Augmentations and Knot Contact Homology

7/27/2019 Augmentations and Knot Contact Homology

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KNOT CONTACT HOMOLOGY AND REPRESENTATIONS OF KNOTGROUPS

CHRISTOPHER R. CORNWELL

Abstract. We study certain linear representations of the knot group that induceaugmentations of knot contact homology. This perspective on augmentationsenhances our understanding of the relationship between the augmentation poly-nomial and the A-polynomial of the knot. For example, we show that for 2-bridgeknots the polynomials agree and that this is never the case for (non-2-bridge) torusknots, nor for a family of 3-bridge pretzel knots. In addition, we obtain a lower

bound on the meridional rank of the knot. As a consequence, our results giveanother proof that torus knots and a family of pretzel knots have meridional rank

equal to their bridge number.

1. Introduction

Let Kbe a knot inR3. The knot contact homology HC(K) ofKis the Legendriancontact homology of a Legendrian torus over K in the unit cotangent bundle ofR3. It appears that HC(K) is both a very powerful invariant and also reasonablycomputable (see [EENS12] and [Ng11]). It was found in [Ng08] that augmentationsare related to theA-polynomial introduced in [CCG+94]: theA-polynomialAK(,)divides the augmentation polynomial AugK(,

2).This paper begins an effort to better understand this relationship through repre-

sentations of the knot group, and particularly to understand factors of AugK(,)failing to appear in the A-polynomial. The results herein present emerging evi-dence that the additional factors may correspond precisely to higher dimensionalknot group representations that satisfy some conditions on the peripheral sub-group. A representation satisfying these conditions is called a KCH representation(see Definition 3.2). We show that the degree of an irreducible KCH representation

bounds the meridional rank from below. In all known cases there is an irreducibleKCH representation in dimension equal to the bridge number (see Section 5).

We work with a specialization of HC(K) obtained from a differential gradedalgebra A|U=1 over the polynomial ring R0 = Z[1, 1] (see Section 2 for a defini-tion, and [Ng12] for a discussion of the more general invariant). An augmentationof HC(K) is an algebra map : HC0(K) C. The augmentation polynomialAugK(,) is defined so that its zero locus is the complex curve that is the closureof points

((), ()) | an augmentation ofHC(K)

.

ThroughoutwedenotebyKthefundamentalgroupofthecomplement R3 \ n(K).It is observed in [Ng12] that KCH representations ofK induce an augmentationofHC(K).

1.1. Results. We first address the dimension of irreducible KCH representationsofK. If

g1, g2, . . . , gr

is a set of generators for K with the property that gi is a

meridian ofKfor each 1 i r, we call the generating set meridional.Theorem 1.1. Let

g1, . . . , gr

be a meridional generating set for K. If : K GLnC

is an irreducible KCH representation ofK then n r.1


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The meridional rank of K, denoted mr(K), is the minimal size of a meridionalgenerating set. Theorem 1.1 produces a lower bound on mr(K). We will see thatthis bound is sharp on torus knots and on the (

2, 3, 2k+ 1) pretzel knots (see

Section 5), and we are lead to wonder if it is sharp generally.The meridional rank is bounded above by the bridge number of a knot. Problem

1.11 of [Kir95], which is attributed to Cappell and Shaneson, asks whether everyknot with meridional rank n is an n-bridge knot. The lower bound from Theorem1.1 suggests that KCH representations might be used to study this problem.

Without much difficulty, KCH representations can be made to have image inSLnC, and from this one sees that theA-polynomial divides AugK(,

2) (see section3; cf.[Ng08]). With this in mind, we will use n-dimensional KCH representationsto define for each n 2 the n-dimensional A-polynomial An

K(, ) ofK.

If is the induced augmentation from an n-dimensional KCH representation ,m is a meridian of K, and a 0-framed longitude, then () and () are certaineigenvalues of(m) and (), respectively. We will define An

K(,) so that its zero

locus is the closure of points in (C)2

given by these eigenvalues.If : K GLnC is a reducible KCH representation, then there is a KCH

representation of lower degree that corresponds to the same point ((), ()) inthe curve of eigenvalues (see Lemma 3.4). Hence Theorem 1.1 implies that thepolynomials An

K(,) stabilize for sufficiently large n. Define the result to be the

stable A-polynomial AK

(,).1 We have that AK(,) divides AugK(,).

Conjecture 1.2. For all knots K in R3, ( 1)AK

(,) = AugK(,).

When K is a 2-bridge knot, Conjecture 1.2 concerns the original A-polynomialAK(,) and was posed in [Ng08]. We provide a proof for this case here.

Theorem 1.3. Let K be a 2-bridge knot and : HC0(K) C an augmentation with() 1. Then is induced from a KCH representation : K

GL2C. Thus,

AugK(,2) = (2 1)AK(,).

On the other hand, torus knots exhibit that there can be irreducible KCH repre-sentations of high degree.

Theorem 1.4. Given 1 p < q, with p, q relatively prime, let T(p, q) denote the (p, q)torus knot and its knot group. For every 1 n p there is a factor (n(n1)pq 1) ofthe stable A-polynomial that arises from an irreducible degree n KCH representation of.In fact,

AT(p,q) =min(p,q)

n=1(n(n1)pq 1).

We remark that the torus knot T(p, q) may be put into bridge position withmin(p, q) bridges. Theorem 1.4 produces irreducible KCH representations in di-mension min(p, q) and, by Theorem 1.1, this implies the meridional rank of T(p, q)is the bridge number (cf. [RZ87]).

1The term stable A-polynomial was coined by Stavros Garoufalidis in private communications.


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Finally, we consider the family of (2, 3, 2k+ 1) pretzel knots, which have aprojection as shown in Figure 1. The bridge number of these pretzel knots is equalto 3 ifk

1, 0.

Theorem 1.5. Let K be the (2, 3, 2k+ 1) pretzel knot, where k 1, 0. Then there is a3-dimensional, irreducible KCH representation ofK. Furthermore, (1 2k+6) divides

A3K

(,).

As a consequence every knot Kin this family has meridional rank at least 3, andso the meridional rank and bridge number of the (2, 3, 2k+ 1) pretzel knot agree(cf. [BZ85]).

The paper will be organized as follows. In Section 2 we provide a quick reviewof the background on knot contact homology, particularly recalling an interpreta-tion of the degree zero part as the cord algebra. In Section 3 we will define theaugmentation polynomial and KCH representations, use the cord algebra to seehow KCH representations induce an augmentation of HC(K) and indicate how

this produces a factor of the augmentation polynomial equal to the A-polynomial.We will also prove the dimension bound in Theorem 1.1. Section 4 will be devotedto the proof of Theorem 1.3. In Section 5 we will prove Theorem 1.4 and Theorem1.5.

Acknowledgments. The author would like to thank Lenny Ng for his guidanceand for many very helpful conversations.

2. Knot contact homology and the cord algebra

2.1. Knot contact homology. The focus of this paper is on KCH representationsand the augmentations they induce on a specialization of knot contact homology.Using results of [EENS12], [Ng08], and [Ng11] augmentations on this specializationmay be understood purely through the cord algebra (using the formulation in

Theorem 2.3) which has a topological definition. The reader who so chooses may,in fact, skipthe review of the definition ofHC(K) via Legendrian contact homologyand begin with the statement of Theorem 2.3, taking it as a definition of HC0(K).

We define knot contact homology through the tool of Legendrian contact ho-mology (LCH), introduced by Eliashberg and Hofer [Eli98]. Many details on LCHare omitted.

Being pertinent to the setting of knot contact homology of knots in R3, weonly discuss contact manifolds P R, where z is the R coordinate, P has an exactsymplectic form d, and the contact 1-form is dz . In fact, we are interested inthe case of the 1-jet space J1(M), topologically equal to TM R, where is thecanonical 1-form. The proof of the invariance of LCH in (P R,ker(dz )) wascarried out in [EES07].

Let be a Legendrian in (V= TM

R,ker(dz

)). A Reeb chord is an integral

curve to the Reeb field that starts and ends on . We assume that has trivialMaslov class and finitely many Reeb chords a1, . . . , an. Let R denote the group ringofH2(V,).

Define A to be the noncommutative unital algebra over R freely generated bya1, . . . , an. The grading on generators is by the Conley-Zehnder index (minus 1),and the base ring has grading 0. Extend the grading to all ofA in the usual way.

The differential on a generator ai is defined by counting holomorphic disks ina moduli space M(ai; aj1 , . . . , ajk) modulo an R-action. After declaring (r) = 0 for


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all r R and defining (ai) for each generator, extend to A via the signed Leibnizrule. The definition of on a generator ai is

(ai) = dimM(ai;aj1 ,...,ajk)/R=0

M/R

sgn()e[]aj1 ajk.

In the above definition, M(ai; aj1 , . . . , ajk) is the moduli space of holomorphic disks in the symplectization ofV(the almost complex structure being compatible andsuitably generic) with the following boundary conditions:

- The boundary of is in the Lagrangian R ;- has k+1 ends, one being asymptotic to ai at +, the other kasymptotic to

aj1 , . . . , ajk at and in that order as determined by the oriented boundaryof the disk;

- sgn() is an orientation sign on , and [] is the homology class of inH2(V,) (using a chosen disk for each Reeb chord in V, with part boundaryon the chord, to cap off ).

By work in [EES07], makes A into a differential graded algebra and H(A, )is a Legendrian invariant of.

Given a knot KinR3, the unit cotangent bundle STR3 with its canonical contactstructure is contactomorphic to J1(S2). Define K to be the unit conormal bundleover K, that is

K =(q,p) STR3 | q Kand p, v = 0 for v TqK .

We obtain a Legendrian torus K in V= STR3. Define the knot DGA (AK, K) tobe the differential graded algebra ofK described above. Knot contact homologyHC(K) is defined as the Legendrian contact homology H(AK, K). Isotopy on Klifts to Legendrian isotopy on K, and so HC(K) is an invariant of the knot type K.

There is an isomorphsism H2(STR3,K) H2(S2) H1(T2), through which we

identify R with the ring Z[U1

, 1

, 1

] where U is a generator of H2(S2

) and (resp. ) corresponds to a longitude (resp. meridian) ofK.This paper focuses on a specialization of (AK, K), where multiplication by U

is trivial. In this specialization, the degree zero homology H0(AK, K) admits auseful topological description called the cord algebra [Ng08]. Were this story to

be paralleled with the full coefficient ring, it would be a hopeful step towardunderstanding recent computations that show coincidences between the three-variable augmentation polynomial, super-A polynomials in string theory, andcolored HOMFLY polynomials.

Specialize the knot DGA so the differential K only accounts for the homologyof H1(K). In other words, we take U= 1 in the base ring. Unless otherwisespecified the specialization to U= 1 is assumed for the remainder of the paper, yetwe leave our notation unaltered.

The (framed) knot DGA in [Ng08] was shown to be an invariant of K inde-pendently of contact homology, and by [EENS12] is a calculation ofHC(K) (withU= 1) that uses a presentation of K as the closure of a braid. We will define thedegree zero part of this DGA in order to arrive at the cord algebra, which in turnwill lead us to KCH representations. See [Ng12] and [Ng11] for a definition of thefull DGA from this perspective.

Given n 1, let An be the noncommutative unital algebra over Z freely gener-ated by n(n 1) elements aij , 1 i j n. Let Bn denote the braid group on n


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strands. Ifk is one of the standard generators, then define : Bn Aut An bydefining it on each generator as

k :

aij aij , i, j k, k+ 1ak+1,i aki, i k, k+ 1ai,k+1 aik, i k, k+ 1ak,k+1 ak+1,k,ak+1,k ak,k+1,aki ak+1,i ak+1,kaki i k, k+ 1aik ai,k+1 aikak,k+1 i k, k+ 1

Under an identification ofAn with a certain algebra of homotopy classes of arcsin a punctured disk, the representation corresponds to the mapping class groupaction (see [Ng12]). Inject : Bn

Bn+1 by letting the last strand not interact and

for B Bn let B = B . We may define two matrices LB,RB Matnn(An) by

B(ai,n+1) =n

j=1

(LB)i jaj,n+1,

B(an+1,i) =n

j=1

an+1,j(RB )ji.

Finally, define the matrices A by

Aij =

ai j i < jai j i > j1 i = j

,

and the diagonal matrix = diag[w, 1, . . . ,1], where w is the writhe of the braidB Bn. The following gives a concrete algebraic interpretation of HC0(K).

Theorem 2.1 (see [Ng12], 4). Let R0 = Z[1, 1] and suppose K is the closure of abraid B Bn. Then

HC0(K) (An R0)I,

where I is the ideal generated by entries in the matrices A B(A) 1

,A LB A,

and A A RB

1.

2.2. The cord algebra. Let K S3 be an oriented knot with a basepoint K. Acord of (K, ) is a path : [0, 1] S3 with 1() = and 1(K) = {0, 1}. Define thecord algebra CK to be the unital tensor algebra over R0 freely generated by homotopyclasses of cords (note that endpoints of cords may move in the homotopy, if theyavoid ), modulo the ideal generated by the relations:


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(1) = 1

;

(2) = and = (3) =

The cord algebra CK is manifestly a knot invariant. Moreover, the isomorphismin Theorem 2.1 was used in [Ng08] to prove that the cord algebra is isomorphic to

HC0(K).

Theorem 2.2 ([Ng08]). CK is isomorphic as an R0 algebra to (An R0)I, and thus to

HC0(K).Finally, it is useful to reformulate the cord algebra CK in terms of the knot group.

Given a cord in CK, and XK = R3 \ n(K) the complement ofK, push the endpointsslightly offofKso the cord is in XK and connect the endpoints via a longitudinalcurve on n(K) to get an element ofK = 1(XK).

Theorem 2.3 ([Ng08]). Let PK denote the underlying set of the knot group K, where wewrite [] PK for K. Let e denote the identity, m a choice of meridian, and the0-framed longitude of K. The cord algebra HC0(K) is isomorphic to the noncommutativeunital algebra freely generated by elements of PK over R0 = Z[1, 1] modulo therelations:

(1) [e] = 1 ;(2) [m] = [m] = [] and [] = [] = [], for any

K;

(3) [12] [1m2] = [1][2] for any 1, 2 K.It is this last interpretation ofHC0(K) that will indicate how KCHrepresentations

induce an augmentation.

3. KCH representations and dimension bounds

3.1. Augmentations of HC0(K) and KCH representations. Given a unital ringS, an S-augmentation of (AK, K) is a graded algebra map : AK S such that = 0. Since AK is supported in non-negative grading, such maps are in

bijective correspondence with algebra maps : HC0(K) S. We fix S = C andrefer to a C-augmentation of (AK, K) simply as an augmentation ofHC(K).

Consider the variety in (C \ 0)2 given by

VK = ((), ()) | is an augmentation ofHC(K) .Conjecture 3.1. The maximum dimensional component of the closure of VK is the van-ishing set of a polynomial for all knots K R3.

Let the augmentation polynomial AugK(,) be the reduced polynomial of Con-jecture 3.1, where by reduced we mean that AugK(,) contains no repeated factors.This only defines AugK(,) up to a unit in C[

1, 1]; however, we may assumeAugK(,) Z[1, 1] (and that the coefficients are collectively coprime), since


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the differential of AK involves only Z coefficients. Further, multiplying by anappropriate power of and of will guarantee the polynomial is not divisible by,, and has no negative powers of,. The result is a polynomial well-definedup to overall sign.

We are now prepared to discuss KCH representations and their relation toaugmentations. We use notation for the knot group and its elements that agreeswith the statement of Theorem 2.3.

Definition 3.2. Let Vbe a complex vector space with dimension n. We say that ahomomorphism : K GL(V) is a KCH representation ofK of degree n if for somemeridian m of K, (m) is diagonalizable and all but one of its eigenvalues equalone. We call a KCH irrep if it is irreducible as a representation.

We often identify VwithCn via a basis {e1, . . . , en} of eigenvectors of(m). In thisnotation e1 always denotes the eigenvector with eigenvalue 0 1. As commuteswith m the vector e1 is an eigenvector for (). Write 0 for the corresponding

eigenvalue.Identify HC0(K) with the cord algebra as described as in Theorem 2.3, generated

over R0 by elements [], K.Theorem 3.3 ([Ng12]). Let : K GL(V) be a KCHrepresentation ofK withnotationas above. There is an induced augmentation with () = 0 and () = 0.

As cited, Theorem 3.3 is outlined in an exercise in [Ng12]. The proof is asfollows.

Proof. Define () = 0 and () = 0, then extend to a ring homomorphism onR0. Define an inner product , on Vso that the basis {e1, . . . , en} is orthonormal.For K define ([]) by ([]) = (1 0)()e1, e1. Extend this to an algebramap on the cord algebra.

We check that is well-defined, referring to the relations of Theorem 2.3.Relation (1) is trivial since (e) is the identity. To check (2), note that, in the basis{e1, . . . , en}, the first row of the matrix for (m)() is 0 times the first row of().Thus for any K,

([m]) = (1 0)(m)()e1, e1= (1 0)0()e1, e1= ([]), by definition.

Moreover, ([m]) = (1 0)()(m)e1, e1 = (1 0)0()e1, e1 since e1 is a0-eigenvector. A similar argument holds for [], [], and [] since e1 is also aneigenvector for ().

Finally, to see that respects (3), let 1, 2 K. Note that (m) = I+ (0 1)E11where Iis the identity andE11 thematrix with 1 in thetop-left entry and0 elsewhere.

As a result (1)(m)(2) = (1)(2) + (0 1)(1)E11(2). From this we seethat

([12] [1m2]) = (1 0)(12) (1m2)

e1, e1

= (1 0)2(1)E11(2)e1, e1= (1 0)(1)e1, e1(1 0)(2)e1, e1= ([1])([2]),


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the penultimate equality coming from the fact that E11(2) has its image in thespan ofe1.

Previous calculations of AugK(,) have been markedly difficult. The situationis much improved upon considering KCH representations ofK as we will see inthe succeeding section.

There is a canonical augmentation with() = 1 where () may be any nonzeronumber [Ng08, Prop 5.6]. KCH representations cannot account for this, since if0 = 1 then the representation is trivial asK is normally generated by m. However,it may be that for any non-canonical augmentation the restriction to R0 agrees withthat of for some KCH representation . Rephrased in terms of the stable A-polynomial (see Corollary 3.5 below), this says the following.

Conjecture 1.2. For all knots in R3, ( 1)AK

(,) = AugK(,).

We recall the relationship found in [Ng08] of Aug K(,) to the A-polynomialAK(,), namely

(1) (2 1)AK(,)AugK(,2).

The A-polynomial was introduced in [CCG+94], and it was shown there that 1 does not divide AK(,). However, 1 does always divide AugK(,) asseen by the canonical augmentation.

Let : K SL2C denote a representation ofK into SL2C. Ifm and denote ameridian and longitude ofKthen (m) and () can be made simultaneously uppertriangular:

(m) =

0 1

() =

0 1

The A-polynomial has zero locus equal to the maximal dimensional component

of the closure of points (,) (C \ 0)2

such that there is a as above with , theupper-left entry of(), (m) respectively.

If we have such a representation : K SL2C, let () := lk(,K)(), wherelk(,K) is the linking number of K with K. The additivity of lk(,K) makes a representation into GL2C; moreover, if 1 then (m) is diagonalizable and

defines a KCH representation with 0 = 2. Thus AK(,) divides AugK(,2).

3.2. Irreducibility and dimension bounds. It can occur that AugK(,) has morefactors than those given in (1). For example, Ng computes extra factors inAugK(,) for the (3, 4)-torus knot and three pretzel knots [Ng08, end of5]. Tosee such factors from KCH representations requires (at least) a KCH representationof degree n > 2.

Similar to the definition of the augmentation variety, consider the variety in

(C \ {0})2

given byUnK =

((), ()) | is induced from a degree n KCH representation .

From UnK

we define the n-dimensional A-polynomial AnK

(,) in similar fashionto AugK(,). We would hope to understand A

nK

(,) by considering only KCHirreps with degree at most n, and may do so by the following result.

Lemma 3.4. Let : K GL(V) be a KCH representation of degree n that is reducibleand corresponds to the point (0, 0) in UnK. Then (0, 0) Un

K

for some n < n.


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Proof. Let V Vbe a proper invariant subspace. Ife1 is not contained in V thentake the quotient representation : K GL(V/V), which is well-defined as Vis invariant. Consider a basis

{e

1+ V

, f

2+ V

, . . . , f

n+ V

}ofV/V

. Suppose that

fi = cijej as a vector in V. Then for fi = fi ci1e1, the set {e1+V, f2+V, . . . , fn+V}is also a basis ofV/V. As (m)(f

i+ V) = f

i+ V for 2 i n, this basis shows

to be a KCH representation.Since e1 + V

is in the eigenvector basis for (m) we get 0, 0 in the inducedaugmentation.

Ife1 is contained in V define a new representation on V by restriction. Then

(m) is diagonalizable in V so this is a KCH representation. Clearly the eigenvalues0, 0 are unchanged.

We now bound the degree of a KCH irrep for a fixed knot K by proving thefollowing.

Theorem 1.1. Let g1, . . . , gr be a meridional generating set for K. If : K GLnCis a KCH irrep ofK then n r.Proof. Let m be the meridian ofK sent by to M := diag[0, 1, . . . ,1] (in the basis{e1, e2, . . . , en}). For each gi, let wi K be such that gi = wimw1i . Write Wi forthe matrix representation of(wi). Now the specific form ofM guarantees that foreach 1 i r,

WiMW1i = I+ (0 1)WiE11W1i .

Now (0 1)WiE11W1i is a rank 1 matrix; let vi be a vector whose span equals theimage. For each 1 j r there is a scalar j

i C for every 1 i r such that

(gi)vj = vj + j

ivi.

This shows that the span of

{v1, v2, . . . ,vr

}is an invariant subspace. We conclude

n r since is irreducible. Corollary 3.5. The sequence of polynomials An

K(,), n 1 stabilizes.

Proof. Combine the result of Lemma 3.4 with Theorem 1.1.

Remark 3.6. One often uses a generating set of K that is not meridional. Ifg1, . . . , gr

is such a generating set, then for each i we may write gi as a product of

pi 1 meridians. An adaptation of the proof above would show that D =r

i=1 piwill be an upper bound for n. However, this adaptation is unnecessary since thereare clearly D meridians that generateK, thoseusedin the products of meridians,each of which gives gi for some 1 i r.

4. The augmentation polynomial of 2-bridge knots

Consider a knot K that has a 2-bridge presentation. Studies of 2-bridge knotshave been carried out in many settings. Of particular interest to our discussion arethe papers [Ril84] and [Sch56].

ThroughoutSection4and5wewillderiveandmakeuseofanumberofidentitieson the values of augmentations : HC0(K) C. In an attempt to avoid a clutteringofs and demarcation symbols, we will therefore adopt the notation

X= Y


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to denote that elements X,Yof the cord algebra satisfy (X) = (Y). At times Ymayinstead be a complex number, in which case we mean that (X) = Y. In addition,we will drop the subscript of

0and simply write both when referring to the

element ofR0 and to the eigenvalue.

Theorem 1.3. Let K be a 2-bridge knot and : HC0(K) C an augmentation with() 1. Then is induced from a KCH representation : K GL2C. Thus,

AugK(,2) = (2 1)AK(,).

Proof. Let p < q < p, wherep, q are coprime and odd and suppose Kis the 2-bridgeknot K(p, q). Setp1 = 2k. From the normal form of Schubert [Sch56], we know thatK has a presentation of the form m, b | wm = bw, where w = m1 b2 . . .m2k1 b2kand i = (1)iq/p. Here m is a meridian ofK(p, q), and will be the meridian used inthe KCH representation we construct.

Given the augmentation : HC0(K(p, q)) C, the relations on the cord algebralisted in the statement of Theorem 2.3 imply [e]

= 1 as well as

(2) [g][h]= [gh] [gmh] and [m1g] = 1[g] = [gm1]

for any g, h K.

Define (m) = M =

00 1

and (b) = B =

([b])1

([b][e])([m][b])(1)2

1 ([e]+[m][b])1

.Using relations (2) we can see that the traces ofB and M agree. In addition, we

calculate

([b] [e])([m] [b]) = [b]([m] + [e] [b]) (1 )[m]= [b]([m] + [e] [b]) (1 )2.

As a consequence, B amd M have the same determinant,

det B=

[b]([e] + [m] [b]) ([b] [e])([m] [b])(1 )2

= .

Upon extending to a group homomorphism, we claim is well-defined and,as a KCHrepresentation, induces . Note that if results from a 1-dimensional (and

thus abelian) representation ofK then (m) = , so we must have [g]= lk(g,K)

for any g K. Particularly, [b] = = [m] implies that B is lower triangular, so that is reducible.

Momentarily suppose that : K GL2C is well-defined; we show that itinduces the augmentation .For this well show that for any word x in {m1, b1}, ifX is the correspondingproduct of matrices M1 and B1, we have

(3) X11 = Xe1, e1 = [x]1 .

We can show this by induction on the length of the word x. Our definition ofB and M guarantees (3) immediately for x = b and for x = m1 upon noting that


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[m]= [e]

= (1 ). In addition, (2) gives

[b] + [b1] = [wmw1] + [wm1w1]= [e] [w][w1] + ([e] + 1[w][w1])= [e] + [m],

and so [b1]= 1([e] + [m] [b]), implying (B1)11 = [b

1]1 . For n > 1,

[bn]= [wmn1w1] n1[w][w1]= [bn1] n1([e] [b])=

1

1 [e][bn1] + [b][mn1] [e][mn1]

=

[bn1][b]1

+([b] [e])([mn1] [bn1])

1

An induction argument shows that (1 )(Bn)11 = [bn] for all n 0. Theargument uses the above equalities and begins with induction hypotheses (Bi)12

=

([b][e])([mi][bi])(1)2 and (1 )(Bi)11

= [bi] for i < n. The hypotheses certainly hold for

i = 1. By a similar argument, using [b1] = 1([e] + [m] [b]), one may check that(1 )(Bn)11 = [bn] for n < 0 as well.

To finish, note that weve shown (3) holds for any x that is a power ofb. Anyother x will have an m1 appear in the word. Assume that (3) holds for any wordin {m1, b1} with length less than x. Consider m, = 1, and x, x such thatx = xmx as words in {m1, b1}.

As in Section 3, let E11 denote the 22 matrix with 1 in the top-left entry and 0

elsewhere. Then M

=I+

(

1)E11 and we find that,X= XMX = XX + ( 1)XE11X,

the (1, 1)-entry of which is (XX)11+ ( 1)(X)11(X)11 = [xx]

1 12 (1) [x

][x]1 , by

induction. But the value of this expression under is [x]1 by (2). Consequentially,if is well-defined, it is a KCH representation that induces .

We now turn to the proof that is well-defined. Following Lemma 1 of [Ril84],we find a U SL2C such that

(4) UMU1 = 10 1

and UBU1 =

0

u 1

for some u in the fieldQ(, ([b])). Indeed, choose a square root z of([m]

[b])C

and define

U=

1z

z1

0 z

.

Then one computes that UMU1 and UBU1 have the desired form, where u =([m] [b]).

Recall the element w = m1 b2 . . .m2k1 b2k appearing in our presentation ofK.We write N = UMU1 and C = UBU1 for the matrices in (4) and write W =


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12

N1 C2 . . .N2k1 C2k. It is shown in [Ril84] that

(5) W11 + (1

)W12 = 0

if and only ifm N, b C determines a well-defined representation of the knotgroup K. This is equivalent to being well-defined.

To demonstrate (5) we will use Lemma 4.1 below. First, let {1, 2, . . . , r} beany set with i {1}. We call W = N1 C2 . . .Nr1 Cr a palindrome ifi = r+1ifor all 1 i r. By considering the transpose ofW, it is shown in [Ril84] thatuW12 = W21 ifWis a palindrome. In our notation, this says ([m] [b])W12 = W21.Lemma 4.1. Suppose W = N1 C2 . . .Nr1 Cr corresponds to a word w in {m1, b1}under the assignment m N, b C, where i = 1 but W is not necessarily a

palindrome. Then for all n Z(6) [bnw]

= [bn]W11 W21.

Proof. The statement is trivially true when w = 1and W= I. Now let W= N1 C2 W

and suppose that (6) holds for W. Define i = 12 (i 1), for 1 i r. Then[bnw]

= [bn+2 w] 11 [bn][b2 w]= [bn+2 ]W11 W21 11 [bn]([b2 ]W11 W21)=[bn+2 ] 11 [bn][b2 ]

W11

1 11 [bn]

W21.

Now, we have that N1 C2 =

1+2 + 121+2([m] [b]) 11

22([m] [b]) 1, and so

W11=1+2 + 12

1+2 ([m] [b])

W11 + 11 W21

andW21

=2

2 ([m] [b])W11 + W21.Thus, to conclude the proof of the lemma, we need to show that

(7) [bn]1+2 + 12

1+2 ([m] [b]) 22 ([m] [b]) = [bn+2 ] 11 [bn][b2 ].

We recall from a previous computation that [bn+1]=

[bn][b]+([b][e])([mn][bn])1 , from

which we derive that

(8) [bn+1]= [bn] ([m] [b])

for all n 0. We show that (7) holds in four cases.Case 1 = 1, 2 = 1: In this case 1 = 2 = 0, so the left side of (7) is

[bn] 2 + [m]

[b] ([m] [b]) which equals [b

n]

[bn][b]

([m]

[b]). Using

(8), we see that this equals the right side of (7).Case 1 = 1, 2 = 1: Here, 1 = 1 and so the left side is [bn]

+ 1[b]

([m] [b]) which equals [bn+1] + 1[bn][b] by another application of (8).Case 1 = 1, 2 = 1:

In this case the left side is [bn] + 1[b]

+ 1([m] [b]). On the right side we

have

[bn1] [bn][b1] = [bn1] 1[bn](1 2 [b]) = [bn] + 1[bn][b] + [bn1] 1[bn].


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13

This provides us the result since [bn1] 1[bn] = 1([m] [b]) by (8).Case 1 = 1, 2 = 1: Finally, the left side in this case is

[bn] 2 + 2([m] [b]) + 1([m] [b]) = [bn] 1 + 1[b1] + 1([m] [b]).The right side is [bn1] + 1[bn][b1]. But (8) says 1[bn] + 1([m] [b]) = [bn1],showing that equality (7) holds. This proves Lemma 4.1.

We now finish the proof of Theorem 1.3. Let W Z[1, ([b])] be the polyno-mial equal to W11 + (1 )W12. In [Ril84], W is called the non-abelian polynomialfor K(p, q). We will use Lemma 4.1 to show that if Wis a palindrome then

[bw] [wm] = ([b] [m])(W).Recall that since W is a palindrome, W21

= ([b] [m])W12. Also note that

[wm]= [w]. By applying the result of Lemma 4.1, with n = 1 and n = 0, we have

[bw] [mw] = [b]W11 W21 ([e]W11 W21)=[b] [e]W11 (1 )W21

= ([b] [m]) W11 + (1 )W12 = ([b] [m])W.

Now since K(p, q) is a 2-bridge knot, we have that bw = wm in K, and so([bw]) ([wm]) = 0 since is well-defined. If([b]) ([m]) (which must be so if([g]) is not the linking number of g with K), this implies that (5) holds, and thus is well-defined by [Ril84].

5. Augmentations and KCH representations in higher dimensions

We advance to knots for which AK

(,) A2K

(,), in which case AugK(,)has more factors than those in the classical A-polynomial. We tackle torus knotsfirst and show that their knot group admits KCH irreps up to degree min(p, q). Wethen consider a family of 3-bridge pretzel knots, and find degree three KCH irreps.These initial results lead us to wonder if the upper bound in Theorem 1.1 is sharpfor all knots.

5.1. Torus knots. Given a relatively prime pair (p, q) of positive integers, let T(p, q)be the corresponding torus knot. Since T(p, q) may be put in bridge positionwith min(p, q) bridges, its knot group has a presentation with min(p, q) meridionalgenerators. By Theorem 1.4 the bound in Theorem 1.1 is realized, giving a newproof that meridional rank equals bridge number for torus knots (cf. [RZ87]).

Moreover, if Conjecture 1.2 holds we obtain an explicit calculation of the aug-mentation polynomial for all torus knots.

Let be the knot group ofT(p, q). We will work with the familiar 2-generatorpresentation for . Let = x, y | xp = yq, where the peripheral elements m and, representing a meridian and longitude of T(p, q) respectively, satisfy = xpmpq

and m = xsyr where r, s are integers such that rp + sq = 1. Without any loss ofgenerality we assume min(p, q) = p 1.Theorem 1.4. Given 1 p < q, with p, q relatively prime, let T(p, q) denote the (p, q)torus knot and its knot group. For every 1 n p there is a factor (n(n1)pq 1) of


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14

the stable A-polynomial that arises from an irreducible degree n KCH representation of.In fact,

AT(p,q) =min(p,q)

n=1

(n(n1)pq 1).

The proof of Theorem 1.4 will proceed from Lemma 5.1 given below. To proveLemma 5.1 we will use Lemma 5.2.

Lemma 5.1. If1 n p < q, there exists X,Y GLnC such that(a) for r, s with rp + sq = 1, XsYr = M, where M is the n n diagonal matrix

diag[0, 1, . . . ,1];(b) Xp = Yq = zI where z is a nth root of

pq

0and I is the identity in GLnC.

Lemma 5.2. Let

Y be an n n complex matrix of the form

Y=

x1

x2

. . . xn1

y01 0 . . . 0 y1

0 1 0 y2...

. . ....

...0 0 1 yn1

.

Denote by a(t) = a0 + a1t+ + an1tn1 + tn the characteristic polynomial ofY. Then forall 1 i n,

ani = xi yni +i1j=1

xjyni+j

where we let xn = 0.

Proof of Theorem 1.4. Considering the knot group for T(p, q), the existence of X,Y

guaranteed by Lemma 5.1 provides a well-defined KCH representation via (m) =M, (x) = X and (y) = Y. It follows () is the diagonal matrix diag[0, , . . . , ]with z

pq0 = 0, proving that A

T(p,q)

has the desired factor.

That this accounts forall factors ofAT(p,q)

is seenby showing that every KCH irrep

of has this form. This is an application of Schurs Lemma. Note xp commuteswith every element of, since xpy = yq+1 = yxp. Now an eigenspace Ez of(xp)must be an invariant subspace, so Ez = C

n and (xp) is zI. Since is a commutator,det(()) = 1, so det((xp)) = det((mpq)) =

pq

0. Thus z is a nth root of

pq

0.

That these representations are irreducible is deduced from Lemma 3.4. Thebound on degree comes from Theorem 1.1.

Proof of Lemma 5.2. To prove the claim, suppose that for a permutation Sn, the

product p = (tI Y)i,(i) is non-zero. If(1) = n then (i) = i 1 for i > 1 andsign()p = (1)n1(1)ny0 = y0. Let (1) = i < n and (j) = n. It must be thati j, for otherwise p = 0.

When i > 1 and j < n, an examination oftI Y(or more precisely, of the minorobtained by eliminating the 1st and jth rows and ith and nth columns) indicates that

p will be zero unless (k) = k 1 when 1 < k i or j < k n, and (k) = kwheni < k< j. This implies p = (1)nji+3xiyj1tj1i. Since the sign of is (1)nji+1,we get that sign()p = xiyj1tj1i.


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15

Ifi = 1 or j = n then a similar analysis shows that sign()p = ijtj1i wherei = (xi i1t) and j = (yj1 nj t), i j denoting the Kronecker-delta function.

Summing the terms corresponding to a fixed power t

n

i

gives the claimed coef-ficients ani.

Proof of Lemma 5.1. Since z is an nth root ofpq

0, we can find numbers 1, 2, . . . , n

that are pairwise distinct qth roots ofz such thatn

j=1 j = p

0. We define cY(t) =n

j=1(t rj ) = cY0 + + cYn1tn1 + tn, and would like to choose entries yi ofYsothat the characteristic polynomial a(t) agrees with cY(t). But Lemma 5.2 shows thatani is linear in yni and independent ofyl ifl < n i. Thus we may set

yn1 = cYn1 x1,then recursively define yni = cYni xi +

i1j=1 xiyni+j for 1 < i n. The resulting

matrix

Ynow has cY(t) as its characteristic polynomial.

Let 1, 2, . . . , n be pairwise distinct pth

roots of z such that nj=1 j = q0 anddefine cX(t) =

nj=1(t sj) = cX0 + + cXn1tn1 + tn.

Letting X= MY with the yni defined as above, consider the formula for yni,1 i n. It is easy to see that yni is linear in xi and is independent ofxl for l > i.Thus, if we replace Xfor Y in Lemma 5.2 we see that the coefficient oftni in thecharacteristic polynomial ofXhas coefficients

bni = (0 1)xi + biwhere b

iis a polynomial in variables xl with l < i. (b1 = c

Yn1) And so, since

0 1, we may solve x1 = (cXn1 + b1)/(0 1) and recursively solve for each xi,for 1 < i n 1. Now observe that the definition of

Ximplies,

det X= 0 detY= 0(1)n( nj=1

rj ) = (1)n1rp0 = (1)nsq0 =n

j=1

sj.

As a result, upon solving for the xi, 1 i n 1, the constant term of the charac-teristic polynomial ofXagrees with cX0 , making cX(t) the characteristic polynomial.

Now, the eigenvalues ofXare pairwise distinct and so Xis diagonalizable. FindP such that X = P1DP, where D = diag[s

1, s2, . . . ,

sn], then define X = P

1DPwhere D = diag[1, 2, . . . , n]. Define Yin a similar manner, so that it is similar to

diag[1, 2, . . . , n] by a similarity matrix Q with Y= Q1diag[r1 , . . . , rn ]Q.Now we have Xs = P1DsP = X= MY, and Yr = Y. Moreover, Xp zI= 0 and

Yq zI= 0 by the Cayley-Hamilton theorem.

5.2. KCH representations for (2, 3, 2k+ 1) pretzel knots. In this section we findirreducible, degree three KCH representations for the (2, 3, 2k+ 1) pretzel knots,which have a projection as shown in Figure 1. Similar to the case of torus knots,the proof of irreducibility relies on Lemma 3.4.

Theorem 1.5. Let K be the (2, 3, 2k+ 1) pretzel knot, where k 1, 0. Then there is a3-dimensional, irreducible KCH representation ofK. Furthermore, (1 2k+6) divides

A3K

(,).


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16

m

w

L

... 2k+ 1

Figure 1. The(

2, 3, 2k+1) pretzel knot with a blackboard framed

longitude L.

Proof. The group K admits a two-generator presentation K = m,w | wkE = Fwk,where m is a meridian, E = mwm1w1m1, F = m1w1mwmw1, and w is a productof two meridians [LT12, 4]. As a consequence of Theorem 1.1, any irreducibleKCH representation that sends m to M = diag[0, 1, . . . ,1] has degree 3 or less.

Let us first observe what would be required of an augmentation : HC0(K) Cthat is induced from : K GL3C, a KCH irrep. We will abuse notation andwrite for () = 0. Then there is a basis for which (m) = diag[, 1, 1] and suchthat (w) has the form

(w) =

x1 x y01 0 y

10 1 y .As induces the augmentation we must have x1

=

[w]1 . Note that y0 x1y1 xy =

det((w)) = 2 since w is the product of two conjugates of m. Furthermore, an

examination of the upper left entry of(w)1 requires that [w1]

1= (w1)11 = y12 in

this basis. Thus, for our supposed representation, we have that in some basis andfor some pair x, y C,

(9) (w) =

([w])1 x

2 2 ([w])([w1])(1)2 + xy1 0 2 ([w1])10 1 y

.

By considering the relations (2) on the values of and that wk = FwkE1 in K,we obtain

[wk]= 2([w1])2[wk+1]21[wk][w1]

1 + 1[w][w1]

+[wk1]

1 + 1[w][w1]

2.

Were to satisfy ([w]) = 1 and ([w1]) = , then this relation would become[wk]

= [wk+1], implying that the (1,1)-entry of(w)k+1 would agree with that of

(w)k.


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17

Motivated by these observations, let us define W to be the matrix in (9) with([w]) replaced by 1 and ([w1]) replaced by . Denote by E0 (and F0) the matrix

corresponding to (E) (and (F) respectively) in our basis.Lemma 5.3. Let W be defined as above, but with y set equal to 1+ + (1 1)x, and let

M = diag[, 1, 1]. Suppose that (Wk)11 = (Wk+1)11 and define R = WkE0 F0Wk. ThenR = 0.

Lemma 5.3 is proved below; let us first use it to complete the proof of Theorem1.5. For a given C, 1, note that ify = 1 + + (1 1)x and k 0,1 thenk= (W

k+1)11 (Wk)11 is a polynomial in x with positive degree. Thus it has a rootin C; set x as a root. Lemma 5.3 implies that from m M and w Wwe obtain awell-defined homomorphism : K GL3C.

To see that is irreducible, we use the following lemma.

Lemma 5.4. If is the augmentation induced from the KCH representation : K

GL3C defined above and K the longitude of K, then () = (2k+6).From Lemma 5.4 we conclude that (1 2k+6) divides A3

K(,). Were this factor

to exist in A2K

(,) there would be a factor of (1 4k+12) in the A-polynomial ofthe (2, 3, 2k+ 1)-pretzel knot. Results in [Mat02, Theorem 1.6] (upon translatingto when n = 2k+ 1) show that the variety has one, or two, components, dependingon whether 3 does not, or does, divide k 1. The geometric component appearingin both cases does not agree with (14k+12) and when 3|(k1) the non-geometricfactor in the A-polynomial is (1 4k+8).

By Lemma 3.4, cannot be reducible. The proof of Lemma 5.4 is presented afterthe proof of Lemma 5.3.

Proof of Lemma 5.3. We begin by noting some identities between the entries of Wk.We will sometimes use y to denote 1 + + (1 1)x.

Since Wk+1 = WkW we see that (Wk+1)11 =1

1 (Wk)11 + (W

k)12. Thus by our

assumption,

(10) (Wk)12 =

1 (Wk)11.

Now, upon noting that E0 = MWM1W1M1 and F0 = M1W1MWMW1, one

may compute that

(11) E0 =

1 1 +

(1)2

1x1

1(1

) x

0 0 1

F0 =

1 0 0

0y

xy12 + 1

0 1 x 12

Now write out the first row ofR in terms of the entries ofWk. In each case, anapplication of (10) shows that R1j = 0 for j = 1, 2, 3.

Using Wk+1 = WWk, and the supposition that (Wk+1)11 = (Wk)11, we also see that

(12)

1 (Wk)11 + x(W

k)21 +

2 +

3

(1 )2 + xy

(Wk)31 = 0


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18

We also have the following two identities:

(13)

1

1 (Wk

)21+

(W

k

)22=

(W

k

W)21=

(WW

k

)21=

(W

k

)11+

3

1 (Wk

)31;

(14)1

1 (Wk)31 + (W

k)32 = (WkW)31 = (WW

k)31 = (Wk)21 + y(W

k)31.

Additionally, note that y = 1 + + (1 1)x implies that 1y(1)21 +1 x = 0.

Now we may use (11), (12), and (13) to calculate that

R21 =

1 (Wk)21 +

1

(Wk)22 y

1

(Wk)21 + x

1 2

(Wk)31

(Wk)31

=

1 (Wk)21 +

1

(Wk)11 (Wk)22

y

1

(Wk)21 + x

1 2

(Wk)31

(Wk)31 1

(Wk)11

=1

1 y(1 ) 2

1 (Wk)21 2

1 (Wk)31 (Wk)31 1

2 1 (Wk)11 + yx(Wk)31=

1

1 y(1 ) 2

1 +1

x

(Wk)21 +

1 +

1 2

1 1

(Wk)31 = 0.

By a similar calculation, that uses (14), we also see that R31 = 0:

R31 =

1 (Wk)31 +

1

(Wk)32 +

(Wk)21

+ x

1 2

(Wk)31

= 1(Wk)21 (Wk)32

+

x

1 2

1

(Wk)31

= 1

1 y(1 )1

(Wk)31

+

x

1 2

1

(Wk)31

= 11 y(1 ) 2

1 +1

x

(Wk)31 = 0.

Similar, though perhaps more extensive, calculations may be carried out for theremaining four entries ofR, showing that each one vanishes. For these calculations,the assiduous reader will want to use identities

(15) x(Wk)31 + (Wk)33 = (W

k+1)32 = (Wk)22 + y(W

k)32 and

(16)

2 +

3

(1 )2 + xy

(Wk)31 +3

1 (Wk)32 + y(W

k)33 = (Wk+1)33 = (W

k)23 + y(Wk)33

in conjunction with (13) and (14) to reduce (Wk)23, (Wk)33, (W

k)22, and (Wk)32 to

expressions in (Wk)11, (Wk)21, and (Wk)31. This, combined with (12) will yield theresult.

Having proved Lemma 5.3 we obtain a well-defined augmentation : HC0(K) C satisfying ([w]) = 1, ([w1]) = , and ([wk]) = ([wk+1]). We will use theseproperties (and others implicit in the proof of Lemma 5.3) to prove Lemma 5.4.

First we require an auxiliary result. Recall that y = 1 + + (1 1)x.Lemma 5.5. Let j 0. Then [wjmw] = 2j+1[wj1mw].


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19

Proof. One can readily use the relations on the cord algebra and the facts [w]= 1

and [w1]= to verify the lemma when j = 0. The claim of the lemma concerns

entries of a linear combination of powers ofW. Indeed, [wj

mw] = [wj+1

] [wj

] and[wj1mw] = [wj] [wj1]. Moreover, by the definition of we know[wj+1] [wj] = (1 )(Wj+1

11 Wj

11) and [wj] [wj1] = (1 )(Wj

11 Wj1

11).

As 1, the statement to be proved is

(17) Aj := Wj+1

11 Wj

11 2j+1(Wj

11 Wj1

11) = 0

We know that A0 = 0. Except for finitely many values of, one checks that Wis

diagonalizable with eigenvalues 1, a

b, a+

b (for some square root ofb). Here

a, b are elements2 ofQ(, x) such that a2 b = 2. Let Sbe the matrix of eigenvectorsso that S1WS = diag[1, a

b, a +

b].

Using the diagonalization, we compute thatthereare numbers3A,B Q(, x,

b),

that are independent of j, such thatAj =

a

bj1

2j+1 (a

b)2j+1A +

a +

bj1

2j+1 (a +

b)2j+1

B.

As 2 = (a +

b)(a

b) we have that

Aj =a

bj

(a +

b)2j1 (a

b)2jA +

a +

bj

(a

b)2j1 (a +

b)2j

B

=a

b

a

bj

2j1 (a

b)2j1A +

a +

bj

2j1 (a +

b)2j1

B

+a

bj

2j1(2

b)A +a +

bj

(2

b)(a +

b)2j1B

= (a

b)Aj1 +a +

bj

1(2

b)A + (2

b)(a +

b)1B.

The numbers A,B, a, b are such that A + (a

b)B = 0, and this implies that

1

(2 b)A+(2 b)(a+ b)1

B = 0. By inductionAj1 = 0andsoAj = 0, concludingthe proof.

We collect some additional observations before proving Lemma 5.4.

Lemma 5.6. The matrix F0 commutes with M and E0W= F10 . We also have, for anyg, h K, the following identities on values of:

[gEwh]= [gF1h];(18)

[F1g]= [g]

= [gF1];(19)

[gE1]= [gw].(20)

Moreover, [gEimw]= i[gmw] and [E1mEig]

= i+1([g] [w1g]) for any g K

and any integer i.Proof. We readily check F0 commutes with M and E0W= F

10 from (11). Identity

(18) is a consequence ofE0W= F10 . That (19) holds is inferred from the block form

ofF0, and (20) from the observation that Wand E10

have the same first column.

2a =x(1)2+(22)

2(1) 123A =

(a1

b)(a

b)(x(1)2(aa+2

b(2

b)))

2

b(x(1)2(21)+2(342+(2a+5)(2a+1)) , B =(a1+

b)(a+

b)(x(1)2(aa+2+

b(2+

b)))

2

b(x(1)2(21)+2(342+(2a+5)(2a+1))


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20

To see that [gEimw]= i[gmw] for any g K, note that Ei = mwmiw1m1

and use that [gmwmi]= i[gmw]. Finally, since [E1]

= [w]

= 1, we know that for

g K[E1mEig] = [Ei1g] [Eig]

= [mw][miw1m1g]= [w](i[w1m1g])= i+1([w1m1][g] + [w1g])= i+1([g] [w1g]).

Proof of Lemma 5.4. In Figure 1 we depict a projection of K and a longitude of Kwith blackboard framing. Denote the longitude there depicted by L; we have

= m(2k+6)L in K. We will show that [L] = 1 . As a result (1 ) = [] =[m(2k+6)L]

= (2k+6)(1 ), and so = (2k+6).

A computation from the projection in Figure 1 allows us to write L in terms ofm,w, and E:

L = (Ew)(E1mE)Ekwk((Ew)1mEw)(E1mE)(Ew)kwk((Ew)1mEw)E1.

Using (18) and that F0 and M commute from Lemma 5.6, we may replace(Ew)1m(Ew) with m in the preceding expression without changing the value underthe map . From this observation and (19) we obtain

[L]= [E1mEk+1wk(mE1mE)FkwkmE1].

Now use the relation wk

E = Fwk

in K and (20) to obtain[L]

= [E1mEk+1wk(mE1mE)wkEkmw].

The facts [gEimw]= i[gmw] and [E1mEig]

= i+1([g] [w1g]) for any g K

inform us that

[L]= 2k([wk(mE1mE)wkmw] [wk1(mE1mE)wkmw]).

By applying relations in (2), (20) and using that [wkmw] = [wk+1] [wk] = 0,[L]

= 2k([wk(mE1mE)wkmw] [wk1(mE1mE)wkmw])= 2k([wkmwkmw] [wkmE1][Ewkmw] [wk1mwkmw] + [wk1mE1][Ewkmw])= 2k([wkmwkmw]

[wk1mwkmw] + [wk1mw][wk1mw])

= 2k1([wkmwkE1mw] [wk1mwkE1mw] + [wk1mw][wk1E1mw])= 2k1([wkmF1wkmw] [wk1mF1wkmw] + [wk1mw][w1F1wkmw])= 2k1([wk+1mwkmw] [wkmwkmw]).

We will calculate that

(21) [wk+1mwkmw]= 2k+2 and [wkmwkmw] = 2k+1.


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21

Our calculation on ([L]) then implies that [L]= 1. We compute (21) by showing

that for any i 0,

(22) i[wk+imw] + i[wkimw] = 0,

Taking i = kwe see that 0= k[w2kmw] + k[mw] which implies [wkmwkmw]

=

[w2kmw]= 2k+1, the first equality here because [wkmw] = 0. By taking i = k+ 1

we get that [w2k+1mw] = 2k+2[w1mw] = 2k+2.Turning to the proof of (22), note that when i = 0, both sides are zero. For the

case i = 1 we have

1[wk+1mw] + [wk1mw]= [wk1E1mw] + 1[wmwk+1]= [wk1E1w] [wk] 1([m1w1][wmwk+1] + [mwk+1]) + [wk+1]= [wk+1]

[wk] + [wk1E1w]

[w1F1wk+1]

= [wk+1] [wk] = 0,

the second to last equality from wkE1 = F1wk. We see that (22) holds for i = 1.Given some i > 1, if (22) holds for j < i then

i[wk+imw] + i[wkimw] = i1[wkiE1mw] + i[wmwk+i]= i1[wkiE1mw] i

[w1m1][wmwk+i] + [mwk+i]

+ i+1[wk+i]

= i1[wkiE1w] i1[wki+1] + i+1[wk+i] i+1[w1F1wk+i]= i1[wiF1wk+1] i1[wki+1] + i+1[wk+i] i+1[wk1E1wi]= i1[wki+1mw] + i+1[wk+i1mw]

+ i

1

[wi+1

m1

w1

m1

][wmwk+1

] + i+1

[wk

1

mw][w1

m1

wi

]= i1[wi+1m1w1m1][wmwk+1] + i+1[wk1mw][w1m1wi],

thelast equality from theinduction hypothesis. Now[wi+1m1w1m1] = 1[wimw]and also [w1m1wi]

= [wi1mw]. But as was shown in Lemma 5.5, [wi1mw] =

2i1[wimw] for all i. And so

i[wk+imw] + i[wkimw]= i1[wimw](1[wk+1mw]) i+1[wk1mw](2i1[wimw])= i1[wimw]

1[wk+1mw] + [wk1mw]

= 0,

since (22) holds when i = 1.

References

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[CCG+94] D. Cooper, M. Culler, H.Gillet, D.D. Long, andP. Shalen. Plane curvesassociated to charactervarieties of 3-manifolds. Invent. Math., 118:4784, 1994.

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Math. Soc., 359(7):33013335 (electronic), 2007. arXiv: math/0505451.[Eli98] Y. Eliashberg. Invariants in contact topology. In Proceeding of the International Congress of

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Mathematics Department, Duke University, Box 90320, Durham, NC 27708E-mail address: [email protected]

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