Mathematical Theory of Knots and Braids (Mathematics Studies 82)

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'


.

(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


.

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
/ ( 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.
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

=


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





where (a,b) = a -1b-1ab. xn = a

,

Put

and xm = b.

Then





.

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

.

2


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)


<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)

=


.

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(&)


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)




,

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




,

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)


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)


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

=


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 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


, 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
/

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
. 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

,..., 5 ; ri, ..., ri-1 ’




.. , (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
.

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,.


However

f o r a l l i and j .

which we can consider t o be


.

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
.

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
'

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


.

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

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 )

=
,

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;')

=


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.