The action of the Kauffman bracket skein algebra of the torus on the Kauffman bracket skein module of the 3-twist knot complement
aa r X i v : . [ m a t h . G T ] F e b THE ACTION OF THE KAUFFMAN BRACKET SKEINALGEBRA OF THE TORUS ON THE KAUFFMANBRACKET SKEIN MODULE OF THE 3-TWIST KNOTCOMPLEMENT
R ˘AZVAN GELCA AND HONGWEI WANG
Abstract.
We determine the action of the Kauffman bracket skeinalgebra of the torus on the Kauffman bracket skein module of the com-plement of the 3-twist knot. The point is to study the relationshipbetween knot complements and their boundary tori, an idea that hasproved very fruitful in knot theory. We place this idea in the contextof Chern-Simons theory, where such actions arose in connection withthe computation of the noncommutative version of the A-polynomialthat was defined in [6], but they can also be interpreted as quantummechanical systems. Our goal is to exhibit a detailed example in a partof Chern-Simons theory where examples are scarce. Introduction
This paper should be viewed as a piece of experimental mathematics. Itdescribes the action of the Kauffman bracket skein algebra of the torus onthe Kauffman bracket skein module of the complement of the 3-twist knot,which is listed as the 5 knot in the knot table. Such computations have beendone before for the trefoil knot [8], the figure-eight knot [11], and (2 , p + 1)torus knots [18] as the main step in the computation of the noncommutativeversion of the A-polynomial defined in [6]. The noncommutative version ofthe A-polynomial has been linked to colored Jones polynomials in [6], [9],[7], and to SL (2 , C )-Chern-Simons theory [13], [4], a difficult area of math-ematics that has yet to be thoroughly understood. The theory of Kauffmanbracket skein modules has been linked to SL (2 , C )-character varieties [2],[20], as deformations of rings of affine characters, and as such they are alsosupposed to be related to SL (2 , C )-Chern-Simons theory, though it is notknown how.The case of the 3-twist knot is probably the most complicated examplethat can still be done by hand; this is why we want to present it to the public.Computational complexity grows very fast in the theory of skein modules,hence there are few examples. A striking feature exhibited in this paper isthe ocurrence of Jones-Wenzl idempotents in the computations within skein Mathematics Subject Classification.
Key words and phrases.
Kauffman bracket, skein modules, Chern-Simons theory. modules. This feature has been observed before by the first author; it seemsthat (arbitrary) skein computations tend to structure themselves in termsof Jones-Wenzl idempotents.In [21], E. Witten has related SU (2)-Chern-Simons theory to quantiza-tions of moduli spaces of flat connections on surfaces, which then leads toquantum mechanical models (see [12] for a complete discussion). Thesequantum mechanical models have a combinatorial version that arises fromquantizing Wilson lines, formulated using reduced skein modules (which arethe building blocks of the topological quantum field theory of Blanchet,Habbegger, Masbaum and Vogel [1]). The action of the Kauffman bracketskein algebra of the torus on the skein module of the knot complement thatmakes the object of this paper is similar to that quantum mechanical model,but here we work with the non-reduced version of skein modules. One mightask what does the model in which we work with the actual skein modulesand not their reduced versions represent? Given such questions, the lack ofexamples, and recent renewed interest in skein modules, we consider worthshowing this particular situation, as it might help clarify the general situa-tion.Given that the structure of the Kauffman bracket skein modules is nowknown for a fairly large family of knot and link complements [16], [17], wehope that the above work will be expanded to the study of structures thatarise from skein modules in other knot complements.We start with some background material. Throughout the paper t is avariable. A framed link in an orientable 3-manifold M is a disjoint unionof embedded annuli. If M is the cylinder over the torus, framed links areidentified with curves on the torus, with the annulus being parallel to thetorus. If we draw a framed link on paper, its framing is parallel to theplane of the paper, unless the link is drawn on the torus, when we use theprevious convention. Let L be the set of isotopy classes of framed linksin the manifold M , including the empty link. Consider the free C [ t, t − ]-module with basis L , and factor it by the smallest subspace containing allexpressions of the form − t − t − and (cid:13) + t + t − , where thelinks in each expression are identical except in a ball in which they look likedepicted. This quotient is denoted by K t ( M ) and is called the Kauffmanbracket skein module of the manifold [19]. The factorization allows us tosmoothen crossings (which we can create at will using isotopy) and to replacetrivial link components by a scalar. Because at each application of thefirst skein relation one term is replaced by two terms, the complexity ofcomputations grows exponentially, and so the computations in this paperare quite involved.For the cylinder over torus, T × I (where I = [0 , K t ( T × I ) is free with basis ( p, q ) T , p, q ∈ Z , p ≥
0, where( p, q ) T = T n (( p, q )) with T n the (normalized) Chebyshev polynomial of first HE ACTION OF THE KAUFFMAN BRACKET SKEIN ALGEBRA 3 kind defined by T ( x ) = 2, T ( x ) = x , and T n +1 ( x ) = xT n ( x ) − T n − ( x ), n = gcd( p, q ), and ( p, q ) is the curve of slope p/q on the torus. We haveproduct-to-sum formula( p, q ) T ∗ ( r, s ) T = t | pqrs | ( p + r, q + s ) T + t −| pqrs | ( p − r, q − s ) T . In this paper we focus on the 3-twist knot K drawn in bold line in Fig-ure 1. When talking about the complement of K we mean the compact yx Figure 1. orientable manifold S \ N ( K ) obtained by removing from the 3-sphere anopen regular neighborhood N ( K ) of K . The operation of gluing the cylinderover ∂N ( K ) = T to S \ N ( K ) induces a K t ( T × I )-left module structureon K t ( S \ N ( K )). In what follows we explicate this module structure.It was shown in [3] that K t ( S \ N ( K )) is a free C [ t, t − ]-module with basis x n y k , n ≥
0, 0 ≤ k ≤
3, where x, y are shown in Figure 1. It suffices tounderstand the action of a set of generators of K t ( T × I ) on the basis, andas generators we have chosen (0 , T , (1 , − T and (1 , − T . The action of(0 , T is (0 , T x n y k = x n +1 y k , so we focus on the other two.Using the fact the x can be pulled back into the cylinder over the boundaryas the skein (0 , T , and using the relations (1 , q ) T (0 , T = t (1 , q + 1) T + t − (1 , q − T , and (0 , T (1 , q ) T = t − (1 , q + 1) T + t (1 , q − T , we see thatthe action of (1 , q ) T on x n y k can be found easily if we know how (1 , q ) T acts on the basis elements 1 = y , y, y , y . It should also be noted that incomputations from this paper x behaves like a scalar .We will change the basis of K t ( S \ N ( K )) to S n ( x ) S k ( y ), where S n is the(normalized) Chebyshev polynomial of the second kind: S ( x ) = 1 , S ( x ) = x, S n +1 ( x ) = xS n ( x ) − S n − ( x ). As such, the basis elements are the curves x and y colored by Jones-Wenzl idempotents . There are two explanations forthis, one is practical: the formulas become simpler. But there is a deeperexplanation for this, namely that the polynomial S n is the character of the n + 1-dimensional irreducible representation of SL (2 , C ), and as such theskein S n ( x ) S k ( y ) consists of two Wilson lines (one for x and one for y )associated to irreducible representations. It is worth pointing out that thecolored Jones polynomials of a knot K are ( − n h S n ( K ) i , where h·i denotesthe Kauffman bracket (of knots and links in S ).In short, the goal of the paper is to find (1 , − T · S k ( y ) and (1 , − T · S k ( y ), k = 0 , , , R ˘AZVAN GELCA AND HONGWEI WANG Formulas in a quotient of the Kauffman bracket skeinmodule of cylinder over the twice punctured disk
It is known that the Kauffman bracket skein module of the cylinder overthe twice punctured disk, i.e. a disk with two disjoint open disks beingremoved, is free with basis x m y n z k , m, n, k ≥
0, where x and z are curvesthat are parallel with the boundaries of the two open disks that have beenremoved, and y is a curve parallel to the boundary of the original disk.In Sections 2 and 3, we make the following convention. We schematicallyrepresent the cylinder over the twice punctured disk sideways, by drawingonly the two curves that trace the punctures in the cylinder. These curveswill either be represented as twisting around each other, such as in the firstdiagram from Figure 2, or as two parallel lines such as in the second and thirddiagram from the same figure. Closed curves in the diagram comprise skeins,taken with the blackboard framing. Whenever a number is written next toa curve, such as the k written next to the y -curve in the first diagram fromFigure 2, that number indicates that the skein contains that many parallelcopies of that curve, as such as in our example there are k parallel copies of y . We factor the Kauffman bracket skein module of the cylinder over thetwice punctured disk by the relation x = z and perform all computations from this section of the paper in this quotient.All computations in this section can be used for general twist knots.In Figure 2, we recall the skeins X i ∗ y k from [10] and define the skeins Y ∗ y k . In the first diagram, the index i counts the crossings of the twostrands that define the genus 2 handlebody. For Y ∗ y k , the undercrossingscan be at the bottom and the overcrossings at the top, as one diagram ismapped into the other by isotopy. ... i i − kk kY ∗ y k : X ∗ y k : X i ∗ y k : Figure 2.
HE ACTION OF THE KAUFFMAN BRACKET SKEIN ALGEBRA 5
Lemma 2.1.
The skeins X ∗ y k and Y ∗ y k , k ≥ , satisfy the recursions: X ∗ y k +1 = t yX ∗ y k + ( t − − t ) Y ∗ y k + 2(1 − t ) x y k , k ≥ ,Y ∗ y k +1 = t − yY ∗ y k + ( t − t − ) X ∗ y k + 2(1 − t − ) x y k , k ≥ ,X ∗ y = X = − t y − t x , Y ∗ y = Y = − t − t − . Proof.
We start computing X ∗ y k +1 as in Figure 3. The first diagram is X ∗ y k +1 = = t + + + t − kk k kk Figure 3. computed as in Figure 4, and is equal to t yX ∗ y k − t Y ∗ y k − t x y k . Sothe first term is t yX ∗ y k − t Y ∗ y k − t x y k . The sum of the other termsis equal to x y k + x y k + t − Y ∗ y k . Adding we obtain the first recursion.Similarly for the second recursion. (cid:3) Applying Lemma 2.1 we obtain X ∗ y = − t S ( y ) − t x S ( y ) + 2(1 − t ) x + ( t − − t − ); Y ∗ y = − ( t + t − ) S ( y ) + (2 − t − t − ) x ; X ∗ y = − t S ( y ) + ( − t ) x S ( y ) + ( − t + 2) x S ( y )+( t − t − t − ) S ( y ) + ( − t + 2 t − − t − ) x ; Y ∗ y = ( − t − t − ) S ( y ) + (2 − t − t − ) x S ( y )+( − t − t − + 2 t + 2 t − ) x + ( t + t − − t − t − );The following result is a straighforward generalization of Lemma 1 in [10]. Lemma 2.2.
The skeins X i ∗ y k , i, k ≥ , satisfy the recursive relation X i +2 ∗ y k = t yX i +1 ∗ y k − t X i ∗ y k − t x y k ,X ∗ y k = t yX ∗ y k − t Y ∗ y k − t x y k . R ˘AZVAN GELCA AND HONGWEI WANG +( − t )= − t ( − t − ) − t x y k = t + − t x y k = t yX ∗ y k + ( − t ) − t x y k = t yX ∗ y k − t − t − t x y k = − t k kk kk k ( − t ) k kk Figure 4.
As a consequence, we obtain X = − t S ( y ) − t S ( x ) S ( y ) − t S ( y ) − t S ( x ) − t ; X = − t S ( y ) − t S ( x ) S ( y ) − t S ( y ) − t S ( x ) S ( y ) − t S ( y ) − t S ( x ) − t ; X = − t S ( y ) − t S ( x ) S ( y ) − t S ( y ) − t S ( x ) S ( y ) − t S ( y ) − t S ( x ) S ( y ) − t S ( y ) − t S ( x ) − t ; HE ACTION OF THE KAUFFMAN BRACKET SKEIN ALGEBRA 7 X ∗ y = − t S ( y ) − t S ( x ) S ( y ) − t S ( y ) − t S ( x ) S ( y )+( − t − t ) S ( y ) + ( − t + 1) S ( x ) + ( − t + 1); X ∗ y = − t S ( y ) − t S ( x ) S ( y ) − t S ( y ) − t S ( x ) S ( y )+( − t − t ) S ( y ) + ( − t − t ) S ( x ) S ( y ) + ( − t − t ) S ( y ) − t S ( x ) + ( − t + 1); X ∗ y = − t S ( y ) − t S ( x ) S ( y ) − t S ( y ) − t S ( x ) S ( y )+( − t − t ) S ( y ) + ( − t − t ) S ( x ) S ( y ) + ( − t − t ) S ( y )+( − t − t ) S ( x ) S ( y ) + ( − t − t ) S ( y ) − t S ( x ) − t ; X ∗ y = − t S ( y ) − t S ( x ) S ( y ) − t − S ( y ) − t S ( x ) S ( y )+( − t − t ) S ( y ) + ( − t − t + 2) S ( x ) S ( y )+( − t − t + 2) S ( y ) + ( − t − t + 2 t − ) S ( x )+( − t + t − − t − ); X ∗ y = − t S ( y ) − t S ( x ) S ( y ) − t S ( y ) + ( − t ) S ( x ) S ( y )+( − t − t ) S ( y ) + ( − t − t ) S ( x ) S ( y ) + ( − t − t ) S ( y )+( − t − t + 2) S ( x ) S ( y ) + ( − t − t + 1) S ( y )+( − t − t + t − ) S ( x ) + ( − t − t + t − ); X ∗ y = − t S ( y ) − t S ( x ) S ( y ) − t S ( y ) − t S ( x ) S ( y )+( − t − t ) S ( y ) + ( − t − t ) S ( x ) S ( y )+( − t − t ) S ( y ) + ( − t − t ) S ( x ) S ( y )+( − t − t − t ) S ( y ) + ( − t − t + 1) S ( x ) S ( y )+( − t − t + 1) S ( y ) + ( − t − t ) S ( x ) + ( − t − t + t − ) . Lemma 2.3.
The skeins X i ∗ y k , i, k ≥ , satisfy the recursive relation X ∗ y k = t − X ∗ y k +1 − t − x y k − t − Y ∗ y k and for i ≥ , X i +2 ∗ y k = t − X i +1 ∗ y k +1 − t − x y k − t − X i ∗ y k . Proof.
To compute X i ∗ y k +1 , we separate a y from y k +1 , slide it so as toproduce two crossings in the link diagram, then solve the crossings using theKauffman bracket skein relation as in Figure 5.In the first term, by sliding the strand to the right we see that this termequals t X i +1 ∗ y k . The second and third terms are each equal to 2 x y .The last term is equal to X i − ∗ y k if i ≥ Y ∗ y k if i = 1. (cid:3) Define the skeins A ∗ y k , A ∗ y k , B ∗ y k , B ∗ y k as in Figure 6. R ˘AZVAN GELCA AND HONGWEI WANG ... i i − k ++ ... i i − k ... i i − k ... i i − k ... i i − k + t − t Figure 5. kA ∗ y k kA ∗ y k k kB ∗ y k B ∗ y k Figure 6.
Lemma 2.4.
The following relations hold A ∗ S k ( y ) = ( − t k +2 − t − k − ) B ∗ S k ( y ) ,A ∗ S k ( y ) = ( − t k +2 − t − k − ) B ∗ S k ( y ) B ∗ y k = t yB ∗ y k − + (1 − t − ) B ∗ y k − ,B ∗ y k = t − yB ∗ y k − + (1 − t ) B ∗ y k − ,B ∗ y = B ∗ y = x. Proof.
The formulas for A ∗ S k ( y ), A ∗ S k ( y ) follow from the standard prop-erties of Jones-Wenzl idempotents.For B ∗ y k (see Figure 7) resolve the crossings specified by the arrow toobtain the first sum in this figure. Perform an isotopy of the first skeinfrom the sum to obtain the first skein on the second row (in the processwe remove and then add a positive twist), then remove a negative twistfrom the second term and perform an isotopy in this term. Then apply theKauffman bracket skein relation in the place specified by the arrow to obtainthe desired relation. HE ACTION OF THE KAUFFMAN BRACKET SKEIN ALGEBRA 9 B ∗ y k is obtained by reflecting B ∗ y k over a horizontal line, and underreflections, in the Kauffman bracket t is replaced by t − . (cid:3) = t + t − ( − t − ) + t − = tk − k − k − k − k − Figure 7.
Corollary 2.5.
The following formulas hold A ∗ y = ( − t − t − ) x,A ∗ y = ( − t − t − ) xS ( y ) + ( − t + 1 − t − + t − ) x,A ∗ y = ( − t − t − ) xS ( y ) + ( − t + 1 − t − + t − ) xS ( y )+( − t − t − − t − + t − ) xA ∗ y = ( − t − t − ) xS ( y ) + ( − t + 1 − t − + t − ) xS ( y )+( − t − t − + t − t − + t − − t ) xS ( y )+( − t − t + 2 + t − − t − + t − ) x. Corollary 2.6.
The following formulas hold A ∗ y = ( − t − t − ) x,A ∗ y = ( − t − t − ) xy + ( t − t + 1 − t − ) x,A ∗ y = ( − t − t − ) xS ( y ) + ( t − t + 1 − t − ) xS ( y )+( − t − t − − t + t ) xA ∗ y = ( − t − t − ) xS ( y ) + ( t − t + 1 − t − ) xS ( y )+( − t − + t − − t − t + t − t − ) xS ( y )+( − t − − t − + 2 + t − t + t ) x. Proof.
The skein A ∗ y k is the reflection of A ∗ y k over a horizontal line. Toget the formulas for A ∗ y k , swap t and t − in the formulas for A ∗ y k . (cid:3) We define the skeins C j ∗ y k , D j ∗ y k , E j , F j as in Figure 8. j jk kC j ∗ y k D j ∗ y k jjE ∗ y j F ∗ y j Figure 8.
Lemma 2.7.
The skeins C j ∗ y k , D j ∗ y k , E ∗ y j , F ∗ y j satisfy the followingrelations C j ∗ y k = t C j +1 ∗ y k − + (1 − t − ) D j ∗ y k − D j ∗ y k = t − D j +1 ∗ y k − + (1 − t ) C j ∗ y k − C j ∗ y = xB ∗ y j , D j ∗ y = t − xB ∗ y j + (1 − t ) E ∗ y j E ∗ S j ( y ) = ( − t j +2 − t − j − ) S j ( y ) , F ∗ S j ( y ) = ( t j +2 + t − j − ) S j ( y ) . Proof.
For C j ∗ y k , separate a strand from y k as in the skein on the left inFigure 9, then resolve the crossings defined by arrows to obtain t C j +1 ∗ y k − + D j ∗ y k − + D j ∗ y k − + t − ( − t − t − ) D j ∗ y k − , which yields the relation. For D j ∗ y k do the same in the skein on the right. k j k j Figure 9.
The skein C j ∗ y is the mirror image of xA j over a vertical line, so, asa skein, equals xA j . Resolving the two crossings in D j ∗ y specified inFigure 10, we get t ( − t − t − ) E j + E j + E j + t − xA j . The formulas for E ∗ S j ( y ) and F ∗ S j ( y ) follow from standard properties of Jones-Wenzlidempotents. (cid:3) HE ACTION OF THE KAUFFMAN BRACKET SKEIN ALGEBRA 11 j Figure 10.
Corollary 2.8.
The following formulas hold C ∗ y = x C ∗ y = x S ( y ) + ( t − + t − t − t − ) x + ( t − t − − t + t − ) C ∗ y = x S ( y ) + ( t + t − − t − t − ) x S ( y ) + (3 − t − t − ) x +( − t − − t + t + t − ) S ( y ) C ∗ y = x S ( y ) + ( t − + t − t − t − ) x S ( y )+( t + t − + 4 − t − t − − t − − t ) x S ( y )+( − t − − t + 4 t − + 5 t + 2 t − t − t − + t − ) x +( − t + t − t − + t − ) S ( y )+( − t − + 3 t − − t + 3 t ) . We define the skeins G ∗ y k , H ∗ y k , and H ∗ y k as in Figure 11. The nextresult has a proof analogous to that to Lemma 2.4. kG ∗ y k k k H ∗ y k J ∗ y k Figure 11.
Lemma 2.9.
The following formulas hold G ∗ S k ( y ) = ( − t k +2 − t − k − ) H ∗ S k ( y ) ,H ∗ y k = t yH ∗ y k − + (1 − t − ) J ∗ y k − , H ∗ y = y,J ∗ S k ( y ) = ( − t k +2 − t − k − ) S k ( y ) . Corollary 2.10.
The following formulas hold G ∗ y = ( − t − t − ) S ( y ) ,G ∗ y = ( − t − t − ) S ( y ) + ( − t − − t − ) ,G ∗ y = ( − t − t − ) S ( y ) + ( − t − t − − t − ) S ( y ) ,G ∗ y = ( − t − t − ) S ( y ) + ( − t − t − − t − ) S ( y ) + ( − t − − t − ) . Formulas in the Kauffman bracket skein module of the3-twist knot complement
For the complement of the 3-twist knot Lemma 6 in [10] gives
Lemma 3.1.
For all k ≥ we have X ∗ y k = − t − X ∗ y k − t − x y k . Lemma 3.1 yields different formulas for X ∗ y k , which, combined withthose in §
2, give relations that successively compute S ( y ), S ( y ), S ( y ): S ( y ) = [ − t − S ( x ) − ( t − + t − )] S ( y ) + [ − (2 t − + t − ) S ( x ) − ( t − + t − )] S ( y ) + [ − t − + t − ) S ( x ) − ( t − + 2 t − )] S ( y )+[ − ( t − + 2 t − ) S ( x ) − ( t − + 2 t − )]; S ( y ) = [( t − S ( x ) + ( t − + t − ) S ( x ) + ( t − + t − )] S ( y )+[(2 t − + t − ) S ( x ) + (3 t − + 2 t − + t − ) S ( x ) + ( t − + t − )] S ( y )+[(2 t − + 2 t − ) S ( x ) + (3 t − + 4 t − + 2 t − ) S ( x )+(2 t − + 2 t − + t − )] S ( y ) + ( t − + 2 t − ) S ( x )+(3 t − + 4 t − + t − ) S ( x ) + (2 t − + 2 t − + 2 t − ); S ( y ) = [ − t − S ( x ) + ( − t − − t − ) S ( x ) + ( − t − − t − ) S ( x )+( − t − − t − )] S ( y ) + [( − t − − t − ) S ( x )+( − t − − t − − t − ) S ( x ) + ( − t − − t − − t − ) S ( x ) ++( − t − − t − − t − )] S ( y ) + [( − t − − t − ) S ( x )+( − t − − t − − t − ) S ( x ) + ( − t − − t − − t − ) S ( x )+( − t − − t − )] S ( y ) + [( − t − − t − ) S ( x )+( − t − − t − − t − ) S ( x ) + ( − t − − t − − t − ) S ( x )+( − t − − t − − t − )] . Also, from Lemma 3.1, we obtain X = t S ( y ) + ( t S ( x ) + t ) S ( y ) + (2 S ( x ) + 1) S ( y ) + t − S ( x ) + t − . Combining Lemma 2.3 and Lemma 3.1 we obtain the following recursivescheme that allows the writing of X j ∗ S k ( y ), 1 ≤ j ≤ , ≤ k ≤ S j ( x ) S k ( y ), 0 ≤ j, ≤ k ≤ X ∗ S k +1 ( y ) = t X ∗ S k ( y ) + t − Y ∗ S k ( y ) − X ∗ S k − ( y ) + (2 S ( x ) + 2) S k ( y ) ,X ∗ S k +1 ( y ) = t X ∗ S k ( y ) + t − X ∗ S k ( y ) − X ∗ S k − ( y ) + (2 S ( x ) + 2) S k ( y ) ,X ∗ S k +1 ( y ) = t X ∗ S k ( y ) + t − X ∗ S k ( y ) − X ∗ S k − ( y ) + (2 S ( x ) + 2) S k ( y ) ,X ∗ S k ( y ) = − t − X ∗ S k ( y ) − t x S k ( y ) . Using also the formulas for X , X , X , X and those that express S ( y ), HE ACTION OF THE KAUFFMAN BRACKET SKEIN ALGEBRA 13 S ( y ), and S ( y ) in terms of the basis we obtain X ∗ S ( y ) = t S ( y ) + t S ( x ) S ( y ) + t S ( x ) S ( y ) + S ( x ) + 2; X ∗ S ( y ) = − t S ( y ) − S ( x ) S ( y ) + ( − t − S ( x ) − t − ) S ( y ) − t − S ( x ) − t − .X ∗ S ( y ) = t S ( y ) + t S ( x ) S ( y ) + [(2 + t ) S ( x ) + 2] S ( y )+( t + 2 t − ) S ( x ) + (2 t + t − − t − ); X ∗ S ( y ) = − t S ( y ) − t S ( x ) S ( y ) − t − ; X ∗ S ( y ) = S ( y ) − t − S ( y ) + t − .X ∗ S ( y ) = − t S ( y ) + ( − t + 1) S ( x ) S ( y ) + 2 S ( y ) + [2 t − S ( x )+( − t − − t − )] S ( y ) + (2 t − − t − ) S ( x ) + ( − t − − t ); X ∗ S ( y ) = t S ( y ) + 2 S ( x ) S ( y ) + S ( y ) + [2 t − S ( x ) + 2 t − ] S ( y ); X ∗ S ( y ) = [ − t − S ( x ) − t − ] S ( y ) + [ − t − S ( x ) − t − ] S ( y )+[ − t − S ( x ) − t − ] S ( y ); X ∗ S ( y ) = [2 S ( x ) + ( t + 2)] S ( y ) + [(2 t + 2 t − ) S ( x )+( t − t − )] S ( y ) + [(2 + 2 t − − t − ) S ( x )+(2 + 2 t − − t − )] S ( y ) + (2 + 2 t − − t − ) S ( x )+(2 t − + t − − t − ); X ∗ S ( y ) = S ( x ) S ( y ) + t − S ( y ) − t − S ( y ) − t − S ( x ); X ∗ S ( y ) = [ t − S ( x ) + (2 t − + t − ) S ( x ) + (2 t − + t − )] S ( y )+[(2 t − + 2 t − ) S ( x ) + (5 t − + 6 t − ) S ( x ) + (2 t − + 4 t − )] S ( y )+[(2 t − + 2 t − ) S ( x ) + (10 t − + 5 t − ) S ( x ) + (3 t − + t − )] S ( y )+(2 t − + t − ) S ( x ) + (6 t − + 3 t − ) S ( x ) + (4 t − + 2 t − ) . The action of the skeins (1 , − T and (1 , − T on the skeinmodule of the 3-twist knot complement As said in the introduction, we compute the action of (1 , − T , (1 , − T from K t ( T × I ) on the basis elements S k ( y ), 0 ≤ k ≤ K t ( S \ N ( K )).The skeins (1 , − T and (1 , − T are depicted in Figure 12, with the cylinder T × [0 ,
1] embedded as a regular neighborhood of the boundary of the knotcomplement. Before starting the computation we prove a lemma.
Lemma 4.1.
The identities from Figure 13 hold. Here the curved strandcan encircle several parallel straight strands.Proof.
Pull the strand in the term on the left until you create a negativetwist in the first identity and a positive twist in the second identity, thenresolve the crossing. (cid:3) (1 , − T (1 , − T Figure 12. = − t − − t − = − t − t Figure 13.
First we find (1 , − T · y k , k = 0 , , ,
3. For that we add y k to the skeinrepresented by the curve on the left side of Figure 12, then push this curveinside the knot complement. There is a small technical detail. The framingthat the curve inherits from the torus does not coincide with the framingdefined by the plane of the paper. The resulting skein (with framing definedby the plane of the paper) is the skein from Figure 14 multiplied by t .We compute this skein from the figure first, then multiply by the adjustingfactor in the end. In Figure 14 we have labeled the 5 crossings in the orderin which they are resolved. We use a boldface curve for y k , and here and insubsequent figures no longer write the label y k next to it.
13 4 5 Figure 14.
We denote by a string of length k consisting of +’s and − ’s inside doublebrackets the skein obtained from (1 , − T · y k by smoothening the first k HE ACTION OF THE KAUFFMAN BRACKET SKEIN ALGEBRA 15 crossings (in the order of labels), horizontally for a plus and vertically for aminus. For example the Kauffman bracket skein relation applied to the firstcrossing reads (1 , − T · y k = t ((+)) + t − (( − )). Applying the Kauffmanbracket skein relation repeatedly we obtain(1 , − T · y k = t ((+ + ++)) + t ((+ + + − +)) + t ((+ + + − − ))+ t ((+ + − )) + ((+ − )) + t − (( − ))where the skeins from this expression are shown in Figure 15. ((+ + ++)) ((+ + + − +)) ((+ + + − − ))((+ + − )) ((+ − )) (( − )) Figure 15.
After removing two negative twists in (( − )) and focusing on the lowerpart of the twist knot only, we obtain that (( − )) is equal to the first skeinin Figure 16 multiplied by t − . This new skein is computed as shown in thefigure by using the skein relation and sliding the strands. In this sum thefirst term is t xy k +1 , the third is xy k , and the fourth is t − xX ∗ y k . = t ++ + t − Figure 16.
Let us focus on the second term in the sum. Applying Lemma 4.1 in theplace specified by the arrow, we can transform it as in Figure 17. The firstterm is just − t xy k +1 . By applying Lemma 4.1 at the two places specifiedby arrows we can transform the first term into − t x y k − t xy k − t xX ∗ y k − t xX ∗ y k . So the term we are computing equals − t xy k +1 − t x y k − t y k − t xX ∗ y k − t xX ∗ y k . = − t − t Figure 17.
Therefore (( − )) = − xX ∗ y k + ( t − − t − ) xX ∗ y k − t − x y k + ( t − − t − ) xy k , which after applying Lemma 3.1 becomes(( − )) = t − xX ∗ y k + ( t − − t − ) xy k . To compute ((+ + − )) we look again at the lower part of the twist knotand apply Lemma 4.1 in the places specified by arrows in Figure 18. Inthe last sum, by resolving the two crossings in each diagram by the skeinrelation we find that the first term is t − xX ∗ y k +1 + t − x y k +1 + t − xX ∗ y k +1 + ( − t − − t − ) xX ∗ y k +1 = ( t − − t − ) xX ∗ y k +1 + t − x y k +1 . − t − = t − + t − − t − = − t − Figure 18.
Resolving the crossings in the second term we obtain that it is equal to t − xy k +1 X + t − A + t − B + t − xX ∗ y k +1 , where skeins A and B are as in Figure 19. Compute B by applying Lemma 4.1at the location specified by arrow to obtain B = − t xX ∗ y k +1 − t xy k +1 .Then transform A by an isotopy, use Lemma 4.1 as shown in Figure 20 toobtain A = t − xX ∗ y k +1 + t xy k +2 + x y k +1 + xy k +1 . Substitute X by HE ACTION OF THE KAUFFMAN BRACKET SKEIN ALGEBRA 17 B : A : Figure 19. − t y − t x to conclude that the second term in the three-term sum fromthe second line in Figure 18 equals( t − − t − ) xX ∗ y k +1 + t − xX ∗ y k +1 + ( t − − xy k +2 +( t − − t − ) x y k +1 + ( t − − t − ) xy k +1 . Similarly, the third term is (1 − t − ) xy k − t − xX ∗ y k . Hence((+ + − )) = − t − xX ∗ y k + t − xX ∗ y k +1 + ( t − − xy k +2 + t − x y k +1 + ( t − − t − ) xy k +1 + (1 − t − ) xy k . After applying Lemma 2.3 this becomes((+ + − )) = t − xX ∗ y k + ( t − − xy k +2 + t − x y k +1 +( t − − t − ) xy k +1 + 2 t − x y k + (1 − t − ) xy k . We turn to ((+ − )) and by working in the lower part of the knot, we apply = ++ t − + t Figure 20.
Lemma 4.1 as specified by an arrow in Figure 21, then remove a negativetwist in each resulting diagram, to obtain the two-term sum in Figure 21.A close examination shows that the last diagram in the figure is the sameas the last diagram in Figure 18. Adjusting for the different coefficient,we deduce that the second term from the sum in Figure 21 is ( − t − + t − ) xy k + t − xX ∗ y k . Computing similarly by resolving both crossingswith the Kauffman bracket skein relation, we find that the first term is t − xX ∗ y k +1 + ( t − − t − ) xy k +1 . Combining, we get((+ − )) = t − xX ∗ y k +1 + t − xX ∗ y k + ( t − − t − ) xy k +1 + ( t − − t − ) xy k . Now turn to the computation of ((+ + + − − )). After an isotopy at thetop part of the twist knot we obtain the first diagram in Figure 22, wherein this diagram we ignore the top part of the twist knot, as we will not useit again. We apply Lemma 4.1 as specified by the arrow to obtain the sum = t − + t − Figure 21. on the right. The first term is computed by applying Lemma 4.1 at the = − t − − t − Figure 22. point specified by the arrow, as in Figure 23. In the sum from Figure 23,the second term is t − xY ∗ y k . For the first term, we perform an isotopyto make it look like in Figure 24, then apply Lemma 4.1 to obtain that itis equal to − t − x y k +1 − t − xyX ∗ y k . Thus the first term of the sum in + t − t − Figure 23.
Figure 22 equals − t − x y k +1 − t − xyX ∗ y k + t − xY ∗ y k . t − Figure 24.
The second term in Figure 22 can be transformed by an isotopy intothe first skein in Figure 25. Then apply Lemma 4.1 as specified by thearrow to obtain the sum on the right. The second term is t − xy k . Thefirst term can be computed by applying the Lemma 4.1 as specified, and is − t − x y k − t − X ∗ y k . Combining the results we obtain((+ + + − − )) = − t − x y k +1 − t − xyX ∗ y k + t − xY ∗ y k − t − x y k − t − xX ∗ y k + t − xy k . Using Lemma 2.2 we write this as((+ + + − − )) = − t − xX ∗ y k − t − xX ∗ y k − t − x y k +1 − t − x y k + t − xy k . HE ACTION OF THE KAUFFMAN BRACKET SKEIN ALGEBRA 19 − t − = t − + t − Figure 25.
To compute the term ((+ + + − +)) we slide the skein to an area wherethe two strands of the twist knot are parallel, as in the first diagram fromFigure 26, then apply Lemma 4.1 at the point specified by the arrow, toobtain the first sum in this figure. = − t − − t − = t − ( y −
1) + t − Figure 26.
Next, apply Lemma 4.1 at the point specified by the second arrow inFigure 26 to obtain the sum on the second row. Resolve the crossings in thefirst diagram to obtain that the first term is t − ( y − t xy k + A ∗ y k + xy k +1 + t − xy k ). The second term is yA ∗ y k . Hence((+ + + − +)) = ( t − y − t − ) A ∗ y k + t − yA ∗ y k + t − xy k +3 +( t − + t − ) xy k +2 − t − xy k +1 + ( − t − − t − ) xy k . Finally, for ((+ + ++)), remove the twist and multiply the skein by − t ,slide the skein over the top of the diagram to get the first skein from Figure 27(again only the bottom of the diagram of the twist knot is shown, and theskein has been moved to the left off the area where the crossings occur).Apply Lemma 4.1 as specified by first arrow to obtain the first equality,then apply the lemma again as specified by second arrow to obtain (afterarranging the terms) the last sum in Figure 27. The second term is − t − yA ∗ y k . After applying Lemma 4.1 as specified by the arrow, the first term isequal to t − ( − y + 1)( − t − yA ∗ y k − t − A ∗ y k ). So((+ + ++)) = t − ( y − y ) A ∗ y k − t − ( − y + 1) A ∗ y k . To simplify the formulas we set u i = S i +1 ( x ) and q = t . − t = t + t − = t − ( − y + 1) − t − Figure 27.
Theorem 4.2.
The action of (1 , − T on K t ( T × I ) is given by (1 , − T · S k ( y ) = X ≤ j ≤ t k +2 j − α kj S j ( y ) where α , = ( − q − u , α , = − u − q + 1) u , α , = − u + ( q − u α , = − u + ( − q − u , α , = qu + (1 + q − ) u ,α , = (2 q + 1 + q − ) u + ( q + q − − q − ) u , α , = ( q + 2) u + ( q + 3) u ,α , = (2 q + 1 + q − ) u + (2 q + 4 + 2 q − ) u , α , = − qu + ( q − u +( − q − q − ) u , α , = ( − q − u + ( − q − − q − − q − ) u +( − q − − q − − q − + q − ) u , α , = ( − q − u + (2 q − − q − ) u +( q − q − q − − q − − q − − q − ) u , α , = ( − q − u +( − q − q − ) u + ( q − q − q + 2 − q − − q − ) u ,α , = qu + u + ( − q − ) u + (2 q − − q − ) u , α , = (2 q + 1) u +( q + 1 + q − ) u − u + ( q − + q − + q − − q − ) u , α , = (2 q + 2) u +(2 q + 1 + 2 q − ) u + ( − q − + 2 q − ) u + ( − q − + 2 q − − q − ) u ,α , = (2 q + 1) u + (4 q + 2 + q − ) u + (6 q − q − ) u +(4 q − − q − + q − − q − + q − ) u . Proof.
Combining the terms computed above, applying Lemmas 2.3 and 3.1,and multiplying with the frame adjusting factor t , we obtain:(1 , − T · y k = t − xX ∗ y k + ( t S ( y ) + t S ( y )) A ∗ y k +( t y + t S ( y )) A ∗ y k + [ t S ( y ) + 2 t S ( y ) + ( − t + 2 t ) S ( y )+( − t + 2 t − )] xy k . Substituting the formulas from § §
3, switching to the basis S j ( x ) S k ( y ),we obtain the formulas from the statement. (cid:3) HE ACTION OF THE KAUFFMAN BRACKET SKEIN ALGEBRA 21
Let us compute (1 , − T · y k , k = 0 , , ,
3. Again we push the (1 , − T skein inside the knot complement and adjust the framing from the plane ofthe torus to the plane of the paper, to get the skein from Figure 28 multipliedby − t . We compute first the skein from the figure, then adjust framing. Inthe figure we label the 5 crossings in the order they are resolved, and use aboldface curve for y k , as before.
13 4 5 Figure 28.
As this skein looks similar to (1 , − T · y k , we expand it in the same way:(1 , − T · y k = t ((+ + ++)) + t ((+ + + − +)) + t ((+ + + − − ))+ t ((+ + − )) + ((+ − )) + t − (( − )) , but now the diagrams of the 6 skeins are different (see Figure 29). ((+ + ++)) ((+ + + − +)) ((+ + + − − ))((+ + − )) ((+ − )) (( − )) Figure 29.
To compute (( − )) we remove the two negative twists (and multiply theskein by t − ), then resolve the two crossings using the Kauffman bracketskein relation. We obtain(( − )) = t − [ t x y k + X ∗ y k + X ∗ y k + t − ( − t − t − ) X ∗ y k ]= t − x y k + ( t − − t − ) X ∗ y k . Next, we focus on ((+ − )). After removing a twist and performing anisotopy, we obtain the first skein from Figure 30. Now use Lemma 4.1 toobtain the sum in the figure. − t − = t − + t − Figure 30.
Resolving the crossings with the skein relation, we find the first term: t − [ t ( − t − t − ) X ∗ y k +1 + X ∗ y k +1 + X ∗ y k +1 + t − x y k +1 ]= ( t − − t − ) X ∗ y k +1 + t − xy k +1 , and the second term: t − ( t x y k + X ∗ y k + X ∗ y k + t − ( − t − t − ) X ∗ y k )= ( t − − t − ) X ∗ y k + t − x y k . Combining we obtain((+ − )) = ( t − − t − ) X ∗ y k +1 + ( t − − t − ) X ∗ y k + t − xy k +1 + t − x y k . Next we compute ((++ − )), which after an isotopy becomes the first skeinin Figure 31. = − t − − t − Figure 31.
Apply Lemma 4.1 to transform this into the sum on the right side ofFigure 31. Apply Lemma 4.1 in the first term on the right, then resolve thecrossings to obtain that this term equals − t − [( − t − )( t x y k +2 + t − ( − t − t − ) X ∗ y k +2 + X ∗ y k +2 + X ∗ y k +2 )) − t − ( t ( − t − t − ) X ∗ y k +1 + t − x y k +1 + X ∗ y k +1 + X ∗ y k +1 )]= ( t − − t − ) X ∗ y k +2 + ( − t − + t − ) X ∗ y k +1 + t − x y k +2 + t − x y k +1 . HE ACTION OF THE KAUFFMAN BRACKET SKEIN ALGEBRA 23
Then resolve the crossings in the second term to obtain that it is equal to − t − [ t x y k + t − ( − t − t − ) X ∗ y k + 2 X ∗ y k ]= ( − t − + t − ) X ∗ y k − t − x y k . Combining, we obtain that((+ + − )) = ( t − − t − ) X ∗ y k +2 + ( − t − + t − ) X ∗ y k +( t − − t − ) X ∗ y k +1 + t − x y k +2 + t − x y k +1 − t − x y k . Which after applying Lemma 2.3 for the first and third terms becomes((+ + − )) = ( − t − + t − ) X ∗ y k + ( t − − t − ) X ∗ y k +1 + t − x y k +2 + (2 t − − t − ) x y k +1 − t − x y k . Compute ((+ + + − − )) by performing an isotopy over the top of the knot toobtain the first skein in Figure 32 (again, ignore the top of the knot), thenapply Lemma 4.1 as specified by arrow to obtain the next sum. Continueapplying Lemma 4.1 to each term as specified, to obtain the sum on thesecond row in Figure 32. Applying Lemma 4.1 three more times yields((+ + + − − )) = − t − x y k +2 − t − x y k +1 + t − x y k − t − x y k +1 − t − x y k − t − x y k − t − Y ∗ y k = − t − Y ∗ y k − t − x y k +2 − t − x y k +1 + ( t − − t − ) x y k . = − t − = t − − t − + t − x y k + t − + t − Figure 32.
To compute ((+ + + − +)), apply Lemma 4.1 as in Figure 33. Thefirst diagram on the right is x times the second diagram in Figure 15 (thatdenoted by ((+ + + − +)) in that figure). So the first term on the right is( − t − y + t − ) xA ∗ y k − t − xyA ∗ y k − t − x y k +2 − t − x y k +2 − t − x y k +3 + t − x y k + t − x y k + t − x y k +1 . The second term on the right can be transformed by an isotopy into thefirst skein in Figure 34. Apply Lemma 4.1 (see arrow) to transform it intothe sum on the right. Now apply Lemma 4.1 in each term as specified bythe arrows to obtain the sum on the second row. = − t − − t − Figure 33. = − t − − t − − t − − t − = t − + t − − t − Figure 34.
The first and third terms can be combined into ( − t − xy − t − x ) Z ∗ y k ,where Z ∗ y k is the skein in Figure 35. We resolve the crossings marked byarrows using the skein relation to obtain that Z ∗ y k = t xy k + A ∗ y k + xy k +1 + t − xy k . The second term from the last sum in Figure 34 can be slidback to the left over the crossing of the twist knot. Then the second term isjust − t − xA ∗ y k , while the last is − t − times the 180 ◦ rotation of C ∗ y k in the plane of the paper, so it is in fact equal to − t − C ∗ y k . Thus((+ + + − +)) = ( − t − y − t − y − t − + t − ) xA ∗ y k +( − t − y − t − ) xA ∗ y k − t − C ∗ y k − t − x y k +3 +( − t − − t − ) x y k +2 − t − x y k +1 + ( t − − t − ) x y k . = Figure 35.
For ((+ + ++)), remove the positive twist, multiply by − t , then slidethrough the top of the knot diagram to obtain the skein in Figure 36. ApplyLemma 4.1 as shown by arrow to obtain the first sum, then apply the lemmaagain in each term, as shown by arrows, to obtain the second sum. Apply HE ACTION OF THE KAUFFMAN BRACKET SKEIN ALGEBRA 25 − t = t + t − = − t − − t − − t − − t − Figure 36. again Lemma 4.1 in each term to obtain((+ + ++)) = t − x yG ∗ y k + t − x F ∗ y k + t − xyA ∗ y k + t − xA ∗ y k + t − x G ∗ y k + t − xA ∗ y k + t − xA ∗ y k + t − F ∗ y k = ( t − y + t − ) x G ∗ y k + ( t − x + t − ) F ∗ y k + ( t − y + 2 t − ) xA ∗ y k + t − xA ∗ y k . We set v i = S i ( x ) and q = t . Theorem 4.3.
The action of (1 , − T on K t ( T × I ) is given by (1 , − · S k ( y ) = X j =0 t − k − j β kj S j ( y ) , where β , = v + qv , β , = ( q + q + 1) v + ( q + 2 q ) v , β , = (2 q + 2 q ) v +(2 q + 2 q ) v , β , = (2 q + 2 q ) v + ( q − − q − ) v ,β , = − q − v + ( q − q + 1 − q − − q − ) v + ( q − q − − q − ) v ,β , = ( − q − − q − ) v + (2 q − q − q − − q − ) v + (2 q + 2 q − ) v ,β , = ( − q − − q − ) v + (3 q − q − − q − − q − ) v +( q − q − q − − q − ) v , β , = ( − q − − q − ) v +( − q − q + 1 − q − − q − ) v + ( − q − q + 2 − q − − q − ) v ,β , = q − v + ( − q + q − q − + q − + q − ) v + ( − q + 1 − q − + q − ) v +( − q − − q − − q − ) v , β , = (2 q − + q − ) v + ( − q + q + q − − q − + 2 q − + q − ) v + ( − q + q − q − + 2 q − ) v + ( − q − q − q − − q − − q − ) v , β , = (2 q − + 2 q − ) v + ( − q + 2 − q − + 2 q − + 3 q − + 2 q − ) v +( − q + 1 + q − − q − − q − + 4 q − ) v + ( − q − q − q − − q − − q − + 2 q − ) v , β , = (2 q − + q − ) v + ( − q + q + 1 − q − +3 q − + 3 q − + q − ) v + ( − q − q + 6 − q − − q − + 3 q − + 4 q − ) v +( − q − q + 1 + 3 q − − q − − q − ) v , β , = − q − v + ( q − q + q − − q − − q − ) v + ( q − q + 2 q − − q − + 2 q − + q − − q − ) v +( q + q − q + 3 − q − + 4 q − − q − + 5 q − + 7 q − − q − − q − − q − ) v ( − q + 3 q + 1 + 5 q − − q − − q − + 2 q − + 6 q − − q − − q − − q − ) v ,β , = ( − q − − q − ) v + (2 q − q − q − − q − − q − − q − ) v +( − q + 10 q − q − q − − q − − q − + 2 q − − q − − q − ) v + ( − q +24 q − q + 4 − q − + 9 q − − q − + q − − q − + 8 q − q − − q − ) v +( − q + 16 q − q + 3 − q − + 6 q − − q − − q − + 6 q − − q − ) v ,β , = ( − q − − q − ) v + (2 q − q − − q − − q − − q − ) v +( − q + 3 q − − q − + 5 q − − q − + 2 q − − q − ) v + ( − q − q +6 + 8 q − + 6 q − − q − − q − + 6 q − − q − − q − ) v + ( − q + 5+12 q − − q − + 3 q − − q − + 4 q − ) v , β , = ( − q − − q − ) v + (2 q − q − − q − + 8 q − + 10 q − − q − − q − + 2 q − − q − − q − ) v + ( − q +5 q + 3 q + 6 + 9 q − + 24 q − + t − − t − + 5 t − ) v + ( − q + 3 q + 3 q + 6+9 q − + 19 q − q − + 3 q − + 3 q − ) v . Proof.
Adding the terms and adjusting framing by − t we obtain(1 , − T · y k = ( t − t ) X ∗ y k +1 + ( t + 2 t − − t ) X ∗ y k +( − t + 1) X ∗ y k +1 + t − Y ∗ y k + [ t y + ( − t + t ) y +( − t + t )] xA ∗ y k + ( t y + t − t ) xA ∗ y k + C ∗ y k +( − t x − t ) F ∗ y k + ( − t y − t ) x G ∗ y k + t x y k +3 +3 t x y k +2 + (4 t − t ) x y k +1 + ( − t + 3) x y k . Then use the formulas in § § S j ( x ) S k ( y ). (cid:3) References [1] Ch. Blanchet, N. Habegger, G. Masbaum, P. Vogel,
Topological quantum field theo-ries derived from the Kauffman bracket , Topology, (1992), 685-699.[2] D. Bullock, Rings of Sl ( C ) -characters and the Kauffman bracket skein module ,Comment. Math. Helv. (1997), no. 4, 521–542.[3] D. Bullock, W. Lofaro, The Kauffman bracket skein module of a twist knot exterior ,Algebraic and Geometric Topology, (2005), 107–118.[4] T. Dimofte, Quantum Riemann surfaces in Chern-Simons theory , Adv. Theor. Math.Phys., (2013), 479–599.[5] Ch. Frohman, R. Gelca, Skein modules and the noncommutative torus , Trans. AMS, (2000), 4877-4888.
HE ACTION OF THE KAUFFMAN BRACKET SKEIN ALGEBRA 27 [6] Ch. Frohman, R. Gelca,W. Lofaro,
The A-polynomial from the noncommutativeviewpoint , Trans. AMS, S (2002)735-747.[7] S. Garoufalidis, T.T.Q. Le,
The colored Jones function is q-holonomic , Geom. Topol., (2005), 1253–1293.[8] R. Gelca, Non-commutative trigonometry and the A-polynomial of the trefoil knot ,Math. Proc. Camb. Phil. Soc., (2002), 311–323.[9] R. Gelca,
On the relationship between the A-polynomial and the Jones polynomial ,Proc. Amer. Math. Soc., (2001), Vol. 130, 1235–1241.[10] R. Gelca, F. Nagasato, Some results about the Kauffman bracket skein module of thetwist knot exterior , J. Knot Theory and Ramif., (2006), 1095–1106.[11] R. Gelca, J. Sain, The computation of the noncommutative A-ideal for the figureeight knot , J. Knot Theory and Ramif., (2004), 1–24.[12] R. Gelca, A. Uribe, Quantum mechanics and non-abelian theta functions for thegauge group SU (2), Fundamenta Math., (2015), 97–137.[13] S. Gukov, Three-Dimensional Quantum Gravity, Chern-Simons Theory, and the A-Polynomial , Commun. in Math. Phys., (2005), 577–627.[14] V.F.R. Jones,
Polynomial invariants of knots via von Neumann algebras , Bull. Amer.Math. Soc., (1995), 103-111.[15] L. Kauffman, State models and the Jones polynomial , Topology 26 no. (1987) 395-401.[16] T.T.Q. Le, The colored Jones polynomial and the A-polynomial of knots , Adv. inMath., (2006), 782–804.[17] T.T.Q. Le, A. Tran,
The skein module of two-bridge links , Proc. Amer. Math. Soc., (2014), 1045-1056.[18] F. Nagasato,
Computing the A-polynomial using noncommutative methods , J. KnotTheory. Ramif., (2005), Vol. 14, 735–749.[19] J.H. Przytycki, Skein modules of 3-manifolds , Bull. Pol. Acad. Sci.Math, (1991) 91-100.[20] J.H. Przytycki, A.S. Sikora,
On Skein Algebras And Sl ( C ) -Character Varieties ,Topology (2000), 115-148.[21] E. Witten, Quantum field theory and the Jones polynomial , Comm. Math. Phys., (1989), 351–399.
Department of Mathematics and Statistics, Texas Tech University, Lub-bock, TX 79409
Email address : [email protected] Department of Mathematics and Physics, Texas A&M International Uni-versity, Laredo, TX 78041
Email address ::