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International Mathematics Research Notices
, Volume 2018 (5) – Mar 1, 2018

15 pages

/lp/ou_press/on-positivity-of-kauffman-bracket-skein-algebras-of-surfaces-FtfoD6nYzx

- Publisher
- Oxford University Press
- Copyright
- © The Author(s) 2016. Published by Oxford University Press. All rights reserved. For permissions, please e-mail: journals.permission@oup.com.
- ISSN
- 1073-7928
- eISSN
- 1687-0247
- D.O.I.
- 10.1093/imrn/rnw280
- Publisher site
- See Article on Publisher Site

Abstract We show that the Chebyshev polynomials form a basic block of any positive basis of the Kauffman bracket skein algebras of surfaces. 1 Introduction 1.1 Positivity of Kauffman bracket skein algebras Let $${\mathbb Z[q^{\pm 1}]}$$ be the ring of Laurent polynomial in an indeterminate $$q$$ with integer coefficients. Suppose $$\Sigma$$ is an oriented surface. The Kauffman bracket skein algebra $$\mathscr S_{\mathbb Z[q^{\pm 1}]}(\Sigma)$$ is a $${\mathbb Z[q^{\pm 1}]}$$-algebra generated by unoriented framed links in $$\Sigma \times [0,1]$$ subject the Kauffman skein relations [6] which are recalled in Section 2. The skein algebras were introduced by Przyticki [12] and independently Turaev [15] in an attempt to generalize the Jones polynomial to links in general three-manifolds, and have connections and applications to many interesting objects like character varieties, topological quantum field theories, quantum invariants, quantum Teichmüller spaces, and many others (see, e.g., [1, 3, 9, 10, 12–14]). The quotient $$\mathscr S_\mathbb Z(\Sigma)= \mathscr S_{\mathbb Z[q^{\pm 1}]}(\Sigma)/(q-1)$$, which is a $$\mathbb Z$$-algebra, is called the skein algebra at $$q=1$$. For $$R=\mathbb Z$$ let $$R_+ = \mathbb Z_+$$, the set of non-negative integers, and for $$R={\mathbb Z[q^{\pm 1}]}$$ let $$R_+= \mathbb Z_+[q^{\pm1}]$$, the set of Laurent polynomials with non-negative coefficients, that is, the set of $$\mathbb Z_+$$-linear combinations of integer powers of $$q$$. An $$R$$-algebra, where $$R={\mathbb Z[q^{\pm 1}]}$$ or $$R=\mathbb Z$$, is said to be positive if it is free as an $$R$$-module and has a basis $$\mathcal B$$ in which the structure constants are in $$R_+$$, that is, for any $$b,b', b''\in \mathcal B$$, the $$b$$-coefficient of the product $$b'b''$$ is in $$R_+$$. Such a basis $$\mathcal B$$ is called a positive basis of the algebra. The important positivity conjecture of Fock and Goncharov [5] states that $$\mathscr S_R(\Sigma)$$ is positive for the case $$R={\mathbb Z[q^{\pm 1}]}$$ and $$R=\mathbb Z$$. The first case obviously implies the second. The second case, on the positivity over $$\mathbb Z$$, was proved by Thurston [14]. Moreover, Thurston makes precise the positivity conjecture by specifying the positive basis as follows. A normalized sequence of polynomials over $$R$$ is a sequence $${\mathbf P}=(P_n(t)_{n\ge 0})$$, such that for each $$n$$, $$P_n(t)\in R[t]$$ is a monic polynomial of degree $$n$$. Every normalized sequence $${\mathbf P}$$ over $$R$$ gives rise in a canonical way a basis $$\mathcal B_{\mathbf P}$$ of the $$R$$-module $$\mathscr S_R(\Sigma)$$, see Section 2. Definition 1. A sequence $${\mathbf P}=(P_n(t)_{n\ge 0})$$ of polynomials in $$R[t]$$ is positive over $$R$$ if it is normalized and the basis $$\mathcal B_{\mathbf P}$$ is a positive basis of $$\mathscr S_R(\Sigma)$$, for any oriented surface $$\Sigma$$. □ Then Thurston proved the following. Theorem 1.1. [14] The sequence $$(T_n)$$ of Chebyshev’s polynomials is positive over $$R=\mathbb Z$$. □ Here in this paper, Chebyshev’s polynomials $$T_n(t)$$ are defined recursively by T0(t)=1, T1(t)=t,T2(t)=t2−2,Tn(t)=tTn−1(t)−Tn−2(t)for n≥3. If $$T_0$$ is the constant polynomial 2, then the above polynomials are Chebyshev’s polynomials of type 1. Here we have to set $$T_0=1$$ since by default, all normalized sequence begins with 1. Thurston suggested the following conjecture, making precise the positivity conjecture. Conjecture 1. The sequence $$(T_n)$$ of Chebyshev’s polynomials is positive over $${\mathbb Z[q^{\pm 1}]}$$. □ We will show Theorem 1.2. Let $$R=\mathbb Z$$ or $$R={\mathbb Z[q^{\pm 1}]}$$ and $${\mathbf P}= (P_n(t)_{n\ge 0})$$ be a sequence of polynomials in $$R[t]$$. If $${\mathbf P}$$ is positive over $$R$$, then $$P_n(t)$$ is an $$R_+$$-linear combination of $$T_0(t), T_1(t), \dots, T_n(t)$$. Besides, $$P_1(t)=T_1(t)=t$$. □ For the case $$R=\mathbb Z$$, this result complements well Theorem 1.1, as they together claim that the sequence of Chebyshev polynomials is the minimal one in the set of positive sequence over $$\mathbb Z$$. For $$R={\mathbb Z[q^{\pm 1}]}$$, our result says that the sequence of Chebyshev polynomial should be the minimal positive sequence. A stronger version of Theorem 1.2 is proved in Section 3. Remark 1.1. The positivity conjecture considered here is different from the one discussed in [11], which claims that every element of a certain basis is a $$\mathbb Z_+$$-linear combination of monomials of cluster variables. □ 1.2 Marked surfaces A marked surface is a pair $$(\Sigma,\mathcal P)$$, where $$\Sigma$$ is a compact-oriented surface with (possibly empty) boundary $$\partial \Sigma$$ and $$\mathcal P$$ is a finite set in $$\partial \Sigma$$. Muller [10] defined the skein algebra $$\mathscr S_R(\Sigma,\mathcal P)$$, extending the definition from surfaces to marked surfaces. When $$R=\mathbb Z$$, this algebra had been known earlier, and actually, the above-mentioned result of Thurston (Theorem 1.1) was proved also for the case of marked surfaces. However, there are two types of basis generators of the skein algebras, namely loops and arcs, and one needs two normalized sequences of polynomials $${\mathbf P}$$ and $${\mathbf Q}$$ to define an $$R$$-basis of the skein algebra $$\mathscr S_R(\Sigma,\mathcal P)$$. Here $${\mathbf P}$$ is applicable to loops, and $${\mathbf Q}$$ is applicable to arcs, see Section 4. A pair of sequences of polynomials $$({\mathbf P},{\mathbf Q})$$ are positive over $$R$$ if they are normalized and the basis they generate is positive for any marked surface. Thurston result says that with $${\mathbf P}= (T_n)$$, the sequence of Chebyshev polynomials, and $${\mathbf Q}=(Q_n)$$ defined by $$Q_n(t)=t^n$$, the pair $$({\mathbf P},{\mathbf Q})$$ are positive over $$\mathbb Z$$. We obtained also an extension of Theorem 1.2 to the case of marked surface as follows. Theorem 1.3. Let $$R=\mathbb Z$$ or $$R={\mathbb Z[q^{\pm 1}]}$$. Suppose a pair $$({\mathbf P},{\mathbf Q})$$ of sequences of polynomials in $$R[t]$$, $${\mathbf P}= (P_n(t)_{n\ge 0})$$ and $${\mathbf Q}= (Q_n(t)_{n\ge 0})$$, are positive over $$R$$. Then $$P_n(t)$$ is an $$R_+$$-linear combination of $$T_0(t), T_1(t), \dots, T_n(t)$$ and $$Q_n(t)$$ is an $$R_+$$-linear combination of $$1, t, \dots, t^n$$. Moreover, $$P_1(t)=Q_1(t)=t$$. □ 1.3 Plan of the paper In Section 2 we recall basic facts about the Kauffman bracket skein algebras of surfaces. We present the proofs of Theorems 1.2 and 1.3 in, respectively, Sections 3 and 4. 2 Skein Algebras We recall here basic notions concerning the Kauffman bracket skein algebra of a surface. 2.1 Ground ring Throughout the paper we work with a ground ring $$R$$ which is more general than $$\mathbb Z$$ and $${\mathbb Z[q^{\pm 1}]}$$. We will assume that $$R$$ is a commutative domain over $$\mathbb Z$$ containing an invertible element $$q$$ and a subset $$R_+$$ such that $$R_+$$ is closed under addition and multiplication, $$q, q^{-1} \in R_+$$, and $$R_+ \cap (- R_+)= \{0\}$$. For example, we can take $$R= \mathbb Z$$ with $$q=1$$ and $$R_+= \mathbb Z_+$$, or $$R={\mathbb Z[q^{\pm 1}]}$$ with $$R_+= \mathbb Z_+[q^{\pm 1}]$$. The reader might think of $$R$$ as one of these two rings. 2.2 Kauffman bracket skein algebra Suppose $$\Sigma$$ is an oriented surface. The Kauffman bracket skein module$$\mathscr S_R(\Sigma)$$ is the $$R$$-module freely spanned by isotopy classes of non-oriented framed links in $$\Sigma \times [0,1]$$ modulo the skein relation and the trivial loop relation described in Figure 1. In all figures, framed links are drawn with blackboard framing. More precisely, the trivial loop relation says if $$L$$ is a loop bounding a disk in $$\Sigma\times[0,1]$$ with framing perpendicular to the disk, then $$L= -q^2 -q^{-2}.$$ And the skein relation says L=qL++q−1L− if $$L, L_+, L_-$$ are identical except in a ball in which they look like in Figure 2. Fig. 1. View largeDownload slide Skein relation (left) and trivial loop relation (right). Fig. 1. View largeDownload slide Skein relation (left) and trivial loop relation (right). Fig. 2. View largeDownload slide From left to right: $$L, L_+, L_-$$. Fig. 2. View largeDownload slide From left to right: $$L, L_+, L_-$$. For future reference, we say that the diagram $$L_+$$ (resp. $$L_-$$) in Figure 2 is obtained from $$L$$ by the positive (resp. negative) resolution of the crossing. The $$R$$-module $$\mathscr S_R(\Sigma)$$ has an algebra structure where the product of two links $$\alpha_1$$ and $$\alpha_2$$ is the result of stacking $$\alpha_1$$ atop $$\alpha_2$$ using the cylinder structure of $$\Sigma \times [0,1]$$. Over $$R=\mathbb Z$$, the skein algebra $$\mathscr S_R(\Sigma)$$ is commutative and is closely related to the $$SL_2(\mathbb C)$$-character variety of $$\Sigma$$ (see [1, 2, 13, 15]). Over $$R={\mathbb Z[q^{\pm 1}]}$$, $$\mathscr S_R(\Sigma)$$ is not commutative in general and is closely related to quantum Teichmüller spaces [4]. 2.3 Bases We will consider $$\Sigma$$ as a subset of $$\Sigma\times [0,1]$$ by identifying $$\Sigma$$ with $$\Sigma \times \{1/2\}$$. As usual, links in $$\Sigma$$, which are closed one-dimensional non-oriented submanifolds of $$\Sigma$$, are considered up to ambient isotopies of $$\Sigma$$. A link $$\alpha$$ in $$\Sigma$$ is an essential if it has no trivial component, that is, a component bounding a disk in $$\Sigma$$. By convention, the empty set is considered an essential link. The framing of a link is vertical if at every point $$x$$, the framing is parallel to $$x \times [0,1]$$, with the direction equal to the positive direction of $$[0,1]$$. By [13, Theorem 5.2], $$\mathscr S_R(\Sigma)$$ is free over $$R$$ with basis the set of all isotopy classes of essential links in $$\Sigma$$ with vertical framing. This basis can be parameterized as follows. An integer lamination of $$\Sigma$$ is an unordered collection $$\mu=(n_i,C_i)_{i=1}^m$$, where each $$n_i$$ is a positive integer, each $$C_i$$ is a non-trivial knot in $$\Sigma$$, no two $$C_i$$ intersect, no two $$C_i$$ are ambient isotopic. For each integer lamination $$\mu$$, define an element $$b_\mu\in \mathscr S_R(\Sigma)$$ by bμ=∏i(Ci)ni. Then the set of all $$b_\mu$$, where $$\mu$$ runs the set of all integer laminations including the empty one, is the above mentioned basis of $$\mathscr S_R(\Sigma)$$. Suppose $${\mathbf P}=(P_n(t)_{n\ge 0})$$ is a normalized sequence of polynomials in $$R[t]$$. Then we can twist the basis element $$b_\mu$$ by $${\mathbf P}$$ as follows. Let bμ,P:=∏iPni(Ci). As $$\{ P_n(t)\} \mid n \in \mathbb Z_+ \}$$, just like $$\{ z^n\}$$, is a basis of $$R[t]$$, the set $$\mathcal B_{\mathbf P}$$ of all $$b_{\mu,{\mathbf P}}$$, when $$\mu$$ runs the set of all integer laminations, is a free $$R$$-basis of $$\mathscr S_R(\Sigma)$$. 3 Theorem 1.2 and Its Stronger Version We present here the proof of a stronger version of Theorem 1.2, using a result from [7] which we recall first. 3.1 Skein algebra of the annulus Let $${\mathbb A} \subset \mathbb R^2$$ be the annulus $${\mathbb A}= \{ x \in \mathbb R^2, 1 \le | x| \le 2\}$$. Let $$z$$ be the core of the annulus defined by $$z = \{ {x}, |x | = 3/2\}$$. It is easy to show that, as an algebra, $$ \mathscr S_R({\mathbb A}) = R[z]$$. Let $$p_1= (0,1)\in \mathbb R^2$$ and $$p_2=(0,2)\in \mathbb R^2$$, which are points in $$\partial {\mathbb A}$$. Then $${\mathbb A_{io}}=({\mathbb A}, \{p_1, p_2\})$$ is an example of a marked surface. See Figure 3, which also depicts the arcs $$\theta_0$$, $$\theta_{-1}$$, $$\theta_{2}$$. Here $$\theta_0$$ is the straight segment $$p_1p_2$$, and $$\theta_n$$, for $$n\in \mathbb Z$$, is an arc properly embedded in $${\mathbb A}$$ beginning at $$p_1$$ and winding clockwise (resp. counterclockwise) $$|n|$$ times if $$n\ge 0$$ (resp. $$n < 0$$) before getting to $$p_2$$. Fig. 3. View largeDownload slide Marked annulus $${\mathbb A_{io}}$$, arcs $$\theta_0$$, $$\theta_{-1}$$, $$\theta_{2}$$, and element $$\theta_0\bullet z.$$ Fig. 3. View largeDownload slide Marked annulus $${\mathbb A_{io}}$$, arcs $$\theta_0$$, $$\theta_{-1}$$, $$\theta_{2}$$, and element $$\theta_0\bullet z.$$ A $${\mathbb A_{io}}$$-arc is a proper embedding of the interval $$[0,1]$$ in $${\mathbb A} \times [0,1]$$ equipped with a framing such that one end point is in $$p_1 \times [0,1]$$ and the other is in $$p_2 \times [0,1]$$, and the framing is vertical at both end points. A $${\mathbb A_{io}}$$-tangle is a disjoint union of a $${\mathbb A_{io}}$$-arc and a (possibly empty) framed link in $${\mathbb A} \times [0,1]$$. Isotopy of $${\mathbb A_{io}}$$-tangles is considered in the class of $${\mathbb A_{io}}$$-tangles. Let $$\mathscr S'_R({\mathbb A_{io}})$$ be the $$R$$-module spanned by isotopy classes of $${\mathbb A_{io}}$$-tangles modulo the usual skein relation and the trivial knot relation. As usual, each $$\theta_n$$ is equipped with the vertical framing, and is considered as an element of $$\mathscr S'_R({\mathbb A_{io}})$$. For $$\alpha\in \mathscr S_R({\mathbb A})$$ let $$\theta_0 \bullet \alpha\in \mathscr S'_R({\mathbb A_{io}})$$ be the element obtained by placing $$\theta_0$$ on top of $$\alpha$$, see Figure 3 for an example. In [7] we proved the following. Proposition 3.1. For any integer $$n \ge 1$$, we have θ0∙Tn(z)=qnθn+q−nθ−n. (1) □ 3.2 Stronger version of Theorem 3.2 We say that a sequence $${\mathbf P}=(P_n)$$ of normalized polynomials in $$R[t]$$ is positive for $$\Sigma$$ over $$R$$ if $$\mathcal B_{\mathbf P}$$ is a positive basis for $$\mathscr S_R(\Sigma)$$. Thus, $${\mathbf P}=(P_n)$$ is positive if and only if it is positive for any oriented surface. Theorem 1.2 follows from the following stronger statement. Theorem 3.2. Suppose $${\mathbf P}=(P_n(t))$$ is positive for a non-planar oriented surface $$\Sigma$$. Then for every $$n \ge 0$$, $$P_n(t)$$ is an $$R_+$$-linear combination of $$T_k(t)$$ with $$k \le n$$. Besides, $$P_1(t)=t$$. □ Proof. Since $$\Sigma$$ is non-planar, there are two simple closed curves $$z,z'\subset \Sigma$$ which intersect transversally at one point. We identify a small tubular neighborhood of $$z$$ with the annulus $${\mathbb A}$$ such that $$z'\cap {\mathbb A}$$ is the segment $$p_1 p_2$$, see Figure 4. Note that $$z$$ and $$z'$$, as homology classes in $$H_1(\Sigma,\mathbb Z)$$, are linearly independent over $$\mathbb Z$$. By definition $$P_1(t)=t+a$$, $$a \in R$$. Fix an integer $$n\ge 1$$. There are $$c_k \in R$$ such that Pn(t)=∑k=0nckTk(t). Let $$z_{1,k}$$ be the curve $$(z' \setminus {\mathbb A})\cup \theta_k$$. We have P1(z′)Pn(z)=aPn(z)+z′∑k=0nckTk(z)=aPn(z)+c0z′+∑k=1nckz′Tk(z)=aPn(z)+c0z′+∑k=1nck(qkz1,k+ckq−kz1,−k), (2) where the last equality follows from Proposition 3.1. Using homology, one sees that in the collection $$\{ z, z', z_{1,k}, z_{1,-k} \mid k \ge 1\}$$ every curve is a non-trivial knot in $$\Sigma$$, and any two of them are non-isotopic. Rewriting $$t = P_1(t) -a$$, we have P1(z′)Pn(z)=aPn(z)+c0(P1(z′)−a)+∑k=1nck[qk(P1(z1,k)−a)+q−k(P1(z1,−k)−a)]=aPn(z)+c0P1(z′)+∑k=1nck[qk(P1(z1,k))+q−k(P1(z1,−k))]−ac0−∑k=1nack(qk+q−k). By the positivity, the coefficients of $$P_1(z')$$, $$P_n(z)$$, $$P_1(z_{1,k})$$ (for every $$k \ge 1$$) and the constant coefficient are in $$R_+$$. This shows that $$a, c_k \in R_+$$ for every $$k$$, and d:=−ac0−∑k=1nack(qk+q−k)∈R+. Note that $$-d = a( c_0 + \sum_{k=1}^n c_k (q^k + q^{-k})) \in R_+$$, since $$a, c_k \in R_+$$. Thus, both $$d$$ and $$-d$$ are in $$R_+$$. Then $$d=0$$ by the assumption on $$R$$. It follows that a(c0+∑k=1nck(qk+q−k))=0. (3) Claim. Suppose $$x,y\in R_+$$ such that $$x+y=0$$. Then $$x=y=0$$. Proof of Claim. We have $$x=-y$$. Thus, $$y$$ and $$-y=x$$ are in $$ R_+$$, implying $$y=0$$. Then $$x=0$$. The claim means that if $$x,y \in R_+$$ and $$x\neq 0$$, then $$x+y \neq 0$$. In particular, $$q^n+q^{-n} \neq 0$$. Note that $$c_n=1$$ because $$P_n(t)$$ is a monic polynomial. All the summands in the sum c0+∑k=1nck(qk+q−k) are in $$R_+$$, and the summand with $$k=n$$ is non-zero. Hence the above sum is non-zero. Since $$R$$ is a domain, from (3) we conclude that $$a=0$$. This completes the proof of the theorem. ■ Fig. 4. View largeDownload slide The union $${\mathbb A} \cup z'$$. The bold arc is $$z'\setminus {\mathbb A}$$. Fig. 4. View largeDownload slide The union $${\mathbb A} \cup z'$$. The bold arc is $$z'\setminus {\mathbb A}$$. 4 Marked Surfaces 4.1 Skein algebra of marked surfaces Suppose $$(\Sigma,\mathcal P)$$ is a marked surface, that is, $$\mathcal P\subset \partial \Sigma$$ is a finite set. A framed 3D $$\mathcal P$$-tangle $$\alpha$$ in $$\Sigma\times [0,1]$$ is a framed proper embedding of a one-dimensional non-oriented compact manifold into $$\Sigma\times [0,1]$$ such that $$\partial \alpha \subset \mathcal P \times [0,1]$$, and the framing at every boundary point of $$\alpha$$ is vertical. Two framed 3D $$\mathcal P$$-tangles are isotopic if they are isotopic through the class of framed 3D $$\mathcal P$$-tangles. Just like the link case, a framed 3D $$\mathcal P$$-tangle $$\alpha$$ is depicted by its diagram on $$\Sigma$$, with vertical framing on the diagram. Here a diagram of $$\alpha$$ is a projection of $$\alpha$$ on to $$\Sigma$$ in general position, with an order of strands at every crossing. Crossings come in two types. If the crossing is in $$\Sigma \setminus \mathcal P$$, then it is a usual double point (with usual over/under information). If the crossing is a point in $$\mathcal P$$, there may be two or more number of strands, which are ordered. The order indicates which strand is above which. When there are two strands, the lower one is depicted by a broken line, see, for example, Figure 5. Fig. 5. View largeDownload slide Trivial arc relation: if a framed 3D $$\mathcal P$$-tangle $$\alpha$$ has a trivial arc, then $$\alpha=0$$. Fig. 5. View largeDownload slide Trivial arc relation: if a framed 3D $$\mathcal P$$-tangle $$\alpha$$ has a trivial arc, then $$\alpha=0$$. Let $$\mathscr S_R(\Sigma,\mathcal P)$$ be the $$R$$-module spanned by the set of isotopy classes of framed 3D $$\mathcal P$$-tangles in $$\Sigma\times [0,1]$$ modulo the skein relation, the trivial knot relation (Figure 1), and the trivial arc relation of Figure 5. The following relation holds, see [8]. Proposition 4.1. In $$\mathscr S(\Sigma,\mathcal P)$$, the reordering relation depicted in Figure 6 holds. □ Fig. 6. View largeDownload slide Reordering relation. Fig. 6. View largeDownload slide Reordering relation. Again one defines the product $$\alpha_1 \alpha_2$$ of two skein elements $$\alpha_1$$ and $$\alpha_2$$ by stacking $$\alpha_1$$ above $$\alpha_2$$. This makes $$\mathscr S_R(\Sigma,\mathcal P)$$ an $$R$$-algebra, which was first defined by Muller [10]. 4.2 Basis for $$\mathscr S_R(\Sigma,\mathcal P)$$ A $$\mathcal P$$-arc is an immersion $$\alpha: [0,1]\to \Sigma$$ such that $$\alpha(0), \alpha(1)\in \mathcal P$$ and the restriction of $$\alpha$$ on to $$(0,1)$$ is an embedding into $$\Sigma \setminus \mathcal P$$. A $$\mathcal P$$-knot is an embedding of $$S^1$$ into $$\Sigma \setminus \mathcal P$$. A $$\mathcal P$$-knot or a $$\mathcal P$$-arc is trivial if it bounds a disk in $$\Sigma$$. Two $$\mathcal P$$-arcs (resp. $$\mathcal P$$-knots) are $$\mathcal P$$-isotopic if they are isotopic in the class of $$\mathcal P$$-arcs (resp. $$\mathcal P$$-knots). A $$\mathcal P$$-arc is called boundary if it can be isotoped into $$\partial \Sigma$$, otherwise it is called inner. We consider every $$\mathcal P$$-arc and every $$\mathcal P$$-knot as an element of $$\mathscr S_R(\Sigma,\mathcal P)$$ by equipping it with the vertical framing. In the case when the two ends of a $$\mathcal P$$-arc are the same point $$p\in\mathcal P$$, we order the left strand to be above the right one. An integer $$\mathcal P$$-lamination of $$\Sigma$$ is an unordered collection $$\mu=(n_i,C_i)_{i=1}^m$$, where each $$n_i$$ is a positive integer, each $$C_i$$, called a component of $$\mu$$, is a non-trivial $$\mathcal P$$-knot or a non-trivial $$\mathcal P$$-arc, no two $$C_i$$ intersect in $$\Sigma\setminus \mathcal P$$, no two $$C_i$$ are $$\mathcal P$$-isotopic. In an integer $$\mathcal P$$-lamination $$\mu=(n_i,C_i)_{i=1}^m$$, a knot component commutes with any other component (in the algebra $$\mathscr S_R(\Sigma,\mathcal P)$$), while a $$\mathcal P$$-arc component $$C$$$$q$$-commutes with any other component $$C'$$ in the sense that $$CC'= q^k C'C$$ for a certain $$k\in \mathbb Z$$. The $$q$$-commutativity follows from Proposition 4.1. Hence the element bμ=∏i(Ci)ni (4) is defined up to a factor $$q^k, k\in \mathbb Z$$. To make $$b_\mu$$ really well-defined, we will fix once and for all a total order on the set of all $$\mathcal P$$-arcs in $$\Sigma$$. Now define $$b_\mu$$ by (4), where the product is taken in this order. To simplify some proofs, we further assume that in the total order any boundary $$\mathcal P$$-arc comes before any inner $$\mathcal P$$-arc, although this is not necessary. It follows from [10, Lemma 4.1] that the set of all $$b_\mu$$, where $$\mu$$ runs the set of all integer $$\mathcal P$$-laminations including the empty one, is the a free $$R$$-basis of $$\mathscr S_R(\Sigma,\mathcal P)$$. Suppose $${{\mathbf P}}=(P_n)$$ and $${\mathbf Q}=(Q_n)$$ are normalized sequences of polynomials in $$R[t]$$. Define bμ,P,Q=∏iFni(Ci), (5) where the product is taken in the above-mentioned order, and $$F_{n_i}=P_{n_i} $$ if $$C_i$$ is a knot, $$F_{n_i}= Q_{n_i}$$ if $$C_i$$ is an inner $$\mathcal P$$-arc, $$F_{n_i}(C_i) = (C_i)^{n_i}$$ if $$C_i$$ is a boundary $$\mathcal P$$-arc. Then the set $$\mathcal B_{{{\mathbf P}}, {\mathbf Q}}$$ of all $$b_{\mu,{{\mathbf P}}, {\mathbf Q}}$$, where $$\mu$$ runs the set of all integer $$\mathcal P$$-laminations, is a free $$R$$-basis of $$\mathscr S_R(\Sigma,\mathcal P)$$. A pair of sequences of polynomials $$({{\mathbf P}},{\mathbf Q})$$ are positive over $$R$$ if they are normalized and the basis $$\mathcal B_{{{\mathbf P}}, {\mathbf Q}}$$ is positive for any marked surface. 4.3 Positive basis in quotients The following follows right away from the definition. Lemma 4.2. Let $$\mathcal B$$ be a positive basis of an $$R$$-algebra $$A$$ and $$I\subset A$$ be an ideal of $$A$$, with $$\pi: A \to A/I$$ the natural projection. Assume that $$I$$ respects the base $$\mathcal B$$ in the sense that $$I$$ is freely $$R$$-spanned by $$I \cap \mathcal B$$. Then $$\pi(\mathcal B\setminus I)$$ is a positive basis of $$A/I$$. □ 4.4 Ideal generated by boundary arcs Lemma 4.3. Suppose $$\alpha_1,\dots,\alpha_l$$ are boundary $$\mathcal P$$-arcs, and $$I=\sum_{i=1}^l \alpha_i \mathscr S_R(\Sigma,\mathcal P)$$. (a) The set $$I$$ is a two-sided ideal of $$\mathscr S_R(\Sigma,\mathcal P)$$. (b) For any normalized sequences $${{\mathbf P}}, {\mathbf Q}$$ of polynomials in $$R[t]$$, $$I$$ respects the basis $$\mathcal B_{{{\mathbf P}},{\mathbf Q}}$$. □ Proof. (a) Suppose $$\alpha$$ is a boundary $$\mathcal P$$-arc. Since $$\alpha$$$$q$$-commutes with any basis element $$b_\mu$$ defined by (4), $$\alpha\mathscr S_R(\Sigma,\mathcal P)$$ is a two-sided ideal of $$\mathscr S_R(\Sigma,\mathcal P)$$. Hence $$I$$ is also a two-sided ideal. (b) In the chosen order, the boundary $$\mathcal P$$-arcs come before any inner arcs $$\mathcal P$$-arc. Hence $$I$$ is freely $$R$$-spanned by all $$b_{\mu,{{\mathbf P}},{\mathbf Q}}$$ such that $$\mu$$ has one of the $$\alpha_i$$ as a component. Thus, $$I$$ respects $$\mathcal B_{{{\mathbf P}},{\mathbf Q}}$$. ■ 4.5 Proof of Theorem 1.3 We will prove the following stronger version of Theorem 1.3. Theorem 4.4. Suppose a pair $$({{\mathbf P}},{\mathbf Q})$$ of sequences of polynomials in $$R[t]$$, $${{\mathbf P}}= (P_n(t)_{n\ge 0})$$ and $${\mathbf Q}= (Q_n(t)_{n\ge 0})$$, are positive over $$R$$. Then $$P_n(t)$$ is an $$R_+$$-linear combination of $$T_0(t), T_1(t), \dots, T_n(t)$$ and $$Q_n(t)$$ is an $$R_+$$-linear combination of $$1, t, \dots, t^n$$. Moreover, $$P_1(t)=Q_1(t)=t$$. □ Lemma 4.5. One has $$Q_1(t)= t$$. □ Proof. Consider the marked surface $$\mathbb D_1$$, which is a disk with four marked points $$p_0, p_1, p_2$$, and $$q_1$$ as in Figure 7. Let $$x$$ be the arc $$p_0p_2$$ and $$y$$ the arc $$ p_1 q_1$$. Suppose $$Q_1(t)= t + a$$, where $$a\in R$$. Using the skein relation to resolve the only crossing of $$xy$$, we get Q1(x)Q1(y)=aQ1(x)+aQ1(y)−a2modI∂. (6) Here $$I_\partial$$ is the ideal of $$\mathscr S_R(\mathbb D_1)$$ generated by all the boundary $$\mathcal P$$-arcs, which respects $$\mathcal B_{{{\mathbf P}},{\mathbf Q}}$$ by Lemma 4.3. Lemma 4.2 and Equation (6) show that $$ a \in R_+$$ and $$- a^2 \in R_+$$. Hence both $$a^2$$ and $$-a^2$$ are $$R_+$$, which implies $$a=0$$. ■ Fig. 7. View largeDownload slide Left: $$\mathbb D_1$$, with product $$xy$$. Middle: $$\mathbb D_5$$, with product $$x^2 y_5$$. Right: Element $$z_{2,5.}$$ Fig. 7. View largeDownload slide Left: $$\mathbb D_1$$, with product $$xy$$. Middle: $$\mathbb D_5$$, with product $$x^2 y_5$$. Right: Element $$z_{2,5.}$$ Now fix an integer number $$n \ge 2$$. Let $$\mathbb D_n$$ be the marked surface, which is a disk with $$2n+2$$ marked points which in clockwise orders are $$p_0,p_1,\dots,p_{n+1}, q_n,\dots,q_1$$. We draw $$\mathbb D_n$$ in the standard plane so that the straight segment $$p_0p_{n+1}$$, denoted by $$x$$, is vertical with $$p_0$$ being lower than $$p_{n+1}$$, and each straight segment $$p_i q_i$$ is horizontal. See Figure 7 for an example of $$\mathbb D_5$$. Let $$y_n$$ be the union of the $$n$$ straight segments $$p_i q_i$$, $$i=1,\dots,n$$. Considering $$x$$ as an element of $$\mathscr S_R(\mathbb D_n)$$, we will present $$x^k$$ by $$k$$ arcs, each going from $$p_0$$ monotonously upwards to $$p_{n+1}$$. Any two of these $$k$$ arcs do not have intersection in the interior of $$\mathbb D_n$$, and the left one is above the right one. See an example in Figure 7 where the diagram of $$x^2 y_5$$ is drawn. Then the diagram of $$x^k y_n$$ has exactly $$kn$$ double points. Let $$z_{k,n}$$ be obtained from this diagram of $$x^k y_n$$ by negatively resolving all the crossings, again see Figure 7. For each $$i=0,\dots,n-1$$ let $$\gamma_i$$ be the arc $$p_i p_{i+1}$$, which is a boundary $$\mathcal P$$-arc. By Lemma 4.3, the set I:=∑i=0n−1γiS(Σ,P) is a two-sided ideal of $$\mathscr S(\Sigma,\mathcal P)$$ respecting $$\mathcal B_{{{\mathbf P}},{\mathbf Q}}$$. It is important that the arc $$p_n p_{n+1}$$ is not in $$I$$. Lemma 4.6. For every $$k \le n$$, one has xkyn=q−knzk,nmodI. (7) □ Proof. Label the $$k$$ arcs of the diagram of $$x^k$$ from left to right by $$1, \dots, k$$. There are $$kn$$ crossings in the diagram of $$x^k y_n$$, and denote by $$E_{l,m}$$ the intersection of the $$l$$th arc of $$x^k$$ and the arc $$p_m q_m$$ of $$y_n$$. Order the set of all $$kn$$ crossings $$E_{lm}$$ by the lexicographic pair $$(l+m,l)$$. Suppose $$\tau$$ is one of the $$2^{kn}$$ ways of resolutions of all the $$kn$$ crossings. Let $$D_\tau$$ be the result of the resolution $$\tau$$, which is a diagram without crossing. Assume $$\tau$$ has at least one positive resolution. We will prove that $$D_\tau\in I$$. Let $$E_{l,m}$$ be the smallest crossing at which the resolution is positive. In a small neighborhood of $$E_{l,m}$$, $$D_\tau$$ has two arcs, with the lower left one denoted by $$\delta$$, see Figure 8. The resolution at any $$E_{l', m'} < E_{l,m}$$ is negative. These data are enough to determine the arc $$\mathfrak u$$ of $$D_\tau$$ containing $$\delta$$, see Figure 9. Namely, if $$m \ge l$$ then $$\mathfrak u$$ is $$\gamma_{m-l}$$, and if $$m < l$$ then $$\mathfrak u$$ is an arc whose two end points are $$p_0$$, which is 0. See Figure 10 for an example. Either case, $$D_\tau \in I$$. Hence, modulo $$I$$, the only element obtained by resolving all the crossings of $$x^k y_n$$ is the all-negative resolution one, which is $$z_{k,n}$$. The corresponding factor coming from the skein relation is $$q^{-kn}$$. This proves Identity (7). ■ Fig. 8. View largeDownload slide Left: Crossing $$E_{l,m}$$. Right: Its positive resolution, and the arc $$\delta$$. Fig. 8. View largeDownload slide Left: Crossing $$E_{l,m}$$. Right: Its positive resolution, and the arc $$\delta$$. Fig. 9. View largeDownload slide The top right corner crossing is $$E_{l,m}$$, with positive resolution. The arc of $$D_\tau$$ containing $$\delta$$ keeps going down the south-west direction. Fig. 9. View largeDownload slide The top right corner crossing is $$E_{l,m}$$, with positive resolution. The arc of $$D_\tau$$ containing $$\delta$$ keeps going down the south-west direction. Fig. 10. View largeDownload slide Left: The arc $$\mathfrak u$$ with $$l=4, m=5$$. In this case $$\mathfrak u=\gamma_1$$. Right: The arc $$\mathfrak u$$ with $$l=4, m=3$$. In this case, both end points of $$\mathfrak u$$ are $$p_0$$. Fig. 10. View largeDownload slide Left: The arc $$\mathfrak u$$ with $$l=4, m=5$$. In this case $$\mathfrak u=\gamma_1$$. Right: The arc $$\mathfrak u$$ with $$l=4, m=3$$. In this case, both end points of $$\mathfrak u$$ are $$p_0$$. Proof of Theorem 4.4. Theorem 3.2 implies that $$P_n$$ is an $$R_+$$-linear combination of $$T_k$$ with $$k \le n$$. Since $$Q_1(t)=t$$, each of $$z_{k,n}$$, $$y_n$$ in Lemma 4.6 is an element of the basis $$\mathcal B_{{{\mathbf P}},{\mathbf Q}}$$. Suppose $$Q_n(t)= \sum_{k=0}^n c_k t^k$$ with $$c_k \in R$$. From (7), we have Qn(x)yn=(∑k=0nckxk)yn=∑k=0nckq−knzk,n+I. Note that $$z_{k,n}\not \in I$$. Since $$I$$ respects the basis $$\mathcal B_{{{\mathbf P}},{\mathbf Q}}$$, Lemma 4.2 shows that $$c_k \in R_+$$ for all $$k$$. This completes the proof of Theorem 4.4. ■ Funding This work was supported in part by the National Science Foundation grant DMS-14-06419. Acknowledgments The author would like to thank D. Thurston for valuable discussions. He also thanks the anonymous referees for useful comments which in particular lead to the current, stronger formulation of Theorem 3.2. References [1] Bullock D. “Rings of $$Sl_2(\mathbb C)$$-characters and the Kauffman bracket skein module.” Commentarii Mathematici Helvetici 72, no. 4 (1997): 521– 42. Google Scholar CrossRef Search ADS [2] Bullock D. Frohman C. and Kania-Bartoszynska. J. “Understanding the Kauffman bracket skein module.” Journal of Knot Theory and Its Ramifications 8, no. 3 (1999): 265– 77. Google Scholar CrossRef Search ADS [3] Bonahon F. and Wong. H. “Quantum traces for representations of surface groups in $$SL_2$$(C).” Geometry & Topology 15, no. 3 (2011): 1569– 615. Google Scholar CrossRef Search ADS [4] Chekhov L. and Fock. V. “Quantum Teichmüller spaces” (Russian) Teoreticheskaya i Matematicheskaya Fizika 120, no. 3 (1999): 511– 28; translation in Theoretical and Mathematical Physics 120, no. 3 (1999): 1245– 59. Google Scholar CrossRef Search ADS [5] Fock V. and Goncharov. A. “Moduli spaces of local systems and higher Teichmüller theory.” Publications Mathématiques. Institut de Hautes études Scientifiques no. 103 (2006): 1– 211. [6] Kauffman L. “States models and the Jones polynomial.” Topology 26 (1987): 395– 407. Google Scholar CrossRef Search ADS [7] Lê T. T. Q. “On Kauffman bracket skein modules at roots of unity.” Algebraic & Geometric Topology 15 (2015): 1093– 117. Google Scholar CrossRef Search ADS [8] Lê T. T. Q. Quantum Teichmuller spaces and quantum trace map . Journal of the Institute of Mathematics of Jussieu (2015): preprint arXiv:1511.06054 (in press). [9] Lê T. T. Q. and Tran. A. “On the AJ conjecture for knots.” Indiana University Mathematics Journal 64 (2015): 1103– 51. Google Scholar CrossRef Search ADS [10] Muller G. “Skein algebras and cluster algebras of marked surfaces.” Quantum Topology , (2012): preprint arXiv:1204.0020 (in press). [11] Musiker G. Schiffler R. and Williams. L. “Bases for cluster algebras from surfaces.” Compositio Mathematica 149 (2013): 217– 63. Google Scholar CrossRef Search ADS [12] Przytycki J. “Fundamentals of Kauffman bracket skein modules.” Kobe journal of mathematics 16 (1999): 45– 66. [13] Przytycki J. and Sikora. 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International Mathematics Research Notices – Oxford University Press

**Published: ** Mar 1, 2018

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