Casson's Invariant for Oriented Homology Three-Spheres: An Exposition. (MN-36) [Course Book ed.] 9781400860623

Table of contents :
Contents
Preface
Introduction
Chapter I: Representation Spaces
Chapter II: Heegard Decompositions and Stable Bquivalence
Chapter III: Representation Spaces Associated to Heegard Decompositions
Chapter IV; Casson’s Invariant for Oriented Homology 3-spheres
Chapter V: Casson’s invariant for knots in homology 3-spheres
Chapter VI: The Topology of the Space of Representations
References

Citation preview

Casson's Invariant for Oriented Homology 3-Spheres An Exposition

Casson's Invariant for Oriented Homology 3-Spheres An Exposition by Selman Akbulut and Jolin D. McCarthy

Mathematical Notes 36

PRINCETON UNIVERSITY PRESS

PRINCETON, NEW JERSEY 1990

Copyright © 1990 by Princeton University Press ALL RIGHTS RESERVED

The Annals of Mathematics Studies are edited by Luis A. CafFarelli, John N. Mather, John Milnor, and Elias M. Stein

Princeton University Press books are printed on acid-free paper, and meet the guidelines for perma­ nence and durability of the Committee on Produc­ tion Guidelines for Book Longevity of the Council on Library Resources

Printed in the United States of America by Princeton University Press, 41 William Street Princeton, New Jersey

Library of Congress Cataloging-in-Publication Data Akbulut, Selman, 1949Casson's invariant for oriented homology 3-spheres: An exposition / by Selman Akbulut and John D. McCarthy, p.

cm.

(Mathematical notes ; 36) Includes bibliographical references. 1. Three-manifolds (Topology) 2. Invariants. I. McCarthy, John D., 1955-

. II. Title. III. Title: Homology 3-spheres.

IV. Series: Mathematical notes (Princeton University Press) ; 36. QA613.A43 1990 ISBN 0-691-08563-3

514\3

89-38532

Contents

Preface Introduction Chapter I:

1.

ix xi

Bepresentation Spaces

The special unitary group

SU(2,£;)

(a) identification with S3

1

S3

(b) the tangent bundle of

TS3

(c) the conjugation actions on

1

S3

and

TS3

3

(d) the trace and argument maps (e) the unit quaternions

8

Sp(I) and

SO3

10

(f) the pure unit quaternions S2

2.

12

The space of representations of G, R(G) (a) the functor

R

(b) the action of SO3

13 on

R(G)

13

(c) the algebraic set associated to a set of generators, S, of

3.

G

(d) induced polynomials maps

18

(e) the real algebraic structure of R(G)

19

Representations of a free group (a) standard model with respect to a basis

20

(b) the homology of R(G)

21

(c) the tangent bundle of

R(G)

(d) induced orientations of

Chapter II:

1.

IV

25

R(G) via a basis of G

28

Heegard Decompositions and Stable Bquivalence

Heegard decompositions and models (a) the standard handlebody

W

33

(b) the Heegard model (W,h)

34

(c) Heegard decompositions of a three-manifold (d) associated presentations of

Πι

and

Hj

M3

34 35

2.

Stable equivalence of Heegard decompositions (a) stable equivalence of Heegard decompositions

40

(b) Singer's theorem on stable equivalence

41

(c) connected sums of Heegard decompositions

42

(d) the genus one decomposition of

Chapter III:

S3

(TljT2)

44

Representation Spaces Associated to

Heegard Decompositions

1.

The diagram of representation spaces (a) the diagram of representation spaces (b) intersection properties of

2.

The boundary map

3

(a) the boundary map

Q1

and

47 Q2

48

and induced orientations 3

53

(b) the singular set S

58

(c) induced orientations

59

Chapter IV;

Casson's Invariant for Oriented Homology 3-spheres

1.

2.

Casson's invariant for Heegard decompositions (a) the intersection products

63

(b) Casson's invariant for Heegard decompositions

65

(c) independence of Heegard decomposition

66

Casson's invariant for oriented homology

3-spheres

(a) Casson's invariant for oriented homology

3-spheres ....

79

Chapter V:

1.

2.

3-spheres

Dehn surgery on knots in homology spheres (a) Dehn surgery on knots in homology spheres

80

(b) preferred Heegard decompositions for

82

Kn

Casson's invariant for framed knots (a) Casson's invariant for framed knots

87

(b) the canonical isotopy

88

central role in the exposition. whether

there

are

other

the difference cycle (It may be of "characteristic"

...»bg

holes with 3 discs.

We obtain

and capping off the resulting

attaching discs to

h-1 (b,),...,h~l(bg)

2

M3

F

and

by attaching discs to

sphere to get

and capping to attach

Wt Wj.

and then This yields

the following presentation:

F

W1

Figure 5 These two presentations, of course, are of the same type. considers

F

as identified with the target of

We shall prefer the latter perspective. Abelianizing, we obtain

H1(M31Z):

h,

The first

the latter with the source.

Since

F

is oriented, we have a well defined intersection pairing on

H.(F,Z):

We assume that the basis for

H,{F,Z) described above is chosen so that this

pairing

matrix

has

the

following

with

respect

to

the

ordered

basis

(c,,...,cg,d,,...,dg):

That is, if we consider the pairing: 38

we It Ccanbining Recall follows havethat the immediately these identity: h identities, is an from orientation the we compute: definition preserving of the homeomorphism intersection product from Fthat: to

-F.

2.

Stable equivalence of Heegard decowpositions

Suppose and

is a closed

3-manifold with Heegard decompositions

We s a y that

ambient isotopy of

H

in an handlebody

W

i s an embedding of

there i s an arc

on

F

such that:

that

a u p

i s the boundary of a n embedded disc

Mote:

Up to ambient isotopy of

H

W.

in

if there is an

taking

An unknotted one handle D* * I

is isotopic to

such B'

with

W there i s exactly one unknotted one handle

Figure 6

40

Given an unknotted one handle

H

in

we obtain a new Heegard

decomposition of

well defined up to isotopy.

Figure 7 We s a y that

is s t a b l y equivalent t

pair of nonnegative i n t e g e r s ,

Note: an

o

i

f

there e x i s t s a

such that

This move of adding an unknotted one handle to increase the g e n u s of

Heegard

decomposition

has

an

alternative

connected sums of Heegard decompositions.

interpretation

in

terms

of

(See sections (c) and (d) below).

(b) Singer's Theorem on Stable Equivalence of Heegard Decompositions of Closed Orientable Three Manifolds fSIN]) Stable

equivalence

is

the

basic

decompositions of a fixed three-manifold.

41

equivalence

relating

the

Heegard

This is the theorem of Singer:

Theorem 2.1 (Singer): 3

Any two Heegard decompositions of an orientable closed

manifold are stably equivalent.

(c) Connected sums of Heegard decompositions S u p p o s e that

i s an Heegard decomposition of

i s an Heegard decomposition of 3-ball

C

in

Choose a

3-ball

and B

in

M'

and a

such that:

Figure 8 If we remove the interiors of and glue

to

B

and

C

from

and

respectively

^ ! by a homeomorphism r e s p e c t i n g the decomposition:

we evidently obtain an Heegard decoiq>osition for the connected sum of

42

and

Pigura 9 We may think of t h i s operation a s t a k i n g

boundary connected

sums of

the

component handle bodies:

Note:

We adopt t h e u s u a l c o n v e n t i o n that

r e s p e c t to t h e i n d u c e d o r i e n t a t i o n s o n The

operation

of

connected

3B

sum

of

9

is orientation r e v e r s i n g

with

and Heegard

decompositions

has

a

particularly simple d e s c r i p t i o n in terms of t h e Heegard models of s e c t i o n 1(b). If

and

M'

if,

i s i d e n t i f i e d with

by

abuse

decompositions b y

of (W,h)

and

notation, and

with

we

(V,g),

denote then:

43

so that:

the

corresponding

Heegard

where the boundary connect sum identifies

(D,0) and

(E,p)

a n d i s

well by assumption. Mote:defined We choose the new disc and base point as in Figure 10.

Figure 10

(d) The genus one decomposition of Of particular interest to u s is the g e n u s one Heegard decomposition of where

is the standard t o r u s embedded in

R^

and

(Actually there is only one g e n u s one Heegard decomposition of [W].)

44

is

Figure 11 It is e v i d e n t from Figure 7 that S i n g e r ' s move of adding an one-handle

is

equivalent

Heegard decomposition of

to forming

the

connect

sum

with

the

unknotted genus-one

S':

The g e n u s one Heegard decomposition of model.

45

has the following Heegard

Figure 12 Choose

a

homeomorphism

of

i n t e r c h a n g e s the simple closed c u r v e s

which a

and

fixes

b:

Clearly, this yields the desired Heegard model:

the genus one Heeg. dec. of

E

pointwise

and

CHAPTER III: REPRESENTATION SPACES ASSOCIATED TO HEEGARD DECOMPOSITIONS 1.

The diagram of representation apaces

(a)

The diagram of representation s p a c e s

Given an Heegard decomposition

of a closed oriented

3-manifold

M^, we have an associated commutative diagram of inclusions:

where:

Note that each inclusion except for cells.

Hence, by applying

n,,

i

is obtained by attaching

2

and

3

we obtain a commutative diagram of groups

where all homomorphisms except

are surjective.

"All the fundamental

group is in the punctured surface".:

Clearly, i* is an inclusion. Finally, by

applying

the

representation

commutative diagram of spaces:

47

functor

R,

we

obtain a

where:

As usual for Horn functors, all the surjections become injections. shown that

(b)

3

(It can be

is surjective. We shall show this later.)

Intersection properties of

and

Prom our previous discussion, we observe:

Since all the injections are natural, we consider them as inclusions. this convention, we conclude that: and

are half-dimensional cycles in

48

Under

Hence, we can consider their intersection product in

R*.

We shall prove the

following proposition.

Proposition 1.1:

Note: We have denoted the top c l a s s of

by

Propositions 1.3.1, 1.3.2 and 1.3.3 thru 9 are the key tools in the proof. We need to develop some background before beginning the proof. Identify

M^

with an Heegard model

(W,h). The diagram of inclusions in

part (a) may be written as follows:

From this, we obtain the corresponding diagram of representation spaces:

49

Considering all the injections as inclusions, we may make the

identifications

below:

In particular, we have fixed orientations on

and

Proof of PropoBition 1.1: Consider the natural cocycles in H'(R*,Z):

If we consider

as having the induced orientation via the ordered basis the dual

cycles are given by:

50

Notes: (1)

is oriented

(2) Similarly,

has the obvious orientation.

Clearly:

Then:

where w

is the top class of

By Proposition 1.3.1,

(h*)* with respect to the ordered basis as the matrix for

with respect to

Using the fact that

51

the matrix for is the same

But, as established in section

),

is

"dual" to w:

Hence, by comparison with section II.1(c):

For the same retison:

We turn now to the proof of part (c).

Using the same models as above, we

may make the identifications: (6) (7) (8) (9) Transversality of

(10) But

id

at

1 is equivalent to the equation:

is the image of the map:

(11) The direct product here is "abstract" and we identify factor as:

52

the second

(12) By Proposition 1.3.3, the map is given by the matrix:

(13) By the above remarks singular.

Clearly, this is equivalent to the

is nonnonsingularity

B. The result follows o Apparently, the fact thatas before. is symplectic is important here.

Note:

The boundary map

3

(a) The boundary map

3

2.

at

and induced orientations

Consider:

This induces a map of representation spaces, the restriction map:

We denote this map as the boundary map

3:

We have the following properties of Proposition 2.1: (2)

(1)

3

{critical p t s of

is s u r j e c t i v e (reducible repa}

53

of

Proof of (1): As in section 1(a):

and

R(n^(3F*,0)) This

latter

is identified with identification

may be chosen

so that

the map

3

may be

written as:

•

Proof of (2): Let:

54

(6) Of course;

(7) By computing derivatives as before and applying appropriate left and right translations to each factor of the given splitting of Tp(R*) and to

we can construct a commutative diagram:

where:

(8) Since the following elements of

are a free basis:

we see that Proposition 2.1(2) is equivalent to the following claim. Claim:

where:

55

Then (9)

(10)

F

is surjective i f and only i f

is irreducible.

Hence, we begin by computing the image of the map:

Suppose:

Then:

Hence:

(11)

In general,

(12)

Hence, let

and:

so:

56

(13)

Since F

we conclude that:

is surjective i f and only i f there exists

j, k

with

and

(14)

Let Suppose

is reducible.

Then there exists

that:

So:

So:

So:

Claim:

Subproof:

is reducible iff

(assuming

We have just shown the necessity.

Then:

57

Now suppose

such

In other words,

(15)

Finally,

is diagonalizable.

it is a straightforward exercise to verify that: Normalizer

The result follows immediately from (13) and ( 1 5 ) .

(t>) T h e singular Bet

S

Let:

By Proposition 1.2.1, every reducible representation is conjugate to a diagonal representation. relation for

Hence, every reducible representation satisfies the generating and we conclude that:

As an immediate corollary to Proposition 2 . 1 3)

(and the definition of

and

we conclude:

Corollary 2 . 2 : (b) is an open smooth manifold of dimension Appealing to Proposition 1.2.1 and the naturality of the deduce the existence of the following free actions:

58

action, we

Let:

From the above observation, we derive:

Proposition 2.3: ( a ) i s (b)

(c)

a smooth open manifold of dimension

is a smooth open manifold of dimension

Induced orientations Consider the following choices of orientation:

induced orientation from induced orientation from induced orientation from induced orientation from

The orientation on

determines a unique

adopt the following convention:

59

generator

of

We

Ra

-

induced orientation from the basis of corresponding to the orientation on

Recall that a short exact sequence of oriented vector spaces:

is said to be compatibly oriented i f we can assume:

Given a short exact sequence of vector spaces with two of there is a unique orientation on oriented.

the third

60

making

U,V,W

the sequence

oriented, compatibly

We shall refer to this orientation as the induced orientation.

By as is Corollary a follows: Next, Now, short given we exact orient 2 . 2an and sequence. arbitrary Proposition byorientation the 2following . 1 , weof conclude convention. we that orient iConsider f these the vector sequence: then: spaces

induced orientation from orientation on induced orientation from induced orientation from

It

is

easy

to

see

that

this

orientations

on

the

tangent

gives

spaces

a

and

well-defined

of

and,

continuous hence,

a

family

of

well-defined

orientation on In a similar manner, we may orient

More precisely, we consider the

sequence:

and the associated short exact sequence:

Since

which

;ts freely on

induces

orientation.

!

an

orientation

we have a natural homeomorphism:

on

(Recall that

is oriented as the quotient

We orient the vector spaces above as follows:

induced orientation from induced orientation from induced orientation from

61

has

a

fixed

Again,

it

is

easy

to

see

that

we

obtain

in

this

manner

a

well

Finally, in the same manner, given an arbitrary orientation

o

defined

orientation on n

w

e

obtain an orientation on

induced orientation from induced orientation from induced orientation from

Note:

Under

our conventions,

and the orbits of

the only and

fixed orientations are those of and

only be compatibly oriented with respect to these. conventions will be seen shortly.

62

and

need

The motivation for these

C H A P T E R IV:

C A S S O N ' S INVARIANT FOR ORIENTED H O M O L O G Y 3-SPHERES

1.

Casaon's invariant for Heegard decompositions

(a)

T h e intersection products Let

be an homology 3-sphere:

Let

be

a

Heegard

decomposition

of

As

an

immediate

consequence of Proposition III. 1.1, we obtain: Proposition 1.1:

By

Proposition

1.2.1,

diagonal representation. factors through

every

reducible

Hence, every

representation

reducible

is

conjugate

representation

to

a

of

We deduce the following corollary:

Corollary 1.2:

is compact. are properly embedded open submanifolds of is compact.

Note: abelian.

If the genus of the Heegard

decomposition is one, then

It follows from Proposition 1.2.3 that:

if if

genus genus

63

is

Choose an isotopy of where

with compact support which carries

is transverse to

Then, b y Proposition III.2.3:

is isotopic to

finite set of points.

Given orientations on of

and

we can define the algebraic intersection

and

By standard arguments, this is well-defined.

Figure 13

64

(b)

Casson's invariant for Heegard decompositions If

is an oriented homology 3-sphere with an Heegard decomposition we define:

where all the orientations are chosen and

S

Notes:

is either

or

1)

1.

S

by

b y the compatibility conventions of

For instance, if we change the orientation of

change that of

111.2(c),

will be determined in section V.5.

is well-defined

section

the conventions of section

and, hence, that of

we must

Hence, by Proposition 1.1, we

can always choose the orientations subject to the convention:

If we change the orientation of

this changes the orientation of

This changes the generator on

to

T h e induced orientations

come, respectively, from the identifications:

Since

inversion

orientation of by 1). hence, on

changes

the orientation

must change.

of

we

see

that

the

We may maintain the given orientation on

In order to maintain compatibility, the orientation on must change.

induced

has a fixed orientation).

65

Therefore:

and,

(c)

Independence of Heegard decomposition We

wish

to

show

decomposition

that

is

independent

of

the

Heegard

It is evident that this invariant is an invariant of

ambient isotopy of Heegard decompositions.

Hence,

from Singer's Theorem, we

conclude that it suffices to show that it is also invariant under

stabilization,

(i.e., under addition of an unknotted one-handle). If

is an Heegard model, let:

Proposition 1 . 3 :

Given an Heegard model

let:

Then: Proof:

Choose a basis

as in section 11.2(d):

66

as in section 11.1(a).

We may make the following identifications:

Then:

1.)

We wish to compare the various terms.

(i)

Comparison of denominators From the above identifications, we observe that:

67

(This last identity follows from the definition of

given in section 11.2(d).)

Consider the Runneth formulae:

As in the proof of Proposition III.l.l, we consider the natural

3-cocycles in

and

Then:

Let

and

be the top-class and coclass of

Then:

68

as in section

1.3(b).

Our conventions on orientations allow us to choose the following orientations:

69

As From Let:in the the above proof observations, of Proposition weI Icompute I . 1 . 1 , we that: conclude:

Let:

Then:

Hence:

2.) (ii)

Comparison of numerators We wish to carry out a similar comparison of numerators.

not a factor of hence,

a

"local

we shall see that factor".

This

will

Although

is a submanifold of be

sufficient,

since

the

is and,

following

considerations imply that all the intersections occuring in the numerators will take place in a regular neighborhood of By an argument similar to section 70

By an obvious abuse of notation:

Since

acts by conjugation on each component, we obtain a well-defined

commutative diagram:

from which we conclude that:

In order to compute the numerators, we need to isotope transverse to

and

and

construct a single isotopy of

and

then count intersections. to accomplish this.

to be

We would

like to

Furthermore, we want

this isotopy to be as simple as possible so that we can compare intersections and

orientations.

neighborhood of

R\S

To

see that this is possible,

in

is a properly embedded subaanifold of In order to see this,

consider

the map:

71

we

investigate

the

regular

Then:

Clearly:

It follows from Proposition I I I . 2 . 1 that bedding.

Moreover,

from the

observations

is a differentiable emabove, this embedding is proper.

Hence, we conclude that: Is a proper embedding is a proper embedding. In addition, we have the following commutative diagram of proper embeddings:

(The latter inclusions follow from the definition of

and

is a properly embedded submanifold of Since all the proper embeddings above are invariant under the we conclude immediately that: ls a proper embedding is a proper embedding

72

S0 3 action,

and that we have a commutative diagram of proper embeddings:

where:

In addition, we observe that:

Identification of oriented tangent spaces From the map

Note:

Considering

we conclude that i f

then:

the given trivialization of

and the associated Hiemannian metric, this splitting can be considered

as

a

natural orthogonal splitting. Recall that our conventions on orientations stipulate that the following exact sequences of oriented vector spaces be compatible:

73

With respect to the given basis, this diagram may be identified with:

where

By the choice of orientations on

and

given in part

the following equality of oriented vector spaces:

(oriented).

From the definition of compatibility, we deduce:

(oriented).

Using the same metric, we may split the tangent space to

Considering, the short exact sequences: we may make the following identifications: 74

we have

Recall again that our conventions stipulate that the compatibly oriented.

above

sequences

be

That is, considering the above identifications:

oriented) (oriented).

From the previous observations,

i f we write:

(oriented)

then we obtain the equation:

Counting dimensions we see that 3.)

(oriented).

By the choice of orientations on

and

given in part

the equations: (oriented)

75

(oriented)

(oriented). Our conventions stipulate the compatibility of:

we have

where

We observe that the first column is orientation preserving:

Having already observed the same for the second column, we conclude that:

(oriented).

Likewise, we must have compatibility of:

Following the argument for equation 3 . ,

write:

(oriented)

to obtain the equation:

Counting dimensions we see that

4.)

76

(oriented).

5.)

(oriented)

Construction of the isotopy Choose a compactly supported ambient isotopy of to

where

theorem

is transverse to

[Hi], extend

assume that this

this isotopy

isotopy

H

respects

By the isotopy extension

to an ambient

the

which carries

orthogonal

isotopy of splitting

We

of the normal

bundle of

Note: isotopy

The existence of such an isotopy follows readily from the proof of the extension

equivariantly in

theorem.

Alternatively,

one

can

construct

H.

Let

next observations follow immediately from the construction of

and

isotopy

using the product structure on the normal bundle of

Finally, extend to an isotopy of

Since

the

are transverse, we conclude that:

and

_

are transverse.

77

The H:

Moreover, by the previous results and the construction of the isotopy:

(oriented) (oriented)

.

Let:

Then we have the equations:

(oriented) (oriented) (oriented).

From this, we derive the equations:

Counting' dimensions, we conclude that:

Since this holds for all

we conclude from the previous

observations that: 6.)

From equations 1 . , 2.

and

6.,

we conclude:

78

2.

Casson's invariant for oriented homology 3-spheres

(a)

Casson's invariant for oriented homology 3-spheres

If

--

is an oriented homology 3-sphere, we define:

where

is an Heegard dec. of

From the results of section If

and

this is well-defined.

are oriented homology 3-spheres and

is an orientation preserving homeomorphism, then decomposition of

to

an

Heegard

h

carries any given Heegard decomposition

of

It is evident from our conventions that:

Hence, we obtain a topological invariant:

Proposition 2 . 1 :

Casson's invariant for oriented homology 3-spheres

is

an homeomorphism invariant.

The obvious question at this point is whether Casson's invariant is computable. from

It is well-known that every homology 3-sphere

by a sequence of

is obtained

surgeries on knots in homology 3-spheres.

Hence, one approach to this question is to study the change in Casson's invariant when we perform

surgery on a-knot in

It turns out that

there is a very simple formula for this change (as we shall see). next chapter we begin the study of this

79

"differential".

In

the

C H A P T E R V:

C A S S O N ' S INVARIANT FOR K N O T S IN H O M O L O G Y

1.

D e h n surgery on knots in homology spheres

(a)

Dehn surgery on knots in homology spheres

L e t b e Let

Let

3-sphere.

be a knot in

Let Let

an homology

3-SPHERES

be a closed regular neighborhood of -

be the boundary of

K.

so that

is a torus.

be

Now:

Consider the inclusions:

80 and the associated surjections:

and them lb yof nontrivial Since Since simple and closed andformcurves a are basis primitive on for elements, which we we we also may may denote assume represent asthat

they intersect transversally in precisely one point.

Each of these curves is

well defined up to isotopy and change in orientation. Let

M3

be oriented.

The orientation on

on

M3(K).

Ta

by the convention that "the orientation on

T3

plus the inward pointing normal vector to

orientations

This orientation on

are

orientations on orientation on

chosen μ

μ

the pair of curves

and

M3(K)

M3

subject λ

induces a well defined orientation on

to

this

M3(K)

is the orientation on

T2".

(We assume that all

convention.)

so that "the orientation on

plus the orientation on (μ,λ)

restricts to an orientation

λ".

We Ta

choose

is equal to the

Subject to these assumptions,

is well defined up to ambient isotopy of

simultaneous change of orientation of standard meridian-longitude pair for

and

μ

the

λ.

Ta

and

The pair is called the

K.

(iO Figure 14 Any simple closed curve

7

on

Ta

is uniquely determined by its

homology class (which is well defined up to a change in sign):

γ = ρμ + ςλ Moreover, the relative primeness of

= 1

ρ

and

q

·

ensure that

ρμ + q*

is

represented

by

obtain a well curve on

a

simple closed

defined

,

Hence,

(up to isotopy)

and vice-versa. is

curve.

for each fraction

homotopically

nontrivial

simple

we closed

(We allow that

surgery on

(

(or just

surgery on

K).

This is defined as follows:

does not depend on the choice of Now

(assuming

homology

Hence

is an

3-sphere precisely w h e n

is

Dehn surgery on

K

.

Mote:

(b)

Preferred Heegard decompositions for

In this section, we construct Heegard decompositions which are compatible with surgery.

These will be useful in comparing

the Casson invariants for

and

Lemma that

K

1.1:

There exists an

Heegard

decomposition for

is a separating curve on

82

such

Figure 15

Proof:

Choose a Seifert surface

(Here,

and

for

K.

Thicken

to

is identified with

Figure 16 Attach

1-handles

handlebody

(from

such that

to is an handlebody

83

to obtain a

new

Figure 17 If we so desire, we may assume that Lemma 1.2: There exists an Heegard decomposition for

Proof:

through

such that

Repeat the proof of Lemma 1.1 to construct an Heegard as in Figure

17.

the

the

effectively

cores

of

adding

Finally, observe that

Then

drill from

l-handles

l-handles

to

which

to were

to obtain

is homeomorphic to

an handlebody

84

decomposition and

attached a

new

to

handlebody and, therefore, is

Figure 18 By a n ambient isotopy, we may assume

Now choose an Heegard decomposition of

1.2):

let:

By an ambient isotopy, we assume that

Figure 19

85

as in Lemma 1.1 (or Lemma

Note:

K

is homologically trivia] in

trivial in Let

(unless W

is a disc, and, hence,

denote

Let

we take to be supported on

Proof:

K

is a trivial knot).

and identify the given Heegard decomposition of

with an Heegard model

Lemma 1.3:

It is not, however, homotopically

denote the Dehn twist about

K

which

A.

is an Heegard model for

Let

B

be

in

Then

is homeomorphic

to the

quotient space:

where:

Identifying

with this quotient, we observe that:

longitude .

Let

be an

arc in

A

from

to

Let

Likewise, w e observe that:

meridian .

Since

is supported on

A

we may consider

Under our conventions on orientations, we have:

86

as a Dehn twist on

(Recall, in general,

S

i

n

c

e

w

e

obtain a

change of sign.) ¥

Figure 20 Let

so that:

In order to obtain off.

we wish to attach a disc to

and cap

From the preceding observation and the same considerations which allow

us to identify

as a "prequotient" of

we conclude, as desired,

that:

2.

Caason's invariant for framed knots

(a)

Casson's invariant for framed knots A framed

integer.

If

knot is a pair K

where

is a knot in an homology

87

K

is a

knot and

n

3-sphere, as above, we define:

is an

(b)

T h e canonical isotopy If we choose Heegard

models for the Dehn

previous section, we compute, as in section

where

is genus(F)

and

is genus

88

surgeries that:

.

That is:

on

K

as in the

We wish to study the difference

Note:

It should be emphasized

here that

is not a cycle in the classical

sense, since it is not compact. We may identify

In these coordinates,

Note:

R

in the following manner:

has a simple form:

We choose the base point for

in

89

As above,

Let

be

abelian and

in

R.

Since

reducible representations

are

is homologous to zero, w e readily deduce that:

Let:

Since

is an homology sphere, we know that:

From the observation concerning

In the complement of element of to

{1}

we conclude:

is given by conjugation of

by an

Hence, we can employ the natural contraction of

which arises from the exponential map,

to construct an isotopy in the complement of 90

(see sections 1.1(b) and which carries

onto

K Figure 21

Note: R_(K)) to

In Figure 21, the shaded region indicates the trace of under the isotopy

R_(K).

H.

It should be observed that

H

Q1

(away from

does not extend

Clearly

is

identity.

We recall from the proof of Proposition 1.3.3 that the tangent space at is spanned space at

Note:

1

to

by curves of cyclic representations S

is

to

tangent

Moreover, for any

In a small neighborhood of

to an open interval containing of

(i.e. the Zariski

1

I.

1,

the isotopy extends in an obvious way

It is with this consideration that w e speak

and

Prom the previous observations w e conclude that:

where

is

projection

onto

the

first

factor.

techniques that there exists an open neighborhood of that:

Figure 22

92

It

follows 1

in

by R,

standard such

We are now able to prove an analogous result for the trace of

Lemma 2.1:

Proof:

There exists a neighborhood of

From the previous observations,

fixes both seta for all

in

Moreover,

-

Hence, w e may choose a neighborhood of

in

Figure 23

•

93

such that:

t.

that:

Finally, let

S:

it follows that:

-

-

near

such

Note:

From

the conclusion, we

invariant.

Moreover,

boundary.

Clearly,

isotopy

H

can

assume

is

defined on

onto

may choose

that

Vn is

-equivariant.

Hence,

we

in the complement of

a

to be manifold

have a

with

canonical

and carrying

Moreover:

Note:

Recall

contained

that

All

in this space,

speak of

(c)

we

see that we

definition).

therefore,

We will, however, for

are

convenience,

H

6

is not a compactly

supported

previous section allow us to employ

with

difference cycle is independent of intersection with

isotopy,

the results of the

to define a compact cycle

"carries the intersection of

As we

(as a genuine cycle in

is independent of

invariant. Let

spaces,

and consider that:

T h e difference cycle Although

(by

quotient

be the compact cycle:

94

n.

which

shall see, this and

its

This will establish Casson's knot

Figure 24 By the discussion in the previous section, w e conclude that:

95

Note:

is a compact cycle in

We assume

that

is disjoint from

manifold neighborhood of

We choose

Since

so that

in

is a manifold.

is a compact boundary in

96 Finally, we define the difference cycle.

. with

Let

Let

be a Let:

compact

It suffices to show, therefore, that the previous

section, we

is homologous to

see that we can assume

regular neighborhood of

that

hence, that

in

Prom

is carried

by

a

is in the image of the

natural map:

We have already seen that trivially on

acts trivially on

This establishes the result.

(indeed,

acts

In summary, we have shown

that:

for

any

Heegard

independent of

decomposition

as

in

Lemma

1.1.

Hence,

is

n.

3.

Casson's invariant for knots

(a)

Casson's invariant for knots Given a knot

K

in an homology

3-sphere, we define Casson's invariant

by:

By the

previous

section, this

is

a

well

defined

topological

invariant.

Moreover: Note: In order to compute this invariant, we investigate the difference cycle

97

4.

T h e difference cycle.

(a)

Identification of representation spaces Using

an

Heegard

model

as

in

Lemmas

1.2

representation spaces as follows:

reducible representations in reducible representations in

regular neighborhood of

in

98

and

1.3,

we

may

identify

Note:

(b)

We shall show that

converges to a cycle in

A n eqinvariant trivialization of We may identify

with

via the diagonal map:

With this identification, we may write:

Figure 25

99

By the results of section III. 1(b), we may choose a trivialization:

For

we have

Now, for sufficiently

and the trivialization restricts to:

small

the exponential map on

projection which is equivariant with respect to the

Being a projection of the normal

bundle of

provides a action:

in

we

obtain

a

diffeomorphism: We may assume that (for sufficiently small

is an equivariant

(c)

Collapsing

6

to a cycle in

trivialization.

B-(K)

As we have already pointed out in section 2(c), (or more precisely by a regular neighborhood of collapse

6

into

S

is carried b y In this section, we

and obtain an alternative description of

100

Let:

Figure 26 We shall require a trivialization of

obtained via the exponential map in

projecting to the "bottom e n d " :

Figure 27

101

where:

We observe that this decomposition of

is preserved under the natural

involution:

The trivialization of

and the involution

are compatible in that:

Clearly, we may write:

The trivialization above allows us to realize

via a map from

Observe that as a cycle: We However, have also in the previously complement remarked of that , 102 weHknow does that: not

extend

across

Hence, we may extend

H

continuously to:

With this extension in mind, we may write:

For the purpose of the next lemma, we define the following actions of on

and on

Lemma 4 . 1 :

There exists a natural

equivariant map:

such that: Note: From the previous observation: All we

need

to

do

is

to construct

Actually, we shall ensure that

as a converges as

way, w e shall realize

as a class in

Proof:

"in pieces".

We construct

103

degree

one

homeomorphism.

tends to zero.

In this

(1)

- Observe that

is the natural

inclusion on the

intersection

and

- Hence,

we may extend

as follows:

(2)

where:

- From the compatibility of

and

we observe that on

iff

- Hence, we may extend

as follows:

(3)

- Clearly,

is an

equivariant, degree one homeomorphism.

104

of

Let:

Then:

Our next goal is to study the limiting

behavior of

With this in mind, we define:

Clearly,

-equivariant.

Lemma 4 . 2 :

Proof:

We check the limit "in pieces"

Case ( 1 ) :

Hence:

105

as

tends to zero.

Case (2):

Hence:

Case ( 3 ) : Choose

sufficiently small such that

Let:

Then:

Clearly:

106

for all

Then,

if

Hence:

Claim:

Note:

We recall that the natural contraction on

S3. Proof of Claim: that Let: Recall

does not extend to Hence:

Now, by definition,

Now:

Moreover:

so that:

Hence:

107

I f we write

we compute that:

That is:

Having established the claim, we conclude that:

From steps ( 1 ) ,

(2) and ( 3 ) , we conclude that:

108

Let: Let:

Consider the natural map:

Corollary 4 . 3 :

The difference cycle

is given as:

with: Proof: we This But Since: conclude the implies From inversion Lemma that that4 map .the 2 , there on following is is a map continuous orientation 109 is orientation map: reversing: reversing:

- Moreover, the natural map is an

- Hence, since

h

is

bundle map:

equivariant,

is also orientation reversing.

- Clearly, - Hence,

(d)

The cycle

as a pullback

In this section, we give an alternative description of a fibre bundle with total space

as a pullback of

We begin by considering the map

of the previous section:

Note:

From hereon [

appropriate

] will

denote

class

action.

Proposition 4 . 4 :

Proof:

equivalence

cover of its image.

It suffices to show that

- But for some fixed - This latter condition, however, implies that:

110

or orbit

under

the

- Since

consists

of

irreducible

representations

and

we

conclude from Proposition 1 . 2 . 3 that:

- Hence,

and

- Thus, - Now

suppose that

Then as above: and

- Once again, by Proposition 1 . 2 . 3 , we conclude that: and

- This is clearly impossible, and hence

The previous proof indicates a natural factorization of cover and an inclusion.

We define the following action of

We have the natural map:

which induces a map of orbit spaces:

111

by

a

2-fold

Finally, we can define a map:

which induces a map of orbit spaces:

Corollary 4 . 5 :

(1)

is an embedding is a

Proof:

2-fold cover

This is an immediate consequence of Proposition 4 . 4 .

Now we have an obvious projection map:

Proposition 28:

Proof:

p

is a fibre bundle with fibre

It suffices to show that the fibre of

- Suppose - Then

and

- Hence, - In other words,

112

p

is

- But - Hence,

Let

be the diagonal in

Proposition 4 . 7 :

Proof: - Hence, - On the other hand, suppose - Then

Hence, we have a pullback diagram:

Corollary 4.8: Proof:

This is an immediate consequence of the previous proposition.

Note:

has a natural orientation inherited

from

via

p.

We

may

113 think of this in two ways (at least). (1) duality Suppose

where

is an oriented submanifold of is a submanifold of

M

N,

M

and

Suppose

N p

are manifolds. is transverse to

with a natural orientation given by:

Suppose Then

where (2)

D

is Poincare duality.

transversality Under the same assumptions as in (1), we may orient the tangent

of

spaces

compatibly by the rules:

where

and

(Compare this with the discussion in

section I V . 1 . ) These two descriptions are, of course, 5.

CaBBon's invariant for fibered knots

(a)

Fibered knots We recall that a knot

K

in a

monodromy

f

(where

is a Seifert surface for

3-manitold

if the knot space

From this fibration of decomposition for

Kn,

equivalent.

M

is a fibered knot with

fibres over and monodromy

with fibre f:

we obtain a strongly preferred

(as in Lemmas 1.2 and 1.3), as follows.

114

Let:

Heegard

Let:

Then:

We consider the following homeomorphism:

115

Then Ithen f wewe we identify see obtain that: an Heegard with modelvia forthe map:

where:

Note:

(b)

This decomposition satisfies Lemmas 1.2 and 1.3.

Casson's invariant for fibered knots in homology

3-spheres

and Lefschetz numbers Given a fibered knot Heegard decomposition for

K

in an homology

3-sphere

M,

as in the previous section.

we may identify representation spaces as follows:

116

let

(W,h)

As in section

be an

reducible representations in

In addition, we have the following objects:

Proposition 5 . 1 :

Proof:

is transverse to

First, we make a dimension count:

- Hence, we see that:

117

in

H.

- Next, we compare tangent spaces

at a point

of

intersection

of

and

- Clearly, - The dimension count and intersection calculation establish transversality in R.

Note:

We did not use the assumption that

establishing

this

last

proposition.

This

M

is an homology

assumption

is

3-sphere in

equivalent

to

the

following condition on the monodromy:

is nonsingular

As in the proof of Proposition III. 1.1, we may interpret this in terms of But we made no assumptions about this derivative.

Proposition 5.2:

Proof:

has an orientable normal bundle in

By Proposition

5.1,

is a submanifold of

All manifolds in question are orientable.

118

(as well as of

We recall from sections 3 and 4 that:

Construct an isotopy of

which carries

transverse to

Extend this isotopy of

and let:

Then: is isotopic to

Furthermore, by Proposition 5 . 1 :

is transverse to

119

to a submanifold

with:

Figure

28

Hence, by Proposition 5.2, we conclude that:

Applying the above observations:

120

Now we wish to apply the projection Proposition 5 . 3 : is a l i f t of

p.

First, we need to observe that:

Proof:

It suffices to construct a section

of the

restriction of

to

We need only to show that the obvious map is well defined:

But this is immediate from the observation:

We need the following simple lemma, which will allow us to compare:

and

Lemma 5.4:

Let

be a map of connected, oriented

Suppose t h a t a n d transverse to

and

are oriented that

submanifolds

lifts to an oriented

of

manifolds.

that

p

is

submanifold of

Then:

Note:

We have already

used

this in "projecting"

to

from

lemma follows readily from the same local transversality argument.

The However,

we shall give an homological argument, via duality, for the sake of variety.

121

Proof:

Note that, under

with

the standard

identifications of

and

is the identity:

- But

is a lift of

That is, there is a section:

~ Hence, - Finally,

Applying this last lemma, we conclude that:

By definition,

Corollary 5 . 5 : sphere

the right hand side is the Lefschetz number of

If

K

is a fibered knot with monodromy

then:

122

f

in an homology

(c)

Fibered knots of genus one We recall that a knot if

Note:

K

K

in a

3-manifold

M

is a fibered knot of genus

has a fibration such that the genus of the fiber is

We do not assume that

Suppose that

g,

is

1.

is minimal. Then:

free group of rank

2

.

Consider the following representation:

Then:

Le—a 5 . 6 :

Note:

If

then

By a dimension count, we know that

is a manifold of dimension

Moreover, we know it is connected, since it is a quotient of a retraction of the connected space

123

and

0. is

We shall, however, give a direct proof.

Proof:

Suppose that

- Let

and

- Hence,

Then:

or

- This implies that - Therefore, at the risk of conjugating

- Now solving the consequence,

we nay assume that:

of the relation above,

that:

- arbitrary

- Hence,

is conjugate to

124

we conclude

Corollary

5.7:

If

K

is

a

fibered

knot

of

genus

one

in

an

homology

3-sphere, then

Proof:

This is an immediate consequence of Corollary 5.5 and Lemma 5.6.

Corollary 5.8:

Note:

Let

Since

K

and

we now choose

S

be

the

trefoil or figure

eight

knot

in

•

Then

have only been defined up to a choice of sign

such that:

(trefoil knot] We have not attempted to calculate 6.

Dehn surgery on unlinks

(a)

Dehn surgery on unlinks Let

S.

be an homology sphere.

in

Then:

If

and

Let

be a link of two components

are standard meridian-longitude pairs for

K

and

L

respectively, then:

Let

denote the manifold obtained surgery to

homology

K

and

surgery to

3-sphere precisely when

L

from

by

performing

(as in section 1(a)).

This is an

(assuming

before, we denote these homology spheres as follows:

125

As

Now suppose that

Now

K

K

and

L

have zero linking number.

is naturally contained in

.

is a standard meridian longitude pair for

Note:

In general,

which depends on

(b)

That is to say:

This assumption implies that K

in

.

I

Hence, it is clear that:

for some surgery coefficient

m, n

and the linking number of

K

and

L.

Preferred Heegard decompositions for surgeries on boundary links Suppose that

disjoint

Seifert

number.)

In a

is a boundary link.

surfaces. manner

(In similar

particular,

K

That is, and

to the construction

L

K

and have

of section

L

bound

zero

linking

1(b), we

may

construct a particularly convenient Heegard decomposition compatible with the various surgeries on Proposition 6.1:

There exists an Heegard decomposition for

that:

and

K

is a separating curve in

126

such

w, Figure 29 Proof: F1.

Choose a pair of disjoint Seifert surfaces for

Thicken

F0

to

F0 * I»

the boundary of

F0 * I

tubing

F1

L,

F2

and

with

disjoint from

and form the connected sum of

together.

we obtain a surface,

F0 * I

K

F",

L = 3F"

and F1. F2

L, Let and

By choosing the tubing away from

F0

and

Fa

be

F1

by

K

and

with the following properties: K

is separating in

F" .

The rest of the proof continues along the lines of the proofs of Lemmas 1.1 and 1.2.

In order to preserve the condition that

need to insure that all component of

F"\K

K

is separating, we only

1-handles are attached in the complement of the

bounded by

the attaching discs off this surface.

K.

This is easily accomplished by sliding •

As in section 1(b), we identify this decomposition with an Heegard model (W,h).

Let

respectively.

and

τχ

denote the Oehn twists about

We obtain the analog of Proposition 1.3.

K

and

L

Proposition 5 . 2 :

7 (a)

is an Heegard model for

Casson's invariant for unlinks CaBson's invariant for unlinks Let

, , )

be an unlink of two components in an homology

That is, assume that observed, Km).

K

K

and

is naturally

L

have zero linking number.

contained

in

(L

3-sphere As previously

is naturally contained in

We denote these various knots as follows:

Similarly, we use the knots

i and

and

to denote Casson's invariant for respectively.

Casson's invariant for the unlink

(K,L)

in

is defined by the rule:

By grouping the terms in two distinct ways, we see:

128

This it (b)is Let Triviality independent shows thatforbe of boundary a mboundary and is links independent n; link in an ofishomology well m and defined. sphere n respectively. Choose Hence, an

Heegard

decomposition

for

as

in

Proposition

6.1

with

corresponding

Heegard models as in Proposition 6.2. Following the argument of section 2(b), we conclude that:

Let

Recall that we have a difference cycle:

Then:

Hence, we conclude that:

Proposition 7.1:

Proof:

Let

(rationally).

Recall that

be

the

restriction of

hg

is carried by

component to

(section

of by

which

contains

Then the action of

given by the rule:

129

Here, we have:

K

and on

denote

the is

In addition we have the projection:

From Corollary 4 . 8 , we know that:

where

is the diagonal of

In terms of Poincare duality,

this

equality may be expressed as follows:

130 From the results of the next chapter (Theorem VI.2.4) acts trivially on the homology of

F',

and the fact that

we conclude that:

Hence, where second and cohomology Since the factor: by following theis Runneth are the dual: obvious diagram formula map commutes: and induced the fact from that

Since acts the rational triviallyhomology on the

we conclude from the previous remarks that:

Corollary 7 . 2 :

If

is a boundary link in an homology

3

sphere

then

Proof:

This follows immediately from the previous identity:

and Proposition 7 . 1 . (c)

Casson's invariant for unlinks. Let

X"(K,C,C')

be an unlink of three components in an homology

That is, assume that every pair of components has zero linking As in previous sections, we adopt notation:

131

3-sphere number.

and the various modified forms. Casson's invariant is then defined by the rule:

By grouping the terms in various ways, we see:

Note:

The fact that all these terms are defined follows from the assumption

that each pair of components is unlinked.

This is an easy exercise which we

leave to the reader.

These

various

independently of

formulations and

show

m.

132

t

h

a

t

i

s

well

defined

Figure 30

8.

Cas8on'B invariant a n d the Alexander polynomial

(a)

T h e Alexander polynomial Let

K

be a knot (or link) in an homology

Seifert surface in

where

is the

M

3

for

K.

"plus p u s h

Alexander polynomial of

K

Let

V

off" of

b y the rule:

133

3

sphere

Let

F

be a

be a matrix for the linking form:

y

([Ro]).

Then

we

can define

the

If we change

the Seifert surface, the polynomial changes

by at most a unit

multiple:

We

shall

denote

any

such

polynomial

as

We

recall

that

is

symmetric up to a unit:

where

r

is the rank of where

In particular, if

is the genus

of

F.

Hence,

K

is a knot,

we obtain a

symmetrized

Alexander polynomial by symmetrizing:

We recall that

is the intersection form:

Hence, we have the following proposition:

Proposition 8 . 1 :

Let

K

be a knot or link.

Then:

if

K

is a knot

if

K

is not a knot

134

Proof:

For a connected surface of one boundary component, the intersection

form is the standard symplectic form which has determinant one.

On the

other hand, a connected surface of more than one boundary component has a degenerate intersection form, since any nontrivial element of the kernel.

Note: In

boundary

component represents a

α

This and subsequent results are well-known for knots and links in

particular,

they

follow

from

results

in

Kaufman

[K]

for

the

S3.

Conway

polynomial. Corollary 8.2:

Let

K

be a knot with Seifert surfaces Ap,(t) = tnAp(t)

Note:

F

Then:

According to the classical definition of the Alexander polynomial as an

S3,

However, for any link

any two Seifert matrices are related by the moves described in [BZ],

pg. 236.

(b)

F'.

η € Z .

elementary divisor, it is only well defined up to a unit. in

and

Hence, the above corollary also holds for links in

The second derivative of

S3.

Air(t)

We have the following identities for links of two components in

Proposition 8.3:

If

K

is a link of two components in

S3

and

S3:

4 (K)

is the

The conclusion is independent of the choice of Seifert surface.

As we

linking number of these components, then: A^(I) = '(K)

Note:

have previously observed, since Seifert surface for

K,

then:

K

·

is a link in

S3,

if

F'

is another

Hence, we compute that:

Proof of Proposition 8.3 The

proof

is

components of

K

components of Hence,

we

Let surfaces for

K

by

induction

on

the

geometric

(i.e. the minimal number required to homotope

first

consider

the

of elementary

where

2-sphere and

respectively

number crossings

of

the

of the

to a splittable link).

situation

be a separating and

K

linking

with

K

is and and

a

splittable

link

be Seifert contained

in

components of

Figure 31 Tubing

and

a Seifert surface for

together along an arc in F.

136

we obtain

Figure 32

The corresponding Seifert matrix has the form:

Hence,

O n the other hand,

T M b establishes the result

for split links. Hence, we may assume, by standard arguments, that there is a projection of

K

with at least one elementary linking of the components of

K,

such

that an elementary crossing at this elementary linking reduces the geometric linking of

K.

137

Figure 33

Using the standard Seifert surface for this projection, a basis for

containing curves

in Figure 34.

Figure 34

138

a

F, and

w e can choose b

as depicted

Note:

If the associated surface for

tube the two components together. we insure that the curve

b

K', F'

By adding this tube as a handle to

V =

has the form:

a

-1

b

-1

*

F1

*

V'

is the Seifert matrix for

(t - 1) tV

K

(F')+

b

V'

F1

exists.

Then the corresponding Seifert matrix for

where

is not connected, then we can

=

F'.

Then we compute that:

-1

0

t

(t - 1)

•

0

*

V - t(V')T

We observe that the matrix obtained by deleting the first row and column is an Alexander matrix for the knot (i .e.

by deleting the generator

a):

L

obtained by slicing

F

transverse to a

Hence, we have the following identity:

Computing derivatives, we conclude that:

From the previous section, we know that:

By induction, we conclude that:

On the other hand, we readily see that;

140

Hence, we have the desired result:

projections Now suppose which differ that at a singleand Figure crossing 36 ofare

links as depicted in below. which

have

Lemma 8.4; the link

Proof:

Let

be a knot in

Let

is a link of two components,

and

be as above.

Then

and:

has two components for homological reasons.

Now choose the Seifert

surfaces as follows:

Figure 37

So we can arrange the corresponding Seifert matrices

as follows:

141

To Hence: Observe obtainthat the symmetrized is homeomorphic Alexandertopolynomials, Hence,we have to multiply b y

Let

Computing derivatives:

Hence, by Propositions 8 . 1 and 8 . 3 : (c)

Casson's invariant and the Alexander polynomial We are now ready to prove the main result relating Casson's invariant to

the classical Alexander polynomial.

Lemma 8.5: exists a knot Proof:

Let

K

L

in

We begin with a simple lemma.

be a knot in an homology such that

It is well known that

components of a link in The proof is by induction on

3-sphere i

Then

surgeries on the

It is obvious for

Now choose Seifert surfaces for

K

and

and

By general

142 one dimensional spines, position, we may assume that there exist for

F

and

neighborhood of

which are disjoint.

We may isotope

F

Hence, we may assume that

and

into a small regular

By a separate isotopy, we may isotope

regular neighborhood of

there

and

may be obtained by

n.

... .

into a small F

and

are

disjoint.

(Of course, this may require crossings of

crossings of

C.)

link in

and

In other words, we may assume that

Hence,

On the other

K

hand,

by Corollary 7.2.

since

K

and

bound

but not self is a boundary

That is:

disjoint Seifert surfaces in

we may also conclude that:

where

is the knot

K

considered as a knot in the surgered

manifold

I f we denote this manifold by

Hence, by induction, there is a knot

L

in

such that:

This establishes the desired result.

Next w e prove an analog of Lemma 8.4.

Lemma 8.6:

Let

be a knot in

Let

is a link of two components and:

143

and

be as above.

Then

Proof:

Suppose that

link in each meeting

such K

assume that

K that

is a knot in C

and

Now suppose that bound disjoint discs,

geometrically twice and algebraically zero times. and

do not separate each other in

is a D

and In addition,

K:

Figure 38 In particular,

is a boundary

bound obvious disjoint punctured tori in

Figure 39

144

link in

for

and

C

Hence,

for all

n.

From this, we conclude that:

This in turn implies that: This means that changing value of change Let

But, by K

K

with a twist across

changing

to the following knot

K

Km: Figure 40

denote the curve depicted below:

Figure 41

145

D

across such

does not change the discs

we

can

We have the following computation:

Figure 42

Hence, from the previous comments:

146

We may assume that

K

is

and

by applying a left twist across the

two

component

link

is the knot obtained from

That is

obtained

from

K

is by

cutting

K

Let

denote

across

and

connecting "the

Figure 43 The move of twisting across the Hence, it clearly leaves

_

disc D invariant.

however, it is evident that

Figure 44

147

only affects one component of For the standard form

It follows that:

As previously observed, we may "normalize"

so that:

For this normalization, we have the identity: We now turn to the main theorem.

Theorem 8.7:

If

K

is a knot in an homology

3

sphere

then:

On the other hand, by induction:

The result follows immediately.

With this result, we are able to establish property (5) be the arf invariant of

Property ( 5 ) :

Proof:

for

Let

K.

(mod 2)

From Theorem 7 of [R],

(after symmetrizing), and Theorem 8 . 7 above, we

conclude that:

for any

knot in

The argument of Lemma 8.5 permits us to deduce the

same fact for any knot in any homology sphere.

(The arf invariant also comes

from a Seifert matrix.)

O n the other hand, Theorem 2 of [G] implies that:

This yields the identity:

149 Since we can write

as a sequence of

and the we property know that: follows by induction on

n

surgeries on a link in

using the identity above.

C H A P T E R VI:

THE TOPOLOGY OF THE SPACE OF

1.

The topology of

(a)

Idenfication of apaees As in previous sections, we identify

inducting on

g

REPRESENTATIONS

with

Since we shall be

at various points, we denote the boundary map as follows:

Likewise, we use the following notation:

Clearly:

The singular set of on

S

is contained in

R+.

Hence,

In particular, we have a fibre bundle with fibre

150

is regular

Hence, there is a trivialization:

with the following properties:

(projection onto second factor) is the usual inclusion

(b)

The topology of

Lemma 1 . 1 :

is connected for simply connected for

Proof:

The restriction of the boundary map:

is a locally trivial fibration with fiber is connected for

It

suffices

to

show

that

and simply connected for

Consider the decomposition of

Let

in

(i.e.

A

is the set of diagonal matrices).

the map:

Then:

151

Consider

Since diagonal matrices commute,

Hence,

-

which

factors to:

implies

Since

the result follows by general position.

We now want to relate the

topology of

to the

topology of

Write:

Let

be the projections onto

Recall the involution of

Hence

and

sending

then

Now construct imbeddings

152

So we can write:

given by:

Then we conclude that:

Consider the diffeomorphism:

Then we get the following commutative diagram:

Hence: Furthermore,

since

153

we

can

write:

X

is the composition of:

Vie can describe the gluing by the diagram:

where

and

are

and

followed by inclusions.

We have a principal fibre bundle with fibre

We have previously shown that

is a point.

The Mayer-Vietoris sequence yields: We conclude that:

154

S03:

Hence,

is homeomorphic to

We can easily check that:

where

denotes the map induced in

homology.

We deduce:

Now, consider the exact sequence:

Since the usual generators of

is onto.

additon, we compute that: Hence:

Also, since

is onto, we have an exact sequence:

This implies that:

155

In

Hence:

Similarly:

By using

coefficients in homology, we get:

Hence: For a fixed

we can define the map:

It follows that:

is an isomorphism.

is an isomorphism.

Let

from the obvious generators of Then:

be the generators of ay the inclusion:

156

induced

This fact will be used later.

Theorem 1 . 2 :

Proof:

The following inclusions are surjective:

(a)

We first prove the case:

Since

is

the

only relevant case is

The

obvious

generators,

and

are contained in

So

they lie in We then show the case:

In this case,

So

the

relevant cases are

Clearly, all the generators of lie in

, and, therefore,

in

and

4

of the 6-generators of

It suffices to prove that given:

and the induced homology homomorphism:

the image of the generator lies in the image of tative diagram:

157

Consider the commu-

It suffices to prove that

By excision,

it suffices to prove that:

is the zero homomorphism. Let

k

be the map given by the commutative diagram:

By the properties of the trivializations of section

(a):

Consider:

Then: (inclusion)

where

is the obvious projection:

Hence,

158

So it suffices to show that the following is zero:

Since

is

and:

is onto, the long exact homology sequences give: So

it

suffices

to

prove

the right hand vertical map is zero.

suffices to prove that:

is zero.

This map factors through the inclusion:

Recall that Therefore,

is

and:

159

In turn,

it

We now assume that

By induction, we assume that (a) is proved

Consider a standard generator of

where each

is

either

or

{1}

and exactly

s

such factors of

are

Case 1:

for some integer

For simplicity, we assume that

By induction,

k

with

the only remaining case to

treat

By assumption:

therefore:

lies in the image of

But the image of the obvious imbedding: clearly lies in Hence, lies in the image of Since at least one factor must be is the following. Case 2: for some integer k

with

160

For simplicity, we

By induction,

assume

that

By

assumption:

therefore: lies in the image of

Therefore, consider the imbedding.

Since:

let:

where: where

can be deformation retracted onto

for some small positive Since: the image of

We now

g

is contained in a regular neighborhood of

deformation retracts into

prove ( b ) .

Suppose

Hence,

Let

Then, for some Then foi

and

be

comes

a

Suppose is homologous to

Consider the imbedding:

161

generator

from

of

with

Then:

Hence,

lies in the image of

Consequently,

and the following commutes:

(a) implies ( b ) .

We introduce the following notation:

Theorem 1 . 3 :

Proof:

we have:

Since:

is onto for

162

when

is onto when

Since:

we conclude that:

when Hence:

From the discussion preceding Theorem 1 . 2 :

163

Theorem Proof: and (b) The hold. 1 . 4 : proof By is Theorem by induction 1 . 3 , for on i g.

For

is

and

(a)

So, after the appropriate restriction on

m,

we have:

We will prove formula (a) by induction on

g.

All the other cases are proved similarly.

From above:

Assume:

This implies:

Hence, by induction, we have:

164

We give the proof when

To prove ( b ) , notice that:

Hence, Poincare duality gives ( b ) .

Corollary 1 . 5 :

Proof:

If

g

and

The first equality comes from

(b)

of Theorem 1 . 4 .

is even:

Hence, Theorem 1 . 4 ( a ) gives

Likewise,

Consider the fibration:

165

If

is odd:

Theorem 1 . 6 :

If

then:

is an injection.

Proof:

Since

is a

®homology

exact sequence: where all cohomologies are taken with

We also

3-sphere, there is a

Thom-Gysin

coefficients and:

make the convention that the rest of the homologies in this proof are

taken with Since:

coefficients.

it suffices to prove that:

is onto for

is injective for starting with the cases

In turn, it is enough to prove that the map:

We prove this by descending induction on and

166

Consider:

By Corollary 1 . 5 ,

the two outside groups are zero,

so

u^

is an isomorphism

Similarly: is onto.

Since

is a Kahler manifold [Nl],

that: is injective for

there is

Hence:

is injective and:

Hence:

is an isomorphism. Now assume

is

injective

and:

for

all

Consider:

It can be written as compositions in 167 two different ways

such

In

the

second

properties of

w

factorization,

both

and by induction.

homoraorphisms

injective

by

the

Hence:

Corollary 1 . 7 :

Proof:

are

is an isomorphism for

and

Consider again the Thom-Gysin sequence:

Of course, the first term is zero.

Hence,

the other hand, by Theorem 1 . 6 , Corollary 1 . 8 : For Proof: By Theorem 1 . 6 , for

is surjective for

_

On (provided

we have a short exact sequence:

Let:

Then: This gives the required formula.

is injective for

168

Consider the maps:

Lemma 1 . 9 :

The maps

and

induce isomorphisms on

also induces an isomorphism.

Proof:

Since, by Theorem 1 . 2 :

is surjective:

is infective.

Hence, Mayer-Vietoris gives:

The map is induced by:

Since the image of:

is the kernel of

is an isomorphism.

Consider the diagram: 169

It commutes except on the middle square which is homotopy commutative,

is homotopic to

(This follows from the proof of Theorem

1.2(a)

by

observing that the homotopy of the map:

to the inclusion:

may be taken to fix the second

factor.

Hence,

it restricts to the

desired

homotopy on Hence, we have the commutative diagram:

So

and

Let

induce isomorphisms on

be the generator of

of lemma.

for Let

be the

Also, let

denote the generators

induced by the

isomorphism of the above

elements of

the inclusions:

170

and

induced by

of

.

Likewise, using Corollary 1 . 7 ,

let these symbols denote classes

in

We have natural

inclusions:

As in the proof of Theorem 1 . 2 ( a ) ,

Case 2, we can deform this map into a map:

(2)1.10:

Proof:

Part (2) is trivial.

To prove part (1), we begin by observing that the

discussion preceding Theorem 5.1 implies:

Hence, we compute that:

We may write the pullback of

as follows:

171

Now we have the following diagrams:

By

choosing

the

deformation

to

diagrams commute up to homotopy.

By definition,

carefully,

we

may

insure

that

Hence, we have commutative diagrams:

and

Hence, we conclude that:

172

these

This implies that

and

Since on

p,

by the proof of Theorem 1 . 3 , we conclude, by induction

that:

Proposition 1 . 1 1 :

The elements

where

and

Proof:

are linearly independent.

The proof is by induction on

g.

Assume that the result

holds for

Recall:

let

_

be a linear combination of

Assume

By induction on

elements

Look at the component of

_

of

)

the from

in:

and Lemma 1.10, we conclude that the only terms in

which can have nonzero coefficients are of one of the following forms: (a) the ones with (b) the ones containing

173

We can find an automorphism of for all

(b')

the

ones

which takes .

containing

Hence, all terms of

at

and

to any

Hence, we can replace (b) by:

least

one

of

for all

i,

satisfy (a) and are in the form:

But the inclusion induces an injection:

for

and such elements are linearly independent in

Hence, they are linearly

independent in

The case of genus

one is

true by inspection. Corollary 1 . 1 2 :

The

element

generates

H'

The

element

generates Proof:

From Theorem 1 . 2 and the

long exact

the short exact sequence:

for

Hence, we deduce that:

174

sequence

for a pair, we

deduce

Consequently, Proposition

has

rank

one

and

the

result

follows

from

1.11.

In a similar manner, we deduce that:

By Theorem 1 . 4 , we conclude that the rank of

is zero in genus one and

one in genus

Again, the result follows from Proposition 1.11.

Proposition 1 . 1 3 :

The elements

are linearly independent in

Proof:

This is obvious f

o

r

L

e

t

elements as above such that

be

a

linear

combination of

As above, the only terms with non-zero

coefficients are in the form

But

implies that there are

no such terms.

Corollary 1 . 1 4 :

The elements

generate

Proof:

By Theorem

we

Hence, Corollary the 1result .15: The follows. element

know

the

rank

generates 175

of

The element

satisfies:

generates

Proof:

Both claims are obvious for

Corollary 1.14

for

The

first

claim

follows

The second claim follows for

On

from the

other hand, by Theorem 1 . 4 :

Hence, the second claim is also true when

Corollary 1 . 1 6 : X

For

is generated by

and

(as a cohomology ring).

Proof:

For

by Theorem 1 . 6 , we have the exact sequence:

where

Hence, the proof follows by induction on

Since

is a

nonsingular

projective

it follows from the Hard

variety of

r.

complex

dimension

lefschetz Theorem

that we have a stronger result:

Corollary 1 . 1 7 :

is generated by

2.

The action of the Torelli group

(a)

The action of the Torelli group

and

(as a ring).

Recall that every base point preserving, homeomorphism, induces an homeomorphism of

where

is the induced homomorphism on

176

I f in addition,

is

orientation preserving, then

f

preserves the boundary of

F*

and we have a

commutative diagram:

T,* H

3

S> S3

id

H*

Note:

If

f

> s3

is orientation reversing,

vertical arrow by the inversion map, By Proposition 7, it follows that

Note:

If

f

parity of

g.

(f-orient. pres.)

A I f*

is orientation reversing,

From the previous

then we must replace the right hand > A

1

.

preserves the orientation of

R*.

this assertion is dependent on the

diagram and remark, we obtain from

preserving, orientation preserving homeomorphism, f,

the

every

base-point

following restric­

tions:

Prom the compatibility with

f# : R^

> R®

f* : Nf

> Nf

3

and the previous remark, it follows that all

these induced maps preserve orientations. Since these maps commute with the preserving maps of orbit spaces:

SO 3

action, we obtain orientation

The mapping class group

where

is the group of orientation preserving

of f

is defined by:

with the compact open topology. by

mapping

f.

We denote the mapping class of

It is evident that the maps induced from class of

actions of

f.

Hence,

we

have

well

homeomorphisms

defined

f

depend only on the

orientation

preserving

on these various spaces.

The Torelli group

is defined by:

The above actions, of course, restrict to

(b)

Triviality on the homology of We wish to show

homology of consider

rational

that the Torelli Since rational cohomology.

suffices to consider dimensions

group acts homology and

On

the

other

2, 3 and 4.

trivially

acts trivially on

178

rational

cohomology are dual, we can hand,

by Corollary

We begin with dimension

homology and cohomology will be assumed to be rational.)

Lemma 2 . 1 :

on the

1.17, 3.

it (All

Proof:

By Proposition 1 . 3 . 1 ,

acts trivially on

is homotopy equivalent to

Since

, Theorem 1 . 2 implies that we have a surjection:

On the other hand, by Theorem 1 . 4 ,

the rank of

this is clearly the rank of

is

Since

the inclusion induces an

isomorphism.

Hence, we have an isomorphism:

By Corollary 1 . 7 ,

the natural projection induces an isomorphism

Since these identifications

are compatible

acts trivially on

(provided

with the actions, we deduce

provided

.

that

The case of genus one is

obvious.

We now consider dimension 2. Lemma 2 . 2 :

Proof:

acts trivially on

As usual, we may assume that

consider

given

in

By Corollary 1 . 7 ,

it suffices to

Tracing through the identification:

the

proof of Corollary 1 . 1 2 , we find that this isomorphism is also

compatible with the

actions.

(The reader should

179

convince

himself

of

this

assertion for the Thorn isomorphism. orientation

preserving.)

Here we use the fact that the actions are

Hence, we may reduce the problem to

By

Corollary 1 . 1 2 , on the other hand, it suffices to consider Again, using the identifications outlined in Proposition 1 . 1 1 , we have:

which reduces the proof to

But certainly

acts trivially on

Finally, we consider dimension 4.

Lemma 2 . 3 : Proof:

,

. ,

acts trivially on

1

From Corollary 1 . 1 7 , Lemma 2 . 1 and Lemma 2 . 2 ,

considering the action

of

the proof is reduced to

on the element

But

x

is the

Euler-class of the bundle:

a rational orientable sphere-bundle. bundle, we conclude that

As observed at the

i

acts orientably on this

preserves

beginning of this part,

result from Corollary 1 . 1 7 ,

Theorem 2 . 4 :

Since

Lemma 2 . 1 ,

2.2,

acts trivially on

180

we can

and 2 . 3 .

deduce the

following

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3-Sphare, Topology, 7