aa r X i v : . [ m a t h . G T ] S e p ON KAUFFMAN BRACKET SKEIN MODULES AT ROOTS OFUNITY
THANG T. Q. LˆE
Abstract.
We reprove and expand results of Bonahon and Wong on central elements ofthe Kauffman bracket skein modules at root of 1 and on the existence of the Chebyshevhomomorphism, using elementary skein methods. Introduction
Kauffman bracket skein modules.
Let us recall the definition of the Kauffmanbracket skein module, which was introduced by J. Przytycki [Pr] and V. Turaev [Tu]. Let R = C [ t ± ]. A framed link in an oriented 3-manifold M is a disjoint union of smoothlyembedded circles, equipped with a non-zero normal vector field. The empty set is alsoconsidered a framed link. The Kauffman bracket skein module S ( M ) is the R -modulespanned by isotopy classes of framed links in M subject to the following relations L = tL + + t − L − (1) L ⊔ U = − ( t + t − ) L, (2)where in the first identity, L, L + , L − are identical except in a ball in which they look likein Figure 1, and in the second identity, the left hand side stands for the union of a link L Figure 1.
The links L , L + , and L − and the trivial framed knot U in a ball disjoint from L . If M = R then S ( R ) = R . Thevalue of a framed link L in S ( R ) = R = C [ t ± ] is a version of Jones polynomial [Kau].For a non-zero complex number ξ , let S ξ ( M ) be the quotient S ( M ) / ( t − ξ ), which is a C -vector space.For an oriented surface Σ, possibly with boundary, we define S (Σ) := S ( M ), where M = Σ × [ − ,
1] is the cylinder over Σ. The skein module S (Σ) has an algebra structureinduced by the operation of gluing one cylinder on top of the other. Supported in part by National Science Foundation.2010
Mathematics Classification:
Primary 57N10. Secondary 57M25.
Key words and phrases: Kauffman bracket skein module, Chebyshev homomorphism.
For a framed knot K in M and a polynomial p ( z ) = P dj =0 a j z j ∈ C [ z ], define p ( K ) by p ( K ) = d X j =0 a j K ( j ) ∈ S ( M ) , where K ( j ) be the link consists of j parallels of K (using the framing of K ) in a smallneighborhood of K . When L is a link, define p ( L ) by applying p to each component of L .More precisely, for a framed link L ⊂ M with m components L , . . . , L m , define p ( L ) = d X j ,...,j m =0 m Y k =1 a j k ! m G k =1 L ( j k ) k ! . Here F mk =1 L ( j k ) k is the link which is the union, over k ∈ { , . . . , m } , of j k parallels of L k . Remark 0.1.
Suppose K ⊂ Σ is a simple closed curve on the surface Σ. Consider K asa framed knot in Σ × [ − ,
1] by identifying Σ = Σ × K with the verticalframing, i.e. the framing where the normal vector is perpendicular to Σ and has directionfrom − K ( j ) = K j , where K j is the power in the algebra S (Σ). Thus, p ( K )has the usual meaning of applying a polynomial to an element of an algebra.But if K is a knot in Σ × [ − , p ( K ) in general is not the result of applying thepolynomial p to the element K using the algebra structure of S (Σ), i.e. p ( K ) = P a j K j .0.2. Bonahon and Wong’s results.Definition 1.
A polynomial p ( z ) ∈ C [ z ] is called central at ξ ∈ C × if for any orientedsurface Σ and any framed link L in Σ × [ − , , p ( L ) is central in the algebra S ξ (Σ) . Bonahon and Wong [BW2] showed that if ξ is root of unity of order 2 N , then T N ( z ) iscentral, where T N ( z ) is the Chebyshev polynomial of type 1 defined recursively by T ( z ) = 2 , T ( z ) = 1 , T n ( z ) = zT n − ( z ) − T n − ( z ) , ∀ n ≥ . We will prove a stronger version, using a different method.
Theorem 1.
A non-constant polynomial p ( z ) ∈ C [ z ] is central at ξ ∈ C × if and only if(i) ξ is a root of unity and(ii) p ( z ) ∈ C [ T N ( z )] , i.e. p is a C -polynomial in T N ( z ) , where N is the order of ξ . Remark 0.2.
We also find a version of “skew-centrality” when ξ N = − Remark 0.3.
Let us call a polynomial p ( z ) ∈ C [ z ] weakly central at ξ ∈ C × if for anyoriented surface Σ and any simple closed curve K on Σ, p ( K ) is central in the algebra S ξ (Σ). Then our proof will also show that Theorem 1 holds true if one replaces “central”by “weakly central”. It follows that being central is equivalent to being weakly central.A remarkable result of Bonahon and Wong is the following. Theorem 2 (Bonahon-Wong [BW2]) . Let M be an oriented 3-manifold, possibly withboundary. Suppose ξ is a root of unity of order N . Let ε = ξ N . There is a unique C -linear map Ch : S ε ( M ) → S ξ ( M ) such that for any framed link L ⊂ M , Ch ( L ) = T n ( L ) . N KAUFFMAN BRACKET SKEIN MODULES AT ROOTS OF UNITY 3 If M = Σ × [ − , Ch is an algebra homomorphism. Actually Bonahonand Wong only consider the case of S (Σ), but their proof works also in the case of skeinmodules of 3-manifolds. In their proof, Bonahon and Wong used the theory of quantumTeichm¨uller space of Chekhov and Fock [CF] and Kashaev [Kas], and the quantum tracehomomorphism developed in earlier work [BW1]. Bonahon and Wong asked for a proofusing elementary skein theory. We will present here a proof of Theorem 2, using onlyelementary skein theory. The main idea is to use central properties (in more generalsetting) and several operators and filtrations on the skein modules defined by arcs.In general, the calculation of S ( M ) is difficult. For some results on knot and linkcomplements in S , see [Le, LT, Mar]. Note that if ξ N = 1, then ε = ξ N is a 4-th root of1. In this case the S ε ( M ) is well-known and is related to character varieties of M . Thismakes Theorem 2 interesting. At t = − S − ( M ) has an algebra structure and, moduloits nilradical, is equal to the ring of regular functions on the SL ( C )-character variety of M , see [Bul, PS, BFK]. For the case when ε is a primitive 4-th root of 1, see [Si].0.3. Plan of the paper.
Section 1 is preliminaries on Chebyshev polynomials and rela-tive skein modules. Section 2 contains the proof of Theorem 1. Section 3 introduces thefiltrations and operators on skein modules, and Sections 4 and 5 contain some calculationswhich are used in Section 6, where the main technical lemma about the skein module ofthe twice punctured torus is proved. Theorem 2 is proved in Section 7.0.4.
Acknowledgements.
The author would like to thank F. Bonahon, whose talk atthe conference “Geometric Topology at Columbia University (August 12-16, 2013) hasprompted the author to work on this project. The author also thanks C. Frohman,A. Sikora, and H. Wong for helpful discussions. The work is supported in part by NSF.1.
Ground ring, Chebyshev polynomials, and relative skein modules
Ground ring.
Let R = C [ t ± ], which is a principal ideal domain. For an R -moduleand a non-zero complex number ξ ∈ C × let V ξ be the R -module V / ( t − ξ ). Then R ξ ∼ = C as C -modules, and V ξ has a natural structure of an R ξ -module.We will often use the constants(3) λ k := − ( t k +2 + t − k − ) ∈ R. For example, λ is the value of the unknot U as a skein element.1.2. Chebyshev polynomials.
Recall that the Chebyshev polynomials of type 1 T n ( z )and type 2 S n ( z ) are defined by T = 2 , T ( z ) = z, T n ( z ) = zT n − ( z ) − T n − ( z ) S = 1 , S ( z ) = z, S n ( z ) = zS n − ( z ) − S n − ( z ) . Here are some well-known facts. We drop the easy proofs.
Lemma 1.1. (a) One has T n ( u + u − ) = u n + u − n (4) T n = S n − S n − (5) (b) For a fixed positive integer N , the C -span of { T Nj , j ≥ } , is C [ T N ( z )] , the ring ofall C -polynomials in T N ( z ) . THANG T. Q. LˆE
Since T n ( z ) has leading term z n , { T n ( z ) , n ≥ } is a C -basis of C [ z ].1.3. Skein module of a surface.
Suppose Σ is a compact connected orientable 2-dimensional manifold with boundary. A knot in Σ is trivial if it bounds a disk in Σ.Recall that S (Σ) is the skein module S (Σ × [ − , ∂ Σ = ∅ , then S (Σ) is a free R -module with basis the set of all links in Σ without trivial components, including theempty link, see [PS]. Here a link in Σ is considered as a framed link in Σ × [ − ,
1] byidentifying Σ with Σ ×
0, and the framing at every point P ∈ Σ × P × [ − , ⊂ Σ × [ − , R -module S (Σ) has a natural R -algebra structure, where L L is obtained byplacing L on top of L .It might happen that Σ × [ − , ∼ = Σ × [ − ,
1] with Σ = Σ . In that case, S (Σ )and S (Σ ) are the same as R -modules, but the algebra structures may be different.1.4. Example: The annulus.
Let A ⊂ R be the annulus A = {−→ x ∈ R , ≤ |−→ x | ≤ } .Let z ∈ S ( A ) be the core of the annulus, z = {−→ x , |−→ x | = 3 / } . Then S ( A ) = R [ z ].1.5. Relative skein modules.
A marked surface (Σ , P ) is a surface Σ together with afinite set P of points on its boundary ∂ Σ. For such a marked surface, a relative framedlink is a 1-dimensional compact framed submanifold X in Σ × [ − ,
1] such that ∂X = P = X ∩ ∂ (Σ × [ − , X is perpendicular to ∂ (Σ × [ − , P ∈ P = ∂X is vertical. The relative skein module S (Σ , P ) is defined as the R -modulespanned by isotopy class of relative framed links modulo the same skein relations (1) and(2). We will use the following fact. Proposition 1.2. (See [PS, Theorem 5.2] ) The R -module S (Σ , P ) is free with basis theset of isotopy classes of relative links embedded in Σ without trivial components. Annulus with two marked points and central elements
Marked annulus.
Recall that A ⊂ R is the annulus A = {−→ x ∈ R | ≤ |−→ x | ≤ } .Let A io be the marked surface ( A , { P , P } ), with two marked points P = (0 , P =(0 , e , u , u − . Figure 2.
The marked annulus A io , and the arcs e , u , u − For L , L ∈ S ( A io ) define the product L L by placing L inside L . Formally thismeans we first shrink A io ⊃ L by factor 1 /
2, we get ( A io ) ⊃ ( L ), where A io is anannulus on the plane whose outer circle is the inner circle of A io . Then L L is ( L ) ∪ L ⊂ ( A io ) ∪ A io . The identity of S ( A io ) is the presented by e , and u − u = e = uu − . Proposition 2.1.
The Kauffman bracket skein modules of A io is S ( A io ) = R [ u ± ] , thering of Laurent R -polynomial in one variable u . In particular, S ( A io ) is commutative. N KAUFFMAN BRACKET SKEIN MODULES AT ROOTS OF UNITY 5
Proof.
Using Proposition 1.2 one can easily show that the set { u k , k ∈ Z } is a free R -basisof S ( A io ). (cid:3) Passing through T k . Recall that S ( A ) = R [ z ]. One defines a left action and aright action of S ( A ) on S ( A io ) as follows. For L ∈ S ( A ) , K ∈ S ( A io ) let L • K be theelement in S ( A io ) obtained by placing L above K , and K • L ∈ S ( A io ) be the element in S ( A io ) obtained by placing K above L . For example, e • z = , z • e = Proposition 2.2.
One has T k ( z ) • e = t k u k + t − k u − k (6) e • T k ( z ) = t k u − k + t − k u k (7) T k ( z ) • e − e • T k ( z ) = ( t k − t − k )( u k − u − k ) . (8) Proof.
It is important to note that the map f : S ( A ) → S ( A io ) given by f ( L ) = L • e isan algebra homomorphism.Resolve the only crossing point, we have z • e = = t + t − = t u + t − u − . Hence, T k ( z ) • e = T k ( t u + t − u − ) because f is an algebra homomorphism= t k u k + t − k u − k by (4) . This proves (6). The proof of (7) is similar, while (8) follows from (6) and (7). (cid:3)
Corollary 2.3.
Suppose ξ N = 1 . Then T N ( z ) is central at ξ .Proof. We have ξ N = ξ − N since ξ N = 1. Then (8) shows that T N ( z ) • e = e • T N ( z ),which easily implies the centrality of T N ( z ). (cid:3) Remark 2.4.
The corollary was first proved by Bonahon and Wong [BW2] using anothermethod.2.3.
Transparent elements.
We say that p ( z ) ∈ C [ z ] is transparent at ξ if for any 3disjoint framed knots K, K , K in any oriented 3-manifold M , p ( K ) ∪ K = p ( K ) ∪ K in S ξ ( M ), provided that K and K are isotopic in M . Note that in general, K and K are not isotopic in M \ K . Proposition 2.5.
The following are equivalent(i) p ( z ) • e = e • p ( z ) in S ξ ( A io ) .(ii) p ( z ) is transparent at ξ .(iii) p ( z ) is central at ξ .Proof. It is clear that ( i ) ⇒ ( ii ) ⇒ ( iii ). Let us prove ( iii ) ⇒ ( i ).By gluing a 1-handle to A we get a punctured torus T punc as in Figure 3. Here the baseof the 1-handle is glued to a small neighborhood of { P ∪ P } in ∂ A , and the core of the1-handle is an arc β connecting P and P . Let ι : S ( A io ) → S ( T punc ) be R -map which isthe closure by β , i.e. ι ( K ) = K ∪ β . Then ι ( u k ) is a knot in T punc for every k ∈ Z , and THANG T. Q. LˆE PP
1 2
Figure 3.
The core β connects P and P in T punc ι ( u k ) is not isotopic to ι ( u l ) if k = l . Since { u k , k ∈ Z } is an R -basis of S ( A io ) and theisotopy classes of links in T punc form an R -basis of S ( T punc ), ι is injective.Assume (iii). Then p ( z ) ι ( e ) = ι ( e ) p ( z ), or ι ( p ( z ) • e ) = ι ( e • p ( z )). Since ι is injective,we have p ( z ) • e = e • p ( z ). (cid:3) Proof of Theorem 1.
The “‘if” part has been proved, see Corollary 2.3. Let usprove the “only if part”. Assume that p ( z ) is central at ξ and having degree k ≥
1. Since { T j ( z ) , j ≥ } is a basis of C [ z ], we can write(9) p ( z ) = k X j =0 c j T j ( z ) , c j ∈ C , c k = 0 . By Proposition 2.5, p ( z ) • e − e • p ( z ) = 0. Using expression (9) for p ( z ) and (8), we get0 = p ( z ) • e − e • p ( z ) = k X j =0 c j ( ξ j − ξ − j )( u j − u − j ) . Because { u j , j ∈ Z } is a basis of S ξ ( A io ), the coefficient of each u j on the right handside is 0. This means,(10) c j = 0 or ξ j = 1 , ∀ j. Since c k = 0, we have ξ k = 1. Since k ≥
1, this shows ξ is a root of unity of some order N . Then (10) shows that c j = 0 unless N | j . Thus, p ( z ) is a C -linear combination of T j with N | j . This completes the proof of Theorem 1.2.5. Skew transparency.
One more consequence of Proposition 2.2 is the following.
Corollary 2.6.
Suppose ξ N = − . Then in S ξ ( A io ) , T N ( z ) • e = − e • T N ( z ) . This means every time we move T N ( K ) passed through a component of a link L , thevalue of the skein gets multiplied by −
1. Following is a precise statement.Suppose K and K are knots in a 3-manifold M . Recall that an isotopy between K and K is a smooth map H : S × [1 , → M such that for each t ∈ [1 , N KAUFFMAN BRACKET SKEIN MODULES AT ROOTS OF UNITY 7 map H t : S → M is an embedding, and the image of H i is K i for i = 1 ,
2. Here H t ( x ) = H ( x, t ). For a knot K ⊂ M let I ( H, K ) be the mod 2 intersection number of H and K . Thus, if H is transversal to K then I ( H, K ) is the number of points in the finiteset H − ( K ) modulo 2. Definition 2.
Suppose µ = ± . A polynomial p ( z ) ∈ C [ z ] is called µ -transparent at ξ ∈ C × if for any 3 disjoint framed knots K, K , K in any oriented 3-manifold M , with K and K connected by an isotopy H , one has the following equality in S ξ ( M ) : p ( K ) ∪ K = µ I ( H,K ) [ p ( K ) ∪ K ] . From Corollary 2.6 we have
Corollary 2.7.
Assume ξ N = 1 . Then µ := ξ N = ± , and T N ( z ) is µ -transparent. A special case is the following. Suppose D ⊂ M is a disk in M with ∂D = K , and aframed link L ⊂ M is disjoint from K . Then, if ξ N = µ = ±
1, one has(11) K ∪ T N ( L ) = µ I ( D,L ) λ T N ( L ) in S ξ ( M ) . Here λ = − ( ξ + ξ − ) is the value of trivial knot in S ξ ( M ).3. Filtrations of skein modules
Suppose Φ is a link in ∂M . We define an R -map Φ : S ( M ) → S ( M ) by Φ( L ) = Φ ∪ L .3.1. Filtration by an arc.
Suppose α is an arc properly embedded in a marked surface(Σ , P ) with ∂ Σ = ∅ . Assume the two boundary points of α , which are on the boundary ofΣ, are disjoint from the marked points. Then D α := α × [ − ,
1] is a disk properly embeddedin Σ × [ − , α = ∂ ( α × [ − , α × {− , } ) ∪ ( ∂α × [ − , F αk = F αk ( S (Σ)) be the R -submodule of S (Σ) spanned by all relative links whichintersect with D α at less than or equal to k points. For L ∈ S (Σ), we define fil α ( L ) = k if L ∈ F αk \ F αk − . The filtration is compatible with the algebra structure, i.e.fil α ( L L ) ≤ fil α ( L ) + fil α ( L ) . Remark 3.1.
A similar filtration was used in [Mar] to calculate the skein module of torusknot complements.A convenient way to count the number of intersection points of a link L with D α is tocount the intersection points of the diagram of L with α : Let D be the vertical projectionof L onto Σ. In general position D has only singular points of type double points, and weassume further that D is transversal to α . In that case, the number of intersection pointsof L with D α is equal to the number of intersections of D with α , where each intersectionpoint of α and D at a double point of D is counted twice.Recall that Φ α ( L ) = L ∪ Φ α , where Φ α is the boundary of the disk D α = α × [ − , F αk is Φ α invariant, i.e. Φ α ( F αk ) ⊂ F αk . It turns out that the action of Φ α on the quotient F αk / F αk − is very simple. Recall that λ k = − ( t k +2 + t − k − ). Proposition 3.2.
For k ≥ , the action of Φ α on F αk / F αk − is λ k times the identity. This is a consequence of Proposition 3.3 proved in the next subsection.
THANG T. Q. LˆE
The Temperley-Lieb algebra and the operator Φ . The well-known Temperley-Lieb algebra
T L k is the skein module of the disk with 2 k marked points on the boundary.We will present the disk as the square Sq = [0 , × [0 ,
1] on the standard plane, with k marked points on the top side and k -marked points on the bottom side. The product L L in T L k is defined as the result of placing T on top of T . The unit ˜ e k of T L k ispresented by k vertical straight arcs, see Figure 4. Figure 4.
The unit ˜ e k , the arc α , and Φ α (˜ e k ). Here k = 4.Let α ⊂ Sq be the the horizontal arc [0 , × (1 / α (˜ e k ) is depicted inFigure 4. In general, Φ α ( L ) is L encircled by one simple closed curve. Proposition 3.3.
With the above notation, one has (12) Φ α (˜ e k ) = λ k ˜ e k (mod F αk − ) . Proof.
A direct proof can be carried out as follows. Using the skein relation (1) oneresolves all the crossings of the diagram of Φ α (˜ e k ), and finds that only a few terms arenot in F αk − , and the sum of these terms is equal to λ k ˜ e k . This is a good exercise for thededicated reader.Here is another proof using more advanced knowledge of the Temperley-Lieb algebra.First we extend the ground ring to field of fraction C ( t ). Then the Temperley-Lieb algebracontains a special element called the Jones-Wenzl idempotent f k (see e.g. [Lic, Chapter13]). We have f k = ˜ e k (mod F αk − ), and f k is an eigenvector of Φ α with eigenvalue λ k .Hence, we have (12). (cid:3) Another annulus with two marked points
Annulus with two marked points on the same boundary.
Let A oo be theannulus A with two marked points Q , Q on the outer boundary as in Figure 5. Let u , u be arcs connecting Q and Q in A oo as in Figure 5. Figure 5.
The marked annulus A oo and arcs u , u Define a left S ( A )-module and a right S ( A )-module on S ( A oo ) as follows. For K ∈S ( A oo ) and L ∈ S ( A ) let KL be the skein in S ( A oo ) obtained by placing K on top of L , N KAUFFMAN BRACKET SKEIN MODULES AT ROOTS OF UNITY 9 and LK ∈ S ( A oo ) obtained by placing L on top of K . It is easy to see that KL = LK .Recall that S ( A ) = R [ z ]. Proposition 4.1.
The module S ( A oo ) is a free S ( A ) -module with basis { u , u } : S ( A oo ) = R [ z ] u ⊕ R [ z ] u . Proof.
Any relative link in A oo is of the form u i z m with i = 0 , m ∈ Z . Theproposition now follows from Proposition 1.2. (cid:3) Framing change and the unknot.
Recall that S k is the k -th Chebyshev poly-nomial of type 2. The values of the unknot colored by S k and the framing change arewell-known (see e.g. [BHMV]): In S ( M ), where M is an oriented 3-manifold, one has L ⊔ S k ( U ) = ( − k t k +2 − t − k − t − t − L (13) S k (cid:18) (cid:19) = ( − k t k +2 k S k ! . (14)Here in (13), U is the trivial knot lying in a ball disjoint from L .4.3. Some elements of S ( A oo ) . Let u k , k ≥ A oo depicted in Figure 6. Theelement u and u are the same as the ones defined in Figure 5. Let v = u and v k , k ≥ A oo depicted in Figure 6. u k = v k = Figure 6. u k and v k , with k = 3 Proposition 4.2.
One has u k = t k − S k − ( z ) u + t k − S k − ( z ) u ∀ k ≥ v k = t − k S k − ( z ) u + t − k S k ( z ) u ∀ k ≥ . (16) Proof.
Suppose k ≥
3. Apply the skein relation to the innermost crossing of u k , we get u k = = t + t − which, after an isotopy and removing a framing crossing, is u k = t u k − z − t u k − , from which one can easily prove (15) by induction.Similarly, using the skein relation to resolve the innermost crossing point of v k , we get v k = t − v k − z − t − v k − , for k ≥ (cid:3) Remark 4.3.
Identity (15) does not hold for k = 0. This is due to a framing change. Operator Ψ . Let Ψ be the arc in ∂ A × [ − ,
1] beginning at Q and ending at Q ,as depicted in Figure 7. Here we draw A × [ − ,
1] as a handlebody. For any element QQ QQ Figure 7.
Arc Ψ connecting Q and Q , and Ψ( z ) α ∈ S ( A ) let Ψ( α ) ∈ S ( A oo ) be the skein Ψ ∪ α . For example, Ψ( z ) is given in Figure 7. Proposition 4.4.
For k ≥ , one has (17) Ψ( T k ( z )) = u (cid:2) t ( t − k − t k ) S k − ( z ) (cid:3) + u (cid:2) t − k S k ( z ) − t k S k − ( z ) (cid:3) . Proof.
Applying Proposition 2.2 to the part in the left rectangle box, we get= t k + t − k . The positive framing crossing in the first term gives a factor − t . Thus,Ψ( T k ( z )) = − t k +3 u k + t − k v k . Plugging in the values of u k , v k given by Proposition 4.2, we get the result. (cid:3) Remark 4.5.
One can use Proposition 4.4 to establish product-to-sum formulas similarto the ones in [FG]. 5.
Twice punctured disk
Skein module of twice punctured disk.
Let
D ⊂ R be the disk of radius 4centered at the origin, D ⊂ R the disk of radius 1 centered ( − , D the disk ofradius 1 centered (2 , D to be D with the interiors of D and D removed. Figure 8.
The twice punctured disk D and the arcs α , α , α , α The horizontal axis intersects D at 3 arcs denoted from left to right by α , α , α , seeFigure 8. The vertical axis of R intersects D at an arc denoted by α . The corresponding N KAUFFMAN BRACKET SKEIN MODULES AT ROOTS OF UNITY 11 curve Φ α i on ∂ D × [ − ,
1] will be denoted simply by Φ i , for i = 0 , , ,
3. If D × [ − , H , which is a thickening of D in R , then the curvesΦ , Φ , Φ , Φ look like in Figure 9. Figure 9.
The curves Φ , Φ , Φ , Φ on the boundary of the handlebodyLet x , x , and y be the closed curves in D : x = , x = , y =It is known that S ( D ) = R [ x , x , y ], the R -polynomial in the variables x , x , y , see[BP]. In particular, S ( D ) is commutative.Let σ be the rotation about the origin of R by 180 . Then σ ( D ) = D . Hence σ inducesan automorphism of S ( D ) = R [ x , x , y ], which is an algebra automorphism. One has σ ( y ) = y, σ ( x ) = x , σ ( x ) = x .5.2. Degrees on S ( D ) = R [ x , x , y ] . Define the left degree, right degree, and doubledegree on R [ x , y, x ] as follows. For a monomial m = x a y b x a define its left degreedeg l ( m ) = a + b , right degree deg r ( m ) = a + b , double degree deg lr ( m ) = deg l ( m ) +deg r ( m ) = a + a + 2 b . One readily finds thatdeg l ( m ) = fil α ( m ) , deg r ( m ) = fil α ( m ) , where fil α is defined in Section 3.1. Using the definition of fil α involving the numbers ofintersection points we get the following. Lemma 5.1.
Suppose L is an embedded link in D and L intersects transversally the arc α i at k i points for i = 1 , , . Then, as an element of S ( D ) , L = x a x a y b , where b ≤ k and deg l ( L ) ≤ k , deg l ( L ) ≡ k (mod 2)deg r ( L ) ≤ k , deg r ( L ) ≡ k (mod 2) . Consequently, deg lr ( L ) ≤ k + k and deg lr ( L ) ≡ k + k (mod 2) .Proof. If L = L ⊔ L is the union of 2 disjoint sub-links, and the statement holds for eachof L i , then it holds for L . Hence we assume L has one component, i.e. L is an embeddedloop in D ⊂ R . Then L is isotopic to either a trivial loop, or x , or x , or y . In eachcase, the statement can be verified easily. For example, suppose L = x . For the mod2 intersection numbers, I ( L, α ) = I ( x , α ) = 1. Hence k , the geometric intersectionnumber between L and α , must be odd and bigger than or equal to 1. Hence, we havedeg l ( L ) ≤ k and deg l ( L ) ≡ k (mod 2). (cid:3) Corollary 5.2.
Suppose L is a link diagram on D which intersects transversally the arc α i at k i points for i = 1 , , . Then, as an element in S ( D ) , deg l ( L ) ≤ k , deg r ( L ) ≤ k , y ( L ) ≤ k , and L is a linear R -combination of monomials whose double degrees are equal to k + k modulo 2. The R -module V n and the skein γ . Let γ and ¯ γ be the following link diagramson D ,(18) γ = , ¯ γ = . Let V n = { p ∈ R [ x , x , y ] | deg l ( p ) ≤ n, deg r ( p ) ≤ n, deg lr ( p ) even } . In other words, V n ⊂ R [ x , x , y ] is the R -submodule spanned by x a x a y b , with a i + b ≤ n for i = 1 ,
2, and a + a even. Lemma 5.3.
One has T n ( γ ) , T n (¯ γ ) ∈ V n .Proof. The diagram γ k has k intersection points with each of α and α . By Corollary5.2, deg l ( γ k ) ≤ k, deg r ( γ k ) ≤ k , and each monomial of γ k has double degree ≡ k + k ≡ γ k ∈ V k for every k ≥
0. Because T n ( γ ) is Z -linear combination of γ k with k ≤ n , we have T n ( γ ) ∈ V n . The proof for ¯ γ is similar. (cid:3) Remark 5.4.
It is an easy exercise to show that T N (¯ γ ) = T N ( γ ) (cid:12)(cid:12) t → t − .6. Skein module of twice puncture disk at root of 1
Recall that γ and ¯ γ are knot diagrams on D defined by (18). The following was provedby Bonahon and Wong, using quantum Teichm¨uller algebras and their representations. Proposition 6.1.
Suppose ξ is a root of 1 of order N . Then in S ξ ( D ) one has T N ( γ ) = ξ − N T N ( y ) + ξ N T N ( x ) T N ( x )(19) T N (¯ γ ) = ξ N T N ( y ) + ξ − N T N ( x ) T N ( x ) . (20)As mentioned above, there was an urge to find a proof using elementary skein theory;one such is presented here. Our proof roughly goes as follows. Using the transparentproperty of T N ( γ ), we show that T N ( γ ) is a common eigenvector of several operators. Wethen prove that the space of common eigenvectors has dimension at most 3, with a simplebasis. We then fix coefficients of T N ( γ ) in this basis using calculations in highest order.Then the result turns out to be the right hand side of (19).Throughout this section we fix a complex number ξ such that ξ is a root of unity oforder N . Define ε = ξ N . We will write V N,ξ simply by V N and λ k for λ k ( ξ ). Thus, in thewhole section, λ k = − ( ξ k +2 + ξ − k − ) . Properties of ξ and λ k . Recall that ξ is a root of 1 of order N . Lemma 6.2.
Suppose ≤ k ≤ N − . Then(a) λ k = λ if and only if k = N − .(b) λ k = ξ N λ implies that k = N − .(c) If N is even then ξ N = − .(d) One has (21) ξ N +2 N = ( − N +1 . N KAUFFMAN BRACKET SKEIN MODULES AT ROOTS OF UNITY 13
Proof. (a) With λ k = − ( ξ k +2 + ξ − k − ), we have λ k − λ = − ξ − − k ( ξ k − ξ k +4 − . Hence, λ k − λ = 0 if and only if either N | k or N | ( k + 1). With 1 ≤ k ≤ N −
1, this isequivalent to k = N − λ k − ξ N λ = − ξ − N − ( ξ N − k − ξ N +2 k +4 − . Either (i) ξ N − k = 1 or (ii) ξ N +2 k +4 = 1. Taking the squares of both identities, we seethat either N | ( N − k ) or N | ( k + 2). With 1 ≤ k ≤ N −
1, we conclude that k = N − N is even. Since ξ has order N , one has ( ξ ) N/ = −
1. Then ξ N =( ξ ) N/ = − (cid:3) Operators Φ i and the vector space W . Recall that Φ i := Φ α i , i = 0 , , ,
3, isdefined in section 5.1. Then Φ i ( V N ) ⊂ V N for i = 0 , , , be the curve on ∂ D × [ − ,
1] depicted in Figure 10. Here we draw H = D × [ − , Figure 10.
The curve Φ and Φ ( x )as a handlebody. We also depict Φ ( x ).We don’t have Φ ( V N ) ⊂ V N , since Φ in general increases the double degree. By count-ing the intersection points with α and α , we have, for every E ∈ S ξ ( D ) = C [ x , x , y ],(22) deg lr (Φ ( E )) ≤ deg lr ( E ) + 1 . Proposition 6.3. If E is one of { T N ( γ ) , T N (¯ γ ) , T N ( y ) , T N ( x ) T N ( x ) } , then one has σ ( E ) = E (23) Φ ( E ) = ξ N λ E (24) Φ i ( E ) = λ E for i = 0 , , (25) Φ ( E ) = ξ N x E. (26) Proof.
The first identity follows from the fact that each of γ, ¯ γ, y, x ∪ x is invariant under σ . The remaining identities follows from the ξ N -transparent property of T N ( z ), Corollary2.7. (cid:3) Remark 6.4.
Note that Φ is C [ x ]-linear and (26) says that E is a ξ N x -eigenvector ofΦ .Let W be the subspace of V N consisting of elements satisfying (23)–(26). This means, W ⊂ V N consists of elements which are at the same time 1-eigenvector of σ , ξ N λ -eigenvector of Φ , λ -eigenvector of Φ and Φ , and ξ N x -eigenvector of Φ .We will show that W is spanned by T N ( y ) , T N ( x ) T N ( x ), and possibly 1. Action of Φ , Φ , and Φ . For an element F ∈ C [ x , x , y ] and a monomial m = x a x a y b let coeff( F, m ) be the coefficient of m in F . Lemma 6.5.
Suppose E ∈ W and coeff( E, y N ) = 0 . Then E ∈ C [ x , x ] .Proof. Let k be the y -degree of E . Since E ∈ W and coeff( E, y N ) = 0, one has k ≤ N − k = 0. Suppose the contrary that 1 ≤ k . Then 1 ≤ k ≤ N − k = N −
1, using the fact that E is a λ -eigenvector of Φ by (25).Recall that fil α is twice the y -degree. One has fil α ( E ) = 2 k . Thus E = 0 ∈ F α k / F α k − .By Proposition 3.2, any non-zero element in F α k / F α k − is an eigenvector of Φ witheigenvalue λ k . But E is an eigenvector of Φ with eigenvalue λ . It follows that λ k = λ .By Lemma 6.2, we have k = N − lr ( E ) is even and ≤ N , we must have E = y N − ( c x x + c ) + O ( y N − ) , c , c ∈ C . We will prove c = 0 by showing that otherwise, Φ will increase the y -degree. Note thatΦ can increase the y -degree by at most 1, and Φ is C [ y ]-linear. We have(27) Φ ( E ) = y N − ( c Φ ( x x ) + c Φ (1)) + O ( y N − ) . The diagram of Φ ( x x ) has 4 crossings, see Figure 11. A simple calculation shows Figure 11. Φ ( x x )Φ ( x x ) = (1 − t )(1 − t − ) y + O ( y ) . Plugging this value in (27), with Φ (1) = λ ∈ C ,(28) Φ ( E ) = y N c (1 − t )(1 − t − ) + O ( y N − ) . If c = 0, then the y -degree of Φ ( E ) is N , strictly bigger than that of E , and E cannotbe an eigenvector of Φ . Thus c = 0.One has now(29) E = c y N − + O ( y N − )Since the y -degree of E is N −
1, one must have c = 0. By counting the intersectionswith α , we see that Φ does not increase the y -degree. We haveΦ ( E ) = c Φ ( y N − ) + O ( y N − )= c λ N − y N − + O ( y N − ) by Proposition 3.2 . Comparing the above identity with (29) and using the fact that E is a ξ N λ -eigenvectorof Φ , we have λ N − = ξ N λ , which is impossible since Lemma 6.2 says that λ k = ξ N λ only when k = N −
2. Thiscompletes the proof of the lemma. (cid:3)
N KAUFFMAN BRACKET SKEIN MODULES AT ROOTS OF UNITY 15
Action of Φ . Recall that Φ is the curve on the boundary of the handlebody H (see Figure 10) which acts on S ξ ( D ) = C [ x , x , y ]. The action of Φ is C [ x ]-linear, andevery element of W is a ξ N x -eigenvector of Φ .Recall that deg r = fil α , and deg r ( x a x a y b ) = a + b . Note that for F ∈ C [ x , x ],deg r ( F ) is exactly the x -degree of F . By looking at the intersection with α , we see thatΦ preserves the α -filtration, i.e. deg r Φ ( F ) ≤ deg r ( F ). We will study actions of Φ onthe associated graded spaces.We will use the notation F + deg r - l.o.t. to mean F + F , where deg r ( F ) < deg r ( F ). Lemma 6.6.
Suppose ≤ k ≤ N − . One has Φ ( a ( x ) T N ( x )) = ξ N x [ a ( x ) T N ( x )](30) Φ ( T k ( x )) = y (cid:2) ξ ( ξ − k − ξ k ) x k − (cid:3) + deg r -l.o.t . mod C [ x , x ] . (31) Proof.
Identity (30) follows from the ξ N -transparency of T N ( z ).Let us prove (31). Apply identity (17) to the dashed box below, we haveΦ ( T k ( x )) == y (cid:2) ξ ( ξ − k − ξ k ) S k − ( x ) (cid:3) + x (cid:2) ξ − k S k ( x ) − ξ k S k − ( x ) (cid:3) , which implies (31). (cid:3) The space W ∩ C [ x , x ] .Lemma 6.7. Suppose E ∈ W ∩ C [ x , x ] and the coefficient of x N x N in E is 0. Then E ∈ C .Proof. Since T k ( x ) is a basis of C [ x ], we can write E uniquely as E = N X k =0 a k ( x ) T k ( x ) , a k ( x ) ∈ C [ x ] . Let j be the x -degree of E ′ := E − a N ( x ) T N ( x ). Then j ≤ N − j = 0. Assume the contrary j ≥
1. Thus 1 ≤ j ≤ N −
1. Notethat E , by assumption, and a N ( x ) T N ( x ), by (30), are eigenvectors of Φ with eigenvalue ξ N x . It follows that E ′ is also an eigenvector of Φ with eigenvalue ξ N x . We have E ′ = j X k =0 a k ( x )Φ ( T k ( x )) = a j ( x ) T j ( x ) + deg r -l.o.t . Using (31) and the fact that Φ does not increase deg r , we haveΦ ( E ′ ) = y (cid:2) a j ( x ) ξ ( ξ − k − ξ k ) x k − (cid:3) + deg r -l.o.t . mod C [ x , x ] . When 1 ≤ j ≤ n −
1, the coefficient of y , which is the element in the square bracket, isnon-zero. Thus Φ ( E ′ ) C [ x , x ], while E ′ ∈ C [ x , x ]. This means E ′ can not be aneigenvector of Φ , a contradiction. This proves j = 0.So we have E = a N ( x ) T N ( x ) + a ( x ) . Because the deg lr ( E ) < N , the x -degree of a N ( x ) is < n . Using the invariance under σ , one sees that E must be of the form(32) E = c ( T N ( x ) + T N ( x )) + c , c , c ∈ C . To finish the proof of the lemma, we need to show that c = 0. Assume that c = 0. Since E has even double degree, N is even. By Lemma 6.2(c), ξ N = − E is λ -eigenvector of Φ . Apply Φ to (32), λ [ c ( T N ( x ) + T N ( x )) + c ] = Φ ( c ( T N ( x ) + T N ( x )) + c ) . Both T N ( x ) and T N ( x ) are eigenvectors of Φ with eigenvalues ξ N λ = − λ , whileΦ (1) = λ . Hence we have λ [ c ( T N ( x ) + T N ( x )) + c ] = λ [ − c ( T N ( x ) − T N ( x )) + c ] . which is impossible since c λ = 0.Hence, we have c = 0, and E ∈ C . (cid:3) Some maximal degree parts of T N ( γ ) .Lemma 6.8. One has coeff( T N ( γ ) , y N ) = ξ − N (33) coeff( T N ( γ ) , x N x N ) = ξ N . (34) Proof.
Since T N ( γ ) = γ N + deg lr - l.o.t. , we havecoeff( T N ( γ ) , y N ) = coeff( γ N , y N ) , coeff( T N ( γ ) , x N x N ) = coeff( γ N , x N x N ) . There are N crossing points in the diagram of γ N . Each crossing can be smoothed intwo ways. The positive smoothing acquires a factor t in the skein relation, and the negativesmoothing acquires a factor t − . The are 2 N smoothings of γ N . Each smoothing s of allthe N crossings gives rise to a link L s embedded in D . Then γ N is a linear combinationof all L s . We will show that the only s for which L s = y N is the all negative smoothing.Consider a crossing point C of γ N . The vertical line passing through C intersect D inan interval α ′ which is isotopic to α , and fil α = fil α ′ . For an embedded link L in D ,as an element of S ( D ) = R [ x , x , y ], L is a monomial whose y -degree is bounded aboveby half the number of intersection points of L with α ′ . The diagram γ N has exactly 2 N intersection points with α ′ , with C contributing two (of the 2 N intersection points). Ifwe positively smooth γ N at C , the result is a link diagram with 2 N − α ′ , and no matter how we smooth other crossings, the resulting link will have lessthan or equal to 2 N − α ′ . Thus we cannot get y N if any ofthe crossing is smoothed positively. The only smoothing which results in y N is the allnegative smoothing. The coefficient of this smoothing is ξ − N .Similarly, one can prove that the only smoothing which results in x N x N is the allpositive smoothing, whose coefficient is ξ N . (cid:3) N KAUFFMAN BRACKET SKEIN MODULES AT ROOTS OF UNITY 17
Proof of Proposition 6.1.
Let E = T N ( γ ) − ξ N T N ( x ) T N ( x ) − ξ − N T N ( y ) . Then E ∈ W . Lemma 6.8 shows that coeff( E, y N ) = 0 = coeff( E, x N x N ). By Lemma6.5, E ∈ C [ x , x ]. Then, by Lemma 6.7, we have E ∈ C , i.e. E is a constant.We will show that E = 0. This is done by using the inclusion of H into R , which givesa C -linear map ι : S ξ ( D ) → S ξ ( R ) = C . Under ι , we have(35) E = ι ( T N ( γ )) − ι ( ξ N T N ( x ) T N ( x )) − ι ( ξ − N T N ( y )) . The right hand side involves the trivial knot and the trivial knot with framing 1, and canbe calculated explicitly as follows. Note that ι ( γ ) is the unknot with framing 1, while ι ( x ) = ι ( x ) = ι ( y ) = U , the trivial knot. With T N = S N − S N − , and the framingchange given by (14), we find(36) T N ( ) = ( − N ξ N +2 N T N ( ) = − ξ − N T N ( ) , where the second identity follows from (21). Similarly, using (13), we have(37) T N ( L ⊔ U ) = 2( − N ξ N T N ( L ) = − ( ξ N + ξ − N ) T N ( L ) . From (36) and (37), we calculate the right hand side of (35), and find that E = 0. Thisproves (19).The proof of (20) is similar. Alternatively, one can get (20) from (19) by noticing thatthe mirror image map on R [ x , x , y ] is the C -algebra map sending t to t − , leaving eachof x , x , y fixed.This completes the proof of Proposition 6.1.7. Proof of Theorem 2
Recall that ε = ξ N , where ξ is a root of 1 of order N . Then ε = 1. The map S ε ( M ) → S ξ ( M ), defined for framed links by L → T N ( L ), is well-defined if and only if itpreserves the skein relations (1) and (2), i.e. in S ξ ( M ), T N ( L ) = εT N ( L + ) + ε − T N ( L − )(38) T N ( L ⊔ U ) = − ( ε + ε − ) T N ( L ) . (39)Here, in (38), L, L + , L − are links appearing in the original skein relation (1), they areidentical everywhere, except in a ball B , where they look like in Figure 12. Figure 12.
The links L , L + , and L − Identity (39) follows from (37). Let us prove (38).Case 1: The two strands of L in the ball B belong to the same component. Then (38)follows from Proposition 6.1, applied to the handlebody which is the union of B and atubular neighborhood of L . Case 2: The two strands of L in B belong to different components. Then the twostrands of L + belong to the same component, and we can apply (38) to the case whenthe left hand side is L + . We have T N ( L + ) = T N ( ) = T N ( )= ε − T N ( ) + ε T N ( )(40) = ε − T N ( ) + ε ( − ε ) T N ( )(41) = ε − T N ( L ) − ε T N ( L − ) , (42)where (40) follows from Case 1, and (41) follows from the framing factor formula (36).Multiplying (42) by ε and using ε = ε − , we get (38) in this case. This completes theproof of Theorem 2. Remark 7.1.
In [BW2], in order to prove Theorem 2, the authors proved, besides Propo-sition 6.1, a similar statement for links in the cylinder over a punctured torus. Here webypass this extra statement by reducing the extra statement to Proposition 6.1. Essen-tially this is due to the fact that the cylinder over a punctured torus is the same as thecylinder over a twice punctured disk.
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