# What are the Jones polynomials for the torus links and the closure of the other braid word below?

I am working on a project to determine the Jones polynomial for the torus links and a class of links which I call tst links. Their braid words are respectively given by $$(\sigma_1 \sigma_2 \cdots \sigma_{p-1})^q \in B_p$$ and $$(\sigma_1^2 \sigma_2 \sigma_3 \cdots \sigma_p)^q \in B_{p+1}$$

I would like to know if anyone familiar with the Ocneanu approach, Chern-Simons Theory approach, and the Burau representations approach can evaluate the Jones polynomials and give a clear, detailed explanation.

# Representation theory approach

First note that the trace of a braid $$b$$ from $$B_p$$ is given by the sum over all irreducible representations of the symmetric group $$S_p$$ given by the Young tableux $$Y$$ of $$p$$ nodes: $$\text{tr}(b) = \sum_Y \tilde W_Y(q,\lambda) \text{tr}(\pi_Y(b))$$ where $$\tilde W_Y(q,\lambda)$$ is the weight of the tableau $$Y$$ and $$\pi_Y(b)$$ is the representation of $$b$$ corresponding to the tableau $$Y$$.

Then it is given that $$\text{tr}(\pi_Y(b)) = 0$$ unless $$Y$$ is of the form with $$(\beta+1)$$ boxes in the first row, then $$1$$ box in each of the $$\gamma$$ rows that follow:

$$Y_{\beta,\gamma}=$$

The weight of the tableau $$Y = Y_{\beta, \gamma}$$ is given by $$\tilde W_Y(q,\lambda) = \left(\dfrac{1-q}{1-\lambda q} \right)^p \dfrac{R_Y(q,\lambda)}{Q_Y(q)}$$ where $$R_Y(q,\lambda) = (1-\lambda q)(q-\lambda q)(q^2-\lambda q) \cdots (q^\beta -\lambda q) \times (1-\lambda q^2)(1-\lambda q^3)\cdots (1-\lambda q^{\gamma+1})$$

$$= \prod_{i=0}^\beta (q^i-\lambda q)\times q^{1+2+\cdots+\gamma}(q^{-1}-\lambda q)(q^{-2}-\lambda q)\cdots(q^{-\gamma}-\lambda q) = q^{1+2+\cdots+\gamma}\prod_{i=-\gamma}^\beta (q^i-\lambda q) = q^{\frac{\gamma(\gamma+1)}{2}}\prod_{i=-\gamma}^\beta (q^i-\lambda q)$$

and $$Q_Y$$ is given by $$Q_Y(q) = (1-q)(1-q^2)\cdots (1-q^\beta) \times (1-q^\gamma)(1-q^{\gamma-1})\cdots (1-q) \times (1-q^{\gamma+\beta+1})$$ Using the notation $$[n] = 1-q^n$$, $$[n]! = [n][n-1]!$$ and $$[0]!=1$$, we can write it as $$Q_Y(q) = [\beta]![\gamma]!(1-q^p)$$ The remaining piece of information needed to calculate the trace is the trace $$\text{tr}(\pi_Y(b))$$ for the Young tableaux $$Y = Y_{\beta,\gamma}$$ specified above and $$b=(\sigma_1\sigma_2\cdots\sigma_{p-1})^m$$.

For coprime $$m$$ and $$p$$, this is given by $$\text{tr}(\pi_Y((\sigma_1\sigma_2\cdots\sigma_{p-1})^m)) = (-1)^\gamma q^{\beta m}$$

This is related to the fact that $$(\sigma_1\sigma_2\cdots\sigma_{p-1})^q$$ is in the center of the braid group $$B_p$$ but I don't know (1) why there is such connection, and (2) why this braid word is in the center.

Putting the pieces of information together, we have $$\text{tr}((\sigma_1\sigma_2\cdots\sigma_{p-1})^m) = \sum_{\begin{matrix} \beta+\gamma+1=p \\ \beta,\gamma \geq 0 \end{matrix}} (-1)^\gamma q^{\beta m} \left(\dfrac{1-q}{1-\lambda q} \right)^p \dfrac{q^{\frac{\gamma(\gamma+1)}{2}}}{[\beta]![\gamma]!(1-q^p)}\prod_{i=-\gamma}^\beta (q^i-\lambda q)$$

Hence the 2-variable Jones polynomial is given by $$X_{\text{Cl}((\sigma_1\sigma_2\cdots\sigma_{p-1})^m)}(q, \lambda) = \left(-\dfrac{1-\lambda q}{\sqrt \lambda (1-q)}\right)^{p-1} \sqrt \lambda^{(p-1)m} \times \sum_{\begin{matrix} \beta+\gamma+1=p \\ \beta,\gamma \geq 0 \end{matrix}}(-1)^\gamma q^{\beta m} \left(\dfrac{1-q}{1-\lambda q} \right)^p \dfrac{q^{\frac{\gamma(\gamma+1)}{2}}}{[\beta]![\gamma]!(1-q^p)}\prod_{i=-\gamma}^\beta (q^i-\lambda q)$$ $$= \left(\dfrac{1-q}{1-\lambda q} \right) \lambda^{\frac{(p-1)(m-1)}{2}} \sum_{\begin{matrix} \beta+\gamma+1=p \\ \beta,\gamma \geq 0 \end{matrix}} (-1)^{p-1-\gamma} q^{\beta m} \dfrac{q^{\frac{\gamma(\gamma+1)}{2}}}{[\beta]![\gamma]!(1-q^p)}\prod_{i=-\gamma}^\beta (q^i-\lambda q)$$ $$= \left(\dfrac{1-q}{1-\lambda q} \right) \lambda^{\frac{(p-1)(m-1)}{2}} \sum_{\begin{matrix} \beta+\gamma+1=p \\ \beta,\gamma \geq 0 \end{matrix}} (-1)^\beta \dfrac{q^{\beta m+\frac{\gamma(\gamma+1)}{2}}}{[\beta]![\gamma]!(1-q^p)}\prod_{i=-\gamma}^\beta (q^i-\lambda q)$$

# Chern-Simons Theory approach

For a general torus link $$K(p,m)$$, it can be viewed as a $$d$$-component link, where $$d=\text{gcd}(p,m)$$, with each component being a $$(P,M)$$-torus knot, where $$\dfrac{P}{M}$$ is the reduced form of the fraction $$\dfrac{p}{m}$$. We can then split the manifold $$S^3$$ into pieces with toroidal boundaries such that each piece contains one $$K(P,M)$$, schematically shown below: Then by the spirit of the partition function, we have $$V_{\text{Cl}((\sigma_1 \sigma_2 \cdots \sigma_{p-1})^m)}(q, \lambda) = \left(V_{\text{Cl}((\sigma_1 \sigma_2 \cdots \sigma_{P-1})^M)}(q, \lambda) \right)^d =\left( \left(\dfrac{1-q}{1-\lambda q} \right) \lambda^{\frac{(p-d)(m-d)}{2d^2}} \sum_{\begin{matrix} \beta+\gamma+1=\frac pd \\ \beta,\gamma \geq 0 \end{matrix}} (-1)^\beta \dfrac{q^{\beta m+\frac{\gamma(\gamma+1)}{2}}}{[\beta]![\gamma]!(1-q^\frac pd)}\prod_{i=-\gamma}^\beta (q^i-\lambda q) \right)^d = \left(\dfrac{1-q}{1-\lambda q} \right)^d \lambda^{\frac{(p-d)(m-d)}{2d}} \left(\sum_{\begin{matrix} \beta+\gamma+1=\frac pd \\ \beta,\gamma \geq 0 \end{matrix}} (-1)^\beta \dfrac{q^{\beta m+\frac{\gamma(\gamma+1)}{2}}}{[\beta]![\gamma]!(1-q^\frac pd)}\prod_{i=-\gamma}^\beta (q^i-\lambda q) \right)^d$$

Similarly, for the tst links, we have the following splitting of $$S^3$$ into $$(d+1)$$ pieces: where the rightmost knot is the unknot $$O$$.

Thus we have $$V_{\text{Cl}((\sigma_1^2 \sigma_2 \cdots \sigma_p)^m)}(q, \lambda) = \left(V_{\text{Cl}((\sigma_1 \sigma_2 \cdots \sigma_{P-1})^M)}(q, \lambda) \right)^d V_{O}(q, \lambda) = \left(V_{\text{Cl}((\sigma_1 \sigma_2 \cdots \sigma_{P-1})^M)}(q, \lambda) \right)^d = \left(\dfrac{1-q}{1-\lambda q} \right)^d \lambda^{\frac{(p-d)(m-d)}{2d}} \left(\sum_{\begin{matrix} \beta+\gamma+1=\frac pd \\ \beta,\gamma \geq 0 \end{matrix}} (-1)^\beta \dfrac{q^{\beta m+\frac{\gamma(\gamma+1)}{2}}}{[\beta]![\gamma]!(1-q^\frac pd)}\prod_{i=-\gamma}^\beta (q^i-\lambda q) \right)^d$$

# Ocneanu trace approach

First consider the torus links. I will start with the easiest case $$\sigma_1 \sigma_2 \cdots \sigma_p$$. The Ocneanu trace is obviously $$z^n$$.

Then I will try $$(\sigma_1 \sigma_2 \cdots \sigma_p)^2$$. Denote $$\beta \sim_O \beta'$$ if they have the same Ocneanu traces, i.e. $$\text{tr}(\beta) = \text{tr}(\beta')$$, and $$\beta \sim_M \beta'$$ if they are related by a sequence of Markov moves. We have $$(\sigma_1 \sigma_2 \cdots \sigma_p)^2$$ $$= \sigma_1 \sigma_2 \cdots \sigma_p\sigma_1 \sigma_2 \cdots \sigma_p$$ $$= \sigma_1 \sigma_2 \cdots \sigma_{p-1}\sigma_1 \sigma_2 \cdots \sigma_p \sigma_{p-1} \sigma_p$$ $$\sim_M \sigma_1 \sigma_2 \cdots \sigma_{p-1}\sigma_1 \sigma_2 \cdots \sigma_{p-1} \sigma_p \sigma_{p-1}$$ $$\sim_O z \sigma_1 \sigma_2 \cdots \sigma_{p-1}\sigma_1 \sigma_2 \cdots \sigma_{p-1}^2$$ $$\sim_O z(q-1) \sigma_1 \sigma_2 \cdots \sigma_{p-1}\sigma_1 \sigma_2 \cdots \sigma_{p-1} + zq \sigma_1 \sigma_2 \cdots \sigma_{p-1}\sigma_1 \sigma_2 \cdots \sigma_{p-2}$$ $$\sim_O z(q-1) (\sigma_1 \sigma_2 \cdots \sigma_{p-1})^2 + z^2q \sigma_1 \sigma_2 \cdots \sigma_{p-2}\sigma_1 \sigma_2 \cdots \sigma_{p-2}$$ $$\sim_O z(q-1) (\sigma_1 \sigma_2 \cdots \sigma_{p-1})^2 + z^2q (\sigma_1 \sigma_2 \cdots \sigma_{p-2})^2$$ Writing $$tr((\sigma_1 \sigma_2 \cdots \sigma_p)^2) = T(2,p)$$, we have the recurrence relation $$T(2,p) = z(q-1) T(2,p-1) + z^2qT(2,p-2)$$ which is second-order, homogeneous and has constant coefficients (no dependance on $$p$$). The characteristic equation is given by $$\lambda^2 - z(q-1) \lambda - z^2q = 0$$ Solving the quadratic equation gives $$\lambda = \dfrac{z(q-1) \pm \sqrt{(z(q-1))^2 - 4(-z^2q)}}{2} = \dfrac{z(q-1) \pm \sqrt{z^2q^2 - 2z^2q + z^2 + 4z^2q}}{2} = \dfrac{z(q-1) \pm \sqrt{z^2(q + 1)^2}}{2} = \dfrac{z(q-1) \pm z(q+1)}{2}$$ We have $$\lambda = zq$$ or $$-z$$. Hence $$T(2,p) = a(zq)^p + b(-z)^p$$ for some $$a$$ and $$b$$.

The initial values are given by $$T(2,1) = \text{tr}(\sigma_1^2) = (q-1)\text{tr}(\sigma_1) + q = (q-1)z + q$$ and $$T(2,2) = \text{tr}(\sigma_1\sigma_2\sigma_1\sigma_2) = \text{tr}(\sigma_1^2\sigma_2\sigma_1) = z\text{tr}(\sigma_1^3) = z(q-1)\text{tr}(\sigma_1^2) + zq \text{tr}(\sigma_1) = z(q-1)((q-1)z + q) + z^2q = z^2(q-1)^2 +zq(q-1) + z^2q$$ Substitution gives $$azq - bz = (q-1)z + q$$ $$a(zq)^2 + bz^2 = z^2(q-1)^2 +zq(q-1) + z^2q$$ Cramer's rule gives $$a = \dfrac{\begin{vmatrix} (q-1)z + q & -z \\ z^2(q-1)^2 +zq(q-1) + z^2q & z^2 \end{vmatrix}} {\begin{vmatrix} zq & -z \\ (zq)^2 + zq & z^2 \end{vmatrix}} =\dfrac{z\begin{vmatrix} (q-1)z + q & -1 \\ z^2(q-1)^2 +zq(q-1) + z^2q & z \end{vmatrix}} {z^2q\begin{vmatrix} 1 & -1 \\ zq + 1 & z \end{vmatrix}} =\dfrac{(q-1)z^2 + qz + z^2(q-1)^2 +zq(q-1) + z^2q}{zq(z+zq+1)} =\dfrac{zq + z^2q(q-1) +zq(q-1) + z^2q}{zq(z+zq+1)} =\dfrac{q(z + 1)}{z+zq+1}$$ and $$b = \dfrac{\begin{vmatrix} zq & (q-1)z + q \\ (zq)^2 & z^2(2q-1) + zq \end{vmatrix}} {\begin{vmatrix} zq & -z \\ (zq)^2 + zq & z^2 \end{vmatrix}} =\dfrac{zq\begin{vmatrix} 1 & (q-1)z + q \\ zq & z^2(q-1)^2 +zq(q-1) + z^2q \end{vmatrix}}{z^2q(z+zq+1)} =\dfrac{zq (z^2(q-1)^2 +zq(q-1) + z^2q - ((q-1)z^2q + zq^2))}{z^2q(z+zq+1)} =\dfrac{z(z-q)}{z(z+zq+1)} =\dfrac{z - q}{z+zq+1}$$ Hence $$\text{tr}((\sigma_1 \sigma_2 \cdots \sigma_p)^2) = \dfrac{q(z + 1)(zq)^p}{z+zq+1} + \dfrac{(z - q)(-z)^p}{z+zq+1} = \dfrac{q(z + 1)(zq)^p + (z - q)(-z)^p}{z+zq+1}$$

Now we try $$(\sigma_1 \sigma_2 \cdots \sigma_p)^3$$. By the same token, we have $$\sigma_1 \sigma_2 \cdots \sigma_p\sigma_1 \sigma_2 \cdots \sigma_p\sigma_1 \sigma_2 \cdots \sigma_p$$ $$=\sigma_1 \sigma_2 \cdots \sigma_{p-1}\sigma_1 \sigma_2 \cdots \sigma_p\sigma_{p-1}\sigma_p\sigma_1 \sigma_2 \cdots \sigma_p$$ $$\sim_M \sigma_1 \sigma_2 \cdots \sigma_{p-1}\sigma_1 \sigma_2 \cdots \sigma_{p-1}\sigma_p\sigma_{p-1}\sigma_1 \sigma_2 \cdots \sigma_p$$ $$\sim_M \sigma_1 \sigma_2 \cdots \sigma_{p-1}\sigma_1 \sigma_2 \cdots \sigma_{p-1}^2\sigma_1 \sigma_2 \cdots \sigma_{p-1}\sigma_p\sigma_{p-1}$$ $$\sim_O z\sigma_1 \sigma_2 \cdots \sigma_{p-1}\sigma_1 \sigma_2 \cdots \sigma_{p-1}^2\sigma_1 \sigma_2 \cdots \sigma_{p-1}^2$$ $$\sim_O z(q-1)(\sigma_1 \sigma_2 \cdots \sigma_{p-1})^3 + zq\sigma_1 \sigma_2 \cdots \sigma_{p-1}\sigma_1 \sigma_2 \cdots \sigma_{p-2}\sigma_1 \sigma_2 \cdots \sigma_{p-1}^2$$ $$\sim_O z(q-1)(\sigma_1 \sigma_2 \cdots \sigma_{p-1})^3 + zq(q-1)\sigma_1 \sigma_2 \cdots \sigma_{p-1}\sigma_1 \sigma_2 \cdots \sigma_{p-2}\sigma_1 \sigma_2 \cdots \sigma_{p-1} + zq^2\sigma_1 \sigma_2 \cdots \sigma_{p-1}\sigma_1 \sigma_2 \cdots \sigma_{p-2}\sigma_1 \sigma_2 \cdots \sigma_{p-2}$$ $$\sim_O z(q-1)(\sigma_1 \sigma_2 \cdots \sigma_{p-1})^3 + zq(q-1)\sigma_1 \sigma_2 \cdots \sigma_{p-1}\sigma_1 \sigma_2 \cdots \sigma_{p-2}\sigma_1 \sigma_2 \cdots \sigma_{p-1} + z^2q^2(\sigma_1 \sigma_2 \cdots \sigma_{p-2})^3$$

But $$\sigma_1 \sigma_2 \cdots \sigma_{p-1}\sigma_1 \sigma_2 \cdots \sigma_{p-2}\sigma_1 \sigma_2 \cdots \sigma_{p-1}$$ $$=\sigma_1 \sigma_2 \cdots \sigma_{p-1}\sigma_1 \sigma_2 \cdots \sigma_{p-3}\sigma_1 \sigma_2 \cdots \sigma_{p-2}\sigma_{p-3}\sigma_{p-2} \sigma_{p-1}$$ $$\sim_M \sigma_1 \sigma_2 \cdots \sigma_{p-1}\sigma_1 \sigma_2 \cdots \sigma_{p-3}\sigma_1 \sigma_2 \cdots \sigma_{p-3}\sigma_{p-2}\sigma_{p-3} \sigma_{p-1}$$ $$= \sigma_1 \sigma_2 \cdots \sigma_{p-2}\sigma_1 \sigma_2 \cdots \sigma_{p-3}\sigma_1 \sigma_2 \cdots \sigma_{p-3}\sigma_{p-1}\sigma_{p-2}\sigma_{p-1}\sigma_{p-3}$$ $$\sim_M \sigma_1 \sigma_2 \cdots \sigma_{p-2}\sigma_1 \sigma_2 \cdots \sigma_{p-3}\sigma_1 \sigma_2 \cdots \sigma_{p-3}\sigma_{p-2}\sigma_{p-1}\sigma_{p-2}\sigma_{p-3}$$ $$\sim_O z\sigma_1 \sigma_2 \cdots \sigma_{p-2}\sigma_1 \sigma_2 \cdots \sigma_{p-3}\sigma_1 \sigma_2 \cdots \sigma_{p-3}\sigma_{p-2}^2\sigma_{p-3}$$ $$\sim_O z ((q-1) \sigma_1 \sigma_2 \cdots \sigma_{p-2}\sigma_1 \sigma_2 \cdots \sigma_{p-3}\sigma_1 \sigma_2 \cdots \sigma_{p-3}\sigma_{p-2}\sigma_{p-3} + q\sigma_1 \sigma_2 \cdots \sigma_{p-2}\sigma_1 \sigma_2 \cdots \sigma_{p-3}\sigma_1 \sigma_2 \cdots \sigma_{p-3}^2)$$ $$\sim_O z ((q-1) \sigma_1 \sigma_2 \cdots \sigma_{p-2}\sigma_1 \sigma_2 \cdots \sigma_{p-3}\sigma_1 \sigma_2 \cdots \sigma_{p-2}\sigma_{p-3}\sigma_{p-2} + qz(\sigma_1 \sigma_2 \cdots \sigma_{p-3})^2\sigma_1 \sigma_2 \cdots \sigma_{p-3}^2)$$ $$\sim_O z ((q-1) \sigma_1 \sigma_2 \cdots \sigma_{p-2}\sigma_1 \sigma_2 \cdots \sigma_{p-3}\sigma_{p-2}\sigma_1 \sigma_2 \cdots \sigma_{p-3}\sigma_{p-2} + qz(\sigma_1 \sigma_2 \cdots \sigma_{p-3})^2\sigma_1 \sigma_2 \cdots \sigma_{p-3}^2)$$ $$= z ((q-1) (\sigma_1 \sigma_2 \cdots \sigma_{p-2})^3 + qz(\sigma_1 \sigma_2 \cdots \sigma_{p-3})^2\sigma_1 \sigma_2 \cdots \sigma_{p-3}^2)$$ $$\sim_O z(q-1) (\sigma_1 \sigma_2 \cdots \sigma_{p-2})^3 + q(q-1)z^2(\sigma_1 \sigma_2 \cdots \sigma_{p-3})^3 +q^2z^2(\sigma_1 \sigma_2 \cdots \sigma_{p-3})^2\sigma_1 \sigma_2 \cdots \sigma_{p-4}$$ $$\sim_M z(q-1) (\sigma_1 \sigma_2 \cdots \sigma_{p-2})^3 + q(q-1)z^2(\sigma_1 \sigma_2 \cdots \sigma_{p-3})^3 +q^2z^2 \sigma_1 \sigma_2 \cdots \sigma_{p-3}\sigma_1 \sigma_2 \cdots \sigma_{p-4} \sigma_1 \sigma_2 \cdots \sigma_{p-3}$$ Writing $$\text{tr}((\sigma_1 \sigma_2 \cdots \sigma_p)^3) = T(3, p)$$ and $$\text{tr}(\sigma_1 \sigma_2 \cdots \sigma_p\sigma_1 \sigma_2 \cdots \sigma_{p-1}\sigma_1 \sigma_2 \cdots \sigma_p) = U(3, p)$$, we have $$U(3, p-1) = z(q-1)T(3,p-2)+q(q-1)z^2T(3,p-3)+q^2z^2U(3,p-3)$$ and $$T(3,p) = z(q-1)T(3,p-1) + zq(q-1)U(3,p-1) + z^2q^2T(3,p-2)$$ Rearrange the last relation to give $$U(3,p-1) = \dfrac{T(3,p)-z(q-1)T(3,p-1)-z^2q^2T(3,p-2)}{zq(q-1)} =\dfrac{T(3,p)}{zq(q-1)}-\dfrac{T(3,p-1)}{q}-\dfrac{zqT(3,p-2)}{q-1}$$ Substitute into the first relation to give $$\dfrac{T(3,p)}{zq(q-1)}-\dfrac{T(3,p-1)}{q}-\dfrac{zqT(3,p-2)}{q-1}= z(q-1)T(3,p-2)+q(q-1)z^2T(3,p-3)+q^2z^2\left(\dfrac{T(3,p-2)}{zq(q-1)}-\dfrac{T(3,p-3)}{q}-\dfrac{zqT(3,p-4)}{q-1}\right)$$

This recurrence relation needs to be solved subject to $$U(3, 2) = \text{tr}(\sigma_2\sigma_1\sigma_2) = \text{tr}(\sigma_1\sigma_2\sigma_1) = z\text{tr}(\sigma_1^2) = z(q-1)\text{tr}(\sigma_1)+zq = z^2(q-1)+zq$$ $$T(3, 1) = \text{tr}(\sigma_1^3) = (q-1)\text{tr}(\sigma_1^2)+q\text{tr}(\sigma_1) = (q-1)^2\text{tr}(\sigma_1)+ (q-1)q + qz = (q-1)^2z + (q-1)q + qz$$ $$T(3, 2) = \text{tr}((\sigma_1\sigma_2)^3) = \text{tr}(\sigma_1\sigma_2\sigma_1\sigma_2\sigma_1\sigma_2) = \text{tr}(\sigma_2\sigma_1\sigma_2^2\sigma_1\sigma_2) = (q-1)\text{tr}(\sigma_2\sigma_1\sigma_2\sigma_1\sigma_2) + q\text{tr}(\sigma_2\sigma_1\sigma_1^2\sigma_2) = (q-1)\text{tr}(\sigma_2\sigma_1\sigma_1\sigma_2\sigma_1) + q(q-1)\text{tr}(\sigma_2\sigma_1\sigma_1\sigma_2) + q^2\text{tr}(\sigma_2\sigma_1\sigma_2) = (q-1)^2\text{tr}(\sigma_2\sigma_1\sigma_1\sigma_2\sigma_1) + (q-1)q\text{tr}(\sigma_2\sigma_2\sigma_1) + q(q-1)\text{tr}(\sigma_2^2\sigma_1\sigma_1) + q^2\text{tr}(\sigma_1\sigma_2\sigma_1) = (q-1)^2\text{tr}(\sigma_1\sigma_1\sigma_2\sigma_1\sigma_2) + (q-1)q\text{tr}(\sigma_2\sigma_2\sigma_1) + q(q-1)((q-1)\text{tr}(\sigma_2\sigma_1^2)+q\text{tr}(\sigma_1\sigma_1))+ q^2z\text{tr}(\sigma_1^2) = (q-1)^2\text{tr}(\sigma_1\sigma_1\sigma_1\sigma_2\sigma_1) + (q-1)q\text{tr}(\sigma_1\sigma_2\sigma_1) + q(q-1)((q-1)z\text{tr}(\sigma_1^2)+qtr(\sigma_1^2))+ q^2z\text{tr}(\sigma_1^2) = (q-1)^2z\text{tr}(\sigma_1^4) + (q-1)qz\text{tr}(\sigma_1^2) + q(q-1)((q-1)z\text{tr}(\sigma_1^2)+qtr(\sigma_1^2))+ q^2z\text{tr}(\sigma_1^2) = (q-1)^2z\text{tr}(\sigma_1^4) + ((q-1)qz + q(q-1)((q-1)z+q)+ q^2z)\text{tr}(\sigma_1^2)$$ But $$\sigma_1^2 \sim_O (q-1)\sigma_1 + q \sim_O (q-1)z + q = qz-z+q$$ and $$\sigma_1^4 \sim_O (q-1)\sigma_1^3 + q\sigma_1^2 \sim_O (q-1)((q-1)\sigma_1^2 + q\sigma_1) + q((q-1)z + q) \sim_O (q-1)((q-1)((q-1)z + q) + qz) + q((q-1)z + q)$$ $$=q^3 z - q^2 z + q z - z + q^3 - q^2 + q$$ Hence $$T(3, 2) = (q-1)^2z(q-1)((q-1)((q-1)z + q) + qz) + q((q-1)z + q) + ((q-1)qz + q(q-1)((q-1)z+q)+ q^2z)((q-1)z + q)$$ $$=q^5 z^2 + q^5 z - 3 q^4 z^2 - 2 q^4 z + q^4 + 6 q^3 z^2 + 4 q^3 z - q^3 - 7 q^2 z^2 - 2 q^2 z + q^2 + 4 q z^2 - z^2$$