On Double-Torus Knots

Peter Clifford Hi

-4 thesis submitted in conformity with the requirements for the deg -ee of Doctor of Philosophy Graduate Depart ment of Mat hemat ics

Cniversity of Toronto

@Copyright by Peter Clifford Hill 1998

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Abstract

On DoubleTorus Knots

Peter Clifford Hill

Doctor of Philosophy

Graduate Depart ment of Mat bematics

L.niversity of Toronto

1998

A well knom-n family of knots is the torus knots. These are the knots

which aiay be embedded in a genus one Heegaard surface in S3. This idea is

generalized in two ways: first. by considering knots which may be embedded

in a genus two Heegaard surface T in S3 (double-torus knots). and second.

by considering knots which are "alrnost- torus knots. i.e. knots whicli live

on the toms except for one -straight" arc. caiied a bridge. This second class

of knots are called genus one bridge one ( g l b l ) knots. In fact it is easily

seen that g l b l knots are double-toms knots.

-4 double-torus knot may be either a separating or a non-separating

curve on T. -4 separating knot. if not trivial, is necessarily genus one. Some

Seifert invariants corning from the 2 x 2 Seifert matris are discussed. and

several infinite families of non-trivial separating knots with trivial AIesander

polynomial are described.

-4rnong the non-separating knots are the g l b l knots. which in turn in-

clude the 2-bridge knots. Formulae for the -Aiexander polynomial are given

in the simplest cases. These show sirnilarities to the torus and Terasaka

formulae for Alexander polynomids.

Acknowledgements

First. 1 tt-ould like to t bank my thesis advisor. Professor Kunio Slurasugi. for introducing me to h o t theory. and for his help and advice over the y a r s . 1 nrould also like to thank the other members of the department. particularly J. McCool and V. Jurdjevic. and the staff. particularly 11. Bachtis and 1. Bulat .

,Le.*. 1 would me to thank Lol for her patience and support. and. of course. Joel and Moira for Lol. ..bd 1 thank Margaret for convincing to return to school in the first place.

Finally, 1 wish to acknowledge financial assistance from the Sational Science and Engineering Research Council. and the Department of Mathe- matics at the Universil of Toronto.

Contents

O Introduction 2

1 Definitions and Notation 7 . 1.1 Double-toms Links . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2 Decompositionof3 . . . . . . . . . . . . . . . . . . . . . . . 9

2 T l T l Knots 30 2.1 E-xistence of T l T l b o t s . . . . . . . . . . . . . . . . . . . . 30 3.2 Seifert Matrices of T lTl Iinots . . . . . . . . . . . . . . . . . 39

. . . . . . . . . . . . . . . . . . . . . . 2.3 Calculation of a, and 3 41 3.3.1 The 1:lCase . . . . . . . . . . . . . . . . . . . . . . . 43

. . . . . . . . . . . . . . . . . . . . . . . 2.3.2 The 1:- Case 4-5 2.3.3 The23Case . . . . . . . . . . . . . . . . . . . . . . . 47

. . . . . . . . . . . . . . . . . . . . . . . . 2.4 Sutured Manifolds 56 . . . . . . . . . . . . . . . . . . . 5 Some typesof TlTl Knots 60

2 . 5 . TlTl knots ntit h 10 or fewer crossings . . . . . . . . . 6-7 . . . . . . . . . . '3.5.2 Genus one knots which are not T l T l 67

3 T2 Knots 69 . . . . . . . . . . . . . . . . . . . . . . . 3.1 Decompositions of 7 69

. . . . . . . . . . . . 3.2 .Al exander Polynomials of 1:l T2 Knots 90 3.3 Genus One Bridge One Knots . . . . . . . . . . . . . . . . . . 93

. . . . . . . . . . . . . . . . . . . . . . 3.3.1 2-bridge knots 103 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Berge Knots 105

Chapter O

Introduction

A p-componeot link is an embedding of y copies of S1 into s3. and a h o t is a 1-cornponent link. One possible way of studying knots is to study simple closed curves in closed orientable surfaces embedded in S3.

The only h o t which may be embedded in a 2-sphere in S3 is the unknot. On the other hand. any h o t may be embedded in a toms embedded in 5" by taking a regular neighbourhood of the knot. which is a solid torus. and moving the h o t by an isotopy to the boundary of the solid torus. This is obviously not very informative. so we restrict the type of embedding of the torus to embeddings which cut S3 into two solid ton. A n embedding of t his End is called a Heegaard surface of genus one. and a simple closed curw which is essential in this surface is called a torus h o t . W e may generaiize this to essential simple closed curves on Heegaard surfaces of higher genus. i.e. surfaces mhich cut S3 into two handlebodies. These knots have not been investigated systematicaily.

In this paper we concern ourselves with knots o n a genus two Heegaard surface in s3. We c d such knots double-torus knots. W e show an esample below :

Figure 0.1

This is a large family of knots which contains other well-linon-n families of knots such as the torus. classical pretzel. twist. ?-bridge. and genus one bridge one knots.However. t here are knots which are not dou ble-torils knots: an example is given in Remark 3.3. The classification of this famiiy is a hard problem: the classification for the simplest type of genus one bridge one knots has been achieved only recently by Bleiler and Jones ([BJ]).

We study these knots by loohng a t the surface that results from cutting the double-torus along them. X torus knot cuts the torus into an annulus. In the case of dou ble-torus knots. cut ting the surface aiong the knot yields a twice-punctured torus or two once-punctured tori In tlie first case n-e say the h o t is non-separating, ot hernrise it is separating. W e first cut along the curve marked O in Figure 0.1 . We assume the knot has proper intersection with 0, otherwise the knot is a torus knot. Cutting along O separates the knot into a collection of disjoint arcs proper1y embedded in tn-O once- punctured tori which we cd1 LHS and RHS. These arcs occur in a t most three parallel classes in each of LHS and RHS. -4 double-torus knot is callecl a j:k knot with 1 5 j. k 5 3 if there are j parallel classes arcs in LHS and k parallel clases of arcs in RHS. -4 double-torus may have a description as a j :k knot and as an m:n h o t with ( j , P ) # (rn.n).

Having cut along O. we then cut LHS and RHS along the arcs. Then we glue the resulting pieces back along the edges coming from O to arrive a t the surface we would have obtained by simply cutting along the h o t . The advantage of making the cut in two stages is that certain symmetries becorne apparent in the order of the arcs met in travelling along the knot. These are used first to give criteria for a knot to be a separating double-torus knot. For example. the number of arcs belonging to each parallel class must be even for the knot to be separating.

It is harder to find sufficient conditioiis for a double-torus knot to be separating, but this can be done if the number of parallel classes of arcs in both LHS and RHS is less than three. For example. if there is only one parallel class of arcs in each of LHS and RHS (the 1:l case). then n-e can specify the h o t by the number n of arcs. the LHS and RHS parallel classes. and a rotation number p which shows how the arcs in LHS match u p n i th those in RHS. The h o t wül be separating if and only if n is even and ( n . p ) = 1. In this case we have a function v of n and p n-hich gives information about the Seifert invariants of the knot. In particular. if v(n. pl = O ie.g. v (8.3) = O ) t hen the -4Iesander polynomial is trivial:

Theorem (Theorem 2.15, Proposition 2.16) If L is a Ir1 l inh n is even. and (n . p ) = 1 . then L is a sepamting h o t with Aleaander polgnomzal

m'th B = -y2 ( n . p)pqp'qf . tuhere p. q (P'. q') are integers detennining the parallel classes of the arcs in LHS (RHS).

This is a special case of a more general theorem: Theorem 2.1 A double-tom link L with ni ( n i ) arcs in each of three

pamllel classes in LHS (RHS). where 1 5 i < 3 und some of the n, ( r i : ) may be O. is a sepamting Anot i f and only i f

(ii) 3 has no annulus components. In particular. this implies

As in the i:1 case. the second condition can be replaced wit h an arith- metic condition involving the integers ni. nz. ns, ni. ni . ni and p. in the 1:2 and 2 2 cases ( s e Theorems 2.20 and '2.23).

We can find many other families of knots with trivial Seifert invariants. in addition t o the 1:l type examples given by v = O (see Esamples 2.1 to 2.5). In order to show t hat at least some of the separating tnots nit h trivial

Seifert invariants are non-trivial. we use Gabai's sutured manifold theory. This leads to the perhaps unexpected result that a separating double-torus knot is trivial if and only if it bounds a disk mhich does not intersect the doubletorus in the interior of the disk (Corollary '2.39). This is in contrast to the case for non-separating knots: for example. the (1.1)-torus h o t is trivial, but any disk it bounds must intersect the toms in its interior. -4 consequence of this result is that almost al1 of the knots in our families of separating double-torus knots with trivial Seifert inm-ïants are in fact non-trivial (T beorem 2.3 1).

Among the separating double-torus knots. which are necessarily genus one. are twist knots and classical (P. Q. R)-pretzel b o t s nrit h P. Q. R odd. In fact. a l alternating genus one knots. and all genus one knots Rith fen-er than eleven crossings are separating double-torus knots (see Theorem 2.32 and $2.5.1). At present we have no esample of a genus one knot n-hich can be shown to have no presentation as a separating double-torus knot.

The separating knots appear as boundaries of genus one surfaces: tlrk is not the case for non-separating knots. which may have arbitrarily high genus. In this case we may use one of seven canonical forms for the surface obtained by cutting the double-torus dong the knot to get a presentation for the group of the knot via the Seifert-van Kampen Theorem. This k applied in the 1:l case to get a formula for the -4lexander polynomial:

Theorem 3.5 Let L be a 1:1 non-sepamting knot. Then L is a band sum of two disjoint torus h o t s . and the Alexander poZynomial of L k of the

where p. q (p', q') are integers determining the parnllel classes of the arcs in LHS (RHS) and A,,, ( t ) is the Alexander polynomial of the ( m. rr )-toms h o t .

This is similar to the formula found bu Terasah for the Alesander poly- nomial of a knot which is a band sum of two disjoint knots. one of n-hich is tnt-ial.

-4 special case of non-separating double-toms knots is genus one bridge one (g lb l ) knots. Here we take a once-punctured torus containing n con- nections, and cap it off with a disk containg n - 1 properly ernbedded arcs. then adding one more arc "outside" the torus:

Figure 0.2

The simplest case of this type of h o t is a satellite knot for which both the pattern and cornpanion knots are torus b o t s (Propositions 3.11 and 3.1'1). In t his case we can write the -Alexander polynomial direct11 The formula for this polynomial leads to a conjecture concerning Alexander polynomials of glbZ knots in general. We test this conjecture by applying it ro the '-bridge knots, which are contained in the g lb l b o t s .

The final section of the paper is a description of some non-separating double-torus knots found by J . Berge which have surgeries yielding lem spaces.

Chapter 1

Definitions and Notation

1.1 Double-torus Links

Let V V and W' be two genustwo handlebodies such that C l r U IV' = S? and W n IV' = 7 is a genus-two ciosed surface. By [Wal68]. T is embedded in S3 in a standard way. i.e. it is not knotted or linked with itself. There is a sphere 5 in S3 such that W n S is a disk. FV'n S is a disic. and II- - X and W' - Z are both disjoint unions of two genus-one handlebodies. Let WL. IVR be genus-one handlebodies with V V - 5 equal to the disjoint union of IVL and WR. Let LHS= LWL - X and RHS= 8F1;R - S. LHS and RHS are once-punct ured tori.

Let X I be a simple closed curve (scc) in the interior of WL representing a generator of Hl(WL) and y1 a scc in the interior of representing a generator of Hl(&). Then [xi] and [yl] generate Hi(W). where [cl is the bomology class represented by the scc c. Ive choose generating scc's +z. y?

for Hi (W') according to the foilowing diagram:

Figure 1.1

Yote that lk(xl . x2) = -lk(yl. y2). where Ik denotes the linking number. Let D = W n S and let Cl = dD. The arrow on 0 indicates the positive direction on O. I f a is a curve on 7 = dVV = d W ' . let a be the image of c\ pushed slightly inside W. Then lk(a. E) is well-defined.

Definition 1.1 A simple properly-ernbedded arc in LHS (RHS) d l be called a connection if it does not separate LHS (RHS). Two connections are parallel if together Nith two arcs on O they form the boundary of a disk in LHS (RHS).

Proposition 1.1 ([Zie63]) Let C be a collection of disjoint connections in LHS. Then C consists of at most three classes of parallei connections { a l . . . .,ai}. {b i , . . .. b,} , { c l , . . . . en} . We assume that if rn = O then n = 0. and if m # O then I # O . Furthemore. there is a point on O auch that travelling along O in the positive direction one encounters the initial points of

a l , . . . ? ai, b i ? . . .. b , , cr . . .. C,

in order. followed by the terminal points of

al.. . . . a l . 6,. . . .. b t . c , . . . . cl

Proof. Complete LHS to a torus LHS' by gluing a disk to its boundary. and complete the connections ai. b,. cr, (see Figure 1.8) to simple closed curves a;. b;. ci on LHS' by arcs in general position on the disk. If E . q are connections, and the arcs completing <. 7 to cm. q' have no proper intersection then c*. q'. and thus (. q, are parallel- If the completing arcs have proper intersection, then they intersect in one point. and <* and q' are scc's on the torus which intersect at one point. Since a t most three non- parallel scc's on a torus can have pairwise proper intersections of one point. there are at most three pa rde l classes of connections on LHS. If < and r,~ are parallel. and w e come to the initial point of another connection ; while moving positively along O from the initial point of f to the initial point of q. then we will come to the terminal points of q . 7 4 . and < in order as n-e continue to move positively along O. since y' is contained in the annulus bounded by <' and q*. This is enough to show that the initial and termina1 points are distributed along O in the order described. CI

1.2 Decomposition of 3

In the diagrams below we have some arcs labelled with numbers. whicli correspond to the '272 segments of O. and other arcs Iabelled by letters. When we refer to labelled arcs we dl a h a y s mean number labels.

Suppose we have on- one class {ai. . . . . a,) of parallel connections in

LHS:

Figure 1.2

Cutting dong al we get:

al

Figure 1.3

Cutting along the rernaining a, we get:

an- i

Figure 1.4

Remark 1.2 In the above diagram we have shown a connection b. alt hough we only have a-connections. Here b is a connection starting in the segment of O labeLled n. and ending in the segment labeiled 2n. Uthough b is not one of the connections we are cutting dong, we use it later ahen we describe canonical curves on subsurfaces of 3. Similady. a b'-connection appears in Figure 1.11.

Yow suppose we have two classes {ai. . . . . a,, } .{bi . . . . . b., }. rr = n 1 + n2.0f parallel connections on LHS:

Figure 1.5

Cutting dong the a; yïeIds:

Figure 1.6

Cutting dong the bj fields:

an, - 1

Figure 1.7

3 0 ~ suppose have t h e classes { a 1. . . . . a , , ) . { b l . . . . . b,, } . {el . . . . . r,, } .

n = ni + n* + n~ of pardel connections on LHS:

Figure 1.8

Cutting dong the a; fields:

Figure 1.9

Cutting dong the bj and ck fields:

Figure 1.10

Definition 1.2 Suppose we have a Iink L embedded in 7 wîth connections { a 1. . . . . an, , bi, . . . . 6 , . ci. . . . . c,,, } in LHS and connections

. b;. . . . . b:;. 4, . . . . c$} in RHS where

We need one more number. the rotation number p. to describe hon- the connections in LHS match up with the connections in RHS. p is defined so that the connection al begins at the beginning of the oriented arc labelled 1 in 0. and a; begins at the beginning of the arc labelled (I + p ) mod 2 n .

Xow we have diagrams for RHS similar to Figures 1.4.1.7. and 1.10. :

Figure 1.11

See Remark 1.2

Figure 1.12

18

Figure 1.13

Xow iet T = T - L. We shall assume that L c ont ains no disjoint trivial components. so 3 has no disli components. FVe can determine 3 from the diagrams 1.2 to 1.13. Since T has EuIer characteristic -2. so does 3. 1Iodulo any annulus components. there are five possibilities for 3:

TlTl : two once-punctured ton T2 : a twice-punctured torus SSTl : a once-punctured toms and a thrice-punctured sphere S3S3 : two thrice-punctured spheres S 4 : a sphere wit h four punctures

Remark 1.3 We will assume that L contains no torus h o t component that h a . no proper intersection with O. Thus 3 consists of one of the five surfaces listed above, plus (possibly) annuii. Since we are most interested in the case where L is a hot , we concentrate on the cases TlTl and T'L. However.

in the following examples we show that all five types of double-torus links eicist.

Notation Given a set {ci. . - . . 6) of connections in LHS (RHS) ahere < can be any Ietter. the letter < without subscript will be used to s tand for any elernent of {cl. . . . . Cm}. or any connection in LHS (RHS) parallel t o the <i - Definition 1.3 Suppose f" is the completion of a connection < in LHS. as in the proof of Proposition 1.1- Then c' is a (p. q)-torus knot. n-here 11 corresponds to the generator X I ? and q t o x?. We say (p. q) is the direction of the connection <. Suppose q has direction (r . s). We define

Given <. r ) with 1 1 = 1. there is an autohomeornorphrn of LHS whicli

maps ,C to a (1. O)-cokection and r ) to a (O. 1)-connection - this follows from the analogous result on the torus. Thus. git-en three sets of parallel connections {ai. . . . . a,, ) .{bi. . . . . b,, }. {ci. . . . . c,, ). ahich are disjoint. n-e may assume

by r e l a b e h g if necessary. and so there is an autohomeomorphism of LHS so that a (1. O ) , b (0. l), and c M (-1.1). Hence the diagranis 1.2 to 1.13 are fuUy general.

In particular. if a and b are (p. q)- and (r. s)-connections respectively. then c is a ( - p + r. -q + s)-connection.

' I I Definit ion 1.4 We let L ( ( n i . n l , ns. n , . R ~ . n3. p ) . ( p . q . r. .s) . (pl . qf . I.'. sf ) ) be the link embedded in T which has rotation number p. n (p. q)-connections. n;! (r. s)-connections, and n~ ( - p + r. -q + s)-connections in LHS. and ni (p'. qt)-connections. n; (r'. s')-connections. and n j (-p' + r'. -qr + .<')- connections in RHS, with

The first (p. q)- connection starts a t the beginning of the arc labelled 1 in O, and the first (p', qf)-connection s ta r t s a t the beginning of t he arc

labeiled (1 + p ) mod 2n in O. For the rest of thii paper we use the symbol L interchangeably wit h L ( ( n 1. nî. nf , n; . n;. ni. p ) . ( p . q. r. s ) . (p'. q'. r f . .sf ) ) .

We c d L a j:k link 1 5 j, k 5 3. if there are 1 classes of parallel connections in LHS. and k classes of parallel connections in RHS.

We d sometimes write an equation

/ 1 / to mean nl = rn1.n.r = m2,ns = ms.nL = m l , n 2 = m;.n; = m;. and cal1 t his the equation of L. Obviously. the equation depends on the given de- scription of L as a double-torus link. and one hli may have many equations.

In the foilowing diagrams we no longer label every edge and connectiori. Example 1.1 .j + O + O = 5 = 2 + 2 + 1. p = 1. The diagrams corre-

sponding to this case are

Figure 1.14

This gives the foIlowing diagram for 3:

Figure 1.15

so L is a T2 knot. Example 1.2 4 + 4 + O = 8 = 6 + 2 + O . p = 3. The diagrams corre-

sponding to this case are

Figure 1.16

This gives the following diagram for 3:

Figure 1.17

so L is an S3S3 Link with four extra parallel components.

Example 1.3 3 + O + O = 3 = 2 + 1 + O. p = 2. The diagrams corre- sponding to this case are

Figure 1.18

This gives the following diagram for F:

Figure 1.19

Example 1.4 4 + O + O = 4 = 4 + O + 0. p = 1. The diagrams corre- sponding to this case are:

Figure 1.20

This gives the following diagram for 3:

Figure 1.21

so L is a TlTl h o t . Example 1.6 2 + 1 + 1 = 2 + 2 + 0, p = O. The diagrams corresponding

to this case are:

Figure 1.22

This gives the following diagram for 3:

Figure 1.23

so L is an S4 iink nrith one extra parallel cornponent.

Definition 1.5 We c d the components of LHS - L (RHS - L) the pzeces of 3. The rectangles in Diagrams 1.2 to 1.13 ndl be called simple pieces. and the cylinders. octagons, and hexagons d l be called cornplez pieces. -4 maximal coilection of contiguous simple pieces is c d e d a band.

Definition 1.6 We now orient the connections. and sa!* t hat a rectang!e in LHS (RHS) with numbered labels (j. k) (in order), 1 5 j. k 5 2 n . represents an on'ented connection < if < starts in the j'th segment of O and ends in the k'th segment of 0, and is disjoint from L. Clearly if a rectangle represents connections and 7, then and q belong to the same class of connections.

Lemma 1.4 (Reversing Iemma) Suppose a band in 3 starts ut label m and tenninates at label ml, and represents connections . . Ei = &1. in order. Then the% ù a band in 3 startang at mi E ( ml + n ) mod 2n and teminating at rn; (m2 + n) mod 2n whzch represents C;" . -=-<;'r

in order. Thus the band which tenninates at m; is the reverse of the band which begins ut m 1. In purticular. if the band begdinning ut rn 1 tenninates at mi. then it às self-reversing.

Proof. Erom inspection of the diagrams 1.2 to 1-13 .we see that if 3 contains a cornplex piece with label m. it &O contains a comples piece with label m' = ( m + n) mod 272. and if 3 contains a rectangle n i th labels (j. k) representing <, then it also contains a rectangle with labels (( j + n) mod 212, (k + n) mod 272) representing E-'. 411

Definition 1.7 Given ni, nz, ns, ni. ni. ni and p we have a famil- of links. Corresponding to this family we have a connection diagnzm. a simplified picture in which O is represented by a vertical line segment. and the con- nections are represent ed by oriented (approximately ) semi-circular arcs. For example. the case 3 + 2 + 3 = 6 + 1 + O. p = 4 has the connection diagram:

Figure 1.24

28

The diagram is misleading in that it hides the fact that paraIlel connec- tions have non-trivial linking in general. and it implies double points that don3 really exkit. Connection diagrams can be used to find the number of components of the link. and the order and orientation of connections met when travelling dong a component of the link.

Definition 1.8 If L is a h o t . then the first h o m o l o ~ group of its comple- ment is infinite cyclic. Define a generator t for this group according to the foUowing diagram :

Figure 1.25

W e dso define the numbers P.Q.Pf and Q' so that x1.x.~.y1 and y- rep- resent tQ. tP. t4' and tP' respective. in the hornolopv of the cornplernent.

Definition 1.9 Given a positive integer a. define r,(m) = rn mod 20 in the range -0 < r 5 a..

1 ' b m a r k 1.5 Let P' = n+p. Then L ( ( n l . RÎ. n3. n l . TI?. n 3 . p ) . ( p . q . r . . ~ ) . (p'. q'. r f . . 2 ) is equivalent to L ( ( n l . n2. n n ni. n;. ni, pl) . ( p . q. r.s). (-JI'. -4'. - r f . - s f ) ) . For this reason we assume

O S p S n - 1

Remark 1.6 It is clear that a j : k link is also a k : j link by a rotation of T through R which maps RHS ont0 LHS. For example. in the 3:l case. L ( ( n i . n*. n j , n. O, 0 . p ) . ( p . q. r . s ) . (pl. q'. *. *)) is equiident t o L ( ( n , O, O, n3, n l , n l . p ) , ( - p l , q'. *. *), ( p - r. -q + S. -r. s ) ) . For this reason we assume j:k is one of 1:l. 12, 1:3, 2 2 , 3 2 , or 3:3.

Remark 1.7 In Figure 1.1 we are observing T from a vantage point located in the handlebody W . We could have chosen a vantage point inside JI-. This Ieads to more equvdences. For example. L ( (nl. O. O. n',. ni. n3. p ) . ( p . q . t. *) . (p'. q'. r'. s') )

is equitdent to L ( ( n l . 0. O , ni, n;. n',, n - p ) . ( p . q , *. *). (-q'fs'. -pl+rl . s'. r ' ) ) . and L ( ( n 1 . n2, ns , a;. nG,O, p ) , ( p , q , r. s ) , (p'. ql , r'. s r ) ) is equivalent to L((n:1 . n-. ni. n'. n;. O . n - P), ( - q + s, - p + r , s , r ) , ( S I , r r , q l , p l ) ) , and so forth.

Chapter 2

TlTl Knots

In this chapter we show that a separating knot on 7 has an even nurnber of connections in each paraIlel class. and. conversely. a knot n l th even numbers of connections cannot be a non-separating h o t .

Xmong the separating knots are the classical (P. Q. RI-pretzel knots n i t h P,Q,and R odd. and various infinite families of non-trivial knots a-ith trivial Alexander polynornial.

2.1 Existence of TlTl Knots

Theorem 2.1 L = L ( ( n l . n2. n3, n i . n i . ni. p ) . ( p . q . r . s ) . (p t . qt. r t . s ' ) ) is a Tl Ti h o t if and only if

(ii) 3 hm no annulus components. In particular. this ingplies gcd(nI. n2. n3. n i ! ni. n 5 . p ) = 1

Proof. The proof is contained in Lemmas 2.2 to 2.11 .

Lemma 2.2 Suppose L is a TlTl h o t . Then either

Proof. Since L separates 7 into two cornponents. as n-e travel dong O we go from one component to the other every time we cross L. Therefore aU the labelled segments in the pieces belonging to one component of 3 are

even, and ail the labels in the pieces belonging to the other component are odd .

If np = n3 = O then the cylinder in LHS has labels n and Zn. and therefore n is even (similarly for RHS). If n;! # 0.n3 = O then the octagon in LHS has labels n, n 1. and Zn. so n and n 1 are even. and t herefore n- is ais0 even (similarly for RHS).

Thus if L is a 1:l. 12. or 2 2 link and L is a T1T1 h o t . then n-e have ( i ) . Suppose L is a 1:3 link. Then n2 = ns = O so n is even. Suppose p is even. Then RHS contains a hexagon aith labek n+p. n:+p. n+n;+n;+p so n; and ni must also be even. and therefore so is ni. and again we have (i). Suppose p is odd. Then RHS contains a hexagon with Labels p. n + ni + p . ni + 4 + p which must all be odd. and so ni and ni must be even. so n3 is also even. and again we have (i).

Suppose L is a 3:2 link. Then ni # O. ni = O. so n is even. and an argument similar to the 1:3 case shows that we have [il.

Now suppose L is a 3:3 liak. If n is even we have a hexagon in LHS n-ith labels n. n 1, n + n 1 + n2, so n 1 and n* are even. and t herefore so is ns. The same argument goes for RBS. so we have (i). If n is odd. the LHS hesagon -5th Iabels n? nl, and n + nl + n2 shows that nl and n;! are odd. and so n:! is odd. Again. RHS is similar. so this tirne we have (ii). CI

Lemma 2.3 Suppose ni = nz = n3 = ni = n; = ni = Z mod 2. Thrn L is s3s3.

Proof. From the diagrams for the three-direction case. in LHS n-e have one hesagon Ho with even labels 2n. n l + nz. n + nl. and one hesagon Hl nit h odd labels n. n+ n 1 + nz . n 1 . In RHS ne have one hesagon Ho nir h even

labels m l , mp, ms, and one hesagon Hi with odd labels n+rnl. nfrn2 . n+mn.

Figure 2.1

Here a. ,3. y E {a'*'. bf*'. c'* l } with [:dl = -[O] + [3] . We may assume without loss of generality that the edge labelled 2n in

Ho is joined to the edge labeiled ml in Ho y - a band .-Io. Then by Lemma 1.1 the edge labelied n + ml in Hi is joined to the edge labelled n in Hi by a band -41 which is the reverse of -Ao.

Suppose the edge labelled n + ni in Ho is joined to the edge labelled

nl + n2 in Ho. Then by Lemma 1.4 we have the foilowing diagram:

Figure 2.2

and so L is S3S3. Suppose then that the edge labelied n + nl in Ho is joined to the eclge

labelled m2 in Ho. Then we have the following diagram:

Figure 2.3

33

so again L is S3S3. Finally, suppose the edge labeiled n + n 1 in Ho is joined to the edge

labelled rns in Ho. By Lemma 1.4 me have the foUowing diagram:

Figure 2.4

The band BI is the reverse of the band Bo. and Cl is the reverse of C;. Now let Vi be the cornponent of 3 containing Hi. and let j; : 1; s 7

be the inclusion map. i = 0.1 - Starting from a base point in the interior of the segment labeiled 2n we can choose canonicd curves e. f for the once- punctured torus Vb with

Similarly. starting a t a base point in the interior of the segment labelled n + rn 1, we can choose canonical curves

for VI.

'iote that since .Ao is the reverse of A l . 10, ([-hl) = jl, ([=ll]). Thus

JO, ([el) = j l t ( [ e l ) and jo# ([A) = jlw ([fl) in Hi ( T h & r) VI = L: in particular Vb r) T.1 is non-empty. and so we can use part

of the Mayer-Vietoris sequence of the pair (Vo. 5) in reduced homology to get the epimorphism

Since Hl (Vo) is generated by {[el. [fl) and Hl(l.i) is generated by { [ e t ] . [fl}. t his implies that HI (T) is generated by the two elements jo, ([el) = j l x ( [ e l ) and jo# ([A) = ji# ([fl) . But Hi (T) is a free abelian group of rank 4. so this is not possible.

Thus n + nl cannot be joined to m3. and so L is S3S3. n

This proves the -onlu if" part of Theorem 2.1. For the rernainder of t his chapter we assume nl n2 n~ ZE n: ni 3 ni I O mod 2 .

Lemma 2.4 Suppose ,- 1

is a Seifert matrix of a h o t . Then 1 b - cl = 1 .

Proof. Since JI is a Seifert matriv for the knot k

Lemma 2.5 Suppose 3 contains a once-punctured toms formed frorn a LHS-octagon and LES and RHS rectangles. Then the rectangles form two bands of odd length. whose rniddle rectangles may not represent parallel con-

nections.

&(t) = n ( 1 - t ) b - c t c - bt d ( 1 - t )

Proof. In order to form a once-punctured torus. the two bands must join opposite edges of the octagon :

Figure 2.5

Here < E {ai1. b*'}, r ) E {{ak1. b*') - {E"}} . We choose canonical curves a. 3 for the once-punctured torus. Since the bands join opposite edges of the hexagon, and opposite edges have labels differing by n mod 'ln. the bands are self-ret-ersing, by Lemma 1.4. Therefore. if the middIe rectangle of one band is parallel t o the connection UT. t hen the band represents a partial worcl xwx' where x' is the reverse of x . Then in homology the band represents to 3[x] + [u'] . Suppose the second band aiso has its middle rectangle parallel to w - the orientation doesn't matter since we can replace 3 by PL. Then the second band represents the partial word ywy' mhere y' is the reverse of y. so in homology the second band represents ?[y] +[w] . Then. in Iiornology. we have

[al = 3bI + KI - h l + LU.']

Suppressing the square brackets and letting z = f + q . we have

Recall that we have defined E. 3 to be the images of a. 3 pushed sliglitly 'inside' T. Then clearly

cannot be a Seifert matris of a knot by Lemma 2.4. But 31 is a Seifert matnu for the component of L that bounds the @\-en once-punctured torus component of 3. So t his is a contradiction. and the middle rectangles of the bands cannot represent parde l connections. O

Lemma 2.6 Suppose F contains a once-punctured toms formed fmrn two LHS hexagons and LES and RHS rectangles. Then the rectangles form .? bands of odd lengtir. no two of which have middle rectangles representing pamllel connectiorzs.

Proof. ThetwohexagonsarelabelledO.n+n~.n~+n~.andn.n~. n + ni + na . Joining O. n + nt. n, + n;! to n. nl. n + ni + n2 respectit-ely by the t hree bands fields a once-punctured torus. No other way of joining the two hexagons by bands results in a once-punctured torus. For esample. suppose n is joined to n + ni . Then by Lemma 1.4. O is joined to nl. so nl + na is joined to n + nl + n*. and we have the follonring diagram:

Figure 2.6

This is a sphere with three punctures, not a once-punctured torus. When we join O, n + nl, nl + nz to n. ni, n + nl + nz re~pecti~ely. we get three self-reversing bands by Lemma 1.4 . The rest of the proof is similar to that of Lemma 2.5 .

Lemma 2.7 No more than two rectangles representing parullel connections can be middle rectangles of bands in a Tl Tl h o t .

Proof. This foIlows from Lemmas '2.5 and 2.6. a

Remark 2.8 Suppose g E {ni. n*. n 3 } . r)' E {ni. ni. ni} . By correspond~nt~ type to q or r)'. we mean for example if q = n 1 then the corresponding type of rectangle to q is a rectangle representing an a-connection. if q' = R B the corresponding type of rectangle is a rectangle representing a et-connect ion. and so on. Then by inspection:

q G 2 mod 4 (q' = O mod 4) e There are an even number of even-labelled rectangles. and an odd number of odd-labeiled rectangles. of the corresponding type.

q O mod 4 (q' G 2 mod 4) There are an odd number of even-labelleci rectangles. and an even number of odd-labellecl rectangles. of the corresponding type.

Lemma 2.0 Suppose L is a TlTl h o t . and w zs a connection belonqing to L . i.e. u* is an arc contained in L . Then 3 contains ezactly one band whose middle rectangle represents UT.

Proof. Suppose there are two bands s i t h middle rectangles repre- senting W . By Lemmas 3.5 to 2.7 there are at most two. and they belong to different components of 3. Then one of the middle rectangles must be odd-labeiled and the other even-labelled. So there are an odd nuniber of odd-labelled rectangles, and an odd number of even-labelled rectangles representing m. since non-middle rectangles occur in pairs in self-reversing bands. But Remark 2.8 shows that this is impossible. u

Corollary 2.10 A j:k Tl Tl h o t has ( j + k ) rectangle types and ( J + k ) bands, and there is ezactly one middle rectangle of each type.

Lemma 2.11 Suppose ni E nl n~ ni ni G n j O mod 2 . and 3 contains no annulus components. Then L is a Tl Tl Lnot.

Proof. Since 3 has no annulus components we have gcd(nl, n l , n ~ , ni, ni, nJ,p) = 1 which implies that p is odd. p odd iniplies that LHS complex pieces have even labels and RHS comples pieces have odd labels. ,iUso. the two labelled edges of each rectangle have the same parity. This shows that 3 is not connected.

An even number of rectangles of a given type must appear in pairs of non-self-reversing bands. and an even number of rectangles of a @en type must appear as non-middIe rectangles of self-reversing bands. Since there are no annuli. all rectangles must occur in bands. There are an odd number of rectangles of each type. so each type must contain at least one rectangle which is the middle of a self-reversing band. In a j:k link t here are ( j + k) bands and ( j + k) rectangle types. so aU bands are self-reversing. and no band is the reverse of any other band.

Therefore. an edge of a complex piece Iabelled rn is joined CO an edge of a complex piece Iabelled (n + m) rnod 2n by Lemma 1.4. and it iç easily seen from diagrams 1.3 - 1.13 that ail components of 3 are once-punctured tori. O

This completes the -if" part of the proof of Theorem 2.1. t hus Theorem 3.1 is proved. a

1 1 Proposition 2.12 Let L = L ((ni. n2. n3. n l . n2. ni. p). (p. q . r. si. (p'. q'. r f . 2') )

be a Tl Tl h o t . Then the number of { n 1, n z , ns} which are congruent to 2 rnod 4 is equal to the number of {n;. R; . ni} which are congruent to 2 r n o d 4 .

Proof- Suppose L is a j:k TlTl h o t . Suppose esactly m even- labelled bands have RHS middle rectangles. O 5 m 5 j . Clearly esactly n, odd-labelled bands have LHS middle rectangles. so m 5 k .

Then there are m LHS rectangles with odd numbers of odd-labelled rectangles. and ( j - m ) LHS rectangle types with odd numbers of even- labelled rectangles. so exactly m of { n t . nz. ns) are congruent to 2 mod 4 by Remark 2.8 .

Similarly. there are m RHS rectangle types with odd numbers of eve~i- labelled rectangles, and (k - m) RHS rectangle types with odd numbers of odd-labeiled rectangles. so exactly m of { n i . n;. n i } are congruent to 2 mod 4. a

Definition 2.1 If L = L ((nl y n* y na, ni. n;. ni! p ) . ( p . q. r. s ) . (p'. y'. r'. a') ) is a TlTl h o t , then p is c d e d regular.

2.2 Seifert Matrices of TITI Knots

In the following we will abuse notation by spealcing of [cl as an element of H1(7). By [t] (or, suppressing the square brachts. just É) we m i l l mean [c'], where <' is the completion of 5 as in the proof of Proposition [Zie63].

SupposeL = L ( ( n l . 122: n & . n ; . n $ . p ) . ( p . q . r . s ) . ( p ' . q f . r f . s f ) ) k a T i T i knot with connections a = (p, q) b = ( r. s ) . c = ( - p + r. -q + s). a' = ( p f . 4'). 15' = (r'. S I ) . 8 = ( -pl + r'. -q' + s') . On one of the punctured tori in 3 n-e can take canonicd curves a and 3 ~ 6 t h

in the homoIogy of 3. ai. Ji E Z (Xgain w e suppress the square bracliets). Let

Then -LI is a Seifert matrix for L . Lemma 2.13

Proof. Define the linking number so that Ik(sl . x S ) = 1. Tlien l k ( y l . y2) = -1. Then we have

- - Ik(a, a) = -pq, lk(b. b) = -rs. Ik(a. b) = -qr. Ik(b. n) = -p.s

- I f Ik(af. 2) = p'q'. Ik(bf. 8) = rfs'. &(af . P) = q'r'. Ik(bt. a') = p -5

and k ( E . 7) = O = lk(qf. c)

where E { a . b} and rlf E {a'. b'}. Thus

0

Let 8 = det (LW) .

If the knot L has a 2 x 2 Seifert matriv JI with det(31) = 6i. ,Aiexander polynornial (t) has the form

1

then its

In particular. AL(t) is trivial. i.e. l ~ ( t ) = 1. precisely n-hen 0 = 0. IfL is a 1:l. 12. or 1:3 Tl T l knot we can take ckl = 1. a- = 03 = mi = O

in which case = 33. 742 = 34, -13 = O = 744 and so

2.3 Calculation of a and B We describe an algorithm for calculating the sequence of connections in a band in a T l T l knot. This can be done by making the appropriate diagrams as in Figures 1.3 to 1.13. and joining similarly labelled edges. W e do this in some of the examples below. Our algorithm is simply the same procediire described in words instead of diagrams. This is then applied to find values for a;, ,di, i = 1,3,3,4 .

1) Define the function d on the set {- n. -n + 1. . . . . n} as follows:

i even. O < i < n l - 8 8 o(i) = n i e v e n . n l < i < n l + n 2 - - - o ( i ) = b ieven, nl + n 2 < i < n - - - O(i) = c iodd. O < i < ni - - ~ ( i ) = EL' iodd. n ; < i < n ; + n ; - - - o ( i ) = b '

, i o d d . n : + n ; < i < n - - - o(i)=c'

and o(0) = O(nl) = o(nl+n2) = o ( n ) = O. For i = 0.1.. ... n. O ( - i ) = -otn-i) .

Thus o takes d u e s in the set {O. ka. ib. f c. fa'. f b'. &c/) .

2) Define the function t.~ on Im(o) as follows: u(0) = O and ~ ( x ) = r ( - X ) for any x in Im(o) :

3) Choose a label m from a comples piece in LHS and define a seqiience -Yl. -fL;, . . . . -YA+L by

-Y1 m - p mod 2n and

where r, is defined as in Definition 1.9. The sequence terminates when d(-X,) = o. Define a second sequence E;. f i . . . . . ki by I.;. = of -Y,).

-4s stated above. we are assuming nl n2 G n3 G n', r n', = n!, n O mod '2. When p = O. a rectangle which includes the edge Iabelled i repre- sents the oriented connection o(i). with the initiai point located in the edge labelled i. For TlTl knots p must be odd. and for odd i. a rectangle n-hich includes the edge labelled (i + p ) represents the oriented connection O( i ) .

For < E { f a . Ib. f c } . the sum of the two labels on a rectangle repre- senting ( is m(<) mod 272. For <' E {ka'. Ib'. id). the sum of the tn-O labels on a recangle representing cf is (v(!.(Ft) + 3p) mod 2n . This can be seen froni the diagrams.

Now choose a label n from a complex piece in LHS. -4ttached to tliis edge %di be a RHS rectangle with label rn which represents O( rn - p) . Let -Y1 = m - p. Let X2 be the other label of this RHS rectangle . Then

-Y2 -(m - p ) + uw(-Yi) +prnod?n

= -XI + wo(Xl) + p mod Zn

and = o( -Y~) = ~ ( m - p) is the oriented connection represented by this rectangle.

If X2 # O. then attached to the edge labelied -Y2 is a LHS rectangle with label -Y2. representing o(X2) . Let (-Yn + p ) be the other label of this rectangle. Then

-Y2 + (-Y3 + p ) = v0(..Y2) rnod Zn

Since the band terminates at a LHS cornplex piece. et-entually n-e have o ( - Y ~ + ~ ) = 0. so the band has length A. If a cuwe in 7 traverses the band for part of its length. the contribution of the band to the curve's homology

2.3.1 The f :l Case

Theorem 2.15 Suppose ni = ni G O mod 9. and nz = nn = n i = na = O. Then p E {1.3. . ... n - 1) is a regular value. i.e. L Ls a TITI h o t . if and only if ( n . p ) = 1.

Proof. Consider the band starting at label O on the LHS cylinder. \Ié are assuming (n . p ) = 1 so the -Yi senes is -p. 2p. -3p. . . . . n p . The lengt h of the band is n - 1. so every even-labeled rectangle is used in the band. By relabelling, i.e by subtracting p from ail labels. it is clear that al1 the odd-labelied rectangles are used in the band attached to the RHS cylinder. so there are no annuli. hence p is regular when ( n . p) = 1. n For canonical curves we take LL parallel to a and 3 as in the folloming diagrarn:

Figure 2.7

43

Xote that 3 contains one (r. s)-connection, as in Figure 1.4. Then .-3? = l.J4 = O since there are no b'- or c'-connections. and J3 = - v ( R . p ) where

mhere a is the sign function.

is a Seifert matrk for L by Lemma 2.13 . But we can replace ( 1 . . s) by ( r - dlp, s - 31q). and so

is also a Seifert matris for L .

Proposition 2.16 If L zs a 1:1 Tl Tl h o t then L has Alezander polynonaial

where 6 = - ~ ~ ~ q p ' ~ ' . In particular. if v ( n . p ) = 0. then lr ( t ) is trivial for al1 valves of p. q . p' and q'.

Remark 2.17 There are (apparentiy infinitely many) d u e s of ( n . p ) for which v(n. p ) = O. For example O = ~ ( 8 . 3 ) = 48.5) = ~(16.7) = - - - = ~ ( 6 4 , 2 5 ) = . - = ~ ( 1 1 2 . 2 3 ) = - - .. For each such value of ( n . p) n-e get an infinite famiiy of knots with trivial Alexander polynomial parametrized by (p, q) and (p'. q'). Later (Theorem 2.31) we show that these knots are never trivial unless one of p. q . p f . q' is equal to O.

Proposition 2.18 The first homology group of the double cover of a 1-1 TITI h o t is cyclic.

Proof. L e t V = M + - M T . ThenYisapresentationmatrkforthe first homology group of the double cover of L (see [RolSO].p. '112). W e have

Now ps -qr = 1 so ps+qr = 1 mod 2. so gcd(2pq.pstqr) = ,ocd(pq. ps+qr) . Suppose m is a prime which divides both pq and ps+qr . .&sume mlp. Then rn /q since gcd (p. q) = 1 and m Jr since gcd (p. r ) = 1. T hus m l(p.5 + q r ) and so gcd(2pq, ps + qr) = gcd(pq, ps + qr) = 1. The greatest common divisor of the entries of V is 1. which shows that the first homology group of the double cover of L is cyclic. O

Remark 2.19 The h o t 946 is a genus one knot but it cannot be a 1:l T l T l knot because the first homology group of its double cover is 2/32 f 2/32. Later we shall see that 94s is a 2 2 T1T1 h o t .

2.3.2 The 1:2 Case

Consider the band starting at label O on the LHS cylinder. Note t hat L - ( U ' )

~(6') r n i mod n. Hence the sequence Si is congruent to

-p. ni - n i - 3p.9~2; ++. . . .. k ( n ; + 'Zp) mod n

with k ( n ; + 2p) O mod n. The length of t his band is 2E - 1. The total number of even-labelleù rectangles is n - 1. Thus if k < n j2. not al1 the even-labelled rectangles have been used so t here exists at least one annulus. Hence a solution for II- of

irnpfies that p is irregular. Conversely. if p is irregular then 3 contains a t least one annulus n-hich

can be labelled

z. -T.Z + n; +?p.. ... z + k(n; + 2 p ) mod n

mhich implies + 2 + k(n; + 2p) rnod n. Thus p irregular implies the existence of a solution for k of (2.2). Hence p is irregular if and only if there is a solution for k of (2.2).

Suppose n = mq where q is an odd prime. Let

1 1 1 p, = q + 5 ( q + l)n; and kq = ~ r n < -n. Then - - 2

so pq is irregular, and so is p = p, + 2rq. r E 2. We wish to show that ail the irregular p's can be found this way. i.e. by going through all the odd primes dividing n.

Theorem 2.20 Suppose n l = ni = ni G O mod 2. n2 = n~ = n i = O . Let

S = ( p : 1 5 p 5 n - l .gcd(nl . n2. n3. ni. ni. = 1.p irregular).

Q, = { q : q zs an odd prime whzch divides n ). Let

S'= U { p : 15 p 5 n - 1. (2q)l c l a n

Then S = Sr.

Proof. It is clear that S' c S. Suppose p E S. Then there esists an integer k 5

(P - p q w

iuch that k and p sa (3.2) . We consider two cases: (i) t here is an odd prime q such t hat qj n and ql (ni + 2p) . (ii) there is no odd prime which divides both n and ( n i + 2p).

(i) ni + 2pq = ni + 2q + qn; + n; = n + q(n; + 2) so

41b; + 3,) = ~ P ( P - P,)

ql(p- p,) since q is odd

(3q) 1 ( p - pq) since p. pq are odd.

a p E S '

(ü) We wnte n = ISqf1 . -q;'< . k(n; + 2p) = M n nhere the qi are tiis- tinct odd primes. and we can assume without loss of generaiity chat (k. W ) = 1. Since none of the qi divide (n i + 2p). (qiL - - qFt ) 1 k. Then n/E = 2' for sorne r 2 2 since Ir < f n and ( B . -11) = 1 . Thus 41 n and therefore 4 n ; by Proposition 2.1'2. But then

But this is impossible since ni O mod 4 and 2p G 2 mod 4 since p is odd. Hence case (ii) does not occur? and the theorem is proved. 0

Corollary 2.22 In the 1.2 case there are a h y s at least two regular p 3 .

Pmof. By the previous coroilary tve can assume there is at Ieast one odd prime dividing n . since n 2 6. Suppose Q, = { q t . . . .. q r } . Then by the Chinese Remaiader Theorem there is a solution to

in therange 1 s p <q l - - -q , . andanother in therange ( q l = - . q , + 1 ) < p < 2ql - - - q r 5 n. O

Example 2.1 Suppose L is a 12 link Mth n = 8. n; = 4 = n!,. By Corollary 2.21, al p's are regular. We caiculate 33. 34 for p = i. 3.

p = 1 Starting at O on the LHS cylinder the -Y; and 1; are

Thus 3 = 3a + b - ?a' - 2b' -.1. 33 = -2. 34 = -2. Since 6i = -pq(,33p'+,34r')(33q'+34s') byequation 2.1. ifme take p' = 1. r' = -1 m p ' = -1. r f = 1 o r q f = 1 . s' = -1 or qf = -1. sf = 1. wesee t hat 8 = O so n-e get infinite families of knots n i t h trivial -liesander polynornial pararnetrized by (p. q ) .

p = 3 Starting at O on the LHS cylinder the -Y; and 1; are

Thus 3 = -3a + b + ?a1 - 2b' 33 = 2 . 4 = -?.and n-e can take pf = 1. r' = 1 etc. to get infinite families of knots n-ith trivial Aiexander polynornial parametrized by (p. q ) .

2.3.3 The 2:2 Case

Consider the band starting at label O on the LHS octagon. Xote that ~ ( a ) G

$ ( b ) a ni mod n and r$(af) to(b') ni mod n . Hence the series Si is congruent to

-p, n;+2p , n l -ni -3p . -n1+2n; ++. . . .. ( k - 1 ) ( - n i ) + k ( n ; +?P) mod n

If 3 contains an annulus it can be Iabelled

I r. - t + n l . r - n l + n l + 2 p . -++hl -ni -3.. . .. - - + k ( - n i +ni +31) mod n

where we begin the labelling at an edge of a RHS piece. The annulus has lengt h 2k and the joining band has Iength 2k - 1. There are ( n - 2) odd- labelled and ( n - 2) even-labelled rectangles in total. Suppose n-e have a solution for k of

1 k ( - n i + ni + 2 p ) G O mod n. 1 <_ k 5 -n

4

Then it follows that we have a solution for k of

1 ( k - l ) ( - n i ) + k ( n : + 2 p ) = O m o d n. 15 k 5 -n

4

and this implies that all of the joining bands on the octagons have length strictly less than ' l ( + n ) - 1 = i n - 1 (relabefing to see this for the odd- Iabelled bands), so the the joining bands account for strictly Iess t han 2 ( a n - 1) = n - 2 even-labelled rectangles and strictly less than ( n - 2 ) odd-lab~lled rectangles. and so there must be annuii.

Conversely. suppose an annulus e-xists. and suppose the shortest an- nulus contains a t least i n rectangies. .&sume this annulus consists of even-labelled rectangles. by relabelling if n e c e s s a . T hen clearly t here can only be one even-labelled annulus. so all the other even-labelled rectangles are contained in joining bands. one of which must have length Iess thari $ [ ( n - 3) - i n ] = i n - 1. which implies the existence of a solution for k of(2.3). But the e.xistence of a solution for (2.3) together n i th esistence of an annulus implies existence of an annuIus Rith length strictly less than

1 2(,-n) = i n . which contradicts our assumption. Therefore esistence of an annulus implies existence of an annulus Nith length strictly less than ( $n).

*

and this implies e-uistence of a solution for k of (2.3). Hence we have shown that p is irregular if and only if there is a solution

for k of (2 .3 ) . Suppose n = mq where q is an odd prime. Let

Then kq(-nt +ni + 2pq) = $mq(-n2 + n; +z) O rnod n. so pq is irregular. and so is p = p, +2rqo r E 2.

Theorem 2.23 Suppose ni = n2 = n; = nî = O mod 2. nn = O = nj. Let

Q, = { q : q is an odd prime which divides n } . Lct

Proof- The proof is simiiar to t hat of Theorem 2.20 0

Corollary 2.24 Under the hypotheses of the pwuiovs theorem. there are always ut Zeast two regular p's. and if n is a power of 8 al1 odd p-s are regular.

Proof. The proof is similar to those of Corollaries 2.21 and 2.2'2 0

Example 2.2 Suppose L is a 2:- ünk with ni = 6 = n\.nz = 2 = n i . p = 1. By Coroilary 2.24 p is regular. Starting at O on the LHS octagon the -Y; and Er are X: -1 S Y: -6' O

Starting at n - 1 on the LHS cyiinder the Si and 1: are X: 5 -6 3 -4 1 -2 Y: a' -a a' -a a' O T hen

If we choose

( p . q. r. s) = (2.311- - 1.3.3k - 1) (do q'. r'. s') = (3.3m - 1.1. rn)

then the second factor above Nill be O, so we have an infinite family of knots with trivial Alexander polynomials parametrized by k and m. Obviously.

t here are infinitely many ot her choices we could make for (p. q. r. S . p'. q'. r'. .sr ) in order to get such a family. for example we could choose

( p . q . r. s ) = (Li. llk + 2.2.5.3.5k + 3)

p'.q'.r'.s') = ( 1 9 . 1 9 r n f 3 . 6 . 6 m f l ) .

Remark 2.25 We do not have theorems analogous to Theorems 2.20 and 3.33 in the 1:3. 3 2 and 3:3 cases. It is not true that alI odd p's are regular if TZ is a power of 2 in these cases. For esample. in the 1:3 case n-ith equation 16 = 8 + 4 + 4. only 3.5.11 and 13 are regular values for p . However. it appears that there are always at least two regular vaiues in the range l S p < n - l .

Example 2.3 Suppose L is a 1:3 Link nrith ni = 10.n; = 4 = 4. nk = 2.p = 5. The diagrams corresponding to this case are

Figure 2.8

This gives the following diagram for 3:

Figure 2.9

This shows that L is indeed a TlTl knot. so p = 5 is regular. l e can reatl the band starting at O on the LHS cylinder from the diagram. or use the algorithm as follows: X: -5 -8 3 -4 -1 -6 1 -2 -3 10

From this we get Y : -bf -a a' -a -8 -a a' -a -6' O

and û = - p q ( 3 p f - 3rf)(3q' - 3s'). Ifwe choose p' = 1. qf = m . rt = 1. sr = rn + 1 for rn E 2. then for any choice of (p. q) ive have 8 = O so AL ( t ) = 1. Later we shall see that al1 but a finite number of these knots are non-t riviai.

Example 2.4 Suppose L is a 3 2 link with n = S. nl = 4. n;, = '2. n~ = 2. ni = 6. n; = 2. p = 1. The diagrams corresponding to this case

Figure 2.10

This gives the following diagram for 3:

Figure 2.11

This shows that L is a TlTl knot. so p = 1 is regular. From the diagrani we can choose canonicai curves a.. 3 with

,3 = -2a - b + 3a' - 6' . thus = -5. y2 = -1. = 2. y4 = -3 - and

8 = -(-5pp' - pr' + 2rp' - 2rr') (-jqq' - qs' + 2sq' - 2s~ ' )

If we choose

( p , q. r. s ) = (1. k - 1.1. Cc)

( p . . r . s ) = (1. m + 2. -1. -rn - 1)

for example. we get 9 = O so ( t ) = 1 for an infinite family of knots parametrized by k and rn . Again. only finitely many of these knots is trivial.

Example 2.5 Suppose L is a 3:3 link Nith n = 6. nl = n;! = n3 = ni = ni = ni = 2. p = 1. The diagrams corresponding to this case are

b'

Figure 2.12

This gives the following diagram for F:

Figure 2.13

This shows that L is a TlTl Lnot. so p = 1 is regular. From the diagrani we can choose canonical curves a. 3 with

If we choose

for example, we get 8 = O so AL ( t ) = 1 for an infinite farnily of b o t s parametrized by k and m . As before, only finitely many of t liese knots is t rivid.

2.4 Sutured Manifolds

The theory of sutured manifolds was developed by D. Gabai [Gab83].[GabST] as part of the t heory of foliations of 3-manifolds. Later 11. Scharlemann [SchS9] gave a defoliated version which is what we s h d use here.

Definition 2.2 Let -bI be a compact oriented irreducible 3-manifolci. For S a connected orientable compact surface define t - ( S ) = ma.s{O. - \ (S) } where y (S) is the Euler characteristic of S. If S is not connected define

Y- (S) = C{'Y- (~ i ) : Si is a component of S}

For ... a subsurface of BM and 2 E H2(-LI. -V) define the Thurston norm x ( z ) to be

X ( Z ) = rnin{y-(S) : (S . aS ) c (LI . . . ' ) . [S.BS] = = in H2(J1.-1-)} .

Definition 2.3 Let S be a surface properly embedded in -11. and let .\;(as) be a regular neighbourhood of 8s in d M . (S. BS) is taut if S is incompress- ible and x([S. 8 S ] ) = k- (S) where x is the Thurston norm on Hz(-Il. .L-(BS) ).

Definition 2.4 .-1 sutured manifold (-LI. 7) is a compact oriented 3-manifold LM together with a partition of 8M = R+ U R- as the union of two surfaces R+ and R- dong their common boundary. a collection 7 of simple closecl curves (scc's) called the sutures . R+ is oriented so that its normal vector points outward, and R, is oriented so that its normal vector points inn-ad.

Let the surface (S. dS) be properly embedded in the sutured manifold (:LIt dM. y). with d S in generd position ni th respect t o ;. The orientation of S induces orientations on S+ and S- where %QP(S) = S+ U S- U (as x 1). Let Mt be the manifold obtained by decomposing .II dong S. -1ny point in 6Mt wis either a point in DM or in S*. Hence any point in 651' n-hich is neither in OS+ or âS- or y has a natural normal orientation. Let y be a point in 8 M t . If y h a . a neighbourhood I: in d M t such that every point in Li - BSI has the same natural normal orientation. give y that orientation. If y has no such neighbourhood, regard y as lying in a suture. Thtis -11' becomes a sutured manifold with sutures y', and we Say

( M . 3 (-IIt. f )

is a satured manifold decomposition .

Definition 2.5 -4 sutured manifold (M. y ) is taut if Ab1 is irreducible and R+ and R- are both tau t surfaces in M. -4 taut sutured manifold decom- position is a sutured manifold decompwition (JI. a,) 4 (W. 7 ') such r hat bot h (M. y ) and ( A l ' . 7') are taut sutured manifolds.

Definition 2.6 Let S be a Seifert surface for a Link L in S3. Let the sutured manifold (~b1. y ) be the exterior of L wit h *{ = 0. Let S' = S n M. and let ( M'. 3 be the su t ured manifold obtained by decomposition dong S'. We c d (Mt. y') the complementary sutured manifold for S.

The main theorems we shall need are the foUowing:

Theorem 2.26 (Gabai,Scharlemann) Suppose ( J I . 7 ) is a sutured man- ifold and ( M . 7 ) 4 (LW. 7') is a sutured manifold decomposition. If ( JI1. ; l )

2s taut. then (M. y ) is taut.

Theorem 2.27 (Gabai,Scharlemann) Let S be a Seifert surface for a h o t in S3. Then the following ore equiualent:

(i) S is a minimal genus Seifert surface.

(ii) The complementary sutured manifold for S is a taut sutured manifiid.

Theorem 2.28 A Tl Tl h o t L ïs non-trivial if and only if ( T I - . L ) and (W'. L ) are both taut sutured manifolds.

Poof . Suppose (. L) and (W. L ) are both taut sutured manifolds. Let = S3 - N(L) where .b*(L) is a regular neighbourhood of L in S3. Af is irreducible since L is a h o t . L cuts 7 into two once-punctured tori. ri, and &. Let Si = I/;. - N ( L ) , i = 1,2. Regard (.JI.:) as a sutiired

s' manifold Nith A, = 0. Decompose ( M y ) by Si. (JI.?) 4 (-\it.:') Sesr decornpose (W. 7') by S2, (LW.-,') -?t (51". y").

If we choose orientations for Si, Sz and & i ( L ) according to the folloming cross-sectional diagram

Figure 2.14

then (&Pt. 7") is the disjoint union of the two genus-two handlebodies II- and W', and the sutures 7" are two copies of L. one on the boundary of each handlebody. Then by Theorem 2.26, (Mt. y') is a taut sutured manifold. and so SI is a minimal genus Seifert surface for L by Theorem 2.37. Since SI has genus one. L is non-trivial.

Conversely. suppose one of (W. L ) ,(W. L) is not a taut sutured manifold. Say (W. L).Then there exkits an incompressibIe surface Q properly ernbedded in W with x,(Q) < y,(Vl) = 1. so k-(Q) = 0. and

Since \- (Q) = O. Q consists of disks. annuii. and tori. But &([QI) = d.([l/i]) = [LI where 8. is the homoiogy boundary homomorphism. and this irnplies that Q has an odd number of boundary components. so Q contains a diik with boundary contained in 3(WIi) nN(L) . Then L is the boundary of a disk properly embedded in W. and therefore trivial. O

Corollary 2.29 A Tl Tl h o t L is teuial if and only if L bounds a disk in or in W.

Proof. Certainly. if L bounds a disk it is trivial. For the converse. take the given disk component of the surface Q in the proof of Theorem 2.28. Ci

Remark 2.30 Let be a trivial knot in S3 which is a separating curve on a Heegaard surface S of genus g. Then for g 5 2. IC bounds a disk either inside o r outside S. Does this hold for g > 2'; I t is not true for non-separating curves. as can be seen from the (1.1)-cume on the torus. for esample.

Theorem 2.31 Suppose any ( j , k)-connectlun belongzng to a TlTl h o t L 2s such that j # O and k # O . Then L is non-triuial.

Proof. If L is trivial then L bounds a dkk in either PT- or TT". by Corollary 3.29. Suppose L bounds the dislc CI in W. Let D be the di& defined in Chapter 1 which decomposes W into the genus-one handlebodies W1 and th. Since D and C- are both d i sk . and dD and 9L- are borh contained in aW-. we can assume their intersection is a collection of properly embcdded arcs. Let < be an outermost arc in C'. and let q be an arc in

= L so tha t f(J Q bounds a disk in CT which does not meet D:

Figure 2.15

Then q is an arc contained completely in LHS or in RHS. i.e. q is a con- nection, say a (2, k)-connection. belonging to L. Then q is a (1. k)-toms knot on the boundary of Wi;. But since j and k are both nonzero. CU ri cannot bound a disk in Fr:. Hence there is no disk in kt; wit h boundary L, cl

In examples 2.1 to 2.5 we obtained famaes of TlTl knots 6 t h tritlal Alexander polynomials. Oniy finitely many of the knots in each family have (j, k)-connections with j = O or k = O. ,JU the other L o t s in these families are non-trivial by Theorern 2.31 .

In particular. say L is a 1:l knot with n = 8 and p = 3. If we choose nonzero values for p. q,p'. q' then L is a non-trivial knot with trivial -llesan- der polynomid.

2.5 Some types of TlTl Knots

Example 2.6 (a generalization of classical pretzel knots with odd numbers of haif-twists) suppose L is a 2:2 iink a;ith ni = 2. n2 = n - '2. n; = n - 2, = 2 and p = 1. The diagrams corresponding to this case are:

- - 1

n - l , ,

n + l r

Figure 2.16

This gives the foliowing diagram for 3:

/ \ - /

n - 3 rectangles

Figure 2.17

We have used 271 - 4 rectangles. which is ail of t hem. so this shows that L is a TITI h o t and p = 1 is regular. From the top component n-e can choose canonical curves CL, ,d with

Suppose p is odd. Choose

I I (p'. q', r'. sr ) = (;(1 - p ) . ,(3 - p). -p. 2 - p )

Then the first f ac t~r in the expression for 0 is O. so we get a family of knotr with trivial Alexander polynomial paramatrized by m E 2. al1 but a finite number of which are non-trivial. by Theorem 2.31 .

Suppose p is even. Choose

Then the first factor in the expression for û is O. so we get a family of knots with trivial -Ilexander polynomial paramatrized by m E 2. al1 but a finite number of which are non-trivial.

Xow choose integers A. A' and let

( p . p. r. s) = (1. A. O. 1)

( p q r ) = (0.1.-1.X').

Then L is a (P.Q. R)-pretzel knot Nith P = 2A - 1. Q = n - 1. and R =

2X' - 1. This can be seen from the fouonring diagram:

Figure 2.18

The vertical dotted line represents the curve O on 7. L intersects (3 in 2n points where n = Q + 1. in the order shonrn. In t his case

p p p ' + P r ' - ~ r p l + p r r ' = -1 and

pqq' + qs' - psqf + pss' = pX + XX' - p + / l X 1

SO O = p ( X + X f - l ) + X X 1

1 = - ( P Q + Q R + R P + l )

4

Thus I lL ( t ) 1 when PQ + Q R + R P = -1. for esample when (P. Q. R) = (7.5, -3). We cannot show that this knot is non-trivial using Theorem 2.31. but this has been shon-n in [Sei311 and [Gab86].

Example 2.7 Let L be the 1:l link Nith n = 2. p = 1 and ( p . q.p'. q f ) = (1, q, 1 ) It is easily seen that this is the turtst h o t with q twists ( see [RolSO].p. 112). We generatize this slightly to let (p .q .pf .q' ; = (1.q. 1.2'). and cal1 this the (q, 9')-double-twist Lnot. It has a genus one Seifert surface as in the foilonring diagram (for q = 2 = 6):

Figure 2.19

Theorem 2.32 An alternating b o t of genus one is a TITI ,hot.

Pmof. It is shown in ([Mur65], see also [hIur96]) that an alternating knot of genus one is either (i) a (P. Q. R)-pretzel h o t a i t h P. Q. R ail odd and the same sign, or (ii) a (q, qf)-double-twist h o t with qq' > 0. O

2.5.1 TlTl knots with 10 or fewer crossings

The following table describes the 14 genus one knots with fen-er than 11 crossings as TlTl knots:

knot - 3 1

41 - 52

61 - - ' 2 - ' 4

- 8 1

83

- 92

9 5

- 935

946

101 103

pp --

Table 2.1

The 1:l presentations in the table are (double-)twists. and the 2:- pre- sentations are pretzels. The twist presentations can be verified direçtly froni a table of knot diagrams. as in [RolgO]. Values of P. Q. R for the pretzel-hot presentations are @en in the table belon;.

knot - ( 4

83 95 935

946

103

Table 2.2

Notes:

a) P.Q. R > O implies that the knot is alternating. An alternating h o t with (P + Q + R) crossings and the given t-alue of 9 must be the given knot .

b) .A h o t with at moût (IPI + IQI + IRI) crosings and the given value of 19 must be the given knot.

c) There are illustrations of the 946 knot as a (3.3. -3)-pretzel h o t in [CFE]. p. 128 and [Ro190]. p.399.

d) The 935 and 943 knots do not have 1:l TlTl descriptions. since the first homology groups of their double covers are not cyclic (Proposition 2.18).

2.5.2 Genus one knots which are not TlTl

Definition 2.7 The canonical genus g,(K) of a knot I\: is the minimal genus of all Seifert surfaces for A- which can be obtained by applying the Seifert algorithm t o diagrams of I<. The put genus g b ( l ï ) is the minimal genus of all Bat Seifert surfaces for K. i.e Seifert surfaces n-hich can be embedded in the plane except in s m d neighbourhoods of crossings. where they appear as in the folloning diagram:

Figure 2.20

It has been shown in [Hir95] that gb(K) = g,(K). Sow let gH ( L ) be the genus of the Heegaard surface of minimal genus in which a minimal Seifert surface for L can be embedded. For exampie, g ~ ( u n k n 0 t ) = g (unknot) = 0. and if L k a TlTl knot, then g(L) = 1 and ~ H ( L ) = '2. Clearly g H ( K ) 2 2g ( K ) .

Given a flat Seifert surface for K' of genus m. it is obvïous how to con- struct a Heegaard surface which contains K. and which A- separates into two once-punctured genus m surfaces. But this does not imply g a ( K ) 5 Zg,(Ii). since for some b o t s g ( K ) < g,(K). If g ( K ) = g,(K) (this is true for alter- nating knots. for example). then we have g H ( K ) = 2g(K).

Question Are there genus one knots for which g~ > 2 . in other n-ords. are there genus one knots which are not T1Tl knots'? More generally. are there b o t s for which g~ > 2g? g~ > 2g," Can g~ - 'Lg be arbitrarily large? [These questions are answered affirmatively in .Cloriah. k:. On the iree genus of knots. P m . Amer. Math. Soc.. 99:.37.3-379. (1988).]

Remark 2.33 Let K o be a trefoil knot. I< = Ko#I~o#I 'o#I~o . a connected sum of four trefoils, and Kcv the Whitehead double of K. The -obvious- genus one Seifert surface for Kcv cannot be embedded in 7 because it carries a copy of K. which we shall see later (Remark 3.3) cannot be embedded in 7. So Krv is a candidate for a genus one knot which is not TITI. but of course K t v may have a different genus one Seifert surface which can be embedded in T.

Chapter 3

T2 Knots

We show that a non-separating knot on 7 has one of seven canonical forms for 3. These can be used to give group presentations for these knots with four generators and t hree relations.

In the 1:l case. a T2 knot has a simple form for its -Alesander polynomial. which is a product of two toms type Alexander polynomials. and a slice type.

Among the T2 knots are the g lb l knots. which include the 2-bridge knots, and Berge's double-primitive knots.

3.1 Decompositions of 3

Lemma 3.1 If L is a T2 h o t , then 3 may be decomposed into piecea nc- cording to one of the following diagranzs:

1:l case

Figure 3.1

4 u and u are bands

# v is the reverse of u

1:2 case

Figure 3.2

L' is the reverse of u

u: is the reverse of itself (self-reversing)

1:3 case

Figure 3.3

0 t. is the reverse of u

w and r are self-reversing

2:2 case

Figure 3.4

v is the reverse of u

0 UT and z are self-reversing

3:2 case

Figure 3.5

c is the reverse of u

4 it.. z and g are self-reversing

Figure 3.6

0 v is the reverse of u

4 z is the reverse of w

h is the reverse of g

Figure 3.7

t? is the reverse of u

u. =. g and h are self-reversing

Proof. We have six cases:

case 1:1 The two labels on the LHS cylinder in 3 differ by n. so t h e band beginning at one of these edges is the reverse of the band terminating at the ot her edge. by the reversing lemma 1.4 . Since 3 is connected. we get the given diagram.

case 1:2 Since 3 is connected. the LHS cylinder is joined to the RHS oc- tagon by two bands which are the reverse of each other. lloreover. these bands meet opposite edges of the RHS octagon. since opposite sides have labek differing by n mod 272. Then the rernaining sides of the RHS octagon. which ais0 differ by n niod 2 n . must be joined by a

self-reversing band.

- For the remaining cases we make use of the following observation: since L does not separate 7. L is not the boundary of a subsurface of 7 . Hence [LI # O in Hi (n.

case 1:3 3 contains the following subsurface:

Figure 3.8

The remaining four labelled edges must be joined by two bands. If we join the edge labelled i to the edge labelled j by one band. and the remaining two edges by another. we get S4, not T2.

Suppose we join i to ( j + n ) [frorn now on it n-ili be understood that

the edges are labelleci mod'>n]:

Figure 3.9

By Lemma 1.4. r is the reverse of W . so [z] = [ w ] in Hi (n . Take the outer boundary of this surface for L. Then

[LI = [el + [r] - [<'l - [il = O in H i ( n

which is a contradiction. Hence we must join i to ( i + n ) . and J to

(.i +n).

case 2:2 3 contains the following subsurface:

n

Figure 3.10

If we join i to j. and the rernaining two labelled edges by another band. we get S4.

Suppose we join i to j + n:

Figure 3.11

.As before, w is the reverse of r so [u*] = [r]. For L take

L = :y-' vgr-l W-lcu-L ,f

Then [LI = ([O] - [ u ] ) + ([r] - [ w ] ) = O. mhich is a contradiction. Hence wemust job i to ( i + n ) , and j to ( j + n ) .

case 3:2 3 contains the fouonring subsurface:

Figure 3.12

If we join i to j we must join i f n to j + n by Lemma 1.4. and ive get S3. Similady. if we join i to h we get S1.

Suppose we join i to j + n:

Figure 3.13

.As before, UT is the reverse of r. If we take the outer boundary of thk surface for L we have

a contradiction.

Suppose we join i to h

Figure 3.14

Here w is the reverse of = and g is self-reversing. We can take L = w<+..-Lz-I f i r - L n <

Then [LI = ([v] - [u]) + ([u] - [z]) = O. a contradiction. Thus n-e must join i to i + n. By the same reasoning we see that j is joined to j +- n . and h is joined to h + n.

case 3:s Since 3 is connected there is a (possibly empty) band joining a LHS hesagon to a RHS hesagon. Thus 3 contains the following subsurface:

Figure 3.15

Suppose f is joined to e. Then f + n is joined to E + n and we already have too many boundary components. Similady. f is not joined to j .

Xow suppose f is joined to i. and so f + n to i + n. If we join j to t

we have TITI. Suppose we join j to j + n. and so E to E + n:

IL* I L' Y

For L take

T hen

Figure 3.16

a contradiction.

Thus if f is joined to i . we have j joined to e + n as in Figure 3.6 .

Next . suppose f is joined to e + n:

Figure 3.17

Ifwenowjoin i to j+n .andso i + n to j . weget S4.sojoin i to i+n:

Figure 3.18

For L we can take L = fh7'-Lg-L

so [LI = [hl - [g] = O so f is not joined to e + R .

Ne-. suppose f is joined to j + n:

Figure 3.19

Figure 3.20

88

so [LI = ( [V I - LU]) + ([g] - [hl) = O. Therefore f is not joined to j + n. and similariy f is not joined to i + n.

Xow we have f joined to f + n. Suppose we joini to E + n :

tr

Figure 3.21

so [LI = ([VI - [II]) + ([w] - [z]) = 0. so i iS not joined to e + n . Similarly. i is not joined to j + n.

Thus we have i joined to i + n and j joined to j + n. as in Figure 3.7 O

Recall the definitions of t , P, Q, Pr and Q' (Definition 1.7). t is a genera- tor of the first homology group of the complement of L' and [xi] = tQ. [x2] = t P , [yi] = tQ', [y2] = tP' in XI (S3 - L).

Remark 3.2 The diagrams of Lemma 3.1 can be used to present the group of a T2 h o t with four generators and three relations. We have S3 - L = (W - L) U(W' - L) and (W - L) n(IV - L) = 3. xl (3) is free on three generators. We can choose scc's on 3 to represent these. For the first one we choose k parailel to one of the boundaries of 3. and so also parallel to L. The other curves which me c d m and 1 will depend on the diagrarn. Having chosen a base point on O - L. for a scc e in 3 let E 1 be the word written in xl and y1 which e represents in rl(VV). and let e;? be the word n-ritten in xz and y2 which e represents in ri ( W') . Then HI (S3 - L ) has the presentation

by the Seifert-van Kampen Theorem.

Remark 3.3 Since the group of a T2 knot L has a presentation wit h four generators and three relations, it follows that the fourt h elementary ideal of L. and therefore its generator &(t ) . is trivial. We can use t his to show that certain knots are not T2. For example. let E be a trefoil knot. and let L = k#k#E#k. Then &(t) G t2 - t + 1. so L is not T2.

Question Can we use the group presentation to find a prime h o t which is not T??

3.2 Alexander Polynomials of 1:l T2 Knots

Lemma 3.4 Let G be n h o t group. so G/G' Z Z 2 ( t ) . Let o be the canon- ical homomorphism G -+ G/Gt extended to an algebra homomorphzsna ZG + Z(G/Gf). Let G = ( S 1 R) be a presentation for G . and let c* be the canonical hornomorphisrn ( S ) -t G extended to an algebra homornorphi,crn Z((S)) + ZG. Suppose x . y E S with ox = ta . oy = t3. cr # 0. 3 # 0.

Let u = z'l y61 - - .xzm y6m and let r be the reverse of u . i. e. c = y6mlcm . . and let p = (rl + . - - + s,)a + (di + . . - + 6,I.j. Let B( t ) . B f ( t ) . C(t ) , Ct(t) be defined so that

iohere & ts the Fox derivative urith respect to x. Then

(i) B'(t) = t W ( t - l ) + t P - 1

(ü) C ( t ) = - B ( t ) + tp - 1

Proof. Let

Sirnilarly. C ' ( t ) = PC(t-') + t P - 1. Sow = f ( x . y)/(y - 1) ahere

so C ( t ) = -B( t ) + t p - 1 which proves (ii).

Finally C t ( t ) = t P c ( t - ' ) + tP - 1 = t ' [ - ~ ( t - l ) +t-'- '- 1 ] + t P - 1 by (ii)

= - t P ~ ( t - l )

which proves (üi) . a

T heorem 3.5 Let L be a I r 1 T2 h o t with ( p . q ) -connections in L HS and (p i , 9')-connections in RHS- Then L is a band surn of two disjoint t o m h o b , and the Alexander poiynornial of L ii of the f o m

uthem A,.,(t) is the Alexander polynornial of the (m. n)- toms h o t .

Proof. TITe can see that L is a band sum by inspection of the connection diagram.

If any of the numbers p,q.pt.q' is O . then L is a torus h o t . and the conclusion is obvious. We assume that none of p. q. pi. q' is O .

In the diagram below

Figure 3.22

92

we may assume that the upper cyiinder is from LHS. and the connections a = ( p , q ) and b = ( r, s ) are oriented as shown. Then

a'= a'. 3'= b&. where d = 3 3 . Take

as canonical curves on 3 with base point in the segment marked 6 t h the double arrow. Then we have

By t-irtue of tht? relation ml = ma, we can wnte

Since [k] = [a] + [a]. we have

Define Bi ( t )? B2 ( t ) so that

We claim Bi ( t ) = B z ( t ) . For u is a word in { a . a'} and

Thus if ..Y is a word in { a . a'). then ,Y" XgW. Moreover. if u = -'Y. then

and our daim follows from induction on the length of u. Let B( t ) = Bi ( t ) . Then we have

Then LW is a presentation matrix for the -4Iesander module of L. Since a, = O = h. we can obtain the -Alexander polynomiai by deleting the am

second colurnn and taking

by section 1.3 of [Tor53], which impües

where

Hence

where f ( t ) = B( t ) - t p . a Remark 3.6 Fox and Miilnor [FM661 showed tha t for a ribbon b o t (nhich is a band surn of disjoint trivial knots). the Xesander polynomial has the form A(t) = f (t) f (t- ')- Terasaka [Ter591 showed that for a h o t which is a band sum of a h o t Iic and a trivial knot. the Alexander polynomial lias the form h ( t ) = rlK(t) f (t) f ( t - l ) . Inspection of the connection diagram shon-s that a 1:l T2 h o t is a band sum of two disjoint torus knots. For esample.

in the connection diagram for the case n = 5.p = 3 belon-. the two disjoint toms knots are shown with dashed lines, and the core of the band is marked with a dotted iine:

Figure 3.23

Remark 3.7 We may also use the connection diagram to find f ( t ) . Take an arc joining the two disjoint torus knots (Le an arc parallel to the dotted

oc' Line) for u. Then since f (t) = B ( t ) - t p and (2) = .%.. w can apply the Fox derivative to find B(t ) . and calculate p directly from u. From the diagram above. for example. we get

tW + t~ '9 ' - t ~ + ~ J 9 ' - 3 and A(p,s) ( t ) A(pt,qt) ( t ) (tPl + tptqt - t ~ + ~ ' ~ J - 9) - p t ~ + p t q ' - - tm - t ~ J q ' + 1)

For example, the 1:l knot wit h n = .5. p = 3 and (p. q. p'. q') = ( 1.1.1.1 ) has Aiexander polynomial

3.3 Genus One Bridge One Knots

Definition 3.1 Let T be a genus one Heegaard surface in S3 containing the distinct points x and y. Let a and ,3 be simple arcs embedded in T . with end points x and y. If we move the interior of a into one component of S3 - T . and the interior of 3 into the other component of S3 - T by isotopies. ne get a knot in S3. -4 knut which cac be presented in this manner is c d e d a genus one bridge one (g lb i ) h o t .

Proposition 3.8 The group of a glbl knot has a presentation with two generators and one relat2on. One of the genemtars as a meridian of the h o t .

Proof. Let To and Tl be pa rde l ton standard- embedded in S3. Assume the knot k consists of one simple arc on each of To. Tl and tn-O %traightY joining arcs between them. Let C be the component of S c Tl containing To . and C' the ot her component of S3 - Tl . Let -4 = C - k. B = - v - k .

Xow B is just a soiid torus. so ;rl(B) = (m) 2 Z . --ln B is a once- punctured torus (punctured by a simple arc). so r i1 (-4 n B) = (L. .If) 2 Z*Z. A is a solid toms minus a properly embedded simple arc. so rl (-4) = (1. a ) 2 Z * 2. where 1 is represented by a longitude of the solid torus. and a is represented by a Ioop around the embedded arc. n is then a rneridian of the h o t .

We may now use the Seifert-van Kampen Theorem to present the group. Suppose that L and -Id are simple closed cun-es whose homotopy classes together generate q ( A n B ) . Suppose also that L is an ( g . hl curve on T (meaning g A-longitudes and h A-meridians). and M is a ( u . c ) curve. and that

Then we get the presentation

where L. :Id are now considered as words in the generators 1 and n . Xow

It is possible to proceed to a one-generator presentation: the folloning argument was pointed out by J. McCool.

Let GvVh = ( L . MILV = M h . LM = ML). 2 Z since it is a pre- sent arion of the first homology group of the cornplernent of the ( c . h )-toms h o t . Since gcd(o, h ) = 1. there is an automorphisrn Q of the free group (L. JI ) with 9 : {L. M) ct { w l ( L . J I ) . w 2 ( L -11)) and

exponent sum of L in w l = c

exponent sum of -11 in W I = h

T hen

w hich implies

Corollary 3.9 glbl Lnots are prime.

PTOOJ Two generator knots are prime. by the theorem of Sor- wood [No~B~] . Dl

Proposition 3.10 A glb l h o t can be embedded on 7 as a j:2 Ti! h o t with ni = n - 1 and ni = 1 and (p'. q'. r'.sf) = (1.0,O. 1 ) . and any double- torus knot with these connections is a glbl kat.

Proof. Let the point y in Definition 3.1 rnove around on T dragging the arcs a and 3 with it phi le x rernains fixed. This WU not change the knot represented by the two arcs. so we can assume that û is a simple line segment. Then if we add a handle to T and embed the arc a on it. we have the glbl knot embedded on a Heegaard surface of genus two. The bfidge becornes an (r', sl)-connection with r' = O and s1 = 1. and the underpasses becorne (p'. 9')-connections with p' = 1 and q' = O. Clearly any double-toms knot with these connections is a glb l h o t . Thus we can embed glb l knots on T as 1 2 2 2 . or 3 2 knots. 0

Proposition 3.11 1 :2 glbl bots are satellite kmots.

Proof. A 1:2 glb l b o t L is specified by the choice of n. p. q . and p. Complete one of the connections to the (p. q)-toms h o t K. Then I ï is the core of a torus U which does not intersect L. Thus L is a satellite n-ith cornpanion h-. 0:

Proposition 3.12 Ifp = 1 then the pattern h o t in the satellite description of L above is an ( n , npq + 1 ) - t m h o t .

Proo f. We have the foUowing connection diagram. omitting the bridge

Figure 3.24

Figure 3.25

Since the LHS connections are al1 pardel to the core K. they can be embed- ded in U. dong with the a'-connections. Then we can complete L nith an arc (the bridge) ernbedded in 24. This would not be possible if. for esample. we had p = 2 and n > 3. For then the diagram above becomes

Figure 3.26

and it is impossible to complete L in U in this way nit hout self-intersections. Now each of the n LHÇ-connections completes to a (1. pq) cun-e on U.

since its iinking number aith the core li is pq. The bridge adds one more turn in the meridian direction, and so the pattern h o t is an ( n . npq + 1)- torus knot. O

Remark 3.13 It is also true that the pattern h o t is a torus knot nhen p = o.

Definition 3.2 We have a presentation for the group of a 9161 h o t as

in Remark 3.2. If we choose the base point so that the single occurrence of b' cornes at the end of the expression for k. we cali this the standard present ation wit h standard relations.

Proposition 3.14 Let (kl = kp) be a standard relation for a 1.2 g l b l h o t L wîth p = 1. Define B ( t ) so that

Pmof. From Corollary 3.12. L has the satellite form of the Alesander

From the connection diagram. if we start from the end of the 6'-connection. we find that P = np.Q = nq,Q' = -1 and

T hen

and B( t ) =

This seems to work in greater generality: Conjecture Let (ki = kÎ) be a standard relation for a J:? glbl knot L .

j E {l. 2.3). p arbitrary. such t hat P # O and Q # O . Define B( t ) so that

3.3.1 2-bridge knots

A canonical diagram of an (o. 3)-2-bridge h o t with O 5 3 5 a. such as the (5.3)-?-bridge h o t below

Figure 3.27

can be converted into a connection diagram for

Figure 3.28

with equation (n - 1) + 1 + O = (n - 1) + 1

a 2 2 glbl knot

+ O . p = 3. n = a. and ( p T q , r T s , p l . ql, r f T s ' ) = (1,O.O. 1.1.0,O. 1) by taking the inside of the rect- angle in Figure 3.27 as LHS. Thus 2-bridge knots are g l b l knots.

.Assume that both a and 3 are odd. This is no loss of generality as far as the -ilexander polynomid is concerned. since an (a. 3)-?-bridge link is a knot if and only if a is odd. and in this case the (a. 3)-%bridge h o t is the rnirror image of the (a. a - (3)-%bridge b o t . There is a well-knoxn algorithm for drawing the canonical diagram for an (a. 3)---bridge link Let Ei = o(r, (id)) for 1 5 i < a - I. where o is the sign function. Letting the underbridges be two line segments containing (a + 1) points labelleci O to cr in order. join the points on different underbridges labelled Ir,(i.j)l and Ir,((i + 1),~?)1 by disjoint arcs. changing direction if c, # z;+i (see Figure 3.37). Converting to the 2 2 g lb l diagram. we get

From t his it follows that Q = 1.Q' = - 1. and

The Iast equality follows from

Lemma 3.15 Given û. and 3 both odd. and the Ti defined as above. s, =

Proof. Let j E 2. TLere are integers m and k. with O < b < n so t hat

=. - - 8 - ( - I I J

u ip = ?mn+ (-1)'k

9 (n - i)p = n ( p - 2 m ) - ( -1)jk

9 (n - i)p = n ( p - Pm - (-1)') + ( - l ) ~ ( n - k) 0 En_; = (- l) j

ff

Remark 3.16 I t follows from Theorem 1 of [Shi761 that P is the signature of the h o t .

Since Q = 1 and Q' = -1

where

and by Lemma 3.4 B( t ) = A(t) - tZP-q t - ' )

where. as usual.

Then our conjecture impiîes

for '-bridge knots with nonzero signature. Now

tP*qt - ' ) = A(t) + 1 - tP

The proof is similar to the proof of Lemma 3.4. using the fact that 51 = z,- i .

T hen

according to the conjecture.

Proposition 3.17 Let K be the (n. p)-%bridge h o t &th n and p odd. let Ei = r n ( i p ) for 1 _< i 5 n - 1, and let

Proof. Let x. y be generators of the group of corresponding to the underbridges. as in Figure 3.2'7 . Here x0"' = t = y*''. We have the standard group presentation

(I. y 1 Yw"-f u*-l)

by the fundamentai formula of the Fox calculus. Then

This gives some evidence for the conjecture. Example 3.1 Let K be the knot r3. which is the (13.9)-?-bridge knot.

In this case we have ( c l . . . .. 5 1 2 ) = ( 1 . -1.1.1. - 1 . 1 . 1 . -1.1.1. -1.1). Thus P = C ci = 1. and

Example 3.2 We now look at a knot mhi ch differs fr orn the glbl form of a %bridge knot onIy in that me take n2 = 2 instead of 1.

Consider the knot L Nith equation 7 + 2 + O = 8 + 1 + O. p = 3. and connections given by ( p ? q. r, S.$, q'. r'. s') = (1.0,O.l. 1.0.0.1). From the connection diagram (first diagram in Figure 3.29). we get

k = aat-la-latb-laIa-l t - L a a t - f a - l t - 1 t a a b aa"'a'-Lnb'

P = - y - 2 r = - l

Q = -q -zs= -2 Q' = -s'= -1

Then the conjecture implies

In the diagram below. we deform a diagram of a projection of L into a projection with 9 crossings. We can use the Wirtinger presentation to

ver@ that t4 - 2t3 + t2 - 3t + 1 really is the .Uexander polynomial of L. Therefore L is the gJ2 knot. This is our first esample of a glbl knot nrhich is not a torus, satellite (because 9~ is a hyperbolic knot nit h volume = 4.057. see [r\HW91]), or P-bridge knot. Another esample is the bot , which we get if we take p = 2 instead of 3 above. leat-ing the xdites RI. n2' n3. ni, n i . ni. p. q . r. S. d. q'. r'. s' unchanged.

Figure 3.29

3.4 Berge Knots

John Berge [Ber] has shown that that if L is a double-torus knot which represents a generator of both iri (W) and al (W'), then there is an integer surgery on L which yields a lens space. These knots are called double- primitive knots by Berge, and Berge knots by C. Gordon in [Kr95]. Since a TITI knot represents the trivial element in both H L ( W ) and H 1 ( W f ) , a Berge knot is a T2 knot. Berge gives several examples which we describe in the notation of this paper.

Choose a base point on O and regard { x i , y*) as generators of K I (W), and (xz, y2) as generators of rri (W'). As usual let a, 6 , c be (p, q)-, (r. s)-, and ( - p + r, -q + s)-connections respectively, and similarly for a', b', c' .

1) Let L be a (torus) knot with 1+0+0 = 1+0+0, p = 1 and p' = 1, q' = 0. Then k = aa', so ki = xyyi and k2 = xz are clearly generators of K I (W) and KI (W') respectively.

II) Let L be a b o t with 2 + O + O = 1 + 1 + O , p = O and (p', q', r'? s') =

(1,0,0,1). From the connection diagram

Figure 3.30

we see that k = ad-'abt-', so kl = ~ 7 ~ ; ~ X: and k2 = zTy;l are again easily seen to be generators of al (W) and irl (W') respectively.

III) Let L be a knot with j+2+0 = (j+l)+l+O,p = 0 and (pl q, r, s , p t , q', r', s f) = (2m + E , &, ml E , 1,0 ,0 ,1) where j 2 O and E = f 1. From the connec- tion diagram

Figure 3.31

we see that L is indeed a knot, and k = b(a'-Laa'-La)Ja'-lbb'-l, so

kl = xy(y;lx~m+r)jy;l 2 1 m kz = z2+*)~ y2 -1 . k2 is clearly a generator of ni (Wf) , and kl can be seen to be a generator of nl (W) by the foilowing Nielsen automorphisms:

TV) Let L be a knot with j + ( j + 1) + O = ( 2 j ) + 1 + O , p = O and (P, P, r, S, dl qf , y', S I ) = (m + E, E, m. E, 1 ,0 ,1 ,1 ) where j > O and E = f 1. F'rom the connection diagram

Figure 3.32

we see that L is indeed a knot, and k = (ba'-laa'-')jbbf-l, so kl = - 1 m+cy;l)jzm -1

(x;"Y* X l (j+l)c -1 yl and k2 = xz y2 . Again, k2 is easily seen

to be generator of I F I (W') , and for k1 we have:

V) Let L be a knot with 1 + ( 2 j f 2 ) + O = ( 2 j + 2 ) + 1 + O , p = O and (p, q, r, s, pl, qf, r', s') = (rn + E, E , rn, E, 1,0,1,1) where j 2 1 and E =

H. R o m the connection diagram

Figure 3.33

we see that L is indeed a knot, and k = (bal-') if ' ad-' (bal-')jbbt-' so

It is clear that both kl and k2 are generators of their groups.

VI) Let L be a knot with ( 2 j ) + 1 + 2 = ( 2 j + 2) + 1 + O , p = O and (pl Q, r, s,p', q', r', si) = (2,1,3,1,1,0,0,1) where j 1 1 . From the

connect ion diagram

Figure 3.34

we see that L is indeed a knot, and k = ~(a' -~a) ja ' -~b(a ' -~a)~a' - 'cb ' - ' . SO

Again, it is clear that both kl and k2 are generators of their groups.

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