THE MATHEMATICAL THEORY OF KNOTS AND BRAIDS An Introduction
This Page Intentionally Left Blank
NORTH-HOLLAND MATHEMATICS STUDIES
82
The Mathematical Theory of Knots and Braids An Introduction SIEGFRIED MORAN University of Kent at Canterbuy
1983
NORTH-HOLLAND - AMSTERDAM
NEW YORK
OXFORD
Elsevier Science Publishers B . V . , 1983
All rights reserved. N o part of this publication may be reproduced, stored in a retrieval system or transmitted in any f o r m or by any means, mechanical, photocopying, recording or otherwise, without the prior permission of the copyright owner.
ISBN: 0 444 86714 7
Publishers:
ELSEVIER SCIENCE PUBLISHERS B.V. P.O.Box 1991 1000 BZ Amsterdam The Netherlands
Sole distributors for the U.S.A.and Canada:
ELSEVIER SCIENCE PUBLISHING COMPANY, INC. 52 Vanderbilt Avenue NewYork,N.Y. 10017 U.S.A.
L i b r a r y of Congress Cataloging in Puhlicaiion D a t a
bran, Siegfried. The mathematical theory of knots and braids. (North-Holland mathematics studies ; v. 82) Bibliography: p. Includes index. 1. Knot theory. 2. Braid theory. I. Title. 11. Series: North-Holland mathemstic6 studies ; 82.
QA612.2.M67 1983 ISBN 0-444-96714-7
514'.224
83-Il430
PRINTED IN THE NETHERLANDS
To the Mathematicians
James Waddell Alexander 1888-1971 Emil Artin 1898-1962 Ralph Hartzler Fox 1913-1973 Christos Demetriou Papakyriakopoulos 1914-1976
and Members of my Family Ruth, Simon, Matthias, h a , Roberta.
vii
"The Incas had another method f o r knowing and calculating the amount of the provisions contributed i n the provinces
... and the method was so
good and s u b t l e , t h a t i n ingenuity it exceeded the carastes which the Mexicans used t o make t h e i r calculations and business transactions : these were the quipus, which are long strands of knotted cords.
Those who were
accountants and knew the combinations of the knots, gave account by means
of them of the disbursements made, o r of other things t h a t might have happened many years before : and on these knots they counted from one t o ten, and from ten t o a hundred, and from a hundred t o a thousand : and i n one of these strands is the count of the one, and i n another of the other :
in such a way t h a t f o r us it i s amusing and blind computation, and f o r them exce 1l e n t .
Pedro de Cieza de LEon, La Cr6nica d e l PerCl S e v i l l a 1553
This Page Intentionally Left Blank
ix
CONTENTS Apologia
xi
Chapter 1 Some Necessary Group Theory
1
Chapter 2 Some Necessary Topology
33
Chapter 3 Knots and Pictures of Knots
63
Chapter 4 Braids and the Braid Group
75
Chapter 5 Some Connections between Braids and Links
111
Chapter 6 The Group of a Link
119
Chapter 7 Group Rings
131
Chapter 8 Derivatives
139
Chapter 9 Alexander Matrices
155
Chapter 10 Elementary Ideal of Alexander Matrix
171
Chapter 11 Alexander Polynomial of a Knot
175
CONTENTS
X
Chapter 12 Alexander Polynomial of a Link
185
Chapter 13 Some Matrix Representations of t h e Braid Group
189
Chapter 14 Operations on Braids and Resulting Links
209
Chapter 15 The Group of a Free Endomorphism
217
Chapter 16 Alexander Polynomials Revisited
223
Chapter 1 7 Meridians and Longitudes
2 31
Chapter 18 Symmetry of Alexander Matrices of h o t s
243
Chapter 19 Symmetry of Alexander Matrices of Links
2 49
Chapter 20 Conjugacy of Group Autamorphisms
261
Chapter 2 1 P l a i t Representations of Links Chapter
281
w
A List of Links
287
Bib 1iogr aphy
289
Index
293
xi
APOLOGIA This book owes its origin t o a s e r i e s of t h i r t y lectures which I have given over a number of years t o Third Year Undergraduate Mathematics Students.
I have now taken the opportunity t o blow these lectures up
i n t o a book.
Hopefully it s t i l l preserves some of the informality of
the original lectures.
Some of the gaps have been f i l l e d i n .
These
were forced on the l e c t u r e r by the necessity f o r paying careful a t t e n t i o n t o the morale of the audience. I have t r i e d hard t o make t h i s account of Knot Theory complete i n a
number of ways. adequate coverage.
However a number of v i t a l topics have not received For instance although the topological product and
the semi-direct product are mentioned a number of times, definitions of these concepts have not been included and general r e s u l t s on these constructions have not been proved. considered i n many other books.
These matters are well and carefully
However the same cannot be said about
some of the other matters which have been sidestepped ( f o r example orientation). Readers of the well known and beautifully written c l a s s i c a l book on Knot Theory by Crowell and Fox w i l l note q u i t e a number of s i m i l a r i t i e s with the account given here.
There is another i n t e r e s t i n g way of
developing the Theory of Knots and Links.
This method was discovered by
E m i l Artin C11 and i n some ways was put on a more sound b a s i s by J.S. B i r m a n
i n her book on Braids, Links and Mapping Class Groups. benefitted greatly from reading her book.
Clearly I have
APOLOGIA
xii
Some of the more important and interesting exercises are given an outline proof.
The reader is invited t o f i l l in the details.
as references go, the following convention i s adopted.
As f a r
W. Magnus C21
refer t o the second paper/book i n Bibliography under W. Magus.
No number
a f t e r a name means tha t t h i s name i s associated only with one reference. Included i n Bibliography i s a selection of papers and books whose aim is t o extend the interested reader’s knowledge of further developments in the subject
.
I am grateful t o Mrs. Dot Fry who took on the arduous task of typing my manuscript
- she did t h i s cheerfully
and e fficiently.
I hope t h a t , by making appropriate selections f r m the pages of t h i s book, it w i l l s t i l l be possible f o r a discerning lecturer t o construct an interesting course i n Knot Theory. Three Mathematicians, whose i d e n t i t i e s are unknown t o me, each pointed out an er r or in the original Mathematics t h e i r help.
0 time!
-
my thanks t o them f o r
Surely other errors remain.
thou must untangZe t h i s , not I ;
I t is too hard a knot f o r me t o untie.
Twelfth Night
Tyler H i l l , Canterbury , Kent. April 1983.
CHAPTER 1
SOME NECESSARY GROUP THEORY
We s t a r t off with an apparently more general definition of a group. In f a c t it i s completely equivalent t o the usual definition.
However it
is more convenient from our point of view. 1.1. DEFINITION.
A group G is a nonempty s e t S of elements together with
an equivalence r e l a t i o n
- and a binary operation
which i s defined f o r a l l
elements x and y of S such t h a t
(i)
x-y belongs t o S f o r a l l x and y of S;
(ii)
(x*y)*z x*(y*z) f o r a l l elements x,y and z of S;
(iii)
there e x i s t s an element e o f S such t h a t e - x
- x f o r a l l x of S;
(iv)
f o r every x i n S there e x i s t s an element
i n S such t h a t
-
- e; i f x - y i n S,
x-’-x (v)
then
Z*X
- z*y and
X*Z
- y-z f o r a l l z i n S.
In the usual definition of a group equality of elements i n a s e t i s taken t o be the equivalence r e l a t i o n 1.2.
NOTE.
-.
(1) To the usual d e f i n i t i o n of a subgroup H o f a group G one
has t o add the condition t h a t i f x belongs t o H and y H.
(2)
So H
- x, then y belongs
consists of complete equivalence classes.
In the usual definition of homomorphism $ : G1 * G2 one i n t e r p r e t s
the condition t h a t $ is single-valued as s t a t i n g t h a t x t h a t $(x)
- $(y) i n
G2.
- y i n G1
implies
to
CHAPTER 1
2
The value of approaching groups i n t h i s way is t h a t it makes p o s s i b l e a more i n t u i t i v e approach t o f a c t o r groups. 1.3.
FACTOR GROWS.
subgroup of G.
Let G = (S,-,.)
be a group and N be a fixed normal
With t h i s data we associate a new group
G/N = (S,!,.)
which has the same s e t S and product In f a c t we define
relation. x
Hy
. but has
a more stringent equivalence
N, by
i f and only i f x-ly belongs t o N.
I t i s now a routine exercise t o verify t h a t (a) x (b)
(c)
-
y implies that x
k! y
f o r a l l x and y i n G;
is an equivalence relation on S ; N ( S , - , * ) s a t i s f i e s the above axioms ( i ) - (v) of a group.
I t i s easy t o see t h a t x
k!
e f o r a l l x i n N and using axiom (v) of a
group t h a t y i f and only i f x-ly
x
N, e .
Hence we have the following useful way o f looking a t the process G
+
G/N.
The group G/N has the same s e t and product as G but i t s equivalence relation
i s obtained from
xN-e f o r a l l x i n N.
- by adding the condition t h a t
3
SOME NECESSARY GROUP THEORY 1.4.
by considering
G = (S,-,*)
where
s i s the s e t of
s cx" =
1.5.
One obtains the usual definition of a group from
REMARK.
;x
E
CONVENTION.
a l l equivalence classes i n S under
- , that is,
S).
In t h i s book we w i l l frequently replace
- and !lby . =
Using t h i s convention, the question as t o what is happening t o the group G when one goes over t o a f a c t o r group G/N has the answer : one i s adding the r e l a t i o n x = e f o r every x i n N. We now construct a special class of groups such t h a t every group i s isomorphic t o a f a c t o r group of a member of t h i s class. f r e e groups.
They a r e called
Hence every group can be obtained from a f r e e group by
adding some relations. 1.6.
FREE GROUPS.
symbols.
Let xa, a
E
M, be an a r b i t r a r y set X of d i s t i n c t
Associated with t h i s set X we have a s e t X - l whose elements
are in one-to-one correspondence with the elements of the s e t X. denote the elements of X - I by xi',
a
E
M.
I t i s important t o note t h a t
we are not assuming t h a t xi1 is the inverse of xa. statement does not have a meaning.
W e
A t the moment t h i s
We take the elements of X u X-'
t o be
an alphabet and consider the s e t S(X) of a l l f i n i t e words i n t h i s alphabet. For example
are words, where a, B ,
y,
6
a l l belong t o M.
An a r b i t r a r y word can be
CHAPTER 1
4
written in the form
€1 x"l
...
xE2
a2
En xa ' n 1 Further xa i s t o be i We also f o r a l l i and n i s a positive integer.
where a l l ai belongs t o M and every interpreted as being xa
=
E~
_+
1.
i take the empty word e t o be an element of S(X).
The product of two words w1 and w2 is defined t o be the word obtained from w1 and w2 by juxtaposing wl in front of w2, t h a t is, w1w2. The word w1 i s defined t o be equivalent t o the word w2 i f and only i f
w2 can be obtained from the word w1 by a f i n i t e number of insertions or deletions of subwords of the form -1
Xa
xa
-1 and xa xa
.
I t i s easy t o verify t h a t t h i s does in f a c t define an equivalence relation
- on S(X).
For example,
x x x -1x a a a
a
-xx
aB'
I t can now be shown t h a t (S(X),.,-) with the set X of free generators.
i s a group F(X)
-
the f r e e group
The empty word e is the unit element
and -E
n n
...
x
-2
X-E1
a2
9
XE2 ...
E x n
.
The associative law can be 1 "2 an established by induction on the length of the middle factor (see below f o r
i s the inverse of xa
the definition of length),
5
SOME NECESSARY GROUP THEORY
I f w1
- w2 and w2 i s obtained from w1 by deleting some subwords of the
form
o r x" xa-1
-1 x" x"
'
then one says t h a t w2 is obtained from w1 by cancellation.
I f w is a word
on which it is n o t possible t o p e r f o m any cancellations, then w is said t o be reduced. I t is easy t o see t h a t every word i s equivalent t o a reduced word.
An a l t e r n a t i v e d e f i n i t i o n of a f r e e group F(X) is the s e t of a l l reduced words with the product being juxtaposition followed by cancellation. The equivalence r e l a t i o n is taken t o be equality. I t is an immediate consequence of the d e f i n i t i o n of a f r e e group and
the law of indices t h a t every reduced word w ( # e) of F(X) has a unique representation of the form xnl xn2 "1 " 2
...
x"k
,
"k
where every ai # ai+l and every n 1. i s a nonzero integer.
The Zength of w
i s defined t o be k
I f w' i s any word and w'
!L(w')
=
- w,
where w is a reduced word, then we define
E(W).
Also we put k(e) = 0. This is the number which is mentioned above i n connection with proving the associative law by induction.
6
CHAPTER 1
1.7.
I f X consists of one element x , then every element of
EXAMPLE.
F ( I x 1 ) has a unique representation of the form
xn, where n is an integer. Hence F({xI) is isomorphic t o the i n f i n i t e cyclic group. 1.8.
EXERCISE.
(1) Show t h a t a f r e e group does not have any elements of
f i n i t e order. (2)
Show t h a t i f F(X) is a f r e e group with X having more than one element,
then F(X) has a t r i v i a l centre. ?HE UNIVERSAL PROPERTY OF FREE GROUPS.
1.9.
of t h e s e t
X i n t o a group G .
Suppose t h a t $ i s a mapping
Then 4 extends i n a natural way t o a
homomorphism 4 o f t h e f r e e group F(X) i n t o G, where
n
"k We may assume t h a t the elements of F(X) a r e reduced words.
PROOF.
Hence
the above mapping 41 : F(X)
G
-+
i s single-valued.
xml "1
...
If
m n x and x 1 ar B1
...
xnS BS
are reduced words, then
m x "1
...
2.
n x
r
r and v
... xn
0s
=
... 2 . xn ... xn "1 %
x"l
U
1 and the element on the right hand side is a reduced
where u
5
word.
I t is also possible f o r t h i s product t o be equal t o
2
,
BS
SOME NECESSARY GROUP THEORY
m
m
...
x 1 "1
"lr
x u or x
%
...
n x s
BV
In a l l these cases one has t h a t
$
or e.
8s
preserves the product and hence
$
is a
homomorphism.
VAN DYCK'S THEOREM.
If a group G has ga' a E M, as a s e t of generators, then G i s isomorphic t o a f act or group o f t he f r e e group F(X), 1.10.
where X = Ix", a
PROOF.
Let
E
MI.
denote the mapping of X i n t o G defined by
$
xa$ = ga f o r a l l a
E
M.
Then, by 1.9. The universal property of f r e e groups, it follows t h a t
$
extends t o the homomorphism $ :
$
F(X)
is onto.
For i f g is an element of G , then
n 1 g = g" 1 since g
,a
G.
+
"k g"k
'"
M, generate G , where we may assume t h a t a.
E
1
# ai+l f o r a l l i.
Hence n [x"1
...
".I
4l = g.
"k
1.11. GENERATORS AND DEFINING RELATIONS.
homomorphism
$
Let K be the kernel of the
from the free group F(X) onto the group G , which is given i n
the proof of the previous theorem.
Then we have t h a t
G 2 F(X)/K,
where K i s a normal subgroup of F(X).
Hence we have t h a t (upto
CHAPTER 1
8
isomorphism) G is obtained from F(X) by putting the elements of K equal (or more precisely equivalent) t o the unit element e . yB, B
E
Now suppose t h a t
N, is a s e t of elements of K which i s such t h a t every element of
K i s a f i n i t e product of conjugates (in F(X)) of the elements y-fl , B
N.
E
B
Such a s e t of elements i s said t o generate K as a normal subgroup.
Then
it follows from 1.1. axiom (v) of a group t h a t the group F(X)/K i s
obtained from the free group F(X) by putting
Y*
-e
for a l l D
E
N.
We now consider the reverse s i t u a t i o n , where we start with an a r b i t r a r y s e t of elements y
B’
B
E
N , which is contained i n the f r e e group F(X).
y belongs t o the normal subgroup of the f r e e group F(X) generated by the elements y
8’
B
E
N , then y i s said t o be a consequence of the elements
I f K denotes the s e t of a l l consequences of
yB, B
E
N, i n F(X).
yB, B
E
N , in F(X), then the group F(X) and the r e l a t i o n s
YB
-e
for a l l 8
determine the group F(X)/K.
E
N
Clearly the group F(X)/K i s uniquely
determined by the s e t of generators.
x a’
EM,
and the defining reZations y = e 8
o r more b r i e f l y y
B
for B
E
N.
This data is usually written in the form
I t follows from 1.10. van Dyck’s Theorem t h a t every group has a
If
SOME NECESSARY GROUP THEORY
9
presentation of t h i s form i n terms of generators and defining r e l a t i o n s .
However one chooses the elements y get a group by the above method.
EXAMPLES.
E
N , in F(X) one w i l l always
I f one chooses two many elements, then
one is l i a b l e t o get the t r i v i a l group 1.12.
6
6'
<
e
>.
(1) Consider the group
G = < x ;x2,x3>. In G, one has x3 = e , x2 = e and hence x = e. (2)
The free group F(X) has a presentation <
(3)
xa,
ci E
M;
-
>
.
Let n be a positive integer.
Then the cyclic group G of order n has
a presentation < x ; x n > . For i f g is a generator of G, then the mapping x morphism I$ of F(Cx3) onto G. F(Cxl)/ker I$
B
+
g defines a homo-
Hence
G.
However xn belongs t o the kernel of
$.
Hence i f N denotes the normal
subgroup of F(Ix1) generated by xn, then N s ker $ and <
x ; xn > / ( k e r $/N)
2 G.
Hence, by Lagrange's Theorem, the group <x ; xn> has a t l e a s t n elements.
On the other hand the group <x ; xn> has a t most the following d i s t i n c t elements e
=
,..., P-',
xo, x 1
10
CHAPTER 1
namely, n of them. <
n x ;x
>I
Hence =
IGI = n ,
which by Lagrange’s Theorem, implies t ha t ker + / N l
=
and ker o C N.
(4)
1
Hence
Let n be a positive integer. A
= <
Then the group
x ~ , . . . , xn ; (x. x.) f o r a l l i 1’
3
<
j
5
n >
,
where the commutator (x.,x.) = x-1 x-lxi x j , i s isomorphic t o the free i j 1 1 abelian group on the fre e generators x~,..., This l a t t e r group can
s.
be defined in a similar way t o a free group. lin ear combinations of the symbols x1 ,, coefficients.
.., xi ,..., )h with integer
The addition of two such expressions is performed by
adding corresponding coefficients.
Two such expressions are equal i f and
only i f corresponding coefficients are equal. t h i s defines an additive group A(Ixly. universal property for abelian groups. commutator calculus (a,b)”
I t consists of a l l formal
= (b,a)
(a-l,b) = a(b,a)a-’ (ab,c)
= b-’(a,c)b(b,c)
(a,bc)
=
(a,c)c-l(a,b)c
th at A is an abelian group.
Now
.., xnl).
Now it can be ver if ied t h a t These groups s a t i s f y the
I t can be verified by means of the
11
SOME NECESSARY GROUP THEORY
A(Ixl
,..., xnl)
2 F(Cxl
,..., xn1)/K,
where a l l ( x . , x . ) belong t o K. 1 1
Fixl,.
.., xn I)
Let N be the normal subgroup of
generated by a l l (x.x.) with i 11
<
j
5
n.
Then N
C
K.
Hence there e x i s t s a homomorphism
e : A+A(Ixl,
..., % I ) ,
which i s onto and xie = x . f o r a l l i. 1
From the universal property of f r e e
abelian groups it follows t h a t e has an inverse and hence e is one-to-one. This shows t h a t A =" A(Ixl,
..., xnI).
We have a l s o shown t h a t A(X)
F(X)/F(X)',
where X = {xl,.
.., xnI
and F(X)
i s t h e commutator subgroup of F(X). Moser i s concerned with giving
The book by H.S.M. Coxeter and W.O.J.
presentations of groups which occur i n some natural ways.
For example
they consider presentations f o r the symmetric groups, t h e alternating groups and some matrix groups. 1.13.
RESULT.
x
= IX1,".,
Suppose t h a t m and n a r e positive integers and
%I , y
Then F(X)
E
F(Y)
i f and only i f m = n.
= cyl,
..., ynI
I
WTER 1
12
(i)
PROOF.
Suppose t ha t m = n.
defined by xi$
=
Then the mapping
yi for a l l i, extends, by 1.9. The universal property f o r
F(X), t o a homomorphism 4 of F[X) onto F(Y).
Similarly there e x i s t s a
homomorphism I$ :F(Y) '-t F(X)
which i s inverse t o (ii)
$.
Suppose t h a t F(X)
Hence F(Y).
$
i s one-to-one and 4 must be an isomorphism. W e consider the number of d i s t i n c t
homomorphisms of F(X) int o the cyclic group
3=
I 0 , l I of order two.
1.9. The universal property of free groups, t h i s number i s equal t o 2
By
m
,
since such a homomorphism i s uniquely determined by the images of the free generators. Zrn
=
2"
Hence which gives tha t m = n.
The number n i n the fre e group F(Ixl,..
., 3))i s called the rank of
the free group. We have shown that the rank of a fre e group uniquely determines the group upto isomorphism. A group is said t o be f i n i t e Z y generated i f and only i f it has a
f i n i t e set of generators.
A group i s said t o be a finiteZy presented
group i f and only i f it has a presentation with a f i n i t e number of generators and a f i n i t e number of defining relations. 1.14.
EXAMPLES.
(1) Every f i n i t e group is a f i n i t e l y presented group.
The elements can be taken t o be the generators and the multiplication table gives the defining relations.
13
SOME NECESSARY GROUP THEORY
(2)
The additive group Q of rationals is not f i n i t e l y generated.
For
suppose contrary t o assertion Q =
Then Q
=
<
., mk/n k > ‘
ml/nl,..
, which is f a l s e since l/(nln2.. .nk+l) does not Hence Q i s not f i n i t e l y generated.
belong t o <1/(nln2.. .nk) >.
I t is
not d i f f i c u l t t o see tha t Q cannot be given by a f i n i t e number of defining relations. (3)
The group -s s G = < x,y ; (x,y xy ) f o r a l l s
L
1>
is f i n i t e l y generated but not a f i n i t e l y presented group. f a c t is proved in A.L.
&elkin.
The l a t t e r
We give an outline of the proof.
Let
Then
K be the normal subgroup G generated by x. rn
K =
II x < y- s xy s > s=o
and K i s abelian.
$=
<
However i n the group
x,y ; (x,y-’xy
S
)
for 1 5 s
5
k >
the normal subgroup generated by x is always nonabelian. A group w i l l in general have many di ffe rent presentations.
In
general it w i l l be a d i f f i c u l t matter t o show that two such presentations are isomorphic.
In f a c t given two f i n i t e l y presented groups, it w i l l in
general be impossible t o decide whether they are isomorphic o r not. We now define some elementary operations on a presentation of a group which give rise t o presentations of an isomorphic group.
They are called
14
CHAPTER 1
1.15. TIETZE TRANSFORMATIONS.
Let
be a finitely presented group.
We define two types of transformation of
this presentation (and their inverses) which give presentations of isomorphic groups. (Con) Let y be a consequence of yl).
..
(C6n) Let yi be a consequence of yl).
yn.
Then form the presentation
.., yi). .. A
yn.
Then form the
presentation
(Gen)
Let z be a new symbol and w be a word belonging to F({xls..
.
Then form the presentation < X1)
...)
; Yl’“.’ Yn, z = W ’
)m9z
*
(Gkn) Suppose that yi is of the form xy’w, where w is a word in the
...
A , symbols x19..., xj
< X1)
1
Then form the presentation
)m.
.)yis ...)x j s . . . ) xm ; y1 (substitute w for x.)).. I ... yn (substitute w for x.)I . h
A
>
1.16. NOTE.
Instead of the presentation
one can also write xas a
E
M ; yB = e, B
E
N
xJ).
15
SOME NECESSARY GROUP THEORY
or < Xa'
a ~ M ; y i = y h ' , B ~ N >
where (yh) -1y;' t h a t a hat
=
?1 yB o r yk'(yA)-l
=
yil for a l l
E
N.
Further we note
over some symbols means remove those symbols.
These four types of operations are called l'ietze transformations-
It
is easy t o see t h a t i f we s t a r t off with a f i n i t e presentation and apply a f i n i t e number of Tietze transformations, then we a r r i v e a t a f i n i t e presentation which is isomorphic t o the i n i t i a l l y given group.
The
converse o f t h i s r e s u l t holds and is the content of the next theorem. THEOREM OF TIETZE.
1.17. <
,..., a,
al
; rl
Suppose t h a t
,..., rk >
and
<
bl
,..., bn
; s1
,...,s 9. >
are two f i n i t e presentations on d i s t i n c t symbols of the same group G . Then one presentation can be obtained from the other by using a f i n i t e number of Tietze transformations. PROOF.
We have t h a t bi
=
w.(a.) f o r 1 2 i 1
7
2
n in G.
Hence using (Gen)
repeatedly we have t h a t G 5 < al,
..., am, bl, ..., bn
; rl,
..., rk, bl
=
Now we how t h a t S 1 (bv ) =
... = s (b ) = e R
G
V
and hence they are consequences of the relations
rI
=
...
=
rk
=
e, bl
=
w (a.) 1
7
,..., bn = wn (a.). J
Hexe applying (Con) repeatedly we have t h a t
wl,
..., bn = wn . >
WTER 1
16 G :< al,
bl
..., a,,,,b l , - . . , bn ; r1(a,,),---, rk(all), = w (a.) ,..., bn w ( a . ) , sl(bu) ,..., sa(bU) . 1 1 n J =
>
Now i n G we have t h a t ai = w!(b.) f o r 1 c i s m. 1
1
consequences o f the above relations i n G.
Hence they are
So applying (Con) repeatedly
we have t h a t
Finally we have t h a t G :< bl
,..., bn
; s1
,..., s a
>
upon applying (C6n) n+k times, since the remaining relations must be consequences of the defining relations
s1,
S2””
sI
for G. The theorem of Tietze can be used t o show t h a t certain constructions which are given i n terms of a f i n i t e l y presented group are invariant under isomorphisms.
Gne has only t o show t h a t these constructions a r e invariant
under Tietze transformations.
We now give some examples t o show how it is
possible t o show t ha t two presentations are isomorphic by means of Tietze
17
SOME NECESSARY GROUP THEORY
transformations. 1.18.
EXAMPLES.
Show t h a t i f m and n a r e coprime integers
(1)
2
2 , then
the cyclic group <
x ; xm >
<
p
a,b ; am, bn, (a,b) >
where (a,b) = a -1b-1ab. xn = a
,
Put
and xm = b.
Then <
x ; xm >
<
x,a,b ; xm, xn = a , xm = b >
.
Now we know, since m and n are coprime, t h a t there e x i s t integers a and
f3
such t h a t 1 = ma + nB.
Hence x = x1 = baaB, am = bn = (a,b) defining relations.
< x ; xm
> 5
=
e a r e consequences of the above
Hence
(b"aB)n = a , (ba aB )m = b, (a,b),am,bn >
Now it is easy t o see t h a t the f i r s t three of the defining r e l a t i o n s are each consequences of the last three of the defining relations. the required r e s u l t follows. (2)
The braid group Bn with generators
a l #'2J"'Y
an- 1
and defining relations
and
Therefore
.
CHAPTER 1
18
f o r every i and k with k # i, i+l
(ui,uk) = e
plays an important role in knot theory.
We reproduce the r e s u l t of
E m i l Artin 111 t h a t i -i = e for 2 (ul,a u 1a
a,ul ; an = (a u,)"-l,
B~ z
I
i
I
n2 -
>
.
... un-1' Then in Bn, we have = u1 ... ui-luiui+lui+2 ... un-1 'i = u1 ... ui-l , uiui+lui . ui+2 ... u n- 1 = u1 ... ui-l . ui+luiui+l . ... un- 1
Put a = u1u2 aui
Ui+2
= ai+l
.a
Hence a u.a-l = u i+l 1 and ai-'u
a-(i-l) 1
= u
i
for 1 2 i
5
n-2
for 2
5
n-1
I
i
.
Now we use (Gen) on a , i n s e r t the l a t t e r consequences which define ui and then apply (Gkn) t o u2,
..., un-1'
This gives t h a t Bn has a presentation
with generators a , ul and defining relations a
i-1 -(i-1) , aiula-i i-1 -(i-1) ula . a ula
=
a iula -i
. ai-lu1a - ( i - l )
, aiula-i f o r 1 I i s n-2
(ai-l -(i-1) k-1, a-(k-ll = ula , a 1
a = ul
. a u 1a-'.
a 2 ula -2
... an-2
f o r k # i, i+l -(n-2) la
These equations are consequences of the following equations, which hold i n Bn :
SOME NECESSARY GROUP THEORY
. a a l a -1 . u1
u1
a iola -1) = e
(ul,
a"
=
(a all
=
a o l a -1
2
for
. ul . a o l a -1 i
2
5
n-2
19
,
,
.
n- 1
A s Bn can be generated by a and a u l , we have t h a t an belongs t o the centre
of Bn i s a consequence of the last of the l a t t e r relations.
Also the same
is true of (01,
-i i an-iu a- (n- i ) 1 1 = (ol, a ola 1
.
= a -i(ul, a iola -i) -lai
So
n-i -(n-i) (al, a ola ) = e is a consequence of t h e following relations
aiu a -i) 1 Finally (ul,
=
e for 2
. a a l a -1 . al
ul
2
i 5 n and an = (a u ~ ) ~ - ' . 2
= a o l a -1
. a1 . a o l a -1
i s a consequence of (ul,
-1 -1 -1 a u 1 a u l a) = e
which i s a consequence of the relations
a
=
u1
. a ola-l . a 2ola -2 . ... an-2 ala-(n-2)
(premultiply by (ala ala i -' (al, a ola
=
-1 -1 )
e for 2
) and 2
i
5
n 2
together with the f a c t t h a t
an belongs t o the centre of Bn. The l a t t e r f a c t , as noted above, is a consequence of
CHAPTER 1
20 an = (a alln-'.
This follows from Example (2) o r can be proved d i r e c t l y by using the equations b = u 1a 2u 1 '
a = u1a2,
(4)
The factor group of the braid group Bn with respect t o the commutator
subgroup BA has generators u l , aiai+lui
f o r 1 s i s n-2
ui+laiai+l
=
..., an- 1 and defining relations
and f o r li
(ai,aj) = e
-
jl z 2.
These relations are consequences of the following relations which hold i n Bn/B;
: Ul
= u2 =
6 .
a
= an-l.
Hence Bn/B;
(5)
P <
al ;
-
>
.
-1 3 > The group < a,b,c ; b(abc-1)2a, c(abc )
i s isomorphic t o the free group <
x,b ;
-
>
, where
x = abc-1
.
where m is a fixed positive integer with
SOME NECESSARY GROUP THEORY
21
m
a = (xy) x and b = xy.
is isomorphic t o the group <
a,b ; (am,b) = e >
,
where m is a fixed positive integer with a = x y and b = y . 1.19.
We now turn t o the following generalisation of a
FREE PRODUCTS.
free group.
Let Ga, a
E
M
,
be an a r b i t r a r y collection of d i s t i n c t
We consider the s e t of a l l elements of the form
groups.
is an a r b i t r a r y element of Ga with ga not equal t o the u n i t i i i element and ai # ai+l f o r a l l i. We a l s o include the empty word e in
where ga
The integer k i s called the length of t h i s element with e
this set.
being considered t o have zero length.
' h o elements are said t o be equal
i f and only i f they a r e of the form
with ga
=
i
gAi in Ga
f o r every i. i
The product of the two elements is given by juxtaposition followed by
replacements of the f o n
and deletions of u n i t elements.
CHAPTER 1
22 For example, the product of
is
I t i s now a straightforward exercise t o prove t h a t t h i s defines a group, except f o r establishing the associative law. the free product of the groups G a ,
a E
This group is called
M, and i s denoted by
In the case of two groups A and B, one writes t h e i r f r e e product a s A*B.
Clearly the f r e e product n*Ga contains a subgroup isomorphic t o Ga, which we denote by Gay f o r a l l
~1
i n M.
Also every element of n*Ga can be
expressed as a f i n i t e product of elements from the various subgroups Gay
(Y.
E
M.
n*Ga
This is usually expressed i n the form = < Ga,
(Y.
E
M
>
and is referred t o as saying that the f r e e product n*Ga of the groups
SOME NECESSARY GROUP THEORY
Gay
M y is generated by its subgroups Ga'
CL E
c1 E
23
M.
This i s obviously a
generalisation of the concept of a group being generated by a s e t of elements.
Another example i s the d i r e c t product
A x B = < A , B > .
1.20.
EXAMPLES.
(1) F(Ixl,
and F(Ix l y . . . , xm}) * F({xm+l e> * A
(2)
<
(3)
< a ; a2 >
..., xn I )
,... , xn})
*
<
b ; b2
>
:F(Ixl
... *
<
,..., x n l ) .
xn '*
-
>
i s called the i n f i n i t e dihedral group and i s
This group has some properties which distinguish it from
a l l other proper f r e e products.
DL
x1 ; - > *
A f o r a l l groups A.
denoted by Dm.
D"m = < e > .
2 <
For example, Dm is soluble.
In f a c t
This follows from the f a c t t h a t the commutator subgroup
= < (a,b)>
is an i n f i n i t e cyclic group.
Every element of D, has a unique represen-
t a t i o n of the form a'bv (a ,b)m, where
u , v = 0 , l and m is an integer.
Note t h a t (a,b)
=
a
-1 -1 b ab
=
(ab)
2
Now it follows e a s i l y , by means of Tietze transformations, t h a t Dm/< (ab)">
Dn
,
which i s the dihedral group of order 2n, f o r a l l n
t
2.
Here we are a l s o
using 1.24. Consequence (1). 1.21.
EXERCISE.
(1) Show t h a t the centre of the proper f r e e product
A*B i s t r i v i a l . (2)
Show t h a t i f g is an element of f i n i t e order i n A*B, then g i s
conjugate t o an element of e i t h e r A o r B.
.
CHAPTER 1
24
Some of the properties of f r e e groups can be generalised t o f r e e products. 1.22.
'THE
We l i s t two such properties.
UNIVERSAL PROPERTY FOR FREE PRODUCTS.
Let Gay a
E
M y be an
arbitrary c o lle ctio n of d i s t i n c t groups and $ a : G a -+ G be a homomorphism of Then there e x i s t s a homoG i n t o a f i x e d group G for every a i n M. morphism $ from the f r ee product i'I*Ga homomorphisms $
1.23.
ay
a
E
M.
i n t o G , which i s an extension of the
In fact
ANALOGUE OF VAN DYCK'S
THEOREM.
Suppose t h a t a group G i s
generated by a s e t of d i s t i n c t subgroups Gay a
E
M.
Then G i s isomorphic
t o a f a cto r group of the f r e e product n*Ga.
1.24.
CONSEQUENCES.
(1) I t i s now not d i f f i c u l t t o show t h a t i f Ga has a
s e t of generators Xa and a s e t of defining relations Ra f o r each a i n M y then the f r e e product n*Ga has a s e t of generators
and a s e t of defining relations
(2)
I t is not d i f f i c u l t t o show t h a t A
xB
(A*B)/(A,B)
where (A,B) is the normal subgroup of A*B generated by a l l commutators of the form (a,b) with a and b varying a r b i t r a r i l y over A and B respectively.
SOME NECESSARY GROUP THEORY
FREE PRODUCT OF TWO GROUPS AMALGAMATING A SUBGROUP.
1.25.
Let G1 and G2
Let H be another d i s t i n c t group which i s isomorphic
be d i s t i n c t groups. under isomorphisms
25
$
1 and $2 with subgroups HI and H2 of G1 and G2
Then we define the following generalisation of the f ree
respectively. product :
8 G2
G1
is the factor group of the f r e e product G1 * G2 modulo
the normal subgroup generated by a l l elements of the form ($lh).($2h)-1 f o r a l l h i n H. We c a l l t h i s product the free product of G and G2 amalgamating the 1 subgroup H. Clearly t h i s product depends not only on the groups G1,
G2
and H but a l s o on the embedding isomorphisms $1 and $2. 1.26.
EXAMPLES. (1)
G1
* G2 <e>
z G1
* G2
(2)
<
(3)
Suppose t h a t G2 contains a subgroup isomorphic t o the group G1.
e > <e> * G2
G1
(4)
* G2 G1
G~
Then
' G2
The group < x,y ; x4, y4, x2
=
y2 >
is isomorphic t o < x ; x 4 > * < y ; y 4 > . i22 The centre of t h i s group i s nontrivi a l f o r x2 belongs t o the centre and x2 $
# e.
The l a t t e r statement i s true, since there e x i s t s a homomorphism
of the group onto the cyclic group < z ; z4> defined by
which maps x2 onto z2 # e.
26
CHAF'TER 1
The braid group Bg
(5)
=
i
a,b ; a 3
b2
=
z
is a l s o an amalgamated free
product amalgamating an i n f i n i t e cyclic group.
Once again the centre is
nontrivial, f o r it contains the element a 3 # e . A unique representation f o r elements of G1
following way.
I; G2
can be obtained i n the
Choose a complete s e t of l e f t coset representatives f o r
$ I ~ Hin G and f o r $2H i n G 2 , so t h a t the coset representatives of $lH and 1 I f gi belongs t o Giy $ I ~are H the u n i t elements of G1 and G2 respectively. w i l l denote the above chosen coset representative of gi($iH)
then i
=
for
Now it is possible t o establish the following
1,2.
1.27. UNIQUENESS OF REPRESENTATION EM. Eoery eZement of G * G2 has a 1H unique representation of the form
where every gi belongs e i t h e r t o G or t o G2, neighbouring terms beZong t o 1 j d i f f e r e n t groups, no gi i s the u n i t element and h belongs t o H. j L
PROOF.
Representation i n the above given form follows e a s i l y from the
(a)
repeated application of the following procedure :
al bl a2 b2
...
, where
a l l ai
al hl
E
G1
bi
E
G2 ,
. bl . a2 b2 ... , where hl $lH al . h2 bl . a2 b2 ... , where hl h2 c $ ~ H - = al . h2 bl . h i . a2 b2 ... , where h2 bl = h2 bl . h i = al . hza, . h i a2 . b2 ... , where h i z h i $lH
=
E
5
=
E
E
and so on, modulo the relations $l(h)
=
$12(h)
for a l l h
E
H.
27
SOME NECESSARY GROUP THEORY
(b)
We consider the s e t r of a l l elements of the form
UNIQUENESS.
where h belongs t o H and k i s any integer G1
;i G2
2
We define an action of
0.
on the r i g h t of r by means of the procedure i n p a r t (a) of t h i s More specifically one has t h a t G1 a c t s on the r i g h t of r as
proof. follows :
-
'
(gi
gi
1
-
(pi
y...,
,..., gi
1
=I
Y
k- 1
hl) if gi -
, g,
E
G1 and
=
hg k
k
with
hgl
hl i n G
k
hl) i f gi
k- 1
1
G1 a n d %
E
k
1
=
1
h1 i n GI
# e
# e
with
We impose a similar definition f o r the action of G 2 on Clearly the unit element of G1 and of G2 leave a l l the elements of r
f o r a l l g1 in G1.
r.
fixed.
Also, by the method of definition of the action of G1 on
r , we have
that
f o r a l l elements gl and g i of G1. of G2 on
r.
A similar r e s u l t holds f o r the action
In f a c t the groups G1 and G2 a c t as groups of permutations on
the elements of
r.
Tnus the mapping
CHAPTER 1
28
gl
+
action of g1 on
defines a homomorphism
r.
elements of
e2 : G2
-t
r
el of G1 into the group S, of permutations of the
There i s a similar homomorphism Sr
.
By 1 . 2 2 . The universal property f o r f r e e products, we have t h a t extend in a natural way t o a homomorphism e:G1*G2+Sr.
Now $ l ( x ) ( ~ 2 ( x ) ) - 1belongs t o the kernel of e when x
E
Hence e induces a group homomorphism
-
* H
e:G1
G2+S,.
I f contrary t o the assertion of the lemma we have t h a t
Y=gi
-
h = e
1 ' ' ' gik
with e i t h e r k
2
1 or k
= i d , .
= 0
and h # e y then
H.
For
el and e2
SOME NECESSARY GROUP THEORY
29
However, if one a ct s with s ( y ) on the element (1) of r one has t h a t
This contradiction establishes the uniqueness of the normal form. I t i s a consequence of t h i s unique representation that G1
G2
contains subgroups naturally isomorphic t o G1 and G2 which we also denote by G1 and G2 respectively.
GI
;1 G2
G1
n G2 =
G1
;I G2.
= < G I ’ G2
Also
’
and
in
H
I t i s not d i f f i c u l t t o give a universal property and an an 1 gue of van Dyck’s Theorem f o r fre e products with amalgamation.
* $2 GROUP G1 $1 H G2 ‘
1.28.
Finally we consider the following generalisation of the preceding product.
Let G1,
G2 and H be d i s t i n c t groups and
$1 : H + G 1 ,
$2 : H + G 2
We define the group
be homomrphisms.
G2 t o be the f ac t o r group of G1 * G2 modulo the normal subgroup generated by a l l elements of the fonn
(tJ1h)-l($,h)
9
where h varies over H.
W T E R 1
30
1.29.
EXAMPLES.
(1) <e>
*
H
"G
2 G
modulo the r e l a t i o n s $,(h) L
=
e for
a l l h in H. (2)
I f $1 and $ 2 are isomorphisms i n t o , then
(3)
Suppose t h a t $1 is not one-to-one, while $ 2 is one-to-one.
Then
there e x i s t elements hl and h2 of H such t h a t $l(hl)
$l(h2)
=
However in G
I
" G2
*
H
we have , t h a t
being onto with non t r i v i a l kernel and i n t o homomorphisms respectively. Then G1'l
;1 $2 G2
= < e > .
I t is possible t o produce a unique representation f o r elements of
similar t o t h a t given €or f r e e products with amalgamation.
However it is
no longer t r u e t h a t
contains natural isomorphic copies of G1 and G2 (see 1.29. Example (1)).
31
SOME NECESSARY GROUP THEORY
There do e x i s t o f course natural homomorphic images of G1 and G2.
REDUCTION LEMMA.
1.30.
which i s a f r e e product with amalgamation. minimal normal subgroups of G1,
Here G1(u),
G2(w) and H(u) are
G and H r es p ect i vel y so t hat the group 2
homomorphisms $* and 4; induced by $ 1 and $ 2 respect i vel y are isomorphisms 1 ( i n t o ) . In f a c t one has t h a t Gi(u)
=
u
nrO
f o r i = 1 , 2 and
Gi(n)
where
-1 G1(n) i s th e norma2 subgroup of G1 generated by $1($2 (G2(n-1))), G2(n) i s -1 t h e normal subgroup o f G2 generated by $2($1 (G1(n-l))), G1(0) = G 2 ( 0 ) = < e > PROOF.
.
F i r s t l y it is easy t o check t h a t Gl(u), G2(u) and H(u) a r e the
minimal normal subgroups so t h a t $1 and $ 2 induce natural isomorphisms
where these subgroups are taken t o be defined i n the statement of the second p a r t of t h e above Reduction Lema.
32
.
CHAPTER 1
Secondly it can be shown by induction on i t h a t H(i) = G l ( i ) in G1 ‘1 * H
” G2
H(w) = Gl(w)
in G1 $1 * $ 2 G2. H
= G2(i) = < e >
f o r a l l i.
= GZ(w)
This implies tha t = < e >
The subgroups mentioned here r e f e r of course t o t h e i r
natural homomorphic images in G1 $1 * $2 G2 H Finally, by Universal Property f o r the group G1 ‘1 * ” G 2 , one has H th at t h i s group can be identified with the free product of G1/G1(w) and G2/G2(w) amalgamating the subgroup H/H(w)
.
This establishes the Reduction
Lemma. I t follows from the above analysis of the situation that
as natural homomorphic images of G1,
G2 and H respectively.
Also
The book by J.P. Serre on Trees i s an interesting account of some topics connected with fre e products with amalgamation.
33
CHAPTER 2
SOME NECESSARY TOPOLOGY
We s h a l l use the r e s u l t s of t h i s chapter almost exclusively i n 3-dimensional Euclidean space IR3, which has the usual distance defined in it.
An open b a l l
i n IR3 w i l l be denoted by
where x is its centre and r i s i t s radius. called an open s e t .
A mapping f : X
+
A union of open b a l l s is
Y of topological spaces i s said
t o be continuous i f and only i f the inverse image of every open s e t i n Y under f i s an open s e t i n X.
I f f is also one-to-one and onto such t h a t
f - l is also continuous, then f is said t o be a homeomorphism. A path i n a topological space X is a continuous mapping p : I
where I = C0,ll.
-+
X,
The points p(0) and p(1) are called the i n i t i a l point
and the end point of the path p respectively. A topological space X i s said t o be p a t b i s e connected i f and only i f
f o r every p a i r of points xo and x1 of X there e x i s t s a path p i n X such that p(0) = xo and p(1) = xl. Let po and p1 be two paths i n a topological space X, which have the same i n i t i a l point and have the same end point.
We s h a l l say t h a t po is
homotopic t o p1 ( r e l a t i v e t o i t s ends) i f and only i f there e x i s t s a
continuous mapping
CHAPTER 2
34
for a l l t
E
I.
P1 Po
- p1
Instead of po is homotopic t o p1 one a.lso says t h a t po can be continuously deformed i n t o p and writes po 1 2.1.
RESULT.
PROOF.
(i) p
- pl.
H ( * , t ) i s a l s o denoted by Ht(*).
Hornotopy is an equivalence r el a t i o n .
- p using the homotopy for a l l s
H(s,t) = p(s)
I and t
E
I.
- p1 by means of the homotopy H(s,t) . gives t h a t p1 - po.
(ii) Suppose that po
homotopy H(s,l-t)
E
Then the
SOME NECESSARY TOPOLOGY
- p1 and p1 - p2 by means of the homotopies F ( s , t ) respectively. Then po - p2 by means of the homotopy
(iii) Suppose that po and G(s,t)
H(s,t)
for all s
E
F(s,2t) =
I.
for 0
G(s,2t-l) for t
5
t
5
4
5
t
5
1
35
CHAPTER 2
36
Suppose that po and p1 are paths i n a topological space X such t h a t Po(1) = P p ) . Then the product path plpo i s defined t o be
i
p0(2s)
(P,P,)
(s) =
for 0 s s
p1(2s-1) f o r
4
2
1
s s s 1.
The product path
A path i n a topological space X whose end point and i n i t i a l point
coincide i s called a Zoop i n X. L ( 0 ) = a(1).
The loop 9. i s s a i d t o be based a t
SOME NECESSARY TOPOLOGY
2.2.
Suppose that to, El, .t2, R
RESULT.
3
are Zoops i n a topozogical
space X based at x and
0
!Lo
-
R1,
- ”.
R2
Then L2k0
-
R3ill.
X
0 Product of loops R2Ro
PROOF.
Suppose t h a t to
and G(s,t) respectively.
- R1 and .t2 -
k3
i s given by the homotopy F ( s , t )
Then
by means of the homotopy
H(s,t) =
F(s,2t)
for 0
2
t
2
4
4
5
t
2
1
G ( s , 2 t - l ) for
and a l l s
E
I.
This homotopy is also denoted by G.F.
37
CHAPTER 2
38
2.3.
Suppose t h a t e is t he t r i v i a 2 loop in a topological space
RESULT.
X a t xoy t h a t is, e(x)
eR
-
xo f o r a l l x i n X.
=
Then
il
f o r a l l loops R in X a t xo.
Also
This follows as a consequence of the following CONTINUOUS MANGE OF PARAMETER LEMMA.
2.4.
topological space X and a : I and "(1) = 1.
-+
Suppose t hat p is a path in a
I is a continuous mapping with a ( 0 )
=
0
Then pa i s homotopic t o p ( r e l a t i v e t o i t s ends).
PROOF. The mapping H(s,t) = p((l-t)a(s) + ts) for all s
P" 2.5.
E
I and a l l t
E
I gives that
- P.
RESULT.
Suppose t h a t R is a loop i n a topological space X a t x
is the loop i n X a t xo defined by
Then
- e.
0
and
39
SOME NECESSARY TOPOLOGY
PROOF.
k(2s)
4
for 0 < s s
=
( A ) ( S )
a(2-2s) f o r
4
s s < 1.
This is homotopic t o e by means of t h e homotopy &(2s(l-t) H(s,t)
a((2-2s) (1-t)) f o r and f o r a l l t
4
4
s
for 0
=
5
s
5
1
I.
E
The above r e s u l t s give t h e following
2.6.
The c o l l e c t i o n of aZl Zoops in a f i x e d topoZogicaZ space
?HEOREM.
X a t a f i x e d base point x0 forms a group
under t h e above defined operation of product and using homotopy a s t h e equivaZence r e l a t i o n .
This group i s c a l l e d the fundamentaZ group of X a t xo.
2.7.
If p i s a path from xo t o x i n a topoZogicaZ space X, 1
RESULT.
then the mapping
defined by
defines a group isomorphism of ~r(X,x ) onto n(X,x )
0
1
.
40
CHAPTER 2
a
PROOF.
(i)
$
is single-valued.
P
For i f Lo
- L1
by means of the homotopy
H, then
P a0p-l
-P
by means of the homotopy idIxI.H.idIxI. (ii) I t follows from (i) t h a t $ is single-valued. P-l (iii)
o $ =~ i d of
$p o
$
=
i d of a(X,x,)
P-I
Hence ( 4 ) - l =
P
(iv)
.(X,x0) and
P-
$
-1 and $
P
P
i s one-to-one and onto.
Suppose t h a t Lo and L1 belongs t o .(X,x0). $
p ( a 1' il 0) = p,a 1.a 0.p
Then
-1
- ~ . i l ~ . e . a ~ . byp - ~a continuous change of parameter 2.4. - p.al.p-l.p.a
.p -1 0
by a homotopy which is somewhat similar t o the one used in the proof of 2.5. Result.
Hence
41
SOME NECESSARY TOPOLOGY
+p(yo) - OP(Rl) s p ( Q . Suppose t h a t X i s pathwise connected.
Then the fundamental group of
X is independent (upto isomorphism) of the choice of the base point.
Hence we denote its fundamental group by n(X) 2.8.
THEOREM.
Suppose t h a t f : X
topological spaces and xo
E
X.
+
.
Y i s a continuous mapping of
Then f gives r i s e i n a natural way t o a
group homomorphism.
which i s defined by
fn(e) = f
0
R
f o r a l l loops R i n X a t xo.
If g : Y
-+
Z i s aZso
a continuous mapping of
topological spaces, then (g
0
fIn
=
gn
0
fn
*
Also
(id$n
=
i d of n(X,xo).
Hence i f f i s a homeomorphism, then fn is an isomorphism.
CHAPTER 2
42
Y f o a
PROOF.
then f o Lo (ii)
f, i s single-valued.
(i)
- f o al
- .tl i n X with homotopy H,
For i f Lo
under the homotopy f o H.
Suppose a. and a1 a r e two loops i n X a t xo. f,(yo)
= f
0
(Rl.i0)
= (f
0
El)
. (f
0
Then
!Lo)
where one i s using the definition of the product of two loops as given a f t e r the proof (iii) Suppose
of 2.1. Result.
is a loop in X a t xo.
Then
SOME NECESSARY TOPOLOGY (iv) Suppose L is a loop in X at xo, (idX)l,(R) Hence ( idx)
= =
idX('l)
43
Then
.
= L
id of ?T (X,xo) .
Suppose that f is a homeomorphism, then f-' exists and is continuous
(v)
and so id.
=
(f-l o f)TI= (f-1) n o fl,.
This implies that (f-l)l,
=
(fn)-' and fn is a group isomorphism.
2.9. NOTE. A s we shall show later 2.11. Example ( 3 ) , if S1 denotes the
unit circle in IR2, then .(S)
= <
x ;-
>
.
There is a natural injection
which is one-to-one and continuous.
which is not one-to-one.
However
Hence f is one-to-one does not in general imply
that f, is one-to-one. 2.10.
X
x
RESULT.
Let (xo,yo) denote a point i n the topological product
Y of t h e topological spaces X and Y. .(X,x,)
Let
.(X,YO)
denote the d i r e c t product of the fundamental groups.
Then
44
CHAPTER 2
Let p1 and p 2 denote the natural projection of X
PROOF.
Suppose J? and II' are loops in X
respectively.
x
x
Y onto X and Y
Y a t (xo,yo).
Then
define the mapping
(i) e i s single-valued.
For i f L
- L' in X
x
Y by means of a homotopy H,
then p1 o R
-
p1 o
II'
and
p2 o
II
-
p2 o
by means of the homotopies P, o H and
(ii) e(a'.a) =
=
( i i i ) If
(p,
o(&'.L),
e(al)
ax and ay
p2 o
R' H respectively.
p 2 o(al.9.))
. e(a).
are loops i n X and Y a t xo and yo respectively, then the
mapping (kx ,Ry) +. 9.
,
where a ( s ) = (ax(s), ay(s)) f o r a l l s i n I , is an inverse mapping t o
8.
Hence e i s one-to-one and onto. 2.11. (2)
EXAMPLES. (1) n(&)
<
e > f o r a l l positive integers n.
Let C be a eonvex subspace of IF?, tha t i s , i f x and y belong t o C,
then,the line segment joining x t o y,tx + (1-t)y belongs t o C f o r a l l r e a l
45
SOME NECESSARY TOPOLOGY
numbers t with 0
For i f
9,
5
t
1.
5
Then
is a loop i n C a t xo, then
!. - e ,
where e i s the t r i v i a l loop at
x0’ under the homotopy H(s,t) = (l-t)R(s) + t e for a l l s
E
I and t
E
I.
The following are examples of convex subspaces: open and closed b a l l s , open and ‘closed cubes, solid cylinders i n IR3. (3) I f S1 denotes unit circle i n EX2, then
Let R be a loop i n S1 a t xO(e S l ) .
Associated with
there is a well
determined integer n ( e ) , which i s the number of times R winds round S’ i n the anti-clockwise direction. number of R .
The number n(a) i s also called the winding
The mapping
gives the above isomorphism.
A. GramainC11 Chapter 1 14 gives a complete
proof of t h i s r e s u l t along these l ine s.
We now give a proof of the f a c t t h a t rr(S1,l) 2 E which i s based on the important concept of covering space. made of t h i s concept.
However no e x p l i c i t use w i l l be
llIR1 is the universal covering space of S’”.
follow the proof given in M. Greenberg Chapter 4. There e x i s t s a continuous mapping IR’ + S1 which is given by x
-t
exphix
We
CHAPTER 2
46
for all x in IR1, where S1 i s taken t o be the unit c i r c l e centre the origin in the complex plane.
{z ; z
E
This mapping has a p a r t i a l inverse
t y I Z I = 1, z # -11
+.
IR1
which i s continuous and is given by
z
+.
1 Logez.
-7
2a1
-1 1 -
(i)
Let a be a loop i n S1 based at 1.
i n IR1 by
where N is a positive integer so t h a t
Then define a corresponding path
47
SOME NECESSARY TOPOLOGY
f o r a l l t i n [ O , l l and n = 1 , 2 continuity of
Q
,..., N-1.
Such an N e x i s t s by the
on C0,ll and the f a c t t h a t l a ( s ) 1 = 1 f o r a l l s in C0,ll.
Now
(ii)
We note t h a t p(') p(')
i s uniquely determined by the following properties:
i s a path i n IRI
,
For suppose t h a t p were some path i n IR1 with the same properties. the continuous mapping p(')t h i r d property above. connected.
p would take only integer values, by the
Hence p(')-
In f a c t ,('I=
Then
p is a constant, since C0,ll is
p, by the second property above.
(iii) Let F be a homotopy of loops keg and g1 i n S1.
Then we define a
corresponding homotopy i n IR' by G(F)(s,t) =
1N-11
L o g e ( F [ y s , N-n t] /F[T N-n-1 s, N-n-1 t)}
,
2 n i n=O
where we use similar conventions t o those given in p a r t ( i ) above.
Now
BIAPTER 2
G(F)(s,l) = p exp 2aiGCF)(s,t)
=
G(F)(O,t)
(a,)
(s),
F(s,t),
= 0,
G ( ~(1, ) t ) is a constant f o r a l l s and t in C0,ll. In order t o see t h a t the l a s t equality holds one proceeds as follows. exp 2aiG(F)(l,t)
=
F(1,t)
= 1
f o r a l l t , which implies t ha t G(F)(l,t)
E
Z2
f o r a l l t.
As 1 x I i s connected, G(F)(lxI) i s connected, which gives t h a t G(F)(lxI)
i s a constant. (iv)
We now define a mapping
by xII = ~ ( ' ~ ( 1 f)o r a l l II in a (S1,l ). By p a r t ( i i i ) above II0
x
-
R1 implies t ha t p
preserves group operations.
with
then the product path
=
p
(a,)
(1) and so
x is
single-valued.
For i f k0 and R1 belong t o a(S1,l)
49
SOME NECESSARY TOPOLOGY
in IR1 has the three basic properties which characterise the path p
as given i n p ar t ( i i ) above.
f o r a l l s in C0,ll.
(ape) 9
So
Further
by the definition of the product of paths. Finally the mapping
x
has an inverse, namely, an integer m is mapped
onto the loop s
Hence (4)
+
exp 211ims
x is
f o r s in L0,lI.
one-to-one and onto.
The surface of a doughnut o r the two-dimensional torus T2
= S1 x S1
has
fundamental group
by the above example and 2.10. Result. We s h a l l find the following r e s u l t useful f o r working out the fundamental group of certain topological spaces. of t h i s r e s u l t until 2.17.
W e postpone the proof
MAPTER 2
50
Let O1 and 0 be pathwise 2 connected open subspaces of a topoZogicaZ space X such t h a t THEOREM OF SEIFERT AND VAN W E N .
2.12.
X = O1 u 0 2
0
=
with
O1 n O2 being pathwise connected, nonempty and xo
E
0.
Then
where f
1 and f 2 denotes t h e natural i n j e c t i o n o f 0 i n t o 01 and 02
respective 2y.
2.13.
REMARKS.
(1)
A s pointed out before i n 2.9. Note, i f f i is one-to-
one €or some i, then it does not necessarily follow t h a t (fi)n is one-toone f o r t h a t i. (2)
I t i s easy t o see t h a t X is pathwise connected and hence n(X,xo) i s
independent (upto isomorphism) of the choice of base point i n 0. (3)
If 0 is simply connected, that i s , pathwise connected with t r i v i a l
fundamental group, then
This follows from 1.28. Example ( 2 ) . (4)
I f O1 i s simply connected, then n(X,xo) is isomorphic t o the factor.
group of n(02,xo) modulo the normal subgroup generated by t h e elements of (f2)n(n(0,xo)).
This follows from 1.28. Example (1).
2.14.
(1) Let X denote the figure eight
EXAMPLES.
embedded as a subspace of IR2
.
X
We take 01,O2 and 0 t o be the open
51
SOME NECESSARY TOPOLOGY
subspaces with open ends
respectively.
I f we take xo t o be the crossing point i n 0, then
Hence IT(X,X,)SF(Ial,a23), by the Theorem of S e i f e r t and van Kampen. (2)
Let n be a positive integer and X be the one-point union of n
triangles (or copies of s ~ ) . Then
by induction on n and using the Theorem of S e i f e r t and van Kampen.
(3)
Let n be a positive integer and X be a closed d i s c with n d i s t i n c t
points removed from its i n t e r i o r (considered as a subspace of R 2 ) . the same procedure as adopted i n Examples (1) and ( 2 ) shows t h a t
Then
CHAPTER 2
52
I t i s a l s o possible t o see t h a t the space in Example ( 2 ) can be naturally injected i n t o the space given in this Example, which induces an isomorphism of t h e i r fundamental groups.
(4)
Similar arguments t o t h a t given i n Example (3) show t h a t the
fundamental group of the following spaces i s isomorphic. t o the f r e e group F({al,.
.., an}) :
IR2 with n d i s t i n c t points removed, IR3 with n d i s t i n c t p a r a l l e l l i n e s removed, open disc 2.15.
EXERCISE
in
IR2)
(1)
w i t h n d i s t i n c t points removed.
Show t h a t the union Yn of n c i r c l e s of the form
(in IR2) has fundamental group isomorphic t o the f r e e group F(Ial,
..., an’).
SOME NECESSARY TOPOLOGY Show t h at the union Zn of n copies of two-dimensional torus T2
(2)
Of
53
Q@
the
.
,
.
= S1x S 1
(Q(Q
(in IR3) has fundamental group isomorphic t o the free product
where every Ai is a fre e abelian group of rank 2 .
Hence deduce t h a t Ym
cannot be homeomorphic t o Zn f o r any p a i r of positive integers m and n. Before giving a proof of the theorem of Seifer t and van Kampen we consider the following concept of LEBESQLE NUMBER OF A COVERING.
2.16.
space and U,,
a
p o s i t i v e in te g e r
Suppose t h a t X i s a compact metric
M, i s an open covering of X .
E E
Then there e x i s t s a
(called t he Lebesque nwnber of t he covering U
a)
a
E
MI
such th a t f o r every element x of X we have t h a t the open b a l l
B(x ; PROOF.
2
E)
some U
.
Let x be an element of X.
open i n X.
Then x belongs t o some U,.
The U, is
Hence there e x i s t s an open b a l l B(x ; r ( x ) ) with r ( x ) being a
positive r e al number such t h a t
where a(x)
E
M.
covering f o r X. b a l l s cover X.
The open b a l l s B(x ; i r ( x ) ) ,
X.E
X, form an open
As X is compact, we have t hat a f i n i t e number o f these Thus
cover X, where xl,...,
x a re elements of X. n
Let
E
be the minimum of the
54
WTER 2
r e a l numbers
Then we a s s e r t t h a t B(x ;
5
E)
some U
(XI
with ~ ( x )E M f o r a l l x i n X.
We now prove t h i s f o r the point z of X.
For since z belongs t o X, we have t h a t z belongs t o B(xi ; $r(xi) f o r some i.
This implies t h a t z belongs t o U
B(z ;
a (Xi) *
Suppose t h a t y belongs t o
Then
E).
d(z,y)
<
E 2
t r ( x 1. 1 .
Hence by the t r i a n g l e inequality d(xi,y)
2
d(xi,z)
+
d(z,y)
2
r(xi).
This shows t h a t
PROOF OF THE THEOREM OF SEIFERT
2.17.
AND VAN W E N .
(An adaption of
the proof given i n A. Gramain C11 Chapitre 11.) (i) Let II be a loop i n X a t xo.
We s h a l l show t h a t
II belongs t o < n (O1,xo), n(02,xo) >
,
where the group equivalence relations are taken t o be homotopy i n X. Since X
=
O1 u O2 i t follows t h a t {t-'(Ol),
of the closed u n i t i n t e r v a l I . covering. that i f
Let
E
I I - ~ ( O ~ ) }is an open covering
be the Lebesque number of t h i s
Then, by the definition of Lebesque number (see 2.16.), we have
55
SOME NECESSARY TOPOLOGY
(a,b)
c
I
with b
-
a s
E
then a((a,b)) 5 O1 o r 0 2 . Let n be a positive integer so that
and denote the r e s t r i c t i o n of the mapping I. = 1
[ $ , q] by p . f o r 0 s i
Then pi maps
<
1
[: , $1
9.
t o the closed subinterval
n.
i n t o e i t h e r O1 or i n t o O2 f o r every i.
For every
i -nw i t h 1 s i < n we now choose a fixed path 6 i which has i n i t i a l point xo and endpoint a(:). I f a(;) belongs t o 0, then we choose t h i s path t o l i e
i n 0.
I f a(;)
does not belong to 0 but belongs t o O1 or 0 2 , then we choose
the path t o l i e i n O1 or O2 respectively. and O2 are each pathwise connected. a = pn-l ! pn-2 ! pn-3 !
This i s possible, since 0, O1
I t now follows that
... ! pi ! ... ! p1
Here we use the following notation. q1 : Ca,bl * X and q2 : Cb,cl
If +
X
are continuous mappings with ql(b) = q 2 ( b ) , then q2 ! q1 : Ca,cl
i s defined by
-+
X
!
Po
*
CHAPTER 2 .
56
This mapping i s continuous (putting together continuous maps Lemma).
Now we note the following connection between putting together two continuous maps and the product of two paths. (q2 ! ql)(s) = ((q2
0
a-1 0 B2)
*
(ql
0
a-1 0 Bl))(t)Y
where i f a s s s b
2 (b-a) a(s) =
Y
1
Bl(t)
=
it
+
s-b
i f b s s s c
for 0 s t
2
1
,
~2 ( t ) = f t +f ol r O S t s 1 . This now, together with 2.4. Continuous Change of Parameter Lemma, gives the required r e s u l t t h a t R belongs t o < a(Ol,xO)
, a(02 ,xo)
>
,
where the equivalence r e l a t i o n i s homotopy i n X. (ii) Suppose t h a t k is a loop i n 0 a t xo.
I f il and i2denote the
natural injection of O1 and O2 i n t o X respectively. then c l e a r l y
Hence together with the f a c t proved i n ( i ) above, we have that .(X,x0) isomorphic t o a f a c t o r group of the group
is
57
SOME NECESSARY TOPOLOGY
Subsequently we s h a l l not distinguish between a loop i n Oi and a loop i n X whose image is contained i n 0. for i 1
=
1,2.
I t remains t o show t h a t there are no other relations holding i n
(iii)
r(X,xo).
Choose a complete s e t of l e f t coset representatives f o r
and for
so that the coset representatives of (fl)s n(0,xg) and (f2),, r(O,%) are the unit elements of a(Ol,xo) and n(O2,x0) respectively. i n 0.1 a t xo, then
If 'li i s a loop
x. w i l l denote the corresponding l e f t coset representative 1
of the l e f t coset
for i
=
1,2.
By 1.29. Uniqueness of representation Lemma, we have t h a t
every element of n(X,xo) has a representation of the form
belongs e i t h e r t o r(O1,xo) o r t o s(Oz ,xo) , neighbouring j terms belong t o different groups, no Ti is the unit element and h belongs
where every 'li
t o s(0,xg).
j By 1 . 2 9 , it remains t o show t h a t t h i s representation is
unique. Suppose t h a t some homotopy H.
i n X by means of xO As O1 and O2 form an open covering for X, we have that
'l
is homotopic t o the t r i v i a l loop e
CHAPTER 2
58
and H-'(O2)
H-l(Ol)
form an open covering f o r I
x
I.
Hence using 2.16
Lebesque number of t h i s covering, we have t h a t there e x i s t s an integer m[> 1) so t h a t f o r every r e a l number so C O1 o r O2
H(so,t)
for j = O,l,...,
for j
=
O,l,
for t
E
I we have t h a t
E
I f we denote the r e s t r i c t i o n of H t o t h e rectangle
m-1.
..., m-1, then we have
the following putting together of maps
where every H(J)(sO,-) maps e i t h e r i n t o O1 o r i n t o 02.
I t i s important t o
note t h a t these maps H(1) do not necessarily preserve the f a c t t h a t
For instance H(O) applied t o the loop
product of k + l loops.
Ti
.t
is a
need no 1
longer be a loop.
We amend the s i t u a t i o n s l i g h t l y .
and denote the deformation path of the point '1
for j = 1,2,..
homotopy. proof. (PIHI
., k and p = O , l , . .., m-1.
i s a genuine homotopy.
1
(1) under the map
by
j Every
i s a s o r t of
We s h a l l i n f a c t c a l l it a homotopy i n the remainder of t h i s For i f we put (p)H(s,
-.ti.
xi
We denote the map
and
fi m t)
=
( p ) H T ( s , t ) f o r 0 s t s 1, then
The r e s u l t of the homotopy
applied t o
(fl).p
w i l l be denoted by a?)
1
and
respectively f o r 1 5 j s k.
Then
59
SOME NECESSARY TOPOLOGY
under a homotopy which is obtained by f i r s t applying
and then
introducing the "feelers"
Now t h i s homotopy gives a loop which i s s t i l l a product of k + l loops.
Hence, by induction, it remains t o show t h a t R cannot be deformed continuously i n t o e F : I
x
xO
under a homotopy
-; 1 , 11 -+x ,
[l
where F(s, 1 - -1)
m
= a(s)
for a l l s
F ( s , l ) = xo f o r a l l s
E
I,
F(0,t) = F(1,t) = xo f o r a l l t F(so,t) 5 e i t h e r O1 o r O2
I,
E
1 [1 - ; , 11 ,
E
f o r a l l so
E
I.
Here we are assuming t h a t e i t h e r k # 0 o r i f k
=
0 then (fl)T(h) i s not
As before we denote i n the given representation f o r R. xO (1) under the homotopy F by y . and the the deformation path of the point 1 j
homotopic t o e
ai
r e s u l t of the homotopy F applied t o
zi
and (fl)T(h) by ai j
respectively f o r 1 2 j s k.
F
= a.
1
i1
a
i2
...
6
a
ik
Then
.
and 6
j
60
CHAPTER 2
I f F1
=
e , then xO
where, because of the f a c t tha t F(so,t)
C
e i t h e r O1 or O2
f o r every fixed so
E
I and a l l t
E
[1 - 51 , 11 , we have that every
There are two cases t o consider.
maps i n t o 0.
which gives t h a t (fl)T(h) is homotopic t o e
.
assumption concerning the representation f o r
11.
xO
k # 0.
y
j
I f k = 0, then ex = 13, 0
This contradicts the Secondly suppose that
Then
CI
il
y l = exo y 1 '
This implies t h at y1 is a loop i n 0 a t xo and hence ai
is a loop i n 0 at 1
xo.
By the above construction,
ail
-
i1 '
This contradicts the assumption concerning the choice of l e f t coset representatives which i s given a t the beginning of p a r t ( i i i ) of t h i s proof. (iv)
By 1.30 Reduction Lemma, it remains t o show t h a t i f kl,
loops i n 01,O2 and 0 with base point xo respectively, then
only when
i 2 , a.
are
SOME NECESSARY TOWLOGY k1 = e , k 2 = e , ko = e
for a l l h
E
Suppose t h a t H : I
{H-’(O1),
x
x
I
-t
X i s the homotopy i n X which deforms kl to
Then, using 2.16 Lebesque number of the covering
H-’(02)I of I
decompose I
respectively
This i s done f o r the loop kl f o r example a s follows.
n(O,x0).
the t r i v i a l loop.
61
x
I one can, as i n pa r t ( i ) of t h i s proof,
I i n t o a grid of rectangles so t h a t H r estricted t o any one
of these rectangles maps it i n t o e it he r O1 or O2 or 0. introducing “feelers” as before, one can see that k1
Finally, by
- e i n X follows from
the fa ct that a loop i n O1 contained i n 0 can also be considered as a loop i n O2 contained i n 0.
f o r a l l h i n s(O,xo).
The l a t t e r i s j u s t the statement
This Page Intentionally Left Blank
63
CHAPTER 3
KNOTS
AND
PICTURES OF KNOTS
We s h a l l develop a mathematical theory of knots which corresponds t o one's usual experience of knots i n a piece of s t r i n g .
I f one considers
the usual concept of a knot on a s t r i n g with loose ends, then one runs i n t o d i f f i c u l t y when studying it topologically as a curve i n IR3.
For then
every such knot can be continuously deformed i n t o a s t r a i g h t l i n e , by pushing an end through the knot.
Hence we w i l l think of a mathematical
knot as being a knot as commonly met with i n ordinary day l i f e with the e x t r a requirement t h a t the two ends are spliced together.
I t also helps
t o think of knots as being loose.
A knot K is the image i n IR3 of a continuous mapping f:S1 u n i t c i r c l e S1, which is one-to-one.
-+
IR3 of the
The l a s t requirement ensures, what
one has i n an ordinary knot, t h a t two d i s t i n c t points on a s t r i n g never coalesce.
The f a c t that a knot is the image s e t rather than the mapping
is another matter which is determined by ordinary knot theory.
Knots as
defined above are very d i f f i c u l t objects t o study, since they can be made up of an i n f i n i t e sequence of simpler knots. ordinary knot theory they are of no i n t e r e s t Topologists find them worthy of study. tame knots.
From the point of view of
-
only some dedicated
We w i l l confine our attention t o
A knot K i s tame i f and only i f it i s the union of a f i n i t e
number of s t r a i g h t l i n e segments i n IR3.
The points where the s t r a i g h t
l i n e segments meet are called the vertices of the knot. r e f e r t o tame knots as poZygona2 knots.
We s h a l l also
I f one takes a tame knot and t i e s
CHAPTER 3
64
such a knot an i n f i n i t e nunber of times successively (on a piece of s t r i n g ) , then one obtains a knot which i s not tame. abbreviate tame knots t o knots.
From now on we w i l l usually
Note t h a t an ordinary knot can be
obtained from the mathematical concept of a knot by c u t t i n g the curve a t a point. 3.1.
EQUALITY FOR KNOTS.
As other books on t h i s subject we w i l l not have
much t o say about t h i s important topic.
Actually the possible definitions
depend t o a large extent on which branch of Topology one i s aiming t o work
in.
From some points of view i t is best t o work with simplicia1 complexes.
A nice account of t h i s theory, which is useful f o r knot theory, can be
found i n the a r t i c l e by W. Graeub (see i n p a r t i c u l a r 17 Satz 111).
We give
three possible t e n t a t i v e definitions. (1)
EQUIVALENCE.
A knot K i s said t o be equivalent t o a knot K1 i f and
0 only i f there e x i s t s a homeomorphism
+
of IR3 onto IR3 such t h a t
+(%) = K1.
This i s the concept we w i l l mainly work with.
(2)
ORIENTATED EQUIVALENCE.
A homeomorphism Q of R3 onto R3 i s said t o
be Orientation preserving i f and only i f it maps the r i g h t handed cork screw
as numbered
onto a r i g h t handed cork screw.
t
x3 = z
For example the i d e n t i t y mapping i d R 3 is
orientation preserving and so are t r a n s l a t i o n and rotation.
However
reflection i n xy-plane, namely
i s not orientation preserving.
I t is true but not obvious t h a t i f Q i s
orientation preserving a t a point, then the same is t r u e a t every point of
KNOTS
AND
65
PICTURES OF KNOTS
Suppose t h a t a homeomorphism of IR3 onto IR3 i s such t h a t when
IR3.
composed with a reflection i n a plane gives an orientation preserving homeomorphism.
Then
i s said t o be an Orientation reversing homeo-
$
morphism. A knot KO i s said t o be orientated equivaZent t o (or the same as) the
knot K1 i f and only i f there e x i s t s an orientation preserving homomorphism $
of IR3 onto IR3 such t h a t $(KO) = K1.
(3)
STRING ISOTOPY.
The knot KO i s said t o be string isotopic t o the
knot K1 i f and only i f there e x i s t s a continuous mapping
such t h a t F(S1,O) =
KO,
0 s t s 1) the mapping s
the knot { F ( s , t ) , s
E
F(S1,l) = K1 and for each fixed value of t (with -+
F ( s , t ) with s
E
S1 i s one-to-one giving l i s e t o
S l l which is polygonal.
Further we require t h a t
associated with the mapping F there is a r e a l number 6 > 0 so t h a t the distance between every p a i r of vertices of the polygonal knot {F(s,t); s
E
S1) is not l e s s than 6 f o r every t
E
I.
This i s t o avoid
the process of pulling a knot t i g h t t o give the t r i v i a l knot. Clearly knots KO and K1 are orientated K1 are equivalent.
equivalent implies t h a t KO and
We now show t h a t i f two s t r i n g isotopic knots are
s u f f i c i e n t l y close together i n IR3 , then they are orientated equivalent. The knots KO and K1, which are defined by mappings fo : S1 f l : S1
a
-+
-+
IR3 and
IR3 respectively, are said t o be E-close together i f and only i f
.e&u . b
d(fo(e), f,(e)) =
E
where d denotes the usual metric i n IR3.
CHAPTER 3
66
3.2.
RESULT.
Suppose t h a t the s t r i n g isotopic knots KO and K1 are
c-close together.
If
c
i s s u f f i c i e n t l y small,thenKo and K1 are
orientated equivalent knots. PROOF.
By further subdivisions of s t r a i g h t l i n e segments and renumbering
i f necessary, we may assume t h a t the vertices of the knots
and K1 both
occur a t the images of the points
under the continuous mappings fo and f l of S1 respectively.
Further it may
be assumed t h a t the s t r i n g isotopy which deforms KO i n t o K1 deforms fo(COi,ei+ll)
1) f o r a l l i. i n t o fl(Ce.,e. 1 1+1
Associated with t h i s
s i t u a t i o n we have the knot K which i s defined by the mapping fl(e)
for
eo
5
e
5
e1
f o r en-2 s
f,(e)
f o r enml s
e
5
en- 1
e s en
In f a c t g(S1) equals fo(Sl) except f o r a c e r t a i n segment which i s pushed aside
67
KNOTS AND PICTURES OF KNOTS
I I
I
Here the second and fourth l i n e s of the definition of the function g(e) represent s t r a i g h t line segments joining fl(e,) fl(en-l)
respectively.
If
E
t o fo(e2) and fo(en-2) t o
i s s u f f i c i e n t l y small, then there e x i s t s an
orientated homeomorphism of IR3 onto IR3 which maps fo(S1) onto g(S1). Repeated application of
Hence knot KO i s orientated equivalent t o knot K. t h i s s o r t of procedure gives the required r e s u l t . RESULT.
3.3.
I f the knots KO and K1 are s t r i n g isotopic knots, then KO
and K1 are orientated equivalent knots. PROOF.
By assumption we have a continuous mapping
such t h a t F(S1,O) = KO, F(S1,l) = K1 f o r a l l s polygonal.
E
I and each knot F(S1 , t ) i s
By 3 . 2 . Result, we have t h a t f o r each to E I there e x i s t s a
positive integer € ( t o ) > 0 such t h a t
F(S1 ,t) and F(S1 ,to) are orientated equivalent knots f o r It
-
to/ 5 €(to).
CHAPTER 3
68
The intervals (to - c ( t O ) , to + ~ ( ) )t with to E I form an open covering 0 of I. As I i s compact, a f i n i t e number of these i n t e r v a l s cover I. Hence KO is orientated equivalent t o K1. Thus we have shown t h a t s t r i n g isotopy is the strongest equality condition t h a t one can impose on knots. Result holds (see G.M.
In f a c t the converse of 3 . 3 .
Fisher and see a l s o E.E. h i s e Chapter 11).
The following problem i s a fundamental m e i n Knot Theory: Given two knots KO and K1,
determine whether they a r e equivalent
or not. A knot K is said t o be unknotted i f and only i f it i s equivalent t o the
trivial h o t
I t is easy t o see t h a t a knot i s orientated equivalent t o the t r i v i a l knot
i f and only i f it can be unknotted. This may be an appropriate time t o consider some s t r i n g games.
Below
we give d e t a i l s of two such games, namely, the t o r t o i s e and the c a t ' s cradle.
The l a t t e r has the reputation of being the oldest s t r i n g game
i n the world.
They are both concerned with s t r i n g isotopies of the t r i v i a l
knot, which give r i s e t o i n t e r e s t i n g shapes.
The book by J. E l f f e r s and
M. Schuyt is an interesting source f o r various s t r i n g games of t h i s s o r t .
KNOTS AND
PICTURES OF KNOTS
69
70
CHAPTER 3 Crusader's Chain bki.1
KNOTS AND PICTURES OF KNOTS 3.4.
71
We s h a l l assume t h a t a l l knots K are given so
KNOT PEU!JEflIONS.
t h a t they s a t i s f y the following two conditions: (i)
the projection of K on xz-plane has no multiple points other than
double points; (ii) a vertex of K does not project onto a double point of the projection
of K on xz-plane. A knot t h a t s a t i s f i e s the above two conditions i s said t o be in
regular position.
Given a knot KO it i s always possible t o give a knot
K1 which is orientated equivalent t o KO and such t h a t K1 i s in regular
position.
I t is only necessary t o make a r b i t r a r i l y small a l t e r a t i o n s i n
the knot KO t o give a knot K1 which is i n regular position. then gives t h a t KO and K1 a r e orientated equivalent.
3.2. Result
R.H. Crowell and
R.H. Fox 3.1. of Chapter I is a relevant reference.
If a knot i s i n regular position then i t s projection on xz-plane gives an unambiguous picture of the knot provided one specifies a t a double point which l i n e segment passes over and which passes under.
This is
specified by using the following convention:
A double point (of the projection of a knot) i s also called a crossing
point. 3.5.
LINKS.
A Zink
(in
polygonal knots i n R3.
IR3)
i s the union of a f i n i t e number of d i s j o i n t
Clearly as we have done f o r knots we can define
the following concepts f o r links:
equivalence, orientated equivalence,
s t r i n g isotopy, projection and regular position.
Similar r e s u l t s t o those
proved and stated above f o r h o t s hold also f o r l i n k s .
CHAPTER 3
72
The varying d i s j o i n t h o t s which make up a link are c a l l e d the components of the link.
If a link L is s t r i n g i s o t o p i c t o the union
L1 u L2 of links L1 and L2 such t h a t L1 and L2 l i e inside d i s j o i n t b a l l s
i n IR3, then L i s said t o be s p l i t t a b l e . then it is said t o be u n s p l i t t a b l e .
If a link is not s p l i t t a b l e ,
The t r i v i a l link of two components
is s p l i t t a b l e , while /\
i s unsplittable. 3.6.
EXAMPLES.
(1) The l e f t handed t r e f o i l knot has the following
projection
(2)
The r i g h t handed t r e f o i l knot has the following projection
\
‘.
KNOTS AND PICTURES OF KNOTS
(3)
The granny knot has the following projection
(4)
The square knot has the following projection
(5) The figure eight knot has the following projection
(6)
The Borromean rings has the following projection
I t i s a link with three components such that any two of them form the t r i v i a l link of two components.
73
CHAPTER 3
74
The standard reference book f o r knots i s Ashley Book of Knots.
In
t h i s book, knots are c l a s s i f i e d by means of the uses they have been put t o . Thus f o r instance the t r e f o i l knot (which is there called the overhand knot) appears a large number of times. granny knot t o be a dangerous h o t .
t o it.
By the way, Ashley considers the He appends the symbol
75
CHAPTER 4
BRAIDS AND THE BRAID
GROUP
The best known example of a braid i s the one t h a t is used when tying a girl's plait.
I t looks as follows:
or
These are examples of 3-braids or braids on three strings. a fixed positive integer.
Suppose n is
We now define precisely what we s h a l l mean by
CHAPTER 4
76
n-BRAID OR A BRAID ON n STRINGS.
4.1.
I t i s given by the following
data: (i) n points Ply Pz,..
., Pn
i n IR3 which have the same z-coordinate, z = a
say, and whose x-coordinate s t r i c t l y increases as one goes from Pi t o Pi+l along the l i n e segment PiPi+l, ( i i ) n points Q,,
z
=
Q2,.
.., a i n IR3 which have the same z-coordinate,
b say, and whose x-coordinate s t r i c t l y increases as one goes from Qi t o
Qi+l
along the l i n e segment QiQi+l,
where
f o r each i ;
For every i there i s a f i n i t e polygonal path joining Pi t o Qi,,
(iii) p
i s a permutation of 1 , 2 , . . . ,
path from Pi t o Q.
111
(iv)
f o r each i ;
n , so t h a t as one t r a v e l s along t h i s
the z-coordinate s t r i c t l y decreases;
a 3-b and no two d i s t i n c t paths i n t e r s e c t . In an n-braid, the path joining Pi t o Q
where 1 5 i
I:
iu
i s called t h e i - t h s t r i n g ,
n. p2
Q1
Q2
pi
Qiu
n'
z=a
a
z=b
BRAIDS AND THE BRAID
77
GROUP
Two n-braids are said t o be equaZ o r string isotopic i f and only i f there e x i s t s a continuous deformation of one n-braid onto the other n-braid which s a t i s f i e s the above conditions ( i ) - ( i v ) throughout the deformation and the distance between two vertices is never l e s s than a fixed r e a l number 6 > 0.
/
are s t r i n g isotopic
and
/ The f i r s t impression given on studying the above definition of an n-braid is t h a t we have l a i d down too many unnecessary conditions. condition t h a t the path joining Pi t o Q.
The
i s made up of a f i n i t e number of
1?J
s t r a i g h t l i n e segments ensures t h a t we do not get involved i n t r i c k y questions of topology i n IR3.
While the conditions t h a t the i n i t i a l points
and the end points of the s t r i n g s are in each case l e v e l together with the f a c t t h a t as one goes along the path from Pi t o Q
iu
the z-coordinate
s t r i c t l y decreases ensure t h a t braids of the s o r t
are not string isotopic.
Otherwise one could use continuous deformations
of the form
which would reduce the subject t o t r i v i a l i t i e s .
78
4.2.
CHAPTER 4
PICTURE OF A BRAID.
W e s h a l l assume t h a t a l l braids are given i n
the form so t h a t t h e i r projections onto xz-plane s a t i s f y the following conditions: (i) a vertex does not project onto a double point;
(ii)
the only multiple points of the projection a r e double points. A braid which s a t i s f i e s these conditions i s said t o be i n reguZar
position.
Given an n-braid u, i t is always possible t o give an n-braid
u' which i s i n regular position and ul i s s t r i n g isotopic t o u.
I f an n-braid i s i n regular position, then it i s possible t o give an unambiguous picture of the braid by projecting it onto xz-plane and making the usual convention about under-passing s t r a i g h t l i n e segments. 4.3.
THE BRAID GROUP Bn OF ALL n-BRAIDS.
integer and Bn be the set of a l l n-braids.
Let n be a fixed positive We now turn Bn i n t o a group
by taking string isotopy as the equivalence r e l a t i o n and t h e following operation as the product.
I f u and u' a r e n-braids, then t h e i r product
u u l i s obtained by f i r s t constructing an n-braid u l ' which i s s t r i n g
isotopic t o u' so t h a t the i n i t i a l points of the s t r i n g s of u r ' coincide with the end points of the s t r i n g s of u and then placing u t r under u. For example the product of the 3-braids
and
BRAIDS AND ME BRAID GROUP
79
is the 3-braid
j!
R
I t i s now straightforward t o verify t h a t the f i v e axioms of a group hold.
The unit element is the n-braid
.
.
I
(n times)
The inverse of a braid u is given i n the following way.
Take a picture of
u and r e f l e c t it i n a l i n e z = ao, where a. is a r e a l number such t h a t u
lies i n the region z u.
c
a . of IR3.
For example, the inverse of
This gives a picture of the inverse of
CHAPTER 4
80
Basic examples of n-braids are given by
x
i
. . .
which is denoted by ui f o r 1 5 i
i+l
5
n-1.
(1) The 2-braid u:
is
Clearly they generate the group
Bn, th at i s , B = < ul, n EXAMPLES
4.4.
..., un- 1 > .
while the 3-braid u i is
So it i s important t o specify which group Bn one is working i n when one
gives a braid as a word i n the generators ul, u2,. (2)
.. .
The g i r l s ' p l a i t s given a t the s t a r t of t h i s section are given by the
words -1
(01
U2lm
and
(01
-1 m 02 1
where m i s a positive integer.
,
BRAIDS AND THE BRAID GROUP (3) The braid A = ulu2
... un-l
*
a1
.. . an-2 . ... .
81
U1U2
. u1 on
n s t r i n g s has the picture
t h a t i s , it represents a t w i s t .
The usual rope on n strands is repres-
ented by the words Am, where m is an integer. fundamental role i n the group Bn.
The braid A plays a
A good account of t h i s can be found
i n J.S. Birman Chapter 2.
(4)
The 2n-braid -1 -1 -1
m
where m i s a positive integer, i s given i n the Ashley Book of Knots no. 2963.
He c a l l s it a punch or wrought mat.
appearance
I t has the following
82
(5)
CHAPTER 4
The n-braid m (a u u
2 4 6”’
where m is a positive integer, i s given i n the Ashley Book of Knots no. 2976.
He c a l l s i t a French sinnet o r t r e s s e Anglaise.
I t has the
following appearance
(6)
The 8-braid
where m is a positive integer, i s given i n the Ashley Book of Knots no. 3001.
He c a l l s it an eight-strand square sinnet.
appearance
I t has the following
BRAIDS AND THE BRAID
(7)
83
GROUP
The 8-braid
where m is a positive integer, i s given i n the Ashley Book of Knots no. 3007.
He c a l l s it an e l l i p s e of eight strands.
appearance
I t has the following
WTER 4
84
Other examplesof braids, which have been used, can be found scattered throughout the Ashley Book of Knots and i n p a r t i c u l a r i n Chapter 38. The group Bn has defining relations
That these relations actually hold in Bn can be seen i n the following pictures :
BRAIDS AND ME BRAID GROUP
,..
85
...
u -1 . u -1 . u.u J 1 J i
I
The f a c t t h a t these are defining r e l a t i o n s f o r Bn is f a r from easy t o see, although E. Artin C11 assumed it t o be t r u e i n h i s o r i g i n a l paper. postpone a consideration of t h i s matter until a l a t e r stage 4.5.
PARTICULAR CASES.
(2)
B2 = < u1 ; - >
(3)
B3 = < ul,u2
by putting a = ulu2
(4)
(1) B1 =
e
<
>
.
.
; u1u2u1
= u u u
2 1 2
>
and b = u u u 1 2 1 '
See 1.18. Example (2) f o r another presentation of Bn.
-
We
4.9. Theorem.
CHAPTER 4
86
We now consider the following important isomorphic embedding of the braid group Bn in the group Aut(Fn) of r i g h t automorphisms of the f r e e By 1.9. The universal property f o r f r e e groups, in order t o
group Fn.
give an endomorphism of a f r e e group it i s s u f f i c i e n t t o specify i t s e f f e c t I t i s important t o note t h a t i n the proof of
on a s e t of f r e e generators.
the following theorem we do not use the f a c t t h a t Bn has any specific s e t of defining relations. 4.6.
ARTIN REPRESENTATION THEOREM.
Let Fn be a f r e e group on a s e t of
f r e e generators x ~ , . . . , xn, where n i s a f i x e d p o s i t i v e i nt eger.
Then Bn
i s isomorphic t o the subgroup of r i g h t automorphisms B of Fn which s a t i s f y the conditions
xi6 = A.1 x i p ' ;A
for 1 s i s n
and
(x1x2
...
x1x2
=
%)f3
... s,
where p i s a permutution of 1 , 2 , . . . ,
n and every A.1 belongs t o Fn.
Under
t h i s correspondence the braid ai goes over t o t h e automorphism
xi
+
xixi+lxi -1
'i+l
+
xi
x j
+ x
f o r all j # i
j
of Fn f o r 1 5 i n. FinaZZy s t r i n g of u goes from Pi t o Q.
1U
4.7.
NOTATION.
If
IS
p
i s defined by the f a c t t h a t t h e i-th f o r a l l i.
belongs t o Bny then the corresponding r i g h t auto-
morphism will be denoted by
;.
When there i s no danger of confusion we
s h a l l also denote the automorphism
;by
a.
BRAIDS AND THE BRAID GROUP
Take a braid a belonging t o Bn and i n the plane z = a (see
PROOF OF 4.6.
4.1.)
choose a point P, which has smaller x-coordinate than the x-coordin-
a t e s of the points P1,
z
=
87
..., Pn
and such t h a t i t s projection Q on the plane
b, has the same property with respect t o the points Q1,
..., Qn.
We now consider IR3 with the s t r i n g s of the braid u removed. the plane z = a becomes the plane p, where the points P1, P z ,
mental group
..., Pn
W e have a similar s i t u a t i o n f o r the plane q.
been removed. 71
Then have
The funda-
.., xn'
(p ; P) i s a f r e e group Fn with f r e e generators xl,.
where xi i s given by the loop
1 .
p1
Pi
'i-1
X
.
*
r
e
n'
i
We use the same notation f o r the funda-
f o r a l l i by 2.14. Example (4). mental group n ( q ; Q)
.
Pi+l
i
We define a mapping 0 of Fn by pushing a loop il (belonging t o n(p ; P)) down (the gaps l e f t by) the n sfrings
-
t h i s w i l l give a loop i n
q a t Q and hence an element of the fundamental group
f o r a l l loops
e . We now have the following collection of propositions.
(i) I f t1 and i12 a r e loops i n p a t P , which are.homotopic ( r e l a t i v e t o P),
then c l e a r l y ill;
and t,; a r e homotopic in q ( r e l a t i v e t o Q).
Hence 0 is
a single-valued mapping. (ii)
I f t1 and ilz a r e loops i n p a t P, then the product loop i12
the property t h a t
. k 1 has
88
R 4
m
(n.2 .P, 1)O
(n. 2 ;).(a,;>.
=
i s a homomorphism.
Hence
G is
(iii)
has an inverse, namely, the
one-to-one and onto, since
pushing up procedure.
Hence we now have t h a t
a
is a r i g h t automorphism
of Fn. (iv) then
I f a ’ denotes a braid on n s t r i n g s and a ’ i s s t r i n g isotopic to a ,
7= 0.
In fact i t i s s u f f i c i e n t t o see t h a t i f a is s t r i n g
a
isotopic t o the unit n-braid, then (v)
If
i s an n-braid, then ? =
a!
definition of the product aa! (vi)
(xlxz
id
Fn
ao
as follows from r e s u l t
7.
This follows from the
.
... xn - = x1x2 .. )a
=
,
xn.
For x1x2
...
loop in p a t P which encircles the points P1, P 2 ,
i s homotopic t o a
..., Pn
once i n the
... s); does the same f o r the points Q,, Q2 ,... , Qn. Hence i t i s homotopic t o x1x2 .. . xn i n q ( r e l a t i v e t o Q). ( v i i ) The evaluation of 5. Clearly clockwise direction.
-
X.U.
1 1
Now (x1x2
for a l l j # i , i+l.
= x.
I
Now
-
xi+lai = x i but -
xiai # x. because 1+1 the loop x.;
1 1
q.
passes in f r o n t of Q
i’
which i s not true of the loop xi+l in
However we how by (vi) that x1
... xi-l(xiq) . xixi+2 ... ‘h
=
xi
... xi-lxixi+lxi+2 ... xn’
BRAIDS AND THE BRAID GROUP which gives t h a t x i y = xixi+lxi
89
.
-1
A l l t h i s holds f o r 1 5 i < n.
(viii)
The evaluation of (q)-'.
A straightforward calculation shows
that --1
-
-
+Ji)
As Bn =
(ix)
<
al,
Xi+l
..., an- 1 > and a
-+
a defines
a homomorphism Bn
-+
Aut(Fn),
i t follows from ( v i i ) and ( v i i i ) t h a t
-
-' f o r a l l i ,
x.a = A. x. 1 1 iu Ai
having the geometric significance described i n the Theorem.
with
r e s u l t can be proved by induction on the length of
This
0 as a word in the
elements
(x)
We have now completed the proof of the Theorem provided we can show
the following lemma, which is of independent i n t e r e s t , holds.
we w i l l have proved t h a t the mapping a
+
For then
0 has an inverse and so i s one-to-
one and onto. 4.8.
LEN.
Suppose t h a t
i s a r i g h t endomorphism of the f r e e group
F(lxl,
..., 51) = Fn s a t i s f y i n g
the conditions
and (x1x2
... xn) B
=
x1x2
... 'n,
CHAPTER 4
90
where A: is an element of F,11 f o r a l l i. J.
Then there e x i s t s an n-braid 0 such t h a t
= 8 and therefore f3
is an
automorphism of Fn. n The proof proceeds by induction on the integer k ( 8 ) =
PROOF.
1
&(Ai),
i=l
where k(Ai) denotes the length of Ai as an element of the free I f R = 1, then
group Fn.
and we define 8' = e.
B = id
Fn Suppose t h a t B0 has been defined for a l l 8 satisfying the above conditions when
(1
S m(> 1).
Now we assume tha t B is such t h a t R
=
m.
We have that the following
holds i n Fn A1 xl,, A i l
. pL2 x2,, 4' ... a s,, s1= x1x2 ... s.
Since the r i g h t hand side of t h i s equality has length n , some cancellations
Two p o s s i b i l i t i e s can occur.
must take place on the l e f t hand side.
some i, as small as possibte, we have e it he r f i r s t l y (a)
... A. x. 5' . x(i+l)p AT11+1 ... ... A. Bi+l'(i+l)p A-i+l .*. 1
=
1lJ
1
t h a t is, Ai+l = Ai xi: Bi+l,
3
where
For
BRAIDS AND ME BRAID GROUP k(Ai)
k(Ai+l)
+
91
1 + k(Ci).
I f neither case (a) nor case (b) occurs, then it follows that A1 =
... = A, = e and hence B = idFn , which is f a l s e .
In case (a) we have that
while i n case (b) we have that
Hence, by the induction hypothesis, we have t h a t : i n case ( a ) , there e x i s t s a uniquely defined n-braid
0 (FB) such t h a t
i n case ( b ) , there e x i s t s a uniquely defined n-braid
(Fi-'s)O such
that (q
-18 ) 0) - =
Ti%.
I f the case (a) occurs f i r s t , then we define 0
-1-0
B = ui (up)
,
while otherwise we define --1
go = Ui(Ui
B) 0
The f a c t t h a t 8' 4.8.
NOTE.
=
.
B now follows from p a r t (v) of the previous proof.
The group with generators
92
CHAPTER 4
and defining relations for li-jl
aiaj = ajai
2
2
and a i ~ i + l ~=i ~ ~ + ~ afor ~ 1as ~ i 2+n-2 ~ can be mapped homomorphically onto the symmetric group Sn.
This is given
by the mapping ai
-+
for 1 s i
(i,i+l)
<
n
and the facts that these transpositions generate Sn and the relations (i,i+l) (j ,j+l)
=
(j,j+l) (i,i+l)
and (i,i+l) (i+l,i+2) (i,i+l) = (i+l,i+2) (i,i+l) (i+l,i+2) hold in Sn.
Hence associated with every element
a
of the above given
group we have a uniquely defined corresponding permutation.
It follows
that the permutation associated with the word U
;1 l1
... a'ikk
is the permutation
We now turn to the problem of showing that the presentation given in 4.8. Note is in fact a presentation of the group of n-braids.
the method given by W.L. Chow.
We will use
The representation of the braid group,
which is given in 4.6. Artin Representation Theorem, is a vital element of this proof. 4.9.
THEOREM. ij=
a.a
a.a j i
The group of n-braids has defining reZations f o r li-jl t 2
BRAIDS AND THE BRAID and aiui+lui
for 1 5 i 5 n-2
= ui+luiui+l
.., un-1'
on the generators u1,u2,. PROOF.
93
GROUP
Let Rn and Bn denote the group with the above given defining
relations and the group of n-braids respectively on the generators
of Rn.
We have already shown t h a t Bn i s a homomorphic image
an-l.
u1,u2,...,
I t remains t o show t h a t the kernel of t h i s homomorphism i s t r i v i a l .
This r e s u l t w i l l be shown t o be t r u e f o r n-1 instead of n. Let Dn be the subgroup of the group
which consists of those
elements whose corresponding permutation (see 4.8.) fixed.
leaves the l e t t e r n
Let
Ni = an-l an-2 and Nn = e.
... a i
for 1 s i < n
Then the permutation corresponding t o Ni maps the l e t t e r n
i < n.
onto the l e t t e r i f o r 1
N1, N2,...,
Ni,
Hence
..., Nn-19 Nn
form a complete s e t of r i g h t coset representatives f o r Dn i n Rn. be an element of
a.
representative of u by
Let
a
Then we denote the corresponding r i g h t coset p (a).
Hence we have t h a t
i f and only i f the permutation corresponding t o a maps the letter n onto the l e t t e r i f o r 1 s i s n.
I t i s worth noting a t t h i s stage t h a t the
above complete set of r i g h t coset representatives (or r i g h t transversal,
as it is a l s o called) has the following so called Schreier p r o p e r t y : Every i n i t i a l segment an-1un-2 an-1un-2
... ai
(with i
5
... u.I of every coset representative
j ) i s a l s o a coset representative.
WTER 4
94
There i s a well known method for obtaining a presentation of a subgroup from a presentation of the whole group.
We apply t h i s method,
which i s known as the Reidemeister Schreier method, t o the subgroup Dn. A more complete and l e i sure ly account of the general method can be found
i n W. Magnus, A. Karnass and D. Sol it a r Section 2 . 3 .
A short account,
which i s similar t o the one given below, can be found i n R.S. Lyndon and P.E. Schupp Chapter 11, Section 4.
F i r s t of a l l we need a s e t of generators for the subgroup Dn.
They
consist of a l l those elements
which are not equal t o the unit element f o r 1 s i s n and 1 s j < n.
By
the definition of right coset representatives we have a t once t h a t a l l the above defined elements y(N. ,u.) are elements of Dn. 1
3
forward calculation shows that i f
is an element of Dn with a l l exponents being +1, then
E.
where Sj-l(a)
=
... u ij1-1- l l1
for a l l j .
Hence using the above notation f o r
y
we have that
Also a straight-
BRAIDS AND ME BRAID GROUP
95
Also
f o r a l l j.
This establishes the f a c t t ha t a l l the elements y(N. u . ) # e 1’ I form a s e t of generators for the subgroup Dn. I t can now be shown t h a t certai n natural preimages of these elements (# e) i n the f ree group F(Iyl,
..., Yn- 1I ) ,
under the mapping defined by yi
generate a f r ee subgroup i n the fre e group. Schreier property.
+
ui f o r a l l i, f reely
This is where one needs the
Details can be found i n R.S. Lyndon and P.E. Schupp
Chapter I Proposition 3 . 7 . or M. Hall Lemma 7.2.4.
The argument goes as
By (*), it is only necessary t o investigate what happens in a
follows.
product of the form
... y(Ni,ui
j
)
. y(Nk ’ukL ) ... .
In f a c t is i s easy t o see (except f o r the above mentioned notational changes) t h a t any cancellation a t the above displayed point cannot e f f e c t
.
This is of course provided they ar e not each equal t o o r uk j II the u ni t element and not inverses of each other.
eith er ui
I f r is one of
Secondly we need a set of defining re la ti ons f o r Dn.
the defining r e l at i ons of the group I$, then every element of the form
r .. where
‘I
However i f we can express
belongs Dn, i s a l so a re l a t ion in I$,.
this element i n terms of the generators y(N.,u.), then t h i s w i l l be one of 1
1
the defining relations of Dn, where we can obviously take
‘I
= 1.
Hence
the defining relations of the group Dn a re obtained by expressing a l l the
Q-IAPTER 4
96
elements
where k = 1 , 2 , . U
u
-’ i j
..., n
ai u
0-l
j
varies over the elements
and r
for li-j
1
2
-1 -1 -1 ~ ai+ ui+l ~ ui ai+l ai
2
with i , j = 1 , 2 , . . . ,
n-1,
i n terms of the generators y(N. ,u.) of Dn.
1 1 We are now faced with the task of actually evaluating these generators
and defining relations i n t h i s p a r t i c u l a r case. i f j = i-1 Ni-l p ( N . a . ) = Ni+l i f j = i
INi
otherwise
and hence
“i u i - 1 Ni ui y(N. u.) = 1’
J
d N i
1-1
Ni;’
ai+l Nf’
N. u. 1 1
Nil
,Ni u. N-1
if j = i a? N - ~ = Ni+l1 i+l = ui
--
‘j
= u
Thus Dn i s generated by the elements a1,u2,
..., un-2
and
where we use the notation
i f j = i+l i f j s i-2 if j
2
i+2
BRAIDS AND THE BRAID GROUP Ai,n = Ni+l for i = 1,2
2 -1 ui Ni+l = un-l
... ui+l
2 -1 ui ui+l
97
... un--11
,..., n-1.
The evaluation of the defining relations of Dn i s a much more tedious task and so we omit most of the d e t a i l s .
To s e t about t h i s task one needs
t o know the following r e l a t i o n s which a r e consequences of the defining relations for N.
1
U;
s: = 'u
k Ni
if k
5
i-2,
E
= kl;
CHAPTER 4
98
Ni
ui:l
= ui
Ni
ui
=
-1 Ni
i s a consequence of the previous relation;
if k N i uE k N i- l ' N i
2
i+2,
E
= +1
Note t h a t these relations were s u f f i c i e n t t o give the form of the generators f o r Dn and these generators are a s e t of f r e e generators of a subgroup of the f r e e group F(Iul,...,
unml1).
Further, from equation (*)
i n the e a r l i e r p a r t of t h i s proof, we have t h a t
which i s given above.
However we a l s o have t h a t
BRAIDS AND THE BRAID
y(Ni,u.-1) = 1
99
GROUP
“i a -1 i - 1 N -1 i - 1 - Ai-l,n 1
i f j =i-l
-1 -1 Ni ui Ni+l
= 1
i f j = i
= u
if j = i+l
d N i
uitl N i l
i
-1 -1 Ni u j Ni
- ‘j
i f j < i-2
-1 Ni u -1 j Ni
= u.
if j
1-1
2
i+2
where we use the f a c t t h a t Ni-l
i f j = i-1 i f j = i i f j # i-l,i
-’.
and the above given r e l a t i o n s f o r N i u j
I t i s now c l e a r t h a t both of
these procedures give the same value f o r y(N. ,TI). 1’
1
We now give a sample of the type of calculation t h a t one can use t o obtain the defining r e l a t i o n s f o r Dn.
The net outcome of a l l these calculations i s t h a t Dn has defining r e l a t i o n s -1 u-l = 1 ‘i j ‘i ‘1 -1
-1 /1 i+lui ui+l
U i+lui
for ( i - j (
ui = 1
2
2
and
for 1 5 i
i,j
5
<
n-2
n-2
100
CHAPTER 4
for i = j + l for i # j , j + I Let Un denote the subgroup of Dn generated by the elements
Then Un is a normal subgroup of Dn, as is shown by the defining relations (**).
Also the above defining relations f o r Dn show t h a t Dn is the
semi-direct product of Rnml and Un. We now consider a homomorphic image of the group Dn. device f o r showing t h a t Un i s a free group of rank n-1. following r i g h t automorphisms of the f r e e group F(txl,
This i s a Consider the
..., s - ~which } ) are
i n one-to-one correspondence with the above 211-3 generators of Dn:
-1 x. a = x. x J j J j + l 'j 'j+l 'j
=
j
xk a j
=
xk
xk A i Y n
=
x i xk xT1 i
-
for k # j , j + l ; f o r i , k = l,...y n-1.
I t can be v e r i f i e d t h a t these 2n-3 automorphisms s a t i s f y relations corres-
ponding t o those given above as the defining r e l a t i o n s f o r Dn.
Hence the
subgroup generated by
i n Aut(Fnml) i s a homomorphic image of D
n'
by
";,n
9
* *
* A-; 1,n
Hence the subgroup generated
BRAIDS AND ME BRAID GROUP
101
i n Aut(Fn-l) is a homomorphic image of
Since <
AiYn,.. .
A i - l , n > i n AUt(Fnml) i s f r e e l y generated by the given
generators, it follows from 1 . 9 . The universal property f o r f r e e groups t h a t the same r e s u l t holds for
Un i s a f r e e group of rank n-1.
We now consider the action of the elements ul,u~,..., This has the form
group Un. -1 uj
for i
=
n- 2 on the f r e e
u
AiYn uj
1,2,...y
=
Aiyn uj by (**)
n-1 and j = l Y 2 , . . . ,
n-2.
Hence, by 4.6. Artin
representation theorem, there e x i s t s an epimorphism
defined by IS$
=
0
for all
ci E
Rn-l,
where 0 is the automorphism of the f r e e group Un corresponding t o u and
The kernel of
$
i s equal t o the c e n t r a l i z e r of Un i n Rn-l.
As 4 preserves
the permutation corresponding t o an element (see 4.1. ( i i i ) and 4.8.),
we
have t h a t the permutation corresponding t o every element of ker 4 is the i d e n t i t y permutation.
Repeated application of the semi-direct product
decomposition of Dn shows t h a t
CHAPTER 4
102
... Un-l
ker 4 5 U2 Us
Un
with representation i n the form of a product being unique, where Ui i s a free group generated f r e e l y by the elements
f o r i = 2 , 3 , . . . , n. A =
A3
Suppose that
... A.1 # e
belongs t o the kernel of i and A . # e with j 3
A-1 A .
3 ~n
A
<
$,
where Ai belongs t o the f r e e group Ui f o r a l l
n, since A belongs t o Rn-l.
Then
= A.
J ,n
which implies that
AT1 A j Y n Aj J
=
A.
1 ~n
since A. i s a word in the generators u ~ - ~ , . . . , u j , while every one of 3 ~n the A 2 , . . . , Aj-l i s a word i n the generators u ~ - ~ , . . . al. , Hence -1 -1 (N.A.N. 1 3 3
. (N.A. J I
NT') , ( N . A . N T ~ =) N. A ~ - 1 Y1 ~ 3 1 1 J j,n j '
NOW N . A N -1 . = A This is a I j ,n I n-l,n which belongs t o the f r e e group Un. consequence of the following relations which are consequences of the
relation (**) : for i = j
A j+I ,n u . A. ' : 0 J l ~ nJ
for i = j + l
=
for i # j , j + l -1 = a
and 'n-1 An-l,n 'n-1
2 -1 n-1 an-1 un-1
=
An-l,n'
BRAIDS AND THE BRAID GROUP
103
Also, by the d e f i n i t i o n of the elements
However, two we have t h a t N . A . N - l belongs t o the f r e e group Un. I I ~ elements of a f r e e group conunute i f and only i f one element i s a power of the other element.
An-l,n
Hence e i t h e r
-1)m
= (N.A.N.
I I 1
f o r some integer m.
-1
Now as we saw above
N . = A. j j,n
N j An-l,n
and hence A. belongs t o Un which implies t h a t j
I
n.
the f a c t t h a t j
4.10.
EXAMPLES.
Hence the kernel of
$
=
n.
This contradicts
must be t r i v i a l and
(1) With every permutation
TT
of 1 , 2 ,
..., n
l e t there be
associated a fixed element N of Bn whose corresponding pennutation is Then the above Let u be an a r b i t r a r y element of the braid group Bn.
TT.
r e s u l t s show t h a t u can be expressed uniquely i n the fonn (with Nid
=
e)
where A. i s an element of the f r e e group Ui f o r every i and 1
permutation corresponding t o a . i s c a l l e d combing the braid.
is the
Putting a braid i n the above given form E. A r t i n C21 i n h i s concluding remarks a s s e r t s
t h a t (he), t h e w r i t e r , i s convinced t h a t any attempts t o carry t h i s out on
a living person would lead t o violent p r o t e s t s and discrimination against mathematics.
He would therefore discourage such an experiment.
104 (2)
CHAPTER 4
The following is a picture of the braid Ai 1
i-1
i
i+l
j-1
,j
j
n
j+l
J . .
7 It is consi-ired to be an --merit of the braid group Bn.
(3) Let Pn be the normal subgroup of Bn which consists of all braids whose
corresponding permutation is the identity permutation.
An
element of Pn
By Example (l), we have that every element of Pn
is called a pure braid.
has a unique representation of the form
where Ai belongs to Ui for a l l i.
Clearly
where Sn is the symmetric group on 1 , 2 , . . . , is the semi-direct product of Pn-l and Un. free group.
n.
It is easy to see that Pn
Hence every Pn is a torsion-
A direct computation shows that
Ar,n Ai,n A-'r,n if s = i
I
if r
<
i
<
s
<
n.
BRAIDS AND THE BRAID GROUP
105
This defines the action of Pn-l on the f r e e group Un. 4.11.
(1)
EXERCISE.
Give a s e t of generators and defining r e l a t i o n s f o r
the pure braid group Pn. (2)
Show t h a t the r i g h t automorphism
%
of the f r e e group 1s
F(Ix l,..., xnl) associated with the pure braid A
TIS
which a r i s e s from the
correspondence defined by 4.6. Artin representation theorem, i s 'X
i
i f i < r or i > s
xr xs x-'r %,s
= '
xr xs
if i = s
. xr . xs-1 xr-1
if i = r
-1 -1 -1 -1 xs . x i . xs xr sx x r
tx r x s x r
i f r < i < s
f o r a l l 1 5 r < s < n. (3)
Show t h a t the centre of the braid group Bn is (u1u2
<
for n
2
(4) Let
... un-2 un-1 )n > Details are given i n J. Birman Proof of Corollary 1.8.4.
3.
A = (ul
A* = (u1u2
(5)
... U ~ - ~ ) ... ( Uun~ 2 ) ... (u 1u 2 ) a 1' ... un-1 )".
Show that t h e r i g h t automorphism
for 1 2 i A-1
I;
n.
ui A =
Hence deduce t h a t 0
n- i
f o r 1 I; i
5
n-1.
is such t h a t
Show t h a t
CHAPTER 4
106
Show t h a t the r i g h t automorphism determined by the element
(6)
... ‘n- 1)htr of Bn i s such t h a t xi (xl ... X y t 1 . . (xl ... xn)-k-l
(a1a2
for i
Xn-+
+.
5
r
and xi
+.
(xl
... xn) k .
Xi-r
. (xl ... xn) -k
r,
for i
where k and r are nonnegative integers with 0
5
r
n.
Show t h a t the r i g h t automorphism determined by the element
(7)
(“i
.*. x
ak-l) +
j
while x . J
+
(xi x
k-itl
of B (n z 2 ) is
n
... xk) . x j . (xi ... xk)-l
for 1 5 i
5
j
5
k
5
n
f o r j < i o r j > k.
J
(8)
Show t h a t Pn.BA/BA i s isomorphic t o the i n f i n i t e cyclic group.
(9)
Show t h a t the commutator subgroup B; is generated by the 3-braids ulail
and a2-1ul.
(10) Suppose t h a t a is an automorphism of the f r e e group F({xl,...,
and
En
xn})
denotes the group of a l l braid automorphisms of t h i s f r e e group,
t h a t is ,
-’
-
x.a = A x 1 i i p *i
for a l l i
and
f o r a l l 0 in
8.
Show t h a t a-lBna is the group of a l l braid automorphisms
of the f r e e group F(ixla,.
., , xna}).
BRAIDS AM) THE BRAID GROUP
107
(11) Show t h a t i n B6 (a1, u u u u u 0 u u u u u u ) = u -1 1 us and 2 3 1 2 4 5 3 4 2 3 1 2 (u1u3u5,
u u u u u u u u u u u u ) 2 3 1 2 4 5 3 4 2 3 1 2
BRAIDS OF BRAIDS.
4.12.
Now
E
= k1
lk
and k = 1 , 2 ,
2
2.
..., n-1.
represents placing each of
lan, lan-l,...,
(k-l)m+l strings
over every one of the lan+l, h + 2 ,
..., (k+l)m s t r i n g s .
For example i n B we have t h a t 4
(2,1)
with
c1
= u u u u 2 3 1 2
e.
Consider the braid group Bm,
fixed positive integers with n
where
=
Define
where m and n are
CHAPTER 4
108
Clearly it i s e a si ly seen by reflection that
f o r a l l k.
We abbreviate
In Bm,
ck(m,l) t o ck f o r k
..., n-1.
= lD2,
the subgroup
is isomorphic t o Bn under the mapping defined by
lk
+
f o r a l l k.
uk
A simple way of seeing t h i s is t o consider Bm a s a group of r ight
automorphisms of the fre e group F((xl,
..., %I)
representation theorem and its proof.
Put
Xk = x (k-l)m+l for k = l , Z ,
..., n.
'(k-l)m+Z
"'
as given i n 4.6. Artin
'lan
Then, it is c l e a r from the Proof of 4.6.,
automorphism of the fre e group F(IX1,.
.., %I)
that the
corresponding t o
ck,
which i s induced by the pushing down process, i s defined by
k'
+ .
5 'k+l
'k+l
+
k'
Xi
+
Xi
f o r i # k,k+l.
This establishes the required isomorphism. Suppose t h at W(cl, given s e t of generators. the braids, t h a t
C2 ,...
,...
u
m- 1) are words in the Then it follows, by looking a t the pictures of In-,)
and w(ul
BRAIDS AND ME BRAID GROUP
WII1, I2,..., w ( y . * , Um-l)
109
commutes with
... w ( ~ ( ~ - l ) m + l , - * ,
We present two examples.
Uh-1)
'.. W("+l"''
Um-l)'
F i r s t l y we consider some braids i n B6.
represents a three-stranded rope, where each strand has two threads. The most c m o n examples occur when
a >
0 and B < 0.
Secondly a four-stranded rope, where each strand has three threads is given by
in B12'
We have taken some of the notation from the Ashley Book of Knots (see there page 23 f o r d e t a i l s ) .
This Page Intentionally Left Blank
111
CHAPTER 5
SOME CONNECTIONS BETWEEN BRAIDS AND LINKS
I f u is an n-braid, then the corresponding l i n k L(a) is obtained from a by identifying Pi with Qi f o r 1 5 i s n.
5.1. EXAMPLES. (1) The l i n k corresponding t o the 1-braid e i s the t r i v i a l knot
(2)
The link corresponding t o the 2-braid e is the t r i v i a l l i n k of two
components
Ql (3) L(ol) and L(o;')
are
Q2
112
QIAPTER 5
respectively, which are both s t r i n g isotopic t o the t r i v i a l knot. (4)
L(ui3) and L(ul)3 are
and
which i s s t r i n g isotopic t o the right handed and the l e f t handed t r e f o i l knots respectively, namely,
and
respectively. BRAID CORRESPONDING TO LINK.
5.2.
W e now aim t o describe the reverse
process, namely, how one can go from a link L t o a corresponding braid u s o th at L(o) i s s t r i ng isotopic t o L. u
-+
We f i r s t analyse the process
L(u)
more precisely.
I t consists f i r s t of a l l i n installing an axis, which is
perpendicular t o yz-plane and Lies behind the braid.
Then Pi i s joined t o
Qi by a f i n i t e polygonal path which passes behind the chosen axis for 1 s i s n.
We then obtain a link which "constantly cir culates around"
the chosen axis i n an anti-clockwise direction as one proceeds down the strin gs and round the back t o the top again
SOME
Here Q,,
CONNECTIONS BETWEEN BRAIDS AND LINKS
113
denotes the end of the f i r s t s t r i n g . Choose an axis R which is perpendicular t o
We s t a r t with a link L.
xz-plane and does not i n t e r s e c t the link L.
Now one can appeal t o an old
Theorem of Alexander (see 5 . 5 . f o r a copy of Alexander's original paper) which says t h a t L i s s t r i n g isotopic t o a l i n k L' which loops the axis II i n an anti-clockwise direction. t h a t s t a r t s from the axis
Finally one "cuts" the link by a fixed plane and such t h a t i t s projection onto xz-plane does
! ,
not pass through any double point.
Then one only needs t o put the braid
i n regular position by straightening out the s t r i n g s . For example i n
A
D
L \ F
E
we need not appeal t o Theorem of Alexander.
We need only cut a t the two
points specified and straighten out the s t r i n g s . the axis i n the following place
A
D
However i f we had chosen
114
CHAPTER 5
then we have to use the procedure given in the proof of Alexander's Theorem to change FA, as it does not loop around R in the anti-clockwise direction
-
Use a string isotopy to replace FA by
the other sides do.
FGA A
D
F
E
We can now cut and straighten to get a braid. A more sophisticated approach to the above described process can be found in J.S. Birman Chapter 2 . In this account we will be mainly concerned with the procedure of going from a braid to the corresponding link. Finally we establish the following result which tells us when the link L(a) is a knot. 5.3.
RESULT.
Suppose that
a
is an n-braid and the corresponding auto-
morphism is given by for 1 where
p
2
is a pennutation of l,Z,
i
2
n,
..., n.
Then the link L(a) has c
components if and only if the permutation 1~. can be expressed as the product of c disjoint cycles. n-cycle PROOF.
In particular L(o) is a knot if.and only if
p
is an
. If one looks at the pushing down process (see 4.6. and its proof)
which defines the automorphism 0 , then one can immediately see that the
SOE
permutation
p
CONNECTIONS BETWEEN BRAIDS AND LINKS
115
occurring above is the same as the pennutation which deter-
mines the end of the i - t h s t r i n g of u f o r 1 s i
5
n.
When one forms L ( u )
by identifying Pi with Qi f o r a l l i , the d i s j o i n t cycles of
11
give one the
components of t h e link L ( u ) . The method given a t the end of 4.8. Note i s on the whole the best way of evaluating the permutation associated with a braid. 5.4.
EXERCISE.
(1) Establish the following r e s u l t s :
Link
Corresponding Braid
Left handed t r e f o i l h o t Right handed t r e f o i l h o t
3
i n B2
-3 u1
i n B2
Granny h o t Square h o t
(2)
Figure eight knot
( a i l a l l 2 i n B~
Borromean rings
(ui' u1)3 i n B3
Show t h a t i f u and
L(uu;')
T
are n-braids, then the links L(T-'uT) and
a r e both s t r i n g isotopic t o the l i n k L ( u ) .
This is the t r i v i a l
p a r t of the deep Theorem of Markov which a s s e r t s t h a t links a r e s t r i n g isotopic i f and only i f they are related t o each other by a f i n i t e number of steps of the above given type.
J.S. Birman Chapter 2 contains the
only generally accessible proof of t h i s theorem. (3)
Show t h a t the links corresponding t o the following braids a r e s t r i n g
isotopic
CHAPTER 5
116
(b) 5.5.
a';
a
a i 2 a1
and
(a 1 a 2a 3a 4a -1 -1a -1 4 )2* 1 -1a3
ARTICLE TAKEN FROM PROCEEDINGS NATIONAL ACADEMY OF SCIENCES
U.S.A.
1923.
A L E N ON SYSTEMS OF KNOTIED CURVES By J.W. Alexander Department of Mathematics , Princeton University Communicated, February 2 , 1923 Consider a system S made up of a f i n i t e number of simple noninteresting closed curves located in r e a l euclidean 3 space.
The curves S may be
a r b i t r a r i l y h o t t e d and linking, but we s h a l l assume, i n order t o simplify matters as much as possible, t h a t each i s composed of a f i n i t e number of s t r a i g h t pieces.
The problem w i l l be t o prove t h a t the system S i s always
topologically equivalent (in the sense of isotopic) t o a simpler system S', where S' i s so related t o some fixed axis i n space t h a t as a point P describes a curve of S' i n a given direction the plane through the axis and the point P never ceases t o r o t a t e i n the same direction about the axis.
An application of t h i s lemma t o the theory of 3-dimensional manifolds w i l l be given a t the end of the communication. I t w i l l be convenient t o visualize the system S by means of its
projection ST upon a plane.
By choosing the center of projection i n
general position, the projection ST w i l l have no other s i n g u l a r i t i e s than isolated double points a t each of which a p a i r of s t r a i g h t pieces actually cross one another.
Wherever a double point occurs, it w i l l be necessary
t o indicate which of the two branches i s t o be thought of as passing behind the other, e i t h e r by removing a l i t t l e segment from the branch i n question or by some equivalent device.
The problem w i l l then be t o trans-
form the figure Sn by legitimate operations into a figure S i which may be
SOME CONNECTIONS BETWEEN BRAIDS AND LINKS
117
thought of as the projection of the desired system S' isotopic with S. Now, l e t L be a point i n the plane of Sn, so chosen as not t o be collinear with any segment of Sv, and l e t LP be a radius vector connecting the point L with a variable point P of ST(.
Then, i f the point P be made
t o describe a broken l i n e of Sv corresponding t o the projection of one of the component curves of S, it w i l l ordinarily happen t h a t a s P moves along certain segments o f the broken l i n e t h e vector LP w i l l turn i n one direction about L, while as P moves along other segments, the vector LP w i l l turn i n the opposite direction.
The figure ST( must be transformed i n such a
manner as t o eliminate segments of the second s o r t .
With t h i s i n view,
l e t us f i x our attention on a segment a of the l a t t e r s o r t .
I f necessary,
we s h a l l cut the segment a up i n t o a f i n i t e number of sub-segments ai such t h a t no sub-segment ai contains more than one crossing point with the r e s t of t h e figure Sn. segments a
i
Then, i f A and B are the extremities of one of the sub-
of a , we may choose a point C such t h a t the t r i a n g l e ABC
encloses the point L and replace ai by the p a i r of segments AC and CB.
Of
course, i f there is a crossing point on ai a t which ai is t o be thought of as passing over (or under) another segment, the new segments AC and CB must be thought of as passing over (or under) such segments of Sn as they may happen t o cross.
I f there i s no crossing point on ai, the segments AC and
CB may be thought of e i t h e r as passing over a l l segments of ST( which they
cross o r under a l l of them, it makes no difference which. ation of figure S.
s= obviously
The transform-
corresponds t o an isotopic transformation of t h e space
Moreover, the transformation replaces the segment ai by a p a i r
of segments f o r which the vector LP turns about L i n the desired direction. By a repetition of the process, the remaining subsegments of ai may be successively eliminated, following which a l l other segments of the type of a may be disposed of.
A t the very end, there w i l l be l e f t a figure : S which
CHAPTER 5
118
may be regarded as the projection of the desired system of curves S'
.
The axis associated with S w i l l be a l i n e through L and the center of projection. I have shown elsewhere (Bull. h e r . Math. SOC., Ser. 2 , 2 6 , No. 8 ,
pp. 369-372.) t h a t every 3-dimensional closed orientable manifold may be mapped upon a 3-space of inversion as an n-sheeted Riemann space ( i n the sense of a generalized Riemann surface) where, instead of branch points as i n the two dimensional case, there e x i s t s a system S of simple closed curves about each of which a p a i r of sheets are permuted.
Since we have
j u s t seen t h a t the system S i s isotopic with a system of the type S ' , we obtain a t once the following theorem: Every 3-dimensional cZosed orientabze manifold may be generated by r o t a t io n about an axi s o f a Riemann surface with a f i x e d number of simple branch points, such t hat no branch point ever crosses the axi s or merges i n t o another.
Thus, the genus of the generating surface remains unchanged
during the rotation.
The branch points of the generating surface trace
out the system S' isotopic with S.
When the surface has completed a
rotation, the branch points w i l l ordinarily be found t o have undergone a permutation. I t i s believed that other applications of the lemma w i l l suggest
themselves i n connection with the c l a s s i f i c a t i o n of knotted and interlacing systems of curves. 5.5.1.
NOTE.
There i s an interesting misprint i n the f i r s t sentence of
the above a r t i c l e .
119
CHAPTER 6
THE GROUP OF A LINK
The group of a l i n k L i s defined t o be
where
C,
(L) denotes the complement of L i n IR3.
In t h i s chapter we w i l l
By the previous chapter, we know
be concerned with evaluating t h i s group.
t h a t L is s t r i n g isotopic t o a link L ( u ) , where u is some n-braid f o r some positive integer n.
We have also seen (3.3. Result i n case of h o t s ) t h a t
L and L(a) are s t r i n g isotopic implies t h a t L and L ( u ) are equivalent links
and hence
G(L)
by 2.8. Theorem, t h a t 2
G(L(u)).
Hence we can concern ourselves e n t i r e l y with the problem of evaluating
-G(L(o)) 6.1.
where a varies over Bn and n i s a positive integer.
THEOREM OF ARTIN AND B I N .
Suppose t h a t u i s an n-braid.
Then
the group G ( L ( o ) ) o f t he l i n k L(u) has a presentation o f t he form < X1'
where
,.., xn
a denotes
determined by u.
-
; x1 = xla
,..., xn
=
-
xnu
>)
the r i g h t automorphism o f the f r e e group F({xl,.
.., xnl)
ConverseZy the group o f every Zink i s given i n t h i s way.
CHAPTER 6
120
(A more sophisticated proof of t h i s theorem
PROOF.
can be found i n J.S. B i r m a n Chapter 2 . )
Let c y l
denote a s o l i d cylinder in IR3, which encloses the braid u and l i e s between z
=
a and z = b (see 4 . 1 . ) .
Let R denote the s t r a i g h t l i n e path joining Q t o P Consider the
(see beginning of Proof of 4.6.). space (&l(u). R
-1
xik
Then
-
-
x.u 1
for 1 5 i
5
n,
by the pushing down procedure as described i n Proof 4.6. Now consider the space T obtained from the s o l i d cylinder c y l by identifying the ends of the cylinder i n the obvious way so t h a t Pi is There e x i s t s a natural continuous
identified with Qi f o r every i. mapping
There e x i s t s , by 2.8.
since _Cr(L(u)) has the i d e n t i f i c a t i o n topology. Theorem, a group homomorphism
and the relations R -1XP..
1
=
-
x.a 1
for 1 5 i
5
n
hold i n t h i s group, since they hold i n T ( $ ~ ~ ( L ( ~ ) ) , Q ) .In the group G
= <
R,xl,
..., xn
; a
-1
x 1R
=
-
x 1u , . . . ,
11
-1
xn %
=
-
xnu
121
THE GROUP OF A LINK
we have t h a t Q
x. Q-l= x.u--1 1
f o r 1 s i s n.
1
This i s because conjugating by Q , as f a r as the elements of the subgroup <
x l , * . . , xn >
of G
are concerned, is equal t o applying the automorphism <
xl,
..., xn >
=
F({xl,
and
..., x n l ) .
We now aim t o show t h a t
G i s usually called the semi-direct product of the groups <
Q
;
-
>
and F({xl,
..., xn}).
I t i s easy t o see t h a t every element of G has a representation of the form I
Qm
. W(Xl,. ..,
Xn)
,
where m i s an integer and w(xl,..
., 5) i s
an element of F(Ixl,.
I t i s not d i f f i c u l t t o see t h a t t h i s representation is unique.
.., xn)). In any
case suppose t h a t Qm
. w(xl, .., xn)
=
e
under the natural homomorphism of G onto n(cT(L(a)), P ) . possible i n $(L(o)) w(xl,.
.. , 5) = e .
i f and only i f m = 0 and
In f a c t the subgroup generated by xl,.
n ( c T ( L ( o ) ) ,P) is isomorphic t o F({xl,.
w(xl,.
.., XJ
.., xnl).
.., xn i n
For a non-trivial word
i n t h i s subgroup could possibly be made equal t o e only by
CHAPTER 6
1 22
pushing it round the torus a number of times. Hence
t o applying a group automorphism t o w.
-
-1
G = ( i l , x i ,,.., xn '* i l x l i l = x1u ,
However t h i s i s equivalent
..., R -1x , R = x un>-
Next we note t h a t i f To denotes the i n t e r i o r of T, then
Now choose a s o l i d torus T' such t h a t To such t h a t T' contains L(a) but P
L
3 T I ,
which i s close t o To, and
TI.
We are now i n a position t o use 2.12. Theorem of S e i f e r t and van Kampen.
s a t i s f i e s a l l the conditions of t h i s theorem.
where $1 and $2 denote the natural injections
Hence
123
THE GROUP OF A LINK
The open s o l i d holed torus +(TI)
is such t h a t
where A denotes an open annulus.
By 2.10. Result and 2.11. Example ( 3 ) ,
we know t h a t
where M and L denote a meridian going once round inside A and a longitude going once round S1 respectively. /
c - - -
-.
Both loops L and M are considered t o be based a t P. We have already determined the group n(gTo(L(u)),P). determine the group IT(C,,(T') ,P) =
TI
(gn3 ( T I ) <
m ;-
,P)
.
I t remains t o
Clearly
>
where m i s a meridian i n C,,(Tt)
based a t P of the form described below
CHAPTER 6
124
Under the continuous mappings cp 1 and $ 2 we have t h a t cpl(L)
-
e
and cp2(L)
qM)-m
* .t
and cpZ(M)
x1x2
as can be seen geometrically.
(cpl)a(L" )@i
=
mB
CY
)h
Hence
and
. (x1x2 ... xn) B
(cp2)lr(L" MB) = a" f o r a l l integers
*'.
9
Now by 2.12. Theorem
and 6, defines (cpl)T and (cp2)a.
of S e i f e r t and van Kampen and 1 . 2 4 . Consequence ( l ) ,we have t h a t G(L(u)) -
<
x1 ,..., \,a,m
-1 ; a x 1R = x,;
,..., R -1SR= xnG,
~ = e , m x1x2 =
... xn
>
which gives the required r e s u l t on using the Tietze transformation (Ggn) twice on II and m. The remaining p a r t of the theorem follows from the f a c t discussed in the l a s t chapter t h a t every link L i s s t r i n g isotopic t o a link of the form L ( a ) , where a i s a braid.
CCNSEQUENCE.
6.2.
-G(L(u))
s <
If a is a n n-braid,
.., xn
X ~ , .
f o r e v e q i w i t h 1 r; i
PROOF.
x1
5
; x1 = x,;,.
-
xla
.., Axi = x . ~ , .. ., Xn = xn ;> 1
n.
We have t o show t h a t xi = xi: =
then
i s a consequence of the relations
,...) P xi = x.u ,...) xn = xna . 1
1 25
THE GROUP OF A LINK
This follows, because we know t h a t (x1x2
... xi ... xn) a = x x2 ... xi ... xn = (X1G)
(x,i)
always holds f o r a l l braids u, since F({xly * 6.2.
*.
%I)
9
0 i s an automorphism of
*
EXAMPLES.
G(L(e))
... (xi;) ... ( x j )
(1)
The t r i v i a l knot L(e), where e
= <
x1 ; x1 = xle >
= <
x1 ; x1 =
E
B1.
x1 >
One of the best known of the deeper r e s u l t s of Knot Theory a s s e r t s t h a t i f the group of a h o t i s the i n f i n i t e cyclic group, then the knot is
I t i s due t o C.D. Papakyriakopoulos.
the t r i v i a l h o t .
D. Rolfson
Chapter 4 gives a readable account of a proof of t h i s r e s u l t . (2)
The r i g h t handed t r e f o i l h o t L(o;’). -3) )
G(L(ol -
5
<
x1,x2 ; x1 = x u - ~> 11
-1 -1 = < x 1 , x 2 ; x 1 = x 2 x1 x2 x1 x 2 > = < X1’X2
E
B3
; x1x2x1 = x 2x 1x 2
>
CHAPTER 6
126
(3) The l e f t handed t r e f o i l knot L(u;).
x1,x2 ; x1x2x1
= <
(4)
=
x 2 x 1x 2 >
-3 -3 The Granny knot L(ul u2 ) -3 -3
-3 -3
G(L(ul u2 ) ) 5 < x1,x2,x3 ; x2 = x2 u 1 u 2 -
x2'1 '3'1
3'
=
'3'1
-3 -3 '2
>
-3 -3 = a- x i a-
-3 . x1 . a x1a , where a = x-1 x-1x x x 3 2 323-X2u2,
-3 -3 '2
. x 2 . x3x2x 3'
1 1 1
u2
= x-lx-lx-l
2
-3 -3 Hence G(L(ul u2 ) ) 5 <
x1,x2,x3,a ; x2
=
a-1x1-1a-1 , x l . a x1a,
x2x3x2 = x 3x 2 x3'
-1 a = x-1 3 x 2 x 3x2x3 "
As a consequence of the defining relations in t h i s group we have that a = x2. Hence, by means of the Tietze transformation (Gan) applied t o a , we have that
gL(u;3u;3)) (5)
5 <
x1,x2,x3 ; x1x2x1
=
x2x1x2, x2x3x2 = x 3x2x3 > '
-3 3 The square h o t L(ul u2). -3 3
G ( L ( q u2))
fi
-3 -3) G(L(Ul u2
.
This can be proved d i r e c t l y or alternatively it follows from 17.8. Consequence.
1 27
THE GROUP OF A LINK -1 2 (6) The figure eight knot L((a2 al) ) .
G ( L ( U ; ~2~ ~E ) < x1,x2,x3 ; x1 = x1(a2-1all 2 , x2 = x (a -10 l 2 2 2 1
>
-
-1
= <
x1,x2 ,x3 ; x1 = x1x2x1
E <
x1,x3 ; x1-1x3
> . x3 . x1x2-1x1-1 , x2= x-1 3 x 1x 3
. x1 . x-1x
=
x3x1-1
*
x3
. x1x3-1 >,
by means of the Tietze transformation (Gh) applied t o x2.
(7)
The t r i v i a l l i n k L(e), where e G(L(e)) -
(8)
xl,
5 <
...,
;
-
E
Bn, of n components.
>.
-1 3 The Borromean rings L((a2 al) ) .
.
< x1,x2,x3 ; x2 = ~ ~ ( a i ~x3a =~X) ~~( ,U ~ ~> U , ) ~
G(L((a2 -10,)')) Now
x2(a2 -1a1)3 = x3(a2 -1a1)2 = x;1x1x3 a2-1a1
=
-1 -1 x3 x1 x3x1
-1 . x2 . x-1 1 x 3 x 1x3 '
=
-1 -1 x3 x1 x3x1
-1 -1 . x2 . x-1 1 x 3 x 1x3 a 2 a 1
=
-1 -1 -1 -1 -1 (x3 x1 x3x1x2 x1 x3 x1x3
x3(6;1a1)3
Now substituting
-1 -1 x2 = x3 x1 x3x1 in the r e l a t i o n
. x2 . x1-1x-1 3 x 1x 3
3 gives ,) x3 = X ~ ( ' J ~ ~ U
-1 -1 x3 = x2 x1x2x1
. x3
*
-1 -1 x 1x 2 x 1 x2 '
. x1x2x;l) . x3 . (. ..)-1.
CHAPTER 6
128
Hence c(L(a,lo1)
3
)
-=
<
-1 . x2 . x-1 1 x 3 x 1x3 ' -1 -1 x3 = x2-1x1x2x1 . x3 x1x-1 2 x 1 x2 "
-1 -1 x1,x2,x3 ; x2 = x3 x1 x3x1
*
Suppose that L and L ' are equivalent l i n k s .
Then there exists a
homeomorphism 4 of IR3 onto itself so that +(L)
=
L'.
This implies that
This homeomorphism induces, by 2.8. Theorem, an isomorphism
G(L)
71(CIR3(L))
7(GIR3(L')
P
G(L').
This establishes the following 6.3.
NONEQLJIVALENCE CRITERION FOR LINKS.
If L and L' are l i n k s so that
t h e i r groups are not isomorphic, then L and L ' are nonequivalent links and hence also L and L' cannot be string isotopic.
Although it is well hown to experts in the subject, it is not obvious that no two of the following groups are isomorphic
Once one has established this fact, one can deduce that no two of the following knots are equivalent trivial hot, trefoil knot, Granny knot and figure eight hot. This will be our concern in the next few chapters.
129
THE GROUP OF A LINK
Finally we note the following nonobvious f a c t s (see Chapter 1 7 . Chapter 8 of D. Rolfsen i s a good reference).
(i) The r i g h t handed and the l e f t handed t r e f o i l knots are equivalent but not s t r i n g isotopic h o t s (the f i r s t r e s u l t i s e a s i l y obtained by means of a reflection
-
look a t it in a mirror).
( i i ) Although the Granny h o t and the square knot have isomorphic groups, they a r e nonequivalent knots. 6.4.
EXERCISE.
Suppose t h a t u i s an n-braid.
Show t h a t the link L ( u ) is
s p l i t t a b l e (see Chapter 3 f o r definition) i f and only i f L(u) i s s t r i n g isotopic t o a link of the form L(wl.w2), where w1 w2 =
W ~ ( U ~ + ~ , . . . ,u n-
=
wl(u
1) f o r some integer i such t h a t 1
2
l,..., i
2
ui) and
n-3.
Show
that
G(L(wpp
= G(L(wl))
* G(L(w2)).
The converse of t h i s r e s u l t a l s o holds, but the proof of it i s not easy (see C.D. Papakyriakopoulos Section 2 7 ) .
In f a c t i f G(L(u)) i s a (proper)
f r e e product, then L(u) is a s p l i t t a b l e link.
This Page Intentionally Left Blank
131
CHAPTER 7
GROUP RINGS
We are interested i n being able t o introduce d i f f e r e n t i a t i o n i n t o a group.
as;(") d t o hold.
In p a r t i c u l a r we w i l l want something l i k e the product rule = d u m v + u -dv-
az
dx
The product given here w i l l correspond t o the group product.
However we w i l l have t o introduce an addition t o make sense of t h i s formula.
We do t h i s by embedding an a r b i t r a r y group G i n a ring called
i t s group ring EG. The elements of ZZG consist of a l l formal f i n i t e l i n e a r combinations
of elements of G with integer coefficients.
Two elements of ZZG are said
t o be equal i f and only i f corresponding coefficients are equal, t h a t is,
i f and only i f n We define
g
=
m f o r a l l g. g
132
CHAPTER 7
One can verify t h a t ZZG with t h i s definition of addition becomes an I f multiplication i n ZZG is defined by
additive abelian group.
then one can verify t h a t ZG becomes an associative ring with u n i t element l . e , where e i s the unit element of G. The ring ZZG has a multiplicative group contained inside i t , which consists of a l l invertibZe elements of ZZG. called units of the ring ZZ G.
These elements a r e also
We denote t h i s group by U(ZZG).
Now there
e x i s t s a natural isomorphism of G i n t o U(ZZG) which i s defined by g
-+
lg
for a l l g
E
G.
We frequently write g instead of l g f o r g i n G. The ring ZZG also contains the subring Ine ; n
E
721,
where e is the unit element of G. the ring ZZ of integers.
Clearly t h i s subring i s isomorphic t o
We frequently write n instead of ne and i n
particular 1 instead of l e . Clearly ZZG is commutative i f and only i f G is commutative. 7.1. (2)
(1) ZZ < e >
EXAMPLES. Z2. < t ;
-
>
act, t-ll,
t-l with integer coefficients.
element of ZZCt, t-'1 ta
. f(t)
ZZ.
which i s the ring of polynomials i n t and I t is easy t o see t h a t every non-zero
can be expressed uniquely i n the form
y
where a i s an integer and f ( t ) i s a polynomial i n t with integer coefficients
GROUP RINGS
and nonzero constant term.
-
133
I t i s c l e a r t h a t the invertible elements of
is an integer.
ZZ
t ;
(3)
Let G be the f r e e abelian group on the f r e e generators
>
are the elements
where
kt',
CY
This group was considered i n 1 . 1 2 Example ( 4 ) . We write G multiplicatively, so t h a t every element of G has a unique representation of the form ml m2
tl
t2
m
... t . i ... tk% , 1
where every mi i s an integer.
Now the group ring ZZG i s isomorphic t o
the ring of polynomials
in
tll,...,
with integer coefficients.
Every nonzero element of ZZG
has a unique representation of the form
ty
t;2
where ml, m2,
... t?
f ( t , , t2
..., mk
a r e integers and f ( t l , t 2 ,
,..., t k ) ,
..., tk with integer
tl, t2,
... , tk)
is a polynomial i n
coefficients and nonzero constant term.
The
invertible elements of t h i s ring Z G a r e the elements of the form ml m2 k
tl
t2
m.
... ti 1 ... tk"k
where every mi is an integer.
, They form a group isomorphic t o the d i r e c t
product <-1> x G. (4)
Suppose t h a t m i s an integer
2
2.
has a unique representation of the form
Then every element of ZZ
< t
; tm>
CHAF'TER 7
134
where tm- e = 0. Given a group homomorphism 6 : G + H
one can define the corresponding ring homomorphism $ZZG : ZZG
+
ZZH
I t i s easy t o verify t h a t $ z G i s single-valued and preserves both the
operations of addition and multiplication. Clearly $EG i s onto i f and only i f
$
i s onto.
the kernels is dealt with in the following r e s u l t . notation:
L(
The r e l a t i o n between Here we use the
) stands f o r the ideal generated by the enclosed elements i n
the ring under consideration. RESULT.
7.2.
If
$ : G +
ker $ z G = I ( k e r PROOF.
$
-
H i s a group homomorphism, then
e ) in ZZG.
Suppose t h a t g belongs t o the kernel of
$,
t h a t i s , $(g) = e.
Then $zG(g
Hence g
-
- e)
=
$(g) - $(e)
=
e
-
e belongs t o the kernel of
e = 0. $ ZZG.
As the kernel of a ring homo-
morphism is an ideal of the r i n g , one has t h a t
-I(ker
$
-
e)
5
ker
+zG
GROUP RINGS
NOW suppose t h a t
$aG'
1 ng g
135
belongs t o the kernel of the ring homomorphism
Then
Let K denote the kernel of the group homomorphism
$.
Then f o r every g i n
G we have t h a t the above equality implies t h a t
Hence, since gk - e = (g
1 ngk(gk)
kcK
-
e) + g( k
=
C ngk(g - e ) kcK
=
n g(k - e ) . kcK gk
+
- e),
we have t h a t
1 ngk g(k - e) + 1 n e keK ktK gk
c
Let R denote a s e t of l e f t coset representatives of K i n G , t h a t is, G is the d i s j o i n t union of the l e f t cosets gK, where g varies over R.
Then we
have t h a t
This gives t h a t
This r e s u l t can be naturally expressed as saying t h a t i n going from group G t o group H we are putting k = e
for a l l k i n K ,
while i n going from group ring ZZG t o group ring ZZH we are putting
136
CHAPTER 7 k - e = 0
f o r a l l k i n K.
Here we are considering the case when $ is a mapping onto H. Subsequently we s h a l l denote the ring homomorphism $ E G by 7.3.
(1) Let G be an a r b i t r a r y group.
EXAMPLES.
homomorphism G
i s given by
E
-t
e > by
<
1n g [gic
1
=
derivative.
1 ng. geG
This homomorphism i s called the
I t plays a v i t a l role i n the definition of a
The kernel of the ring homomorphism
i s called the augmentation i d e a l . AG =
L(G -
Denote the t r i v i a l
The corresponding ring homomorphism
E.
augmentation homomorphism.
$I.
E
is denoted by AG and
By the above r e s u l t we have t h a t
e).
In f a c t it i s easy t o see t h a t every element of AG has a unique representation as a f i n i t e sum of the form
Let G be an a r b i t r a r y group. G onto G / G ' ,
Then there is a natural homomorphism
where G ' i s the commutator subgroup of G.
The reason
f o r applying the ring homomorphism 2 is t h a t it maps the group r i n g BG, which i n general i s non-commutative, onto the group ring ZZ (G/G')
is a commutative ring.
-I ( G ' - e) (3)
, which
The kernel of the ring homomorphism 2 i s
i n EG.
Let G be an a r b i t r a r y f i n i t e l y generated group.
homomorphism e from the f r e e group F({xl,
Then there e x i s t s a
..., x j * * * *xn)) y
onto the group G.
GROUP RINGS
137
This gives the ring homomorphism onto
e : Z Z F ( I X ~ . . . . ~xnl)
ZZG.
-+
I t follows from the above r e s u l t s t h a t i f G has a group presentation
(4)
of the form
xl....,
<
xn ; rg(8 E N) >.
then the kernel of the ring homomorphism 8 : ZZF({xl.....
xnl)
+
ZZG
is the ideal generated by a l l elements of the form
rg
-
with B
e
E
N.
To see t h i s one has t o make use of the e q u a l i t i e s -1
x - e =-x xy
-
e = (x
-
-1
e)(y
(x
-
- el
.
e)
x
+
-
e + y
-
f o r a l l elements x and y i n the
e
free group.
W e note the following information concerning the group r i n g of the group G/G' when G i s the group of a link. RESULT.
7.4.
Suppose t h a t the link L(u) has c components and G is the
group G(L(o)) of the link L ( u ) .
By 6.1. Theorem of Artin and Birman, we know t h a t
PROOF.
G = G(L(o))
where
Then
G
<
x19..
., xn
-
; x1 = x1uy.
is a braid with n s t r i n g s and
..
xn
=
-
xnu
>
.
138
CHAPTER 7
with Ai being an element of the free group F({xl,..
..., n.
is a pennutation of l,Z,
., %I)
G/G'
<
2 <
il
3
; x1 = xlp
)..., x
-'
;u v iC
uv
xy ( a l l x and y) >
for a l l u and v >
from the c d i s j o i n t cycles (one from each).
on group rings.
is a product of c
,..., xn = sp, x-ly'l
by means of Tietze transformations, where i l , iz,.
group of rank c.
p
Hence
xl,..., x
p
Now we have seen t h a t , since L(o) i s a
link with c components, by 5.3. Result, we have that d isjo i n t cycles.
for a l l i and
,
.., ic are integers taken
Hence G/G' is a free abelian
The remainder of the proof i s given i n 7.1. Example (3)
139
W T E R 8
DER I VAT I VES
Let G be an arbitrary group.
A mapping D : Z Z G
+
E G i s said t o be
a d e r i v a t i v e i f and only i f (1) D(f+h) = Df + Dh and
(2)
D(fh) = (Df).(Eh) + f.(Dh) (product r u l e ) .
where
E
i s the augmentation homomorphism, f o r a l l f and h i n E G as given i n
7.3. Example (1).
Note t h a t i f h belongs t o G , then ( 2 ) reads D(fh) = Df + f(Dh). Besides the ttzero" derivative, the following mapping i s
EXAMPLE.
8.1.
always a derivative Df
=
f
-
~ ( f ) for a l l f in ZZG.
For
f+h -
D(f+h) = f+h =
since
E
E(f+h) ~ ( f )- ~ ( h ) ,
is an additive homomorphism.
D(f+h) = Df + Dh. Also
D ( f h ) = fh =
- ~(fh)
fh - s ( f ) . a ( h ) ,
Hence
CHAPTER 8
140
since
E
i s a multiplicative homomorphism.
Further
(Df) (Eh) + f(Dh) = (f-E ( f ) ) (Eh) + f (h-E (h)) fh - ~ ( f ~ ) (h).
=
Hence axiom ( 2 ) f o r derivatives follows. CONSEQUENCES OF AXIOMS.
8.2.
f o r a l l integers n ;
(ii)
D(n e ) = 0
(iii)
D(g-I) = -g-I(Dg)
PROOF.
(i)
f o r a l l g i n G.
i s a well hown consequence of the f a c t t h a t D i s an additive
homomorphism. (ii)
D(e) = D(e.e) = (De) + (De) by the product rule.
Hence De = 0.
This r e s u l t together with ( i ) shows t h a t (ii) holds. ( i i i ) 0 = De = D(g-'g) follows 8.3.
=
Dg-I + g-l(Dg) by the product rule.
Hence ( i i i )
.
EXAMPLES.
Here we assume t h a t we are working i n a group ring ZZG
with a given derivative D.
where x and y are elements of the group G. (2)
Let the group G be generated by the elements g l y g 2 y , , . y
normal subgroup of G generated by the elements
and K be the
DERIVATIVES
141
for a l l i < j .
gig;'
Then i n the group G/K we have t h a t 81 Let
@
=
82
=
=
- * *
gn.
denote the natural homomorphism from ZZG onto ZZ (G/K)
$ ( D x-1y-1x y )
= 0
f o r x,y
E
.
Then
G I .
This follows from the previous example.
One a l s o writes
f o r a l l x and y i n the commutator subgroup
GI.
Hence one has by the
product rule, t h a t (h)gl=
... =&
=o
f o r a l l elements u belonging t o the subgroup G" (3)
By repeated use of the product rule one gets t h a t xn-1 Dxn = yq-
where
.
x- 1
Y
is an abbreviation f o r
1 + x + x2 + -x-l - .-2
0
-
... + xn-l i f n is a positive integer ... - x" i f n i s a negative integer
ifn=O
Here x i s an element of G and hence x-1 need not be i n v e r t i b l e i n ZZG. the above formula is not t o be interpreted as involving division by x-1.
So
142
CHAPTER 8
(4)
Let r and s be elements of a group G.
r
=
s
If we impose the relation
on G ,
then we go over t o another group H.
-1 (D(rs l ) r = s = (Dr
-
rs DS),,~
= (Dr
-
Ds)~,~
Now i n ZZH we have that
-1
= (D(r-s) 1r=s.
(5)
Let G be a group and RG denote the s e t of a l l derivatives of G.
know that
CG
i s nonempty.
We turn Ilc, into a r i g h t ZZG-module
EG
We
by means
of the following definitions (D1 + D2)f = Dlf ($)f
=
+
(Df)h
for a l l D1, D 2 , D i n
CG and a l l f , h i n
ZZG.
The verification of t h i s i s
We only verify t h a t Dh i s a derivative.
simple but tedious. $(f1
D2F
+ f 2 ) = D(fl + f 2 ) . h =
(Dfl + Df2).h
=
(Dfl).h + (Df2).h
=
$fl
+ Dhf 2
Dh (flf2) = D(fl.f2).h = ( ( D f l I ( ~ f 2 )+ fl(Df2))h =
(Dfl).h.(€f2
=
(Dhf l ) . (Ef2) + fl(Dhf2)
for a l l f l and f 2 i n ZZG.
+ fl(Df2).h
DER1VAT1 VES
8.4.
FREE DERIVATIVES.
143
We w i l l construct some specially simple derivat-
ives on the group ring ZF({xl,.
. ., x j ,. .., xn))
of the free group F(Cxl,
..., x j p * * ' 9% I ) ,
which we w i l l a l s o denote by
ZZF and F respectively. k~ Let II xis be an a r b i t r a r y (not necessarily reduced) word i n the f r e e s=l s group F, where every
for a l l j .
E
We define the f r e e d e r i v a t i v e s by
= kl.
For example
a (x-1x x x x x
_.
axl
a ax
S
1 2 3 1 1 2
) = -x-l + x;1x2x3 + x1-1x2x3x1
1
.
can be extended t o ZZF by defining
j
LEIvNA.
8.5.
a ax
The mapping
i s a single-vaZued d e r i v a t i v e on TZF such
j
that
ax. 2= ax
-1 axi and - = -x-l b i j ax i j
bij
j
PROOF.
(i)
af + a (f+h) = ah ax ax ax j
j
for a22 i and j .
for all j
j
and a l l f , h i n ZZF, by definition.
I f gl and g2 are elements of F, then
CHAPTER 8
144
ax. a
by definition of
we have t h a t
I
a ax. (g1g2) 1
=
Ql ax. 1
81
+
Q2 ax j
(see also (iii) below).
’
Hence
which establishes the product rule. (ii)
2 ax
is single-valued.
For i f axixilb is an element of F, then
j
a (axixi-1b) ax
=
j
aa -1 ab ax + ax.x i i
aa + ax j
a
+
a - a x . x ~ ’ + ax.x.-1 ab 11 1 1 ax
- -a x , (a b) I
if
j
j
j
for a l l j
if i
=
j
DERIVATIVES
145
Similarly we have t h a t
a (ax:'x.b) ax 1
j
axa
Hence
=
1
a (ah) -
for all j .
ax
j
i s single-valued.
j
(iii) Clearly
-1 ax.
axi = ax.
6.
ij
I
8.6.
EXAMPLES
(1)
2 ax.
and
3
axa (X-1y-1x y )
=
by 8.3. Example (1), i n ZZF({x,yl).
=
-x-l 6 . . f o r a l l i and j . i ij
x-1(y-1- e ) g + x -1y-1(x - e ) Hence
Further a2 -1 -1 - (x y x y )
=
ax2
a2 -1 -1 ayax (x Y x Y)
a2
-(x
-2+ x-1
= -x
= 0
-1 -1 Y
-1 -1
-1 -1 -1 -1 = y xy) = - x y + x y
aY2 a2 axay (x-1y-1x y )
(2)
=
-x -1
+
.
x -1y -1 + x -1 = x -1y -1
I f f is an element of ZZF and x . does not occur i n f , then I
-a f - - 0. ax j
2,
146
CHAPTER 8
Let h. be an element of ZZF f o r j = l,Z, 1 unique derivative D on ZZF such that (3)
Dx. J
=
h
..., n.
Then there e x i s t s a
for every j .
j
For define
for a l l elements f of ZZF. n Dxi=
Then
axi
1ax hJ.
j=1
=
j
h.1
for a l l i and D i s a derivative by 8.3. Example (5).
Hence we have shown
that every derivative on ZZF can be obtained i n the above simple way from the free derivatives
)...., -a .
a
a -
axn
axl
We now explain how an element of EF can be recovered from its free derivatives.
THEOREM (Fundamental Formula).
8.7.
EF = EF({xl,
..., xn}). n
f-e(f) =
c
j=1
PROOF.
af
Let f be an eZement of
Then
(Xj
-1).
J
€S
Let f be a group element g = rl x. s=l
of F.
Then
DERIVATIVES
9(xj
Hence
12 ( X j J
[n
1
-1) =
- n
x’:
i = j ssr
r
j
s
s
XI’]
147
and cancellation gives t h a t
s
-1) = g - 1 .
J
Thus i f
f =
1
n g , then
grF
af
J =
c ng ( g - 1 )
g =
c n g g - 1 ng
g = f
8.8.
EXAMPLES.
-
€(f).
af -
(1)
0 f o r a l l j i f and only i f f = n e f o r some
I
integer n.
For i f f = n e , then we already have, by 8.2. (ii) , t h a t
every derivative of f i s zero.
-a-f - o ax
I f , on the other hand,
f o r every j ,
j
then, by the Fundamental Formula, f = E(f) which is an integer multiple of e. (2)
h e can apply the Fundamental Formula t o af axk
.
This gives t h a t
n
f o r every k.
Hence i f one s u b s t i t u t e s i n the Fundamental Formula one gets
CHAPTER 8
148
that
f or a l l elements f of Z F . - -
0
-
In p a r t i c u l a r one can show t h a t
f o r a l l j and k
a x jaxk
i f and only i f n
1
f = m +
mkxk
k=1
f o r some integers m,m1,m2,...,
mn.
I t i s obviously possible t o give a
general version of t h e analogue of Taylor's Theorem.
Hence it i s possible
t o give necessary and s u f f i c i e n t conditions f o r an element f of Z F t o be such t h a t a l l i t s fr e e derivatives of order
i s an integer (3)
2
2
q a r e equal t o zero, where q
1.
I f u is an element of the comutator subgroup F' of the f r e e group F,
then u = l +
1
a2u (x. - 1)(Xk - 1 ) . J
j ,k=l axjaxk
For i f u belongs t o F ' , then
where every gr and hr belongs t o F.
by the product formula, and also
Now i f D i s a derivative on Z F , then
149
DER I VAT1VES
Hence E(DU) = 0, since €(D(gs,hS)) = 0 f o r a l l s.
Hence, by the previous
example, we obtain the required r e s u l t . I f the a r b i t r a r y group G be generated by the elements ga,
(4)
is, G
=
fa, a
E
gay ~1
E
ci
E
M y that
M >, then f o r every element f i n ZZG there e x i s t elements
M y i n ZZG such t h a t a l l but a f i n i t e number of the elements f a are
equal t o 0 and f = E(f)
1 fa
+
(g, -1).
a
For the element f the form g
-
-
E(f) is a f i n i t e l i n e a r combination of elements of
1 with integer coefficients, where g belongs t o G.
kl k2 g = ga ga
1
kS
*"
2
where every kl
NOW
ga
Y
S
,..., ks
is an integer.
Hence i f one applies the following
i d e n t i t i e s , which hold i n ZZG, a f i n i t e number of times a-'
-
1 = -a -1( a - 1 )
a b - 1 = (a-l)(b-1) + (a-1) + (b-1), where a and b belong t o G , then one obtains the required r e s u l t . (5)
Suppose t h a t h j (xj
- 1) =
0
I
i n the group ring ZZF of the free group F({xl,.. every h . belongs t o ZZF. 1
., x j y * * *%y I ) ,
where
Then h . = 0 f o r a l l j . 3
Suppose contrary t o the above assertion, not a l l h . are equal t o 0. I
CHAPTER 8
150
For every element h which i s not equal t o 0, choose a word w . (belonging j' 3 t o F) of maximum length so t h a t w actually occurs (has nonzero j coefficient) i n h . when it i s written as a linear combination of elements I of F. Let R denote the maximum of the lengths of a l l the words w If j' there e x i s t s an integer j such t h a t the length
R(w.) = R and w. does not terminate i n xT1 when written a s a reduced I I I word i n the f r e e generators of F, then ~ ( w . x . =) R + l . This makes it impossible f o r the given equality t o 3 1 hold i n ZZF. So suppose t h a t f o r a l l integers j such t h a t the length .t(w.) = R we have t h a t w. terminates i n xT1 when written as a 1 3 1 reduced word i n the free generators of F. Then when one multiplies out, one gets (hjXj
-
h j ) = 0,
3
where now the words of maximum length t occur (have nonzero coefficient) i n the sum
Hence they cannot cancel out and we a r r i v e a t a contradiction. (6)
The above Examples (4) and ( 5 ) enable one t o give an alternative
definition o f the free derivatives
ax a
f o r 1 s j s n.
I f f is an
j element of Z F , then f
-
E(f) =
1 f.(Xj - l ) , j=1 J
where the elements f . are uniquely determined f o r 1 5 j 1
5
n.
We define
151
DERIVATIVES af ax = j
fj
for every j.
It is now a straightforward exercise to see that these partial derivatives are in fact derivatives and that
This gives us at once the Fundamental FoMrmla and also a statement of uniqueness of this representation.
(7) The Fundamental Formula can be used to give the partial derivatives of
an element of ZZF, by the uniqueness which was established in above Example (6) and using the identities given at end of above Example (4). We illustrate this by working out the partial derivatives of
x-2y-2x2y2 -1
-2 -2
-2 -2
(x y
=
(x-Zy-2 -1) C(X2 -1) (y2 -1) + (x2 -1) + (y2 -1)1
=
-1) (x2y2 -1)
(x y
=
+
-1) + (x2y2 -1)
+
(x-2 -1) (y-2 -1) + (x'2
-1) + (y-2 +1)
+
(x2 -1) (y2 -1) + (x2 -1) + (y2 -1)
L(x-Zy-2 -l)(x +1) - x-Z(x +1) +
[(x
-2 -2
y
-l)(x2-l)(y+l)
- (x-2 -l)y-Z(y +1) +
-
-2 -2 (x y -l)(y+l)
(x2 -l)(y +1) + (y+1)1 (y -1).
aax (x-2y-2x2y2) = (x-2y-2 - 1 q - 2 + 1) (x + 1) x-2 (y-2 - 1) (x + 1).
+
x +11 (x -1)
y-Z(y +1) +
Hence
=
+
+
+
+
-
152
CHAPTER 8
Also
5a (x -2 y -2 x2y2 )
C(x-Zy-2 -1) (x2 -1) + (x-Zy-2 -1)
=
-y-2 + (x2 = x-2y-2 (x2
-
-
(x-2 -1) y-2
1) + 11 (y +1)
- 1)( y + 1),
One can check the answers are correct by the i n i t i a l l y given technique of evaluating f r e e derivatives. 8.9.
(1) Determine a and
EXERCISE.
axl
a
of the words
ax2
where m i s an integer. (2)
Determine a2 -a 2 axi ’ ax2axl
a2 Y-
’
axlax2
a2 -
ax;
-2 x2 -2 x1x2. 2 2 of the word x1 (3)
Determine, by means of the Fundamental Formula, the element w of the
f r e e group F({xl,x21) such t h a t
aw =
_.
axl
aw aX2
=
-1 -1 x-i - i X x-z -xl - x1 2 1 - x1 2 1
-1 x1
+
-1 -2 x1 x2x1
- i Xx-z 2 1 x2
x1
are a r b i t r a r y elements of the group ring ZZFn n of the free group Fn on the s e t of f r e e generators xl,.. ., 3. Show t h a t (4)
Suppose t h a t fly...,
+
and
f
the s e t of simultaneous d i f f e r e n t i a l equations
153
DERIVATIVES
2Y
)...) 2axnY
fl
=
axl
= f
n
Further show t h a t i f y = f and y = h are two
has a solution i n ZZF,.
solutions, then f - h i s a constant. (5)
.., yn))
Let M({yl,.
=
y
denote the ring of a l l formal power s e r i e s i n
the noncommuting indeterminants yl,...,
Show
yn with integer coefficients.
t h a t the mapping x1 -+ 1 + y l , . * . , xn
-+
1 + y,
defines a ring isomorphism of the group ring ZZ F({xl,..
-M.
y
The ring
., xn I )
= ZZ Fn
into
is called the Magnus ring of formal power s e r i e s (see f o r
instance W. Magnus, A. Karnass and S. S o l i t a r 8 5.5.).
Show t h a t the image
of Z F n under t h i s isomorphism contains the ring of a l l "polynomials" i n the noncomuting indeterminants yl,...,
y with integer coefficients. n Let Myi denote the l e f t ideal of y generated by yi, t h a t i s , ;y
y y i = Iy.yi
E
y1
Show t h a t the following d i r e c t sum
f o r a l l i. n
zz e e
yyi
i=1
which contains the image of ZZFn under the above given
i s a subringof
isomorphism into.
Hence it follows t h a t i f f i s an element of t h i s
subring then n f = c +
1fiYi
i=l
uniquely, where c is an integer. af
~a(ri-1)=
fi
f o r a l l i.
One can now adopt the d e f i n i t i o n
154
CHAPTER 8
This is another way of introducing free der ivat i v e s.
of giving a clearer insight into 8.7.
I t has the advantage
Fundamental Fonnula and the analogue
t o Taylor's Theorem as given i n 8.8. Example 2 .
155
CHAPTER 9
ALEXANDER MATRICES
Let <
xl,
..., 5
; rl,
...' rm > Then the Alexander matrix assoc-
be a f i n i t e presentation of a group G.
iated with t h i s presentation is the matrix
where 8 denotes the natural ring homomorphism ZZF(Cxl
,..., xnl)
-+
ZZ < x l ,
..., xn
; rl
,..., rm>
given i n 7 . 3 . Example (4), and 5 denotes the natural ring homomorphism ZZ <xl,
..., xn
; r1
,..., rm>
-+
ZZ ( <xl
,..., 5
; rl
,...
/ <xl
,..., xn
; r1
,..., rm > ' )
rm> /
given in 7 . 3 . Example ( 2 ) . Clearly t h i s matrix depends on the p a r t i c u l a r group presentation which has been chosen f o r - t h e group G. 9.1.
EXAMPLE.
Let k be a positive integer > 1 and kl and k2 be coprime
positive integers > 1 such t h a t k = kl < 2 ; 2%
5 <
x,y ; xkl,
rk2,
.k2.
x-ly'lxy
Then > = G
156
CHAPTER 9
as we saw i n 1.18. Example (1).
The Alexander matrix of the f i r s t group
presentation is the 1 x 1 matrix
[I
... + zk-lJ
z + z2 +
+
,
k where the entry belongs t o the ring 22 < z ; z > with zk -.l = 0.
Alexander matrix of the second group presentation i s the 3
I
1+ x
.t
... + xkl-1 0
1+ y +
-1 -1 - x-l
X
0
... +
:2-j
x
The
2 matrix
)
x-1y -1x - x -1y-1
Y
where the en t r i e s belong t o the ring ZZ ( G / G ' ) which is isomorphic t o the I t i s not c l e a r how these two matrices are related. ring ZZ< z ; z k >. We now investigate t h i s matter. We aim t o define an equivalence o f Alexander matrices so that isomorphic group presentations have equivalent Alexander matrices.
Using 1.17. Theorem
of Tietze, we can reduce t h i s question t o a consideration of the following four cases. (Con) Let e and ZZF({x19..
., xnl)
8'
denote the natural ring homomorphism from
onto E < x1
,..., 5
; rl,..., rm> and ZF({xl
,..., 31)
t o ZZ< x ~ , . . . , 5 ; r l ) . . . , rm,r> respectively, where r i s a consequence of
r l , . . . , rm. r
=
Hence
9 "k JI u ~ ' r. u.
k=l
'k
k'
,
k'
.
where every u. i s an element of F(Ixl,.. , 51)and every ak is an integer. 'k Let A denote the Alexander matrix of the f i r s t presentation
157
ALEXANDER M4TRICES
which is an m
x
n matrix.
second group presentation.
which i s (m + 1) matrix.
x
We now calculate the Alexander matrix of the I t is of the form
n matrix.
We now calculate the (m + l ) - t h row of t h i s
I t i s of the form given below for j = 1 , 2 , . .
9 = ae
l
1
ufl u. p = l k=l k 'k k'
9 =
1 2 el
p=l Now
and
for a l l p.
(.I1 rCIP u
a j
1,
P i~ i~
., n.
-1
Op
P
P
P
since e t ( r i ) = 1 f o r a l l k. k
158
WTER 9
so
Hence we have t h a t the Alexander matrix of the second group presentation i s obtained from the AZexander matrix of the f i r s t group presentation by adding a new row which i s a linear combination of the other rows.
Here we are
making the following obvious identification (which we also use subsequently) : i f f is an element of the group ring ZF({xlY..., % I ) , then a ( e ( f ) ) i s identified with (C&)
a(et ( f ) ) .
Let e and 8' denote the natural ring homomorphism from
ZF({xlY..., xnI) onto Z Z < xlY..., onto ZZ < xl,..
., \ ; rly..., rm-l
5 >
; r l J . . . rm>and ZZF({xlJ..., xnl)
respectively, where rm is a consequence
By the above case of (Con) we have that the AZexander of rl,..., rm-1' matrix of the second group presentation i s obtained from the AZexander matrix of the f i r s t group presentation by removing the Zast row, which i s a linear combination of the preceding rows.
(Gen) Let e and
8'
denote the natural ring homomorphism from
,...
ZZF({xlY..., x I ) onto ZZ< xl,.. ., 5 ;rl,..., rm> and ZF(Cxl , %I) n onto ZZ< xl,. , %, y ; r l y . rmyy-l w > respectively, where y is a new
..
..
symbol and w is an element of the fre e group F({xlY..., % I ) . Alexander matrix of the f i r s t group presentation is
The Alexander matrix of the second group presentation i s
Suppose the
ALEXANDER MATRICES
[ [ $ y-l w]]
since 2 8 ’
=
.
-y -1
159
Hence we have t h a t the Alexander matrix
of the second group presentation i s obtained from the AZexander matrix of
the f i r s t group presentation by adding a row and column of t h e form 0
.... *
*
0
u
,
where u i s an i n v e r t i b l e element of the group ring.
(Gh)
Suppose t h a t < x1
,..., xn,y
; rl
,..., rm >
is a group presentation
such t h a t rm is of t h e form y-lw, where w i s a word i n the f r e e group F({xl,.
.., 5)).Then the resulting Tietze transformation
(Ggn) applied
t o remove the generator y can be expressed as the product of a f i n i t e number of Tietze transformations (Con) and (C6n) t o give the presentation
..., xn,y ; ri, ..., rmt >, where ri, ..., r;-l a r e words i n F({xl, ..., xn)) and rm= rm’ as before, followed by t h e Tietze tranformation <
xl,
of the form x1
,..., xn,y ; ri ,..., r’m >
<xl
,..., 5 ; ri, ..., ri-1 ’
<
goes t o
9
where rmt is of form y-lw with w being a word in the f r e e group F({xl,
..., xnl).
In t h i s special case of (Gb) we have t h a t (as shown i n
case (Gen) above). t h e Alexander matrix of the second group presentation i s obtained from the Alexander matrix of the f i r s t group presentation by renmving the l a s t row and column, which i s of the form
WTER 9
160
0
0
*
.... * u
,
where u i s an i n v e r t i b l e element of the group ring. 9.2.
Let A1 and
EQUIVALENCE OF MATRICES.
% be matrices with entries
from a fixed commutative ring R with unit element 1. be equivatent to
if and only if
Then A1 is said to
can be obtained from A1 by a finite
number of elementary opemtions of the following form or their inverses: (1) Permute the order of the rows; (2)
Permute the order of the columns;
(3) Add to any row a linear combination of the remaining rows; (4)
Add to any column a linear combination of the remaining columns;
(5)
Insert a row of zeros;
(6)
Insert a new border of the form 0
A+
A
' 0
*
.... * u
where u is an invertible element of R. Obviously the above defined equivalence relation is a genuine equivalence relation.
It is clear that a l l the requirements we need to ensure
that isomorphic group presentations have equivalent Alexander matrices occur in this definition,
They assert that the following operations on the group
...,
presentation < xl,
..., rm > each lead to an equivalent
; rl,
161
ALEXANDER MATR I CES
Alexander matrix: (1) Permute the order of the r e l a t i o n s ; (2)
Permute the order of the generators;
(3)
Replace ri by rir, where r is a consequence of rl, r i + l , . * * 9
..., ri-1,
A
'is
rm ;
Replace x . by x.x o r x x . f o r k # j ; I I k k l ( 5 ) Insert the r e l a t i o n e ; (4)
(6)
Insert the new generator y and the new r e l a t i o n u y, where u is a word
i n the f r e e group F(Ixl,
..., % I ) .
There is a b i t of a mystery here.
Where did condition (4) come from?
First of a l l we note t h a t we can add any other conditions we care t o lay down which make calculations with Alexander matrices more easy t o carry out.
In f a c t we have already made use of t h i s principle.
For we a r e only
going t o use equivalence of Alexander matrices i n a negative way, t h a t is, i f the Alexander matrices of two links groups a r e not equivalent, then the corresponding l i n k s a r e not equivalent.
Secondly condition (4) (together
with the other conditions) ensures t h a t the Alexander matrix of a f i n i t e l y presented group F,/K
does not depend on the s e t of f r e e generators which one
chooses f o r the f r e e group Fn. f r e e generators f o r Fn and rl,
In other words, i f xl,
..., rmgenerate
..., xn
i s a s e t of
K a s a normal subgroup of Fn,
then the Alexander matrices
are equivalent, where
CY
i s an automorphism of Fn.
I t i s well known t h a t
the group of automorphisms of the f r e e group Fn i s generated by the following elementary automorphisms of Fn:
162
CHAPTER 9
xj ; xi
(i)
X.
+
Xk, "k
(ii)
xj
+
xj
+
xjxk, xi
+
xi
for i f j ;
+
xkxj , xi
+
xi
for i # j .
1
(iii) x
j
xj
(iv)
-1
+
, xi
+
xi
-+
xi
f o r i # j ,k;
for i # j ;
See for instance W. Magnus, A. Karnass and D. S o l i t a r Chapter 3.
We now
verify t h a t these automorphisms do i n f a c t give r i s e t o equivalent Alexander matrices. In the case of the automorphism ( i i ) above, we have
ari
ax
(Xj
-1) = -
j
ari
ax j
x . ( x -1 -1) J J
and hence
ari
ari
7 = -ax a(xj 1 j
X
j
'
by the definition of free derivatives given i n 8.8. Example ( 6 ) .
Now one
needs t o appeal t o 9.5. Example (1).
In the case of the automorphism ( i i i ) above we have that
and hence
ari -
- ari
_
a(x.x ) I k
ari
ar.
ari
and axk= L axk - ax 1 j
I
ax.
X
j
by the definition of free derivatives given i n 8.8. Example ( 6 ) . Automorphisms given by ( i ) and (iv) have now a l s o been d e a l t with.
ALEXANDER MATR ICES
163
THEOREM (Uniqueness of Alexander matrix).
9.3.
two f i n i t e presentations.
Suppose t h a t a group has
Then t h e i r Alexander matrices are equivalent.
We may assume without l o s s of generality t h a t the two group
PROOF.
presentations are of the form <
x1 ,..., xn ; r1 ,..., rm
where {xl <
and < yl,
..., yq
..., y4 3 are d i s t i n c t
,..., xnl
and {ylY
..., xn
; r l Y . . . ,r
xl,
>
m
>
<
yl,
..., yq
; sl,.
., sp >
s e t s of symbols. ; sl,
NOW
..., sP >
and it follows from 1.17. Theorem of Tietze t h a t the second group present a t i o n can be obtained from the f i r s t group presentation by a f i n i t e number Hence, by the above given calculations and
of Tietze transformations.
definition, we have t h a t the corresponding Alexander matrices are equivalent. I t i s worth noting t h a t we have used some i d e n t i f i c a t i o n s here which
may cause trouble t o the uninitiated. given groups and
@
Suppose t h a t GI and G2 are the two
i s the isomorphism between them.
The Alexander matrices
of G1 and G2 have e n t r i e s i n the rings ZZ(G1/Gi) and ZZ(G2/Gi) respectively. @
induces a ring isomorphism between these two rings.
go from one ring t o the other. 9.4.
EXAMPLE.
This i s best i l l u s t r a t e d by an example.
Consider the group
G = < x1,x2 ; x1x2x1
=
x2 x 1 x 2 > ’
which we have met a number of times already. the groups
This enables one t o
The group G i s isomorphic t o
CHAPTER 9
164 under the mappings defined by x1
3
, x2
y2y1
-+
y2 and x1
-+
z1-1
, x2 -, z2
respectively.
Now G/G'
= <
,
xl,x2 ; x1 = x2 >
G1/Gi
<
~ 1 , ~; YI 2 = e >
G2/Gi
<
z1,z2 ; z1-1
and =
z2
>
.
I f we denote x2,y2 and z1 by t , s , and u respectively, then we have t h a t
and the isomorphisms G/G'
G1/Gi
-f
and G/G'
-+
G2/G2
are given by t
and t
-+ s
respectively
.
We now go on to consider some further operations on Alexander matrices which give equivalent Alexander matrices. 9.5.
EXAMPLES.
(1)
[
... a m ]
A
=
Let A be an Alexander matrix.
If
B
aml
and u i s an invertible element of the r i n g over which A is defined, then B
A-[
uaml
J.
... ua
ALEXANDER MATRICES
165
For
B
A-
a ml 0
B mn
"'
...
0
- am l
1
uaml
amn ua
...
mn
1
B
:::
u
[u:ml
-
B
- [ u a ml ...
ua
1
m n .
Hence i f any row of A i s multiplied by an invertible element, then the resulting matrix i s equivalent t o A. (2)
Let A be an Alexander matrix.
If
and u i s an invertible element of the ring over which A i s defined, then a
In
A-
B
0
a
B
.
In
o
u
In
.
0
o
ua
uamn 0
0
u
CHAPTER 9
166
N
Hence i f any column of A i s multiplied by an invertible element, then the resulting matrix i s equivalent t o A. (3)
If G
= <
x ~ , . . .x~n ; r l , . . . , rm > i s a f i n i t e l y presented group and
u i s an a r b i t r a r y element of the second commutator subgroup F({xl,..
. , xnl)
I ,
then the Alexander matrix of G i s equivalent t o the
Alexander matrix of the group
This is a consequence of 8.3. Example ( 2 ) . 9.6.
THE ALEXANDER MATRIX OF A LINK.
equivalent l i n k s .
Suppose t h a t L(o) and L ( o ' ) are
Then we have shown t h a t there e x i s t s an isomorphism
$T
G(L(u')) (see statements p r i o r t o 6.3.). between the groups G(L(o)) and Now t h i s isomorphism induces an automorphism $*
o f 7Z
[ti', ..., t:']
c is the number of components of the l i n k L(u) by 7.4. Result.
invertible elements of the ring ZZ
*
ml m2 t2
tl
...
mC
tc
[ti',. .., tf']
, where
Since the
a r e a l l of the form
,
where a l l mi are integers, it follows t h a t $ * must be of the form
ALEXANDER MATRICES c $,(ti) = for 1 s i
5
m t ij
with m . .
j =1 j
13
n and detCm..] = 11
f
167
E
1, since every t i i s a group element i n
Hence i n p a r t i c u l a r we have t h a t i n the case when L(o)
QL(u))/&(L(u))'.
and L(a') are equivalent knots the corresponding automorphism +* of ZZCt,t-ll i s of the form
.
$ * ( t ) = t-+1
Thus i f A(L(u) ; t l , .
.., tc) and A(L(o');
tl,.
.., tc) are the Alexander
matrices of the equivalent links L(a) and L ( o ' ) respectively, then the above 9.3. Theorem t e l l s us t h a t
In the p a r t i c u l a r case when L(a) and L ( o ' ) are equivalent knots we have t h a t A(L(a) ; t )
- A(L(a') ; t'l).
We s h a l l l a t e r show t h a t
- A(L(a) ; t ) '
A(L(u) ; t-l)
f o r a l l h o t s L ( u ) (see 18.1. Theorem). substitution t
-+
Hence, except possibly f o r the
t-l, every p a i r of equivalent knots have the r'same'r
Alexander matrices.
The general case of a link i s more complicated.
been proved (see f o r instance G. Torres and R.H. A(L(u) ; tl,..
Fox o r Chapter 1 9 ) t h a t
., tc) - A(L(0) ; tl-1,. .. , tc-1) ' -
We s h a l l i n s i s t t h a t each link has i t s group given i n a fixed way <xl
,..., ' h ; x l = x l u- ,..., xn
-
= xn u > ,
I t has
CHAPTER 9
168
where a i s an n-braid which determines the l i n k , with the elements xl,
..., xn always being determined
the automorphism
0
i n the way given i n the definition of
of Fn (see Proof of 4.6. Theorem).
determines an Alexander matrix f o r each link L(o)
This uniquely
, where
a
i s an n-braid.
Hence i f two links are equivalent, then t h e i r Alexander matrices are related by an equivalence of the form given i n (*). 9.7.
Let B be a r i g h t automorphism of the free group Fn.
EXAMPLE.
Then
the link group G(L(o)) = < xl,
..., xn
;X
~ U . ~ ~ ~ , . x. n.a,x-' , n
>
.. , (xnu.xil)B
>
i s isomorphic t o the group
.., xn
H = < xl,.
*
(xla.xl-1) B , .
'
under an isomorphism induced on the factor group by B , which we w i l l also The above derived theory t e l l s us t h a t the Alexander matrices
denote by B .
of the two group presentations are related as follows:
where c i s the number of components o f the link L ( o ) , can be any element of Aut(Fn). i f Cm..I i s an c 13
ant i s equal t o c
tiB*
=
x
The automorphism B
One can verify by a detailed argument t h a t
c matrix with integer coefficients so t h a t its determin-
t 1, then there e x i s t s a B belonging t o Aut(Fn)
so t h a t
m
II t i j j=1 j
f o r a l l i.
This follows from the well known f a c t t h a t GL(n ; ZZ) i s isomorphic t o the group of automorphisms induced by Aut(Fn) on the free abelian group Fn/FA. Hence one can see t h a t in general it w i l l be extremely d i f f i c u l t t o deter-
ALEXANDER M T R ICES
mine whether two Alexander matrices are equivalent o r not.
169
For t h i s
reason we s h a l l introduce an invariant which i s constructed from an Alexander matrix. 9.8.
EXERCISE.
(1)
< XYY ; X 2 , Y 2 ,
Determine the Alexander matrix of the dihedral group (XYIzn>
,
where n is a positive integer.
(2)
Describe the Alexander matrix of the f r e e product of two f i n i t e l y
presented groups.
This Page Intentionally Left Blank
171
CHAPTER 10 ELEMENTARY IDEAL OF ALEXANDER
MATRIX
Let A be an Alexander matrix which has e n t r i e s from a commutative ring R with u n i t element 1.
I f A is an m
x
n matrix, then we define the
e2ernentary idea2 E(A) i n R t o be the i d e a l generated i n R by a l l determin-
ants of every (n -1)
x
(n -1) submatrix of A.
An (n -1)
x
(n -1)
submatrix of A is a matrix obtained from A by choosing n -1 rows of A and from these rows choosing n -1 columns.
There are a number of special
cases which are not covered d i r e c t l y by the above definition.
However,as
shown below we have: I f n = 1, then E(A) = R = (1); I f m < n - 1, then E(A) = (0).
10.1. EXAMPLE.
(1) Let k be a positive integer.
Then the Alexander
matrix of the group presentation
i s equivalent t o the Alexander matrix
r
l+Xl+
... +XIk-l
01
0
which has elementary ideal (1). < x l ; - > .
A similar argument works f o r the group
CHAPTER 10
172
(2)
Suppose that the group G has a presentation o f the form <
xl,
..., xn
where m < n
-
1.
; rl,
..., rm '
9
Then the Alexander matrix is an m
x
n matrix A.
Now A
is equivalent t o the Alexander matrix
[
'n-m- 1 ,n
]
and i t s elementary ideal i s (0).
THEOREM (Invariance of Elementary Ideals).
10.2.
A are equivaZent Alexander matrices.
2
PROOF.
Suppose that A1 and
Then
We have t o show t h a t each of the operations (1) - ( 6 ) , which
define the equivalence of Alexander matrices, does not change the element a r y ideal of an Alexander matrix.
This follows a t once from properties
of determinants f o r the operations (1) - ( 5 ) . (6) which can be considered t o lead from m
x
This leaves the operation n matrix A t o the
(m +1) x (n +1) matrix
-
0
-
0
B =
A 0
oo...
0u
,
where u i s an invertible element of the ring R. B are of form
The n
x
n submatrices of
ELEMENTARY IDEAL OF ALEXANDER MATRIX
-
r
0
173
01
L1,n-l
-0
...
0
0 u
where An-l,n-l’ An-l,n and *n,n-l are submatrices of A of the appropriate dimension. Now it is easy to see that E(B) = E(A) in this case, since u is an invertible element.
This completes the proof of the theorem.
Once again we have that in general it is extremely difficult to decide whether two elementary ideals are equal or not. a more simple minded invariant.
So
we move on to consider
This Page Intentionally Left Blank
175
W T E R 11
ALEXANDER
POLYNOMIAL OF A KNOT
The elementary ideal of a knot group is p a r t i c u l a r l y easy t o calculate. For suppose t h a t the l i n k associated with an n-braid u i s a knot K ( u ) . Then, by 6.1. Theorem of Artin and Birman and 6 . 2 . Consequence, we have t h a t <
G(K(o))
xl,
..., xn
-
; xlu.xl
- -l ,..., xn-lu.xn-l
-1
Hence the Alexander matrix of t h i s group i s the (n -1) A = [a.
> .
x
n matrix
.],
11
where
- -1 a(x.0.x. 11
x1
for 1 S i s n - 1
=
... = xn
and 1 s j s n.
n
-
-1
j=l =
o
= t
Now, by 8.7. Fundamental Formula,
... = xn = t
-1
in Z Z C ~ , ~ - ' I ,
As ZZCt,t-'l has no divisors i x i p Ai f o r a l l i , by 4.6. n So the Alexander matrix of G(K(u)) i s of zero, we have t h a t 1 aij = 0. since xi:
= A
j=1
equivalent t o the matrix
W ' I E R 11
176
x1 =
. I .
=
= t
On,1
i,j < n-1
This shows that the elementary ideal of G(K(u)) is a principal ideal, namely, the ideal generated by the element
-
-1
det"
L
]
i,j
i n ZZ[t,t-ll.
... = x n = t
x1 = 5
n-1
1
A generator of t h i s ideal i s of the form f ( t ) , where f ( t ) is
a polynomial i n t with integer coefficients and nonzero positive constant term when f ( t ) # 0. of the k n o t K ( u ) .
Such a polynomial i s called the AZexander poZynomia2 Note t h a t the Alexander polynomials of knots are
equivalent i f and only i f they generate the same ideal i n the ring ECt,t
-1 1,
t h a t i s , i f and only i f one i s a multiple of the other by an invertible element i n ZZ[t,t-ll. ZZ[t,t-'I
By 7 . 1 . Example (2), an invertible element of
i s of the form t t", where a is an integer.
I t follows from the
invariance of elementary ideals 10.2. t h a t the AZexander poZynomiaZs of equivalent knots are equivaZent.
Hence if two knots have nonequivazent
Atexaander poZynomiaZs, then they cannot be equiuaZent. 11.1. EXAMPLE.
(1) The t r i v i a l knot is given by the 1-braid e .
Its
group i s
while i t s Alexander matrix is the 1 x 1 matrix COI.
The corresponding
elementary ideal i s (1) and so the Alexander polynomial i s 1.
177
ALEXANDER POLYNOMIAL OF A KNOT (2)
The r i g h t handed t r e f o i l knot i s given by the 2-braid
and, by
6.2. Example (2) i t s group i s <
x1,x2 ; xlul-3 =
XI >
-1 -1 -1 = < x 1 , x 2 ; x1 x 2 x 1 x2 x1 x2 ” Now
a (x-lx-lx-lx x x ) 1 2 1 2 1 2 axl
=
-x-1
-
x-1x -1x-1 1 2 1
-1 -1 -1 +
x1 x2 x1 x2
and -1 -1 -1 -1 -1 a ax2 (x1 x 2 x 1 x2 x1 x2) = - x 1 x 2
+
x-l -1 -1 1 x 2 x1
+
x-lx-lx-lx 1 2 1 2 1 ’
Hence the Alexander matrix i s
[-
t-1, t - 2 - t-3
t-1 - t - 2 + t-3
3,
The Alexander polynomial i s 1 - t + t‘.
(3)
-3 -3 The Granny Knot is, by 6.2. Example (4), given by the 3-braid u1 u2
and i t s group is <
x1,x2,x3 ; x1x2x1 = x2x1x2, x2x3x2 = x3x2x3
>
.
Now we have, by the calculations i n the previous example and by 8.6. Example ( Z ) ,
t h a t the Alexander matrix is
C I M E R 11
178
r" 1
-(1 - t + t 2 )
1 - t + t
-(1
O
-
O1
.
01
t + t2)
The Alexander polynomial i s (1 (4)
-
t +
$I2.
The figure eight knot i s , by 6.2. Example (6) given by the 3-braid and i t s group i s
(ail
x1,x2 ; x1-1x2.x1.x2-1x1 = x2x;l.x
<
x x-1 2' 1 2
>
Now
a (x-1x x -1x -1x x x -1x x x -1) 1 2 1 2 1 2 1 2 1 2 axl
= -x-l
-
x1-1x2x;l + x;1x2x;1x;1
-
x1-1x2x1-1x2-1x1x2x;1 +
+x;1x2x1-1x2 -1x1x2x1-1x2 and
a
-1x x-1x -1x x x -1x x x-1) 1 2 1 2 1 2 1 2 1 2
-(x ax2 =
x1-1 - x -1 x x -1x -1 1 2 1 2
+
x-l -1 -1 1 x 2 x 1 x 2 x 1 - x -1x x -1x -1x x x -1 1 2 1 2 1 2 1
- x -1 x x -1x -1x x x-1x x x -1 1 2 1 2 1 2 1 2 1 2 ' Hence the Alexander matrix i s
[1-
3t-l
+
t-2
The Alexander polynomial i s 1 - 3 t + t2
.
-
1 + 3t-l
-
t-2 ]
.
-
ALEXANDER
POLYNOMIAL OF A KNOT
179
The net r e s u l t of a l l these calculations i s t h a t the following h o t s a r e nonequivalent h o t s : r i g h t handed t r e f o i l knot, Granny knot and figure eight h o t .
Also not one of these three knots can be u n h o t t e d . 11.2.
NOTE.
When working out the Alexander matrices of the above h o t s
one can e a s i l y see t h a t any one of these Alexander matrices i s equivalent t o an Alexander matrix where any column can be assumed t o have a column of This is t r u e i n general for the Alexander matrix of any knot group.
zeros.
The proof of t h i s f a c t proceeds i n a similar way t o the calculation given a t the beginning of t h i s chapter.
I t is a consequence of 8.7. Fundamental
Formula. 11.3.
EXERCISE.
ring Z [ t , t - ’ I .
(1)
Let A be an m
x
n matrix with e n t r i e s taken from the
A r e s t r i c t e d elementary operation (see 9.2. and 9.5. f o r
a more general version of elementary operation) on a matrix A is an operation of one of the following types: (i)
Interchange any two rows o r columns;
(ii)
Add any l i n e a r combination of rows (columns) t o another row (column);
(iii)
Multiply any row o r column by f ta f o r some integer a .
Show t h a t A can be transformed by a f i n i t e number of r e s t r i c t e d elementary operations t o a matrix of the form
0
180
CHAPTER 11
where every di is a polynomial of the form
with a l l coefficien t s being integers and di
divides di+l
Show t h a t dld2 every i
x
... d.
1
f o r a l l i.
is g r e a t e s t common divisor of a l l determinants of
i submatrix of A, f o r a l l i.
ar e called the eZernentary divisors of A.
The polynomials dl,d2,
..., dr
Many books on Linear Algebra
have sections dealing with elementary divisors.
Note t h a t i f A i s n
x
n
Alexander matrix of a knot group <
x1
,..., xn
; s1
,..., sn > ,
then the Alexander polynomial of the knot i s the product dld2 the elementary divisors of the matrix A.
Here r = n -1.
... dnml of
The study of
the products of elementary divisors i n t h i s case was introduced by R.H. Fox. This associates some more polynomials with a knot, which i s of some help i n the nonequivalence problem f o r knots. R.H. Crowell and R.H. Fox.
Examples can be found i n the book by
A l l of these new polynomials a r e of course
f a ctors of the Alexander polynomial i n the ring ZZ Ct,t-ll. (2)
Verify the following r e s u l t s which give another i n t e r e s t i n g way of
looking a t the Alexander polynomial of a knot. ( i ) The normal subgroup N of the f r e e group F(Ixl,. generated by the elements
i s f r eely generated by t h e elements
.., x n l ) , which is
181
ALEXANDER POLYNOMIAL OF A KNOT
m -1 x .x.x n i n with i
-m *
n'
n and m = 0, 2 1,
<
group (F(Ixl,.
... .
Clearly N contains the commutator sub-
. ., xn})) '
(ii) Let a be an n-braid and the l i n k L ( o ) be a h o t .
Let
e denote the
natural homomorphism of the group r i n g of the f r e e group F({x l,..., xnl) onto the group ring EG(L(o)) of the h o t group G(L(a)).
Then
ON = (G(L(u)))'.
(iii)
.xf'
xi:
belongs t o N f o r a l l i and xi;
-
c
-1
a(x.0.x. 1 1 a (x~.xjx~'.x;l")
m,j
.
.xi1
-
1 equals
-1. x-m (x; . x . x - 11 I n n
j zn
i n the group ring Z N , where only a f i n i t e number of the terms w i l l be nonzero, f o r a l l i. (iv)
Let A denote the augmentation ideal of the group ring ZZ
where G
=
G(L(a)) and
= (GI)!.
(G'/G'
') ,
So A i s the kernel of the augmentation
homomorphism
and
(There should be no danger of confusion with the other meaning of A certain type of braid.)
Then the multiplicative abelian group
isomorphic t o the additive abelian group A / A ~ under the mapping
,'/,'I
-
a
is
CHAPTER 11
182
(Note t h a t t h i s r e s u l t holds i f /,"'
i s replaced by any abelian group.)
Here A2 denotes the ideal generated by product of a l l p a i r s of elements of A.
(v)
Let 2' and 6 denote the natural r i n g homomorphisms
A
+
A/A~
respectively.
m,1 1,
€ l a cI i ,m )]
Then f o r a l l i we have
.
I
1=o
x; . x ]. xn . x n - 1 -m
where
f o r a l l i , j (# n) and m.
f o r a l l i , j (# n) ,m.
Here 2 is the usual ring homomorphism of group rings
which make them commutative.
ALEXANDER POLYNOMIAL OF A KNOT
183
( v i i ) Using 13.1. Chain Rule gives t h a t
f o r a l l i , j ( # n),m. G ' / G-" ) Taking SO(%) = t and considering A / A ~ ( : -
(viii)
t o be a ZZ[t,t-'] -
module under the action
for a l l j
<
n and a l l integers m, one has t h a t
- -1 1
j=1
1
for a l l i
5
n -1.
(ix) Let R be a commutative ring with u n i t element and M be a R-module which i s given by generators yl,
P
1
j=1
b . . y . = 0 with i 11 I
and every bij belongs t o R. represented i n the form P k=1
ak yk
=
1,2
..., yP
and defining relations
,..., 9
This means t h a t every element of M can be
CHAPTER 11
184
and the above given r e l a t i o n s determine a l l the r e l a t i o n s which hold i n M - make repeated use of "scalar" multiplication by elements of R and of addition.
The p
x
q matrix [b. .I i s c a l l e d the reLation matrix of the 13
module M ( r e l a t i v e t o the set of generators yl, t h a t the (n -1)
x
Now it follows
(n -1) matrix
is the r e l a t i o n matrix of the ZZ [ t , t - l I
(x)
..., YP' .
- module
A / A ~ :G T /-G t l .
I f f ( t ) is the Alexander polynomial of the knot L(o), then f(t)
. (A/A~) =
~2
o r equivalently f(t)
.
(,/,,)
=
,,I.
See W. Magnus, A. Karnass and D. S o l i t a r Section 3.4. f o r a d i f f e r e n t approach t o the r e s u l t s of t h i s Exercise.
185
CHAPTER 1 2
ALEXANDER POLYNOMIAL OF A L I N K
We now consider the general case of a link.
Let u be an n-braid and
L(u) be the corresponding link with more than one component.
Then, by 6.1.
Theorem of Artin and Birman and 6.2. Consequence, we have t h a t <
G(L(o))
xl,...,
5
-
; xlu.xl
,...,
-1
- -1
Xn-l".Xn-l
'*
Suppose t h a t under the group homomorphism
a e : F({xl,..., xnl)
+
G(L(u)
we have, by 7 . 4 . , t h a t a e x
-
j
= t S(j1
forls j s n
where 1 < 6(j) s c > 1 with c being the number of components of the l i n k Then the Alexander matrix of the group G(L(u)) i s the (n -1)
L(o). matrix
A = [a. . I , 11
where
Now, by 8.7. Fundamental Formula,
x
n
186
CHAPTER 1 2
-1 since x 1. 0 = A.1 x ip Ai , by 4.6., and 6 (ip)
= 6 (i)
f o r a l l i.
In order t o
evaluate the elementary ideal E ( A ) , we have f i r s t of a l l t o evaluate the determinant A . of the submatrix of A, which is obtained from A by deleting 1
the j - t h column of A, for j = 1 , 2 ,
i n ZZ
[ti',. .. , ti+11 by (*).
j and k.
Hence t
So t6(k)-l divides
f o r a l l j and k i n ZZ
.
*
9
6
Now
w-1
nk for a l l k.
[ti', ..., t f l l .
the ideal generated i n ZZ Ct;l,.
Y
...,n .
divides Ak(t6(j)-l)
for a l l
Hence
The elementary ideal E(A) of A is
.. , ti+11 by the elements
'n.
Hence i f we put
then E(A) i s the ideal generated i n ZZ
A.(tl-l),
A.(t2-l),
The element A of ZZCt;',
[ti', ..., ti+11 by the elements
..., A.(tc-l).
..., ti+1I i s called the AZexander poZynomiaZ of
the
Zink L(o) u i t h more than one component. 12.1.
EXAMPLE.
the c-braid e.
(1) The t r i v i a l l i n k with c ( > 1) components i s given by
I t s group is
ALEXANDER POLYNOMIAL OF A LINK
,..., xc
< x1,x2
187
; e >,
while its Alexander matrix i s the 1
x
c matrix Ol,c.
The corresponding
elementary ideal i s (0) and so the Alexander polynomial i s 0. (2)
The group associated with the 2-braid u12 is <
’.
x1,x2 ; x2 = x x x-l
1 2 1
Now
a ( x x x-1x -1) 2 1 2 1 axl
= x 2
a (x2xlx;1x;1)
= 1
-
-1 -1 x2x1x2 x1
and
-
.
x2x1x2 -1
aX2 Hence the Alexander matrix is
It2
-
1
1 - tll.
The elementary ideal is the ideal generated by tl -1 and t2 -1 i n
+1 +1 ZCti , ti 1. (3)
The Alexander polynomial i s 1.
The group associated with the Borromean rings is, by 6.2. Example ( 8 ) ,
isomorphic t o <
-1 -1 x1,x2,x3 ; x2 = x3 x1 x3x1 , x2
x3
-1
-1
= x2 X1X2XL
-1 . x3 . x 1x-1 2 x 1 x2
The Alexander matrix i s equivalent t o
1
- t ; l ( l - t 2 ) (1-t3)
0
x -1x x . x-1 1 3 13’ ”
CHAPTER 1 2
188 A1 = - ( l - t l ) 2 (1-t2)(1-t3) A2 =
-(l-tl)(l-t2) 2 (l-t3)
-1
2
A3 = -tl ( l - t l ) ( l - t 2 ) (l-t3) ,
Hence the Alexander polynomial i s ( l - t + (l-tZ) W 3 ) 12.2.
EXERCISE.
(1) Show t h a t the Borromean rings l i n k i s n o t equivalent
t o a t r i v a l link or t o (2)
~(0:).
State a general c r i t e r i o n €or the nonequivalence of l i n k s with more
than one component i n terms of Alexander polynomials.
189
CHAPTER 13
SOME MATRIX REPRESENTATIONS OF THE
BRAID
GROUP
I t would seem t o be desirable t o be able t o construct the Alexander
matrix of a link d i r e c t l y i n terms of a corresponding braid without f i r s t This w i l l be the aim of the present
constructing the group of the link. chapter. 13.1.
RESULT
Suppose that
(Chain Rule).
are words i n the f r e e group F({xl,
a
ax W(V1(XI'...Y
Xn),.*-,
[[
xn)
j
=
aw(Xl,...,
k=1
for j = 1,2
PROOF.
axk
..., xnl).
5))
VnCX1""'
1 Xk =
Then
Xn)
aV"X1,.",
Vk(X1,.
.., 'n)
ax
j
1
,..., n.
W e use the d e f i n i t i o n of f r e e derivatives given i n 8.8. Example (6). aw(xl,.
j ,k=l
.., 'n)
1
[ xk=vk(xl,. . . n) ,x
aVk(X1,"" ax. I
xn)
CHAPTER 13
190
..., 'n)
[
n aw(xl, = 2 1 axk
n =
1
j =1
('k
[ -
.*
W(Vl(X1,.
-l)]xk=vk(xl,...,x n )
, XJ
,. .., vn(xl,. .., xn,)
1
-1),
(Xj
which gives the required r e s u l t .
NOTATION.
13.2. Then
p
Suppose t h a t
and
p
v
are permutations on 1 , 2 ,
..., n.
and v are said t o have the same p a t t e r n i f and only i f when
and
p
v
are decomposed i n t o products of d i s j o i n t cycles one has t h a t every p a i r of integers belong t o the same d i s j o i n t cycle of belong t o the same d i s j o i n t cycle of v.
p
i f and only i f they also
The pattern of v i s said t o
embrace the pattern of p i f and only i f when u and v are decomposed i n t o
d i s j o i n t cycles one has t h a t every p a i r of integers which belongs t o the same d i s j o i n t cycle of
p
a l s o belongs t o the same d i s j o i n t cycle of
V.
For example (34125) (67) and (15432) (67) have the same pattern, while the pattern of (3156) (24) embraces the pattern of (1635) (2) (4). permutation always has the same pattern as its inverse. any permutation on the numbers 1 , 2 ,
..., n.
A
Suppose t h a t p i s
Then the pattern of p always
... (n) , while the
embraces the pattern of the identity permutation (1)(2) pattern of any n-cycle always embraces the pattern of p. 13.3.
NOTATION.
Suppose t h a t p i s a permutation of the numbers 1 , 2 ,
and t h a t p = (il
... i
)(im +1 ml 1
... i
m2
)
... ( imc- 1+1 ... i
mC
)
..., n
SOME MATRIX REPRESENTATIONS OF THE BRAID GROUP
191
i s the decomposition of p i n t o d i s j o i n t cycles such t h a t every im.+l (take 1 i mo+l = i 1) i s the smallest number i n i t s cycle and
Then putting 5 = t w i l l denote the application of the ring homomorphism P ZZF({xl,.
..
xn})
-+
22
[ti’,...
t:1]
defined by X1 .
+
ml+l
x. lm1
-+
Xi
tl,
-+
t2’
t2’
m2
X
im c-1+1
X
i
-+
-+
tC’
tc.
*c For example i f n = 6 and p = (1635) ( 2 ) (4)
then putting 5 =
P
t means
apply the ring homomorphism defined by
13.4.
MATRIX REPRESENTATION OF n-BRAIDS.
with corresponding permutation
lY2’...’
p.
Suppose t h a t u is an n-braid
I f p is a permutation of t h e numbers
n such t h a t the pattern of p embraces the pattern of
p,
then u @P
CHAPTER 13
192
is defined t o be the n
"-IxJ
x
n matrix
-P-
with coefficients i n the r i n g ZZ
bi',. .., ti'], where c i s t h e number of
d i s j o i n t cycles occurring i n p.
THEOREM.
13.5.
Suppose t h a t u and
permutations p and v respectivezy. 1,2,.
T
are n-braids with corresponding
I f p i s a permutation of t he numbers
.., n so t h a t the pattern of p embraces the patterns
of u,v and p v ,
then =
(U T ) "
U"
m
TJlp
.
PROOF.
x = t
-Pax.:;
by 4.6.
-PHence i f x.0 = wi(xl, 1
and
-
X.T
1
=
..., %)
v.1 (xl,..
., 'n)
f o r a l l i , then r
7
193
SOME M4TRIX REPRESENTATIONS OF THE BRAID GROUP
by means of 13.1.Chain Rule.
By 4.6. Theorem and the assumption concerning
the patternsof the permutations, one has t h a t
13.6.
(1) Suppose t h a t p i s any n-cycle.
EXAMPLE.
Then the above
representation gives the h r a u representation +B : Bn
-+
G L(n ; ZZCt,t-'I).
By the above theorem, qB i s determined by i t s e f f e c t on a1,a2,
-
f o r k = l,Z,
'k-1
0
0
0 -
0
1-t
t
0
0
1
0
0
0
0
0
0
..., n-1.
..., un-1'
This is how W. Burau o r i g i n a l l y defined the Burau
representation. (2)
Suppose t h a t p is the i d e n t i t y permuation ( 1 ) ( 2 )
... (n).
Then one
obtains the following Gassner representation due t o B.J. Gassner
where Pn denotes the normal subgroup of Bn which consists of a l l pure braids
194
CHAPTER 13
as given i n 4.10. Example (3).
By the above theorem, bG i s determined
by i t s e f f e c t on the s e t of generators
‘Ar,s of Pn,
with 1 5 r < s s n l
Nowusing 4.11. Exercise (2) we have t h a t
0
0
0
0
0
0
I
0
Ir-1
0
... 0 0 0
0
0
s-r-1
0
0
0
In-s
Note t h a t I;-1
0
0
0
1-t+t2
t(1-t)
0
0
(l-t)2
-(1-t)2
0
0
(l-t)2
-(1-t)2
0
0
(1-t)2
-(1-t)2
0
1-t
0
t
0
0
0
0
0
In-s
...
...
...
We now show t h a t the image of
u
of the link L(o) corresponding t o u.
under
J,
..* 0
i s almost the Alexander matrix
SOME MATRIX REPRESENTATIONS OF THE BRAID GROUP THEOREM.
13.7.
permutation.
Suppose t h a t u i s an n-braid and
195
i s t h e corresponding
Then the Alexander matrix of the group of t h e l i n k L ( u ) i s
By 6.1. Theorem of Artin and B i r m a n , we have t h a t the group of the
PROOF.
l i n k corresponding t o u i s given by < x1
- -1 ,..., x 0.x-l n n
,...,
; xlu.xl
>.
Hence t h e Alexander matrix i s
Now
a x.;.x:l ax 1 1
=
ax.; 1 ax
j
(xi.)
xi1 6ij
and, by 7.4. Result (and i t s proof)
j
and 4.6. Theorem, we have t h a t the Alexander matrix i s of t h e required form.
We w i l l a l s o r e f e r t o u”
NOTATION.
13.8.
matrix of t h e braid
-
In as being the Alexander
0.
The reduced Alexander matrix of the braid u is defined t o be the matrix u
$B
- In.
The Alexander matrix of an n-braid
(I,where
L(u) i s a knot, can now be
evaluated by means of 13.6. Example ( l ) , 13.5. Theorem and 13.7. Theorem. The Alexander matrix of an n-braid u , where L(a) i s a l i n k with more than one component, can be evaluated by f i r s t combing the braid (4.10. Example (1)) and then using 13.5. Theorem, 13.6. Example (2) and 13.7. Theorem.
13.9.
of B2.
EXAMPLE.
(1) Let m be an integer and u1 be considered as an element
Then it is easy t o prove by induction t h a t
CHAPTER 13
196
($B
L
1-(-t)m+l
t(l-(-t)m)
1-(-t)m
t (1- (-t)m-l
= (l+t)-l
[
1-t t
since (u:)'~
= (u:)~
and u:B
=
Alexander matrix of the link L(u:)
c
-t(l-(-t)m)
(l+t)-l
l-(-t)m
0]
is
.
1.
This gives t h a t the reduced
I.
t(l-(-t)m
-l+(-t)m
In p a r t i c u l a r i f m is an odd integer, then,by 5.3. Result, L(oy) i s a knot, the above matrix i s i t s Alexander matrix and l-(-t)m l+t i s i t s Alexander polynomial.
When we put m =
f
3, then we get the well
known case of the t r e f o i l knot (11.1. Example ( 2 ) ) .
The knot L(ol) 5 has a
projection of the form
I t i s called a pentacle and was considered i n the Middle Ages t o have great mystical powers.
SOME MATRIX REPRESENTATIONS OF THE
(2)
Let m be an integer and
197
be considered as an element of B2.
=
0:"
BRAID GROUP
Then
!
l-t1+t1t2
A+G = 192
1
(l-tl)tl
1-t2
tl
by 13.6. Example (2).
Now a somewhat tedious calculation enables one t o
show t h a t (l-tl)
Cl-tX)
+
t; t;,
tl ( l - t l ) (1-t;t;)
l-t1t2
l-t1t2
(1-tX) ty t;
(1-t2) (l-tyt;)
t1(l-t2)
-k
¶
l-t1t2
l-t1t2
Hence the Alexander matrix of the link L(Ay,2) i s
1-tl t2
1
1-t2
tl-1
and i t s Alexander polynomial is
m t2 m 1-tl 1-tl t 2
(3)
Let ml amd m2 be integers and
0:
:a
Then, by above Example (1), we have t h a t
be considered as an element of B3.
CHAPTER 13
198
1-(- t ) m l + l
t(l-(-t)ml) l+t
l+t
1- ( - t ) m l
t(1-( - t ) m l + l )
l+t
l+t
0
0
1
0
0
0
m2+1 1-(-t) l+t
t(1-(-t)"2) l+t
1- (-t)m2
t(1- ( - t ) m Z - l
l+t
1-(-t)ml +I
l+t
*
l+t
*
*
0
*
Hence the reduced Alexander matrix of the link L ( u y l u?) the matrix (see beginning of Chapter 11)
*
i s equivalent t o
199
SOME MATRIX REPRESENTATIONS OF THE BRAID GROUP
m
I f ml and m2 are both odd i n t e g e r s , then L(ull u?)
t s Alexander polynomial i s
Result, &d
{ l-(;$+l
-
} { t(l-(-t)m2-1 l+t
N
i s a knot, by 5.3.
{ 1-(-tlrn1 } . { l+t
1-(-t)m2 l+t
-
1
*
The Granny knot and the square knot are examples of such knots. (4)
The t r u e lover's knot is given by the 3-braid (ul3 u 2 ) 2
has a picture of the form
l-t+t2-t3
(u?)
=
@ :
t (l-t+t2)
l-t+t2
t(1-t)
0
0
.
This braid
W T E R 13
200
rl
'1
o
0
1-t
0
1
0
1-t+t2-t3
t(1-t)
t2(1-t+t2)
(l-t+t2)
1-t+t2
t(l-t)2
0
1
t2(1-t)
0
*
*
*
(1-t+t2) (1-t2)
*
t 2 (1-t+t3)
l-t+t2
*
t2
(1-t)
1
1
1
1
Hence the Alexander matrix is equivalent t o the matrix (see beginning of Chapter 11) ( 1 4 ) (1-t+t2) 1-t+t2
t 2 (1-t+t3) t2
(1-t) -1
and the Alexander polynomial i s (1-t+t2)
[(1-t) (1-@)t2-(1-t2)-t2 (1-t+t3)]
which is equivalent t o
3 -1 2 ( 5 ) The f a l s e lover's knot i s given by the 3-braid (al a2 )
has a picture of the form
.
This braid
SOME MATRIX REPRESENTATIONS OF THE BRAID GROUP
3 $B
(0,)
=
[
l-t+t2-t3
t (1-t+t2)
l-t+t2
t(1-t)
201
0
0
1
0
0
0
0
t-1
O1
l I
1-t-q 0
t (l-t+tZ)
1-t+t2
0
t(1-t)
0
t-l
1-t-1
l-t+tZ-t 3
*
:. *
(1-t+t2) (l-t+tZ-t3)
1-t
t-l ( l - t + t Z )
t-+ 1-t-I)
11
CHAPTER 13
202
Hence the Alexander matrix is equivalent t o the matrix (see beginning of Chapter 11)
c
(1-t+t2) (l-t+t2-t3)
-t t -1(1- t-1)
t-l(1 - t + t 2 )
0 OI
and the Alexander polynomial i s (1-t+t2) [ t - l ( l - t - l )
(l-t+t2-t3)
+1]
which is equivalent t o (1-t+t2) ( 1 - 2 t + t 2 - 2 t 3 + t 4 ) . (6) The three-lead four bight Turk's head knot i s given by the 3-braid -1 4 This braid has a picture of the form (u2 ul)
.
SOME MATRIX REPRESENTATIONS
(02
1
-lQB
OF THE BRAID GROUP
203
=I! 0
0
0
t-I O
1-t-l
( l - t ) 4 + l - t - t (1-t t-+ 1-t)2+
t(l-t)3+t
)
(t-l-lp
2-t-t-1
*
*
rI
Hence the Alexander matrix is equivalent t o the matrix (see beginning of Chapter 11)
c
(1-t) 4 - t - t ( l - t - y t-+l-t)?+(t-1-1)
and the Alexander polynomial i s (t2-t+l)' (t2-3 t + l )
t ( 1 - t ) 3+t 1-t-t-I
01
0
CHAPTER 13
204 13.10. 0
I f the n-braid u can be decomposed in the form
RESULT. = WI(U1Y.*.Y
Oil
W2"i+l,""
i n Bn, then the reduced Alexander poZynomiaZ of u i s equaZ t o the reduced
Alexander poZynomiaZ of w 1 (considered as an (i+l)-braid) times the reduced AZexander po ZynomiaZ of w2 (considered as an (n-i) -braid). PROOF.
Suppose t h a t i n i t i a l l y we consider w1 and w2 as braids on (i+l)
and (n-i) strings respectively.
Then *
L*
...*I
Hence
The Alexander matrix of u i s equal t o
.
a
,
*
% *
SOME MATRIX REPRESENTATIONS OF THE BRAID GROUP
*
*
205
-
* . , . * *
%-I
-
which is equivalent t o the matrix (see beginning of Chapter 11)
Therefore the reduced Alexander polynomial of a i s equal t o det (Al-I)
. det (%-I).
However d e t ( A1-I)
and det(%-I) a r e the reduced Alexander polynomials of
the braids w1 and w2 (considered as elements of Bi+l
and Bnmi respectively).
This gives the required r e s u l t . Suppose t h a t a h o t i s such t h a t it is s t r i n g isotopic t o a h o t corresponding t o an n-braid of the form
where the links corresponding t o the (i+l)-braid w1 and the (n-i)-braid w2 are knots K1 and K2 respectively. the knots K1 and K2.
Then K i s s a i d t o be the composition of
One a l s o says t h a t K is obtained by t y i n g the h o t
K2 i n the knot K1.
The granny and square knots are examples of h o t s which can be obtained
in this way from t r e f o i l h o t s .
CHAPTER 13
206
A h o t K i s always string isotopic t o the composition of t h e t r i v i a l
knot S1 and K.
If t h i s i s the only composition of knots t o which K is
s t r i n g isotopic, then K is said t o be a prime h o t .
Such a table of knots can be found
usually enumerate only prime knots. a t the end of t h i s book i n Chapter 13.11.
EXERCISE.
Tables of h o t s
0.
(1) The bowline knot has a picture of the form
Determine a braid whose corresponding link is the bowline h o t and evaluate the corresponding Alexander polynomial.
(2)
Draw a picture of the 3-braid (ula2)
3
.
Determine the corresponding
link and evaluate the Alexander polynomial. (3)
Determine the Alexander polynomial of the knot corresponding t o the
u-braid
SOME MATRIX REPRESENTATIONS OF M E
(4)
BRAID
207
GROUP
Show t h a t the Alexander polynomial of the link L ( 4 , s ) i s zero f o r a l l Y
r and s with s
2
3.
( 5 ) Establish the following generalisation of 13.10. Result.
Let u be an
n-braid of the form u = w1(u 11.. with
p, v
., Ui) . w2(ui+l,. .., un- 1)
and p denoting the permutations corresponding t o wl, w2 and u Then the p a t t e r n of p embraces both the pattern of
respectively.
the pattern of w .
p
and
If
denote the Alexander polynomials of w1 and w2 respectively, then the Alexander polynomial of u is
except when both c1 ahd c2 are not equal t o 1.
In the l a t t e r case it i s
equal t o
(6)
Suppose t h a t
.., xnl).
F({xl,.
a
and B are r i g h t endomorphisms of the f r e e group Then i n the group ring of t h i s f r e e group one has t h a t
and n
(X
-1)B
P
=
1 b .(xj-l) j=1 PJ
,
CHAPTER 13
208
for i , j , p = l , Z ,
..., n.
Hence show t h a t the following matrix equation
holds
This i s a p a r t i c u l a r case of 13.1. Chain Rule and can be used t o give another proof of 13.5. Theorem. (7)
Define the concept of the reduced Alexander polynomial of a l i n k .
209
CHAPTER 1 4
OPERATIONS ON BRAIDS AND RESULTING LINKS
Let w(ui) denote an element of the braid group Bn.
We now perform
some geometric operations on the n-braid w(oi) and investigate the resulting e f f e c t on the corresponding link L(w(ui)). 14.1.
Rotate the braid w(ui) through an angle of
gives the n-braid A-' A = (al
TI
about the z-axis.
This
w(ui) A = w ( u ~ - ~where )
... un)(ul ... un- 1) ... (u 2 u 1)u 1'
by 4.11. Exercise 5 .
Clearly L(w(ui)) and L(A-' w(ui)A) a r e s t r i n g
isotopic. 14.2.
Take a plane a t r i g h t angles t o the ends of the braid w(ui).
instance z = a w i l l do i f the braid l i e s between z = a and z = b. the braid i n t h i s plane. L(W(ui)) and L(w(ui)-')
For Reflect
This w i l l give the n-braid w(ui)-'.
The links
w i l l not i n general be s t r i n g isotopic.
For the
above given homeomorphism i s orientation reversing. L(w(ui)-l) are equivalent links.
The link L(w(ui)-')
mirror image Zink of the link L(w(ui)).
However L(w(ui)) and
i s called the
A link which i s s t r i n g isotopic
t o i t s mirror image i s called Qmphicheiral.
For example, the l e f t handed
t r e f o i l knot i s not amphicheiral (see f o r instance D. Rolfsen Chapter 8 or below 17.6. Example), while the figure eight knot is amphicheiral. , ) ~ figure eight knot i s given by the 3-braid ( U ~ ~ Uand
The
CHAPTER 14
2 10
( a ; ' ~ ~ ) - =~ (a;1a2)2
= A
-1 -1 2 (a2 al) A ,
where A = a 1a 2 a 1' 14.3.
We now consider what happens when we go over from the n-braid W(..)1 =
€1 € 2 ai2
a
.'.
'k ik
t o the n-braid
a
'k ik
... a €i22 a €1 , il
which we denote by Rev w(ai). have the same appearance directions.
-
Clearly the links L(w(ai)) and L(Rev w ( a i ) )
t h i s i s so i f one reads them in opposite
These links w i l l not i n general be orientated equivalent.
I f they are orientated equivalent, then the link L(w(ai)) i s said t o be invertible.
H.F. Trotter and C. Kearton have each given examples of knots
which are not invertible.
The t r e f o i l h o t L(a;) is an invertible h o t .
I f one r o t a t e s
through an angle of
71
about the x-axis, then one obtains the braid
OPERATIONS ON
BRAIDS AND RESULTING LINKS
In the past ( f o r instance i n the Proof of 4.6.),
211
we assumed a preferred
direction along the s t r i n g s of a braid, namely, the downward direction. This gives a fixed orientation along a link.
I f we now a l t e r t h i s
convention and thus "read" a braid upwards, then instead of w(ui) we have Rev w(ui).
The problem we now address ourselves t o i s : What happens t o
the Alexander matrix? Let
denote the Magnus ring of formal power s e r i e s in the non-
commuting variables y1,y2a...a yn.
This implies t h a t every element of
has a unique representation of the form
where the coefficients are integers.
We need the r e s u l t s of 8.9. Exercise
These r e s u l t s can a l s o be found i n the book by W. Magnus, A. Karnass
(5).
and D. S o l i t a r Chapter 5 .
In p a r t i c u l a r it is a well known c l a s s i c a l
r e s u l t of W. Magnus t h a t the group ring Z Z F ( I X ~ , . . . ~xn)) of the f r e e group F({xl,.
.., xn))
under the mapping defined by' x1
-+
l+yl,
..., xn
-+
l+yn.
is ring isomorphic t o a subring of
CHAPTER 1 4
212
E has
a subring hll of those formal power s e r i e s which have zero constant
term. We go on t o consider the dual ring h$
. ., xn,
x1,x2,. <
which are defined on
*.. yi
yi
, xj P
1
1 ~j
ril
... 'jq
.. ,>
plicative unit element of
E.
LEMMA.
MT
comuting variables
PROOF.
e
i f ip = j q
otherwise
and the assumption t h a t <
14.3.1.
(with values taken i n Ml) by
>
yjq-l
= tlo
M1
of power s e r i e s i n the functions
is "bilinear".
Here e denotes the multi-
i s t h e subring bll of the Magnus ring i n t h e non-
xl,
..., Xn'
Suppose t h a t
1 Bjl
... j q Xjl'*'
xjq = 0 (zero function).
Then
This implies t h a t a l l '1,
... jq = 0.
In order t o avoid needless complicati.ons we s h a l l identify ZZF(IXl,..*,
xn1) = 2ZF
213
OPERATIONS ON BRAIDS AM) RESULTING LINKS
E.
with i t s image i n the Magnus ring
This means t h a t we shall i d e n t i f y
yi with xi -1 f o r a l l i. Let A be the augmentation ideal of the group ring as defined i n 7.3. Example (1).
So
l n g;lng=o}. {g
A =
Then A
=M1
and ZZF = Z Z 8 A .
i s the topological closure
I t is not d i f f i c u l t t o see t h a t
of ZZF,
where A
3
A
2
3
...
3
Am
3
...
are taken t o be the neighbourhoods of 0 in ZZF.
Let
$
Also
be a r i g h t endomorphism of the f r e e group F(Ixl,
define the corresponding l e f t ring endomorphism $* of <
xi-l, $*X. > = < 3
(xi-l)$,
xj
..., xn}).
9by
>
f o r a l l i and j with the added condition t h a t $* i s continuous on Finally $* extends t o a r i n g endomorphism of t h e dual ring
when we s t i p u l a t e t h a t $*(e) = e . 14.3.2.
LFMA.
Suppose t h a t
n (xi-l)$ = j=1d1'. . (x) (x.-1) 7
for 1 s i
5
n
MT.
We
CHAPTER 1 4
214
i n Z F , where every d.11.(x) i s an element of ZZF.
in
Then
El.
PROOF.
First we note t h a t
f o r a l l j and some
x' and x" i n
$I
implies that
Now
As these e q u a l i t i e s hold f o r a l l i and j , the required r e s u l t follows. 14.3.3.
THEOREM.
If A i s the Alexander matrix of the link L(a)
, then
the
transpose A' i s t h e Alexander matrix of the Zink L(Rev a ) , where a i s an n-braid. PROOF.
According t o 13.7. Theorem, the Alexander matrix of the link L(a) is
OPERATIONS ON
where
215
BRAIDS AM) RESULTING LINKS
0 i s the automorphism of the free group F({xl, ..., 51) corresponding
t o the n-braid u and
i s the corresponding permutation.
Now, as we have
already pointed out i n 8.9. Exercise ( 5 ) , i f (xi-1); =
1 d.11.(x)(x.-1) J
j
f o r every i , then
a ( (xi-i)O)
a (xi;) =--
ax j
ax. J
- d . . (x)
= < (Xi+,
11
-
xj
>
However
f o r a l l i and j .
which we can consider t o be <
(xi-1)
,
f o r a l l i and j
-
(Rev u)xj >
.
Hence, using the above given way of looking a t
a , we axk
have t h a t the Alexander matrix of the link L(Rev (I) i s
=
[ dji(x)J
by 14.3.2.
Lemma.
-
In = A ' ,
Here we are also using the f a c t t h a t
morphism of the Magnus ring
-
p
(see 14.3.1. Lemma).
6* i s
a ring endo-
Finally we note t h a t
u* is taken t o be the mapping corresponding t o the generator n-braid ui i
2 16
CHAP'IER 1 4
with the direction along the s t r i n g s being taken i n the upwards direction for 1 5 i
5
n-1.
R.C. Blanchfield has a proof of 14.3.3. Theorem i n the case of knots,
which uses the relation matrix approach t o Alexander matrices of knots as given i n 11.3. Exercise ( 2 ) .
217
CHAPTER 15 THE GROUP OF A FREE ENDOMORPHISM
Suppose t h a t
i s an endomorphism of a free group Fn on the s e t of f r e e
a
generators x ~ , . . .xn. ~ we have t h a t
a
Then, by 1.9. Universal Property of f r e e groups,
i s determined by i t s e f f e c t on xl1..
a f r e e endomorphism.
., xn’
We also c a l l
a
Associated with every free endomorphism a we have
the group of the p e e endomorphism Gn(a) = < xl,.
.., xn
; x1 = xla,.
.., xn = xna > .
We have already considered such groups.
I f u i s an n-braid, then the group
of the link L(u) is the group of the free endomorphism
G,
by 6.1. Theorem
of Artin and Birman. Suppose t h a t a i s a f r e e endomorphism of the f r e e group Fn.
EM.
15.1.
Then the relations w a “W hold i n Gn(a) f o r a l l words w i n Fn.
Hence a induces the i d e n t i t y auto-
morphism on Gn(a) and also any group on n generators which has the same property i s isomorphic t o a factor group of Gn(a) PROOF.
We know t h a t i n
xi1 = (xia)-’
=
%(a)
.
we have t h a t xi = xia and hence
x i l a f o r a l l i, since
u i s an endomorphism of Fn.
t h a t w1 and w2 are two words belonging t o Fn.
Suppose
By induction we assume t h a t
CHAPTER 15
218
w1
= wla
and w2
= w2a
are relations holding i n Gn(a). WlW2
= w2a.w2c1 =
Hence
(W1W2)"
i s a relation i n %(a), since a i s an endomorphism of Fn.
w = wa
Thus
holds i n %(a)
for a l l w belonging t o Fn. Further suppose tha t G i s a group on generators glyg2,.
.., & such
th at ga = g fo r a l l elements g i n G.
Here g a is defined by first writing g as a word
w(gi) i n the generators gl,.
..,
and then putting
NOW it i s c l ea r t h at G i s isomorphic t o a factor group of %(a). 15.2.
COROLLARY.
If yl,
..., yn is a s e t of
free generators f o r the free
group F({x ly..., % I ) , then
= < X1
Hence
,..., $ ; X1
%(a)
y...,
Xn = %a >
.
i s independent of the choice of the s e t of free generators for
the free group F({xl,. 15.3.
= Xla
EM.
..
x,,}).
Suppose t h a t 6 is an endomorphism of the free group Fn whiZe
a is an autornorphism of Fn.
Then
219
THE GROUP OF A FREE ENDOMORPHISM Gn(a 6a-l) PROOF.
Gn(B)
and Gn(a-')
=
Gn(a).
The group Gn(a 6a-l) can be considered to be the factor group of
F, modulo the relations
Apply the automorphism a.
This induces an isomorphism of Gn(a8a-')
with
the factor group of Fna modulo the relations (xia)8 =
(xis)
for i
=
1,2
,..., n,
which is the required isomorphism. The automorphism a induces an isomorphism
a
of the group Fn modulo the
relations x.a-' 1
=
xi
for all i
onto the group Fna modulo the relations xi
=
x.a 1
for all i.
This says that l maps Gn(a-l) onto Gn(a). this proof, one also has that
However, by the first part of
a maps
which gives the required equality. 15.4.
COROLLARY.
Suppose t h a t u and
'I
are n-braids.
Then we have t h e
foZZowing r e l a t i o n between t h e groups of t h e Zinks L(Tu'I-'), L(0) :
L ( 0 - l ) and
CHAPTER 15
2 20
5 . 4 . Exercise 2 and 1 4 . 2 . give another way of deriving these isomorphisms.
EXERCISE. (1) Let x be a fixed element of the free group Fn.
15.5.
Determine the group of the inner automorphism xi
x x.1xel
for all i.
Suppose that a and 8 are endomorphisms of the free group Fn such that
(2) a
-+
is an automorphism.
Show that the group Gn(8 a-l) is isomorphic to the
factor group of Fn modulo the relations =wa
WB
for all elements w in the free group Fn' %(a Ba'') 5 $(B) and Gn(a-') 2 %(a). (3)
Hence deduce that
Show that if a is an automorphism of the free group Fn, then in
general one cannot assume that the collection of all elements o f the form xa.x-',
where x varies over the free group Fn, forms a normal subgroup of
(4) Show that if 8 is an endomorphism of the free group Fn, then $(8)
is
isomorphic t o the group Fn/ (ker 8) modulo the relations
y (5)
=
y
for all elements y of Fn/(ker 8 ) .
Show that if B is an endomorphism of the free group Fn and k is a
fixed positive integer,then there exists a natural homomorphism of
(6) Show that every group which has a presentation on n generators and n
defining relations is isomorphic to the group of some free endomorphism, where n is a fixed positive integer.
THE GROUP OF A FREE ENDOMORPHISM
221
(7) Show that if B is an endomorphism and a is an automorphism of the free group Fn so that a 8 Gn(B)
*
= 8 a,
then
ci
induces an automorphism of the group
This Page Intentionally Left Blank
223
CHAPTER 16 ALEXANDER POLYNOMIALS REVISITED
The aim of t h i s chapter is t o produce the Alexander polynomial d i r e c t l y from the braid a. REDUCED MATRIX REPRESENTATION OF n-BRAIDS.
16.1.
n-braid (n
2
2) with corresponding permutation
Let p be a permutation
.., n such t h a t the pattern of p embraces the pattern of
of the numbers 1 ’ 2 , . p.
p.
Suppose t h a t a i s an
We take yi
=
x1x2
... xi
Then we form the n
x
for i
=
1,2,...,
n.
n matrix
where we are using the notation of Chapter 13.
The l a s t row of t h i s matrix
is 0
... 0 1 ,
W e denote the reduced matrix obtained from the above since yn 0 = yn. matrix on deleting the l a s t row and the last column by
224
CHAPTER 16 Suppose t h a t a is an n-braid (n
RESULT.
16.2.
I f p i s a permutation of1,2,.
ponding permutation.
embraces t h e pattern of
where C = C(xl,.
2
p,
2 ) and p i s the corres-
... n whose pattern
then
... xn)& =
and
P
C(Xl
PROOF.
Now
..... 3)
=
1
...... 0 0 0 ...... 0 y2 0 ...... 0 .................
1 1 1
0
0
By 13.1. Chain Rule, we have t h a t
[ w-Ix a [2 1 (Yi3
=
=
ayi -axk
{
f o r a l l i and k.
axk x = t
-P Also
0 y1 y1
-P
i f k s i if
where we take y-l = 1.
,
’
Hence we have t h a t
ALEXANDER POLYNOMIALS REVISITED
225
. C = C . u $P
which gives the required r e s u l t . 16.3.
Suppose t h a t u and
COROLLARY.
'I
are n-braids (n
ponding permutations p and v r es pect i vel y. nwnbers 1 , 2 ,
..., n so t hat
2
2) wi t h corres-
I f p i s a pem ut at i on of t h e
t h e pat t er n of p embraces the pat t erns of 11, v
and ~ v , then $ ((I
T)'P
ur
6
$ 'Ir
p.
This r e s u l t follows from 13.5. Theorem and 16.2. Result.
PROOF. 16.4.
=
(1) If p i s an n-cycle, then the resulting mapping
EXAMPLE.
r 6p : Bn
+
GL(n-1 ; ZzCt,t-'l)
is denoted by r J,B and i s called the reduced Burau representation. follows from 16.3. Corollary t h a t r 6B is a group homomorphism. calculation shows t h a t -1 Y ~ yi+ ~yiWl
if j = i
i f j = i+l otherwise and hence
It A simple
CHAPTER 16
226
rB '
with ui
=
-
0
0
0
0
1
0
0
0
0
t
-t
1
0
0
0
0
1
0
0
0
0
0
In-i-2
Ii-2 0
-
for 2
5
i
n.
-
Thus we have t h a t these n-1 matrices determine the reduced Burau representation of Bn.
If p i s the i d e n t i t y permutation, then the resulting mapping
(2)
r J,p : Bn
GL(n-1 ; art;',
-+
..., ti+11)
i s denoted by r$G and i s called the reduced Gassner representation.
It
follows from 16.3. Corollary t h a t the r e s t r i c t i o n of r J, G t o the subgroup Pn of pure braids i s a group homomorphism of the group Pno
THEOREM.
16.5.
Suppose t h a t the permutation corresponding t o an element
u o f the braid group Bn(n
2) is
2
I f p i s not an n-cycle,
p.
then the
AZexander poZynomiaZ o f the Zink L(u) is J,
(x1x2
If
p
...
1 Xn-l)x = t
'
det(ar
-
In-1).
-u-
i s an n-cycle, then the Alexander polynomial o f the knot L(u) i s t-1 -
tn-1
PROOF. matrix
.
det(J
J,
-
InJ.
By 13.7. Theorem, The Alexander matrix of the l i n k L(a) i s the
22 7
ALEXANDER POLYNOMIALS REVISITED
where we have taken the group of the l i n k L(o) t o have generators
..., xn and defining relations xlu . x1-1,..., x -u . a? n
x1 ,
By 15.2. Corollary, we may take
where y . = x1x2 1
... xI. for j
= 1,2,.
.. , n.
By a similar procedure as t h a t
used i n the proof of 1 3 . 7 . Theorem, we have t h a t the Alexander matrix is equivalent t o the matrix
Lo.. .o
01
Using the r e s u l t s of Chapter 1 2 , we have t h a t i f An denotes the determinant of the submatrix obtained from the above matrix by deleting the n-th row and the n-th colunn, then e i t h e r
...Anxn-1) x
(x1x2
= t
-u
is the Alexander polynomial of L(o) according as 16.6.
EXAMPLE.
p
is not or i s an n-cycle.
(1)
We consider the pure 2-braid A1,2 = ul.2
y2
and ArJIG 1,2 =
I t i s easy
t o see t h a t
,2 aY1
=
Ctlt21, which is 1
x
1 matrix.
CHARER 16
228
m 1 for every integer m. Also (AY,2)r'G = Ctm, t 2
Hence the Alexander poly-
nomial of the link L(Ay,2) i s
ty t; - 1 tl t2 1
-
-
See 13.9. Example ( 2 ) f o r a more tedious way of obtaining t h i s r e s u l t . (2)
We determine the Alexander polynomial of the link
The pelmutation associated with t h i s 3-braid is p = (1)(23).
By 16.3.
Corollary, we have t h a t
Hence
Thus the Alexander polynomial i s
1 tlt;
-
1
det
[
I
t1t2+t2(1-t1) (1-t2+t2) 2 -1,
-t2(1-t1) 3
t 2 (l-t2+t2) 2 ,
3 -1 -t2
tlt2 -1
ALEXAMIER POLYNOMIALS REVISITED
I
1-t2+t2 tlt2 -1
-
- tl t2 2 -1
1-tl
-t2 -1
t2
-
229
1 - t 2 + 2t 2 '
Another way of obtaining t h i s r e s u l t i s t o use 13.11.Exercise (5). 16.7.
EXERCISE.
(1) Show that the Alexander polynomial of the n-braid u
i s zero i f and only i f 1 i s an eigenvalue of the matrix a r
ii,
',
where u i s
the permutation corresponding t o a . (2)
Show t h a t i f u i s a 3-braid, then the Alexander polynomial of u is
either
or
where (3)
p
i s the permutation corresponding t o u .
Show d i r e c t l y t h a t u, a
-1
and
T
-1
UT
have equivalent Alexander polynomials, where a and (4)
are n-braids.
Show t h a t
ro
0
... 0 -t1
............... 0 0 ... t -t
1 (5)
T
1
Show t h a t i f the link L(o) associated with t h e n-braid u is a knot and
f ( t ) is i t s Alexander polynomial, then
CHAPTER 16
2 30 f(1)
=
f 1.
There are two ways of proving t h i s r e s u l t . above Exercise (4).
F i r s t l y it can be proved using
See J.S. Birman Corollary 3.11.2.
Another method of
proof, which uses the f a c t t h a t the Alexander polynomial of t h e t r i v i a l group i s 1, can be found i n R.H. Crowell and R.H. (6)
Fox Chapter I X .
Show t h a t (see 4.11. Exercise ( 2 ) )
-1 -1 -1 -1 Y i Ys-1 Ys Yr-1 Yr Y s Ys-1 Yr Yr-1
yi
for i
2
for r s i
s.
s
Hence determine the reduced Gassner representations of the 3-braids and the 4-braids
(7)
A1,3’ *1,4’
%,4’
%,3 *3,4‘
Show t h a t i f the Alexander polynomial of t h e n-braid u i s zero, then
the Alexander polynomial of the n-braid urn i s zero f o r every nonzero integer
m.
231
W T E R 17
MERIDIANS AND LONGITUDES
We examine more closely the proofs of 6.1. Theorem of Artin and B i r m a n and of 4.6. Artin Representation Theorem. n-braid u . xl,.
We were there concerned with an
We chose loops
.., xi,. ..
x n
as follows
I ;(
. . . . Pn. 'i+l
P
X. 1
xly,.., xi,
..., xn
a r e c a l l e d meridians of the l i n k L ( u ) .
Now i f we take
the loop xi and push x . down the braid u , then we get a loop 1
x.0 1
=
x
iu
A;'
for 1 s i
5
n.
Here the word Ai represents a loop which starts a t P , goes along the path
t r a v e l s along a polygon path very close t o the i - t h s t r i n g of u and goes back t o
Q
along t h e path
CHAPTER 1 7
232
Q'
'Qip
Now l e t mi denote the order of the d i s j o i n t cycle of i
=
..., n.
l,Z,
m
x.
1
a
=
p
which contains i f o r
Then
-1 Bi x.1 Bi '
where Bi is a word belonging t o the f r e e group F(Ixl
for 1 5 i
,..., xi ,..., 5
n.
I f we assume t h a t every Bi xi
every Bi i s uniquely determined.
Bil
i s a reduced word, then
Now Bi represents a loop which s t a r t s a t
the base point P, follows closely the i - t h s t r i n g of ,"i u n t i l the l a t t e r comes back t o Pi and then goes back t o the base point.
This is w h a t is
called a longitude of L(a) corresponding t o the meridian xi f o r 1 2 i
2
n.
I t is considered t o be an element of the group
Suppose t h a t the sum of the exponents of the element Bi written as a reduced word i n the free group F({xl,.
.., xi,. .., %I)
i s ~ ( i ) . Then
can also be considered as a longitude of L(a) corresponding t o the meridian We c a l l t h i s the Zongitude of L(a) corresponding t o the meridian xi i' and denote it by long (a) f o r 1 5 i 5 n. Once again every long (a) i s i xi considered t o be an element of the group of the link L(a).
x
17.1.
RESULT.
Let
CI
be an n-braid w i t h
MERIDIANS AND LONGITUDES
233
x.0 = A x A;', 1 i iu
Ail
where Ai xiu every i.
i s a reduced word i n the f r e e group F(Cxly..., %I) for
Then i n G(L(a))
long 'i (u)
=
Ai A.11!
... Aipm(i)-l
-E
(i)
xi
a
where m(i) i s the order of the d i s j o i n t cycZe of 1-1 which contains i and
E(i) i s the i n t e g e r which makes the exponent s m equaZ t o 0 , f o r every i.
If i and j ( i # j ) belong t o the same d i s j o i n t cycZe of
and m ( i , j ) i s the
p
smaZZest p o s i t i v e i n t e g e r so t h a t
then i n G(L(o)) we have t h a t
xi = (Ai Aiu
... Aium(iyj)-l)
xj
*
(Ai A i u
* a -
A m ( i y j ) - l1-l
iu
and
long
(cr) = (Ai Ail,
xi for every i.
... A
ium(i,j)-l
AZso xi and long
) .longx ( u ) . (A. A.
j
1
1u
... A ium(i, j ) - 1 1-l
(u) are commuting eZements of G(L(o)) for
xi every i.
PROOF.
By the d e f i n i t i o n of the longitude corresponding t o the meridian
xi, we have t h a t
By 6.1. Theorem of A r t i n and Binnan and the notation of Chapter 15, we have t h a t long
(u) i s an element of the group
xi
q:)
CHAPTER 17
234
and the above equality is considered t o be an equality i n t h a t group. 15.1. Lemma, the automorphism
group.
By
a c t s as the i d e n t i t y automorphism on t h i s
This gives the required f i r s t equality.
By 6.1. Theorem of Artin and B i r m a n , we have the following relations holding i n the group G(L(u)) :
.................. Substitution gives the required equation relating xi and x
j'
By definition, we have t h a t long x. (u) = Ai 1
where E(j) = E ( i ) .
... Ai um(i,j)-l* A.J
A j,
... A
-E
iu
(i)
m(i)-1 * x i
Appropriate calculation gives the required relation
between long x , (u) and long
(u)
j
1
.
The relations
.................. hold i n G(L(u)) f o r every i. xi = long x. (u) 1
17.2.
DEFINITION.
Hence on substitution one has the consequence
. xi . (long
i
(u))-l.
The above r e s u l t tells us t h a t every component of a link
MERIDIANS AND LONGITUDES
235
L(u) has associated with it a c l a s s of conjugate abelian subgroups. an abelian subgroup i s called a peripheral subgroup of G(L(u)). words, every < x
long (u) i' xi subgroups of G(L(u)).
In other
and a l l i t s conjugates are the peripheral
>
17.3. EXAMPLE. (1) We determine the longitudes of the t r e f o i l knots L ( u l-3 ) and L(u:).
xlul-3
=
x2 u-3 1
=
x2-1x1-1
- x2 x l x 2 x2-1x1-1x2-1 . x1 . x 2 x 1x 2 ' *
Hence -3 -1 -1 -1 -1 -1 long x l ( ~ l ) = x2 x1 x2 x1 x2 -3 (ul )
long
=
-1 -1 -2 -1 x2 x1 x2 x1
x2
. x15
and
. x25
4 -1 -2 -1 = x 2 . x1 x 2 x 1 '
Also -1 -1 . x2 . x-1 1 x2 x 1 x -1 . x1 . x-1 2 1 '
xlul3 = x1x2x1 3 = x1x2 x2u1 Hence
. xi5 = x-41 x2 x12 x 2
3 2 (ul) = x1x2x1x2 x1 3 long (u2) = x x x x x 1 2 1 2 1 x2
long
- x2-5
*
By 6.2. Examples ( 2 ) and ( 3 ) , we have t h a t
z where a = x1x2
c
a,b ; a3 = b2 >
and b = x1x2x1
Such
=
, x 2x 1x2 '
CHAPTER 1 7
2 36
Therefore, under the above given isomorphism, the elements long
(ui3) and long (ul)3 x1 x1
go over t o the elements
b-2(a-1b)6
and b2 (a-1b)-6 respectively.
They determine peripheral subgroups
(2)
-1 3 The Borromean rings i s given by the 3-braid (u2 al)
6 . 2 . Example (8)
long
)
x3
According t o
we have t h a t
((uilul) 3
=
x3 -1x-1x x 1 3 1 '
((uilul) 3
=
x3 -1x1-1x3x1
x2
long
.
. xi'.
x-1x-1x x 1 3 13
*
-1 x1x2x1
-1 = x-1 2 x 1x2 x 1 '
THEOREM. Suppose t h a t u and u' are n-braids and there e x i s t s an orientation preserving homeomorphism $ of IR3 onto i t s e l f such t h a t 17.4.
Then there e x i s t s a group isomorphism $r of G(L(u)) onto G(L(o')) which s e t s up a one-to-one correspondence between complete oonjugacg classes of peripheraZ subgroups of G(L(o)) and o f G ( L ( a ' ) ) . Further i f x. i s an arbitrarg meridian of u, then 1
4axi = wi
.
-1
Xie
. wi '
where e is a mapping of { 1 ' 2 ,
...)n l i n t o i t s e l f ,
and
237
MERIDIANS AND LONGITUDES (a)) = wi xi
$,(long
with
E
. (long
. wi-1 'ie
having t h e same f i x e d value 1 or -1 for a l l i.
OLITLINE OF PROOF.
The existence of the isomorphism
demonstrated i n Chapter 6.
$T
Let xi be a meridian of u.
has already been Then, by using
linking numbers (see D. Rolfsen Chapter 5 ) , it follows t h a t it is possible t o consider
$,
t o be such t h a t
$,(Xi) = wi
. x i o . wi-1
where wi is an element of G(L(o')) f o r every i and {1,2,.
.., n l
wi. (long X
ie
8
is a mapping of
The longitude l o n g x (u) is mapped by i ( u ' ) ) ~ . w ~ 'which , belongs t o the conjugate subgroup into i t s e l f .
$,
onto
of the peripheral subgroup <
xie, long
'i e
(at) >
of the group G(L(u')).
The required r e s u l t now follows from the f a c t t h a t
there is a one-to-one correspondence between complete conjugacy classes of peripheral subgroups of G(L(u)) and components of the link L(u). 17.5.
NON ORIENTATED EQUIVALENCE CRITERION FOR LINKS.
Suppose t h a t L(u)
and L(u') have isomorphic groups such t h a t t he resuZting isomo2rphism is such t h a t
$*(Xi) = w.1 x i e wi but $*(long
(u)) # wi xi
. long
(a')
'ie
. wi-1
CHAPTER 1 7
238
Then L(a) is not oriented equivaZent t o L ( o ' ) .
for some i i n G ( L ( o ' ) ) .
This i s a d i r e c t consequence of 17.4. Theorem. EXAMPLE.
17.6.
We show t h a t the l e f t handed t r e f o i l knot and the right
handed t r e f o i l knot are not orientated equivalent knots.
By 17.3.
Example (1), we know t h a t long a-'b long
a-lb
(03)
=
b-2(a-1b)6
=
2 -1 -6 b (a b)
and
which gives the peripheral subgroup
We now show t h a t there does not e x i s t an automorphism of G which maps b2(a-1b)-6
onto b-'(a -1b) 6
and which induces the i d e n t i t y automorphism on G/Gt = < a-'b
; - >.
centre z(G) of G is the (normal) subgroup generated by a' in G. morphisms of a group map the centre onto i t s e l f .
.
Every automorphism of G/z(G) i s of the form a + a o r a2 o r b a b b
+
b
o r a-'b
a
by 1.21. Exercise (2). (b a)6
onto
or b a2 b
o r aba-' This automorphism i s intended t o map
(a 2 b) 6 .
Auto-
So we can now consider
the automorphism of the f r e e product G/z(G) = < a ; a3 > * < b ; b2 >
The
MERIDIANS AND LONGITUDES
239
There are only two automorphisms of G/z(G) which w i l l do t h i s , namely, a + b a 2 b,
b-tb
2 a + a
b + a
and -1
,
ba.
However, neither of these automorphisms induce the i d e n t i t y mapping on
’*
(G/z(G))/(G/z(G)) RESULT.
17.7.
Suppose t h a t t he n-braid u is such t h a t
which i s a f r e e product w i t h amalgamation.
Here w2 i s a l s o taken t o be an
automorphism of t h e f r e e group
long x. ( 0 ) = long 1
PROOF.
(w,) xi
. long yi(w2) .
Apply the automorphism
wil
of the f r e e group
t o t h e group presentation <xl
,..., xn
; x 1u = x l
,..., xn u = xn ’ .
This gives the isomorphic group presentation on generators xl,. defining r e l a t i o n s
.. , xn and
CHAPTER 1 7
2 40
XIWl
=
xlw;l,
..., x w
=
sw;?
These relations are of the form:
for j
'jWl = 'j XiWl
=
-1 x.w 1 2
x j
=
x.w-l 1 2
< i
for j > i
.
By 6.2. Consequence, we have t h a t
I f we now apply the automorphism w2 of the free group F(Ixl,.
. .,
X i - 1 ~Y
iJi+l,*.*y
\I)
3
then we get back t o the original group presentation.
We further g e t , by
an argument similar t o the one given i n Proof of 15.3. Lemma, t h a t
The required decomposition is now obtained by noting t h a t L(wi1w1w2) and L(wl) are s t r i n g isotopic links (see 5.4. Exercise ( 2 ) ) .
The r e s u l t
concerning the longitude of the composition of two knots follows now from the definition of longitudes.
CONSEQUENCE.
17.8. Ul
= Wi(Ul
Suppose t h a t t h e n-braid u' is such t h a t
)...)q l ) . w p i )...) n- 1) (I
and so t h a t the Zinks L(wi) and L(w2) respectiveZy.
Then
L(wi)
are equivaZent t o the Zinks L(wl) and
MERIDIANS AND LONGITUDES
241
Thus, in particular,
EXERCISE.
17.9.
(1) I t follows from 17.7. Result and 17.3. Example (1)
that
and long x2
(ul-3u2) 3 = (x2x1x2)-2. x;. (x2x3x2)2
By 6 . 2 . Examples (4) and (5)
, the
. x2-6 .
granny knot and the square knot have
groups isomorphic t o <
x1,x2,x3 ; x1x2x1
=
x2x1x2
y
x2x3x2
=
x x x 3 2 3 > '
Simple calculations show t h a t (u;3u23)
long
=
-1 x1-1x2
x2
x-1 3 . x1 . x-1x -1 . x 23 . x 3-1x 2-1 . x 3 . x -1 2 3 'x2
while long x2
-3 3 (ul u2) = x-lx-' 1 2
. x1 . x2-1x1-1 . x3x2 . x3-1 . x2x3 .
Now follow the account of R.H. Fox [11 t o show t h a t the granny knot and the square h o t are not orientated equivalent knots. (2)
Let a be an automorphism of the f r e e group F({xl,. E
x.a = w..x iA.wi 1
Show t h a t
G(L(o))
such t h a t
-1 f o r every i , where every wi i s an element of the f r e e
group, A is a permutation of ly2y...y n and 1.
.., xnl)
Gn(a-lua)
f o r every n-braid u.
E
i s always e i t h e r -1 o r always
242
(3)
CHAPTER 1 7
Show t h a t the assumption i n 1 7 . 4 . Theorem t h a t u and
0'
a r e both
braids on the same number of s t r i n g s i s not r e a l l y a r e s t r i c t i o n i n the context of t h i s theorem. 17.10.
FURTHER RESULTS.
Deep r e s u l t s of F. Waldhausen C11 show t h a t a
knot i s determined by i t s group, a meridian and t h e corresponding longitude. J.H. Conway and C.
McA. Gordon have used t h i s r e s u l t t o construct a group
for an a r b i t r a r y knot which c l a s s i f i e s a l l knots.
This group contains the
group of the knot as a subgroup and contains two e x t r a generators which a r e associated with a meridian and the corresponding longitude.
24 3
CHAPTER 18 S Y M T R Y OF ALEXANDER MATRICES OF KNOTS
We begin by recalling some of the basic r e s u l t s which we w i l l use i n t h i s chapter.
Associated with every n-braid u o f the form
Ql
Qn
we have a s t r i n g isotopy c l a s s of links which i s determined by identifying Pi with Qi f o r every i.
In p a r t i c u l a r t h i s can be done by employing the
construct ion
Ql
Qn
244
W T E R 18
where Pi is joined t o Qi for every i by a polygonal path lying behind the braid u with no further crossings being allowed. by L(u)
.
Such a link i s denoted
By a r e s u l t of Alexander (see 5.2. and end of Chapter 5 ) , every
link is s t r i n g isotopic t o a link of the form L ( u ) . Every n-braid u has associated with it the n loops X1'
..., xi,
xn
. . . I
a t the base point P I where
for a l l i. There a r e two ways of inserting arrows along the braid str ings a l l up o r a l l down.
-
either
We adopt the convention t h a t once the arrows have been
fixed along the braid s t r i n g s , then the arrows along the loops xl,...,
xn
are chosen so that t h i s gives a right handed corkscrew. Finally we r e c a l l t h a t the group G ( L ( u ) ) of the link L(o) i s the fundamental group of the space G(L(o)) =
C,
(L(u)) and
-
Here u i s an n-braid,
0 is the automorphism of the f r e e group F(I xl,..
associated with u and Gn(a) is the group of
0 (see Chapter
., xnl)
15).
18.1. THEOREM.
Let K be a knot and A(K,t) denote t h e correspoding -1 Alexander matrix of K. Then A(K,t) i s equivalent t o A ( K , t ) ' . PROOF. u
=
w(ui)
The knot K i s s t r i n g isotopic t o a knot of the form L(a), where is an n-braid and the corresponding permutation 11. is an n-cycle
245
SYWTRY OF ALEXANDER MATRICES OF KNOTS
(see 5.3. Result).
A(K,t)
=
By 13.6. Example (1) and 13.7. Theorem,
w ( c ~ ) " ( ~-) In 9
where + ( t ) is the Burau representation of the b r a i d group Bn.
for i
=
1,2,
..., n-1. I_ Ii-1
Now 0
0
O
l
Then it is easy t o v e r i f y t h a t D2
Let
for 1 s i
<
n.
So
As + ( t ) and +(t-')
=
In and
a r e group homomorphisms, t h i s gives t h a t
Now = )Rev( ( A -1w(ci)a)-'), W ( -1 U~-~
where we use the same notation a s used i n Chapter 14.
The required r e s u l t
2 46
CHAPTER 18
now follows from the f i r s t given equality of t h i s proof and 14.3.3. Theorem i f we can show t h a t the group of the knot
i s isomorphic t o the group of the knot L(w(ai)) under an isomorphism which induces the i d e n t i t y automorphism on the group
t ;
-
>.
Here we are
also using the f a c t t h a t i f one multiplies an Alexander matrix by the matrix D , then one only permutes i t s rows o r columns and hence obtains an We r e c a l l t h a t every knot group modulo its commutator
equivalent matrix.
subgroup i s naturally isomorphic t o the group < t ; - > . (plus first p a r t of proof o r 14.1.) results
.
18.2.
COROLLARY.
Now 15.3. Lemma
gives us the required necessary
I f f ( t ) i s the AZexander polynomial of a knot, then
f ( t ) = td f(t-1) f o r some nonnegative i n t e g e r d.
I t i s a l s o necessary t o use the f a c t t h a t f(1) # 0 (see 16.7. Exercise ( 5 ) )* 18.3.
Similar r e s u l t s t o the ones given above hold f o r reduced
NOTE.
Alexander matrices and reduced Alexander polynomials of a r b i t r a r y links. The method of proof i s the same. I t i s interesting t o compare the methods and r e s u l t s of t h i s chapter with those of Fox and Torres, Chapter 8 (C7).
I t is an old r e s u l t of S e i f e r t t h a t the properties
1
f(1) =
k
f(t)
td f(t-1)
=
Crowell and Fox Chapter I X and D. Rolfsen
(see 16.7. Example ( 5 ) )
247
S W E T R Y OF ALEXANDER MATRICES OF KNOTS
characterise the Alexander polynomials of h o t s (see for instance D. Rolfsen Chapter 7 (CS)) 18.4.
.
EXERCISE.
(1) Suppose t h a t the Alexander polynomial of a h o t can
be expressed i n the form
antn + an- 1tn-l +
... + a 1t
with an and a. being nonzero.
+ a.
(every ai is an integer)
Show t h a t n i s even and al i s an odd 2n
integer. (2)
Show t h a t the Alexander polynomial of a h o t can be expressed i n the
f om th + h 1 ci t h - i ( l - t ) 2 i i=l
... , ch.
f o r integers c1 ,c2 ,
Show t h a t t h i s polynomial i s d i v i s i b l e by the
Alexander polynomial t + q ( l - t ) '
i f and only i f
where q i s an integer. (3)
Show t h a t i f
i s the Alexander polynomial of a h o t , where a2h # 0, then there e x i s t s a matrix B i n the group SL(2h;Q) so t h a t f ( t ) = aZh det(tIZh-B). One can proceed as follows.
F i r s t l y suppose t h a t a;;
f ( t ) is an
irreducible polynomial i n Q C t l , where Q i s t h e f i e l d of r a t i o n a l numbers.
CHAPTER 18
248
Then
-1 f ( a ) a2h
=
0
f o r some complex number a .
This enables one t o define a linear mapping 6
of the algebraic number f i e l d Q ( a ) i n t o i t s e l f by + ( & ) = aj+'
for a l l j <2h,
mapping t h a t a;:
+
..
,a2h-1 as a l i n e a r space over Q. The l i n e a r -1 has aZh f ( t ) as i t s c h a r a c t e r i s t i c polynomial. Secondly suppose
since Q ( a ) has a basis l , a , .
f ( t ) i s reducible over Q C t l .
irreducible factors over Q C t l .
Then it s p l i t s i n t o a product of
For each irreducible f a c t o r we proceed as
above and construct an algebraic number f i e l d and a corresponding l i n e a r mapping.
Now take the d i r e c t sum of these algebraic number f i e l d s and the
d i r e c t sum of t h e i r linear mappings. This is the l i n e a r mapping which has -1 a2h f ( t ) as i t s c h a r a c t e r i s t i c polynomial and a l s o its determinant i s equal t o 1. Note t h a t the c h a r a c t e r i s t i c polynomial g ( t ) of a matrix belonging t o SL(2h) has the symmetric property g ( t ) = t2hg(t'1).
249
CHAPTER 19
S Y M T R Y OF ALEXANDER MATRICES OF LINKS
The r e s u l t s and methods of proof i n t h i s chapter are similar i n s p i r i t t o the previous chapter except t h a t the detailed calculations are more For t h i s reason some of the d e t a i l s of proof w i l l be omitted
unpleasant.
interested readers can find them i n S. Moran 131. and A . r 9s can be found i n Proof of Theorem 4.9.
The next lemma deals with simple r e s u l t s on the n-braids A Relevant r e s u l t s concerned with A
r,s
and i n 4.10. Examples.
19.1.
LEMd4.
Let A denote a generator o f t he pure braid group Pn. r,s
Then
where w(A
) denotes an ar bi t r ar y element i n t he pure braid group Pn.
Here r <
r,s s 5 n and r,s
PROOF.
One needs t o know t h a t
=
1,2,
..., n.
and A-'
ui A = u
n-i
f o r a l l i.
-
CHAPTER 19
2 50
The l a t t e r r e s u l t can be found in 4.11. Exercises (4) and ( 5 ) . EM.
19.2.
Suppose t h a t p and q are integers s o t h a t
1sp s q
5
n.
Then
*-1(Rev ap,q)* = un-qtl,n-p+lS where u
PI9
is short f o r the n-braid
up opt1 " ' uq-1 with u
= 1.
PIP
This is a simple exercise on the formula
PROOF. A
-1 u i A = u n- i
f o r a l l i.
We r e c a l l some of the notation given i n 13.2. Notation and 13.3. Notation.
The l a t t e r w i l l be amended somewhat i n order t o s u i t our
present s i t u a t i o n .
NOTATIm.
19.3.
Let p and
recollect t h a t the pattern of only i f when p and
1.1
be permutations of 1 , 2 , .
p p
.., n.
Then we
i s s a i d t o embrace the pattern of p i f and
are decomposed i n t o d i s j o i n t cycles one has t h a t every
p a i r of integers which belong t o the same d i s j o i n t cycle of p a l s o belongs t o the same d i s j o i n t cycle of
p.
Amending s l i g h t l y and extending the
notation used in Chapter 13 we proceed as follows. and
and v be permutations of 1 , 2 , .
p
. . , n so t h a t
the pattern of the permutation associated with u . $&, a
--
9
tn,,
Let u be an n-braid t h e pattern of Then
p
embraces
SYWETRY OF ALEXANDER MATRICES OF LINKS
denotes the n
x
251
n matrix
Here xi = ti“ -lJ-
denotes the following ring homomorphism which maps t h e group r i n g
onto the “polynomial” ring
with the additional conditions t h a t ti = t . i f and only i f i and j belong t o the same d i s j o i n t cycle of 1
p
The actual ring homomorphism i s defined by the data xi The n
x
+
tiu
f o r a l l i.
n matrix
i s defined i n a similar way with the corresponding ring homomorphism being defined by the data
In the p a r t i c u l a r case when t h a t the ring homomorphism
l~
is the permutation associated with a , one has
.
CHAPTER 19
2 52
x = t
-v-
(which i s short for xi = t i ) i s given by f i r s t going over t o the group ring -1-I-
Zi!
G ( L ( u ) ) and then making t h i s ring commutative.
In t h i s new notation 13.5. Theorem and 13.7. Theorem can be restated
as given below.
THEOREM.
19.4.
permutations SO
p
Suppose that a and
and v respectively.
are n-braids u i t h corresponding
T
If
is a permutation of l,Z,..., n
that the pattern of X embraces the pattern o f p , v and pv, then
Further
i s the AZexander matrix of the Zink L ( o ) . We come now t o the main r e s u l t of t h i s chapter,
..
Let L be a Zink with c components and A(L ; tl,. , tc) -1 be i t s AZexander matrix.. Then A(L ; t, ) ’ i s equivaZent t o TIEOFEM.
19.5.
ti’, ...,
A(L ; t l , .
.. , tc).
.
Hence i f f ( t l y . . , tc) i s the Alexander poZynomiaZ of
L, then
f(tl,
..., t ) = 2 tla1 ... t,ac f ( t ,-1,..., tc-1) C
f o r some integers al,. PROOF.
.,
a
C‘
As was implied in 4.8. Note
S W E T R Y OF
253
ALEXANDER MATRICES OF LINKS
where Pn i s the normal subgroup of a l l pure n-braids and Sn i s the symmetric group on 1 , 2 , p
..., n
(see also 4.10. Example (1)).
For each permutation
belonging t o Sn we choose an n-braid M whose corresponding permutation
is
P
P.
Then every element of Bn has a unique representation of the form
where w(A ) is a word i n the generators of t h e pure braid group Pn. By r,s 5.4. Exercise (2), conjugate braids give r i s e t o s t r i n g isotopic links. Hence we need only consider a collection of elements, where M is chosen P
so t h a t u comes from a complete s e t of representatives of the conjugacy classes of Sn.
In f a c t we take
p
t o be a product of d i s j o i n t cycles of
the form p =
(1, il-l, i l - Z , . .
., 2)
(il, i2-1, i 2 - 2 ,
..., i l + l )
... (ik-l,ik-1, ik- 2 ,..., ik-l+l), where a l l the integers belong t o { 1 , 2 , . . . ,
n).
Also the cycles are such
t h a t they decrease in length as one goes from l e f t t o r i g h t .
The
corresponding n-braid we take t o be
We consider two p a r t i c u l a r cases before we deal with the general case. (i)
Suppose t h a t L i s the l i n k associated with the pure braid w(ArYs).
Then, by 19.4. Theorem, the Alexander matrix of L is J'id (tl,' * ' Y tn) , = w(ArYs) A(L ; t i , . * *tn)
-
In
where qid is the Gassner representation qG as considered i n 13.6. Example (2).
There one can find the actual value of the matrix representation
CHAPTER 19
254
Now a s l i g h t l y tedious calculation shows t h a t
..., tn)
= D(tl,
-1 $id(t,l,.", t l 1
%-s+l ,n-r+l
. D(tl,,..,
tn) -18
where
0
t2
D(tl,.
.., tn)
=
tn'
0
On using 19 1. Lemma and the product formula i n 19.4. Theorem, one gets that "('r,
s
. D ( t l , ..., tJ-1 . This i s the r e s u l t we w i l l need. (ii)
Suppose t h a t L is the link associated with the braid M
19.4. Theorem, the Alexander matrix of L is
Now it is easy t o see t h a t
P
.
Then, by
255
S W E T R Y OF ALEXANDER MATRICES OF LINKS
k- 1
M)
.
$?J = j= n o (u.i j , ij+l-l)
So it is necessary t o consider the p a r t i c u l a r case of
where v is the permutation (p q q-1
... p + l ) .
A straightforward calculation shows t h a t
r
IP-1 0 0
0
0
1-t P 1
t - t2 P P 0
... ... *..
0
0
t9-Pqt-P P P
tq-P P
0
0
1
0 0 0
lo
1 0
......
0
I
0
n-q
We now introduce two further b i t s of notation. D(t l,..., tn ; A ) ,
where X is a permutation of 1 , 2 , . .
F i r s t l y we l e t
., n,
be the matrix
tn) a s given i n p a r t ( i ) of t h i s proof together with the identi-
D(tl,..., fications
ti = t . i f and only i f i and j belong t o the same d i s j o i n t cycle of
3
Secondly i f =
[ ii]
then the dual A ' of 1 is defined t o be the permutation
a.
CHAPTER 19
256
' 1
=
[
n-i+l n-ai+l
]
This a r i s e s i n braid theory because A' i s the permutation associated with A-1
u A when X i s the permutation associated with u. A further calculation shows t h a t using 19.2. Lemma we have t h a t
= D(tly..
., tn ; v ) .
-1 ( A (Rev u
-1 Jlvt(tn S ' . ' Y
)-'a) P ,9
-1 tl 1
. D ( t l y ..., tn ; v) -1 . So f i n a l l y one has t h a t
(iii)
Suppose t h a t L i s the link associated with t h e n-braid
"(%,s) ' where w(A
TYS
)
(with 11 as given above),
MP
# e and
p
# id.
By 19.4. Theorem, one has t h a t the
Alexander matrix of L i s
where cases ( i ) and ( i i ) of the above proof give the form of each term of the product.
So altogether for every s t r i n g isotopy c l a s s of links there
is an n-braid u such t h a t the Alexander matrix of a link in t h i s c l a s s i s
257
SYMMETRY OF ALEXANDER MATRICES OF LINKS
with
F in ally it remains t o i nves t i gat e how t he l i n k s corresponding t o A
-1
Rev u ,
u A,
u
-1
are r e l a t e d t o the l i n k L(o).
This i s s e t t l e d i n 14.1, 14.3 and 14.2
with 15.3 Lemma. 19.6.
EXERCISE. (1) Verify the s t e p s i n t he following argument which
shows : Let f ( t k
, 1 tk2
components K1,
,...) t
) be the Alexander polynomial of a l i n k L w i t h c
kc
KZy...,
and L' be the l i n k obtained from L by removing
Kc
the component Kc, where c
2
2.
Then f ( t k
,..., tkc-l, 1)
i s equal t o
1
R1 k 2 Itkl tkz
of L ' .
times the Alezander polynomial
*"
Here a
m
denotes the linking number of the component Kc w i t h the component Km, f o r
m
= 1,2,...y
c-1.
( i ) L is the l i n k associated with an n-braid u y where t he permutation associated with u can be taken t o be a product of d i s j o i n t cycles
u
= (1 2
which i s 1 2 3
... kl) ... n-1
(kl+l
..'. kz) ... (kc-1+1 *
kc)
Y
n with t he above given d i s t r i b u t i o n of brackets.
CHAPTER 19
258
I t can be assumed t h a t the braid u can be expressed i n the form
(ii)
u = u
. a" . u'
y
where u i s a pure n-braid which belongs t o the free group
and which i s freely generated by these generators with s = kcml+l.
a'
'
Further
and u' are n-braids whose 1, 2 , .
.., s-1 s t r i n g s ,..., n
and s , s + l
strings
are undisturbed respectively. (iii)
The Zinking number km is equal t o the sum of the exponents of Ails,
where i varies over the set of indices
when u i s expressed as an element of the free group Us, €or m = 1 , 2 , See D. Rolfsen p.132 €or the concept of Zinking numbers.
. I , I=
Is-1 0
where T = tk 5 tka2 1 2
... tk:: .
Also
*
o
T
O
...,C-1.
2 59
SYMMETRY OF ALEXANDER MATRICES OF LINKS
0
0
...
0
1
... 0 0 0 1 ... 0 0 . . ... . . 0 0 ... 1 0 1
0
0
Finally
A
*
0
-1
0
T
0
..a
... 0 0 0 1 ... 0 0 . . ... . . 1 -1
0
0
0
0
0
-1
...
0
1 -1,
where A is the Alexander matrix of the l i n k L’ which is associated with the (s-1)-braid u t (2)
.
Use the r e s u l t proved in Exercise (1) t o derive the following
additional information t h a t i s proved in the reference given below. f(tk
,...) tkc) =
1
41-1 tkl
(-1)
qc-l ... tkc
,..., tkc-11 ,
-1 f(tk 1
where qi is congruent moduZo 2 t o the Zinking number of the knot Ki w i t h the l i n k K1 u
... u Kiml
u Ki+l
u
... u Kc
for i = 1 , 2 , .
..,
C.
(Details of proof can be found i n G. Torres and R.H. Fox 17). (3)
Produce a short proof of, 19.5. Theorem which uses 14.3.3. Theorem and
20.18. Lema Part ( 5 ) .
This Page Intentionally Left Blank
26 1
W T E R 20
CONJUGACY OF GROUP A U T W R P H I SMS
We consider some generalisations of concepts introduced i n Chapter 15. I f 8 and
y
are automorphisms of an a r b i t r a r y group G , then we show t h a t they
are conjugate in ht G (the automorphism group of G) exactly when there e x i s t s a certain type of isomorphism between the HNN R.C.
-
extensions (see
Lyndon and P.E. Schupp f o r the general theory of HNN <
G,x ; x -1 g x
= g 8
for a l l g i n G >
<
G,x ; x -1 g x = g y
for a l l g in G >
-
extensions)
and
.
We then apply t h i s r e s u l t t o examples i n the braid group Bn 20.1.
DEFINITION.
(c
Aut F,).
Let 8 be an endomorphism of an a r b i t r a r y group G.
Then R(8) w i l l denote the normal subgroup of G generated by a l l elements
of the form (g 8 ) . g-l, where g varies over G. G/R(8) by
G G (8)
We denote the f a c t o r group
a
The following two lemmas are analogues of 15.1. Lemma and 15.2. Lemma. Their proofs are similar t o t h e i r analogues and so we omit them. 20.2.
E M V I A .
Let 8 be an endomorphism of an a r b i t r a r y group G.
induces the i d e n t i t y automorphism on
G
(8).
Then 8
Any f a c t o r group of G on
which 8 induces the i d e n t i t y automorphism i s isomorphic t o a f a c t o r group of
@).
I f G is generated by the elements gi, i
E
I , then
G
(8) i s
CHAPTER 20
262
given by the group G modulo the relations
Suppose t h a t a i s an automorphism and B i s an endomorphism
EhRvl.4.
20.3.
Then
of an a r b i t r a r y group G.
- G ( a -1 and
=
gG(a)
induces an isomorphism of
01
DEFINITION.
20.4.
Let
f3
G G(B)
-1
onto
G(a
Ba).
be an endomorphism of an arbitrary group G.
We
take a new symbol x and consider the f r e e product
of G with the i n f i n i t e cyclic group < x ; - >.
Then B can a l s o be taken
t o be an endomorphism of the free product when we define =x.
X B
We w i l l consider the group
where
Lx
h
denotes the inner automorphism
+
x-lh x
f o r a l l h i n the above given free product. denote the above defined group by G(B
We usually abbreviate and
L;').
We omit the proof of the following r e s u l t which i s easy. 20.5.
L E N .
Suppose t h a t B is an automorphsim of an a r b i t r a r y group G.
263
CONJUGACY OF GROUP AUTOMORPHISMS
is the semi-direct product of
Then G(6 J);.'
x ; - > and the normal
<
subgroup G with
x-1g x
=
Suppose t h a t B and y are automorphisms of an arbitrary
THEOREM.
20.6.
f o r a l l g in G.
g8
Then the following are necessary and s u f f i c i e n t conditions f o r
group G.
there t o e x i s t an automorphism a-li3
CI
=
CI
of G so t h a t
y.
There e x i s t s an autornorphism 6 of G *
<
x ;
-
> which induces an i so-
morphism of
so t h at
(i) 6 r e s t r i c t e d t o the norma2 subgroup G of
-1 G(B 1, )
gives an auto-
morphism of G;
(ii) 6 induces th e i d e n t i t y automorphism on G.
In fact
PROOF.
(a)
&IG =
CI
and
6 =
Suppose that a-lg
the automorphism 6 of
Now
x
CI
x ;-
> modulo the centre o f
x. = y
f o r some automorphism
CI
of G.
Define
MAPTER 20
264
Hence, by 20.3. Lemma, we have t h a t 6 induces an isomorphism of
Clearly the above conditions (i) and ( i i ) hold, by 20.5. Lemma. (b)
Suppose t h a t the isomorphism 6 e x i s t s and s a t i s f i e s conditions (i) and
(ii).
Then
x
6
x.u,
=
where u i s an element of the centre Z(G) i n
G(y 1;').
Now i n G(g i i 1 ) 6
we have t h a t
g B 6
=
(x-lg x)6
=
u-lx-l.
=
(g 6 ) y
(g 6 ) . xu f o r a l l g in G ,
since u belongs t o Z (G)
.-I$
a =
.
Put 6 I
= a.
Then we have t h a t
y
i n Aut G. 20.7.
NOTE. G(Y
-1 1, 1
I f u is an element of G , then B G(Y
-1 ixu)
This isomorphism i s induced by the following automorphism of G x
-+
xu
and g
+
g
for a l l g i n G.
*
x ;-
>
265
CONJUGACY OF GROUP AUOMORPHISMS
This enables one t o e s t a b l i s h , as i n the proof of 20.6. Theorem, the following Suppose t h a t B and y a r e automorphisms o f an a r b i t r a r y
THEOREM.
20.8.
group G.
Then t h e f o l l o w i n g are necessary and s u f f i c i e n t c o n d i t i o n s f o r
there t o e x i s t an automorphism =
y
of G so t h a t
c1
modulo ~ n nG,
where Inn G denotes t h e normal subgroup of i n n e r automorphisms o f G i n There e x i s t s an automorphism 6 of
Aut G .
* < x ; - >
G
which induces an isomorphism o f G(R I-I) -X
onto G(y
Lil)
so t h a t
(i)
6 r e s t r i c t e d t o t h e normal subgroup G o f G(B
Jx-1) g i v e s an auto-
morphism of G ;
(ii) 20.9.
6 induces t h e i d e n t i t y automorphism on < x ;
-
>
modulo t h e group G.
EXAMPLE. (1) Suppose t h a t
Then the mapping R(m) such t h a t a ~ ( m ) = a b m and b B(m)
=
b,
where m i s a nonzero integer, gives an automorphism of G. G(m) = G(B(m) J i l )
= <
-1
a,b,x ; a - l b-la b =1, x-la x = a bm, x bx = h
>,
266
CHAPTER 20
Hence G(ml) # G(m2)
lmll # lm21.
, when
So B(ml)
ml and m2 are nonzero integers such t h a t
i s not conjugate t o B(m2) i n Aut G f o r lmll # I m 2 ( .
On the other hand G(m) i s isomorphic t o G(-m)
under an isomorphism
which sends a
-+
a,
b
-f
b-l,
x
-+
x.
A l l of t h i s can be expressed i n Matrix Theory as saying t h a t
are not conjugate i n GL(2,ZZ) when lmll # Im21.
(2)
Suppose that
Then the mappings B and
ay = a
y
defined by
and b y = a - ' b a
give automorphisms of G.
Now
While
CY
261
CONJUGACY OF GROUP ALJTOYORPHISMS
G(6L;')
= <
a,b,x ; a
3
= b
3
= 1, x - l a x = a',
-1 3 -1 G(yLx ) = < a,b,x ; a3 = b = 1, x a x = a ,
x -1b x
=
b >,
-1 bx
=
-1 a ba >
x
.
Their factor groups with respect t o t h e i r commutator subgroups are obviously not isomorphic.
Hence the groups are not isomorphic, which
gives t h a t B i s not conjugate t o
y
i n Aut G (or even i n Aut G modulo Inn G ) .
W e now apply the above theory t o braids.
With every braid a of the
braid group Bn we associate a braid u p n of Bn+l which i s obtained from u by looping a new s t r i n g around it i n the following way. 20.10. DEFINITION.
pn =
an
... u 2 u 1
a
2
*'.
n-1 'n
The following properties of the (n+l)-braid
pn
a r e e a s i l y verified.
CI-WTER 20
268 -1 -1 x i Pn = %+I
(2)
-1 X n + 1 P,
(3)
p
=
(XI
' xi
' 0 '
for i s n
xn+1
sxn+J
1
.
Xn+l
(XI
*
*
'
xnxn+l)
n belongs t o the centraliser of B, i n Bn+l.
According t o 6.1. Theorem of Artin and Birman and 20.1. Definition, the group G(L(a
p,))
of the l i n k L(a
is G
)p,
(a pn).
-Fn+ 1
We now put t h i s
group i n t o familiar form. 20.12.
IEWA.
Let a be an n-braid.
= <
xl,
Then
..., 3,xn+l ; s+l -1 xi xn+l
In t h i s group the c e n t r a l i s e r of xn+l contains x 1x 2 PROOF.
= x1 .a
for i s n >
'.. 5.
The f i r s t equality follows from 20.11. Lemma p a r t ( 3 ) , while the
second equality follows from 20.3. Lemma.
Now
-1 and R(a pn)pn i s the normal subgroup generated by a l l elements of the form
as g varies over Fn+l.
I t is easy t o v e r i f y t h a t R(a
pn)pil
the normal subgroup generated by a l l elements of the form
is i n f a c t
269
CONJUGACY OF GROUP AUTOMORPHI SMS
Now one can use 20.11. Lema p a r t ( 2 ) t o obtain the t h i r d equality.
Note
t h a t the l a s t statement of t h i s Lemma is a consequence of the defining relations for a l l i
2
n
and the f a c t t h a t
20.13.
CONSEQUENE OF 20.6. THEOREM.
Suppose t h a t u and u ' a r e n-braids.
Then there e x i s t s an automorphism a of the free group Fn such t h a t a-1 u a
=
u'
i f and only i f there e x i s t s an automorphism 6 of the f r e e group Fntl which induces an isomorphism of
so t h a t (i)
6 r e s t r i c t e d t o the normal subgroup Fn of
sF
(u pn) gives an
n+l
automorphism of Fn; (ii)
6 induces the i d e n t i t y automorphism on <
In f a c t 6 1
and ~
= a
~ = x+
n + l~'
Fn 20.14.
r-lu
T
=
T
>.
6
CONEQUENCE OF 20.6. THEOREM.
Then there e x i s t s an n-braid
x ;-
Suppose t h a t u and u 1 a r e n-braids.
such t h a t
u'
i f and only i f the links L(u P,)
and L(u'pn) a r e equivalent under an
orientation preserving homeomorphism which maps the directed loop given by
2 70
CHAPTER 20
the (n+l)-th s t r i n g onto the directed loop given by the (n+l)-th s t r i n g . For a more detailed proof the reader i s asked t o consult
OUTLINE PRDOF.
H.R. Morton C11, Proof of Theorem 1.
A meridian (see Chapter 17) i s
mapped onto a meridian under the given homeomorphism. xi
-1 Ai xiu Ai
+
So
for a l l i
and n
n
xi i=l
n xi i=l
-+
Hence the corresponding automorphism of Fn i s
by the l a s t condition. given by an n-braid. DEFINITION.
20.15.
Let Fn be a f r e e group with X = {xl,
s e t of free generators.
An automorphism
and only i f there e x i s t elements A1, 1,2,
...,
c1
...,
as a
of Fn i s said t o be Xinn i f of Fn and a permutation p of
..., n such t h a t x.
= A. 1
1
20.16.
x;+1 A;'
EM.
1u
f o r a l l i.
The collection of a l l Xinn automorphisms of the f r e e group
Fn forms a subgroup Xinn Fn of Aut Fn which contains Inn Fn.
We introduce an Xinn automorphism which is important f o r Braid Theory. 20.17.
DEFINITION.
generators xl,.
x 1. r
= X
The automorphism
r
of the f r e e group Fn on free
.., xn is defined by
-1
n-i+l
for a l l i.
The following properties of
r
are e a s i l y verified.
271
CONJUGACY OF GROUP AUTOMORPHISMS
20.18.
LEIMA.
(4)
r -1A -1
(5)
A
-'r -'.
ui
. rA
u-'
. I'A
-1
= ui
=
Rev u
for 1 s i < n f o r a l l n-braids u
Property (3) above a s s e r t s t h a t conjugating a braid by
r is equivalent
t o looking a t the braid from the other side. By considering the action of a on Fn/F;,
one can obtain the following
simple extension of a well known r e s u l t of E. Artin C21 (see 4.8. Lemma). 20.19.
Suppose t h a t a belongs t o X i n n Fn.
LEWA.
Then
a
i s a braid
i f and only i f
20.20.
EXAMPLE.
B1 =
uil
The Burau representations of the braids
u f uY2 u 2
and
-2 u 2 B2 = u 2 u1
-1 2 ul
have the same c h a r a c t e r i s t i c polynomial, but B1 and 2 a r e not conjugate i n B3 (as i s shown i n J.S. Birman Corollary 3.11.3).
272
CHAPTER 20
20.18.
Lemma p a r t (3) gives t h a t
and hence
where 0
k]
0
D=[Y
1 0
0
and 11, denotes the Burau representation of Bj.
THEOREM.
20.21,
The n o m a l i s e r of Bn in Xinn Fn is the semi-direct
product
r
<
>
for n
2
2.
Suppose that a belongs t o the normaliser of Bn i n Xim Fn.
PFUOF. A
. Bn
2
= (ul
... un-1) n
Then, by 1.18. Example ( 2 ) and 4.11. Exercises (4), (S), A centre of Bn and x. 'A
=
1
f o r a l l i. (xl
(xl
As u
...
XJ
. xi . (xl ...
~ ' ~ 1belongs -l
... 5 ) a A2 u -1 -- x1
..a
t o Bn, it follows t h a t
5 ,
which implies t h a t (xl
Let
... XJ
[(xl
... ]&x
(xl
...
XJ
-1
= (XI
* *
2
belongs t o the
CONJUGACY OF
... %)a
... xn, m
= (xl
f o r some integer m, i n Fn.
As
Y
CI
belongs t o Xinn Fn, the induced auto-
inorphism on Fn/FA gives t h a t m = t 1.
By 20.19. Lemma, t h i s shows t h a t
by 20.18. Lemma ( 2 ) .
. Bn.
>
Suppose t h a t a and a! are n-braids and t he l i nk L(a) is I f there e x i s t s an Xinn automorphism
u n s p l i t ta b le.
F on the s e t of f r e e generators X = Ixl,
n
a%
belongs t o
The reverse inclusion is a consequence of 20.18. Lemma (3).
THEOREM.
20.22.
CI
Hence the normaliser of Bn i n Xinn Fn i s contained in
< r > . Bn. r
I f m = 1, then by 20.19. Lemma, a
On the other hand i f m = -1, then
belongs t o Bn.
<
273
So
holds i n Fn. (xl
GROUP AUTOMORPHISMS
a
=
..., xnl
CI
of the f r e e group
s o t hat
a',
then L ( u ) and L ( a ' ) are equivalent l i n k s .
By 20.3. Lemma,
PROOF.
G (u) -F n
CI
induces an isomorphism of
onto G ( a ' ) . -Fn
If x. u 1
=
-1 Ai xiu Ai
for a l l i
then, by 17.1. Result,
Now x.au! 1
=
x.aa 1
=
(Ai")
. (xipa) . (Aia)-'
f o r a l l i, and
CHAPTER 20
2 74
x 1 a,...,
Xn"
are each meridians of L ( u ' ) , since a i s an X
h automorphism of Fn.
Once
again 17.1. Result gives t h a t (Aia)(Aiva)
f o r a l l i.
... . (xis)
-E
(i)
Clearly a maps longitude onto corresponding longitude.
So,
by F. Waldhausen C11 Corollary 6.5, the complement of a closed tubular neighbourhood of the unsplittable link L(o) i s homeomorphic t o the complement of a closed tubular neighbourhood of the link L(u').
Now it
i s well known t h a t t h i s gives the equivalence of the links L(o) and L ( u ' ) (see for instance D. Rolfsen Chapter 4 C3). We now determine the reduced Alexander matrix and the reduced Alexander polynomial of the link L(apn) i n terms of the Burau representation u ' ~ of u.
This w i l l enable one t o show that certain n-braids a r e not
conjugate i n Aut Fn. 20.23.
THEOREM.
Suppose t hat a i s an n-braid.
Then t he reduced
Alexander matrix of the l i n k L(apn) is equivalent t o the matrix
where $B i s t h e Burau representation of the braid group Bn. Alexander polynomial of t h e link L(apn) i s
det
(2B- t-lIn)
which i s d i v i s i b l e by t-1. PROOF.
The group of the link L(upn) i s
The reduced
275
CONJUGACY OF GROUP A U W R P H I S M S
by 6.1. Theorem of Artin and Biman together with 20.12. Lemma.
The
( i , j ) - t h entry i n the reduced Alexander matrix of t h i s group is
for j = n+l.
[ l - t
Hence the reduced Alexander matrix of the link L(opn) i s
where E
is the column matrix whose every entry is 1. Suppose t h a t aij n,l denotes the ( i , j ) - t h entry i n the above matrix. Then, by 8.7. Fundamental
Formula, me has t h a t
which implies t h a t n+l
1
j=1
a.. = O 1J
for 1 5
i
This gives the required r e s u l t s . In a somewhat similar way one can establish the following r e s u l t .
2 76
CHAPTER 20
THEOREM.
20.24.
Suppose t h a t u beZongs t o t h e group Pn o f pure
n-braids.
Then t h e AZexander matrix of t h e pure Zink L(opn) is equivaZent t o t h e matrix
I_ U
+G
-
L+l In
(l-'i)tn+l-1
,
I where $
G
denotes the Gassner r e p r e s e n t a t i o n of t h e pure braid group Pn.
The AZexander poZynomia2 o f t h e pure link L(upn) i s
20.25.
(1) The links L(01u2p3) and L(ulu;lp3)
EXAMPLE.
have reduced
Alexander polynomials (l-t3) (1-t+t2) respectively.
and
(1-t) (2-t+2t2)
By 20.13. Consequence of 20.6. Theorem, there does not e x i s t
an automorphism a of the free group F(Cx1,x2,x3I) so t h a t a-1cllu2a
=
U1U2
-1
.
Also t h i s equality does not hold modulo Inn F({x1,x2,x3I).
s p i t e of the f a c t t h a t the links L(o u ) and L(o om') 1 2 1 2 (2)
The links L(030102u;~up4~ and
L(o;101u203u2~4)
have reduced Alexander polynomials
This i s in
are s t r i n g isotopic.
277
CONJUGACY OF GROUP AUTOMORPHISMS
(l-t)(l+t3)’
and
( 1 - t ) ( l + t 2 +t4 +t6 )
By 20.13. Consequence of 20.6. Theorem, there does not
respectively.
e x i s t an automorphism a of the f r e e group F(Ix1,x2,x3,x41) s o t h a t a -1
. u31u1u2u3u2 .a
=
u u u u -1
3123‘2‘
Also this equality does not hold modulo Inn F((x1,x2,x3,x41).
This is i n
s p i t e of the f a c t t h a t the links L(u -1 u u u u ) 3 1 2 3 2
and L(03ulu2u3-1u2)
are s t r i n g isotopic. (3)
gF3(u1Lik)
i s isomorphic t o G (uY3 I - l ) , but G (ul) i s not iso-F3 -x3 -F2 The f i r s t r e s u l t can be (ui3). The second r e s u l t i s easy.
morphic t o G -F2 found i n D. Rolfsen page 49, where it is proved that (i)
C,
(ii)
(L(u1p3))
L( 0 1 ~3 )
is homeomorphic t o
C,
(L(ui3p3)) ;
and L(u-’p ) are not equivalent links. 1 3
Actually the isomorphism is easy t o establish d i r e c t l y as follows.
G -F3
-1 x3
(ul
-1 ) has generators x1,x2,x3 and defining relations
Lx
3
. x1x2 . x3
=
x1x2,
x-1 3 x2 x 3 = xl’
-3 -1 G (ul Lx3) has generators x1,x2,x3 and defining relations -F3
-1 x3
. x1x2 . x3
=
x1x2,
s
. x2 . s-l
=
xl, where s = x 3x2-1x -1 1’
Now the required isomorphism is induced by the following automorphism of the f r e e group F( {xl ,x2 ,x31)
2 78
CHAPTER 20
x1
+
x2
X1$
+
x2,
x3
+
s
A.
Finally it i s interesting t o note t h a t
- u1
u3 I 1 -x1x2
20.26.
The above method of disproving conjugacy of automorphisms
NOTE.
of a group G using reduced Alexander polynomials can be applied only t o groups G which have a presentation of the form <
xl,
xn ; r l $ ' " $ m
...$
where every r
x1 =
..*
'
3
is congruent t o 1 modulo the relations
j =
'n.
Furthermore the automorphisms must be of the form
x.1
+
x.i p .ui
for 1 s i s n ,
where every ui i s congruent t o 1 i n G modulo the relations
20.27.
EXERCISE.
(1) Give a very short proof of 20.6. Theorem using the
concept of a group with operators. (2)
Let a denote an automorphism of the free group F({x19.
i s defined by x.a 1
=
%a
=
'i+l x1
*
for i
n
..
%I) which
2 79
CONJUGACY OF GROUP AUTOMORPHISMS
Then show t h a t ui ak
ui+k
=
f o r a l l i+k s n-1, where k i s a positive integer. (3)
Let
Q
i,j
be the autmorphism of the f r e e group F(Ixl,.
.., %I)
which
i s defined by x x
i - + xj j
-t
xi
xi+l+‘j+l
‘j+l+
’i+l
’k
‘k
-f
f o r k # i , i+l,j , j + l
with { i , i + I }n { j , j + l l =
@.
Show t h a t
(4)
The proof of 20.21. Theorem shows t h a t Q A’Q-’
belongs t o Bn
f o r an X i n n automorphism of the f r e e group F(Ixl,. a belongs t o <
(5)
r
>
. Bn.
What i s the n o m a l i s e r of Bn in Aut Fn?
unsolved problem.
.., \ I )
Warning:
i f and only if
This is an
This Page Intentionally Left Blank
28 1
CHAPTER 2 1 PLAIT REPRESENTATIONS OF LINKS
As was noted already i n the c l a s s i c a l book of K. Reidemeister Chapter I 5 6 there i s another useful way of associating a link with a braid which i s d i f f e r e n t from the way considered e a r l i e r i n t h i s book of u goes t o L ( u ) .
The aim of t h i s chapter i s t o give a summary of the more
elementary r e s u l t s i n t h i s new type of association.
A f u l l e r account can
be found i n the book by J.S. Birman Chapter 5. 21.1.
Let u be an 2n-braid of the form
I I
I I
.. .. I I
.--
I 1
...
..
I I
I I
Then one can associated with u the link L (u) i n the following way P
.. ..
...
v u 21.2.
..
v
Let 1-1 be the permutation associated with the 2n-braid u and 6 be the
permutation on 1 , 2 , .
.. , 2n-1,
2n of the form
CHAPTER 2 1
282
1 2
...
2i-1
2i
2 1
...
2i
2i-1
I
... ...
i
with n
2
2.
Then L (a) is a knot i f and only i f the permutation p 6p-’6 when expressed P as a product of d i s j o i n t cycles contains an n-cycle which does not move two successive integers of the form 2 i - 1 , Z i
.
21.3.
If u i s a 2-braid, then L (u) i s s t r i n g isotopic t o the t r i v i a l knot. P
21.4.
The 4-braid
U ; ~ ~ U ~ U ;where ~ ,
f o r interesting examples.
q i s an integer, i s already a source
The links $ ( ~ ; ~ qulu;l)
are a l l knots and when
q i s p o s i t i v e can be pictured as follows:
with q twists where 1 t w i s t i s
PLAIT REPRESENTATIONS OF LINKS
283
The following are well known examples of such h o t s
21.5.
q = -1
gives t r e f o i l knot
q = 0
gives t r i v i a l knot
q
=
1
gives figure eight h o t
q
=
2
gives stevedore's h o t .
Let a be an n-braid and denote also by u the 2n-braid obtained from
the n-braid a by considering it as an element of B2n.
21.6.
One can s e t up some s o r t of plausible elementary argument f o r
showing t h a t t h e group of the l i n k $(u)
namely, G(L ( u ) ) , where a i s a P
2n-braid, has generators
x1,x2,
.*
Y
'2n
and defining relations
. (x 2 1.a) for i = 1,2
,..., n.
(X2n-p)
(XZn4
=
x2i-1
. xZi
=
e
Clearly the relation =
e
is a consequence of the other defining relations. 21.7.
Then
The group of the l i n k
CHAPTER 2 1
284 has a presentation of the form <
x2,x3 ; (x2,x3)-q
. x2 . (X2,X3)9 . x2-1x3-1x2 ' .
Its Alexander polynomial i s qt2 21.8.
-
(2q+l)t + q.
Let a be an endomorphism of the f r e e group F(Ixl,x2,.
. ., ~
~ ~ 1 ) .
Then it i s of some i n t e r e s t t o study the group G(L (a)) which is generated - P by X2n-1'x2n
X1,X2,"'Y
and has defining relations
for i (i)
=
1,2,
..., n.
In p a r t i c u l a r one has t h a t
I f a i s an automorphism, then there e x i s t s an isomorphism onto
(ii) If B i s an automorphism of the f r e e group F(Ixl,.
.., xzn1), then
B
is an isomorphism onto, where the generators on t h e right hand side a r e taken t o be
21.9.
Let u be an n-braid.
Then one can give a d i r e c t group theoretical
argument which establishes 6.1. Theorem of Artin and Birman as a consequence
PLAIT REPRESENTATIONS OF LINKS of 2.15, 21.8 and 21.6.
285
This shows t h a t 21.6 i s of more importance than
6.1. Theorem of Artin and Birman
-
it i s a consequence of the former
theorem. Suppose t h a t u and h a r e Zn-braids with h belonging t o the normal
21.10.
subgroup of BZn generated by
2 A
and A1,2n %,2n
Then the links $(u) 22.11.
EXERCISE.
%n-1,2n'
and L (oh) a r e s t r i n g isotopic. P Let u be a 2n-braid of the form
. . ... . . Determine a presentation of the group of the l i n k obtained from u by t h e identifications of the form
This Page Intentionally Left Blank
207
CHAPTER w
A LIST OF LINKS
We will give a list of nonequivalent links which have up to six crossing points. this property.
In the case of knots we will list only prime knots with In t h e case of links with more than one component we will
list only nonsplittable links with this property. PRIME KNOTS
Alexander & Briggs Notation
Braid giving knot
e
-lo2O3U u1 12 -lu u-l ZU3 3 3
(92
-1 2 -2 ul u2'1 u2
Braid belongs to
CHAPTER
288
o
NONSPLITTABLE LINKS (with more than one component)
Alexander & Briggs Notation
Braid giving link
Braid belongs t o
The method f o r obtaining a link from a braid i s described i n Chapter 5 above. More extensive tables of prime knots and nonsplittable links can be found i n the a r t i c l e by J. Conway and a t the end of the book by D. Rolfsen. They also give more information about these links. Alexander polynomials are given.
For instance t h e i r
The book by M. Murosugi also has some
interesting tables on links t h a t a r i s e from 3-braids.
BIBLIOGRAPHY
289
Alexander, J.W., A Lemma on Systems of Knotted Curves. Proc. Nat. Acad. Sciences U.S.A. 9 (1923), 93-95. Armstrong, M.A., Basic TopoZogy. McGraw (1979). Artin, E., Theorie der Z6pfe. Hamburg Abh. 4 (1925), 47-72. Artin, E., Theory of Braids. Ann. of Maths. 48 (1947), 101-126. Ashley, C.W., The Ashley Book o f Knots. Faber & Faber (1979). Blanchfield, R.C., Intersection theory of manifolds with operators with applications to knot theory. Ann. of Math. 65 (1957), 340-356. Princeton Univ. Birman, J.S., Braids, Links and Mapping Class Groups. Press (1974). Brieskorn, E., Sur les groupes de tresses Cd'aprss V.I. Arnold]. Seminaire Bourbaki no. 401 (1971). Burau, W. , &er Ztpfgruppen und gleichsinnig verdrillte Verkettungen. Abh. Math. Sem. Hamburg Univ. 11 (1935)) 179-186. Chow, W.L., On the Algebraic Braid Group. Ann. of Maths. 49 (1948)) 654-658. Conway, J., An enumeration of hots and links and some of their abstract properties. Computational ProbZems in Abstract Algebra. Edited by J. Leech. Pergamon (1970). Conway, J.& Gordon, C. McA., A group to classify knots. Bull. London Math. SOC. 7 (1975), 84-86. Coxeter, H.S.M. & Moser, W.O.J., Generators and Relations for Discrete Groups (4th Edition). Springer (1980). Craggs, R., On finite presentations for groups. Proc. American Math. SOC. 78 (1980)) 170-174. Crowell, R.H. & Fox, R.H., Introduction t o Knot Theory. Springer (1977). Dyer, J.L., The Algebraic Braid Groups are torsion-free: an algebraic proof. Math. Zeit. 172 (1980), 157-160. Dyer, J.L. & Grossman, E.K., The automorphism groups of the braid groups. American J. of Maths. 103 (1981), 1151-1169. Elffers, J. & Schuyt, M., Cat's CradZes and other String Games. Penguin (1979) Feustel, C.D. & Whitten, W., Groups and Complements of knots. Canadian J. of Maths. 30 (1978), 1284-1295. Fisher, G.M., On the group of all homeomorphisms of a manifold. Trans. American Math. SOC. 97 (1960), 193-212.
.
2 90
B I BLIOGRAF'HY
Fox, R.H., On the complementary domains of a certain pair of inequivalent knots. Ned. Akad. Wetensch. Indag. Math. 1 4 (1952)) 37-40. Topology of 3-manifoZds Fox, R.H. , Quick t r i p through knot theory. and related topics. Edited by M.K. Fort. Prentice Hall (1962). Abh. Math. Sem. Hamburg Univ. 25 Gassner, B.J., On braid groups. (1961) , 19-22. Goldsmith, D.L., Homotopy of braids - in answer t o a question of E. Artin. Topology Conference. Edited by R.F. Dichan Jr. and Springer Lecture Notes in Maths. Vol. 375 (1974). P. Fletcher. Knot Theory. Gordon, C. McA., Some aspects of c l a s s i c a l knot theory. Springer Lecture Notes in Maths. Vol. 685 Edited by J.C. Hausmann. (1978). Gorin, E.A. & L i n , V. Ja., Algebraic equations with continuous coefficients and some problems of the algebraic theory o f braids. Mat. Sbornik 78 (120) (1969) , 579-610. Translation in Math. U.S.S.R. 7 (1969)) 569-596. Sitsungsberichte Graeub, W., Die semilinearen Abbildungen. der Heidelberger Akademie der Wissenschaften. Springer (1950). Press Univ. de France (1971). Gramain, A., Topologie des Surfaces. Gramain, A. , Rapport sur l a thdorie classique des noeuds. SCminaire Bourbaki no. 485 (1976). Benj amin (1967) Greenberg, M. , Lectures on AZgebraic Top0 logy Maanillan (1959). Hall, M. , The Theory of Groups. Hempel, J., 3-rnanifoZds. Princeton (1976). Hillman, J.A., AZexander Ideals of Links. Springer Lecture Notes i n Maths. Vol. 895 (1981). Math. Proc. Hillman, J.A., The Torres conditions are insufficient. Cambridge Phil. SOC. 89 (1981) 19-22 , Proc. American Math. SOC. 78 Johnson, D, , Homomorphs of knot groups. (1980) , 135-138. Proc. American Kearton, C . , Noninvertible knots of codimensian 2. Math. SOC. 40 (1973)) 274-276. Kosniowski, C. , A First Course i n AZgebraic Topology. Cambridge Univ. Press (1980). Lyndon, R.C. & Schupp, P.E., CombinatoriaZ Group Theory. Springer (1977).
.
.
BIBLIOGRAPHY
291
Magnus, W. , Karnass, A. & S o l i t a r , S. , CombinatoriaZ Group Theory, Interscience (1966). Magnus, W., Braid groups: a survey. Proceedings Second International Conference Theory of Groups Canberra 1973. Edited by M.F. Newman. Springer Lecture Notes i n Maths. Vol. 372 (1974). Magnus, W . , Rings of Fricke Characters and automorphism groups of f r e e groups. Math. Z e i t . 170 (1980), 91-103. Harcourt (1967). Massey, W.S. , Algebraic TopoZogy: An Introduction.
Proc. Massey, W.S. & Traldi, L . , Links with f r e e groups are t r i v i a l . American Math. SOC. 82 (1981) , 155-156. Mehta, M.L., On a r e l a t i o n between torsion numbers and Alexander matrix of a knot. Bull. SOC. Math. France 108 (1980), 81-94. Moise, E.E., Geometric TopoZogy i n Dimensions 2 and 3. Springer (1977). Moran, S., Matrix representation f o r the braid group B4. Archiv der Math. 34 (1980), 496-501. Moran, S. , The Alexander matrix of a knot. Archiv der Math. 36 (1981) , 125-132. Moran, S . , The Alexander matrix of a link. Forthcoming publication. Morton, H.R., I n f i n i t e l y many fibred knots having the same Alexander polynomial. Topology 1 7 (1978) , 101-104. Morton, H.R., Closed braids which are not prime h o t s . Math. Proc. Cambridge P h i l . SOC. 86 (1979), 421-426. Memoirs h e r . Math. SOC. No. 151 Murosugi, K., On Closed 3-braids. (1974). Neumann, B.H., An essay on f r e e products of groups with amalgamations. Phil. Trans. Royal SOC. A 246 (1954), 503-554. Princeton (1965). Neuwirth, L.P. , Knot Groups. Neuwirth, L.P., The s t a t u s of some problems r e l a t e d t o h o t groups. Topology Conference. Edited by R.F. Diclanan Jr. and P. Fletcher. Springer Lecture Notes i n Maths. Vol. 375 (1974). Papakyriakopoulos, C.D., On Dehn's lemma and the asphericity of h o t s . Ann. of Maths. 66 (1957), 1-26. Chelsea (1948). Reidemeister, K. , Knotentheorie. Publish o r Perish (1976). Rolfsen, D., Knots and Links. Math. Ann. 110 (1935), S e i f e r t , H., k e r das Geschlecht von Knoten. 571-592. S e i f e r t , H. & Threfall, W., Old and new r e s u l t s on Knot Theory. Canadian J. of Maths. 2 (1950), 2-15.
f
292
BIBLIOGWHY Seifert, H. & Threfall, W., A Textbook of Topology. Translated by M.A. Goldman. Academic (1980) Serre, J.-P., Trees. Springer (1980). Simon, J., Roots and centralizers of peripheral elements in knot groups. Math. h.222 (1976), 205-209. &elkin, A.L., On soluble products of groups. Sibirsk Mat. 2. 6 (1965), 212-220. Smythe, N.F., The Burau representation of the braid group is pairwise free. Archiv der Math. 32 (1979), 309-317. Stallings, J.R., Constructions of fibred knots and links. Symposiwn i n Pure Maths. Algebraic and Geometric Topology. Vol. 32, Part 2 . Edited by R.J. Milgram. American Math. SOC. (1978). Thurston, W.P., Three dimensional manifolds, Kleinian groups and hyperbolic geometry. Bull. American Math. SOC. 6 (1982), 357-381. Torres, G. & Fox, R.H., Dual presentations of the group of a knot. Ann. Of Maths. 59 (1954), 211-218. Trotter, H.F., Non-invertible knots exist. Topology 2 (1964), 275-280. Waldhausen, F. , On irreducible 3-manifolds which are sufficiently large. Ann. of Maths. 87 (1968), 56-88. Waldhausen, F., Recent results on sufficiently large 3-manifolds Symposium i n Pure Maths. AZgebraic and Geometric Topology. V01.32, Part 2. Edited by R.J. Milgram. American Math. SOC. (1978). Whitten, W., A Classification of unsplittable-link complements. Michigan Math. J. 23(1976), 261-266.
.
293
INDEX
Alexander matrix 155, 166 of a braid 195 Alexander polynomial of knot 176, 180 of link 186 Amphicheiral link 209 Augmentation homomorphism 136 Augmentation i d e a l 136 Automorphism corresponding t o braid 86
-
Braid Braid Braid Braid Burau
76 corresponding t o link group 78 of braids 107 representation 193
Cancellation 5 Chain rule 189, 208 Combing a braid 103 Component of a link 72 Composition of knots 205 Consequence 8 Continuous mapping 33 Crossing point 7 1 Defining r e l a t i o n s Derivative 139 Dual permutation Dual ring 2 1 2
8
255
Elementary divisors 180 Elementary ideal 1 7 1 Elementary operations on matrices 160 - Restricted 179 Embrace
190, 250
112
Equivalence of Alexander polynomials 176 Equivalence of knots 64 Equivalence of matrices 160 Factor group 2 F i n i t e l y generated group 1 2 F i n i t e l y presented group 1 2 Free abelian group 10 Free (or p a r t i a l ) derivative 143, 150, 154 Free endomorphism 217 Free group 4 Free product 2 1 Free product amalgamating subgroup 25 Fundamental fonnula 146, 1 5 1 Fundamental group 39 Gassner representation 193 Generated by subgroups 23 Group 1 Group of f r e e endomorphism 2 1 7 Group of l i n k (knot) 119 Group ring 131 HN-extension 2 6 1 Homeomorphism 33 Homotopic paths 33 Invertible element 132 Invertible l i n k 210 Knot 63 Knot bowline 206 f a l s e lover's
200
294
INDEX
Knot figure eight 73, 115, 1 2 7 , 178, 209, 283 granny 73, 74, 115, 126, 1 2 9 , 1 7 7 , 199, 205, 2 4 1 pentacle 196 square 73, 115, 126, 129, 199, 205, 2 4 1 stevedore's 283 three-lead four bight Turk's head 202 t r e f o i l 72, 1 1 2 , 115, 125, 126, 129, 1 7 7 , 205, 209, 210, 235, 238 t r i v i a l 111, 125, 176, 283 Knot projection 7 1 Lebesque number 53 Length of word 5 , 2 1 Link 7 1 Link Borromean rings 7 2 , 115, 1 2 7 , 187, 236 t r i v i a l 186 Link corresponding t o braid 111 Linking number 237, 258 Longitude 232 Loop 36
Magnus r i n g 153 Meridian 231 Mirror image link 209 n-braid 76 Nonsplittable link Open b a l l
33
7 2 , 288
Open s e t 33 Orientated equivalence 64 Orientation preserving 64 Orientation reversing 65 Path 33 Pathwise connected 33 Pattern of permutation 190 Peripheral subgroup 235 Permutation corresponding t o braid 92 Polygonal knot 63 Presentation 9 Prime knot 206, 287 Product path 36 Product rule 139 Pure braid 104 Rank 1 2 Reduced Alexander matrix 195 Reduced Alexander polynomial 208 Reduced Burau representation 225 Reduced Gassner representation 226 Reduced word 5 Regular position 71, 78 Relation matrix 184 Same pattern 190 S p l i t t a b l e l i n k 72, 129 String of braid 76 String isotopy 65, 77 Tame knot 63 Tietze transformation Tying knots 205 Unknotted
68
14
INDEX
Unsplittable link 72 Vertices of knot 63 X i n n automorphism 270
295
This Page Intentionally Left Blank
