Following arXiv:1211.6075v1 I want to calculate colored Jones Polynomial for trefoil knot. I have the formulas:
$ J_{\oplus R_i} = \sum_i J_{R_i} (K,q)$
$J_{R} (K^n, q) = J_{R^{\otimes n}} (K,q)$
I can decompose my representation $2 \otimes 2=1 \oplus 3$ and use $J_{1 \oplus 3} = J_1 \oplus J_3$ and $J_{2 \otimes 2} = J(K^2,q)-1$. And here's my problem. I have no idea what $J(K^2,q)$ is and how to calculate it. I don't understand the statement:" $K_n$ is the n-cabling of the knot K, obtained by taking the path of K and tracing it with a cable of n strands."
How can I calculate the polynomial of $K^2$ to get $J_3$ ?
They are referring to the operation of cabling a knot. Take a diagram with writhe $=0$, and replace the knot with $n$ parallel strands plus a twist to make sure they join up into a knot. Below is a $2$-cable of the figure $8$. To do a cable of the trefoil you need to add some kinks in the usual diagram first so that it has writhe $0$.