Quantum traces and embeddings of stated skein algebras into quantum tori

Abstract

The stated skein algebra of a punctured bordered surface (or equivalently, a marked surface) is a generalization of the well-known Kauffman bracket skein algebra of unmarked surfaces and can be considered as an extension of the quantum special linear group ${\mathcal {O}}_{q^2}(SL_2)$ from a bigon to general surfaces. We show that the stated skein algebra of a punctured bordered surface with non-empty boundary can be embedded into quantum tori in two different ways. The first embedding can be considered as a quantization of the map expressing the trace of a closed curve in terms of the shear coordinates of the enhanced Teichmüller space, and is a lift of Bonahon-Wong’s quantum trace map. The second embedding can be considered as a quantization of the map expressing the trace of a closed curve in terms of the lambda length coordinates of the decorated Teichmüller space, and is an extension of Muller’s quantum trace map. We explain the relation between the two quantum trace maps. We also show that the quantum cluster algebra of Muller is equal to a reduced version of the stated skein algebra. As applications we show that the stated skein algebra is an orderly finitely generated Noetherian domain and calculate its Gelfand-Kirillov dimension.

Triangulation and quasitriangulation

1.1 Face and vertex matrices

Here we give a more formal definition of the face and vertex matrices.

Suppose $\Delta$ is an ideal triangulation of the punctured bordered surface ${\mathfrak {S}}$. For each non-self-folded face ${\mathfrak {T}}\in {\mathcal {F}}$, the edges of ${\mathfrak {T}}$ are cyclically ordered. Given $a,b\in \Delta$, if a and b are distinct edges of ${\mathfrak {T}}$, define

$$\begin{aligned} Q_{\mathfrak {T}}(a,b)={\left\{ \begin{array}{ll} 1,&{}b\text { is clockwise to }a,\\ -1,&{}b\text { is counterclockwise }a.\\ \end{array}\right. } \end{aligned}$$

Otherwise, let $Q_{\mathfrak {T}}(a,b)=0$. For a self-folded face ${\mathfrak {T}}$, set $Q_{\mathfrak {T}}=0$. Then the face matrix is

$$\begin{aligned} Q=\sum _{{\mathfrak {T}}\in {\mathcal {F}}}Q_{\mathfrak {T}}. \end{aligned}$$

(75)

The vertex matrix is defined for a quasitriangulation ${\mathcal {E}}$.

For each boundary puncture $v\in {\mathcal {P}}_\partial$ and half-edges $a',b'$, let $P_{+,v}(a',b')=1$ if $a'\ne b'$ and $b'$ is counterclockwise to $a'$ at v. Otherwise let $P_{+,v}(a',b')=0$. Define the “positive part" $P_+$ of the vertex matrix by

$$\begin{aligned} P_+(a,b)=\sum P_{+,v}(a',b'), \end{aligned}$$

where the sum is over punctures $v\in \mathring{{\mathcal {P}}}$ and half-edges $a',b'$ of arcs $a,b\in {\mathcal {E}}$. For this definition to make sense, an isotopy might be necessary to make a and b disjoint, which is important for the diagonal elements. The vertex matrix is then $P=P_+-P_+^T$. Alternatively, let $P_v(a',b')=P_{+,v}(a',b')-P_{+,v}(b',a')$. Then $P(a,b)=\sum P_v(a',b')$.

1.2 Extended matrices

Define the projection matrix $J:\Delta \times \Delta _\partial \rightarrow {\mathbb {Z}}$ or $J:{\mathcal {E}}\times {\mathcal {E}}_\partial \rightarrow {\mathbb {Z}}$ (depending on the context) by

$$\begin{aligned} J(a,b)={\left\{ \begin{array}{ll}1,&{}a=b,\\ 0,&{}a\ne b.\end{array}\right. } \end{aligned}$$

The extended matrices $\bar{Q}:\bar{\Delta }\times \bar{\Delta }\rightarrow {\mathbb {Z}}$ and $\bar{P},\bar{P}_+:\bar{{\mathcal {E}}}\times \bar{{\mathcal {E}}}\rightarrow {\mathbb {Z}}$ can be written in block matrix form

$$\begin{aligned} \bar{Q}&=\begin{pmatrix}Q&{}\quad J\\ -J^T&{}\quad 0\end{pmatrix}, \end{aligned}$$

(76)

$$\begin{aligned} \bar{P}&=\begin{pmatrix}P&{}\quad -(P_++P_+^T)J\\ J^T(P_++P_+^T)&{}\quad -J^T PJ\end{pmatrix}, \end{aligned}$$

(77)

$$\begin{aligned} \bar{P}_+&=\begin{pmatrix}P_+&{}\quad P_+J\\ J^T P_+&{}\quad J^T P_+J-2I\end{pmatrix}. \end{aligned}$$

(78)

Since J is a projection, it picks out blocks of matrices with the correct dimension.

1.3 Face-vertex matrix duality

For a punctured bordered surface ${\mathfrak {S}}$ with at least one boundary puncture, there is a relation between the face matrix and the vertex matrix. Consider a quasitriangulation ${\mathcal {E}}$ and its completion $\Delta$. Let $H=I_\partial -Q_{\mathcal {E}}$, where $Q_{\mathcal {E}}$ is the restriction to ${\mathcal {E}}\times {\mathcal {E}}$, and $I_\partial =JJ^T$. The restriction to ${\mathcal {E}}$ does not lose any information, since by definition, all entries involving a self-folded edge is 0, so the full matrix can be recovered by a 0-extension, i.e., $Q=Q_{\mathcal {E}}\oplus 0$. The extended matrix $\bar{H}$ is defined by, using the block matrix form,

$$\begin{aligned} \bar{H}=\begin{pmatrix}-Q_{\mathcal {E}}&{}\quad J\\ J^T&{}\quad -I\end{pmatrix}. \end{aligned}$$

Lemma A.1

$HP_+=2I$, $\bar{H}\bar{P}_+=2I$.

Proof

The extended case follows from the first part by a routine calculation.

The punctured monogon is a special case we can directly verify. Here $P_+=(2)$ and $H=(1)$. Thus we focus on other surfaces from now on. In particular, all edges that bound self-folded faces are interior.

There are three cases to consider.

Case 1. Suppose $i\in \mathring{{\mathcal {E}}}$ does not bound a self-folded face. By definition,

$$\begin{aligned} (HP_+)(i,j)=\sum _{k\in {\mathcal {E}}}H(i,k)P_+(k,j)=P_+(a,j)-P_+(b,j)+P_+(c,j)-P_+(d,j), \end{aligned}$$

where a, b, c, d are as in Fig. 12a. Some of the sides might coincide, but using the sum-of-faces definition (75) of Q, the result is unaffected. We further split the sum into half-edges, and group the half-edges into pairs by corners of the quadrilateral.

First we consider $j\notin \{i,a,b,c,d\}$. If a half-edge $j'$ ends on one of the boundary punctures of the quadrilateral, then every corner incident to that puncture is entirely on one side of $j'$. Thus every such corner contributes $1-1$ or $0-0$ to the sum, so the total is 0.

If $j\in \{a,b,c,d\}$, isotope j into the interior of the quadrilateral. j goes through two consecutive corners, and they contribute $1-1$ to the sum. In the two corners that j does not go through, the calculation is as before and they cancel out.

If $j=i$, then a and c both contribute $+1$ in the local picture, and any other corner incident to either end of i contributes 0 as before. Thus the total is 2.

Case 2. Suppose $i\in \mathring{{\mathcal {E}}}$ bounds a self-folded face. We can use the same picture Fig. 12a and identify $c=d$. In this case

$$\begin{aligned} (HP_+)(i,j)=P_+(a,j)-P_+(b,j). \end{aligned}$$

a and b form a (punctured) bigon. We can group half-edges of a and b into corners and repeat the calculation as in Case 1.

Case 3. Suppose $i\in {\mathcal {E}}_\partial$, then

$$\begin{aligned} (HP_+)(i,j)=P_+(i,j)+P_+(a,j)-P_+(b,j), \end{aligned}$$

where a, b are as in Fig. 12b. The calculation is again analogous to the first case. Note even though the i and a terms have the same sign, there is nothing counterclockwise to the i, a corner. $\square$

Fig. 12

Local pictures of the computation

Proof of (73)

This is a matrix equation, where $\sigma ,K:\bar{{\mathcal {E}}}\times \bar{\Delta }\rightarrow {\mathbb {Z}}$ are given by

$$\begin{aligned} \sigma (a,c)=\sigma _a(c),\quad K(a,c)=K_a(c). \end{aligned}$$

Then (73) can be written as $\sigma =\bar{H}K$. By Lemma A.1, we just need to show $\bar{P}_+\sigma =2K$, i.e.

$$\begin{aligned} \sum _{b\in \bar{{\mathcal {E}}}}\bar{P}_+(a,b)\sigma _b(c)=2K_a(c),\quad a\in \bar{{\mathcal {E}}},c\in \bar{\Delta }. \end{aligned}$$

(79)

Also note $\bar{P}_+$ is the restriction of K to $\bar{{\mathcal {E}}}\times \bar{{\mathcal {E}}}$ by definition. Compare (70) with (78).

If $c\in \bar{{\mathcal {E}}}$, then $\sigma _b(c)$ is nonzero only when $b=c$, and $\sigma _c(c)=2$. Thus the equation is satisfied.

If $c=e_v$ for $v\in \mathring{{\mathcal {P}}}$, then $\sigma _b(e_v)$ is nonzero only when $b=b_v$ is the arc bounding the monogon containing v. In this case, $\sigma _{b_v}(e_v)=1$. (79) is then equivalent to

$$\begin{aligned} \bar{P}_+(a,b_v)=2K_a(e_v). \end{aligned}$$

This is true by checking the local picture at v. We can always isotope a to be outside of the monogon. Then for each half edge of a, either $b_v$ counts twice and $e_v$ counts once, or none of them counts. Thus we have the desired equality. $\square$