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 of this form 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 $b=(sigma_1sigma_2cdotssigma_{p-1})^m$.

For coprime $m$ and $p$, this is given by
$$text{tr}(pi_Y((sigma_1sigma_2cdotssigma_{p-1})^m) = (-1)^gamma q^{beta m}$$

This is related to the fact that $(sigma_1sigma_2cdotssigma_{p-1})^q$ 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
$$text{tr}((sigma_1sigma_2cdotssigma_{p-1})^m)
= sum_{beta+gamma+1=p; beta,gamma geq 0} (-1)^gamma q^{beta m}
left(dfrac{1-q}{1-lambda q} right)^p dfrac{q^{frac{gamma(gamma+1)}{2}}}{[beta]![gamma]!(1-q^p)}prod_{i=-gamma}^beta (q^i-lambda q) $$

Hence the 2-variable Jones polynomial is given by
$$X_{text{Cl}((sigma_1sigma_2cdotssigma_{p-1})^m)}(q, lambda)
= left(-dfrac{1-lambda q}{sqrt lambda (1-q)}right)^{p-1} sqrt lambda^{(p-1)m} times
sum_{beta+gamma+1=p; beta,gamma geq 0} (-1)^gamma q^{beta m}
left(dfrac{1-q}{1-lambda q} right)^p dfrac{q^{frac{gamma(gamma+1)}{2}}}{[beta]![gamma]!(1-q^p)}prod_{i=-gamma}^beta (q^i-lambda q)$$
$$= left(dfrac{1-q}{1-lambda q} right) lambda^{frac{(p-1)(m-1)}{2}}
sum_{beta+gamma+1=p; beta,gamma geq 0} (-1)^{p-1-gamma} q^{beta m}
dfrac{q^{frac{gamma(gamma+1)}{2}}}{[beta]![gamma]!(1-q^p)}prod_{i=-gamma}^beta (q^i-lambda q) $$
$$= left(dfrac{1-q}{1-lambda q} right) lambda^{frac{(p-1)(m-1)}{2}}
sum_{beta+gamma+1=p; beta,gamma geq 0} (-1)^beta
dfrac{q^{beta m+frac{gamma(gamma+1)}{2}}}{[beta]![gamma]!(1-q^p)}prod_{i=-gamma}^beta (q^i-lambda q) $$

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

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

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
$$(sigma_1 sigma_2 cdots sigma_p)^2 $$
$$= sigma_1 sigma_2 cdots sigma_psigma_1 sigma_2 cdots sigma_p $$
$$= sigma_1 sigma_2 cdots sigma_{p-1}sigma_1 sigma_2 cdots sigma_p sigma_{p-1} sigma_p $$
$$sim_M sigma_1 sigma_2 cdots sigma_{p-1}sigma_1 sigma_2 cdots sigma_{p-1} sigma_p sigma_{p-1} $$
$$sim_O z sigma_1 sigma_2 cdots sigma_{p-1}sigma_1 sigma_2 cdots sigma_{p-1}^2 $$
$$sim_O z(q-1) sigma_1 sigma_2 cdots sigma_{p-1}sigma_1 sigma_2 cdots sigma_{p-1} + zq sigma_1 sigma_2 cdots sigma_{p-1}sigma_1 sigma_2 cdots sigma_{p-2}$$
$$sim_O z(q-1) (sigma_1 sigma_2 cdots sigma_{p-1})^2 + z^2q sigma_1 sigma_2 cdots sigma_{p-2}sigma_1 sigma_2 cdots sigma_{p-2}$$
$$sim_O z(q-1) (sigma_1 sigma_2 cdots sigma_{p-1})^2 + z^2q (sigma_1 sigma_2 cdots sigma_{p-2})^2$$
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
$$T(2,2) = text{tr}(sigma_1sigma_2sigma_1sigma_2) = text{tr}(sigma_1^2sigma_2sigma_1) = ztext{tr}(sigma_1^3) = z(q-1)text{tr}(sigma_1^2) + zq text{tr}(sigma_1) = z(q-1)((q-1)z + q) + z^2q = z^2(q-1)^2 +zq(q-1) + z^2q$$
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
$$a =
dfrac{begin{vmatrix}
(q-1)z + q & -z \
z^2(q-1)^2 +zq(q-1) + z^2q & z^2
end{vmatrix}}
{begin{vmatrix}
zq & -z \
(zq)^2 + zq & z^2
end{vmatrix}}
=dfrac{zbegin{vmatrix}
(q-1)z + q & -1 \
z^2(q-1)^2 +zq(q-1) + z^2q & z
end{vmatrix}}
{z^2qbegin{vmatrix}
1 & -1 \
zq + 1 & z
end{vmatrix}}
=dfrac{(q-1)z^2 + qz + z^2(q-1)^2 +zq(q-1) + z^2q}{zq(z+zq+1)}
=dfrac{zq + z^2q(q-1) +zq(q-1) + z^2q}{zq(z+zq+1)}
=dfrac{q(z + 1)}{z+zq+1}
$$
and
$$b =
dfrac{begin{vmatrix}
zq & (q-1)z + q \
(zq)^2 & z^2(2q-1) + zq
end{vmatrix}}
{begin{vmatrix}
zq & -z \
(zq)^2 + zq & z^2
end{vmatrix}}
=dfrac{zqbegin{vmatrix}
1 & (q-1)z + q \
zq & z^2(q-1)^2 +zq(q-1) + z^2q
end{vmatrix}}{z^2q(z+zq+1)}
=dfrac{zq (z^2(q-1)^2 +zq(q-1) + z^2q - ((q-1)z^2q + zq^2))}{z^2q(z+zq+1)}
=dfrac{z(z-q)}{z(z+zq+1)}
=dfrac{z - q}{z+zq+1}
$$
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
$$sigma_1 sigma_2 cdots sigma_psigma_1 sigma_2 cdots sigma_psigma_1 sigma_2 cdots sigma_p$$
$$=sigma_1 sigma_2 cdots sigma_{p-1}sigma_1 sigma_2 cdots sigma_psigma_{p-1}sigma_psigma_1 sigma_2 cdots sigma_p$$
$$sim_M sigma_1 sigma_2 cdots sigma_{p-1}sigma_1 sigma_2 cdots sigma_{p-1}sigma_psigma_{p-1}sigma_1 sigma_2 cdots sigma_p$$
$$sim_M sigma_1 sigma_2 cdots sigma_{p-1}sigma_1 sigma_2 cdots sigma_{p-1}^2sigma_1 sigma_2 cdots sigma_{p-1}sigma_psigma_{p-1}$$
$$sim_O zsigma_1 sigma_2 cdots sigma_{p-1}sigma_1 sigma_2 cdots sigma_{p-1}^2sigma_1 sigma_2 cdots sigma_{p-1}^2$$
$$sim_O z(q-1)(sigma_1 sigma_2 cdots sigma_{p-1})^3 + zqsigma_1 sigma_2 cdots sigma_{p-1}sigma_1 sigma_2 cdots sigma_{p-2}sigma_1 sigma_2 cdots sigma_{p-1}^2$$
$$sim_O z(q-1)(sigma_1 sigma_2 cdots sigma_{p-1})^3 + zq(q-1)sigma_1 sigma_2 cdots sigma_{p-1}sigma_1 sigma_2 cdots sigma_{p-2}sigma_1 sigma_2 cdots sigma_{p-1} + zq^2sigma_1 sigma_2 cdots sigma_{p-1}sigma_1 sigma_2 cdots sigma_{p-2}sigma_1 sigma_2 cdots sigma_{p-2}$$
$$sim_O z(q-1)(sigma_1 sigma_2 cdots sigma_{p-1})^3 + zq(q-1)sigma_1 sigma_2 cdots sigma_{p-1}sigma_1 sigma_2 cdots sigma_{p-2}sigma_1 sigma_2 cdots sigma_{p-1} + z^2q^2(sigma_1 sigma_2 cdots sigma_{p-2})^3$$

But
$$sigma_1 sigma_2 cdots sigma_{p-1}sigma_1 sigma_2 cdots sigma_{p-2}sigma_1 sigma_2 cdots sigma_{p-1} $$
$$=sigma_1 sigma_2 cdots sigma_{p-1}sigma_1 sigma_2 cdots sigma_{p-3}sigma_1 sigma_2 cdots sigma_{p-2}sigma_{p-3}sigma_{p-2} sigma_{p-1}$$
$$sim_M sigma_1 sigma_2 cdots sigma_{p-1}sigma_1 sigma_2 cdots sigma_{p-3}sigma_1 sigma_2 cdots sigma_{p-3}sigma_{p-2}sigma_{p-3} sigma_{p-1}$$
$$= sigma_1 sigma_2 cdots sigma_{p-2}sigma_1 sigma_2 cdots sigma_{p-3}sigma_1 sigma_2 cdots sigma_{p-3}sigma_{p-1}sigma_{p-2}sigma_{p-1}sigma_{p-3}$$
$$sim_M sigma_1 sigma_2 cdots sigma_{p-2}sigma_1 sigma_2 cdots sigma_{p-3}sigma_1 sigma_2 cdots sigma_{p-3}sigma_{p-2}sigma_{p-1}sigma_{p-2}sigma_{p-3}$$
$$sim_O zsigma_1 sigma_2 cdots sigma_{p-2}sigma_1 sigma_2 cdots sigma_{p-3}sigma_1 sigma_2 cdots sigma_{p-3}sigma_{p-2}^2sigma_{p-3}$$
$$sim_O z ((q-1) sigma_1 sigma_2 cdots sigma_{p-2}sigma_1 sigma_2 cdots sigma_{p-3}sigma_1 sigma_2 cdots sigma_{p-3}sigma_{p-2}sigma_{p-3} + qsigma_1 sigma_2 cdots sigma_{p-2}sigma_1 sigma_2 cdots sigma_{p-3}sigma_1 sigma_2 cdots sigma_{p-3}^2)$$
$$sim_O z ((q-1) sigma_1 sigma_2 cdots sigma_{p-2}sigma_1 sigma_2 cdots sigma_{p-3}sigma_1 sigma_2 cdots sigma_{p-2}sigma_{p-3}sigma_{p-2} + qz(sigma_1 sigma_2 cdots sigma_{p-3})^2sigma_1 sigma_2 cdots sigma_{p-3}^2)$$
$$sim_O z ((q-1) sigma_1 sigma_2 cdots sigma_{p-2}sigma_1 sigma_2 cdots sigma_{p-3}sigma_{p-2}sigma_1 sigma_2 cdots sigma_{p-3}sigma_{p-2} + qz(sigma_1 sigma_2 cdots sigma_{p-3})^2sigma_1 sigma_2 cdots sigma_{p-3}^2)$$
$$= z ((q-1) (sigma_1 sigma_2 cdots sigma_{p-2})^3 + qz(sigma_1 sigma_2 cdots sigma_{p-3})^2sigma_1 sigma_2 cdots sigma_{p-3}^2)$$
$$sim_O z(q-1) (sigma_1 sigma_2 cdots sigma_{p-2})^3
+ q(q-1)z^2(sigma_1 sigma_2 cdots sigma_{p-3})^3
+q^2z^2(sigma_1 sigma_2 cdots sigma_{p-3})^2sigma_1 sigma_2 cdots sigma_{p-4}$$
$$sim_M z(q-1) (sigma_1 sigma_2 cdots sigma_{p-2})^3
+ q(q-1)z^2(sigma_1 sigma_2 cdots sigma_{p-3})^3
+q^2z^2 sigma_1 sigma_2 cdots sigma_{p-3}sigma_1 sigma_2 cdots sigma_{p-4} sigma_1 sigma_2 cdots sigma_{p-3}$$
Writing $text{tr}((sigma_1 sigma_2 cdots sigma_p)^3) = T(3, p)$ and
$text{tr}(sigma_1 sigma_2 cdots sigma_psigma_1 sigma_2 cdots sigma_{p-1}sigma_1 sigma_2 cdots sigma_p) = U(3, p)$, 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
$$dfrac{T(3,p)}{zq(q-1)}-dfrac{T(3,p-1)}{q}-dfrac{zqT(3,p-2)}{q-1}= z(q-1)T(3,p-2)+q(q-1)z^2T(3,p-3)+q^2z^2left(dfrac{T(3,p-2)}{zq(q-1)}-dfrac{T(3,p-3)}{q}-dfrac{zqT(3,p-4)}{q-1}right)$$

This recurrence relation needs to be solved subject to
$$U(3, 2) = text{tr}(sigma_2sigma_1sigma_2) = text{tr}(sigma_1sigma_2sigma_1) = ztext{tr}(sigma_1^2) = z(q-1)text{tr}(sigma_1)+zq = z^2(q-1)+zq$$
$$T(3, 1) = text{tr}(sigma_1^3) = (q-1)text{tr}(sigma_1^2)+qtext{tr}(sigma_1) = (q-1)^2text{tr}(sigma_1)+ (q-1)q + qz
= (q-1)^2z + (q-1)q + qz$$
$$T(3, 2) = text{tr}((sigma_1sigma_2)^3) = text{tr}(sigma_1sigma_2sigma_1sigma_2sigma_1sigma_2) = text{tr}(sigma_2sigma_1sigma_2^2sigma_1sigma_2)
= (q-1)text{tr}(sigma_2sigma_1sigma_2sigma_1sigma_2) + qtext{tr}(sigma_2sigma_1sigma_1^2sigma_2)
= (q-1)text{tr}(sigma_2sigma_1sigma_1sigma_2sigma_1) + q(q-1)text{tr}(sigma_2sigma_1sigma_1sigma_2) + q^2text{tr}(sigma_2sigma_1sigma_2)
= (q-1)^2text{tr}(sigma_2sigma_1sigma_1sigma_2sigma_1) + (q-1)qtext{tr}(sigma_2sigma_2sigma_1)
+ q(q-1)text{tr}(sigma_2^2sigma_1sigma_1) + q^2text{tr}(sigma_1sigma_2sigma_1)
= (q-1)^2text{tr}(sigma_1sigma_1sigma_2sigma_1sigma_2) + (q-1)qtext{tr}(sigma_2sigma_2sigma_1)
+ q(q-1)((q-1)text{tr}(sigma_2sigma_1^2)+qtext{tr}(sigma_1sigma_1))+ q^2ztext{tr}(sigma_1^2)
= (q-1)^2text{tr}(sigma_1sigma_1sigma_1sigma_2sigma_1) + (q-1)qtext{tr}(sigma_1sigma_2sigma_1)
+ q(q-1)((q-1)ztext{tr}(sigma_1^2)+qtr(sigma_1^2))+ q^2ztext{tr}(sigma_1^2)
= (q-1)^2ztext{tr}(sigma_1^4) + (q-1)qztext{tr}(sigma_1^2)
+ q(q-1)((q-1)ztext{tr}(sigma_1^2)+qtr(sigma_1^2))+ q^2ztext{tr}(sigma_1^2)
= (q-1)^2ztext{tr}(sigma_1^4) + ((q-1)qz + q(q-1)((q-1)z+q)+ q^2z)text{tr}(sigma_1^2)$$
But
$$sigma_1^2 sim_O (q-1)sigma_1 + q sim_O (q-1)z + q = qz-z+q$$
and
$$sigma_1^4 sim_O (q-1)sigma_1^3 + qsigma_1^2
sim_O (q-1)((q-1)sigma_1^2 + qsigma_1) + q((q-1)z + q)
sim_O (q-1)((q-1)((q-1)z + q) + qz) + q((q-1)z + q)$$
$$=q^3 z - q^2 z + q z - z + q^3 - q^2 + q$$
Hence
$$T(3, 2) = (q-1)^2z(q-1)((q-1)((q-1)z + q) + qz) + q((q-1)z + q) + ((q-1)qz + q(q-1)((q-1)z+q)+ q^2z)((q-1)z + q)$$
$$=q^5 z^2 + q^5 z - 3 q^4 z^2 - 2 q^4 z + q^4 + 6 q^3 z^2 + 4 q^3 z - q^3 - 7 q^2 z^2 - 2 q^2 z + q^2 + 4 q z^2 - z^2$$