to a ribbon presentation through Andrews-Curtis moves [2] ..... Andrews-Curtis equivalent to the trivial group: ...... J. A. Hillman: Aspherical four-manifolds and the.
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
H·
(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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