Formal Knot Theory (Mathematical Notes, No. 30)

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CONTENTS

lUI L

Introduction •..•...•...•.••....•.••.••....•....••••

:L

2.

States, Trails, and the Clock Theorem •...•.•......•

11

3.

State Polynomials and the Duality Conjecture ...•..•

4.

Knots and Links .....•................•..•••.....••.

5' 67

5. 6.

Axiomatic Link Calculations .......•.•...•..•.•...•.

78

Curliness and the Alexander Polynomial .....••....••

95

7.

The Coat of Many Colors ••....•..•.•..........•.....

105

8.

Spanning Surfaces ...........•...........•........•.

114

9.

The Genus of Alternative Links .....•...•.•....•.•..

125

10. Ribbon Knots and the Arf Invariant ................•.

143

Appendix. The Classical Alexander Polynomial ........•..

156

References. • • • • . . . . . . . • . • . • . . • . • . . • • . • . • • . • . • • • . . . . • • . .

165

1

1.

Introduction These notes constitute an exploration in comblnatortol

and knot theory.

Knots and links in three-dimensional

may be understood through their planar projections.

.,ao.

A knot

is usually drawn as a schematic snapshot, with crossings indicated by broken line segments. trefoil knot. diagram.

ThUS~

represents the

We shall refer to such a picture &s & ~

The projection corresponding to .uch & diagram

forms a (directed) multi-graph in the plane, with tour ed,e. incident to each vertex.

We shall study these graphs .epa-

rately, and for their own sake. In order to do this work it is very important to underline key concepts by adopting terminology and conventions that are easy to remember.

For this reason I have taken a

perhaps startling. but certainly memorable, set of terms for the graph - theoretic side of the ledger. A

(directed) planar graph with four edges incident to

each vertex will be termed a universe. trefoil universe.

Thus

~

18 a

These universes have singularities (the

crossings); they will a180 have

~, ~~, ~

~, ~nd ~.

A ~ of a universe is an assignment of one vertex in the forms

+ +

~

++

per

2

so that

each~~o~ in

marker.

ThUS'

the graph receives no more than one

~

is a state of the trefoil

*~

universe.

Two regions of the state will be free of markers

(since the number of regions exceeds the number of vertices by two in a connected universe). habited by

These free regions are in-

(*).

the~.

It should be mentioned at once that each universe has states.

1n

In fact,

~ ~

!!!£! in adjacent regions

one-to-one correspondence with Jordan trails

~.

~

~ ~ ~­

A Jordan trail is an (unor1ented) path that traverses

every edge of the univlrse oncl and forms a simple closed curve in thl plane.

This correspondence is obtained by re-

garding each state marker al an instruction to split its crossing according to the tollowing schema:

Xr-+><.

By splitting all crossings in a state, the Jordan trail automatically appears.

Conversely, a choice of stars at the

Jordan trail determines a specific state. for more details.) trefoil.

The process is illustrated below for the

~~

f.)U

(See section 2.

1

3

This correspondence underlines the importance ot .tate. with adjacent stars; further reference to state. will

a•• unt

star adjacency unless otherwise specified.

ma,k.,.

are classified into the categories black ~, ~

x

Ib1iI ~,

£r

~ ~

or••• ~nl

black hole

x x ~

~,

according to placement with reapect to tne

orient~tion:

The

The atat.

white hole

up

~,

down

a(s), is defined by the formula

a(S) = (_l)b where b denotes the number of black holes in S.

Just as the sign of a permutation changes under single

transpositions of its elements, so does the sign of a state change under a state transposition.

A state transposition is

a move, from one state to another that is obtained by switching a pair of state markers as indicated below.

H

transpose

1

~

>7' /

4

Note that in a state transposition both state markers rotate by one.quarter turn in the same clock-direction.

A state

transposition in which the markers turn clockwise (counterclockwise) will be termed a clockwise (counter-clockwise) move.

A .tat. is said to be clocked if it admits only clock-

wise mov•• ,

~

it it admits both clockwise and counter-

clockwi" MOve., and counter-clocked if it admits only counterclockWi •• MOV••• Tbl kl, oombinatorial result in our study is the followin, ualOl'lm.

ZbI ~

Theorem (~).

Let U be a universe and S the

'1' of .tates ot U for a given chOice ot adjacent fixed .tal".

Then

$ has a unique clocked state and a unique

oounter-clocked state.

Any state in 0 can be reached from

the aloaked (aounter-aloaked) .tate by a series of clockwise (aounter-olookWi •• ) move..

Hence any two states in S are

connected by a •• rie. ot .tate transpositions. By defining 8

< 8'

whenever there is a series of clock-

wise moves connecting the state 8'

to the state 8, the

collection of states becomes a lattice whose top is the clocked state, and whose bottom is the counter-clocked state. This result is illustrated for a particular universe in Figure 1.

><

An oriented universe has directed edges so that

each crossing has the form

In section 2. we shall give an algorithm tor constructing the clocked and counter-clocked states.

This algorithm, to-

5

clocked

0

I

.~ tt

,

/1 \

/ counterclocked

Lattice of states

6

gether with the Clock Theorem, gives an efficient method for enumerating all the states of a universe. Patterns of black and white holes in the states give rise to a Duality Conjecture and to a series of results bridging combinatorics and the topology of knots and links. Duality Conjecture.

Let

C

be the collection of states of an

oriented universe U with a choice ot tixed adjacent stars. Let

denote the number of states in 8

N(r,s,l) - N(r,s)

with

r

black holes and

N(r,s) • N(s,r)

for all

s

white holes.

r , I.

As w••hall ••• in .eotion ,.

!1!t

plag.ment.

I conjecture that

That iI, it

I'

~

11 independent £!

is another state collection

ariling trom a 4ifterent choice of tixed stars, then NCr,I,I) • NCr,.,I')

tor all

r . s.

Thi. independenoe result depends crucially and subt1ey on the Clock Theorem.

It is verified by interpreting the ~ polynomial F(O). t (-l)rN(r,s,O)BrW s (belonging to the r,s polynomial ring in variables B and W over the integers: Z[B,W]) as a determinant of a matrix associated with the universe and with

I.

The signs of the permutations that occur

in the expansion of this determinant coincide with the signs (_l)r, and these are the signs of the states being enumerated! A generalization of the state polynomial marks the transition into the theory of knots.

A knot-diagram is a universe

with extra structure at the crossings.

To create a knot-

diagram from a given universe entails a two-fold choice at

7

each crossing. with

c

Hence

crossings.

2c

knot-diagrams projeot to •

univlr••

It is convenient to designate thl •• Ih.ioo.

by.placing a code at each crossing.

OUr codes take thl

'0~~1W.

ing torms:

standard

~

reverse code

Thus a knot or link diagram is an oriented universe with standard or reverse codes at each crossing. In standard code the labels

B and W hover over poten-

tial black and white holes respectively. in the reverse code.

The knot or link

Labels are flipped

obtain~d

by labelling

a universe entirely with standard (reverse) code will be called a standard

(reverse)~.

A reverse knot is the

mirror image of the corresponding standard knot. is standard.

The trefoil

,8

Let

K be a knot and S

~':r1ying

uni ver&e

U.

We

.. state, both sharing the saae inner-product

~&n

and a ~ po.l\lJM!!lial


,

so that whenK 1.s standariii th1.s new 'Polynomial coincides

tor 8. In order to do this the inner ~et is detined as tOllows: Superimpose K and S on the univlrae U. Let x denote the number ,ot coincidenoe. ot W· labels in K with state ma:r~.rs in S. :WIt "I ,dlnote thl number of coincidences ot I · l&_elB with It&tl marklrB. Then wi:th the original .state :pol7tl0mial

When K t, ,'-adard, X i. the numblr ot white holes and y i. the

num~.r

at

~l&ak

holel in

S.

Por example, if K is

the tretoil knot, then

= '

and

cons'itleratiml

'<xjS>

= '132

-0

Tbe,se state

+

'ctf

two other trefoil stat,es shows that

W2.

po1~ls

give rise at onee to topological.

inv,arlants of the 'knot or link X. thatios a topological invariant,

To obta1.n a poJ.,ynom1al

~

i l l WI:! =.!

~ ~

9 .1 =

!i -~.

and

VK(z)



Then

becomes a po-lyn
is a topologtee.l invarirmt of:

K.

z" 'X(.),

In tb

tS'IIoU

VIC "" 1 + zZ.

example we have

The polynomial

is identical with whe.t I" call,e41 tb

VK

Alexander-Conway polynomial.

ar !.

It is'

classical Alexander polynom1.al ([I},.

[51~

II

refinement of' the'

and [21]): and :ts '

characterized by the following three axioms: 1.

To each oriented knot or link

K

.,

ated a polynoudal

""K{z} (Z[zl

,.,",

there 1s &ssocl-

...

such that amtI'.fe!t"

isotopic links receive identica.! polyrtOllldals,. is an unknotted circle· then

2.

If

IC

3.

If

IC, ie, and

the site

L

or .2!!!

VIC", 1.

are three l1nu that d1ft~r at,

croas1ng as 1nd1.cate4 beJ.o\iIII' .. tben

VIC - ViC "" ZVL •

IC These axioms alone sutf:iae to calcula.te the polynomial, W'lthout reference to the underlying state polynomial or to any other model.

g·ee section 5. of these notes far a. sel:f'-'eon-

ta1ned treatment via the axioms.

The Alexander-Conway poly-

nomial is a true refinement of the Alexander pokJDOm1al. Because it is def'ined absolutely (rather. tban up to sign and powers of the variable) it is capable of distinga1sh1ng many links from their mirror images - a eapab111ty not available to the Alexander polynomial.

10

The model using the state polynomial is a reVision of Alexander's original combinatorial appraoch.

The Clock Theorem

provides the underpinning that makes this Sign - precision possible - by allowing the identification of state signs with permutation signs (see section 3).

In section 6. we detail

the connection with the Alexander polynomial and show how it contains a hidden calculation of the Whitney degree ([14), (38)

of the knot projection seen al a plane curve immersion.

In this section a combinatorial vlrlion ot the Whitney degree, dubbed curliness, il given tor unlver.e •• Section 7. d.velopi tne via Itat. pol,nomlal modell,

m~ltl·variable

link polynomials

Here the power of our method

rlall,. oeme. into pl,,. .inoe there 11, at present, no axiomatil'tlon tor tn••• P01,nOMi'11,

The .tate summation method

then Ilvel 11mpl1 proot. ot 30hn Oonway'l identities for the "polfonroml Iklln",

alation 8. oonltitutea a necessary digression into the topololY ot orientable spanning surtaces for knots and links. We discuss genus, Seifert surface, Seifert pairing, and another model ot the Alexander-Conway polynomial. Section 9. contains our main topological theorem, a wide generalization of the Crowell-Murasugi Theorem on the genus of alternating knots ([8J, (29). to the class of alternative

I generalize this theorem

~ (~).

Alternatives in-

clude alternating knots and links, knots that arise as the links of plane algebroid singularities ([27), the Lorenz links of Birman and Williams ([3), and many others.

Our

11

proof of the Theorem is quite simple· thanks going ena. 'I,ln to the Clock Theorem for control of signs! Section 10. discusses the Arf invariant of a knot.

We

show that the Arf invariant is the mod-2 reduction of the degree-2 coefficient of the Alexander-Conway polynomial. with the help of

Then,

J=i, we obtain a quick proof of J. Levine's

Theorem relating the Art invariant and the value of the Alexander polynomial at

-1

taken modulo eight.

The appendix is an exposition of a classical approach to the Alexander polynomial via the Dehn presentation of the knot group. I would like to take this opportunity to thank the many people with whom I have worked and conversed over these many years.

Particular thanks go to my collaborators Tom

Banchoff, Alan Durfee, Deborah Goldsmith, Walter Neumann, Larry Taylor, and Francisco Varela; to George Spencer-Brown for conversation, inspiration and insight into formal mathematics; to Kyoko Inoue and David Solzman for innumerable informal conversations; and to Milton Kerker, William Browder, and Ralph Fox for introducing me to research and to the study of knots.

Finally, my thanks to Ms. Shirley Roper, Head Math

Typist 'at the University of Illinois at Chicago, for her excellent typing job.

June 9, 1982 Chicago, Illinois

12

i. §tates, Trails,

~ ~

Clock Theorem.

In this section I shall delineate the states - trails oorrespondence, give an algorithm tor constructing clocked (oounterclocked) states, and prove the Clock Theorem.

It is

assumed that the reader is tamiliar with these concepts from the introduction, but, for the sake ot completeness. exact detinitions will be given below.

Betore embarking on this

formal development, ! would like to make a tew remarks that highlight the key ingredients in the Theorem: called a

A

~root

ot the Clock

universe in the torm llven ln Figure

~

compo.ltlpn.

2

will be

It 1. qUlte clear that any shell

e

compolition ha. & unlque clocked Itat. where each shell recelves marker' ln the torm

•

••• --~-----i--t-------1-t-----tt-------------···

13 Note that the markers within this elementary shell st.., wUhin the

she~l

when they rotate

is elaborated to a more

e~ockwise.

comp~ieated

If' a shell compol1Uon

universe. then markers

in-

terior to the original. shells may have the opportunity to rotate outwards.

If' the ,original shell is

c~ocked.

new rotations are necessarily cOWlter clockwise.

Interior

clocking leads to the potential for exterior counter

~

then these

~-

,. e1aborate< to

The new universe bas a state

derived from an interior shell marker rotating outward.

On the other

hand~

there are elaborations of

shel~

compo-

sitions such as

0: where the new state 1.8 stUl. clocked.

We shall see that these

properties of shell compositions are central to the Clock Theorem .. The unique clocked and counter-clocked states ot a universe are found by derlv1ngan appropriately related shell composition. States ,and Trails Definition

£d. A universe is a connected planar .(multl-)

graph with 4-valent verti:ces.

A univene is oriented 11' each

14 edge in the graph is directed

80

that the vertices take the

torm ot an oriented crossing ot two line segments:

x

We distinguish two methods of traversing a universe in the neighborhood

o~

one

o~

its vertices.

A path that proceeds

across a vertex along one of the atorementioned line segments will be said to £!2!! that vertex.

A path that meets a ver-

tex but does not cross it will be 1&14 to call that vertex. Note that paths may move against the orientation directions of the universe.

There are two possible local calling con-

figurations at each vertex.

x ><

x

sites

croll in,

Each of the.e oonsist. in two segments of the path meeting at the vertex. form a!i!!.

It is convenient to separate these segments to As

ind~cated

segment has a cusp point.

above, a site is drawn so that each If these cusps are brought together

the site becomes a crOSSing.

><~X

An exchange of sites

will be called a re-assembly.

A 'path that traverses the entire universe, using each edge once and calling every vertex, will be called a

~

!£!!!. A Jordan trail becomes a Jordan curve in the plane if we separate the path at each vertex to form a site. Jordan curves with sites will always be used to indicate

Such

15 Jordan trails. ~

Proof.

2.2.

Every universe admits at least one Jordan tratl.

It follows from the Jordan curve theorem that a uni-

verse cannot be disconnected by both of the possible splittin,1 of a given vertex.

Therefore, choose a sequence of vertex

splits that retain connectivity at every stage, and so that every vertex is split.

The graph with sites that results

from this process is a simple closed curve.

Hence it corres-

ponds to a Jordan trail on the universe. The method of this proof is illustrated in Figure

In closing the sites of a trail we obtain forget how it was brought about.

a

1.

universe, but

In order to remember how to

split the vertices and regain the trail, place a state marker of the form

X

at the vertex, and split marked

tlces according to the schema

)(

ver~

16

A ,tate of a universe U is an assignment of

Detinition~.

state markers to vertices so that no region of U contains more than one state marker.

Unless otherwise specified, it is

assumed that the (two) unoccupied regions are adjacent and marked

by

(*).

stars

Theorem 2.4.

Let I

denote the collection of all states of

a universe U tnat ahar. a fixed choice of adjacent stars.

r

Let

be the ,oll.,tion ot &11 Jordan trails on U.

8 and

r ar.

!IaI£.

GiVln & .tate 8, let

Then

1n on.-to-one oorrespondence.

.pl1"1nl all v.r'10"

al

&(S)

.~'01ti'd

denote the result of by their state markers.

In the atep-by-'t.p ,rooess ot vertex

s~lit8

we can never

arrive at a situation where the marked and unmarked (m and n

~

respectively in

the same region.

) sides of a vertex are part of

For if this were so, then this region would

separate the stars (and the stars are adjacent). This guaran-,' tees the loss of one region after each vertex is split.

Hence

there must be exactly two regions that remain after all vertices are split.

Thus

a(s)

is a Jordan trail, and we have

a:$->1'. To obtain b: l'

->

is. grow two trees, each rooted at one

star as indicated in Figure 4.

These trees then determine a

collection of state markers so that maps.

This completes the proof.

ab

and

ba are identity

17

Tree growth creates state

b(T}

from trail

T.

18 ~.

A checkerboard coloring ot the universe divides the

regions into those that lie inside and those that lie outside any Jordan trail.

Thinking ot states as region - vertex assignments, hence as permutations ot the vertioel relative to an ordering of the regions, it is natural to oonlider transposition ot permutations.

The geometrio ver8ion ot a transposition is indicated

in Figure 5.

In thi8 tilUre the regions X and Yare dis-

tinct, and the dia.rama 8how part of a state with adjacent Itarl.

~n.

relionl Inare oommon boundary that abuts to two

Itate markerl, on. trom each region. it il

~ollible

Under these conditions,

to Iwitch the markers as shown in Figure 5.

A new Itate il formed by this switch.

Call such a move a

(state) transposition.

..·J:c~~:Jr-'" . ... )I'

~

19 Figure 5 also shows sites that correspond to the trine. position.

Dotted lines display how the rest ot the trill

connects with this local situation.

While there will in

general be many other sites in such a trail, the topology of the connection to the transposition sites will always be 'as indicated. For states with adjacent stars it is easy to verity that both markers in the transposition rotate in the same direction.

clo<:,'~

Hence transpositions may be labelled clockwise

or counterclockwise accordingly. A state is said to be clocked if it admits only clockwise transpositions and counterclocked it'it admits only counterclockwise transpositions. Clock Theorem~.

Let

We aim to prove the

U be a universe and

£ the set of

states ot U tor a given choice ot adjacent tixed stars.

C has a unique clocked sta.te and~ ~ _~~!:\UI!I, _:,?~nter.: clocked sta.te. Any state in £ can be reached from the

Then

clocked (counterclocked) state by a series ot clockwise ,---.------- ------ . . ------(counterclockwise) moves. Hence any two states in £ are ~--

connected by a series ot state transpositions. By defining

S

~

S'

whenever there is a series ot clock-

wise moves connecting the state -S'

to the state

S, the

collection ot states becomes a lattice whose top is the clocked state, and whose bottom is the counterclocked state. An important consequence ot the Clock Theorem is that each state

S

has an intrinsically defined Sign, a{S), that

may be identified in a coherent way with the sign ot an asso-

20

eiated

permutation.

(In order to see this we

~irst

the state markers into the categories black holes,

Ya. and

~

~

h2!!!.

as in the introduction.)

Definition 2.6. Let

classity

b = b(S)

Let

S be a state

o~

an oriented universe U.

denote the number ot black holes in S.

the .!!.!.Em o~ 1m! ~ S

by

De~ine

the tormula a(s) = (_l)b(S).

Let U be an oriented universe with regions

Detinition -. 2.7.

Rl ,R2 ,···,Rn , ~+l' ~2 and vertices Vl ,V2 ,···,Vn • Let be the collection ot state. ot U with stars in Rn+l and Rn+2

(letting the.e denote

ad~acent

regions).

S

Choose the

ordering ot the vert ice. .0 that the region - vertex assignment

(R t • Vi

I, and

10

I 1 • 1, •• t,n) corresponds to a state So 1n

that thil Itate haa aign equal to 1.

denote the .et ot permutations ot the letters any permutation p Recall that

p.

in

Sen). let

.gn(p)

.gn(p)

il equal to

(_l)t

Let

Sen)

1,2, •.• ,n.

denote the Sign where

t

For o~

is the

number ot moves in a sequence ot transpositions that transtorms

p

to the identity permutation

Since each state

S in

S

is completely determined by some

region - vertex aSSignment, there is a P:

e = 123 ••• n.

well-de~ined

injection

£ -> sen) where Vp{S)(i) has its marker in region Ri

(i ,.. 1, ... n).

!9J:.

!.

Call P: $

->

Sen)

a permutation assignment

21

Let

Lemma~. •0

that 5'

Then 8 Proot.

is obtained

and

8'

be states at an oriented uniy.,••

S and 5' trOlll

S

differ in sign:

contemplate Figure. 6.

lw

..

tw

t

lw lw

• at • 8t

by.!m!

0'(8')

state tranl,oliUon. =

-0'(8).

This completes the proot. (Wand B reter to possible white or black holes.)

ai' ,t

=>

a{S')

= -0-(8).

Proposition 2.8.

Let 0 be a state collection tor an oriented

universe U. and

P:

8.

S -> Sen) a permutation assignment tor

Then the signs ot the states agree with the signs ot their

corresponding permutations. all

8

Proot.

in

That is, o'(S) = sgn(p(S»

tor

e.

Use the notation ot Detinition 2.7.

Thus

So is a

.tate with 0'(5 0 ) = sgn(p(So»· - 1. Let e = P(80 ). Let 8 be any state in S. Then, by the Clock Theorem, 8 can be obtained trom t

80

by

a sequence ot state transpositions.

be the number ot transpositions in some such sequence.

Then, since

P transtorms state transpositions into trans-

Let

22 positions o~ permutations, sgn(p(S» ~(S) = (_l)t

by

Lemma 2.7.

= (_l)t,

while

Thus a(S) = sgn(P(S», completin,

the proot. This agreement between state and permutation signs will be ot great use to our

~her

investigation ot states in

section 3., and to the ascent into knot theory that begins in section 4.

In order to prove the Clock Theorem we first

discuss an algorithm for conltructing clocked and counterclocked states. Constructing Extr!!!l St.t••

B.v an .xtrll"

~

count.ro1ooked 1'.'1.

% m.an either a clocked state or a

It will b. most convenient to phrase

the di.eullion in tlrm. of universes in string torm; these will b. 01111d ,trin.l.

A .tring il obtained

o,ln in'lrior .re from an .dg. of • universe.

~

il a trefoil string.

by

deleting an

Thus

It will be assumed

that anJ state ot a string corresponds to a state whose stars are in the two regions adjacent to the deleted edge.

Thus

~iS a state of the tretoil string. Drawing in this manner, the starred regions are alW&Js above and below the string.

:~

23

Here we indicate the three trefoil states clockwise moves on each state).

(an~

available

The corresponding trail is

drawn to the right of each state.

_'@

\. Before giving formal details of the clocking algorithm, he.. 18'"

U

In order to make the outer trail boundary appear in the (clocked) form

-B-- '

trace U as follows:

-(if) Tracing the outer shell leaves a Similar configuration on the outside.

Repeat the procedure on the inner form (recursively)

24

)

PUtting this all together, obtain the trail

and .,.,.

Here

S

T

•

is weli wound up, with only one available clockwise

move. Det'init1on &.,2.

fine

Ae B

line ot'

B

Let

A

and

B

be strings as below, and de-

by splicing the right line of'

A to the left

(respecting orientations if theB. are present}.

;l

A string

C is said to be irreducible

if

:it cannot be written:

25 in the form A $ B unless

A

notes the trivial string.

=E

or B

=E

whIr. I

.,-

We shall refer to 'ftl 11"

and right lines of the string as the input line and .21IlIIl line respectively. Remark.

Any string determines two universes with speci.t1c

starred regions.

These universes are obtained

by

connecting

the lnput to the output above or below the body of the str1ns. We may. if we 11ke, regard the input line ot the string as extending indefinitely to the left, and the output as extend· ing indefinitely to the right.

Then this (infinite) string

divides the plane into" two unbounded regions that correspond to the starred regions in either of the related universes. With this convention, we shall refer to the bounded and gnbounded regions .2f the string. peflnition 2.10. of e

A if e

An

edge in a string A is an interior edge

separates two bounded regions in

A.

An edge

is a connecting edge if it separates the two unbounded

regions. pefinition~.

Let A and

B be given strings, and

an interior point of a non-connecting edge of A.

Let

A $ [B,p] • A $ {B] = C denote the string obtained ing the trivial string at p

by a copy of

p

by

replac-

B (that does not

tntersect the rest of A).

AetaJ

-i . ~.t-

26 We shall say that and a

~

B.

C = A $ [BJ

decomposes into a carrier A

This definition generalizes to a collection

{Bl ,B2 , ••• ,Bn J and a set of pOints The resulting composition is denoted

ot riders in A.

A .. [Bl'P1J

{Pl,P2, •.. ,Pn

e ... $ [Bn,Pn J .

Note that this composition is not associative. since (A

$

[B])

$

[B']

may have B'

that is unrelated to B, or Definition 2.12.

may ride on B.

e

~

riders of the same torm.

has no r1derl.

B'

A ~ composition is a string that is ob-

tained from the shell

Det1n1t1gn~.

riding on some part of A

(The riders

A Itr1nl 11

~

m~

by adding have riders.)

it it is irreducible and

'or example, the tretoil string is atomic, as

11 the Ihell Itrinl. In an atomi0 Itr1nl A one ot the following two forms

--0will always be obtained upon removing all interior edges (retaining small interior arcs where an input or output line crosses into a bounded region). and the second a

~.

Call the first form a curl

27

pefinition 2.14.

We shall define the boundary of & .'r~nl.

It will be a graph

,"nt'

~A

composed of curls and shell•••

inductively as folloWs:

1. If A is atomic, then aA is the single curl or shell obtained (as above) by deleting interior edges.

2. 3.

a(A a(A

= o(A)

$

B)

$

[B,p]) = &(A)

$

&(B) $

[&B,p]

When p

belongs

&A. o(A. [B,p]) ~ &(A) when p does not belong to aA. (The small interior arcs of the boundary to

do not contain composition points 4.

o(

p.)

) =

!1ixa.mple.

,,/(B

~

.-D

~

C = A $

oC

=

oA

-C~

Remark.

$

[B,p] • [D,q] [aB,p] ~ [&D,q)

Since the decomposition of a string into atomic

Itrings is unique, the string boundary is well defined. With this definition, a composition of curls is its own boundary.

The reason for this choice is related to the fact

that a curl composition has only one Jordan trail (hence only one state).

Since this state has no available state trans-

28

pOlitions, it satisfies the Clock Theorem vacuously.

Curl

oompo.itlons are the only universes with one state. In order to give a procedure ot constructing extremal

.

Itate. we must consider the collection ~ With sltes.

ot string universes

This contains the ordinary strings

also contains the trails ~.

~,

and it

(I .hall use the same notation

.

tor string trails and Jordan trails with sites.) We shall now \define a deri vati ve D t rt

-> rt.

By

applying the deri va-

tive iteratively to a given etrinl we arrive at a shell composition with sltel.

Thi. Ihlll oompolition is then used to

construct extremal It&tl'. Pttlolt101)~.

Itrin,l.

WI tlr.t dlttne the derivative tor atomic

In thl oa.. ot • OUl'l,

D( -e...

A 1. atomi0 with a ehlll boundary, let ot .,11tt1nl th. vlrt1ol' 00 1o~.otl

lnto DA.

Note that

~oundary

)• ~ .

DA

denote the result'

edges at A so that

This uniquely specifies

is also a string. Example.

'aA

OA

aA

DA for atoms.

DA is a string with sites, and that DA -

A

If

~A

=

IA

29 Having defined the derivative for the atomic str1nl' w•••,.in it generally via:

2.

e B) = D(A) e DCB) DCA e (B]) = DCA) • [D(B)]

3.

D( ---+- ) =

l.

D(A

~

]

..

rf1: 1£ -> 1£ n = 1,2,... are defined

~ derivatives

a8

..

tollows.

Let 1£* stand for _ strings with sites, regarded!!

.trings.

That is, 1£ and ft* have the same information but

"

matters of atomicity and irreducibility are referred soley to

. -> 1t*

Let F: ft

the underlying string structure.

be the

mapping that associates to a string with sites its corresponding string, and

p-l: 1£*

->

i

the inverse map.

With this

understanding, we have an immediate extension of the derivative to D:

~ ->

the right hand

it

via the formula

D(X) = p-1DF (X)

D our original derivative on strings.

with The

higher derivatives now exist via iteration of this operation. Lemma 2.16.

Let A be a string.

tive integer N such that DrlA

Then there exists a posi-

= riA.

Let

DA

denote this

Itring with sites, and call it the dissection of !.

Then

the dissection of a string A is a shell composition with Iltes. l~ot.

This to1lows easily from the definitions.

In a composition ot shells the vertices on each shell can be split in either clockwise or counterclockwise fashion," as illustrated below:

"30

CX

(clockwise)

C'X

(counterclockwise)

If Y is a composition of shelll, we let

CY

denote the

result of performing a clockw11e Iplit on each shell (similarly for

C'Y) •

Theorem 2.17.

K'A

= CIDA.

Let A be & .trln,. Then the tra11.

KA

Let

and

K'A

...

KA = CDA

and

correspond to

clocked and counterolooke4 .t.te. of A respectively. Thl1 theorem oompletel cur statement of the algorithm tor oonltruotlng extremal Itates.

In order to prove it we

Ihall analyze the sort of sites that may be added to a shell composition to obtain a string with sites of the form By specifying allowed sites the proof will emerge.

...

DA.

Figure 7

illustrates this algorithm as it applies to the example just prior to Definition 2.9. Allowed and Forbidden Interactions It is often convenient. when viewing a string with sites to see it decomposed into the same atoms as the pure string obtained by closing all the sites. the map

. J: ft - ) 1t

To this end we define

that closes all the sites.

A string

with sites will be said to be J-atomic or J-irreducible when

31

u

DU

Deriving

~

Clocked

Figure 7

~

32 its image under

is atomic or irreducible.

J

A pure string

(without sites) divides the plane into regions (with two unbounded regions by our conventional.

By these same conven-

tions a trail T divides the plane into two regions, but the regions for the corresponding string, J(T), are apparent from the diagram, Since they are bounded by edges and sites. sites correspond to

doorw~.

The

in rooms - one can identify the

interior of a room even when the doors are open.

Therefore A

we define the !22m! at a Itring with sites regions of the pure .tr1ns J(X).

X E~

Since J(J(X»

to be the = J(X), the

regions ot a pure Itrlnl are identical with its rooms (and some roOllls

m~

haVe no doors).

Conlider a oomposition ot ehells Y and a single circle a

in Y.

Then

lower arc a_, and q Let

00

a

11 divided into an upper arc Q+

Here

a-a+Ua_,o+no_=(p,qj

are the intersection points of denote the intersection of

(bounded) region determined by a, and 00 -may each have riders. U1'U2 ••• · ,Uri 0_

A

°

and a

where

with a line

p t.

with the interior

In Y. the lines

Suppose that

0+

0+, 0_

has riders

Ll •••• ,Ls; a o has riders Ml ••••• Mt • Each rider is itselt a shell composition, and the riders are all disjoint except for their connections along 0+, 0_

°

has riders

and

and its riders.

°0 ,

Fo denote this composition of See Figure 8. Let

33

Figure 8 Let

Fa

plus its

and Fa

= riders,

denote 0

respectively.

a=

plus its riders, and

Thus

Fa = (input and output lines) U Fa U Fa U Fa . Ipecify how sites may be lites

F~, so that

ao

We wish to

+ 0 added to Fa' torming a string with

DJ(F~)

= F~. A point

of

p

a.

FI

will

be said to be a pure boundary point it there exist paths, confined to single rooms of output strands of of

F~,

F~,

from

p

to the input and

and also a path from

p

to some point

Using this concept, the following interaction rules o are adopted: F~.

Interaction rules. 1.

•

Let

denote a string with sites •

Then - ......-

may be replaced with either ot

the two torms: 2.

It .0. FI

is a string with sites that has been obtained

trom a shell composition ot the torm Fa' then further sites may be added between

and F~ as long as o :I:: the point (cusp point) of the site com1ng trom F' F~

Cl.:l:

is a pure boundary pOint. }.

Rules 1. and 2. may be app11ed to any J-rider on a liven 'trins with 'ite,.

With "Iard to

rul. ", note that a J-rider 1s simply a sub-

.,.1ns wlth 11t•• that ha. no .1te interactions with its canta1ft1nl .tr1n. (h.nce it corre.ponds to a decomposable part under

J,).

Propos1tion 2.18.

Let

SH

denote all strings with sites ob-

tained by elaborating shell compositions according to the ~.

interaction rules 1., 2., and 3.

Let

t10n of s1 te - free strings, and

J: SH

denote the collec-

->

~

the mapping

that closes all sites. J

0

Then J

is a one-to-one correspondence, "and D 0 J

D=~.

Thus

SH

= lSH' .

is exactly the set of elaborated shell

compositions obtained by dissecting strings via

D.

35 Using the Interaction Rules Here are a tew examples of the application of the in' •• action rules:

~)')F'

_, 1.

.-/

't:::Y' '--

!lJF' = F'

~ boundary point G

l-..;:;.:...2._~

G'

roo'

.a

G'

Proof' of'~.

D.

J = lSH'

Since

J

D = ltt'

0

it suff'ices to show that

Since this property is preserved under applica

t10n of' the interaction rules, the result follows by induetion. ~.

It is worth observing how things go wrong when the

interaction rules are violated. site to J(F') =

--&

in

~

DJ(Ff) ! F'

(and

DJ(F')

-

=

-@-.

Then Thus

violates rule 2.).

Before proving Theorem placement ot atate

~'.

the f'orm

and F'

For example, if we add a

ma~ke~l.

2.17.

we need a lemma about the

Conlider an atomic string.

g!ft

Take ita tirat derivative and start putt1ng 1n the markers f'or the clocked state that will result from the algorithm of' Theorem 2.17.

Notice that markers at sites between the top of the shell (the shell produced by taking the derivative) and the middle .tring go to tbe right in tbe 'oro

~

where the +

sign labels the cusp from the top part

ot tbe shell.

• ......

larly, the markers tor sites between the middle and the "',.. part ot the shell point to the left.

This phencmenon ,ropa-

gates, and we obtain the Lemma 2.19.

Let

X E SH

be a sheU composition with s1te.

al.lowed by the interaction rules. clocking all shells in

Let ex, the result ot

C. be decorated with state markers·

according to the procedure ot Theorem 2.4 (states trails correspondence).

Let Fa

be a shell configuration within

X, as depicted in Figure 8.

Denote a site with contributing

and a site and Fa by the notation ~ ~ + 0 V with contributing cusps trom Fa and Fa by A o Then markers tor these sites will appear in ex on the

cusps from Fa

right and lett, respectively.

That is, they will have the

y

y

foms

Proot of 2.19.

A

and

A

Combine the observation just prior to the

statement of this lemma with Proposition 2.18. ~ .2!~.

the trail

cbA

Let A be a string.

= KA

We wish to show that

corresponds to a clocked state.

it is easy to see that

ex

Since

is clocked whenever X 1s a pure

shell composition (without sites), it will suftice (by 2.18) to show that if XI action rules, and clocked.

is obtained trom X E SH via the inter-

ex

is clocked, then

ex'

is also

We must show that the addition of an allowed site

38

cannot create a counterclockwise move. We can limit our considerations to interactions on a form Fa as depicted in Figure. 8. figure.

Terminology will refer to this

Call Fa the top, Fa the middle, and Fa_ the + 0 Then a new site may be between middle and top,

~.

middle and bottom, or it may be a self-interaction of one of these forms.

Using the notation of Lemma 2.19, we may assume

that all middle - top or middle - bottom sites receive markers in the forms

~

A

~ ~

and

By interaction rule 2 ••

the cusps from the top and from the bottom form parts of·the boundary between the interior and the exterior of the curve Q,

Theretore & counterclockwi.e move involving either of

tho •• markerl would nlce"arlly move the marker into the ext.r10r ot

oan

Q,

Therlfore no interior counterclockwise moves

arl,. trom '1t., ot thi. type. Pina11Yt con'ider the introduction of a top-top, bottom-

~ottomt

or middle-middle site.

If this occurs along Fa • o

-e-

then it will have the form

A counterclockwise move then entails an interaction of the form

y

Y ~

~-.~Since the marker at the

or (x,I3)

A

site is on the left and

13

is part of the middle portion, x must also be part of the

39 middle portion (since top - middle sites havemarkeZ'. right).

But, this is a forbidden interaction.

Eaoh t1m. a

self-interaction occurs, a curve or composition off.

Further self-interactions must come from

on th.

~ ~

18 .,llt

itselt.

Thus no counterclockwise moves can come from self-interaction. of the middle. Similar arguments apply to the top and bottom. have shown that

CF

Thus we

admits only clockwise internal moves.

Since any composition in

SH

can be decomposed into forms

of this type, we have shown that counterclockwise moves.

KA

does not admit any

It remains to prove the existence of

clockwise moves in KA. As we observed after Lemma 2.16, a composition of curls has no available moves.

This may occur when the shell com-

position underlying

is trivial.

DA

composition is non-triVial then move.

KA

If, however, this shell does admit a clockwise

Such moves can be located by searching for a deepest

!h!!! in

DA.

A deepest shell in a shell composition is a

shell whose middle, top, and bottom have no riders.

Such

shells exist since there are a finite number of shells in the composition.

It is then easy to see that an elaboration of a

deepest shell Via the interaction rules will always admit a clockwise move. Remark.

This completes the proof.

Figure 9 illustrates some elaborations of riderless

shells, and the available clockwise moves.

40

Fi~re

9

We are now readr to approach the proct ot the Clock Theorem.

The cor. ot the proot rests on a procedure for

loina trom one .tate to anf other state bf a series of transposition..

Once this methOd i8 clear, the theorem will

tollow, and we will be able to show that the extremal states constructed by Theorem 2.17 are unique.

In order to·create

a series of transpositions between two states, we first show that there is a series ot exchanges between any two trails. Each exchange then factorizes into a series ot clockwise or counterclockwise moves between the corresponding states. Definition 2.20.

A trail T'

trail T by an exchange it T' two sites of T.

is said to be obtained from a is the result of reassembling

41 Note that upon reassembling a single site, a tratl will break up into two components.

If these two componente

in'.r-

act at another Site, then a second reassembly at this Ittl will constitute an exchange.

Using string form, the first

reassembly produces an extra component that is homeomorphio to a circle.

Thus the generic form of the exchange (ignorinl

the presence of other sites) is as illustrated below.

If there are no other sites between the top and middle, or between the middle and bottom, then the exchange 1s accomplished by a single transposition of the corresponding states. With intervening sites, a series of transpositions can do the job (as we shall prove).

Examine Figure 10.

Since the generic

form of the exchange replaces a clocked form with a counterclocked form (or vice-versa), we shall refer to clockwise

~

counterclockwise exchanges where a clockwise exchange replaces a clocked form by a countercloc"ked form.

In Figure 10 we see

that a clockwise exchange corresponds to a series of clockwise transpositions. Proposition 2.21. U. T' •

This is always the case. Let

T and

T'

be trails on a universe

Then there exists a sequence of exchanges taking T to

42

'exchange

Factorizing

~

Exchange

~

Figure 10

Transpositions

Proot

~

2.21.

Two trails on the same

at an even number

01'

u~iverse

mu.t dirt.r

sites (since one trail can be tranl 8

tormed into the other by reassembling all the sites at whiDh they ditter, and an odd number connected torm).

These sites may be paired ott with each

other to give the desired set Proposition 2.22. U so that T' change.

Let

reassemblies leaves a 411-

01'

01'

exchanges.

Let T and T'

be trails on a universe

is obtained trom T by one S and S'

be states

01'

U (With the same

star placement) corresponding to T and T' Then S'

may be ob.tained trom S

transpositions.

clockwise~ex­

respectively.

by a sequence

01'

clockwise

Except tor the state markers at the exchange

sites (which must each turn through 90°), any state marker involved in these transpositions will turn through a total 01'

either 1800

or 3600

•

Thus clockwise (counterclockwise) exchanges factorize into clockwise (counterclockwise) sequences of transpOSitions. Proof'

~

2.22.

We shall prove this result by induetion on

the number of vertices in the universe U.

In order to do

this induction it is necessary to state the details of the factorization procedure more precisely.

The generic f'orm of

the clockwise exchange is that of' a clocked shell.

As such,

it has a top, bottom, and midline as indicated below.

44 midI E€t~ ~

in,

-bottom

In practice, the top, middle and bottom will all have extra sites.

The midline itself is a trail with sites (self-

interactions) and cusps (places where the midline has sites with top and bottom).

Let

~

denote this midline trail with

its sites and cusps.

Then

~

may be written as a sum,

~

= ~l e

~2

e ...e

~n'

of J-irreducible trails with sites and

cusps, where we extend the notion of J-irreducible and J.atomic (see discussion atter 2.17) ( --""'--- or

~)

by

taking an isolated cusp

as J-atomic.

Recall that a

trail is J-atomic it the string obtained by clOSing allot its sites is atomic.

For the rest ot this proof, atomic

(irreducible) will always be used in place of the term Jatomic (J-irreducible). Each

~K

is then a composition of atoms, and the atoms

are partially ordered is a rider on B.

by

the relation:

If A < B and

that atoms on different

~K

B

A < B whenever A


are unrelated

then A < C. by

Note

this partial

order. Induction Hypothesis:

The clockwise exchange factors into a

sequence of transpositions so that 1.

Only markers at the exchange sites and at sites along the midline trail are utilized.

45 clockwise~,

to',

2.

Each exchange site marker rotates

3.

For markers that move, those at sites between the midline and the top or bottom will turn a total ot 180· clockwise, while those markers at selfinteraction sites of the midline will turn a total of 360· clockwise.

4.

If A is an atom in the decomposition of the midline, and it A has a cusp rider, then every marker on A will turn.

5.

Call an atom involved it all of its markers turn in the tactorization. It A is involved, and A < B, then B is involved.

6. Rules 1. through 5. specify exactly the markers (hence the sites) involved in the factorization. This technical induction hypothesis constitutes an exact statement ot Proposition 2.22.

Note that it is clearly true

tor the.tirst tew examples, such as that shown in Figure 10. The proposition is also easy to verify tor the case where the midline trail has no. selt-interactions (hence it consists entirely of· cusps).

We leave this case as an exercise for the

reader. Thus we may assume given an exchange situation on a universe with N vertices, so that the proposition is known tor all universes of tewer vertices.

It the midline trail has no

selt-interactions, then we are done by the above remarks. Therefore, it may be assumed that there exists a selfinteraction site s ..

>-<

s

on the midline.

Remove

s

, obtaining a smaller universe UJ.

all other state markers on U'

via.

>-< .... )

Note that

are the saae as those on

46 U, and that an otherwise identical exchange problem is presented tor U'. Its

The induction hypothesis applies to U'.

is a site on an uninvolved atom, then no new in-

volvement is created by its removal. the tactorization tor U'

Hence, by induction,

extends to a factorization tor U

that still satisties the induction hypothesis. Suppose that

s

is on an involved atom, and that all the

crossings nearest to

s

remain involved when

s

is removed.

Then, by induction, these nearby markers undergo, rotations as described by statements 1. through 5. these rotations in U' at

Upon replacing

360·

induce a

s,

rotation of the marker

The geometry ot this induction is illustrated in

s.

Figure 11. Finally, suppose that when

s

s

is on an involved atom and that

is removed, one or both of the segments from the site

belong to uninvolved atoms in U'.

In this case we are pre-

sented with a situation in the torm

.,,--t¥-...

...--q;r-...

.

8" .: - ... ~ ... \

:

....~ ...

~

1.1.'

...

Ci

The site B.

s

is an interaction site between two atoms

These ride on the larger atom

When

s

involved.

is removed, the atoms

A and.

C, which is involved in U.

A and

B may no longer be

We have illustrated how B could lose involvement:

47 In U' B

a direct transposition at

need not be utilized.

x and y is &v..UI~lI, and

Note that by induction, thil

position,does occur in the factorization for U' larger atom

'rift••

(s1nce 'he

We are now pre.

C is still involved in U').

sented with a small factorization problem of the same typel Namely:

\ - - - -. .1-::::-- ••• ";J

Since this occurs on a smaller universe, the induction hypothesis applies, and we obtain a partial factorization in U where the marker at

s

turns through

180".

Now apply the

same reasoning to A, and obtain the full rotation of 3600 tor the marker at

s.

This completes the induction, and hence the proot of Proposition 2.22.

3600 Induction Figure 11

48 Remark.

Here is a concrete example ot the last part ot the in-

duction argument tor 2.22. Thus

U'

Let U and

UI

be as shown below.

is obtained trom U by deleting the site at

(indicated in state torm).

The atoms

A and

B in U'

uninvolved, but become part ot a larger involved atom U.

s are

C in

Contained within the larger factorization for U are the

small tactorizations with A or B and the site

.__A

u:

··

.

•

s•

49 The clocked state is unique. We are now prepared to show that the clocked stat. il

In

unique (and that the counterclocked state is unique).

It

order to do this, consider the form of an atomic trail.

it is a curl form, then no exchanges are available, and it is the only trail on its universe; thus there is nothing to prove.

Thus we may assume that the trail

an atomic,string A that is not a curl.

T corresponds to This means that

T

can be obtained from the schema

by

1.

splitting the vertices at

x

and

y,

2.

connecting {a,b,c} and (a',br,c') 1. and :2. produce a c'onnected curve,

}.

adding extra sites except at the input and output lines.

so that

In Figure 12 we have enumerated the three basic possibilities for such an atomic string.

Type

a

illustrates a clocked

shell upon which extra sites may be added. include a counterclocked shell type

~

the sites

x

and

y

a'

We should also

under this case.

In

have been split in a clocked

(or counterclocked) manner, but the pitchfork connections have been made in the one other possible manner. trated is

~o'

Also illus-

the form with the least number of sites that

is in this category.

The same remarks apply to

'Y

and

'Yo'

Any trail in the categories

~

extra sites to the prototypes Since

~o

and

~o

.or

~

and

~o

is obtained by adding ~o.

admit both clocked and counterclocked

exchanges, no trail in these categories can correspond to a clocked or counterclocked state (using Proposition 2.22). Thus a clocked atomic trail is in type clocked atomic trail is in type outer form of the trail.

at.

a, and a counterThis determines the

By repeating this criterion on the

resulting midline trail (recursively) we arrive at exactly the description of the clocked state that is summarized in Theorem 2.17. This completes the proof that the extremal states are unique.

Since any state can be transformed into a clocked

state by performing successive counterclockwise moves, this shows that any state is connected to the clocked (counterclocked) state by a sequence of transpositions.

Figure 12

51 The collection of states is a lattice. A specific marker in a state can be involved in no more than one transposition at a time.

Consequently, it 1s

~Ossi­

ble to label all state transpositions trom the clocked .tate on one diagram in the pa.ttern

For example:

In this example the moves labelled stricted by the heirarchy

and

b2

are re-

~

/"

bl

That is, bl

a, bl , b2 , c

b,

\c

cannot be performed before

a, and

c

must be done after and then do

b2 . If we write rs to indicate "do r S", then rs = sr whenever this makes sense. ab2 c # aCb2 since cb2 is Here equality means that~the resulting states

That is, abl b2 = ab2bl meaningless. are identical.

but

ThUS, trom the heirarcby ot operations we

52 sen.rate the collection of states

(Here

1

denotes the

clocked state.)s

Recall that a lattice is a partially ordered set that every pair ot elements

X and

Y of

S

S

sue~

have a well-

defined infimum X ~ Y and a well-defined supremum X V Y. If we indicate states by clockwise move sequences as above, then

X V Y

is the move sequence corresponding to the in-

tersection ot the set of moves for this detines a unique state.).

X and Y (Observe t~i

Similarly, X "Y

to the union of the moves tor X and Y.

corresporia.."

With the partiaJ.: '

order X < Y whenever there is a series ot clockwise movei ' trom Y to X, this gives a lattice structure to the set of states ot a string. The proot ot the Clock Theorem is now complete.

53 ,1.

~

Polynomials

~

the Duality Conjecture

Suppose that labels have been placed in each ot the four corners ot every vertex in a universe kth

% k

D

k

and suppose that the universe has ~

K and a state

K denote

vertex, Vk • are

B

Define the

Let

For purpose of discussion suppo,.

this labelled universe. that the labels at the

U.



product

W k n

vertices

(k. 1, •.•• n).

between the labelled universe

S of this universe by the formula

-

a(S)Vl (S)V2 (S) ••• Vn (s) where Vk(S) is the label touched by the state marker of S at this vertex. when the state and the labelling forms are superimposed.

Vk(S) = ~, Wk' Uk' or state marker of of the state We regard

Dk according to the position of the

S at the given vertex.

a(S)

is the sign

S.



as an element of the polynomial ring R

whose generators are the collection of labels of coefficients).

Thus

K (integer

The state polynom1al for a labelled universe

K is then defined by the formula

The state polynomial, appropriate matrix:

Let


is the determinant of an

Rl.R2' ..•• Rn.Rn+l'Rn+2

ing of the regions of U so that if' of U with stars in

~+l

and

}t

be an order-

is the set of' states

Rn+2' then this ordering

54 satisfies Definition 2.1, giving a permutation assignment for

d. Define the (seneralized) Alexander ~ (see (1]) A(K) = (A ij ) ot a labelled universe K to be the n x (D+2) matrix with columna corresponding to the ordered set of regions and rows to the ordered set of vertices, and entries as indicated below, if Rj

does not touch Vi' 1 1 , Wi' Ui or Di if Rj touches Vi in the corner corresponding to 'bhi. label, Bi + Wi or Ui + Di 0

Aij •

it Rj

touches two corners at the

vertex Vi' Let

A(K,~)

(the reduced Alexander matrix) denote the

matrix obtained trom and

A(K)

n Xn

by striking out the columns Rn+l

Rn+2'

Proposition~.

The state polynomial is the determinant ot

the reduced Alexander matrix,

Thus

= Det A(K,8). ~,

The terms

= a(s)VI(S) ... vn(S)

of the state

polynomial correspond exactly to the non-zero terms in the expansion of the determinant of

2.8) a(S)

= sgn(p(S»

A(K,C).

Since (proposition

for this permutation assignment

P : C -> S(n), we see that this propos.ition follows directl, from the definition Det A(K,l')

t

pES(n)

of

the determinant

sgn(p) Ap lAp 2" .Ap n' I 2 n

55 The state polynomial will be specialized to various sorts ,

of labelling.

Tne reader should compare this formal develop-

ment with the discussion in the introduction. The first labelling will be the standard

~

~

which we shall abbreviate as siMply

~

A blank corner indicates the label 1. Proposition~.

Let

oriented universe U label).

K be the standard labelling ot an (i.e., each vertex receives a standard

Let 8 be the set ot states tor U with a given

choice ot adjacent fixed stars.

Then tor a state

S in 8,

= (_l)b(S)Bb(S)WW(S) where b(S) denotes the number of black holes in S, and w(S) holes in S. N(b,w,8)

Hence



=

t

denotes the number of white

(-1)~(b,w,8)Bbww where

b,w is the number of states in 8 with

b black holes

and w white holes. Proof. Remark.

This follows directly trom definitions. Propositions 3.1 and

,.2

taken together show that

the states of a universe may be enumerated by taking the determinant of a matrix.

This is an analog of the well-

known Matrix - Tree Theorem (see [9]).

The Matrix - Tree

Theorem enumerates rooted trees in a graph.

In

tact, the

states of a universe are iri:one-to-one corr~spondence with rooted trees in a graph associated with the universe. iraph, G(U), is obtained as follows:

This

Checkerboard color the

regions of U with the colors black and white (say that the unbounded region receives black).

The vertices ot G(U)

are

in one-to-one correspondence with the white regions of

Two vertice.'in

are cormected by an edge in

G(U)

U. when~

G(U}

ever the corre.ponding white regions share a crossing.

For

example:

It is ea., to ... t:rOlll the proof of Theorem 2.4 that maximal:' rooted 1EUJ. in

A.aU

m s: JiUU..u

1l.

l8e this rl.uU, let

are.!!! one-one correspondenee ~

m.! given .£h2!£!

of

fixed~.

.:l1!!!

To

denote the (dual) graph obtained.

G' (U)

by thl saml con.tJ'\\ction on the black regions.

mal rooted tree on either of the graphs

Then a maxi-

G(U) or G'(U)

uniquely determines a tree on the other once the roots are specified--together the two trees exhaust the set of crossings in U.

A pair of trees specifies a state as in Figure.'t :'"::1":1

The Matrix - Tree Theorem has been applied in knot theorY via the graphs

G(U)

and

G'(U)

(see [8], [9], [12J. [3:H);'2' "

To the best of my knowledge, the cormection between trees in>;\ G(U)

and the classical Alexander matrix has not been noted

betore. The main result in this section is Theorem hl.

Let

K be a standard labelling for an oriented

uni verse

Let

J and J.

U.

be two collections of state.

for different choices of' adjacent fixed stars.

Then

. Hence, for each pair of non-negative integerl .

57

(r,s), NCr,s,l) = NCr,s,O") = N(r,s).

l'!Uh .!: black the choice .Qf.

~!:!l!!

~

.!

The number of mi!!

whi te holes II independent 2t

.!1W..

At this point it is worth stating the Duality Cgnjecture:

Let NCr.s)

3.2 •• then NCr.s) = N(s.r)

be defined as in Theorem

for all non-negative integers

.!:.t....!. There is good computational evidence for this conjecture. We shall compute one example at the end of the section. In order to prove Theorem 3.3 we need an indexing of regions due to Alexander «(1]).

Each region is assigned an

integer index so that adjacent regions by one.

hav~

indices differing

The increase or decrease of index from region to

region depends upon the orientation of the intervening boundary as in the schema below.

~

hl.

l!22!. pattern

Every universe has an Alexander indeXing.

By splitting each vertex in U according to the

X ... ~

, we obtain a disjoint collection

of Jordan curves in the plane, the Seifert circles for U,

([12). (36).

Regions that coallesce under this splitting

have the same Alexander index.

Hence it suffices to index

the regions of a universe that consists of a disjoint collection of oriented Jordan curves. 13. )

This is clear.

(See Figure

58

Alexander Indexins Figure 13 We shall use the Alexander indexing to obtain linear dependence relations among the columns of' the Alexander matr1~ The strategy i8 as tollow8:

Combinations that sum to zero.at

each vertex imply global combinations that sum to zero. More 'pecifically, Since the columns of' the Alexander matrix are in one-one correspondence with the regions of' the:

,~,

universe (see the definition prior to 3.1), we may speak of' the columns of index p

tor an indexed universe.

denote the sum ot the columns of' index p.

Let

Cp Then, by e~1ri!q

the f'orm ot a single corssing, we see that if x satisfies the quadratic equation x2 + x(B + W) + 1 = 0, then xP+2 + xP+Le + xP+lw + xl' = 0 have the global relation

t xpC

p

tor any integer p, and we P

= O.

See Figure 14.

59

Figure 14 of

~

x2 + aa

px

Let

+ 1 = 0, denoted

=1

have

Theorem~.

and a +

a

a= -p.

~ aPcp = 0

=B

p

and

+ W.

a,

Then the roots of

satisty the equations

Thus, with notation as above, we ~ (iPcp = O.

and

Hence, letting

[a] = a - a-I, we have

tor any index Let

k

(multiply both sums by

ak

and subtract).

K be the standard labelling of U, and let A = A(K)

be the Alexander matrix. trom A(K) of index regions

by

s.

and that

A(k,s)

be a matrix obtained

deleting one column of index k

Let

~l

Let

F(k,s)

and Rn+2

A(l,O)

Det A(k,s).

(*) yields

[ak-r]C r =

the scalar

[a k - r ]

We may suppose that

have indices 1 and 0 respectively,

is obtained

Hence, by 3.1, F(l,O) =

and one column

by

deleting these two columns.

. For s

~

k

~

r, the relation

t [aP-k]cp and therefore pot,k,r (**) [ak-r]F(k,S) _ [ak-S]F(k,r). To see this identity bring

into the determinant

lng a column of index

r.

F(k,s) by multiply-

Use (*) in the form

g1 ven

above

60 to replace the relult1ng column of index not equal to can see only a

k

l~ecit1c

the column ot index a column ot index

r I

or

r.

(k,s,r)

k

~ I

and

By linearity this determinant

column of index

s

in this sum.

Thus

is effectively deleted and replaced by multiplied by

From this we ealily obtain that whenever

a sum involving all columns

by

r

and once tor

~

t.

[ak-S).

Thus proves (**).

[ar-t]F(k,s). [ak-s]F(r,t)

(Apply (**) tWice, once for

(s,r,t).)

The symbol

s

denotes

equality up to .iRlfl. and

In particular, it F(k,s) • '(r,t).

Since

•

r - t = 1, then

Det A(r,t)

for some choice:;! .. A'

r - t . 1, we conclude that =
ot columns 10 that

these polynom1als is determined intrinsically as

(_l)b.

Tht$l

completes the proof of the theorem. Remark.

We have crucially used the identity between the in7"

,

tr1nsically defined state signs and the signs of the permut';r, ',i

tions that may be associated with the states.

This, in

Hence this proof used

~

',',

"

depends upon the Clock Theorem.

<,

turn~, evelT~1

'"

thing we have built up to this point. Digression on Star-Separated States We use the notation of the proof of 3.3. ship

[ar-t]F(k,s). [ak-s)F(r,t)

states with widely spaced stars. standard state polynomial, and let d = k - s

The relation-

gives information about Let Fd

F

denote

denote

is the index spacing of the stars.

F(l,O), F(k,s)

Then

tbl

wh'''<~'

61

[a)Fd • [adJF

and hence

Let

Yd - [ad]/[a] quadratic x2 + xp + 1 Yd

Proposition~.

Fd • ([adJ/[a]}F.

where

=0

a

and

and

a

are the roots of the

p = B + W.

satisfies the recursion' Y1 - 1, Y2 • -p Yd+l = -PYd - Yd - l

Proof. a +

This follows directly from the formulas

a = -P,

aa. 1,

Yd = [ad]/[a) = (ad-ad)/(a-a).

Thus we have the table Y1 = 1, Y3

= p2

Y2 = -p - 1

Y4 = _p3 + 2p

Y5 - ,,4 - 3p2 + 1 Y6 = _p5 + 4p3 - 3p

Y7

42 = p6 - 5p + 6p

- 1

The sum of the absolute values of the coefficients of Yk is the kth Fibonacci number (The Fibonacci numbers are the members of the sequence fk + f k+1 •••• ). coefficients.

1, 1, 2, 3, 5, 8 ••••• f k , t k+ l ,

The coefficients themselves are binomial

Consider the simplest case of this phenomenon.

Each uni-

verse in the sequence shown in Figure 15 has

F = 1

has only one state for adjacent fixed stars.

If we give the

universe

Ud

its maximal star placement of

since it

d, and let

Fd = Fd(Ud ) with thil placement, then- Fd = ~dF = ~d' Since there is no cancellation ot terms in ~d when expanded as a polynomial in B and W,

~d

actually lists the states tor

this star placement. Jor example, ~3 = (B + W)2 - 1 = B2 + w2 + 2BW - 1. Thul we expect U3 to have five states, one with two blaok holes, one with two white holes, two with a black hole and a white hole, and one state with no black or white holes.

This il born out by the enumeration in

Figure 16.

u, Figure 15

Figure 16

To obtain a purely geometric view of the same let rd = ES~ Bb(s)WW(s)

where

d

,,'u."on

8d denotes tpe .tt 0' ••••••

of Ud with farthest star placement as in Figure 15. Shaw. using only the geometry and an induction argument, that r l = 1, r 2 = p and r d+l • prd + r d_l where p = B + W. In the general case the polynomial Fd will not explicitly list all the states.

This is due to cancellation ot

terms in the determinant expansion.

Also, signs for star -

separated states cannot be computed simply from a black hole count.

We leave it as an adventure for the reader to look at

further properties of states with non-adjacent stars!' The Duality Conjecture We end this section by computing an example that illustrates the Duality Conjecture. universe U shown in Figure 18.

The example is based on the Here we have drawn the clocked

state and illustrated the operation heirarchy (see the example at the end of section 2.).

Also listed is the effec,t that

each operation has on the black-white hole count: For example, BW- l next to the operation b2 indicates that this operation adds a black hole and subtracts a white hole. Figure 19 shows the state lattice generated from the hierarchy.

Finally, Figure 20 shows the black-white hole counts

computed from this lattice. firmed for U.

The Duality Conjecture is con-

64

U with operations avaIIible trom its clocked state. -

a-

SW

h. ba

S-'W

c,

BW-'

Col

a-'W-1

ell

8-'W- '

J~

BW

Operation Hierarcny

e

Figure 18

8W·'

BW- '

:L

I

a,

State lattice LeU). generated!l2m -----operation heirarchY. Figure 19

~

66

Black ~ - White ~ Classitication

~ .. ~ g! ~ in!2!:!!! BrW s Duality COnjecture:

l!.L!:..a..!l = !'!!.!...l:l..

Figure 20

9!. Iill!l

67

We generalize the state polynomial from and links.

unive~sl'

to knots

The key to this ascent into a third d1men.1on is

entailed in the code for crossings:

x·~

(standard)

(revlrse) A knot or link (diagram) is an oriented universe with standard or reverse labels at each crossing. Given a diagram K with underlying universe U, and a stat~

S of U, we have defined (section 3) the inner pro-

E Z[B,W]. In this case is given by formula - a(s)sXwY where x denotes the number duct

the of

coincidences of B-1abe1s with state markers' of S that occur when the diagrams for Similarly. y

X and

S are superimposed.

denotes the number of W-coincidences.

It is instructive to compute the state polynomial by superposition. and also via the Alexander matrix.

Such a

computation is shown for the trefoil knot in Figure 21. Theorem 4.1. verse

U.

Let

Let

X be a knot or link with underlying uni-

0 be the set of states of U for a given

choice of adjacent stars. nomial



more. if X.

Let

~

.

Then the poly-

is independent of the choice of stars.

K.

Further-

L are links that differ at the site of

~

68 crossing as indicated below

><

~

XR

K.

....----:t.

L

04(R') K"> In other worda, <~) then

~.

.c:::

= (W-S)
- =(w-a)<~

It may happen that the universe

U corresponding to

K and K is connected, while the universe U' ing to. L is disconnected.

('

In this case

correspond-

K,

K,

and

L must

have the form indicated below

This observation justifies setting lying universe is disconnected.



= 0

when the under-

For it is not possible to

have adjacent stars in a state of the form Hence, when U'

is disconnected then all

states of U will have either up or down markers at the vertex in question. so -

Therefore

=

and

= 0 = .

We reformulate this observation by stating: cal dotted line between the two strands of a region of U', then

= O.

~ ,,

~L

If the verti-

L does not cleave

69 Now assume that this dotted line does cut a reSion. denote the states of UI. we see that

$1

denotes the states of 8 1

£1 =

Letting

lui where l

with a marker to the lett ot the

dotted line, and

!

the dotted line.

(If the region in UI

the states with marker to the r1sbt ot between the two

strands is occupied by a star, let the star reside on one side or the other of the dotted line.

!

is empty.)

In this case either l

or

By closing the site at the dotted line, and

adding a marker on the unoccupied side, we obtain one-to-one correspondences of states of UI U.

!l

where

with a subset of states of

is in one-one correspondence with ~, and ! ~

and

~

with !

are the states of U with a white hole or

a black hole respectively at. this crossing

z<->~:

"><.

<->~:

)(

! Thus



=


W

<x1t1> ... B

-B

<XI!> ... -W •

(The black holes contribute the minus signs via the formula for the sign of a state S: cr(S)

= (_l)b(S).)

= (W-B) .

Since

Hence -

can-

not see up - states or down - states, - <'K> =

-



=> - <'K>

= (W-B).

This completes the proof

of the exchange identity. We use the exchange identity to prove independence of star

70

placement.

Since

L

has fewer crossings than

may assume by induction that of stars. if



K,

we

is independent of the choice

Hence, by the exchange identity,

is independent.

K or



is independent

Since any two links with the same

underlying universe can be connected by a sequence of crossing exchanges, it suffices to produce one is independent of the star choice. Theorem 3.3 for

K so that



This is provided by

K the standard labelling.

The proof of 4.1

is now complete.

*~ +W~ J .~ -we + 8' R :1..

::1.

A(K) .. ~ 0

.

8

:1. 'W'

A(K~.Lr

/

Se.l

Fi~e

-- Det A(K.,4} \ 21

Topological Inyariance We now investigate how

changes under elementary de-

formations of the link diagram.

The three basic deformations

(the Reidemeister moves [33]) are as shown in Figure 22.

Elementary M2!!! Figure 22 Figure 22 shows representative s1tuations for each move.

The

diagram must have the local forms as shown 1n this figure; the move is performed without disturbing the rest Defini tion 4.2. ~

(K

~

K')

Two link d1agrams

K and

o~

K'

the diagram. are egu1 va-

if there is a sequence of elementary moves

carry1ng K to K'.

(Two diagrams whose underlying uni-

verses are 1somorphic as planar maps are regarded as identical.

,

Thus topological deformations of the underlying universe

are placed in the background of th1s discussion.

Such defor-

mations have no effect on the set of states or on the state polynomials.)

72 Equivalence ot link diagrams corresponds to ambient

~.

isotopy ot the corresponding links in three-dimensional space (see [33]).



We shall see that

becomes an invariant ot the equi-

valence class ot K if we set WB = 1. this collapse, note that if I rated

BW - 1, then

by

Z[B,B- l ].

homomorphism.

is the ideal in

Z{B,W]/I

--->

Let. :Z{B,W]

Z{B,W]/I

and let 1.

2. 3.

gene-

denote this quotient

We then define the Alexander-Conway Polynomial by

We shall see that

vK denote when BW = 1 z = W - B. Then

Theorem.!L2.

Z[B,W]

is isomorphic to the ring

the formula vK = *. VK is a polynomial in z = W - B. of .! link K

In order to formalize

Let

=> K ... 0 => vK - "K =

K ... K'

(as

above~.

VK = VK , VK

=1

ZVL

(0

denotes the trivial knot)

when K, Ie, and L are related

as in Theorem 4.1. FUrthermore, these three properties sutfice to calculate

vK without any reference to its definition as a state polynomial.

Remark. ~,

In section 5 we will take

1., 2., and 3. above as

arid show that they define a unique invariant of knots

and links.

Consequently, the last part ot 4.3 (calculation

from the axioms) is deterred to the next section.

73

Proot

£!~.

Since

is independent ot the star

~l&oem.nt

(by 4.1). we shall choose those star locations that are mOlt

convenient in the calculations to tollow. For elementary moves of type I. we assume that the curl is star-tree and hence must contain a marker.

Thus there is

one-to-one

Since the marker in the curl is either an up or down type (depending upon the orientation ot this string). this correspondence ot states shows that



is invariant under moves

of type I. For moves of type II. we distinguish two cases:

In Case

A

we have

Choosing star location as shown below, the states tall into the types

So' Sl' 82 :

+~+

74 We remark that the reg10ns

X and Y above and below

these d1agrams may be assumed d1st1nct.

For 1f X and Y

connote the same region. the situat10n 1s of type

Hence there are no states of type 81 and clude

82

80

when

X = Y.

Since

contribute equally w1th oppos1te s1gn. we con-

o

=

X # Y.

Now return to Case A when

Then there 1s a one-to-

one corre'TYT~ l3 C} . It follows that

<~

'>

= BW<~ ~>.

In Case B'we have the cod1ng

Aga1n. states of type

Sl

and

S2

tr1bution to the state polynom1al.

cancel each other's conFrom the code above and

the correspondence of states, it then follows that invariant under the move of type II B. verif1cation that



is

This completes the

VK is 1nvariant under moves of type I.

and II. Analysis of the cases for the moves of type III. will proceed by listing state types for the star placements

75

" . , ' " A ."

..

~

These indicate corresponding regions before and after the

move

(compare with Figure 22). For example, suppose we consider a state of the form.

Here the

•

signs occupy regions that have state markers

other than at this triangle.

The state depicted above is the

only type for the horizontal line in the upper position. ing the horizontal line downward, we find three types

Slid-

So'

~¥~~

In order to catalog the contributions of these states, suppose that

K and KI

(differing by a type III move) are given

locally by the diagrams

K,:

"

Za

X y

/ /"

,,-

=

76 Le~

a

denote the contribution to (Kls> from outside the

(K'IS~=

_w2 a, 2 a. Hence (K' Is~ USi U Sp = a. This (K' lSi> = a, (K' 182> - W given triangle.

Then = BWa while

proves that . - t
IS~

U8i U8p.

Identical sorts of

calculations go through in all the remaining cases (The cases are listed in Figure 23.).

This proves the invariance of vK

under type III. equivalence. Since part 3. of this theorem has been proved in Theorem

4.1, this completes the proof of parts 1., 2., and 3. of Theorem 4.3. Discussion.

In Section 5 we shall discuss how to calculate

directly from the axioms.

Here it is appropriate to record

a few examples via state summation: 1.

If K is a trefoil knot, then we see from Figure 21 that 2

vK = 1 + z , since

- W2 - WB + B2

=

· (w - B) 2 + WB.

2.

If L is the link of two unknotted circles of linking

3.

If K'

is the standard knot corresponding to the universe

of Figure· 1, then K'

is as shown below and it is easy to see

from Figure 1 that the state polynomial is - 3WB

= 2(W-B)2 +

WB.

Hence

vK1

=1

~K'

+ 2z2.

(K'> =

2w'2 + 2B2

77

~

: ,••• 1

;

•f

A

I

I

I

J•

~lAAA I

V- lA t

YV . 1AA. · •• 1

YYY1~ ,

Y

-~

••

A

«

••I

1 Tri~e

states

Figure 23

78

2.

Axiomatic Link Calculations Except for the definition of equivalence of link (diagrams)

via elementary moves (See Figure 22.), this section is selfcontained.

We discuss, with many examples, the subJect of cal-

culating the Alexander-Conway polynomial, vK' solely :from the axioms.

Henceforth, this polynomial will be referred to as

the Conway Polynomial.

It's axioms are a partial list of

properties given by Conway in [5] (With a slightly different normalization).

Conway's paper discusses calculations of the

kind we are about to do; it does not give models for the axioms or discuss consistency.

Here, of course, we have already

proved the axioms in Theorem 4.3.

Nevertheless, this section

will end with a brief discussion of Ball and that the axioms are self-consistent.

Meh~ah's

proof

They provide their own

model! Axioms for the Conway Polynomial 1.

To each oriented knot or link K there is associated a polynomial

VK(Z) E Z[z]

such that equiva-

lent (ambient isotopic) links receive identical polynomials. 2.

If K is an unknotted circle, then

3.

If K,

K,

and

vK = 1.

L are three links that differ

at the site of one crossing as indicated below, then

79

When links in the form

A, J

and

>
Care (diagrammatiC&ll,) "~I".

A

we shall write

A.

C and

B = A e C.

Thus, by axiom 3.,

vASB ~ vA + ZV B and vA8C = vA - ZVC' Regard e and e as non-associative, non-commutative binary operations defined on appropriately related pairs of links.

(Warning:

These opera-

tions should not be confused with the string combinations of section 2.

Since we do not deal with strings outside of

section 2, I have taken the liberty of using the same notation for these different operations.) While we work with knots and links via planar diagrams tor theoretical purposes, it is worth noting that they can just as well be seen as collections ot embedded and oriented circles in three-dimensional space.

For example, here is a

standard procedure tor producing an unknotted circle from a given diagram: ~

.!2 !h!&. you

Choose.! point

!!!:!! draw

:e. .2!!.!ill.!

diagram

~

!:!l over-crossing· l1n.e !:1

!!!:!! encounter !!!Jill .! crossing, undercross !! .!ill.! encounter, .!:!!£. continue until you ~ 12.'2,

draw.!

.!ill.!

~

It is\ easy to see that this procedure produces an unknot, but quite a task to show it via the elementary moves. Figure 24 tor an example of this unknotting method.

See

80

Producing

~ ~

Figure 24 A l1nk L

1•• &1d to be split if it is equivalent to a

union ot non-.mptr .ublinks

Ll

and

L2

that are contained

in disjoint regions ot the planar diagram (or in disjoint three - ball. in three dimensional space). Lemma~. ~.

vL • 0 when

We may assume that

L is a split link. L has a diagram in the form

given below:

Since K is ambient isotopic to

K

via a

2~

rotation,

vK = Vi' By axiom 3., vK - ~ = ZVL . Hence 0 = ZV L • Therefore vL = O. We are now prepared to calculate the trefoil knot. axiom 1. implies that

81

Figure 25 Refer to Figure 25.

is an unknot and U is an unknot. a split link,

~

= LeU.

We have K = iC e L, L

= O.

Since

L

VK = 1 +

Theretore

where it

is equivalent to

ZVL ,

VL

=Z

and

:. vK = 1 + z2.

In this example we can write K ~ iCe (L e u). a

~

decomposition of K.

decomposition are unknots

This is

The individual parts of a skein

(0)

and unlinks

(0 0 0 ••. 0).

We refer to these as the generators of the skein.

(This

terminology is due to Conway (6].) Lemma~.

generators.

Every link L has a skein decomposition into Hence the ConwS¥ polynomial may be calculated

via the axioms. Proof.

If L is a knot then there exists a sequence ot

crossing changes

(><

~

X)

that will unknot it.

L is a link, then some sequence of crossing changes will produce a collection of unknotted, unlinked components

If

82

Hence there is a sequence is obtained

Li

~rom

a skein generator.

Xe

where from

~-l

by switching one crossing, and

Thus we have

-y = X e y and

Lk

and

6k

=

.1.

Here

Lk

is obtained

by splicing out the switched crossing.

This gives a skein decomposition of L {Li,L2, .•• ,~J.

Li +1 Ln is

such that

L = LQ, L1 •••• ,Ln

into

~

and

Since the links in this latter set have

diagrams with one fewer crossing than the diagram L, we may assume inductively that each has a decomposition into generators.

This completes the proof of the lemma.

~.

Note that the method of proof of this lemma is the

way we computed the polynomial for the trefoil knot.

Follow-

ing the notation of the proof, we may conclude that vL = VLn + z~=l knot, and

Lk ·

6 kV

Hence

VL

=1

+ zD when L is a

VL = zD when L is a link with more than one

component. A connected

~

of links

K and

K', denoted

K f K', is

defined by splicing two strands as shown in Figure 26. may depend upon the choice of strands.

This

83

Figure 26 proposition .5..:2,.

* K'

1.

If K K and

denotes a connected sum of the links K', then VK K' = VKVK,·

2.

If K* is the result of reversing all orientations on all strands of K. then vK* = vK •

3.

Let K' be the mirror image of K (obtained by switch~ng every crossing in K). Then VK!(z) = VK(-Z).

4.

If L is a link with A compopents, then VL(-Z) • (-l)A+lvL(z). Hence VL ! = (-l)~+lVL'

*

,

The proof of this proposition will be omitted.

All these

results follow easily by induction and the fact that the skein is generated by trivial knots and links.

Note that part 4.

implies that a link with an even number of strands is inequivalent to its mirror image whenever it has a non-zero Conway polynomial.

L = t • K, ~

:. vL

= -z,

vK = 1 + z2

= -z + Z(1+z2) = z3.

a4

2.

Let

M be the knot shown below.

Then M = Me Q where

Q is isotopic to two unlinked circles, and

to K

*K

where

M is

isotopic

K is a trefoil knot.

Q:Y)mJ05) M

M

The knots

M and

Q,

K f K are distinct (as can be verified by,

other means) but they have the same polynomial. 3.

Let

below.

~

Let

vn+l - v n _l

= 1,2,3 •..• be the Vn = V~. Then Vl = = ZVn (by axiom 3). n

sequence of links indicated 1, V2 = Z

and

. .. Hence

Vl

= 1,

V4 = z3

V5

4

v2

= z,

V3 = z2 + 1

+ 2z 2

+ 1 3 V6 = z5 + 4z + 3z v7 = z6 + 5z 4 + 6z 2 + 1 Va = z7 + 6z 5 + lOz3 + 4z2 = ,z

+ 3z

85 4.

Let

K be the standard k.not (all crOssings;':!~, "",

associated with the univer8e of Figure 1.

Then

~;,lt

>4

..

..

~l

C@

gram

We leave it to the reader to show

VK = 1 + 2z2. This is a good example to compute via the axioms in more than one way.

It can also be computed from the lattice of states given in Figure 1.

5.

Let

K'

be the standard knot associated with the universe

of Figure 18.

Then K'

and we see that K' Hence

has diagram

is equivalent to K5 4' 2 VK1 = Z + 3z + 1.

From Figure 20 it is easy to deduce that

ot example 3.

K'

has state poly-

nomial (see section 4)

=



Hence (via

z = w - B and WE = 1 under t) we see that

t = VK,. as predicted 6.

a Weave:

Figure 27. Claim:

(W-B) 4 + (WB+2) (W_B)2 + 1.

Let,

1\ f+

,by

RI\

Theorem 4.3.

be the operation indicated in

-n -

Then, as shown in this Figure. R"-".

v

n R "

=

2

"1\- z v n-l R

"

86 ~.

Let "~ • RnA.

Thus Xn

N

A.

~~~ L""

-zvXn-l Q.E.D.

Figure 27 In Figure 27 a knot

A is given with unknown tying within

the black box, and some particular starting configuration. The transform RA

is a knot of the same torm.

The claim

shows that we may recursively calculate the polynomials for these knots .• For example, let

"

be a trefoil knot.

;; is an unknot. and we have (letting fn tn

=1

2

- z f n _l •

= v~)

fo

Then

=1 +

z2,

87

... fO = 1

+ Z2

f

.. 11m f •

n-

=

246 1 - z + z - z :!:".

n

fl = 1 - z2 - z4 f2 = 1 - z2 + z4 + z6

It 1s tempting to try to make sense out ot ing

f.

such that

f.

by assign-

as an extended invariant for an infinite knot

RK.

idea, see [20).

~

K.,

K.

For a possible formalization of this

88 7.

Tangle

.!hee ".

This)~l' 18 a quick introduction to Conway's theory

ot tangles. ways.

The material here can be generalized in various

See [5] and [15J for more inf'ormation.

A tangle is a

piece ot a knot diagram with two input strings and two output strins., oriented as in

Figur~

28.

Each input is connected

to one output, and there may be any other knotting or linking (Without tree ends) inside the tangle box. and

I, the tangle

output.

61

Given tangles A

A + B is defined by connecting inputs to

in Figure 28.

Also, there are two ways to form a

knot or link from a given tangle A. (numerator) and D(A)

These are denoted N(A)

(denominator) as N

Given a tangle A, let A

= VN(A)

in

Figure 28. D

and let A = vD(A)'

The traction of the tangle A is then defined by the formula F(A) _ AN/AD, Theorem!2.d,. + F(B),

Let A and B be tangles.

That is, (A+B)N _ ANBD + ADBN

Then F(A+B) ... F(A)

and

For example, consider the tangles Q and Q=

~:

-.. --;~

~.~

Q+Q=:t=o Q+ ~

(A+B)D = ADrf,

= ::;> c!::

=..

~ + !: ...

::t> c:!::> c:r: = ~l

¥= ¥+ ¥. . F(Q) + F(Q) F(Q+~) = F(~) = ~ = ¥+ ~ = F(Q) + F(~)

:. F(Q+ Q) ...

F(!!+!!) ... F(!!l)"'~ = ~ + ~ ... F(!!) + F(!!).

89

= 1/0 +

Note that the identity % formal addition of fractions

1/0

tollows from

alb + c/d = (ad + bc)/bd.

we have verified the theorem for the tangles .Q and !!.

Thus It

will be left as an exercise for the reader to show that this actually suffices to prove the theorem!

(compare with Lemma

5.2. )

As an application of this addition theorem, let Ln note the link illustrated in Figure 29. Vtn = Num(Z + ~ + ~ +.•• +~)

5.4 that

denotes a continued fraction with n

It then follows from where this formula

terms.

t A+B

N(A)~ tj

B

P

D(B)

geoo N(A)

Figure 28

de-

0 (A)

90

Figure 29 8.

The Coefficients Let

~K(Z) = ao(K) + al(K)Z + ~(K)z2 + a,(K)z3 + •••

Then the coefficients of the polynomial are themselves invariants of link type. that

ao(K)

= {l o

It follows immediately from the axioms

if K is one component otherwise.

For certain calculations it is useful to adopt a notation to indicate the operations of switching and eliminating a crossing.

Accordingly, let

E(X) =

E(X.)

=

S(~)

~(X) = ~ and let EiK

and

operations to the

SiK

='X

denote the result of applying these

ith crossing of K.

We have implicitly, up until now, aSSigned indices to the crossings according to the formulas:

IeX'):: Let

+i

IiK denote the index of the

)

I(~)

=-i.

ith crossing of K.

Then

91 the exchange identity of axiom 3 may be expressed by the formula

It follows that the coefficients of the Conway polynomial obey the series of identiti•• :

In particular, it is now easy to ••• that terpretation in terms of linking Definition~.

al(K)

has an in-

num~.r.1

Let K be a two-component link.

Define

the linking number, Lk(K), by the formula Lk(K) = i tilT(K)Ii(K).

Here the set T(K)

of crossings of different components of K. do not contribute to tne linking number.

Lk((9)) = Lemma~.

Proof.

denotes the set Self-crossings

For example

+ -1- •

a1(K) = {Lk(K) if K has two components, o otherwise.

This follows at once from the definition of the link-

ing number, and from the exchange identity for Remark,

al ,

It is obvious from its definition that the linking

number is invarfant under elementary moves·, The problem of direct interpretation for

~(K)

and for

the higher coefficients of the Conway polynomial is more mysterious! ~(K)

We shall now give an interesting property of

taken modulo-2, and in section 10 we shall show that

a2 (K) (mod-2)

is the Arf invariant «(34]) of the knot

K,

lat ed two kn ots th at are re Le t K and K' be am below: ac co rdi ng to the d1agr

TheOre!!,.~.

oth erw ise 1d en tic al fo r K and K' are Th at 1s , the diagrams (m od ulo -2) . Then ~(K). ~(K') ld cro ss1 ng as in rti ce s at the to ur -fo Pr oo t. La be l the ve dic ate d below. = E4S3S2SlK~ = ~S2S1KJ an d X4 a E2S1K, ~ , ElK l· Le t X pli es th at ca tio n ot axiom :3 im pli ap ted ea rep a en Th cu lar , + Vx - "'X 4 ). In pa rti vK - VK, .. z(vX 1 - Vx 2 :3 a1 (X 4 ). Th is Xl) - al(~) + a l ~( .. ) KI ~( ~(K) of a 2 to a problem of th e change the es uc red la mu for so cia ted lin ks mbers to r the to ur as nu ng ki lin of rn tte pa as an ex erot the pr oo f 1s le tt st re e Th • X X" 4 , Xl ' X2 ms of co nn ec tio n e are thr ee ba sic for er th at. th te No e. cis e component}. ng to form a kn ot {on ssi cro ld -fo ur fo the to r schamas: They co rre sp on d to the

X3

(X3) -

•

93 Remark:

Theorem 5.7 also holds for a four-fol.d ClZ'OIl:Lnl

switch with reversed orientations on one pair strands.

Thus

K and K'

or

~ar&11.1

forml

• In general, all that is required is that each pair of parallel strands have opposite orientation.

Call two knots pass-

equivalent it one can be obtained trom the other by a combination of ambient isotopy and four-fold switches of this type. For example, the trefoil and the figure eight knot are passequivalent:

~ trefoil In section 10, when we discuss the Arf invariant, we shall prove the following theorem. Theorem. only if'

Two 90ts

K and

~(K). ~(K')

K'

are pass-equivalent if and

(modulo-2).

In particular, any knot

is pass-equivalent to either the trefoil knot or the unknot. This theorem provides & direct geometric interpretation for the

mod-2

reduction of

~(K).

94

o

9,

Co ns ist en cy ,

8, le t no tat ion of example Us ing the sw itc hin g . ori~nted lin ks ere K and K' are wh lK "'S _l Sr Sr = K' id en tit y shows ati on of the exchange Then rep ea ted ap p1 1c

so th at K' is un a sw itc hin g seq ue nc e Sin ce we can ch oo se s th at it sh ou ld be th is for mu la su gg est kn ott ed or un lin ke d, ter ms .of an ' In ely de fin e an +l in tiv uc ind to ble ssi po the Conway po lytak e th is ap pro ac h to [2J Ba ll and Mehtah ry pr oo f of the ele ga nt and ele me nta no mi al, ob tai nin g an eir ap pro ach is the iom s. The ke y to th ax the of cy en ist ns co d in Fig ure 24 , pro ce du re ill us tra te ing ott kn un the of e us r pa pe r fo r fu rth er r is re fe rre d to th ei de rea ted es ter in e Th de tai ls.

95 §.

Curliness

~

!n! Alexander Polynomial

In this section we discuss the structure of a univt,.. viewed as a plane curve immersion.

The total tuminl numbe,

of the tangent vector to such an immersion will be given a combinatorial torm called here curliness.

We then show that

Alexander's original algorithm for calculating the Alexander polynomial contains a hidden curliness calculation.

It is

this calculation that make. the Alexander polynomial dependent up to factors' of the form • tk, upon the particular choice of knot projection for a given knot.

Thus we show how to

normalize the classical polynomial to obtain the state polynomial model for the Conway Polynomial. prove the translation formula the Alexander polynomial,

~(t)

In the process, we

= vK(Jt - l/Jt)

relating

~(t),

(Here

=

and the Conway polynomial. denotes equality up to factors of the form ctK

where

k

is an integer.)

Immersions If

ex: 81

->

R~

is a differentiable mapping of the

circle into the plane with non-vanishing differential

da,

then

is

a

is said:to be an immersion.

said to be normal if the pre-image p E R2

ex-l(p}

ex

of any point

is either empty. a Bingle point, ,or two pOints

(Pl,P2) = a-l(p) pendent.

An immersion

with

dex(Pl}

and

da(P2)

linearly inde-

Thus a normal immersion is a locally one-to-one

mapping with normally crossing singularities in the image,. The same remarks and definitions apply when the single circle

96

is replaced with a collection of disjoint quently, ,g, universe !! may

~

c~cles

A.

Conse-

represented .!!. the image

a(A)

!2t..!: ~ immersion a: A - ) R2. where II ll!: diS-

joint

co~ection

of Circles.

a,~:

Two immersions

1\ - )

R2

are said to be regularly

homotopic i t there is a family of immersions

o~ t

~

1

so that to = a, fl =

continuously wi tb t.

and

dtt vary Just as ambient isotopy tor knots and ~

ft

t t : " - ) R2 • and

links is discretized by the three elementary moves (Rei demeister moves) of sections 4 and 5, we obtain a discrete version of regular homotopy via the moves of type A and B illustrated below:

A.

B.

Definition 6.1. homotopic

Two universes

(U .. UI

)

if U'

U and U'

are regul,ar1y

can be obtained from U by a)

sequence of moves of type A and

B.

G1ven an immersion a: 81 - ) R2

of an oriented circle

into the plane, Whitney [}8] defined a degree D(a) E Z.

The

degree measures the total. number of times (counted with sign) that the image unit tangent vector turns through 2".

as the

97

curve is traversed onoe. abo~t

clockwise circle

By convention, we tate & counte,-

the origin to have degree

and Graustein proved the fundamental result:

ex. ~ : Sl - )

R2

are rell-\l&:rl.y homotopic

if

one.

Whitnlf

Two curves

and only it they

have the same degree. Using calculus, the desr.e 18 detined as follows:

ex is an immersion, !la'il

+0

coordinate on sl) • Let

~. a t III II I

vector to a.

where da

II

= aldO

Ce

Since is the

be the unit tangent

Then 1 J211' DCa) • ~ ~d8.

o

We now give a combinatorial interpretation of' the Whitney degree, and relate this interpretation to the states of a universe. Definition 6.2.

When

C is an oriented Jordan curve in the

plane, define the curliness ~(C) - =1 &ecording as the curve is oriented clockwise (+1) or counterclockwise (-1). If U is any uniVerse, let

U be

the collection o~ Seifert

circl.es for U (see Lemma 3.4 and Figure l}). obtained from b

.

Thus U is

by splitting all vertices in oriented fash-

ion (compare Figure 30).

Define the curliness of U by the

formula: ~ (U)

= 8' (U) .. I:cEO ff (C).

98

Figure 30 Lemma.§.:2. U s a(Sl)

Let

a: sl

->

R2

be a normal immersion.

be the corresponding universe.

degree of a

Let

Then the Whitney

coincides with the curliness of U; that is,

D(a) = tr (U).

The (easy) proof of the lemma will be omitted.

Following

our combinatorial theme, it is of interest to give a proof that ~(U)

is invariant under regular homotopy.

As we see ~(U)

is

related to the topology of Jordan curves in the plane.

As

from the next lemma, this analysis will show how

usual, a configuration such as

denotes a universe

containing this local pattern, and a formula containing more than one such pattern refers to a single universe

th~

has

been changed locally to conform to the indicated patterns. Lemma 6.4.

---.;r

eo(

.rc--

~(

!.- ......" )

)

~(

J.\

- ft( 'L

) =

1

)

-1.

99 It suffices to Oheck these formulas for UftiYl•••• • t

Proof.

Jordan curves.

The lemma then follows from the char, ••~IW

and the chart we have not drawn. obtained trom the vlei_1, chart by reversing all the arrows.

J.t

~ ~

8 @ ~ Proposition~.

~(U)

e( ~t)

((~)

g

:1. -:1. -:2.

::2..

@ 00

0

-:1.

If U is regularly homotopic to U', then

= ff(u').

Proof.

=ff(~4)

==ff(;14) =

~(:x=?)

==

~(~)

-1

(lr ) +1-1 = f1 (;>C) .

.. eo

:. ff Thus

~

(~)

(6.4)

is invariant under moves of type A.

ff

(A)

== ff~~) =~\~~) = r,

The final case for the type next page.

\-)t-) B

move is illustrated on the

100

This completes the proof of Proposition 6.5.

Any universe is regularly homotopic to a disjoint union of 'the standard torms indicated in Figure 31.

For the reader

interested in constructing a combinatorial proof of this fact, the ''Whitney triCk". as illustrated in Figure· 32, will ,

be usef'Ul.. /

101

CJJOc)~ -a

-1..

1

0

standard

.

'"

~

Figure 31

Yo--~~~~.

~1l~1l Whitney Trick Figure 32 In order to relate the curliness of a universe to its states, it is convenient to use the string form.

Upon de-

composing a string (see section 2) into Seifert Circles, there will be a single string without self-crossings, and a collection of Jordan curves.

Assign curliness zero to the string

without self-crossings, and the usual plus or minus one to each Jordan curve.

Thus

102

a te s trom. an y st th e cu rU n es t ac tr ex to how We now show . o f a u n iv er se o f th e el is a la b el b la g in ss o a l cr .6 . A v e rt ic D ef in it io n 6 se U . V (U ) F or & un iv er . 1 .. AD where p la ci n g form ~ ob ta in ed by U f o g n la b el li th e v er ti ca l w il l de no te o ss in g . el a t ea ch cr b la l ca ti er e a v and V (U ) th n g un iv er se ri st a be hen L et U te o t U. T 'l'I!eorem §.:.1. be an y st a 8 t le ; U el li n g o t v er ti ca l la b

D is cu ss io n. ch U fo r whi

of is t st at es S ex e er th at rv e th We fi rs t ob se example: i8 tr u e. F or .7 6 : ot a ul th e fo rm

X

~

A l _ A~ (U)
->2-

103

In order to construct a state for whi:ch first form

U,

th'

'l1l\I&1 ..... ,

the set of Seifert circles for U.

a trail by the following procedure: with empty interior.

Locate

.... ......

.i,.i.

a a.1t."

Reassemble one of its sites

resulting configuration has one fewer circle.

10

'ho

Repeat

process, never using a given site more than once.

th.

th~.

The

re.~'

is a trail whose corresponding state has the required prope,',. (The proof of this fact will be omitted.)

For example:

Trail with state markers.

Seifert circles with indicated re-assemblies.

The State S. Proof of 6.7.

By using the relation AD = 1

see that when S and

S'

it is easy to

are related by a sequence of state

transpositiO~S, then .. .

Thus the theorem

follows from the Clock Theorem and the existence of a state S that satisfies the formula.

10 4

mial The Al ex an d!r PolYno the ap pe nd ix) is de r po lyn om ial (se e The cla ss ica l Al ex an tio n by powers of t. sig n and mu lti pl ica we ll- de fin ed up to co de : de r us es the fol low ing In ou r ter ms , Al ex an

is of a kn ot or lin k K po lyn om ial ~K{t), r de an ex Al the us Th where ~(t) is la ~K(t) ~ Det(~(t» mu for the by ed fin de sthe lab ell in g co rre (se e se cti on 3) fo r x tri ma r de an ex Al an lbol, ~. de no tes eq ua an de r co de . The sym ex Al the to ing nd po in teg er . n the form :'t • n an of rs cto fa to up ity itt ed . xt lemma wi ll be om The pr oo f of th e ne 2 is giv en by the de r po lyn om ial in t an ex Al e Th . 6.8 Lemma Al ex an de r where B.K(t) is th e .K(t» t(B De ~ ) t2 ~( la for mu ma tri x fo r the code

Theorem .2,&.

~ t-f (U )v K( t) and 6.8 th at ~(t2) 6.7 m tro s low fol It Pr oo f. l = W - B ~d the l (ta ke W = t, B = t- , z ) tt z.= where let es mp co is Th ). 4.3 ed in Theorem fin de as ial om lyn po Conway the pr oo f.

105

1.

The Coat .2!

~

Colore

We now generalize to link polynomia1e in ~ .,...,.~•• I A different variable is assigned to each component' of 'hi link.

For illustrations we sha11 use three different " ,••

of arrow-head to denote three distinct strands. are indicated in Figure 33.

The ftlW .....

Note that it is the under-orOIi.

ing line that determines the code variables.

x\x x\x.

• )(

Definition

"x.

yy

~i4 xx

)(.

---t>y ~z:

Crossing Codes

r~=i) Y!' = j" ZZ

-1-

~»

Ll. Let

a link L.

44

t>

Xl'~,

Assume that

••• ,~

~~

=1

be the code variables for for each index k.

(Thus

these codes are the modified Alexander codes of Lemma 6.8.)

[x{- ~~ ... X:nJ

Let

[JS.

~

~J

.)o.~

denote the difference al an n:.:al _an =Xl ... ~ + (-l).lI.l ... ~

As in the single variable case, we have inner products

and state polynomia1s



for a given link L

whose strands have been labelled with these code variables. However, it is now necessary to give a norma1ization factor since the unadorned state polynom1a1 is no longer independent

106 ac co mp lis h th is sta rs . In or de r to ot the ch oic e ot tix ed the un iex fo r the reg ion s ot ind ple lti mu a e us en d, we sh all ve cto r ind ex of U is as sig ne d a ve rse U. Each reg ion se s ,or de ) The kth ind ex , Pk , inc rea 'p = (P l'P 2" •• ,P n' ind ica ted in Fig ure the kt h str an d as cre as es upon cro ssi ng (0 ,0, ••• ,0 ). ion is as sig ne d ind ex 34 . The unbounded reg

p•

(~,...,1\.•...

x" Po)

r

p' • ( ... ... . I\. +'-•.••• p.l

M ult ipl e Ind ex ing Fig ure 34 sind ex ed un ive rse co rre Le t U be a mu lti ply cti on of sta tes to r Le t $ be th e co lle L. k lin a to ing nd po ion s wi th ind ice s are in a pa ir of reg rs sta ed fix ose wh U and pI = p = (P l,P 2, ••• ,Pn )

. Proposition~

kth pla ce . Le t sta rs . ot the ch oic e of fix ed /I$1 is ind ep en de nt

on ly in the Then

itt ed . It tol low s op os iti on wi ll be om The pr oo f ot th is pr pr oo f of Theorem 3. 3. the same' tor ma t as the we must tak e olO gic al inv ari an ce , In or de r to ob tai n top ven L as above, 6) in to ac co un t. Gi on cti se e (se ss ne cu rli lab ell ed su ,U Ln ' th e se pa rat ely ••• U L3 U ~ U Ll = le t L ell in g di sti nc t ve the op tio n of lab ha we at th ote (N . lin ks e va ria bl e.) Then str an ds wi th the sam

107

is a topological invariant of potential function of the Example

liD!

L.

We refer to

&.

U.

([5], [17]).

x

t =

1

(by convention)

It I • X(X-~), ~(L) • X- l • X :. DL = l/(X-X). Example

1.d.. p'S 0

181

= (X2 (-1»X(X-X)

DL

= (~(L)/ltl> = x/x2x- 1 (x-X)

• DL = l/(X-X).

DL u tne

loa

,_,. uK~ aD K.. 1K

l=i®,$)

= -If +

It/ _

(f2 'Oy2'(-1»i(y_i) = Y(Y-i)

OK = =

XY = x(y-i)

(X~(Kl\'t(~) lit/ ) (xy/y(y-y»x(y-i)

:. ,OK = 1.

Example L.§..

Thus we ha.ve

O( 0 ) = 1/(X-X) D( b) = 1/(Y-f)

D(ct)) = 1 and

by

a similar calcula.tion,

D(~)

= -1.

In general. it L is a single variable link (with variable

X assigned to all strands), then

109

where the Conway polynomial is regarded as a function ot z = X-X.

In ([5]) Conway uses the notation for the poten-

tial function that we have reserved for the Conway polynomial With this caveat, our normalizations coincide with his. The potential function still satisfies the basic identity

X

(X]

K

OK - OK

=

=X- X

[XJO L

for links related at crossings involving identically labelled strands. In addition to the basic identity there are also identities for crossings of differently labelled strands: Theorem Ll.

[XY] .. XY + Xi.

Let

Then

1)

D(~)

+

O(:>C
2)

O(~)

+

D(;Y-~)" [xf]D(~ )

Let

Kl'~'

is.,

~

~

)

be related as shown below.

" t?Y ~

= (XYJO(

V- -A j( /X IS ~

Then These weaves can take any consistent orientations, and may be

111

These identities do not yet provide an axioma'l ••'lOft

Remark.

It appears to be neces'sary to ift-

for the potential function.

clude the values ot the potential fUnction on a large class of links (calculated via state enumeration or by other methods) in order to d'etermine the potential for all links.

This must

be contrasted with the situation for the Conway polynomial. The

Conw~

polynomial is determined for all knots and links

by its axioms once we know the value of the unknot.

How

large a collection of links must we specify in order to determine the potential recursively from the identities and the fact that what

~

D

vanishes for split links??

the generators .Q!

Proposition 7.8.

Let

L'

~

polychrome

In Conway's terms,

~?

be obtained by linking a strand of

the link L with an unknotted circle of linking number >. =:1:1 where

D L , = >.[Yl DL = :try] DL Y is the variable labelling the strand of L that as shown below.

Then

is linked by the new component.

rI

X Y

Proof.

Apply the definitions to the states and labels indi-

cated below.

\

The details of the calculation are omitted.

112

Example

1.:2.

str an d wi th

W ad din g a new be ob ta. ine d fro m L be low . number n, as shown

Ln

~t

J,.~k1ns

y

x.

t t (LO

Th en

OL = 0

o

OL

1

Dr.-xl Th us

(7 .a)

= [Y JO L

(7.7)

n '-'Ln _2

n _l = [x y]'-'L n

D~ = (xy ] 4 1

is sp lit )

...

[XY ](Y )

Dr.

-l) [Y JD r. O L - «(x y)2••• :3

Example 7.1 0.

-=>

L

O L = [xYJ - [X]{YJ.

De mo ns tra tio n.

w' -O L + O f: = (X ]O Sim il.a .rl y, O f: =

:. O L

= O f:

Ow = [YJ

(by 7.a )

[xY];

- [XHYJ

= {xYJ - [X ][Y ].

"

/

1l.3

Example 7.11.

B

B

DB

is known a.s the Ballantine rings.

ot its potential function.

= [X] [y}[Z] •

We omit the calcuJ.a.tion

(There are sixteen states in the

Ballantine Universe.) Example 7.12.

B'

Let this tamily inherit color and orientation trom the Ballantine rings in

~le

c:J B

=

7.11.

-[XYZ]

(calculation omitted).

DB' = 0

(split).

DB + OBI'" DB + OBI (Theorem 7.7) •

.. D B .=

-([XYZ]

+ [XJ[Y][Z}).

11 4

§.

Sp an nin g

S~tt&ces

al .ph ere . nte d) thr ee -di me ns ion rie (o the ote d~n Le t S3 is sa id to be a F embedded in s3 An or ien ted sll rfa ce K is id en tic al a kn ot or lin k K if sp an nin g su rfa ce fo r n on F ind uc es F, and the or ien tat io wi th the bo un da ry of n on K. the giv en or ien tat io r co ns tru cti ng a sp ec ifi c method fo In [36 J Se ife rt gave m. His th a giv en kn ot dia gra wi ted cia so as ce rfa a sp an nin g su rfa ce of the dia gra m. na ted the Se ife rt su su rfa ce wi ll be de sig s alg ori thm , re la te all de sc rib e Se ife rt' In th is se cti on we sh genus ot the Se inway po lyn om ial to the the de gre e of the Co tio n of the se re lacu ss the ge ne ral iza fe rt su rfa ce , and dis s vi a the Se ife rt ry sp an nin g su rfa ce tio ns hip s to ar bi tra pa iri ng . Se ife rt' s Al go rit hm

) on the kn ot di acle s (se e Fi gu re 13 Form th e Se ife rt cir as ind ica ted be low . arc s at ea ch cro ssi ng o tw ng wi dra by m gra

::K

ve rsi ng the di athe n ob tai ne d by tra are s cle cir rt ife Se The s when ar riv in g moving alo ng th e arc and s ge ed its ng alo gram (in the pla ne ) ch ci rc le 1s embedded ea us Th . ng ssi cro at a rt su rfa ce is cro ssi ng s. The Se ife the of nt me ple com in the Se ife rt cir cle s ac hin g dis cs to the att ) by d cte tru ns co the n ee thr the in tly in are embedded di sjo so th at the se dis cs ten din g above ot the dia gra m and. ex ne pla the ing lud inc sp ac e in a tw ist ed band co mp let ed by til lin g it. The su rfa ce is

110

single, dual, or, tri-colored (that is, there may be one, two, or three variables involved in the weave). Proof.

Since the potential function is a normalized state

summation, these identities can be demonstrated by the same method as used in Theorem 4.1.

In Figure 35 we have shown all

of the local state configurations with their contributions to the terms of state summations for the first identity (1). Four configurations cancel in pairs, leaving the pattern of this identity. ion.

The second identity follows in the same fash-

The verification of the third identity is a long calcu-

lation involving the triangle states (Figure 23) and is

+. ,.....-

- - T.........

L..

L\

~.&._.&._

YX

Xy

x:x.

YX

XY

~

YX

x9

::><:A:.

>'~

Xy

~

YX

Xy

YX

Xy

:::><x:.. " " :x::>c

Figure 35

~ Co

~ c

'\ /

L

116 are Note that the Seifer t circle s constr ucted in this proces s the same as those produc ed by splitti ng sites to preserv e orient ation. ThUS, in the example above, the c1.re.les drawn by site-s plittin g appear as

u

ClD

Seifer t Circle s

ted Each twisted band has· the same homotopy type as its projec image in the plane, a tilled -in crossin g:

1

pro' ••'

~ ::::I::

Conseq uently, the Seifer t surfac e has the homotopy type of the 2-cell complex obtain ed, from the univer se U underlying the knot diagram , by adding disjoi nt 2-ce11 s to the Seifer t circle s.

(Regard these cells as autom aticall y tilling

in the crossin gs.)

~

complex

117

Lemma 8.1.

Let K be a knot or link (diagram) with underlyinl

universe U.

Let

K, and let X{F) p{F)

F

be the Seifert surface}itor the diagram

denote the Euler characterl$tic of F, and

denote the rank of the first homology group

Hl{F).

Then X(F)

= s(u) -

R(U) + 2,

P(F) = R(U) - s(u) - 1, where R(U) = the number of regions in U. and number of Seifert circles in U.

s(u)

= the

Note that the Euler char-

acteristic and rank are functions of the universe alone. Proof. and

E

V = V(U)

Let 2

R = R(U)

EeU)

and

denote the number of vertices of U,

denote the number of edges ot U, while Then V - E + R = 2

S = S(U).

since the

plane has Euler characteristic 2, while the Euler characteristic ot the Seifert surface is given by

~(F) =

V- E+S

since the surface has the homotopy type of the two-cell plex constructed on the universe.

ooa-

(Recall that the Euler

characteristic of a two-complex is equal to NO - Nl + N. where

Ni

is the number of

i-cells in the

com~lex.

Rlaall

also that for a connected surface with boundary, the Eullr characteristic equals

2 - P where

first homologylgroup.)

p is the rank ot the

The lemma follows at once from thiS'

formulas. Remark.

A surface with boundary is said to have genus

g

if

it becomes a sphere with

g handles upon adding disks to all

the boundary components.

This gives rise to the relation

118

p = 2g + ponents.

- 1

~

tor a connected surface with we have the

Henc.~

Corollary 8.2.

Let

connected universe

K be a link diagram with U, R regions, and

Then the Seifert surface the fonnula ~.

boundary com-

~

g = i (R - S -

F ~)

for

1.1

oomponents,

S Seifert circles.

K has genus

g

liven by

•

It is of interest to note that a surface of relatively

low genus can arise from a diagram with many regions if these regions are balanced off by a goodly circles.

of Seifert

The simplest case is an unknot of the fonn

R

10

S

9

1.1 ~.

collec~ion

=

g

= i (10 - 9 - 1)

o.

1

Lest we give the impression that the Seifert surface

is easy to draw in perspective, please note that nested Seifert circles give rise to nested diSKS.

It is a well-known

bit of folklore that any knot has a diagram in its ambient isotopy class without nested Seifert circles.

I leave it to

the reader to make this concept of nesting preCise, but point out that of the two equivalent diagrams shown below tor the figure eight knot, only the second has a Seifert surface that can be drawn without overlapping disks (allowing one diSK to planar region) .

cv

119

cOPway

The Degree of the

<

PolYnomial

Call a site of the torm of the form '}

a

X

an active .!!i!. and a site

pIIUve~.

We adopt this terminology

because, in the state oorre.ponding to a trail, the black and white holes of the state are in one-to-one correspondence with the active sites of the trail. In order to determine an upper bound for the degree of the Conway polynomial, we shall USI the state polynomial model for

VK(z).

and

Z

VK(Z) '"

Thus

=B -

W.



(See section 4.)

BW = 1

with the caveat that Thus the degree of

~K(z)

is equal to the maximal power of B in the state polynomial



reduced

by the relation

BW

= 1.

By the definition

of the state polynomial, this maximal power can be no more than the largest number of active sites possible for a trail on the universe underlying K. ~~.

Let

U be a universe with

Proof.

S

Then any trail on U has no more than

Seifert circles. R - S - 1

R regions and

active sites.

Split every vertex of U to form an active site.

This creates the collection of Seifert circles ~. be converted into a trail in no less than S - 1 assemblies.

~

~

can

re-

using this many reassemblies, the trail has

the maximal number, A, of active sites.

Thus

where V is the number of vertices of U.

A = (R-2) - (S-l).

A .. V-

(S-l)

Since V'" R - 2, Q.E.D.

120

Proposition..§.:.!t.

Let

K

a knot or link (diagram) with con-

be

nected underlying universe U. for K, and

p = rank H1(F).

Let

F be the Seifert surface

Then

deg VK(Z) 5. p. Proof.

By

Lemma 8.3 and the remarks preceding it,

deg VK(Z) 5. A = R - S - 1

~

p

by

Lemma 8.1.

This completes

the proof. ~.

When

deg"K = P, then we know that the given Seifert

surtace has minimal genus among all Seifert surfaces spanning other diagrams for the knot. knot shown below.

Then

the Seifert surface for

For example, let K be the

4 p = 4, and "K = 1 - z2 - z.

K has minimal genus.

2

"K' = 1 - z v

L

-V K

= z3

+ vK '

(Section 5)

= z"L

:. vK = 1 - z2 - z 4 •

Thus

121

Drawing the Seifert Surtaoe Each Seifert circle diVides the plane into two regions. Call a Seifert circle ot

lZR!. 1. if one of these regions is

void of other Seifert circl•• J otherwise call the Seifert circle of

~

II.

In order to understand the structure of Seifert surfaces involving type II circles, it 1. convenient to have a picture of the surface near its boundary.

The difficulty is illustr-

ated by trying to see this for the t1gure eight knot:

The dotted Circle will have a disk attached to form the surface.

This disk is out ot the plane, attached perpendicularly

along the dotted curve.

We do not yet see a collar of the

boundary ot the surface along segments

a, b, c, d.

To

rectify this consider the following diagram (Figure 36).

~.....'., = .. Figure 36

122

In Figure 36 the dotted circle has been replaced so that it forms a new component in the diagram that over-crosses any old component that it meets.

If the old diagram is

denote this new diagram.

As Figure 36 demonstrates,

only type I Seifert circles.

The Seifert

A

K, let

s~rface

t

K has

FK is

obta1ned from Fi by adding d1sks to the dotted circles (the new components) in Fi'

(Note that each new over-crossing

component is endowed with an orientation opposite to that of its Seifert circle of origination.) Fi

Thus we have an embedding

FK. The surface Fi depicts the planar portion of FK in a graphic and useful manner. C

Example.

Here is a sequence of drawings leading to a depic-

tion of the (minimal) Seifert surface for the knot discussed just after Proposition 8.4.

In the final drawing the Seifert

surface is shown with one of the disks occupying the unbounded portion of the plane. by

It is obtained trom the previous picture

swinging the upper (twisted) band underneath the rest ot

the diagram.

,§

K

-

"~ '/2)

,

K \

~-'';

I

II

123

Arbitrary Spannins Surtaces Proposition 8.4 can be generalized to arbitrarY spanning surfaces.

That is, fOr any connected, oriented surface F c: 83 ~

spanning a link K, deg VK(z)

p(F)

where

p(F)

is the rank

of Hl (F). , In order to see this generalization, we need another model, of the Conway polynomial. in [21].

I will outline the result here.

Given a spanning surface (The Seifert pairing) formula

S(a,b)

into

a+

F c: S3, there is a pairing

e: Hl(F) x Hl(F) -->

= Lk(a+,b)

number in S3, and a

This model is discussed

where

Z defined by the

Lk( , )

denotes linking

is the result of translating the cycle

s3 - F along the positive normal to

F.

The Seifert

pairing is an invariant of the ambient,isotopy class of the embedding of the surface in the three-sphere.

It can be used

to create invariants of the embedding of the boundary of this surface. Let I

~(X) = D(XI - X-lIT)

where

D denotes determinant,

denotes a matrix of the Seifert pairing with respect to a

basis for

Hl(F). and T denotes matrix transpose. Then, letting z = X - x- l , one has that VK(Z) = 0K(x). This means that the Conway polynomial can be computed via the Seifert pairing from any spanning surface for the link. the Seifert matrix has size

Since

P(F) X P(F), the rank inequality

follows immediately. Example.

Let

see that

K, the boundary of F, is isotopic to the trefoil

knot.

Hl(F)

F

be the surface shown below.

is generated by a

and

b

It is easy to

so that the Seifert

124

matrix ls I(a,a)

[-~:-~J w1th respect to this basis. That ls,

= 9(b,b)' -

-1

and

I(a,b)

= 0,

'(b.a)

= 1.

b

(_X+X-1)2 + 1

(X_X- 1 )2 + 1

Remark.

It ls curious to note that with

X = z + l/X, hence

X = z + _1_ _ Z + 1

z +

and

1

=z-+:---

I7K(Z) = OK(Z + ; + '1 z + 1 z +

In particular

17K(l)

z = X - X- 1 we have

=

nK(1 ~

J5).

125

.2. The.9!ru!!,gt. Alt'NUY' ~. In this section we dlt1ne a class of alternative prove that

~

~

and

Seit.rt 'wraee of .!!:!! alternative link diagram

!!!! minimal genus !m2!!I connected spanning surf'aces f'or the ~

in three-dimensional.lR.!£t..

This theorem generalizes a

corresponding result b.Y Crow.ll and Murasugi for alternating links [32].

It is also related to the work of'Murasugi and

Mayland on pseudo-alternating links [8].

Alternative links

are pseudo-alternating, and we conjecture that these two classes of' links are identical to one another. An alternating link has a weaving pattern that is instantly

recognizable f'rom a suitable diagram.

In such a diagram the

over and under crossings alternate as one traverses the components of' the link.

For example, the tref'oil knot. the Ballantine

rings. and the example just after Proposition 8.4 are all alternating. An alternative diagram also has a recognizable weaving pat-

tern, but this pattern does not become apparent until the diagram is translated into the form of the Seifert circles with an appropriate coding.

For this reason, I first consider codes

and designations f'or crossings in a knot diagram. lists the dif'ferent notations that we shall use. the diagrammatic or pictorial representation. the well-known [1] double-dot convention.

Figure 37 Row (a) gives

Row (B) shows

Two dots are placed

to the left of' the under-crossing line, one on each side of' the over-crossing line.

Row (C) is a variant of Row (B)

The crossing has been split to form an active site (see section 8 for the definition of active and passive

126

sites) and one dot the Site.

&~~e&rs

just to the left or to the right of

I Call thie the site marking code.

Row (D) exhi-

bits the label code that we have used for calculating state polynomials. .

CA)

(B)

(e)

CD)

X X

~

~

»(

diagram

dot-convention

site-marking

label-code Figure 37

X X ~

~

X

By labelling a diagram with the dot convention it is easy

to translate to the site-marking code.

Note that the site-

marking code appears as the collection of Seifert Circles decorated with dots at the sites.

For example:

6D & ~

127

Observe that the Seitert circles divide the plane into connected sets that we call spaces (as opposed to the regions of the knot diagram).

Thus the trefoil diagram has five regions,

while the corresponding set ot Seifert circles has three spaces.

In a checkerboard coloring of the knot diagram each

space -in the diagram of Seitert circles receives a solid color.

Spaces that receive the same color will be said to

have the same parity. spa.ces £t parity have same color-.--

Example.

~

checkerboard Definition~.

universe, and

Let CK

K be a link with connected under-lying

be the dia.gram of Seifert circles for

K

decorated according to the site marking convention (Figure 37).

K is said to be an alternative .ll!:!! i f all marks in any s

given space of

CK have the same type.

When speaking ot

links up to topological equivalence, we shall say that & link is alternative if there is an alternative diagram in it' topological equivalence class. In Figure 38 we give an example of an alternating knot and show that it is alternative.

On the other hand, there

are knots and links that are not alternating but nevertheless alternative.

The same figure illustrates this for the

(3,4)

128

torus knot.

K-alternating

K(3,4)-torus .\m91 Figure 38 Alternating links are alternative links with the following special property: Lemma.2.:£.

A link diagram is alternating if and only if it is

alternative and spaces in the diagram of Seifert circles receive the same or different marking type according to whether they have the same or different parity. The evidence for this lemma can be seen from the example in Figure 38.

Figure 39 illustrates the key geometry that is

relevant for the proof of the lemma. omitted.

Further details are

129

alterna.ting ~

site-marking code opposite

same marks with 'Sa:iiie"" parity

~

J!!1b. opposite parity

Figure 39 We are now prepared to state the main theorem of this section. Theorem 2.:2.

Let L be an alternative link diagram, FL the

Seifert surface for surface for

L.

L.

Then

In fact. deg

FL ~L

is a minimal genus spanning

= P(FL)

where

P(FL )

is the

rank of the first homology group of FL' This is the promised result about alternative links. order to prove it, we must locate those states verse underlying L that yield highest ducts ~

In

S of the uni-

B-power inner pro-

. The key observation is due to Ivan Handler [16]: the site-marking £Q!!!

~ ~

the dot always falls .Qn the !!.

label £Q!!!

~

See Figure 40.

superimposed,

130

site-marking

label code Figure 40 ~

Moral:

sites on

To obtain a maXimal CL

~ ~

X

B-power state

S, reassemble

(the diagram of Seifert circles) so that there

are a maximal number of active sites, and in the associated state the state marker falls on the site marking dot at each active site. Each such state marker will then contribute a .

The state

when there are

S will have

SL-l

deg

B to

= P(FL )

exactly

reassemblies, and every state marker at

an active site (i.e., a site that has not been assembled) falls on a dot. We now describe an algorithm that creates high power states, and give examples of its application. Alternative Tree Algorithm (ATA)

1.

Let

L be an alternative projection.

Seifert circles of that code for

L.

CL

denote the

decorated with the site marking dots

L

By definition, each space of

markings of a single type. adjacent regions of

Let

L.

CL

Choose stars in a pair of

has

A \1

i

:~

131

Grow a tree rooted at a star.

The tree can branch

from one reglon 'ot L to another if and only if a)

The secqnd rel~on is unoccupied by tree branche ••

b)

There is a .ite that opens from the first region to the second.

c)

At an active site, the marking dot must be in the second region.

Grow trees from each star until a), b), and c) can no longer be satisfied. There will now be a collection of sites (at boundaries of regions occupied by tree-branches) without any branohes passing through them.

(Further branching

being forbidden by the restrictions of 2.) sites will be reassembled.

These

Denote the reassembly by

placing a circle around the site.

~><-

The trees now have new access.

Continue growing them

/

according to part 2., until no further growth is possible.

Apply 3. again and continue in this manner until

each site has a branch passing through it.

132

4.

Reassemble the boxed sites and use the two trees to create a state

S;

This state will satis:ty

There will be

SL - 1

1'(:Pi) '.., deg(Lls>.

reassemblies, and everY state marker

will fall 'on a dot. In this algorithm each stage of growth involves branching choices.

Thus many different states can appear as the end

product.

However, 14: .§. and .§.!..

~

states obtained El ATA

from the ~ link diagram It" then .§. ~.!im:

Reason.

and

.§.!..

have the

a(S) = a(S').

Different branching choices always occur, within the

same space of

CL

and will therefore leave the total number

of black and white holes in the corresponding states invariant, since all markers in a given space are of the same type. By the dint of this sign constancy, the polynomial

~K

p(p' }

has a term

:l:

N•B

L

where

N is the number of dif'f'erent

states produced by the algorithm ATA. once it is shown that

Theorem 9.3 follows

N is not equal to zero, and that the

algorithm produces all of' the maximal power states.

This is

true, and will be verified after we give a series of' examples that illustrate how the algorithm works. Exam)2le 2.d.

Let

K

be the figure-eight knot.

Q)K (DCK *tt

d

Then K is alternating, and the diagram for

CK

shows the

dots of opposite type in spaces of oppOSite parity.

By

133 choosing stars as shown below, the .first stage of ATA

yields

unique trees and reassemblies:

In this case, one-more pass through ATA

complete. the pro-

cess;

Here is the resulting trail:

If S denotes the state corresponding to this trail, then S has one black hole and one white hole.

Hence a(S) = -1.

The

marker at each of the active sites falls on a dot, and hence on a

~ in the label diagram for

K.

S is a maximal, B-power state for

Hence


= B2,

and

K.

Note that in computing this example the end result is actually obtained (just before the last pass) when all the reJ

assemblies have been indicated.

In tuture computations we

shall stop the process at this point.

134 Example 9.2.

Let L be the alternating link shown below.

In this case, the first pass through ATA produces two maximal states:

Example~.

Let L be the

(3,3) torus link as shown below.

This diagram is alternative, but not alternating.

L

In this case, the algorithm produces a unique state in two passes:

135

Next pass:

Corresponding trail/state:

Example.2...:1.

Let K be the alternating knot shown below.

In this case, one can see that ATA states.

constructs 11 maximal

1,6

If we let K* be the knot obtained by switching crossings

a

and b

(labelled above), then the resuiting pro-

jection is not alternative.

The algorithm can still be used

to enumerate the maximal states of K*.

Of these states, 7

have negative signs and 4 have positive signs. cancellation there is a coefficient of power term, and degree

~K*

= P(FK*).

-,

Hence atter

on the highest

We conjecture that

K*

is not alternative in any projection. This example shows how these methods may be used to show that apparently non-alternative knots may achieve minimal genus on a Seifert surface.

The problem of showing that such

a knot does not possess an alternative prOjection is very difticul t.

Wh.y ATA produces all the high power states. We now explain why the algorithm ATA

produces all the

high power B-states for an alternative link L. if

CL has

s

Recall that

Seifert circles, then each maximal trail/state

is obtained trom

CL

by reassembling

s-l

sites (not all

such reassemblies yield maximal states.). In order to understand the geometry of the reassembly it is helpful to phrase it more generally:

The set of Seifert

circles is a special case of an arbitrary collection of dis-

137

Given two circles, 1" • ,..

joint circles in the plane.

n'

assembly between them be denoted by a straight line ..... drawn from one to the other.

----G;1.ven

s

Thus

circles in the plane, we can ask how many re-

assembly lines need be drawn to create a simple closed curve. The answer, of course, is of Seifert circles.

s-1

just as for the special case

In the general case we are tree to re-

assemble wherever we please, rather than at specified sites. For example:

@Q)C@)@ s =4

s-1

=3

reassemblies

Where there are a minimal number of reassemblies s-1

reassemblies for

connected

~

s

(i.e.,

circles) then each space remains

deletion of the reassembly lines.

For

ex~­

ple, in the diagrams above, one of the spaces is a disk with two holes -- atter cutting it along the reassembly lines it becomes a disk:

, This transformation of spaces to disks must occur if the re-

138

assembly is to produoe a Jordan curve in the end.

To obtain

such a curv~ w1th the least number of reassemblies it suffices to choose a .et ot reassemblies in each space that will cut that space to & disk. Let TA

denote the algorithm ATA

(2c) (d1reotedness of the tree).

without stipulation

Then TA

chooses a set of

reassembl1ea in each space by growing a tree in that space -if we translate to the language of circles and reassembly lines it becomes clear that this produces a set of reassembly lines that cut the space to a disk.

Thus

TA

will produce every

!!!!! that corresponds to s-l reassemblies. The algorithm ATA

is designed to select exactly those

states produced by TA where every marker at an active (not reassembled) site falls on a dot.

In order to do this

ATA

must grow a tree in each space so that the reassemblies cut the space to a disk. ~aph

That this can be done follows from a

theoretic lemma:

Definition~.

A graph is said to be directed if each edge

has an assigned orientation. graph is said to be 1) and 2)

~

A finite, connected, directed

if:

Each vertex touches an even number of edges. At a given vertex, half of the edges are outwardly directed, and half are inwardly directed (that is, away from, and toward the vertex respectively).

A vertex v

in a graph G is said to be the root of an

oriented maximal tree in tree in

G if v

G and every vertex in

is contained in a maximal

G can be reached from

v

139 by an outwardly oriented path in the tree. Lemma~.

Then

v

Let

G be an even graph and

a vertex of O.

v

is the root of an oriented maximal tree

T

~

G (ori-

entation induced by G). Proot.

Since an even graph cannot be itselt a tree, it must

contain cycles. cycles.

Let

Hence. by the definition. it contains oriented

A ~ G be an oriented cycle.

Let

G'

the graph obtained by collapsing A to a pOint.

= G/A

be

That is, G'

is obtained by removing all edges in A and identifying all vertices in

A to a single vertex. Note that

G'

still satis-

ties the hypotheses of the lemma (It may be a single vertex.). We say that . G is obtained by blowing up a vertex of may assume by induction that the lemma is true for T'

G',

G'.

We

It

is a maximal oriented tree tor G'. then we obtain a tree

T tor

G by lifting T'

to

G, and extending it by traveling

around the cycle (See Figure 41). the lemma.

" ~ '~

: .....

...

--

~

Figure 41

This completes the proot of

140

We apply this lemma as follows:

Let

L be an alternative

ct be the collection of Seitert cir-

link diagram, and let

Let

cles of L marked by the dot-marking convention. a space of

CL.

Then

Assign a graph G(p) 1)

is divided into a number ot regions,

P

accessible to each other to

The vertices in

by

passing through the cusps of sites.

P via: G(p)

are in one-to-one correspond-

ence with the regions of 2)

P be

P.

Two vertices are connected by an edge of G(p) whenever the corresponding regions are accessible to one another via a site.

The edge is

directed toward the region that contains the dot that marks the site.

It is easy to see that when all the dots in

P

are of

the same type (as they must be for an alternative diagram) then the graph G(p)

is even.

This follows from the fact

that interacting Seifert circles are oriented coherently. Hence the lemma applies to tree-growth demanded by ATA Thus we have proved that of states.

G(p).

This guarantees that the

is actually possible. ATA

produces a non-empty set

This completes the proof of Theorem 9.3.

141 Examples and Comments 819 of the knot tables ["l i. standard, hence alternative, but it is not alternating [9].

Example~.

The knot

This is the simplest example ot an alternative knot that 1. neither a torus knot nor an alternating knot.

Example~.

['1

The Lorenz links studied by Birman and Williams

are standard, and hence alternative.

These links arise

as knotted periodic orbits in dynamical systems on three space.

Lorenz links include torus knots and algebraic knots

(an algebraic knot is obtained as the link of a plane curve singularity [27]). Example

~,:

Murasugi has proved that the coefficients of

the Alexander polynomial of an alternating knot are all nonzero

(i.e., there are no degree gaps) and that they alter-

nate in sign (See [30]).

It is a good exercise to prove

this theorem using our methods.

The alternation is then

seen to rest in the change in state signs under transpositions.

l~

Figure 42

Exercise~.

Is the alternative knot shown in Figure

alternating in some prejection?

4~

143 10.

Ribbon Knots

~

tbe Arf Invariant

In this section we examine the notion of pall-equivalence (section 5. remark after Theorem 5.7) more caretully, and show its connection with the detection of ribbon knot. and with the Arf invariant.

We conclude with a short proof of

the theorem of J. Levine, relating the Arf invariant of a knot to the value of its Alexander polynomial at minus one (taken modulo eight). Definition 10.1.

A ribbon knot

an immersed disk

6 c: 83

K c: S3

is a knot that bounds

having only ribbon singularities.

A ribbon singularity is a transverse self-intersection of the immersed disk

6 along an arc

is represented by a map a-l(s)

s

so that if the immersion

a: D2 - )

consists of two arcs on

8'

(a(n2) = 6)

then

D2, one contained in the

interior of n 2 , the other touching the boundary of D2 transversely at its endpoints. embedded in n2 .

Each 'arc is nonsingularly

A typical ribbon singularity is illustrated in Figure 43.

Figure 43

144 Figure 44 shows that the connected sum of a tretoil and its mirror image. ;1s ribbon. ribbon disk for this knot. to prove that

~he

I have given two examples ot a The first should enable the reader

connected sum ot

~

knot with its mirror

image is ribbon. Remark.

A ribbon knot is an example ot a slice knot, where

this term refers to a knot

K c s3

that bounds a smoothly

embedded disk in the four-dimensional ball, D4, whose boundary is

s3.

This notion was introduced in the paper by Fox

and Milnor [13). ribbon (see (12).

It is conjectured that all slice knots are In accordance with our combinatorial

theme, we shall restrict our attention here to ribbon knots. They will be viewed as knots whose projections admit a special weaving pattern described by the ribbon disk.

, = K# K:

Figure 44

145 Here is a formal Itatement of the result about mirror images: Lellll/l& 10.2.

Let g

,

be a knot, and let g.

be its mirror

image (obtained trom a d1alram for K by reversing all

,

crossings) • Then the connected sum, K f K·, of K and its mirror image is a ribbon knot. Proot.

The ribbon disk il obtained by joining points on the

knot and its mirror companion al shown in Figure 44. Remark.

This result is due to Fox and Milnor [13J.

It forms

the basis for the construction ot the knot concordance group (See [13], [24], [4J). The rest of this chapter is devoted to the study of the Arf invariant ot a knot.

This invariant was discovered by

Robertello in [34], and has been examined and reformulated by many authors (See particularly the paper [25] by J • . Levine).

our approach to the Arf invariant is seometrica.l,

based on certain constructions in Robertello's o:r1linal paper, We show that under pass equivalence all knot. tall into two distinct classes, and that ribbon knot. tall 1nto a cla•• containing the trivial knot.

The Art invariant identU'ie.

the pass-class of the knot, and thus can be used to shOW that certain knots are not ribbon.

A more sophisticated (and more

algebraic) formulation of the Arf invariant, using the Seifert pairing, shows that it vanishes on any slice knot.

146 Pass Equivalence' Recall that two knota equivalent

(Kl'~

K2)

Kl

and K2

are said to be pass-

it one can be obtained trom the other

by ambient isotopy combined with a sequence ot pass-moves.

A

pass-move is a local operation on the diagram in one of the following two f'orms:

t I

1.

2.

Notation. while

p

The symbol

= will be used for ambient isotopy

will denote pass-moves.

The next lemma is the key ingredient in showing that ribbon knots are pass-equivalent to the unknot.

It is due

to Bob Brandt, Thaddeus' Olzyk, and Steve Winker (all students in a knot theory-seminar held at the University of Illinois at Chicago in 1982).

r-a J&.g. .) .

b)

~ ~~-+

~~-~

Single strand pass between parallel strands causes twist.

360·

147

)

Theorem 10.3.

If K is a ribbon knot, then

equivalent to the unltnot: Proof.

K is pass-

K till O.

By a suitable ambient isotopy the ribbon singulari-

ties for a ribbon disk spanning K may be exhibited in the local forms:

•

J

. (The proof of this statement is omitted.) With this presentation of the ribbon disk, use the Lemma to pass

K to a diagram free from ribbon singularities:

~t-I

lJ,

Then K till K' K'

~~--=\=l~

,

where

K'

bounds a non-singular disk.

is ambient isotopic to the unltnot.

pass-equivalent to the unltnot.

Therefore

Hence

K is Q.E.D

148

Figure 45 illustrates an ambient isotopy of'

the:.~ev.dore's

knot to a ribbon knot, and a passage of' this ribbon to the unknot.

This is an example of' a ribbon knot that

connected sum ot a knot and its mirror image.

i~

not the

(The qonway

polynomial of' the stevedore is irreducible, while the polynomial of any non-trivial connected sum is a non-trivial product.)

This example should underline the extraordinary sub-

tlety in deciding whether a given knot is ribbon.

-0-

Figure 45

149

The trefoil is not ribbon.

Corollary 10.4.

,

Proot.

"';':

According to Theorem 5.7 of these notes, the seicond

degree coefficient,

~(K).

of the Conway polynomial of a knot

K is an invariant of pass-equivalence when taken modulo-two. Since

a 2 (unknot)

~(trefoil)

equals zero, while

equals

one, we conclude that the trefoil has no passage to the unknot.

If the trefoil were ribbon, it would be pass-

equivalent to the unknot. Definition 10.5. the value of

This completes the proof.

The Arf invariant, A(K), of a knot

~(K)

K, is

(see above) taken modulo-two.

This is not the historical definition of the Art invariant.

It is the most convenient definition to give at this

point in these notes.

We shall make connections with ordin-

ary mathematical reality shortly.

Just as in Corollary 10.4,

we now know that no knot with A(K) Theorem 10.6.

Two knots

and" only if A(K)

= A(K').

K and

K'

=1

can be ribbon.

are pass-equivalent if

In particular, any knot is pass-

equivalent either to the unknot or to the trefoil knot.

Any

knot is pass-equivalent to its mirror image. Proof.

An orientable spanning surface for a knot

represented as ~ embedding of

a standard

K can be

surface of genus

seen as a disk with attached bands (see Figure 46).

g,

up to

ambient isotopy, the embedding can be so arranged that the disk is embedded in standard form while the bands may be twisted, knotted, and linked with one another.

Since the edges

of a given band receive opposite orientation, passing bands

.+'.

(see below) result in a boundaries.

,

150 ~ass-equivalence

of the surtace

1';a

..

"". -:=:

By passing bands in this manner we can disentangle the surface until it is a boundary connected sum ,of embeddings of surfaces of genus one.

Further disentanglement through band passing then

reduces each genus one surface to a surface whose boundary is either a trivial knot or a trefoil knot.

The crucial move for

this process is the twist cancellation:

~ ~ ~~om1llI. The surface can be represented so that all twists are accomplished by curls of the form follows from orientability).

~

(this

In such a representation, the

mirror image knot is the boundary of the surface formed by

-"n'M:OO~ Thus any knot is pass-equivalent to its mirror image. So far we have shown that any knot is pass-equivalent to a sum of trefoils,. I

Let""K

denote the trefoil knot.

K ,. K ... K" K· ... 0, the first pass because

Then

K passes to its

mirror image, the second because the connected sum of a knot and its mirror image is ribbon, hence passes to the unknot.

151

Thus any sum of trltoils is reduced through passing to, a sinai, trefoil or to thl unknot. knot is

pass~equ1V&1.nt

This completes the proof that any

to either a trefoil or an unknot.

Since these two cla •••• are distinguished by A(K), this completes the proof of the theorem. Remark.

Part of the dis.ntanglement process of this proof is

illustrated in Figure 47.

p

pi

~=3 Figure 46

Figure 47

152

Seifert Pair1n;s

1'_ k'

Invariant

The Arf invar&ift' .t & knot is usually (se;, (34)

defined

as the Arf'invarilft' .t & mod-2 quadratic form that is associated with the knO'll,

WI now give a brief outline of this ver-

sion.

",:t.o torm is a function

A mod-Ii! CJ,UIod..

q: V -> Z/2Z

where

V is a tin:!.tl d:l.mln.1onal vector space over the field of two

(1/11), and q satisfies the formula q(x+ y) • IICX) + 11(,.) + x.y where x·y is a symmetric bilinear tom dltined on V. The quadratiC form is said to be elements

non-desenerate :l.t this bilinear form is non-degenerate.

This,

in turn, mean. that a matrix for the bilinear form with respect to a basi. tor V 1s non-s1ngular.

When V has even

d1mension the non-degenerate bilinear form is particularly simple in a symplectic basis ei·e j

~

fi·f j

=0

for all

B - (el,fl, ••• ,eg,fg} where i,

j

ei·e.., ~ 5i.1 • Here

5i.1

is the Kronecker delta,

Any non-degenerate form

has a symplectic basis. When q

is a non-degenerate quadratic form on a vector

space of even dimension 2g, then it is not possible for an equal numbef of elements of V to go to q. 1

0 and to 1 under '-

The Arf invariant of q, ARF(q), is defined to be

0 or

according as the majority of elements of V' go to 0 or

to 1 under q.

153

Let

K bea knot in the three-dimensional sphere

let F c s3

be

an oriented spanninevsurface'tor K.

s3, and Let

8 : Hl (F) x Hl (F) - ) Z be the Seifert pair:l,ng as defined in section 8 of these notes. q : V -> Z/2Z

Let V = Hl (K;Z/2Z), and define

b1 the formula q(x) = 8(x,x) (mod-2).

one can verify that q non-degenerate.

Then

is a mod-2 quadratiC :form that is

It is associated with the mod-2 intersection

form for cycles on the spanning surface.

One well-known

definition for the Arf invariant of the knot K is: ARF(K) = ARF(q)

with q

Theorem 10.7 (Folklore).

defined as above. Let K, and it be two knots that

differ by switching one crOSSing, and let L be the link of two components obtained by splicing this crOSSing.

Let Lk(L)

denote the linking number of this two-component link. the Arf invariants of K and

K

are related by the formula

ARF(K) - ARF(R) • Lk(L) Proof.

By

Then

(modulo 2).

using Seifert's algorithm for constructing a spann-

ing surface (see section 8) plus the remarks in the proof ot Theorem 10.6, we may assume the crossing being respondS to a ing surface.

1800

switc~ed

twist in one of the bands on the spann-

The link L is obtained by cutting this band

and taking the boundary of the resulting surface. 48.

cor-

Let the band under discussion be labelled P.

assume that it is aSSOCiated with a band pI

See Figure We may

one of whose

feet stand between the teet of P as in the normal form for an oriented surface (Figure 46).

Let

a

and

b be cycles

154 on the

surfa.c~,

Z'\IIIII:Lft. th:rough the bands

P and P I

re-

""

spectively (aee

':L~,.t

46 and 48.). Then one can verify

e(b,b). Lk(L).

that

We use th:!..

'ao,

to prove the Theorem:

ant of a qu&4:ra,:La fol'm

The Art invari-

q

can be computed from a symplectic

a

and

basis by the formula

(Se. [,4]).

The aUI'Ve'

b can be placed in a sym-

plecti0 ba.:!.t tor H1 ('JZ!2Z) so that e l · = a and e2 = b. He:re , denot •• the .pann1ng surface for the knot K. It is then ea.y to .ee that the difference of the Arf invariants for K and

R

18 equal to

&(b,b) (modulo 2)

Seifert pairing tor F.

Hence,

by

where

e is the

the above remarks,

ARF(K) - ARF(i) = Lk(L) This completes the proof.

F

Figure 48

(modulo 2).

155 Corollary.J&&. mod-2

Let

reductionot

K be a knot, and let." A(K) '2(K)

denote the

as in Definition 10.5.

Then

A{K) '" ARF(K). ~.

R be as in Theorem -10;7. From Lemma

Let K and

5.6 and the Identities tor the Conway polynOmial we have the ~(K)

formula

- a2 (R) a Lk(L) (assuming that K has the positive crossing). Hence A(K) - A(K) = Lk(L) (modulo 2). Since this identity suffices to calculate A(K) value on the unknot, and since ARF(K)

trom its

satisties an identi-

cal identity by Theorem 10.7, it suffices to show that ARF(K) and A(K)

agree on the unknot.

This is true.

Q.E.D.

This completes our brief description ot the standard approach to the Arf invariant of a knot.

We end with a short

proof of Levine I s Theorem (see [25]). Theorem 10.9. sphere.

Let

K be a knot in the three-dimensional

~(t)

Let

denote the Alexander polynomial of K•

Then

.ARF(K) '" 1 ~ ~(-l) = =3 ARF(K) = 0 ~ ~(-l) = :1:1

~.

Let ,;. 'denote equality up to sign and powers of

Then we have the identity Hence

~(t),;.

(modUlo 8). (modulo 8).

VK(Jt'-l/Jt"),

t.

(Theorem 6.9)

"le( -1)

,;. VK(2 J=I). Since K is a knot, the Conway polynomial has only even powers of Z in vK(z). Therefore I

vK(2 J=I} = 1 + a2 (K)(2 J=I)2 + a4 (K)(2 J=I) 4 + ... Since A(K),= ARF(K) = a2 (K) (modulo 2), we have ~(-l)

,;. 1 + 4ARF(K)

This completes the proof.

(modulo 8).

156 APPENDIX.

The 91...819&1 Alexander Polynomial

In this appendix we shall sk.etch one approaoh";$O the Alexander polynomial.

This material is standard, and is

based upon Alexander's original paper [1]. Let

G be a finitely presented, tinitely related group

that is equipped with a surjective homomorphism t : G -> Z where Z denotes the group ot additive integers.

Kernel(t) = G', the commutator subgroup ot G.

that

H = GI/G" It

Assume

s

Let

be the abelianization of this commutator subgroup.

is an element of G such that

f(s) - 1

and x

de-

notes the (mUltiplicative) generator ot the group ring r = Z[Z] = Z[x,x- l ], then H becomes a r-module via the action on G':

x(g) _ sgs-l.

We say that the pair

(G,t)

is an indexed &:Q!!P..

Two in-

(Gl,t l ), (G2 ,t2 ) are isomorphic it there is a group isomorphism h: Gl -> G2 such that f2 h = fl' It

dexed groups

0

H(G,f)

denotes

GI/G"

with modUle structure as above, then

H(G,f)

is an isomorphism invariant of the indexed group

(G,t). As we shall see, H(G,f) over

r.

is f1nitelypresented and related

Suppose that there exists such a presentation with

an equal number

n

of generators and relations.

be the case tor the knot group.)

(This will

Then there is an exact

sequence

rD ~> rD -> where

H(G,t)

->

0

A is an n X n matrix with entries in

r.

157 Let

D(A)

denote the determinant of A.

ture on H(G,f)

tn.

Let

AA

= D(A)I

is indu¢ed by scalar

where

the image of A. D(A)

module.

(r)

r-module struc-

~ltiplieation

on

A denote the adjoint matrix to A so that I

shows that for all Thus

The

is the A,

n xn

identity matrix.

D(A)a • A(Aa), and hence D(A)[a] = a

Therefore

This

D(A)a

fo,r all

is in

[a] f H(G,f)

is an ann1,hilating element for ···H(G,f)

r-

as a

We shall see that the Alexander polynomial takes the

form of D(A) =

~(x)

for an approrpriatc§; matrix A.

Before doing more algebra, let's turn to the geometry. G.= ~1(S3_K)

Let

plement.

be the fundamental group of the knot com-

This group is finitely presented and related, with

a particularly useful presentation known as the Dehn presentation.

In the

Denn presentation

each region of the knot

diagram corresponds to an element of .1(S3_K) following conventions:

Replace S3_K by R3_K and assume

that the knot lies in the and under-crossings.

(x,y,O)

p = (0,0,1)

plane, except for over

These crossings deviate in the

direction (third variable) by 0 point

via the

< Izi

«1.

Let the base-

be a point above the knot diagram.

ciate to each region R a loop that starts at

zAsso-

p. descends

to pierce R once, and then returns by piercing the unbounded region once.

See Figure 49.

In this form1each region of the knot diagram. corresponds to a generator of the fundamental group, except tor the unbounded region, which corresponds to the identity element.

158

..

r~cl.)( (A) =Lk (K,~)

~ 7(C AS'CD': i Crossing relation in

~

Figure 49

presentation

159 Each crossing in the knot diagram corresponds to a relation in the fWldamental grOUlI) (as illustrated in Figure 49). gives a complete set 01' relations for the group. f : "1 (S'-K)

This

The mapping

-> Z exists lince a fundamental group of a knot

complement abelianizes to

Z.

The map can be specifically

described by linking numberel

f([g]) - Lk(K,g)

denotes linking numbers 01' curve. in R'.

where

Lk

With thi. inter-

pretation we see that, when we let elements 01' the group correspond to regions above, then

R in the knot diagram as described

feR) = Index(R)

ander index of the region

where Index denotes the Alex-

R (See. Lemma 3.4.).

bounded region is assigned index zero. Remark. tion.

The r-module

RCG.f)

Let q: X -> S'-K

ing to the representation

(The Wl-

See Figure 49.)

has the following interpreta-

be the covering space correspondf: G -> Z.

The space X is the

infinite cyclic cover of the knot complement (see [26]). first homology group, Hl(X;Z), is a

r-module via the action

of the group of covering translations of X. ture, Hl(XjZ)

and

H(G.f)

The

are isomorphic

With this strucr-modules.

Th1s

interpretation is very important. but will not be pursued here. Returning to algebra, we wish to describe a presentation for

GI. and thence compute

presentation of the form: with

GI IG".

Suppose that

G has a

G = (s'gl'g2' ...• 9nfRl' •.. ,Rm)

160 1.

n .. "mi" ('1'ru. tor the Dehn presentation since there are' two

mo•••,,1ons

than crossings, and ,one region

corre.,ond. to the identity element.) 2.

or· m ~ n.

o.

f(e).~. t(ll) • f(g2) = ••• = f(~) =

The seoond aondttton 1s accomplished from an arbitrary presentation bJ ohooltns an the other ••n••,tol"

s

(f(s) = 1), and re-defining

via multiplication by approrpriate

power. ot ., to tnlure that they all hit zero under f. Iloall that r. Z[x,x- l ] acts on G via xg = SgS-l. It is eal, to .e. that

G'

is generated by the set

CxkS1/k • Z, i . l, .•• ,n}.

In particular, each relation Rk

can be rewritten in terms of these generators.

Let

p(Rk)

denote th"'s rewr1tlns of Rk • Then (~P(Rk)/k E Z, i - l, .•• ,m) is a Bet of relations tor G' (proof via covering spaces or combinatorial group theory). G'

= «(xkgi}/(xkp(Rj»)).

By

Thus

abelianizing these generators

and relations, and writing them additively, we obtain the structure of H(G,f). Consider the Dehn presentation. region of index 1

(or

-1

Let

s

if necessary).

correspond to a Suppose A, B,

C, D are the regions around a crossing with in
= feD) = p+l,

fCC) = p+2

as in Figu~ 50.

(sPa)(sP+lb)-l(sp+2 c ) (sP+ld)-l = sPab-lscd-ls-p-~.

=p,

Then we have

a, b, c, d with A = BPa~ B = sp+1a, sP+2 c , D = sP+ld. The relation R = AB-1CD- 1 becomes R = AB-1CD- l

new generators C=

f(A)

161

The next calculation rewrites xKa, x~, ~c, xkd

R in terms of the generators

of G'.

R = sPas-PsPb-ls-psp+lcs-p-lsP+ld-ls-P-l (xPa)(xPb)-l(xp+lc) (Xp+ld)-l "p(R). Upon abel1anizing G'

to form

the additive relation

x'P(a-b+xc-Xd).

over the group ring r. H{n,f) a,b.c.d, •••

H(G.f), this''i<elation'becomes Thus. as a module

is generated by symbols

corresponding to the regions of the knot diagram.

with generating relations. one per crQ6sing. of the form a - b + xc - xd.

These relations can be remembered by plac-

ing :l.:x around the crossing in 'the regions that they correspond with

(as shown in Figure 50).

P

P+l

,., AID • SIC

T-+ -il ~ .i.

p+-.

~

Fim!:re 50 Earlier in the notes we have referred to this labelling as the Alexander code.

The symbol corresponding to the un-

bounded region is set equal to zero, and an adjacent region corresponds to

s

(f(s) = 1)

and is also eliminated.

The

resulting square relation matrix is exactly what we have described in sectiOn 3 as the Alex~der Matrix for this code. Its determinant is the Alexander polynomial,
and that it is a topological invariant of the

knot

K can be verified by examining the comb1natorial group

theory that we have sketched. Alexander apparently felt that the algorithm tibould stand on its own right, and he wrote his first paper on the polynomial from a combinatorial standpOint with slight mention of the background group theory and topology. Remark.on Determinants In these notes we have used a tormulation of'certain determinants (ot Alexander matrices) as state summations over the states of a universe (knot or link graph).

The signs in

the determinant expansion come trom the geometry of the universe.

I wish to point out here that the formula for any

determinant tollows a Similar pattern. The sign ot a permutation is determined geometrically by the tollowing prescription.

List the numbers

l, •.• ,n

order, and the permutation ot them on a line below.

in

Connect

corresponding numbers by arcs so that all arcs intersect transverselyat double ,points. is

(_l)c

arcs.

whe~e

c

Then the sign of the permutation

is the number of intersections ot the

For example let

P

= (~ ~ ~ ~ '~ ~):

1·.~:(p) ~ ~5

12345

6

+1

The combinatorics underlying the determinant expansion of an n x n matrix conSists of the grid.

n!

A grid state is a pattern of n

grid

~

of an

rooks on the

chessboard so that no two rooks attack each other.

n xn

n xn (Rooks

move only on hori·zontal and vertical files. tains more than one rook.) a

.Thus no file con-

For example, the grid states for

2x2

Given a grid state

S

and matrix

be the product of the entries of ponding to rooks of

S

A, define



to

A that are 1n boxes corres-

(when the matrix and grid state are

superimposed). The sign of a grid state

S, denoted a(S), is the sign

of the permutation of the rows that.produces this state from the diagonal state (all rooks on the main diagonal) .. To obtain this sign directly, draw arcs outside the grid from top row positions to left column positions so that each arc marks the row and column position of' a corresponding rook.

a(S) arcs.

~ (_l)c

Then

where

c

Then

is the number of crossings of the

Det(A) = t a(s)
of all grid states for the

where

.

n xn

grid.

~

is the collection

164

+

+

+ A S

C

I)

E F

CD

H ,
+AE\r

- 8D~

-AFH

+

+

CDH

C.E(!)

Il

De.t(M)

=:.

~rr(s)<~/S>

.
~

Figure 51

BF
- Summation

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J.W. Alexander. Topological invariants of knots and l1nks. Trans. Am. Math. Soc., 2Q (1923). 275-306.

2.

R. Ball and M.L. Mehta. Sequence of invariants for knots and links. (To appear in Journal de Physique.)

3.

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4.

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5.

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6.

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7.

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9.

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12.

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13.

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14.

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Ann. of

Ill. J. Math. 1

166 16.

I. Handler.

17.

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L.H. Kautf'man. 33-44.

19. 20.

(Private communication.)

Link manifolds.

Mich. Math. J. l i (1974),

L.H. Kauffman. Planar surface immersions. (1979). 648-665.

~

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Ill. J. Math.

Topology;, 20

L.H. Kauff'man and T.F. Banchoft.' Immersions and mod-2 quadratic forms. Amer. Math. Monthly 84 (1977), 168-185. L.H. Kauffman and L. Taylor. Signature of links. Amer. Math. Soc. 216 (1976), 351-365.

Trans.

24.

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25.

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26.

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29.

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30.

K. Murasugi. nating knot.

31.

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the Alexander polynomial of the alterOsaka J. Math. 10 (1958), 181-189. On

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~

e:nd Links.

Comm.

Publish or Perish Press

~

Math. Ann.

Combinatorial Group

Department of Mathematics, Statistics e:nd Computer Science The University or Illinois at Chicago Chicago, Illinois 60680

Compo