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HOMOTOPY TYPE INVARIANTS OF FOUR-DIMENSIONAL KNOT COMPLEMENTS

Alexandru Ion Suciu

Submitted in partial fulfillment of the requirements of the degree of Doctor of Philosophy in the Graduate School of Arts and Sciences

COLUMBIA UNIVERSITY 1984

@

1984

ALEXANDRU ION SUCIU

ALL RIGHTS RESERVED

Abstract

HOMOTOPY TYPE INVARIANTS OF FOUR-DIMENSIONAL KNOT COMPLEMENTS

Alexandru Ion Suciu

This thesis studies the homotopy type of smooth four dimensional knot complements.

In contrast with the classi-

cal case, high-dimensional knot complements with fundamental group different from

2

are never aspherical.

The sec-

ond homotopy group already provides examples of the way in which a knot in

s4

can fail to be determined by its funda-

mental group (C. MeA. Gordon, S. P. Plotnick). A natural class of knots to investigate is ribbon knots. They bound immersed disks with "ribbon singularities". method is given for computing

n

2

of such knot complements.

s4

I show that there are infinitely many ribbon knots 1n with isomorphic ules).

n

1

but distinct

A

(viewed as

ZTI

1

-mod-

They appear as boundaries of distinct ribbon disk

pairs with the same exterior.

These knots have the funda-

mental group of the spun trefoil, but none is a spun knot. To a four-dimensional knot complement, one can associate a certain cohomology class, the first k-invariant of Eilenberg, MacLane and Whitehead.

In a joint paper, Plotnick

and I showed that there are arbitrarily many knots in

s4

whose complements have isomorphic

n

modules), but distinct k-invariants.

1

and

(as

2n

1

-

Here I prove this

result using examples which are somewhat more natural and easier to produce.

They are constructed from a fibered

knot with fiber a punctured lens space and a ribbon knot by surgery. The proofs involve writing down explicit cell complexes, computing twisted cohomology groups, combinatorial group theory and calculations in group rings.

Contents Page Abstract Acknowledgments

ll

Chapter I.

II.

INTRODUCTION

1

RIBBON KNOTS AND DISKS

8

1.

Ribbon knots and

2.

n

n

8

1

2 of a ribbon 2-knot 3. Meridians and ribbon disks 4. Ribbon knots with different

I I I.

k-INVARIANTS OF KNOTS IN

16 23 n

s4

2

27

34

1. Lens spaces

34

2. 2-twist spun 2-bridge knots

38

3. The knots ~,q

45

2 (xp,q) 5. A cell complex for Xp,q

50

6. Computation of the k-invariant

63

7. Proof of the Theorem

67

4. Computation of

n

References

55

87

l

Acknowledgments

It 1s a great pleasure to thank Professor Steven Plotnick for his constant and invaluable guidance during the writing of this thesis.

His ideas and the cooperation leading to our

joint paper were decisive in the shaping of this work.

I wish to thank also Professor John W. Morgan, who introduced me to low-dimensional topology; Professor Michael Davis, who taught me many skills; and Professor John Harer, from whom I learned all I know about four-manifolds.

And last but not least, I am grateful to my wife, Anda, for her patience and encouragement.

ii

1

I.

INTRODUCTION

The study of the homotopy type of knot complements started around 1910, when Dehn and Wirtinger gave a method for presenting the fundamental groups of knots in

s3.

In

1956, Papakyriakopoulos [41] proved that classical knot complements are aspherical; and hence, that their homotopy type is completely determined by their "algebraic 2-type" (i.e. the fundamental group).

The situation is completely

different in higher dimensions.

Andrews and Curtis [1] s 4 - the spun trefoil

first gave an example of a knot in whose complement has nontrivial

n

2

.

D. B. A. Epstein [13]

showed that the exterior of a spun knot is aspherical if and only if the knot is trivial.

More generally, Dyer and

Vasquez [11] proved that an aspherical knot complement in sn

( n ~ 4) has fundamental group

2' •

Combined with re-

sults of J. Stallings [50], J. Levine [34], J. Shaneson and C. T. C. Wall [51], this implies that the knot is trivial if n for n

~

5.

M. H. Freedman [16] shows that this is also true

= 4,

in the topological category.

The natural question to ask next is which homotopy type invariants can be used to distinguish among high-dimensional knot complements with the same fundamental group. Let

K

let

X

=

(Sn+ 2 , Sn)

be a (smooth) n-dimensional knot, and

be its exterior, that is, the closure of the com-

2

plement of a tubular neighborhood of

sn

sn+ 2 .

in

By the

usual abuse of language, we will call the homotopy type lnvariants of In [20],

X

the homotopy type invariants of the knot

c. MeA. Gordon gave examples of knots in

isomorphic n

but different n

1

2

~n

(viewed as

s4

K.

with

1 - modules).

S. P. Plotnick (43] generalizes this to arbitrarily many knots.

In [44], he produces infinitely many examples in

the TOP category, using the results of Freedman [15].

We

somewhat improve Plotnick's result and get examples in the DIFF category (Theorem 1.3). s4

Now suppose we're given two knots in phic n

1

and n

2

(as

their complements?

~n

1

-modules).

A knot exterior

with isomor-

How can we distinguish X

=

s4 - s2

x

o2

has the homotopy type of a 3-dimensional CW-complex. any 3-complex

X, one can associate an element

To

k(X) E

3 H (n X,n X), the first k-invariant of Eilenberg, MacLane 1 2 and Whitehead [12], a retraction

K(n x, 1

[39].

It is the first obstruction to

1)------~x.

augmented chain complex of so that

Ci(X)

is a free

the number of i-cells of exact at

..v

c 2 (X). Add to

To identify it, let

"' the un1versal . X, cover of

X,

~ x-module, with rank equal to

1

X. This complex may fail to be N

c 3 (X) a free

~

1 (X)-module

and map so as to kill

1T

2 X·.

c3

3

This is now a partial free resolution of

2'

over

::l''ITl X, and the map

k = pa : c3 -+ TI X determines a well3 2 3 defined class k(X) = [k] E H ( Til X, TI X) . 2 The triple ( n x, n X,k(X)) is called the "algebraic 32 1 ~"

of

X [39]. A map ( n X, n X,k(X))----+ 1 2

consists of a homomorphism morphism S :

a3 S

TI

2

X

TI

-+

2

a :n X

n X'

-+

1

1

( n X', n X' ,k(X')) 1 2 and an

a-homo-

* a(k(X'))=

X' satisfying

( n1 X, ( n x' ) a>. When this condition holds, we say a and 2 preserve k-invariants. The complexes X and X' have

the same algebraic 3-type when

a

and

S

are isomorphisms.

A theorem of MacLane and Whitehead [39] asserts that there exists a map a

and

S

f:X

-+

X' inducing

preserve k-invariants. .,._

morphisms, and

a

and If

a

S and

precisely when S

are 1so-

N

H (X) = H (X') = 0, the Hurewicz and White3 3

head theorems show that f 1s a homotopy equivalence. ln

with exterior

quasi-aspherical.

X

satisfying

Knots

are called

The reason, observed by S. J. Lomonaco,

Jr., is that, in analogy with the classical case, the algebraic 3-type of a quasi-aspherical knot determines its homotopy type.

4

In [37], Lomonaco provides a way for computing the algebraic 3-type of a 2-knot from a motion picture.

He

asks (Problem 16) whether there are knots which are distinguished by their k-invariants.

Plotnick [43] argues

that the only reasonable candidates among fibered knots are, given the present status of 3-manifold theory, the 5-twist spin trefoil and its 2-fold cover.

The question

whether their k-invariants correspond reduces to a difficult problem about

Z [SL(2,5)].

Therefore, one should

look at non-fibered knots for examples.

The main theorem

in [45] is:

s4

Theorem 1.1: There are arbitrarily many knots in complements have isomorphic

n

1

and

n

2

(as

whose

Zn -modules),

1

but distinct k-invariants.

We will prove Theorem 1.1 using examples which are somewhat more natural and easier to produce.

In our paper

[45], we relied on the fact that for

I.1

closed, orientable, aspherical 3-manifolds ad-

mitting no orientation reversing homotopy equivalence. Instead, we will use here J. H. C. Whitehead's classification of lens spaces up to homotopy type. The knots constructed 1n [45] and here are quasiaspherical, so the k-invariant is the last obstruction to a homotopy equivalence. cal [18],

[46].

But not all knots are quasiaspheri-

Hence one might ask (Problem 1 in [37]):

5

does the algebraic 3-type of a knot exterior its homotopy type? of

X

to look at are n

i (X 2 ) = 0

determine

The invariants in the Postnikov tower 3

x

and the second k-invariant

4 k 2 (X) E H (X 2 ,n 3 x), where n i (X 2 ) = n i (X) n

X

for i > 2.

for i

~ 2 and

To answer this question seems a

hard task. Another question which arises 1s whether the exterior of a knot determines the knot. swer is not known.

For classical knots the an-

For multi-dimensional knots, a given

exterior corresponds to at most two distinct knots [17], [6], [34].

Examples of inequivalent n-knots with the same comple-

ment were given by Cappell-Shaneson [9] for n

c. MeA. Gordon [22] for n tion about disk knots

= 2.

=

3, 4 and

One can ask the same ques-

Dn CDn+ 2 .

Hitt-Sumners [30], [31]

construct arbitrarily many examples of distinct disk knots Dn c Dn+ 2

with the same exterior for

amples for n

=

4.

many examples for n

n ~ 5, and three ex-

s. P. Plotnick [44] produces infinitely ~

3.

For n

=

3, his proof requires

Freedman's solution to the four-dimensional Poincare conjecture, so he gets results in TOP.

We give our own exam-

ples, which are somewhat simpler and also work in DIFF for

n

=

3.

Ribbon disks constitute a natural class of examples to work with, one which is interesting 1n its own right. o d b y pus ho1ng 1n Th ey are o b ta1ne 0

sn+ 1

Dn+2

a

Dn

with singularities of a certain type.

can be built with just 0-, 1- and 2-handles.

1mmerse d 1n 0

0

Their exterior Using a con-

6

struction similar to that of Hitt-Sumners, we prove:

Theorem 1.2: disks

There exist infinitely many distinct ribbon Dn c Dn+ 2 , n ~ 3, with the same exterior.

A nice feature of these disk knots is that trefoil knot group.

n

1

is the

The difference comes from the fact

that their meridians are not equivalent under any automorphism of

n

1

ribbon knot.

.

The boundary of a ribbon disk is called a Analyzing the boundaries of the examples pro-

vided by Theorem 1.2, we prove:

Theorem 1.3: There exist infinitely many ribbon knots in s 4 with isomorphic n 1 but distinct n 2 (as zn1 -modules). The proof involves the study of the of 2

x

2n -module structure 1

2 , and the reduction of the problem to a question about 2 matrices.

n

The simplest examples of ribbon knots are spun knots. We show in II.1 that, for most knots in s 3 (including torus knots), the fundamental group of the knot determines the spun knot.

Combining this with the above theorem yields:

Corollary 1.4: There are infinitely many distinct knots 1n s 4 which are not spun but have the fundamental group of the spun trefoil.

7

This thesis is organized as follows. studies ribbon disks and knots. definitions and look at computing quences.

n

Chapter II

In §1 we discuss several

1 . In §2 we give a method for of a ribbon 2-knot and derive some conse-

2

n

We prove Theorem 1.2 in §3 and Theorem 1.3 in

§4. Chapter III is devoted to the study of k-invariants, leading to the proof of Theorem 1.1.

§1 interprets

Whitehead's homotopy classification of lens spaces in terms of the k-invariant.

§2 describes certain fibered knots

with fiber punctured lens spaces.

In §3, a construction

based on surgery yields knots

Kp,q , our examples for In §4 we compute n 2 of the knot exteriors

Theorem 1.1.

§5 describes a cell complex for X p,q and identifies p,q k(Xp,q ) on the cochain level. In §6 we compute

X

and locate the k-invariant.

§7 completes the proof of

Theorem 1.1.

1 and n 2 and A long computation in group

It studies automorphisms of

their action on

3 H ( Til' TI 2) .

n

r1ngs wraps up the proof. All stated.

~TI

-modules are left-modules, unless otherwise

An element

u : 2 n ~ 2 TI

u E

2n

induces the

via right multiplication.

2n

Vectors 1n

are row vectors and matrices with entries in the right.

-module map

2TI

(2 n )n

act on

For any unexplained fact about cohomology of

groups, see Brown's book [7].

8

II.

RIBBON DISKS AND KNOTS

§1. Ribbon knots and

n

1

In this section we introduce ribbon knots and disks and discuss several definitions. computing

n

1

We recall a method for

of a ribbon knot and show that a spun knot

is determined by its fundamental group. Ribbon n-knots were first defined by Fox [14], for n

= 1,

and Yajima [53], for n

1s a ribbon knot if sn+ 2

sn

= 2.

A knot K

=

(Sn+ 2 ,sn)

bounds an immersed disk

Dn+ 1~0~---.~

with no triple points and such that the components

of the singular set are n-disks whose boundary (n-1)spheres either lie on

sn

or are disjoint from

I

Figure 1

sn.

9

Dn+ 1

Pushing

into

Dn+ 3

produces a ribbon disk

(Dn+ 3 , Dn+ 1 ), with the ribbon knot boundary.

(Sn+ 2 , Sn)

on its

It can be pictured via a motion picture, as

shown in Figure 2.

Figure 2

The double of a ribbon (n+1)-disk is an (n+1)-ribbon knot. Every (n+1)-ribbon knot is obtained in this manner. Given a knot lS

the

standard

(n+1)-knot

d?

I

!)1).

K

=

(S3,

a n (K) For n

s 1 ), the n-s2in of

= a (K = 1, we

X

Dn+1)' where K

~

1,

=K -

get the usual sp1n of K.

It is folklore that every n-spun knot is ribbon. see this from Figure 3.

K, n

One can

10

Figure 3

The spun trefoil is the double of the disk in Figure 2. The fundamental group of a ribbon n-knot can be computed from a motion picture [14] S

n-1

TT 1

x . ) joined together by (1 ~ i

S. l

_ Dn ) (D n+2 + +

where

w. l

m

l

n-1

to

=

(

=

~

x0 x1 1

m) .

1



••

The equatorial

s n-1 0

consists of disjoint spheres

meridians

sg- 1

[54].

1

I

•••

1

n-1 sm

(with

bands running from

Then xm

x

0

= w.x .w. l

l

l

-1

1

~

~

i

E: . .

lJ II xk. J

+1

1

if the i-th band goes over xk . J

and

E:. .

lJ

=

-1

1

if the i-th band goes under xk . J

0

1

otherwise.

m)

I

11

We call such a presentation of For example,

rr 1 a "ribbon presentation".

the spun trefoil, with equatorial section the

square knot (Figure 4), has rr1 (

s4

-

s2

)

=

(t,x

t

I

=

1 (xt) x (xt)- )

t

=

(t,x

1

txt

=

xtx) .

X

cs

_r

.

-1~

Figure 4

Does the fundamental group of a ribbon knot determine the knot?

We will see in §4 that this is not the case.

For spun knots, however,

it is reasonable to conjecture an

affirmative answer to this question.

We can prove this for

a large class of knots:

Theorem 1 . 1:

~ rr 1 x2 ·

Let

Assume

non-trivial knot .

and

Kl K. 1

K2

are not

Then

on(Kl)

be knots in (p,q)-cables,

=

an ( K2) .

$3

with

IP I

~

2,

7T

1x1

of a

12

Proof:

Results of Johannsen,

Feustel,

Whitten, Burde and

Zieschang (see [23, p. 9-10]), imply that either (i) K. are l

prime knots, with

x1

=

x2

or (ii) Ki are composite knots,

with the prime factors equal, up to orientations. ( i)

1

= a n (X 2 ),

an(Xl)

Cappell [8], for n

;>

1,

and by Gluck

[17],

In case

for n = 1 and In case (ii), the

a n (K1) = a n (K2).

argument in Gordon [21] yields the equivalence of

a (K . ) . n

1

0 Remark: spun

The

knots

aspherici ty of classical knots implies that

with

isomorphic

'IT

1

have

homotopy equivalent

exteriors.

Not every ribbon 2-knot is spun .

For example,

the

ribbon 2-knot with cross section the stevedore knot (Figure 4 - s 2 ) = (t,a I t = (at- 1 ) a (at- 1 )- 1 ) 5) has G = 'IT1 (s 2 (t,x I txt- 1 = x ) (compare [14, p. 136]). --1 x=at

a

t

Figure 5

13

But G is not a 3-manifold group [25],

[26], so the knot

is not spun.

This knot will play an important role in

Chapter I I I .

Even more striking, if we let the band on

Figure 5 wrap

k

times around

( s4 1 (t,a It= (at- 1 )k a (at- 1 )-k)

and

s2 )

a ribbon 2-knot with n

a

a

t

( k > 1), we get

=

Baumslag-Solitar non-residually finite group.

But,

according to Thurston, 3-manifold groups are residually finite. Ribbon knots are, by definition, slice knots.

Wheth-

er the converse is true 1s a famous question in classical knot theory, but in higher dimensions counterexamples abound [29],

[ 48],

[10] .

All 2-knots are slice [31], but

nom-twist spun 2-knot (m > 1) is ribbon [10].

This fact

follows easily from [27, Cor. to Thm. 1], but unfortunately there is a gap in that paper. Here is another construction of ribbon disks and knots. (Dn+ 3 ,Dn+ 1 ), the standard disk pair, with meri-

Start with dian

circles r. l

h~

t. Add 1-handles x. l

n Dn+ 1 =

l

, and 2-handles

If

and

r. l

(1

/ 1·

~

h~l

,
).

=

If we perform equivariant surgery on the lifts of

"" X.

M, we get

the cosets

TI

/Z(

IT/2'(X)

1

in

IRX D3, indexed

M consists of copies of

by the cosets

r

s3, indexed by tubed together by "connectors" s 3 x I

t)

I

and copies of

IR

X

indexed by IT .

Figure 8 depicts the cover, together with three lifts of the surgery curve r = txt- 1 x- 2 . The lifts of the "fiber" s 3 , gs 3 = s! , are indexed by IT The lifts of

r

are indexed by their basepoint

g E-'IT.

I

17

~

f-..

1'---

!

..........-

1"--.

../

tt ...._

~---

.

--

~

_./

s3

t

~--"--

tt

s3 xi "--- ----:.::~-

X

- ....... r-~

0

'

I'

~------- ......

ft

x1

--. - . s3 ..-f-.~ 1

__.,




3 D X R

ic:""".----,

3 S X R

Figure 8

b

1

3 D X R

~----":> -

18

Let

M

= Mo

u u

S

1

u X

0 , the exact sequence defining

e

= rr

2

gives

36

3 .

5

O~H (Zp'

3~

we have that

=

d k ( 1)

ok

4

Z:lp/N)-"""-~H (Zp, Z)-O •

~

k (N)

s s(q-1) N • ( l+a + ••• +a )

q· e

zz--~---~~z

is

homomorphism 4

3

.

takes (k]

5

H (Zp' Z)

generates

=

~

From the diagram

=

q • N , so

This means that the conn ecting to

q. [ e ] •

Since

[e]

ZP , we conclude 0

0

0

4

H (TTl L(p,q), rr 2L (p, q))

H (i!p,!) """ Zp """'

w

UJ

~

k(L(p,q))

q

(Compare with Plotnick [42j). 0

The pun ctured lens spaces 2-ske leton, independent of valent for all 0

L(p,q ' )

q •

(rel o )

extends to

f:

L(p,q}

(n,p )

induces

=

L(p,q') •

:::>.
= Hf1.

I

( soo

X

Rl Z!:)

=

0

'T"

for

i>l •

(Since

Shapiro's lemma:

[H 1 Z]

Hi(H 1 ZH)

=

p

=

< (X)

I

this also follows from

Hi(lt 1 1ZZ))

cohomology with coefficients in



l(H/Zp)

To compute the 1

first note that

42

l(H/Z) p

=

L

has one cell in each dimension, the cohomology of

~

Z

=

(p)

$ ~,

~z -modules.

as

p

z

z~p

oo

Z p

ZH ®

commutes with direct sums.

Since

K(Z ,1) p

=

Hence EBZ ;i even>O l

i

= H (Z , p

$ ~

p

Z)

0; i odd •

The Wang sequence for the fibration with coefficients

Since

$z ~

x

Z(H/Z ) yields p

acts trivially on

4

H

(~

p

,Z), the action of

is just the permutation of the cosets

p

Hence

4 H (H,Z(H/Z )) -=0 p

and

Z

=

H/Z

x

p

5 H (H,Z(H/Z )) ""' Z p

p

The exact sequence from Lemma 2.1 breaks into the short exact sequences

o~

A

~zH ____.TI

2---o

(*)

and (-N, x-1) o-Z(H/Z'> )----..?Z:B EB Z(H/Z )---.A~O p p

These two coefficient sequences yield :

(**)

on

43

3

H

(H,n

2

5

}~H

4

(H,A)

:::>




~:m-z----o

c1 (B')

lo/( x~1)

=

-1

-t

-1

2

lrr(P)EBl:rr(e )EB(Ilrr& c

\

01

o2 = ZH(b)

I

a3

(i3))

02

zH

1

o2 =

ker

l

=

2

~

a2

llm~ c (i3) 1 ZH

(B))

al



e

,.~rr ~z-o

lrr(P)EBzrr(b)

I~

Zrr

~

(x-1) (1+x -1 -t -1~ ) N ( 1+x

-1

-t

-1

) 0'1 ~

62

This shows that the cocycle representing to the cocycle representing • ( l+x

-1

-t

-1

):

TT

k(B

2 ---"rr X • 2 p,q 2 p,q B

k(X

p,q

)

restricts

2 ) , followed by the map p,q

Computation of the k-invariant

§6.

We now identify

k(X -p,q )

as an element of

In addition to the facts about the cohomology of

H

mentioned

in §2 we need the following:

(1) The cohomological dimension of

G

is

2

by Lyndon's

theorem or from t he Mayer-Vietoris sequence for the extension

HNN

(2) H = ~ ~Z p

G = ~z';

has a

K(H,l)

in each dimension, so

with finitely many cells

H* (H,

~

M.) = 1.

~

H* (H,M.) • 1.

The Mayer-Vietoris sequence for the amalgamation

1T

= G*H z

i > 3

yields, for i H ( n ,Zrr )

=

i

H (H,Zrr)

=

0 •

i

We also have to compute

H

(IT,Z(rr/Zp}), fori= 4,5.

In

order to do that, we need the following lemma from [45], the proof of which we include for completeness:

Lemma 6.1:

A*B

,

let

-1

E c

Given a free product with amalgamation a E A

be such that there is no

a' E A

with

c a'aa'

Proof: form

.

Then

waw

-1

Recall that each

E A implies

w E A*B

c

w = cd •• • dn , where the di 1

w E A

.

has a uniqu e normal are chosen alternately

64

c"'. A

from fixed coset representatives for We say that any of d

n

A

and

; -A

I

w

have length

-1

-1 -1 -1 = cd ••• dna dn ••• d c 1 1

a contradiction.

Thus

2n-l •

d

n

E A

p

rr--+rr/z

g

-1

Suppose

hg E Z

aj E - z

aj

As

p

h

hE~ Z [hg]

~

p

~

(

I

.

As for

$

waw

-1

E A •

n

Zrr ~l:Z Z

~(rr/~)

D The projec-

via

g E rr , h E - H •

Hence, for g E H

I

g E

h·s = Then

~-H·l

1T

-

H

X

I

I

h~HZ[hg] ~ Z(H/~)

thus get the direct sum decomposition of

Z(rr/Z )

If

2n+l > 1

cannot be conjugated into a power of

.

ZH

~- c

.

is a torsion element and thus equals

g E H

Lemma 6.1 gives

E A

-1 d a d n n

Since

acts on

h·g = g , for some

, so

p

-1

p induces a homomorphism H

hs .

.

waw



Elements

[ 40].

has length

Hence, n=l , and w=cd

g~g

,

p

n

Suppose

is the induced module

z(rr/Z )

tion

< 1.

B

waw

has length

as above has length

c"'. B

and

.

We

ZH-modules

ZH) ffi Z(H/l: ) • p

The Mayer-Vietoris sequence for the amalgamation

rr = G*zH •

together with the above direct sum decomposition, facts (1) and (2), and the cohomology computations from §2, yield (for i=4,5)

65

i H ( rr, Z ( rr/Z· ) ) p

i H (Fh Z hr hz ]) ~ EB Hi (H, ZH) E9 Hi ( H, Z ( H/z. ) ) p ~ p -H·l

2!!

Hi(H,Z(Hf:l )) p

2!!

~{~

;

i = 4

;

i = 5

Recall from § 4 the short exact sequences of Zrr-modules

(-N,x-1)

The long exact sequences for these coefficient

se~~ences

yield 3



(rr,rr

2

0

)---~H

""'

4

(rr,H (M).) 3

As in the proof of Proposition 2. 2, we see that 5

the zero map on 3 H (TT,TT2)

~

H (tr,z(rr/z ) ) p 4

5

·(x-1)

is

, and so

H (rr,H (M)) == H ( rr, Z ( rr/Z ) ) 3 p

::::..

zp

This, together with Proposition 2.2, Remark 1 from §4, and the results from

§s, proves

66

Proposition 6.2: n (X ) 1 p,q 1

3

2

"""'

3

n Bp,q )~H (rr p,qJ 2

x

1 p,q

1

0

(L{p,q))~rr 0

.

1

2

(Bp,q).._c_~~

0

"""'

H (rr L{p,q), rr L{p,q)).______ 1 2

induce isomorphisms

3 2 H (rr B

rr

The inclusions

, rr 2xp,q )

under which

k-invariants correspond, namely 3 ::::.. '>H ( zP, ZZp/N) ~ zP

w k{X

p,q

w )-----':!ook(B

w

U)

2 )------~k(L{p,q))~q p,q

0

67

§7.

Proof of the Theorem

We now prove Theorem 3.2. -. ---'~X

f: X

and an

inducing an isomorphism

p,q 1

p,q

S: rr

a.-isomorphism

S

preserve k-invariants.

X

~x

p,q

p,q

1T

H = ~ p ,q Z = (a , x

! ap

Proposition 7.1: has the form

Proof:

a. :

X ~rr X 2 p,q 2 p,q ,

by a map

f

and

= G* oz!I ,

=

G

where

a.

E Aut ,

G

First no t e that

~ 11 X

11 X

1

1 p,q

Then



f

p,q 1

a.

and

by a map

~X

p,q

p,q

,

,

the

still preserve the k-invariants.

a~an±l { x--.x

only subgroup of

X

H = ~pX\ ~

TI

(n,p)

I

(t,x

= 1, xax -l = a -l)

Let

a. :

If we precede

, or follow

composed maps on Recall that

Assume there is a map

txt

-1

= x 2)

and



Up to conjugation,



=

a.

1 .

is torsion-free and that the ~

isomorphic to

is

Zp (a)

From the structure theorem for subgroups of amalgamated a. (~)

products [40, p. 243], Thus, up to conjugation,

irp· ,

a. (x) • a • a. (x)

that

a. (x) E H , so

a.

-1

-1

is a conjugate of

a. ( Zp)

E Zp •

=

ZP •

Since

x

lp . normalizes

Applying Lemma 6.1, we see

a. (H) c H •

The same argument gives

(H) c H • Suppose the r e is no

Then, since

h

~ H with

a. (t)' a. (x) • a. (t) -l

Lemma 6.1 implies

O. (t) E -H.

=

h O. (x)h-l E -Z(x)

a. (txt -l)

Since

=

H (n) 1

2 a. (x ) E' -H ,

=

Z(t)

and

68

H c

, this is a contradiction.

[TI,TI]

tion by an element

a

I HE -Aut

and

H

f

Following

a{x) E z(x)

h E H ,

= !{x)

H (H) 1

by a map

Hence, up to conjugaSince

a (x)

,

X



D which realizes

~X

p,q•

p,q•

conjugauon by a suitable element, we may assume above f orm.

I

=

a (x )

f

a

has the

. x -1 , construct a f.~ber-preserv~ng

diffeomorphism

F (x)

=

x-l

(This is possible since

* the monodromy

is an involution).

n3 x s 1

preserves a

s 1 x n3

T

F:X

~X

p,q

us to assume that

a

a~a

F

, in which we take connected sum with

and do surgery, so that

equivalence

We may assume

F

extends to a homotopy

Replacing

p,q

by

f

f oF

permits

has the form n

a: x~x

Note that

a(t)

of such.

(n,p)

=

1 .

can be quite complicated.

a(t)

might have

,

or

m a(t) = a t

Later on we will further modify

projects to

id

G

For example, we , or a composition

a

until it

69

Now we examine k-invariants. a(~

p

Since

a(H) = H

and

) = l:p , we get the following commuting diagram of

isomorphisms:

k(X

p,q

,) E

l

As

is not a

:qq'

~~adratic

residue mod p , all we have to

show in order to derive a contradiction is

13

*

=

±1 •

Before proceeding with the proof, let us collect some facts about ( i)

ZIT

that will be needed.

·(g-1) l'IT --=--~Zn

is a monomorphism, far

g E

TT

of

infinite order (Lemma II 2.1). ~~) ( ~·

7TT

~

•(l+x

-1

-t

-1

)

Proposition 4.1).

~TT

~

is a monomorphism (proof of

70

(iii)

Z~-resolution

Tensoring the standard left (right) ~

of

with

on the left (right)

Z TI

induces the free left (right) Zrr-resolution of ~

a-1

N

a-1 ·

::ZTI -~ZiT-~"> ~TI

ker N

=- Im(a-1)

e

This gives

-~') llTI ~z-0

=

and ker (a-1)

z

ImN

1

where the

maps are multiplication on the right (left). (iv)

has

The augmentation ideal presentation

(l~ox2

:~:=i

(:~~)

x+Noa -1)

a-1 ( ""~-" )3- - - - - -0 - - - - ----=.;.(Zrr )3 --'---~I rr ~o

where the map

( il'fT )

3

~

( :lTI )

3

i s the "Jacobian

matrix" of Fox der ivatives associated to the presentation of

(see [7 1 p. 45-46]

TI

we now start the study of the action of cohomology.

The

a -map

S

lifts to an

1

or look

S* on

a -map

'e :

Zrr~

ln

-1 ) ( 1 .

We now determine what s ort of satisfy {16).

(l+x

u

E

=

~G

u

1

u E ZG'

can

=

Equation (16) yields

-t n

g

+

n

xg

- n

tg

=

;

-1

g = 1 or

t

n

gt \...0

g = t

-r

+ n

gxt

+ n -r

gt

-r+l

g=ht

r-1

X

-1

or htr

; otherwise

1

81

where

n

g

For

g

=

0

=

t

if i

~

g

G'.

Since only a finite number of the

an integer = 0

n

I

such that

L for

Let

= -n .

.

t~h

>

L

Hence



. -n i+l = 0 • t h

xt~h

> r+l such that there is + n

Then

=

.

0 , and so

xtJh

=

=

.

Ii I

for

n . + n

be the largest integer

h E G' with n . f 0 • t]h

are nonzero, there is

g

tih

i > r+l , we again have

j

n

= 0

n

.

/ ..__ -1

~

tih For

n . + n i -n i+l = 0 • t~h xt h t h

h , i < -2 , we have

(-l)s

tJt-JxtJh

x

This is an infinite seq-uence of equalities, since infinite order. i

> r+l

For

g

for

Hence

n . = 0 , t]h

have

n

th

which proves

~

= t -1 h

, we have

n _ t

Hence

has

nl = t

and

~

nh + '11xh - nth

= 0

1

+ n h

for

={ ~

xt

h

;h=l,x i

_

h;el, X

1

-nh

={--to

;h=l i

h

~ 1

hfl •

g = h

For

I

we

-1

which gives

-1

= 0 •

For

1


0 0 •

such that there is an For such

h , equation

htr (20)

gives n

= -n

htr

which implies

x

2

r+l

r =

ht

= 0 •

...

= { -1)

Hence

s

n

x2

r+s

= 0 ,

htr

for

=







I

X

k

83

Let n . xJtr

be the largest positive integer such that

j

i

0 •

Equation (19) gives

= 0 •

= -n j+l r + n '2j r + n 2r+l+2' t

X

Hence

n k

x tr

=0

t

X

, for

Jtr

X

k > 0 •

The conclusion of these

computations is k

u=.t+(I: nk x)t k 0 •

r

( 21)

can be, we go back to equation (17),

=

0 , for

h

=

t

-i k i

x

t

,

i

> 2 or

Combining (17) and (18) gives

n -1 + n -1 • t ht t xht S e~' t '~g

h -- t-lxkt , k odd, 1n · th e a b ove equa t'1on y1e · ld s

n -1 k + n -1 k 2r t X t t X tx

X t

which implies

n -1 k = 0 • t

u

=

This gives

n -1 k 2r = t x tx

n -1 k t

=0

~

k O X 0 2::

m (

k x )

r

e (u) = m · 2

In particular,

r

But



k==-2 +1

€(u)

=

0 (mod p)

and

p

is odd, hence

m

=

pm

1

.

We now

have

u

0 2::

=

X

r

2k r t

k=-2 +1

=

,r., + m (

0 2:

k=-2

r-1

2k r -l 0 = ,(.,+p(m . ~ x t ) (l+x ) 1 k=-2r-l+l

+1

"---v----1 \)1

=

t

+ pv (l+x 1

-1

-t

-1

) + m •

°i: r-1

k=-2

x +1

2k r 1 t -

85 .

Induction on

r

~

produces an element

u = t +

p~(l+x

-1

-t

-1

E

~G

such that

) + m ,

thus proving lemma 7.2.

0

Why does this proof work for these examples, but not for the 2-twist spun knots? X

p,q

'f

X

p,q

,

"""'B2

but

,

That is, why is it that We can see this

?

p,q' .

Recall that

algebraically as follows.

rr B

2

1 p,q

=

H ,

and

is given by the exact sequence

S is given

a = id

Take by

z.q

(x;l) EB ZH

la~(:

}~"f:---0

:)

z_q _______,. TT

ZH EB ZH

This imposes the conditions

2

S

N

is an isomorphism, we also want a

analogous conditions and

uu'

=

=

pick

1, q'

=

2

we need a

I

u =a+ a'

uu' = N + 1 •

, u'

=

u

u' satisfying

(u (x-l) + v 1 N) + 1 • 1

conditions are not hard to satisfy. q

= (x-1) a + Nb u = (x-l)c + Nd

(x-1) u {

Since

----.-0

with

1 + a +a

-1

For example, if

s* =

~ (U)

Then

p = 5 , We may

= 2

9

These

=(u 0)

0 u

and

With the more complicated module structure on

86

'li

2

(X ) , these p,q

·u' s

are no longer available.

One might also ask, why not project X

2

( la)

~D

, and work in

I

1

(

lb)

I

(

2)

1

(

2I

3

,

G-n

as in [45}.

-1

-t

-1

uu' u'u

I txt -l

=

Equations

project to

)

-

-1

-

-1

(a(x-1) + pb) (l+x p(l+x

=(t,x

3

)u = (c(x-1) + pd) (l+x

-t -t

-1 -1

-1

) ) -1

= (u (x-l) + pu ) (l+x -t ) + 1 0 1 -1 -1 = (u (x-l) + pui> (l+x -t ) + 1

0

Unfortunately, one can produce examples which satisfy these equations and have the required augmentations. if

p = 5, pick

u = 2 + tx •

For example,

We get

1 (x-1) (l+x -l-t- ) ) (

Zn EBZn 3

3

5 ( l+x

-1

-t

-1

)

3

---------------~ZD ------» ~D / (~)

3

u~2+tx

0 ) -1 2+tx

j

--o

S

Zn EBzo ~--------------~zn ~~D /(*)~0

3

Pick also

3

u' = 2tx + 1 •

We have

uu' = (2+tx) (2tx+l) = 5tx + 4 = (x+tx

So it really seems necessary to work in order to prove the theorem.

3

3

-1

ZG

) • 5 ( l+x

-1

-t

-1

as we did in

) -1

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