symbolism in a natural way to obtain finer in-
The underlying idea is maybe best understood by a detailed study
of an example.
174
is only a very crude invariant of a
(For instance any singular point of a smooth function
show how to extend the variants.
E1
Consider the smooth mapping ]R2 -, IR2 given (in complex numbers) by we consider the deformed map
Given a (small)
E > 0
z "
where the bar denotes complex conjugation.
z2
+ 2cz,
(x, y) H
is defined by
f
u
=
(u, v)
f
:
IR2
-+ IR2
z - z2.
given by
In real numbers
where
x2 - y2 + 2Ex
v
=
2xy - 2Ey
and has Jacobian matrix
(2x + 2E 2y
which has rank < 2 2
X
+ y
2
2
=
E
-2y 2x - 2E
when its determinant vanishes,
this then is its singular set.
:
i.e.
on the circle
The bifurcation set is soon
If we parametrize the singular set by putting
found as well.
x= E cos a
y=
E sine
then we obtain a parametrization of the bifurcation set in the form
u
=
E2 (cos 26 + 2cos e)
v
=
E2 (sin 2e
- 2 sine)
which is a standard representation of a tricuspidal hypocycloid, the curve traced by a fixed point on a circle rolling inside another circle of three times its radius.
175
In fact our circle set
Elf,
x2 + y2
=
E2
is precisely the first-order singularity
since clearly the Jacobian matrix cannot have rank
We are
0.
therefore unable to distinguish one point on the circle from another by just looking at the symbol
On the other hand there are three points on the
Z1.
circle (the complex cube roots of
E3
in fact) which very clearly need to be
distinguished from the others in that they map to the cusps on the hypocycloid.
A clue as to how we should distinguish these three points is obtained by further analysis.
Let us concentrate our attention on the way in which onto the hypocycloid.
maps the circle
f
fjz1f.
We look therefore at the restriction
compute the rank of the restriction at a point
(x, y)
on the circle.
Let us Recall
that the differential of the restriction is the restriction of the differential of
Now the tangent line to the circle
to the tangent line to the circle.
f
at the point vector.
(x, y)
is the line through the origin perpendicular to this
And a unit tangent vector will be
under the differential of
f
(x, y)
at
(-y/E,
The image of this
x/e).
will be obtained by applying the
Jacobian matrix to it, yielding the vector
Zx + 2E
-2y
-y/E
-2xy - Ey
Zy
2x - 2E
X/E
-y2 + x2 - EX
and it has rank
The differential of the restriction certainly has rank . 1;
zero only when this last vector vanishes, which happens precisely at the cube roots of
c3.
In other words our three points are distinguished precisely
by the fact that they are
E1
other points on the circle are
points for the restriction EO
:
176
]Rn -+ P.
whilst all
points for the restriction.
The next step in the theory becomes clear.
f
fIE1f,
Given a smooth mapping
we have the first-order singularity sets Zlf.
If these are
E1'3f = E3(flzlf).
submanifolds we can introduce second-order singularity sets
If these sets are submanifolds we can
And this process can be continued.
introduce third-order singularity sets E1'j'kf
=
Ek(ffEi'Jf).
And so on.
The sets obtained in this way are the higher order Thom singularity sets of f. As Thom observed when he introduced these sets there is an unsatisfactory element here in that the definitions only make sense as long as we continue to However, as we saw in Chapter II, there is a way out of
obtain submanifolds.
this difficulty submanifolds
Z
of the jet-space
for which the inverse images under
J1(n, p)
are precisely the required sets
j1f
and then for a generic mapping
Elf:
these sets will be smooth manifolds.
Thom proposed the problem of imitating
this procedure for the kth order singularity sets, manifolds
One defines
at least for the first-order singularity sets.
-
k
in the jet-space
E
J (n, p)
i.e.
of defining sub-
such that for a generic
k the pull-back under j f are smooth manifolds, and precisely 11,...,lk th f. The case k = 2 was solved by the k order singularity sets E
f
:
]Rn
-+
]Rp
H. Levine, but the general case waited till 1967 when it was solved by We cannot hope to give a full account of Boardman's solution in a
Boardman.
what we can do however is to describe the construction 1,...,ik is to be attached to the k-jet whereby one decides which symbol E
book of this nature:
of a given map-germ, since it is an
entirely practical one.
We start with an algebraic idea. in the algebra
s
let
f1, ..., fp
>.
1.
We define
ideal generated by all
A I
s x s
s
I
be a finitely generated ideal
be generators for
be a system of co-ordinates in
y1' , yn integer
dn,
Let
&n.
to be the ideal
I.
and let
Suppose one is given an I + I'
where
I'
is the
minors of the Jacobian matrix
177
aft
aft
ayI
ayn
aft ay1
The ideal
(5.1)
AsI
so obtained depends neither on the choice of gen-
erators, nor the choice of co-ordinates, tors for
I,
z1, ..., zn
and
i.e.
g1, ..., gq
if
are genera-
is a system of co-ordinates, then
cides with the ideal generated by
and the
I
s x s
E
I
coin-
minors of the Jacobian
matrix
ag1
ag1 8z1
azn 1
.....
\agg az1
ate/ azn
Proof
Clearly, it suffices to show that any
in psI.
Each
gi
s x s
lies
minor of
can be written as a linear combination of the
fk,
with
ag. coefficients in
fin.
Thus each
afk
2zl
j
bination of the az plus an element of
can be written as the same linear comI.
Using the multilinearity of the
J
determinant we see that any
by
178
I
and the
s x s
s x s
minor of
*=
lies in the ideal generated
minors of the Jacobian matrix
af
of I
..... az n
az1
az
.....
az1
Thus it will be enough to show that any
az azn
***
minor of
s x s
For this observe that the Chain Rule allows us to write each
lies in & I. as a linear
az J
combination of the
with coefficients in
,
The result now follows
&n
ayk
on appealing again to the multilinearity of the determinant.
In practice we shall work with the standard system of co-ordinates x1, ..., xn but it will be important for us to know that any system will do.
A I
Note that
when
I
=
s
C A nI C An-1I C
I
Let
Example I
L I
=
=
I
I
and the
<xk-1>.
And A
generated by
I
Let
Example 2 ated by
so
A1I
minors;
I,
=
and the
I
<xk> 1 x I
s
&1
C
...
with
I.
k > 1.
All
minors of the Jacobian matrix <xk>
for
=
=
<xy, x2 + y2>
A2I =
in
I
I x I
<x, y>.
clearly A2I
also that one has the inclusions of ideals
s > n,
is the ideal k-1 (10C
),
so
s > 2.
in
32.
is the ideal gener-
All
minors of the Jacobian matrix
is the ideal generated by
<x2, xy, y2>.
And
I
and the
A I = <xy, x2 + y2>
2 x 2
for
s
s>.3. 179
I C gn we shall adopt the notation
Given an ideal
s
0 I
and refer to the ideal
AII,
A2I,
dnI
An-s+1I as the successive Jacobian extensions of
In view of the sequence of inclusions above we have
I.
I Let us call
...,
=
I
=
...
AOI C All C A2I C
proper when
I = gn
(Note:
I A en.
C A nI
saying that if we take a finite set of generators for generator has constant term Jacobian extension of is proper.
I
Suppose
0.)
is the last ideal
0
1
I
then at least one
I
of successive critical Jacobian extension of
The critical
in the sequence
It has in turn a critical Jacobian extension A
In this way we obtain an ascending sequence
on.
*
is the same thing as
is proper.
I i
.
px1I,
A 20i1I,
and we say that
I.
And so
I.
2L
which
*
...
I
has
Boardman symbol
(i1,
Example 3
One checks easily that the Boardman symbol of the ideal
I
<xk>
=
(k - 1)
in
d2
in
g1
repeated
i2,
mentioned in Example I is Its.
(1,
1),
..., 1, 0, ...)
with
And the Boardman symbol of the ideal I = <xy, x2 (2, 0, 0, ...).
mentioned in Example 2 is
The Boardman symbol of a germ
+y
f
:
(]Rn, 0)
-+
(]RP,
0)
is defined to be
that of the ideal If generated by the components f1, ..., f p. (5.2)
invariant,
The Boardman symbol_ of a germ i.e.
(]Rn, 0)
-j
(]RR, 0)
is a contact
if two such germs are X -equivalent they have the same
Boardman symbol.
Proof 180
Suppose the germs are
f
with components
f1, ..., fp
and
2 >
fi, ..., f'.
with components
f'
Suppose
Step 1
f, f'
coincide, so the Boardman symbols of
If,
Step 2
Suppose
invertible germ
h
x1, ..., xn
Jacobian matrix of
yield a system of co-ordinates.
h
of
ft
If we
relative to the standard system
we get the same ideal as if we compute the relative to the system of co-ordinates
f1, ..., fp
Again it follows from (5.1) that
h1, ..., hn.
there exists an
i.e.
(]Rn, 0) - (IRn, 0) for which f o h, f' coincide.
:
compute the Jacobian matrix of of co-ordinates
If,
coincide, by (5.1).
are right-equivalent,
f, f'
his ..., hn
The components
By (2.1) the ideals
are g-equivalent.
f, f'
f, f'
have the same Boardman
symbol.
The required result is immediate from the two preceding steps.
Step 3
Now we can extend our definition.
is X -equivalent to a germ f0 symbol of
to be that of
f
:
f0.
Certainly any germ
(]Rn, 0) - (IRP, 0).
f :
(]Rr, x)-+(]R, y)
We define the Boardman
In view of (5.2) this definition is
unambiguous.
The first
(5.3)
f
k
integers in the Boardman symbol of a germ
(]Rn, x) - (IRp, y) depend only on the k-jet of
:
Clearly, we can suppose
Proof
x = 0,
f. Let
and let
(i1, i2, ...)
generated by the components
f1, ..., fp
the Boardman symbol of
It is evident, by induction on
i
Lsa k-1
I.
of
f,
be the ideal
y = 0.
k,
I
be
that an ideal
i
... A 1I
f1, ..., fp.
is generated by partial derivatives of order
. k
of
And whether this ideal is proper or not depends only on the
181
values of all these derivatives at
0:
thus
depends only on the k-jet
ik
f.
of
We return now to our objective of partitioning the jet-space
Given k
into kth order singularity sets. that a germ
f
(IRn, x) -+ (IRp, y)
:
Boardman symbol has the form
Jk(n, p)
i
('1,
integers i1, ..., ik we say ik,...,ik when its is of type E
.. ., ik;
...).
We define
E
11'"''lk
to
be the subset of the jet-space comprising those jets having a representative in view of (5.3) this definition is unambiguous.
germ of type
In the case
Example 4-
k = I
the above definition recovers the first-
order singularity sets studied in Chapter II. Consider a jet in
J1(n, p)
and target, and let Now
A3I
One sees this as follows.
having a representative germ with zero source
be the ideal generated by the components of the germ.
I
is generated by
I
and the minors of order
(n - s + 1)
of the
Jacobian matrix, and will be proper if and only if all the minors of order(n - s + 1)
are zero, S
it follows that rank precisely
A I s
-
i.e.
if and only if the Jacobian has kernel rank >.
will be critical if and only if the Jacobian has kernel which is the same thing as saying that the jet lies in
the first-order singularity set
Es,
as defined in Chapter II.
Before we turn to further examples we shall determine just when the kth k k in The answer order singularity set E J (n, p) is non-empty. is provided by
(5.4)
A necessary and sufficient condition for the set
i1,...,ik E
C Jk(n, p)
should be satisfied:
182
s:
to be non-empty is that the following conditions
n>i1i2>....>ik> 0
(i)
(ii)
i1
(iii)
if
Proof
n -p i1 = n - p
i1 = i2 _ ... = ik.
then
Note first that we need only concern ourselves with jets
having zero source and target.
Necessity
(i)
i1, i2, ..., 'k
and that
n > i1,
It is an immediate consequence of the definitions that
Take
we proceed as follows.
I
are all
to be the ideal in
components of a representative of some jet in Alt that
A1I lis generated by
...
To see that
> 0.
generated by the
8
Z
.
We can suppose
say, with
g1, ..., got
s
A A J... A 1I
Suppose
(n - s + 1)
is proper.
ii > ij+1
<.
a1
a2 s ...
.
This implies that all the minors of order
of the Jacobian matrix of
vanish, and therefore
g1, ..., gc J
in particular all the minors of order
vanish, so that the ideal
g1, ..., go j-1
hence
( n - s + 1)
s = ij+1
Taking
s 5 ij.
(ii)
of the Jacobian matrix of
AsAJ-1 1 ... AI 1 is proper, and
we obtain
ij+1
s
ii.
As we observed above in Example 4 the first index
the kernel rank of the jet, so certainly
is
i1
> n - p. i
(iii)
immediately that
Sufficiency
i1 = n - p
If
i1 = i2 =
then
A 1I = Ap+1I = I,
and it follows
..
Suppose conditions (i), (ii) and (iii) are satisfied.
have to produce a germ say, which is of type
f Z
:
(]Rn, 0) -+ (]RR, 0)
with components
We
f1, ..., fp
We consider cases, leaving the computa-
tions as good exercises for the reader.
183
Case when
i1 = n - p
Case when
i1
In this case we choose
> n - p
f1 = x1, ..., fp = xp.
In this case the following choice will do.
(1 5 i <, n - i1) n-i3
n--i1
fn-i1+1
J x
-
+
n-i1+1
fi
+
X.
n-i2+1
(n - i1 + 2 , i ,
0
=
One immediate consequence of this result is that the partition of k is actually finite. by the non-empty E
Example 5
The only non-empty singularity sets in the jet-space
22,2,
22,1,
22'0,
are
21,1,
and
beginning of this section, Cu, v)
at any point
u = x2
with
the mapping
i.e.
- y2 + 2ex,
in the plane.
(x0, y0)
Boardman symbol of any germ f0 to the germ of
f
(x, y) H (u0, v0)
184
I
J2(2, 2)
E0'0
f
at
(x0, y0).
:
:
]R2
-+ IR2
v = 2xy - 2cy
We shall compute the second-order Boardman symbol
Here,
Jk(n, p)
Let us return to the example we studied in detail at the
Example 6
(x, y) -
21,0
p)
defined by and
e > 0.
for the germ of
21'j
f
By definition, we need to compute the
(IR2, 0) _, (IR2, 0)
which is X equivalent
An obvious choice for
f0
is given by
where
fuo(x, Y)
=
u(x + x0, Y + YO) - u(x0, YO)
vO(x, Y)
=
v(x + x0, Y + YO) - v(xO, YO)
is the ideal generated by
u0, v0
and
All
is the ideal generated
by
u0, v0
and the minors of order
2 The ideal
is generated by
I
of their Jacobian matrix
(3 - i)
au0
au-0
ax
ay
8v0
8v0
ax
ay
u0, v0
and the entries in the Jacobian, and au0
cannot be proper as two of its generators The ideal
A1I
is generated by
av0
ax
have constant term pL 0.
ay
,
and the determinant
u0, v0
D
of the above
Jacobian, and will be proper (hence critical) when the constant term 2
2
x0 + y0 - e
2
the germ of and the
in
D
vanishes:
has type E.
f
2 x 2
this then, as we saw before, is the set where The ideal
A1A1I is generated by
minors of their Jacobian:
u0, v0, D
a line or two of computation will
verify that the constant terms in these generators are
x0 + Y0 - E2;
And the ideal
A1A1I
vanish simultaneously,
2
Y0(2x0 + E);
2
Y0 - x0 + x0e.
will be critical exactly when the last three expressions i.e.
exactly at the three complex cube roots of
Thus the three exceptional points on the circle cisely by the fact that the germ of
f
21f
E3
are distinguished pre-
at these points has type
whereas at all other points on the circle it has type
E1'1'
E1,0
In Chapter II we proved that the first-order singularity sets
E1 C J1(n, p)
were smooth manifolds, and computed their codimensions.
We shall only state the much harder result of Boardman.
(5.5)
If the k
th
11$ order singularity set
E
...,lk
in
k
J (n, p)
is non-
empty then it is a smooth submanif old of codimension
185
(p - n + 11)µ(i1, ..., ik) - (11 - i2)µ(12, ..., ik) - ... - (ik-1 - ik)µ(ik) where
µ(i1, ..., ik)
denotes the number of sequences
(j
,
-
..., jk )
of
1
integers which satisfy the following conditions
j1 ' j2 ' ... % jk > 0
(1)
is 3 js
(ii)
Example 7
for all
In the case
dimension of
Z
i
in
J1
(3Rn ,
I
k = I
< s < k
and
we have
µ(i) = i
and hence the co-
which agrees with the
(p - n + i)i,
is
IRp)
j1 > 0.
formula we obtained in Chapter II.
Suppose
Example 8 µ(1, ...,
1)
=
k,
Clearly we have
E1'...$1
so the codimension of
Note that in the equidimensional case
(p - n + 1)k. just
k,
i1 = i2 = ... = ik = 1.
the number of repeated
In the case
Example 9
so the codimension of
El's
Jk(n, p)
p = n
will be
the answer is
1's.
k = 2 in
in
one has
J2(n, p)
µ(i, j)
=
i(j + 1) - J
-
1
2
'
is given by the formula
(p - n + i)i + 12{ (p - n + i) (2i - j + 1) - 2i + 2j] . k In view of (5.5) the singularity sets submanifolds of
Jk(n, p).
are called the Boardman
Z
The Thom Transversality Lemma, proved in §4 of
Chapter II, yields the following.
(5.6)
The set of all smooth mappings
f
:
IRn -+ IRp
for which
i1,...,ik transverse to all the Boardman submanifolds
C O° (IRn, RP) .
186
Z
is dense in
jkf
is
We shall call a smooth mapping f : ]Rn - ]Rp generic in the sense of k i1,...,zk Boardman when j f is transverse to all the Boardman submanifolds E , for every integer k 3 1.
For such a mapping the set
1 ,...,1 1
E
i ,...,i
kf
(jkf)-1(E
1
k)
=
i1, ..., ik
n will be a smooth submanifold of
having the same codimension as
]R
E
.
Boardman showed that
Let
(5.7)
f
k'1k+1f
]Rn - ]Rp
be generic in the sense of Boardman:
then
Ek+1(fiEkf)
=
In other words we have the following.
Any smooth mapping
f
:
]Rn -* ]Rp
can be forced to be generic in the sense of Boardman by an arbitrarily small i 1,..., i
perturbation:
moreover, for such a mapping the sets
k f
E
manifolds and coincide precisely with the Thom singularity sets.
are smooth
Note inci-
dentally one trivial consequence of (5.7), namely that
E11
Let
Example 10
f 2) f
:
2
E11
E11.41 2 13
f 2) ]R3
]R3 ->
be generic in the sense of Boardman. By Example 7 the codimension
We ask which Thom singularity sets can occur. of
J1(3, 3)
in
E1
clearly then
Elf
is
E1,1,1f
1, 2
hence
with codimension
ity set which can occur. dimensions
i2,
Elf
f 2 ...
Elf I
has codimension
2, 3
splits into
21'0 f and E1'1f with co-
respectively;
k,
so will not appear for
E1,1,0
f and
no further splitting can
take place since the kth order Thom singularity set sion
in X23:
is the only first-order Thom singular-
respectively, and E1'1f splits into
with codimensions
i2
E1' " '1f
has codimen-
k > 4.
187
A pleasant illustration is provided by the dovetail mapping
Example 11
f
: ]R3 -> ]R3
(x, y, z) H (u, v, w)
given by
u=x
v=y
w=z
where
-uz -vz2.
The reader will easily check that the possible Thom singularity sets
E1'" '1f are given by the equations below. aw as
E1f
0
aw az
E1,1f
=
0
Z= 0 E111,11f
:
1
Lw
a2w
and
=
E1,1,1f
aw 3
2 and
as
aZ
is the folded surface illustrated below:
E f
0
az2
=
0
and
=
0.
aZ
E1'1f is the fold curve, and
is the origin.
In succeeding sections it will be useful for us to know that the Boardman symbols
are invariant under unfolding in the following precise
E
sense.
(5.8)
Let
F
:
(]Rr
unfolding of the germ f Boardman symbol. 188
x ]Rn00) - (]Rr x ]RP, 0) be an r-parameter :
(]Rn, 0) - (]RR, 0): then f, F have the same
Proof
x1, ..., xn
Let
u1, ..., ur
those in ]R'.
u1, ..., ur,
f1, ..., fp
be the standard co-ordinates in ]Rn,
We take where
and
to be the germ whose components are
f'
f1, ..., fp
are the components of
f.
The proof proceeds in two steps.
We claim first that
Step I
Suppose
has Boardman symbol
f
i
A A
that the ideal
k,
on i
have the same Boardman symbol.
f, f'
(i1, i2, ...).
k-1 ...A 10
We shall show, by induction
is generated by
If,
and
u1, ....4 ur
i
A s A k-1 ... A O If.
Here we put
9n with a subset of
i0 = 0
where
n,
for convenience, and tacitly identify
n' = n + r.
It will follow immediately
A k
that
... A ° If,
is critical, so establishing the claim.
the assertion is trivial.
Suppose it holds for
When
Take
k.
J
k = 0
to be the
Alk Jacobian matrix of some fixed set of generators for to the standard co-ordinates
x1, ..., xn,
of the same set of generators augmented by
generated by
u1, ..., ur,
minors of order and the identity order
u1, ..., ur.
x1, ..., xn,
ordinates
(n' - s + 1) (n - s + 1 )
u1, ..., ur, (n - s + 1)
of P.
Now
of
010
If ,
k ... A 0 If'
and the
is the direct sum of
J'
J
coincides with the ideal generated by the minors
J'
of
J.
i.e.
k s Thus A A A k A
i
.. A 0 If,
J,
Step 2
We claim that
...
f', F
is generated by
1
it is generated by
i k ... AiOIf AA
valent.
...
s
A A
matrix, so the ideal generated by the minors of
the generators for of
relative to the co-
Consider the ideal k
If relative
to be the Jacobian matrix
u1, ..., ur
i
of order
J'
the generators for A
(n' - s + 1)
r x r
and
... 010
If, and the minors of order u1, ..., ur
and
are g-equivalent, and hence .gi'-equi-
It will then follow from (5.2) that
f', F have the same Board-man
symbol, which fact combined with Step 1 will clinch the result.
Since
F 189
unfolds
it has components
f
F1, ..., Fp
u1, ..., ur,
and for
I
s i 6 p
we have Fi(x1,
..., xn,
identically in
x1, ..., xn.
write each
=
by
F.
fi + Thus
u1, ..., ur.
=
fi(x1, ..., xn)
It follows from the Hadamard Lemma that we can where
Ci
u1, ..., ur,
generate the same ideal in
F1, ..., Fp f', F
i
0, ..., 0)
lies in the ideal in f1, ..., fp Sn
and
&n,
generated
u1, ..., ur,
so the corresponding germs
are P,-equivalent by (2.1).
The germ F
Example 12
(IRn, 0) -, (Mn, 0) with the components
:
F1, ..., Fn where
(1, isn-1)
xi
F.
n-1 F
_
xnn +1
n
x xl
+
i
n
i=I
is an unfolding of the germ f
(IR, 0) - (IR, 0) given by f(x) = xn+1
:
has type L1,...,1,0 with n repeated 1's using Example 3. Example 13
The germ F
:
(IRk, 0)
-
(IR , 0) with the components
F1, F2, F3, F4 where x1
F1
190
F2
=
x2
F3
=
x3x4
F4
=
x3 + x 4 + x1x3 + x2x4
2
so
is an unfolding of the germ f
f(x, y)
:
(IR2, 0)
(v, x2 + y2) so has type
=
-4
E2'0
(IR? 0)
given by
by Example 3.
Classifying Germs under JP-equivalence
§6.
In this section we shall consider the very simplest situations where it is possible to obtain explicit lists of germs under the relation of -Y-equivalence. As in Chapter IV the whole thing turns on the idea of "determinacy". a germ
g
f
(]Rn, 0) ,
:
(IRn, 0)
:
-+
(IRR, 0)
(IRR, 0) with jkf
We call
. -k-determined when any germ =
f.
jkg is X -equivalent to
By
analogy with (IV.3.1) one might reasonably expect the following result.
(6.1)
A sufficient condition for a germ
f
0) - (IRP, 0)
to be
-k-determined is that .,1V n+1. &n'p C Tf
more than
In fact the result is correct, and its proof turns out to be no a slightly complicated version of the proof of (IV.3.1): shall omit the proof. finitely
for that reason we f
By further analogy with Chapter IV we call
7,'-determined when it is
,5'-k-determined for some
k >.
follows immediately from (2.3) and (6.1) that a germ of finite
must be finitely -Y-determined.
1.
It
X-codimension
In fact that statement is the only appli-
cation we shall make of (6.1) in this book.
As a starting point, let us look again at the case of germs of functions, only this time under
-T'-equivalence rather than R-equivalence.
The first
step is the Splitting Lemma.
191
Let
(6.2)
f e (n be a germ of corank
c
and finite
f -codimension:
then f is X-equivalent to a germ g(x1P ..., xc)
with g
+
xc+1
-
...
-
xn
e,,o,13
The normal
The proof is exactly the same as that given in Chapter IV.
forms for germs of corank 0 are given by the Morse Lemma. of corank
then
one obtains almost exactly the same classification as before.
I
Let
(6.3) f
And for germs
f r =,//2n
have corank
I
and finite ,5'-codimension
is ,'Y'-equivalent to a germ of the form
xk+1
±
k >
1:
x2 + ... + xn 2
Again, the proof is exactly the same as that of the corresponding result in Chapter IV.
One can continue in this way, just as we did in Chapter IV,
and it turns out that up to a certain point the two classifications are more or less identical, but then they begin to diverge.
;Ye shall not pursue the
point further.
Let us now turn our attention to germs of smooth mappings, as opposed to germs of smooth functions.
we are now in a much more complicated situation,
and can only hope to obtain results in very special cases.
It is in just
this situation that the Boardman symbol of a germ proves to be useful, in that it allows us to distinguish special cases.
The first fact one ought
to be aware of is
(6.4)
Let
f
:
(IRn, 0 ) -- (]RP, 0)
its Boardman symbol must have the form integer
192
k >
1.
be a germ of finite -codimension: (i1,
..., ik, 01 0, ...)
for some
Proof
As we have remarked already
so X-equivalent to
g
:
(7Rn, 0)
given by a polynomial of degree Boardman symbol of
(IRR, 0)
-+
p
>>
E
1 .
2
is the case when
and
has the same form.
f
(]Rn, 0) - (IRP, 0)
The first order Boardman symbol is
n = 1.
Now suppose the germ has finite
We ask for the classification up to ,5
1
Let f
(6.5) .
Clearly, the
with Z0
or
In the former case the germ is non-singular and a normal form is pro-
vided by (I.1.4). E
case to study for germs
k 3 1.
..., uk, 0, ..., 0),
(u1,
since the Boardman symbol is a .5r-invariant, that of
Perhaps the simplest
is
each of whose components
for some integer
. k,
has the required form
g
must be finitely X -determined
f
(IR, 0)
:
Then
-codimension.
equivalence.
(]RP, 0) be a germ of type Z1,
-+
is necessarily of type
f
, '-codimension, and type
E(with
k
and in that case isX-equivalent to
k 3 1,
repetitions) for some integer
and finite
k+1).
(0, 0,
..., 0,
Proof
That
f
has the type indicated is an immediate consequence
Next,-we claim that
of (5.1+) and (6.1+).
f
E1'...'1'0
(with
k
Indeed by induction on
k
has type
repetitions) if and only if the following conditions are satisfied. (i)
ajf
ajf.
=
1 (0) axj
...,
0,
a
(0)
=
0
(j
k)
aJf. (ii)
some
0
(0)
axj
where
f1, ..., fp
denote the components of
one readily checks that ajfl tives
with
Al ...
1 , i s p,
AIIf 1
f.
is generated by
5 j 5 k.
If,
and the deriva-
The claim follows immediately.
ox" 193
In view of (IV.2.4) the conditions (i) and (ii) are equivalent to saying that If
=
k+1
<X
and then (2.1) tells us that this is the same as
g-equivalent to
So far so good. p f
..., xk+1).
(0, 0,
being
f
The result follows.
Let us press on further.
If we insist on both
n
and
being > 2, then the next simplest case to study is that of germs (IR2, 0)
:
E0
E1,
(]R2, 0).
-.
Here the possible first-order Boardman symbols are
In the first case the germ is non-singular, and a normal form
E2.
is provided by (1.1.3).
Let
(6.6)
f
:
For germs of type
(]R2, 0)
Then
'-codimension.
f
.
E1
we have the following result.
be a germ of type
(IR2, 0)
and finite
E1,
is necessarily of type E(with k
repetitions) for some integer k 3 1,
and in that case is .9i-equivalent to
k+1
(x, y
).
Proof f
As in the proof of the preceding proposition, the fact that
has the type indicated is an immediate consequence of (5.4) and (6.4).
As
f
has rank
I
we know from (4.1) that
parameter unfolding of a germ f0 21,...,1,0
(5.8) also has type
then X -equivalent to J '-equivalent to
:
(IR, 0) -
(with
(x, f0(y))
yk+1,
and hence
f
k
can be assumed to be a 1-
of rank 0, which by
(IR, 0)
repetitions).
And clearly
But (6.5) tells us that f
We have still to treat germs (IR2, 0)
is - -equivalent to
(]R2, 0)
f
f0(y)
(x)
is
is
yk+1
At this
of type E2.
point we are entering a decidedly more complex situation, and must proceed cautiously. E2,2
in order of increasing degeneracy.
plete result
194
The possible second-order Boardman symbols are
(due to J. Mather).
2210,
E2,1
For the first we can obtain a com-
(6.7)
Any germ
f
:
-
(]R2, 0)
(]R2, 0)
of type E2'0 and finite
,
'-
codimension is -Y-equivalent to one of the germs listed below.
Iab,
b
IIa,b
(xy, xa - y
IVa
(x2 + y2, xa)
Note
these germs.
b3a>2
(xy,xa+y)
b t a 3 2,
)
even
a 3 3
We have kept to Mather's notation
Ia b,
IIa b,
IVa
for
His list, which was for a more general situation, included in
addition certain germs denoted
IIIa ,
Proof
a
b'
Va'
The first thing to note is that the components of
no linear terms, so the 2-jet of
f
f
have
can be thought of as a pair of binary
quadratic forms
(a1x2 + 2b1xy + aIy2,
a x 2
2
+ 2b2xy + c2y2)
In Chapter III we saw that by applying linear changes of co-ordinates at source and target we can suppose that the 2-jet has one of the normal forms
written out in the table below. second-order Boardman symbol.
Beside each normal form we have written the The computations are very easy, and left to
the reader.
195
Boardman symbol
normal form
for pencil
E2,0
y2)
(xY, x2 + (xy,
x
E2,0
2
E2,0 (xy, 0) (x2 + y2,
E2,0
0)
(x2
0)
(0,
0)
E2,1
E2,2
Since the Boardman symbol is a Y -invariant we can discard the last two normal forms in the table.
xy
2-jet
The
xy
or
to
x2 + y2.
xy.
will have
f
We consider these cases separately.
The first component of
case
of corank
Certainly then the first component of
f
is a smooth germ
(]R2, 0) -i (7R, 0)
It follows from the Morse Lemma that this germ is R-equivalent
0.
Applying the same change of co-ordinates to (xy, c(x, y))
.5 -equivalent to a germ
is of finite
have supposed
f
and therefore
C(x, y)
where
C
f
we see that it is
has no linear terms.
, '-codimension, so finitely
can be supposed to be a polynomial.
We
-determined, It follows from
(2.1) that we shall not change the . 5'--gquivalence class by subtracting from a multiple of
xy
(xy, a(x) + Q(y))
in
92:
thus we can suppose our germ has the form
with a, (3 polynomials.
At this point a couple of remarks are in order. has order
a >>
2
(i.e.
xa
is the lowest power of
Suppose
196
becomes
+ xa.
And similarly, if R
0
a
0,
and
x which appears in a):
then certainly we can produce a change of co-ordinates a
5
x F-
and has order
X under which b > 2
we can
find a change of co-ordinates x. X
here that
under which
generate the same ideal in 91.
the same ideal in
&1,
becomes
(3
+ yb.
y, Y
and likewise
Note generate
Now we consider the possibilities regarding a, p.
are both zero
a, /3
y - Y
This case yields the germ
which is not
(xy, 0),
of finite .Z'codimension, so can be discarded.
Just one of
is zero
a, 0
apply in the case Q
Suppose
(A similar argument will
0.
a ,
x H X, y - y
A change of co-ordinates
0.)
brings the germ to the form
Xy, xy generate the
and since
+ xa),
(Xy,
same ideal this is X-equivalent by (2.1) to
(xy, xa).
But
(xy, xa)
is
also not of finite . -codimension, so this case can likewise be discarded.
A change of co-ordinates
are both non-zero
a, $
brings the germ to the form
(XY, + xa + y
point that we are at liberty to suppose either component by
-1.
Also, if
by a change of co-ordinates
xa
I
and since
a , b;
ab
xy, XY
generate the
(xy, + xa + yb).
Note at this
and of course we can multiply
is odd we can change the sign in front
a
x H -x,
is odd we can change the sign in front of a germ of type
)
X-equivalent by (2.1) to
same ideal this is
of
b
y H Y
x H X,
yb.
y H y:
and similarly if
b
Thus in every case we obtain
save when the signs are different, and
a, b
are both
s
even, and in that case we obtain a germ of type
The
x2 + y2
the
xy
where
S
f
is
T,-equivalent to a germ (x2 + y2, C(x, y))
is a polynomial with no terms of degree s 2.
change the .-equivalence class by subtracting from in f
92 :
b.
Following the same initial reasoning used in
Case
case we see that
IIa
in particular we can suppose
has the form
(x2 + y2,
C
a(x) + y/3(x))
C
By (2.1) we do not
a multiple of
has no terms with factor with
a, (3 polynomials.
x2 + y2
y2,
i.e.
Note that
197
the second component cannot be identically zero, as
5-codimension. order term is
Denote by
a > 3
pxq + gyxa-1,
we can suppose
p / 0,
the order of
(x2 + y2, 0)
a(x) + YP(x),
so the lowest
p, q ' 0.
say, with at least one of
q = 0.
has infinite
I claim
To this end consider a linear change of co-
ordinates
X= x cos 0+ y sin 6 y
Notice that ulo
x2 + y2
X2 + Y2,
=
=
-x sin e + y cos B
+ Y.
X2
Using the fact that
Y23 =(-1)jX2j
mod-
a straightforward computation yields
gyxa-1
pxa +
=
QYXa-1
PXa +
modulo
X2 + Y2
where
P
=
p cos ae - q sin ae
Q
=
p sin ae + q cos ae
and the claim follows an observing that we can choose P / 0,
Q = 0.
It follows that our germ is
which is the desired normal form of type
g
is such a way that
9l-equivalent to
(x2
+Y2 ,
xa),
IVa.
Of course one could go further, and take up the next case of germs of type E2,1
and finite ,Y-codimension.
At the time of writing no complete list is
available, but one could certainly work with increasing codimension and gradually generate a list.
However the reader can probaoly see for himself by
now that such computations will become increasingly complicated and uninteresting.
And that is as far as we shall pursue the problem of listing germs
under the relation of
The next step in our programme is to
spell out just how all this enables us to list some of the simplest types of stable germs under A-equivalence. 198
Some Examples of Classifying Stable Germs
§7.
We are now in a. position to write out explicit lists of stable germs (IRn, 0) -+
(IR',
Boardman symbol.
under certain restrictions on the dimensions and the
0)
Let us start with the case
situation is that of a non-singular germ.
n s p.
Such a germ is necessarily of type
automatically stable, and has normal form
E0,
x
x , ...
nj,
0
by
0)
We can reasonably expect the next simplest case to be stable germs
(1.1.4).
of type E.
Here we have a complete result, due to B. Morin.
Let n 5 p,
(7.1) of type
The simplest possible
Then
E1.
F
and let
F
(IRP, 0)
be a stable germ
is necessarily of type E(with k
titions) for some integer in that case
(IRn, 0) -
F :
with
k
q = p - n + 1.
And
(IRn, 0) - (IRP, 0)
with
5 k . n/q where
1
G
is A-equivalent to the germ
:
repe
components
1 G1
=
(1 s i 5 n - 1)
u. k
G
n+i
jk-1
Gp
i
(0
uik+jx3
=
q - 2)
=1
j
=
u(q-1)k+jx
+ x
k+1
j=1
where we write
Proof (6.4).
u1,
... J,
un-1'
The initial statement follows immediately from (5.4) and
The theory of §4 tells us that
parameter unfolding of a germ sion, and also of type f
is
for the standard co-ordinates on
x
-equivalent to
f
:
E1'...,1,0 (0,
F
(IR, 0)
must be A-equivalent to an (n-I)->
by (5.8).
..., 0, xk+1),
(Iaq, 0)
of finite 5
codimen-
In view of (6.5) we know that
of 1'-codimension
qk + q - I
by 199
The theory now tells us that
Example 7 in §2.
The deformation in question was computed in Example 2 of
xk+1).
..., 0,
is A-equivalent to the
, -versal deformation of the germ
stable germ associated to a (0,
F
and the germ written out above is clearly the associated stable germ.
§3,
Finally, the theory tells us that we need only consider those
the X-codimension
qk + q - 1
is
5 p,
for which
i.e.
k
for which
k < n/q. 1-1
It is probably worthwhile isolating the equidimensional case of this result, the case
i.e.
Let
(7.2)
F
F
:
1
: k E n,
(]Etn, 0)
:
-+
(]R1', 0)
(]Rn, 0)
-'
E1' " .'1'0
must be of type
with G
n = p.
(with
k
be a stable germ of type
E1.
repetitions) for some integer
k
and in that case is A-equivalent to the germ
(itn' 0) with components
(1 4i 4n-1) k -1 Gn
uxi +
=
xk+1
.
j=1
Of course, this is the germ which we have previously dubbed the "generalized" Whitney mapping, and which in the special case
n = 2, k = 2
yields the Whitney cusp mapping of the plane.
Let us now reverse the emphasis by taking up the case of stable germs n
(]R '0)
-+
(IR , 0)
with
n 3 p.
is that of a non-singular germ.
200
Again, the simplest possible situation Such a germ is necessarily of type
En-p
automatically stable, and has normal form
(x1,
..., x
by
)
(I.1.3).
We
p
can reasonably expect the next simplest case to be that of stable germs of type
n-p+1
The starting point here, as always, is that the theory of §1+
E
tells us that such a germ is A-equivalent to a (p - 1)-parameter unfolding of
a germ
f
(lip, 0)
:
degree of complication in
singular).
Let
1
f
:
Then
OmIf
' , axm
Notice that
c.
f
Indeed that is the case.
(]Rm, 0)
- OR, 0)
has type
be a germ of type
if and only if
Em'c
We start with the condition for
Proof The ideal
depends at root on its corank
It is therefore a good guess that the two invariants will
f.
be closely connected.
(7.3)
in = n - p + 1,
where
and the second order Boardman symbol,-both depend solely on
c,
the 2-jet of
f
n-p+1.
Our experience in Chapter IV tells us that the
and of finite X-codimension.
the corank
also of type
(7R, 0)
-+
is generated by
f
f
f
Em
i.e.
has corank
to be of type
c.
Em,c
and the partial derivatives
and is to be critical.
The ideal
the same list, together with the minors of order
AsAmIf
(m - s + 1)
is generated by of their
Jacobian of
...
ax1
a2f ...
of
axm
a2f axloxm
ax2
a2f
a2f ...
ax2
axin ax 1
201
The condition for the ideal
m
s
A A I
should have zero constant term, (m - s + 1)
should vanish at
row, since all the
to be proper is that all its generators
f
i.e.
In this matrix we can disregard the first
0.
vanish at
ax
that all the minors of order
so are left precisely with the Hessian
0,
J
matrix of
sdmIf
Thus the condition for the ideal
f.
the Hessian should have corank
and the condition for
> s,
critical is that the Hessian should have corank exactly we see that
f
has type
Em'c
if and only if
The simplest situation is when
to be proper is that
f
f
AsAmIf Taking
s.
has corank
has corank
0.
(IRn, 0)
(IRP, 0)
to be s = c
c.
This yields another
result of B. Morin. Let n > p,
(7.4) of type
En-p+1,0
:
and let
then
F
F
:
->
be a stable germ
is A-equivalent to a germ
(]Rn, 0)
(Pp, 0)
>
given by
(1 s i s p- I)
U.
Gp
Proof
=
2
+ xP +
...
+
xn-1
As was pointed out above,
(p - 1)-parameter unfolding of a germ f and finite k-codimension, where
2
+ xn
F :
is certainly A-equivalent to a
(IRm, 0) - (IR, 0)
m = n - p + 1.
By (7.3)
f
of type
Em'0
has corank
0,
so by the Morse Lemma is R-equivalent, hence X -equivalent, to a germ
+ x2
p
+
...
+
2 xn.
The theory of §4 now tells us that
F
is A-equivalent
to the stable germ associated to a p-parameter , -versal deformation of this germ.
It is a trivial computation to verify that the germ written out above
is the stable germ in question.
202
The next simplest case is when
has corank 1, giving rise to the foll-
f
owing result, also due to B. Morin.
Let
(7.5)
En-p+1,1
of type
k
(with
F
case
n
p,
and let
Then
F
F
:
(IRn, 0)
-+
(]RP, 0)
En-p+1,1,...,1,0
is necessarily of type
repetitions) for some integer k with is A-equivalent to a germ
G
:
(]Rn, 0)
be a stable germ
s k s q:
1
(]RR, 0)
-+
and in that given by
k 2
_ ±xp2
+ ... + xn-1 +
-
xn+2 + )
u in
J=1 Proof
The initial statement follows immediately from (5.1+) and
The rest of the proof follows exactly the same lines as that of
(6.4).
(7.4), save that this time by (6.3).
f
is X-equivalent to
+2
+ xP + ... + xn -1 +
The computation of the deformation is very straightforward, and
can safely be left to the reader.
Of course, one could push these techniques further, by systematically classifying germs of functions with increasing ,5-dimension.
However the
point of the last two results has already been made in that we have indicated the simplest complete results which can be obtained. in the equidimensional case
n = p
the last two results yield (7.2).
the next case to study is that of
E2
Let
Having dealt with germs of type
us pursue the equidimensional case further. E1,
Note incidentally that
germs.
In view of the discussion
of §6 it is clear that we can only expect a complete result in the case of E2'0
germs.
The result is due to J. Mather.
203
Let
(7.6)
F
:
(Mn, 0) - (]Rn, 0)
be a stable germ of type
then F is A-equivalent to one of the following germs G :
Types
a-l+j
E2'0:
(]R", 0) - (:u
U.
(1 5 i < a-1)
=
vj
(I s j
=
xy
=
xa ± yb + a-1 E ui x
0).
<< b - 1)
Ia,b
and Ila,b
a+b-1 a+b
b-1 vjyJ
i=1
j=1
(I si.
(1 s j s a - 1)
vj
Ga-1+j
-x
iC-2I
+ y
2
a-1 =
G2a
+ Z ulxi
xa
+ E
i=1
parameter unfolding of a germ f finite types
j=1
By the theory of §4 the germ
Proof
Y-codimension. Ia,b'
IIa,b'
IVa
:
(]R2, 0)
-->
F
F
is A-equivalent to an (n -2)-
(7R2, 0)
By (6.7) this germ is .
And then
vixj-1y
of type
E2'0
and
-equivalent to one of the
will be A-equivalent to the stable
germ associated to a 9F-versal deformation of these germs.
The deformations
were computed in Examples 3, 4 of §3, and the corresponding stable germs are evidently the germs written out above.
With these results we have gone as far as is possible in a book of this nature, and have achieved our objective of indicating how one goes about classifying singular points of smooth mappings.
We shall however pursue
the matter a little further by looking briefly at certain mappings whose singular points are necessarily stable. 204
Singular Points of Stable Mappings
§8.
We have so far discussed only stable germs of smooth mappings
f
]Rn - ]RP.
:
But the underlying ideas can be introduced equally well for mappings themselves.
By analogy with germs there is a natural notion of "equivalence"
for smooth mappings; e
when there exist diffeomorphisms
,.lent
-
namely, we call two smooth mappings fV f2 g, h
:
]Rn -,]Rp
for which the following
diagram commutes
Now ]Rn,
So the pair
(g, h)
Diff(]Rn) x Diff(]RP) to be
(g, h).f
Diff(]R n)
is an element of the group
h
and
]RP.
is an element of the group
g
Diff(]RP)
of all diffeomorphisms of of all diffeomorphisms of
is an element of the product group
and this group acts on g-1.
h o f o
cf°(]Rn, ]RP)
Of course this action lies outside the frame-
work discussed in Chapter III because neither the group nor the set
C°O(]R', ]RP)
if we define
Diff(]Rn) x Diff(]RP)
are in any way finite-dimensional.
But for all
that there is no harm in proceeding by analogy, just as we did with germs. Recall that in Chapter III we introduced the equivalent notions of "stability" and "infinitesimal stability, the former being geometric in nature, and the latter algebraic.
When dealing with germs we interpreted stability in terms
of unfoldings, but for mappings it will be easier if we stick to the geometric idea.
We should therefore call
close"
g
:
f
: e-+ ]RP "stable" when all "sufficiently
]Rn - ]gP are equivalent to
f.
On this basis we introduce the
following formal definition.
205
A smooth mapping
f
IRn - ]Rp is stable when there exists a real number
:
e > 0
such that every smooth mapping
of
is equivalent to
f
g
is the e-neighbourhood
7Rn -+ ]Rp
:
Our first result gives us some idea of just how
f.
nice stable mappings are.
(8.1)
Any stable mapping
f
]Rn -r IR13
:
is generic in the sense of
Boardman. Proof
Let
k >>
I
be an integer.
transverse to all the Boardman submanifolds
We have to show that jkf is i1,...,ik To this end recall Z
from (5.6) that the set of all smooth mappings ]Rn - ]Rp is dense in
C O°(it , ]Rp):
thus, given any
with this property
we can find a
e > 0
]Rn - ]Rp in the e-neighbourhood of f for which jkg is transverse to i1, ...,ik And since f is stable we can choose a so small that all the E g
:
every map in the e-neighbourhood of to observe that if
f
ity condition, then f
is equivalent to
f
is equivalent to does as well
g,
and
g
It remains only
f.
satisfies a transversal-
which fact we leave as an exercise
-
7
for the reader.
(8.2)
Let
f
:
]Rn - ]Rp
be stable:
then the germ of
f
at any point
is stable.
If we now combine these results
We shall omit the proof of this result.
with the classifications obtained in the previous paragraphs we obtain some aesthetically very pleasing theorems describing completely all possible singu-
lar points of a stable mapping f
:
]Rn - IRp
for certain values of
Let us start by looking at the equidimensional case stable mapping
f
:
]Rn -, in.
first-order Boardman submanifolds
206
First of all, El
-
n = p.
n, p.
Consider a
f must be transverse to the
which by (5.1) in Chapter II have
codimension Elf
-
In particular, if
1 2.
n t 3
which have the same codimension
-
the first-order singularity sets
must be empty for
only possible singular points are those of type
f
(8.3)
In fact we can be more
E1.
k repetitions) has codimension
(with
thus the only singularity sets with
and the
Recall that in the equidimensional case the singularity set
explicit. E1'
i >> 2,
k 5 n.
Let
If we take
which can be non-void are those
Z
we obtain immediately from (7.2)
n = I
f :1R -+]R
k, by Example 8 of §5:
be a stable mapping:
its germ at any point is
equivalent to one of the following
E0
:
y1
=
E1,0
Taking the next case
x2
=
y
(regular)
x1
(simple minimum).
1
we recover a famous result of H. Whitney des-
n = 2
cribing the possible singularities of stable mappings from the plane to itself.
(8.4)
Let
f
]R2
:
-
be a stable mapping:
]R2
its germ at any point is
equivalent to one of the following
0 E
=
x2
y1
=
x1
Y2
_ =
x2
y1
=
x1
Y2
=
Y2
E1'0
E1,1,0
x1
y1
2
(regular)
(fold)
(cusp).
x2 + x1x2
207
In order to obtain geometric insight into these germs it is best to think of them respectively as composites
(
xI, x 2 )I--->
(x
(x Is
, ---) (x', x2 )
(xx2
F
)I
O
(x 1,
proj
x2, x2 )
x2, x2) x2, x
x2 )
;
(x 1, x2)
prod
+ x 1 x2 )
j
p Toj
(x
,
x
+ x 1 x2 )
2
2
where in each case the second mapping is (the restriction to the image of the first of) the projection
(x1, x2, x3)
ualize our maps as follows. x2
x1
A x2
xl
x2
x1
208
-*
(xV x3).
In this way we can vis-
Take for instance the situation of Example 1+ in §4 of Chapter II, namely the projection of a torus onto a plane.
Locally, such a mapping is from the
plane to itself, and generically such a mapping is stable shall have to ask our reader to accept points of the above three types.
-
-
a fact which we
so we should only see singular
Indeed this is the case.
Recall that the
set of critical values looks like this.
Here the curves which make up the picture are the images under the projection f
of the singularity sets
torus folding over.
E1'0 f:
at a point in such a set we simply see the
But the four exceptional points are the images of the
E1'1f: here we have the more complicated situation of two
singularity sets
folds coming together to form a "pleat".
We can of course gain one more result by taking the final case
(8.5)
Let
f
:
7R3
-
7R3
be a stable mapping:
n = 3.
its germ at any point is
equivalent to one of the following.
y1
x1
y2
=
x2
y3
=
x3
(regular)
209
E1,0
(fold)
x2
y2
2
y3
=
x3
E1,1,0
(cusp)
y3
1,1,1'0
x3 + x1x3
y1
=
x1
y2
=
x
y3
=
(dovetail)
2
X4
+ x 1 x 3 +
x2x3
Of course, when we go up to the next dimension n = 4 we no longer obtain only
As we have already pointed out
points.
E1
Elf
has oodimension
in the equidimensional case so when n = 4 we can'certainly have but avoid E2'2f
E1
points for
with codimensions
i 3 3.
4, 7, 10
three sets can be non-void. germs of type
Let
(8.6)
E2'0
f
:
Now
splits into
respectively:
points,
E2'0f, E2'1f,
thus only the first of these
Appealing to the classification of stable
given in §7 we obtain the following:
]R4 -* ]R4
be a stable mapping:
equivalent to one of the following.
{yi =
xi
EO
Y4 =
210
E2f
E2
i2
x4
its germ at any point is
(1 E i (3)
yi
f Y4 X.
yi
51,1,0
{
51,1,1, 0
Y4-
yi
{ Y4
xi
=
x4 + xx4 + x2x
f
(1 <.is3)
xx + x1x_ + x2x
Y4.
,c is3)
2
xi .
yi
51,1,1,1,0
(1
=
2
+ x3x43
y1
52,0 Y2
x2
Y4
x3 + xlf + x1x3
I2?32
2
2
52'0
Y2 1
=
x2
=
x3x1f
=
x3 - xL + xlx3 + x2x4
y3
2,2
Y,
2
2
6I-+
The reader is invited to continue the listing process for stable mappings
5
I, 5
6 IR I-+
That is as far as he will get by just dipping into
the results of the previous sections. germs of type
52,1,
Stable mappings ]R7 -+ 7R7
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
We ask if there are pairs of integers
n << p.
having the property that a stable mapping Mn --' 1Rp
points of type
need
We know that
i(p - n + i) . n in the case
E1
i = 1,
i.e.
then be looking at pairs
i 3 2,
in the range
(n, p)
in this range:
(n, p)
Ecan arise. repetitions) is k - I
we have
of having 21,...'1'0 i.e.
E1'0
points.
Suppose we
E1,...,1,0
p < 2n
We shall certainly avoid singular points of type
ural range to consider then is given by times called the metastable range.
if
n' for k >.
(p - n + 1)k >
3n - I
s
2p <
4-n,
2,
A nat-
which is some-
Combining these arguments with the
results of §7 we deduce the following result, which in the case
212
k
ensures that for k- I
imposing only a very slight further restriction.
is due to H. Whitney.
(with
so we are not excluding the possibility
(with two or more repetitions)
2p > 3n - 1,
so
we ask which singular points of type
The condition
(p - n + 1)k 4 n,
E1,
3n - 4 < 2p < 4n.
Recall that the codimension of
(p - n + 1)k.
we
We should certainly
It is worthwhile analysing the situation a little further. have a pair
E1
But for
p < 2n.
2p > 3n - 4.
i.e.
(n, p)
has codimension
we do not wish our mapping to have singular points of type
i(p - n + i) > n for
n < p
admits only singular
If our mapping is to admit singular points of type
i(p - n + i).
i > 2
with
(n, p)
We can determine a useful range of such pairs
E1.
by a simple combinatoric exercise.
must have
Take for instance
p = 2n - 1
(8.7)
Let
f
:
]Rn -+ 1Rp
the metastable range:
then
f
be a stable mapping with the pair
(n, p)
only admits singular points of type
in
E1,0
and its germ at such a point is equivalent to the germ
(14isn-1) yn-1+i
-
(1 si.p - n)
xixn
2
yP
The case
n = 1
=
xn
of this result is simply (8.3).
something new, namely that a stable mapping lar points of type
E1,0
f
:
But when n = 2
7R2 - 7R3
we get
only admits singu-
and that the germ at such a point is equivalent to
the germ
whose image is the so-called Whitney umbrella depicted'below.
213
What about the case pairs
(n, p)
n 3 p?
Here again one could construct ranges of
for which a stable mapping
points of a very simple type.
]Rn
-+ RP assumes only singular
However the combinatorics are rather more com-
plex, and it is probably more illuminating to take specific pairs write out a complete list of possibilities.
(8.8)
Let
f
:
]R3 -+
One of the simplest cases is
be a stable mapping:
1R.2
(n, p) and
its germ at any point is
equivalent to one of the following.
y1
E1
'
32
x1
x2
=
x1
y1
E2'0 2
2
y2
=
+ x2 + x3
y1
_
x1
E2,1,0 Y2
Proof
In this situation Elf
be empty for
Now Elf
±
i > 3,
splits into
i.e.
x2 ± x3 + x1x3 2
has codimension
i(i - 1)
so must
we can only have singular points of type
E2'Of,
E2'1f,
E2'2f
with codimensions
E2.
2, 3, 5
respectively, so we need only consider the first two of these sets. singular point of type
E2'0
normal forms are provided by (7.4).
remaining possibilities are singular points of type repetitions):
this has codimension k + 2,
so we need only consider the cases (7.5).
214
k = 0, 1
E2,1,---,0
At a The only
(with
k
as an easy computation verifies,
with normal forms provided by
Appendix A
The theorem of Sard
The key idea here is borrowed from measure theory, and is that of a "null set"; this will provide our starting point.
in Mn when, given any real number
A set
duct of
n
V
e > 0,
union of cubes the sum of whose volumes is
V C ]Rn
is called a null set
is contained in a countable (A cube in ]Rn
< c.
is a pro-
open intervals, called its sides, and its volume is the product The reader will readily check that in using this defi-
of their lengths.)
nition one can assume that each cube has all its sides of equal length.
Note that when V
is compact the countable union of cubes can be assumed
finite.
The first fact we need is
(Al)
Any countable union V
of null sets
V.
in ]Rn
is again a null
set in ]an. Let
Proof
< e/2J:
of total volume
C J.,k
We can cover
e > 0.
of total volume
V.
by countably many cubes
Ci,k
hence we can cover V by the countably many cubes
<
--r
e/2J < e
El
It seems worthwhile spelling out a simple consequence of (Al) which we shall use several times in the proof of Sard's Theorem.
Let
(A2) point
x e X
set in ]Rp
f
: X -+ Y
be a mapping with X C ]Rn, Y C MP .
has a neighbourhood then
f(X)
UX
in ]Rn
for which
f(UX n X)
If each is a null
is a null set in ]RP.
215
Proof
(Ux)X for
From the cover
subcover, so
X
one can extract a countable
is a countable union of null sets in IRp,
f(X)
hence a null
set in ]Rp, by (AI). 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
(A3)
if
V C N
f : N -+ P
be a smooth mapping with
is a null set in ]Rn
then likewise
N. P
f(V) C P
open sets in ]Rn:
is a null set in
]Rn
Proof
By (A2) it will suffice to show that for each cube
in IRn
C
in ]Rn
is a null set
f(C (1 V)
whose closure is contained in
N.
It
follows from the Mean Value Theorem that there exists a real number K > 0 such that for any
x, y in
C
<
If(x) - f(y)I
E > 0.
Choose
Since
countably many cubes
C (1 V
Ci C C,
and of total volume
ci,
i
KIx -
yI
is a null set in ]Rn each
having all its sides of equal length
Ci
It follows that each
ci < E.
tained in a cube having all its sides of equal length that
f(C 0 V)
(Kn2)nci
<
it can be covered by
f(Ci)
(Kn2)ci,
and hence
is covered by a countable family of cubes of total volume (Kn2)ne.
It follows that
f(C 0 V)
is a null set in IRn.
The next proposition is a very special case of Fubini's Theorem. it we need a little notation.
Suppose
V C in = IR x ]Rn-1.
For
define
Vs
216
=
is con-
{(x1, ..., xn-1) E
]Rn-1
.
(s, x1, ..., xn-1) E V1-
To state
s s IR
Let V C ]Rn
(A)+)
Vs C ]Rn-1
the set
s e ]R
be countable union of compact sets such that for all is a null set in ]R.n-1.
then V
is a null set
in ]Rn. Suppose first that
Step 1
interval
V
is compact, so there exists a closed
for which V C I x ]Rn-1.
I
By hypothesis
V
total volume <
c> 0.
and let
s e I
can be covered by finitely many cubes
s
in ]Rn-1
C
of
s,J
indeed it follows from the compactness of
e:
is an open interval
with mid-point
Is,
same cubes cover all the open intervals
Now let
for
Vt
s,
in
t
that there
having the property that the From this cover of
Is.
we can extract a finite subcover
Is
V
Is
,
1
certainly suppose that this subcover is minimal,
i.e.
..., Is p
by
I
:
one can
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
1
volume is
It follows that V
s 2eL.
Step 2
Now suppose
V
covering V, and their total i,J is a null set in ]Rn.
X Cs
Is
is a countable union of compact sets.
follows immediately from Step 1 and (Al) that
So much for technical preliminaries.
V
is a null set in ]Rn.
The next step is to extend the basic
notion of a null set to subsets of smooth manifolds. smooth manifold. chart
V C N
A set
0 : A n N ->
are open sets in ]R5,
B
one has ]Rn
It
Let
Nn C ]Rs
be a
is called a null set in N when for every O(A 0 V)
respectively.
a null set in ]Rn:
of course A, B
Sard's Theorem is the following
proposition.
(A5)
Let
f : N - P
be a smooth mapping with
The set of critical values of
f
is a null set in
Nn, Pp
smooth manifolds.
P.
217
Proof
One proceeds by induction on
(It is a convention that IRO
that (A5) holds for
The case
n.
comprises a single point.)
and will deduce that it holds for
(n - 1),
N, P
One easily checks that
that its image under this mapping, namely
f
f(N).
is precisely
Proof
N x 0
n < p
Of course when n > p
is
given
is a null set in ]Rp, so
is a null set in ]Rp
f(N),
It remains only to observe that when
and one needs a more subtle argument: n > p,
n < p
Consider indeed the smooth mapping N x RP -n -+ P
(x, y) - f(x).
of
to be open
It is probably worthwhile pointing out that the case
relatively trivial.
(A3).
It fol-
n.
IRp.
Remark
by
is trivial.
We suppose therefore
lows immediately from the definitions that we can suppose
sets in ]Rn,
n = 0
by
the set of critical values
this reasoning breaks down
this shows that we may as well assume
so the critical points are precisely the singular points.
Write
(continued)
for the set of points
E
for the singular set of
f.
And write
Ek
x E N where all the partial derivatives of order
. k
Thus
are zero.
E 2 E 1 2 Z2 2.. We have to show that
f(E)
is a null set in IRp.
In view of (Al) it will
suffice to establish the following steps.
Step 1
:
f(E- E1)
Step 2
:
f(Ek
Step 3
:
f(Ek)
218
Ek+1)
is a null set in IR.p, for all k.
is a null set in IRp, for some k.
Note first that we can assume
Step 1
have
-
is a null set in ]Rp.
E = E1,
and there is nothing to prove.
p > 2
since when p = 1
we
By (A2) it suffices to show
that each point x image of
in
under the restriction flu
E - E1
in N for which the
has a neighbourhood U
E - E1
is a null set in ]Rp.
And
using (A3) it is clear that we can replace this restriction by any equivalent mapping, which we shall also denote
f.
f(x) = 0:
at
and since the rank of
f
In particular we can assume
x
is
>.
and
I
(by choosing linearly adapted local co-ordinates) that where
s
and each f
E ]R,
s
x = 0,
we can suppose
< p
f(s, x) = (
fs(x))
,
3n-1
is a smooth mapping from an open set in
to an open set in 3Rp-1.
V
We have to show that the set
of critical values of
Notice that the set of critical points of
]RP.
a countable union of compact sets:
of critical points under
With the notation of values of
fs,
f,
(Au-)
it follows that
is a closed set,
f
the image of the set
V,
observe that
Vs
is precisely the set of critical by the induction hypothesis for
V is a null set.
the theorem.
It follows from (A4) that
Step 2
By (A2) it suffices to show that each point
in N for which the image of
f
is a null set in ]RP.
f
is
/ 0,
axg- with
say
hence
is likewise a countable union of compact sets.
which is a null set in ]RP-1,
has a neighbourhood U
is a null set in
f
At x g
x in
U (1 (Ek
E1 - Ek+1
Ek+1)
-
under
(k + 1) th order partial derivative of
some
a kth order partial derivative of
f,
so there
J
certainly exists a neighbourhood set defined by
Ek C G :
g = 0
=
so its image under
U fl Ek
f
of
x whose intersection
is a smooth manifold of dimension
in particular x E G.
u n (Ek - Ek+1)
U
Clearly if
U
G
(n - 1).
with the Note that
is small enough
is contained in the set of critical points of fIG,
is a null set in ]Rp,
by the induction hypothesis of
the theorem.
Step 3
Let
a null set in ]Rp.
k be an integer
> n - 1.
We shall show that
By (A2) it is enough to show that
f(C r Ek)
f(Ek) is is a null 219
set in ]Rp tained in
ber
for any cube
in ]Rn,
C
for which If(x) - f(y)
divide
x, y in C
say, whose closure is con-
c
As a consequence of Taylor's Theorem there exists a real num-
N.
K > 0
for all
of side
above inequality 2K(o/r)k+1,
Introduce an arbitrary integer
C rl Ek.
R = rn
into
equal cubes
C1, ..., CR
r > 1,
of side
f(C n) Ek)
is contained in the union of
and sub-
By the
c/r.
is contained in a cube in e of side
f(C n Ek)
and hence
KIx _ ylk+1
s
rn
2
cubes in
of total volume
]Rp
rn.(2K)p(c/r)p(k+1)
=
Note that n - pk - p < 0 as k > P - 1,
so that this last expression -- 0
It follows immediately that f(C nzk) is a null set in ]RP.
r -oo.
as
r(n-pk-p).
constant x
In order to establish the consequence of Sard's Theorem used in Chapter II to prove the Basic Transversality Lemma we need
Let
(A6)
lement of
V
be a smooth manifold and V C P
is dense in
P = ]RP.
P.
Suppose (A6) were false, so we could find a cube
is a subset of
V.
V C U CI
e > 0,
(C1)
Vol (C)
Vol (C.1 ) 1=1
220
C.
a finite subcover
-
whose closure Since
find a countable family of cubes
and with Y vol(C.) <
from the open cover
C
One derives a contradiction as follows.
a null set one can, given
with
then the com-
a null set:
It follows immediately from the definitions that we can assume
Proof
C
P
5
As
C
V is
(Ci)
is compact we can extract say;
Cl,, ..., Ct,
Vol (C. )
<
e
then
and
e
being arbitrary one deduces that
vol (C) = 0,
which is the desired
contradiction.
t
Note that we assume in the above proof that if t
cally reader.
(A7)
Ni, P
vol (C)
vol (Ci) :
C C U C.
1=1 1
then automati-
we leave this fact as an exercise for the
Finally, we deduce
Let
fi
:
Ni - P be a countable family of smooth mappings with the
smooth manifolds:
the set of common regular values of the
fi
is
dense in P.
Proof
The set
C.
of critical values of
by Sard's Theorem, so the union
C = U C i
fi
is a null set in P,
is a null set by
of common regular values is the complement of
C,
(Al).
The set
so dense in P by (A6).
221
Appendix B
Let fold
:
Semialgebraic group actions
G x M - M be a smooth action of a Lie group
M.
G
on a smooth mani-
The objective of this appendix is to show that if the action is
"semialgebraic", 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 At the time of writing the only reference for the theory of semi-
to them.
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 C ]Rn is algebraic when it can be obtained by finitely many applications of the operation of intersection, starting from sets of the form
{x a in
:
f(x) = O
with
f
a polynomial function on ]Rn.
a little linear algebra shows that any linear subspace of ]Rn
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 in 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 ix E 1Rn
222
:
f(x) > 0;
with
f
a polynomial function on ]Rn.
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
of all real
M(s)
obvious way with a Euclidean space: subset of
plement
M(s),
s x s
matrices, identified in an
the singular matrices form an algebraic
given by the vanishing of the determinant, so that the com-
The idea can be extended to mappings in an obvious way.
with A C ]Rn
is semialgebraic when graph
f
A mapping is semialgebraic
Linear projections provide simple examples of semialgebraic
in 1Rn x IRp.
mappings, since their graphs are linear spaces. arises from considering rational mappings which each component on IRn,
This is a sub-
GL(s).
is semialgebraic.
GL(s)
f : A - 1Rp
A
with 0i
fi
=
where
Oi/Oi
nowhere zero on
A.
A wider class of examples i.e.
f : A -+ IRp, chi, 0i
those for
are polynomial functions
It seems worthwhile spelling out the
following proposition. (B1)
If A C Mn is semialgebraic, and f : A - IRp is rational, then
f is semialgebraic. Proof
With the above notation define polynomial functions 0? ..., Bp
on Mn x MP by O (x, y)
where
x = (xI)
The vanishing of
=
Oi(x) - yioi(x)
..., xn ) is in ]Rn, and y = 8i
defines an algebraic subset
(yl, H.
...,
yp)
is in IRp.
of IRn x MP.
The
proposition follows on noting that graph f
is necessarily semialgebraic.
=
(A x ]RP) () HI fl ... n H p
o 223
One of the basic facts about semialgebraic mappings is the Tarski-Seidenberg
theorem. Let
(B2)
algebraic with
X C IRn be semialgebraic, and let then the image
A C IRn :
f
:
A -, IRp
be semi-
is semialgebraic in IRp.
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
semialgebraic then so too are the factors ested primarily in group actions
x e A
(B3)
Let
M
which are semialgebraic
U0A
Let
First, a preliminary definition.
ar (of dimension
is re
in in for which
G x M -
have to be semialgebraic.
G, M
We need one more fact. A point
:
We are going to be inter-
A, B.
It follows from the preceding remarks that
in the sense just described. then automatically
A x B is
And by the same argument, if a product
under a linear projection.
d)
when x
A C IRn.
has a neighbourhood U
is a smooth submanifold of IRn (of dimension d). then
A C in be a non-void semialgebraic set:
A
has at
least one regular point.
Now we can put the bits together to obtain the main result.
Let
: G x M
smooth manifold M.
-,
M
And suppose that the action is semialgebraic.
all the orbits are smooth submanifolds of
Let
Proof ,
G
be a smooth action of a Lie group
x0 E M.
of the semialgebraic set
The orbit
G x xOJ,
on a
Then
M.
G.x0
through x
0
is the image under
hence semialgebraic by the Tarski-
Seidenberg theorem.
By (B3) the orbit has at least one regular point, of
say.
But the homogeneity property for orbits implies that
dimension
d,
then every point on the orbit is regular, of dimension 221+
d,
i.e.
the orbit
is a smooth submanifold of
M
of that dimension.
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 =
h o f o g
1
We have
GL(n) x GL(p)
on
Hd(n, p)
given by
is semialgebraic, and hence all the orbits are smooth
manifolds.
Proof
GL(n),
product,
i.e.
and
GL(p)
Hd(n, p)
semialgebraic, so their
are all
the domain of the action, is likewise semialgebraic.
of (BI) it will now suffice to show that the action is rational.
will be convenient to think of our linear mappings (gig), (hid)
respectively.
component of
h o f o g1
f.
where
F
h o f o
g-1
denote the components
F1, ..., FP
Certainly the entries
f.
are rational functions of the entries
denominator.
ith component of
is a rational function of the entries in
and the coefficients of components of 1
Now the
g, h and
it will therefore be sufficient to show that each coeffi-
cient of each component of
g
as square matrices
is a rational function of the entries in
is precisely hi1F1 + ... + hipFp,
F = f o g 1;
For this it
We have to show that each coefficient of each
the coefficients of components of
of
g, h
In view
And the claim follows since
gig
of
g,
gig
g,
of
with nowhere-zero are obtained respec-
Fi, ..., F P
tively from the components
gijx, + ... + ginxn
f1, ..., fp
for the variable
of
f
by substituting
xi.
225
Appendix C
Real algebras
V together with a bilinear
By a real algebra is meant a real vector space
mapping V x V -,
written
V,
(x, y) - 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
&n
of all germs
(]Rn, 0) - (IR, y) of smooth functions.
This is endowed with the
algebra product induced from that on the reals. The real vector space
Example 2
f = > fax in n x
=
(x1,
..., x ) n
real indeterminates and
tion for the expression follows. h
Given f
hx' and
=
9n
a
=
x1, ..., xn.
x1 1 x2 2 ... xnn
.
The algebra product is defined as
= > g
g
Here, given
we write x as an abbrevia-
..., an)
(a1,
= >2 fax ,
of all formal power series
xR
we set
f.g
=
h where
= > fa.g
hY
NN
V. W an
Given real algebras
mapping with O(x.y)
=
p(x).O(y)
The natural mapping &n
(C1)
algebra homomorphism for all
-+
do
: V- W is a linear
x, y in V.
given by f
is an algebra
homomorphism.
Here we keep to the notation of Chapter IV:
Proof
n that
226
n
denote the unique maximal ideals in
(f.g)
d, n .
in particular
We have to show
= f.g; it suffices to show that these are equal modulo an
n+1,
element in
f = fk,
g
Then, modulo
for all
gk modulo .,ll A// n+1
,
k >. 0.
k+
wi th
Choose
fk
k.
gk
By (IV.2.li-) we can write
polynomials of degree < k.
we have
(f.g) bearing in mind that the Taylor series of a polynomial function is precisely that polynomial. 1-1
227
Appendix D
The Borel lemma
Our starting point is the explicit construction of smooth functions having very special properties.
There exists a smooth function
(D1)
that
0(t)
=
t
for
5 1/2
ItJ
and
:
(P(t)
]R -+]R with the property
=
for
0
Certainly there exists a smooth function
Proof 0 .< 0(t)
4,
1
for all
for which
t,
0(t) = 0
(0
-1/t2
The required function 0
1.
>.
Iti
B
:
]R -+]R with
precisely when
when
t 5 0
when
t > 0
t 5 0, namely
.
is obtained by setting
t6(1-t) 2 Y
(t)
=
0(1-t
2)
+ O(t2-
)
Now we come to the Borel Lemma itself, as it was stated in Chapter IV.
The natural algebra homomorphism do _, do
(D2)
given by f -+
is
surjective.
x
=
(x1,
..., x ) n
integers 3 0,
of real numbers, and a sequence
we write
a
a1
=
(a1,
..., an)
of
an
x as an abbreviation for x1...xn.
have to show is that given an element
228
Given a sequence
We adopt the following notation.
Proof
What we
x in n there exists an
in Sn
f
element
for which
standard co-ordinates
?
is the Taylor series of
x1, ..., xn
i.e.
a, where we write a: = a I .... an: , Daf
fa = I Daf(0)
alai
=
axiI
To this end let
0
:
>Rn -+ IRn
the formula Cx1, ..., xn) trary sequence limit
Ek
=
f
... axnn
(
to be the smooth mapping defined by
(xJ), ...,
of real numbers with
(Ek) 0,
=
and
be a smooth function having the properties
IR -+ IR.
listed in (D1) and take
relative to the
for all choices of
+ ... + an
= a1
la l
f
O(xn)). 0 < Ek <
and the corresponding function
f
We consider an arbiI
for which
: Mn -+
IR
defined by
k -+oo
f(x)
fk(x
_
where
k=0
fk(x)
Ek?
_ I
Note first that finitely many terms
f
(k)
a I =k
is well-defined, since for a given value of fk(x)
are non-zero.
Note also that
only
fk(x) = 7 fax
Ijk
near the origin, and hence that 0 Dafk(0)
x
if
k
lal
_
alfa if k = lal We claim that it is possible to choose the sequence that, on some neighbourhood of uniformly for any choice of
0
in such a way
the series > Dafk
converges
k=0 It then follows from standard theorems in
a.
calculus that the sum-function
in IRn,
(Ek)
f
is smooth, and that
229
CO
_ X Dafk(0)
Daf(0)
a:fa
=
k=0 just as we required.
By the usual comparison test it will suffice to choose 00
(Ek)
in such a way that
ving
a)
5'
supn 'Dafkl
k=0
which converges for any choice of
an upper bound on D fk.
is dominated by a series (involClearly, we need to produce
a.
First of all, note that (by sheer differentiation)
we have
Dafk(X)
Ek'aI fR''(a, 9, k, X)
=
k where
as
n
(a, f3' k, x)
=H j=1
As the function 0
a.
CC
ax.J
k)
vanishes outside a compact set we have a well-defined
bound
B(a, g, k)
=
II'(a, 9, k, x)J
sup XE]Rn
so setting
B(a, k)
fR B(a, 6, k)
=
I/I=k we see that
sup . in and hence that the series
I Dafkl
5
Ek Ia I B(a, k)
sup IDaffi
is dominated by
k=0 All we need to do now is to choose the sequence latter series converges for all choices of
230
a:
(Ek)
(a, k).
k=0 in such a way that the
and that can be done by
choosing
ek
in such a way that
e.
>
2k
7,
B(a, k).
231
Appendix E
Guide to further reading
This guide is addressed largely to those who have read at least part of the present book and wish to pursue their interests, but who do not have the advantage of expert advice.
Singularity theory, in common with most areas of
mathematics, has an extensive technical literature, much of which is not readily available.
In preparing this guide I decided only to quote those
sources which I know to be available and which, in my view, a student can hope to gain from reading.
I have made no attempt to compile a comprehensive list
of references, on the grounds that such a list would add little to the value of the book.
In particular, virtually all the mathematics in the last two
chapters has its genesis in papers of J. Mather, listed for instance in [LS1].
(I)
Material Directly Related to this Book
On a general level I feel that anyone starting off in this subject should look at the classic survey paper [Al] by Arnol'd.
[LSI] still provides a very
good reference, and gives one a fair idea of the state of the subject at the turn of the decade.
As far as more systematic texts are concerned, [L] com-
plements the material of this volume in several respects, whilst [GG] provides a fairly formal treatment of much of the global theory. Here are some more detailed suggestions. topology has several good expositions. excellent little volume [M1].
The beginnings of differential
First and foremost I recommend the
The reader who wishes to graduate to the
abstract idea of a smooth manifold will find good accounts in [GP] and [BJ].
These two volumes also contain rather more material on the subject of transversality, as does the more advanced [GG]. 232
The classification of function
germs has now been taken far beyond the beginning indicated in Chapter IV,
though good accounts of this fascinating area of the subject have yet to appear.
Certainly the next step in this direction would be the first few
pages of [A2], whilst some idea of how the subject develops can be found in [A3].
For a lucid exposition of the connexions with other areas of mathema-
tics try [A4].
The reader who wishes to fill in the gaps in Chapter V should
refer to [M], though it is not all easy going.
For this one certainly needs
to be familiar with the Preparation Theorem.
Here I recommend Wall's
"Introduction to the Preparation Theorem" in [LS1] followed by the appropriate chapter of [BL].
Further information on the classification of stable germs
can only be obtained by dipping into the research literature on the subject.
(II)
Material Not Directly Related to this Book
Beyond the differentiable theory of smooth mappings, where the changes of coordinates are diffeomorphisms, lies the important topological theory where the changes of co-ordinates are just homeomorphisms.
Some ideas concerning the
local theory can be found in the latter chapters of [L], whilst an exposition of part of the global theory can be found in [GWPL].
leans heavily on the idea of a "stratification":
The topological theory
unfortunately, there is as
yet no real introduction to the theory of stratifications, but some idea of what it is all about can be gleaned from the references just quoted.
Another
aspect of the topological theory is the subject matter of the classic [M2].
(III)
Applications
In Chapter IV we indicated how singularities of smooth mappings arose naturally in both differential geometry and algebraic geometry.
And it is prin-
cipally to these areas that one looks for applications of the theory within
233
mathematics.
As far as differential geometry is concerned I suggest looking
at [P], whilst the reader who wishes to see further ahead will find solid reading in [W].
The applications of singularity theory to algebraic geometry
are still at a rudimentary level, and no account of this is likely to appear
for some timne. The applications of singularity theory to the physical sciences have centred around the catastrophe theory of Thom.
Those with a serious interest in such
matters will want to look at Thom's own book [T].
Quick introductions to the
ideas can be found in [Z] and [S) while [SP] provides a comprehensive account of the state of the subject at the time of writing.
For the mathematics of
the matter, the account in [BL] extends the material in Chapter IV of the
present volume, whilst more appears in [ TZ] .
I recommend [LS2] for further
ideas relating singularity theory to problems in the physical sciences.
References [Al]
Arnol'd, V. I. Surveys
Singularities of Smooth Mappings.
(1968) 1 - 43.
Russian Math.
Translated from Uspehi Math. Nauk 23 (1968)
3 - 44. [A2]
Arnol'd, V. I. Points, larities.
[A3]
Normal Forms for Functions near Degenerate Critical
the Weyl Groups of
Ak, Dk
and
F.k,
Functional Anal. Appl. 6, 254-272,
Arnol'd, V. I.
and Lagrangian singu1972.
Normal Forms of Functions in Neighbourhoods of
Degenerate Critical Points.
Russian Math. Surveys
29,
10-50,
1974.
[Al
Arnol'd, V. I.
Critical Points of Smooth Functions.
ernat. Congr. Math.
234
Vancouver
1974,
19-39.
Proc. Int-
[BL]
BrBcker, T. & Lander, L.
Differential Germs and Catastrophes.
London Mathematical Society Lecture Notes 17. Press, Cambridge. [BJ]
1975
BrBcker, T. & JAnich, K.
Einftlhrung in die Differentialtopologie.
Springer, Berlin & New York. [GG]
New York, [GP]
1970
Golubitsky, M. & Guillemin, V. larities.
Cambridge University
Stable Mappings and their Singu-
Graduate Texts in Mathematics 14.
Springer, Berlin &
1973.
Guillemin, V. & Pollack, A.
Differential Topology.
Prentice Hall,
1974.
[GWPL]
Gibson, C. G.,
E. J. N.
Wirthmialler, K.,
du Plessis, A. A.,
Looijenga,
Topological Stability of Smooth Mappings.
Lecture Notes in Mathematics
552.
Springer,
Berlin & New York,
1977. [L]
Lu, Y. C. Theory.
[LS1]
Singularity Theory and An Introduction to Catastrophe Springer, Berlin & New York,
I.
and New York, [LS2]
Wall, C. T. C. Symposium
II.
Lecture Notes in Mathematics 192.
Martinet, J.
Springer, Berlin
1971.
Proceedings of Liverpool Singularities
(Ed.)
Lecture Notes in Mathematics 209.
Berlin & New York, [M]
Proceedings of Liverpool Singularities
Wall, C. T. C. (Ed.) Symposium
1976.
Springer,
1971.
Deploiements Versels des Applications Differen-
tiables et Classifications des Applications Stable.
Notes In Mathematics
535
Springer,
Lecture
Berlin & New York,
1975.
235
[M1]
Topology
Milnor, J.
From the Differentiable Viewpoint.
-
The University Press of Virginia. [M2]
Milnor, J.
Princeton University Press.
61
Stewart, I. N. 68,
[SP]
447-454,
[T]
London
Wall, C. T. C. Manifolds.
Lecture Notes Zeeman, E. C. 65-83, [TZ]
1978.
Benjamin-Addison Wesley,
from Geometry and Topology III
New York.
1975
Proceedings of
pp. 707-774
Springer
no. 597.
Catastrophe Theory.
Scientific American
234,
1976.
Catastrophes of Codimension s 5.
236
(trans-
Geometric Properties of Generic Differentiable
Trotman, D. J. A. & Zeeman, E. C.
525
New Scientist
Catastrophe Theory and its Applica-
Symp. at IMPA, Rio de Janeiro, July 1976.
[Z]
543-564.
Structural Stability and Morphogenesis
lated by D. H. Fowler) [W]
(1971)
1975.
Pitman.
Thom, R.
Vol. 5,
The Seven Elementary Catastrophes.
Stewart, I. N. & Poston, T. tions.
1968
The Normal Singularities of a Submanifold.
Porteous, Z. R.
Journal of Differential Geometry. [S]
Annals
Singular Points of Complex Hypersurfaces.
of Maths. Studies [P]
Charlottesville, 1965.
Springer.
Berlin & New York.
Classification of Elementary Lecture Notes in Mathematics. 1976
pp. 263-327.
Index
Action of a group
61
Critical value
Algebra
226
Cubic curves
Algebra product
226
Cuspoids
homomorphism
226
222
Algebraic set
Bifurcation set
40 68-70 130
Curve
Deformation
25
of germ
140
induced
163
44
Binary cubic form
linear
65, 66
Boardman symbol
180,
submanifold
181
186 101, 228
Borel Lemma g-equivalence
144
Chain Rule
9, 21, 23
Change of basis
72
of co-ordinates
72
of parameter
90
Chart
13
Codimension of function germ
98
56
transversal
160
universal
164
Dense
52
Descendant
110
116, 191
Determinacy finite
Determinacy,
120, 191
Diffeomorphism
8, 12
Differential
8, 20
Double point
130
Equivalence of deformations
163
152
of germs
of orbit
77
of maps
16, 205
of submanifold
15
of unfoldings
89, 142
of map germ
Co-ordinates,
local
13
Corank of function germ Critical point Critical point,
139
125 29, 40
degenerate
59
Fold curve
46
Folded handkerchief mapping
55
Flow line
26
237
Fundamental neighbourhood
General linear g roup
Generic Germ,
51
Jets,
equivalence of
62
of germ
37
63 187
-r- e quivalence
143 ,
144
equivalence of
35
immersive
35
Lie group
73
invertible
34
Linear systems
71
rank of
35
representative of
34
Local Existence and Uniqueness Theorem
28
singular
35
source of
34
submersive
35
target of
34
Graph
28
flow
Manifold
12
Metastable range
212
Morphism of deformation
164
Multiplicity
130
Nakayama Lemma
102
Normal bundle space
133
Null set
215
21
Hadamard Lemma
100
Hessian
59, 67
Homogeneity Homomorphism,
62
induced
Infinitesimally stable point Inverse function theorem Isomorphism of ideals
Isotropy subgroup
Jacobian extension ideal
module
112
78
Orbit
61
Parametrization
12
Pencil
71
Product structure
80
Quadratic form
64
10
149 80
180
97 151
Quadratic form,
index of
65
Jet extension
37
rank of
65
space
37
semi-index of
65
Jets 238
37
R -equivalence
94
Transversality Lemma, Basic
49 47
Rational mapping
223
Transversality Theorem
Regular point
224
Transversality Theorem,
value
40
Sard's Theorem
1+9, 217
elementary
51
of Thom
53
Transverse
39 38
Semialgebraic set
222
Transverse intersections
Semi-direct product
150
Triple point
Singular point
44
Singular point,
non-degenerate
123
of type Ak
130
Singularity set
130
44, 54
Singularity set,
first order
higher order Slice
55 177
79
Smooth mapping
Umbilic bracelet
68
Unfolding
81
Unfolding
of germ
141
induced
89
minimal
81
morphism of
90
transverse
81
universal
90
versal
90
8, 12
Splitting Lemma
125
Vector field
Stable germ
142
Vector field,
map
point
206 77
Submanifold
15
Submersion
40
24
equivalence of
30
time dependent
32
time independent
33
Whitney cusp mapping Whitney umbrella
Tangent bundle
87, 213
24
bundle space
22
mapping
22
space
16
Tarski-Seidenberg Theorem
44
224
239
