Prove that two variable Jones polynomial can be expressed by Finite type invariant

Question

I have this question that says: Prove that two variable Jones polynomial can be expressed by Finite type invariant. Can somebody explain how this is done? many thanks in advance!

Answer 1

Birman and Lin showed in 1993 that the Jones, HOMFLY, and Kauffman polynomials are determined by the set of finite type invariants.

Birman, Joan S.; Lin, Xiao-Song, Knot polynomials and Vassiliev’s invariants, Invent. Math. 111, No. 2, 225-270 (1993). ZBL0812.57011. (PDF)

For the Jones polynomial $J_L(t)$ of a link $L$, if you substitute $J_L(e^x)$ and do a power series expansion $J_L(e^x)=\sum_{i=0}^\infty U_i(L)x^i$, it turns out $U_i(L)$ is a finite-type invariant of links. Then since the collection of all these $U_i(L)$ determines $J_L(e^x)$ and thus $J_L(t)$, the Jones polynomial (while not being a finite-type invariant itself) is determined by finite-type invariants. In case the substitution seems unmotivated: from the perspective of quantum groups, the Jones polynomial is already a polynomial in $e^x$.

Section 4 of the paper describes the HOMFLY polynomial. One point of view is that the HOMFLY polynomial is a sequence of polynomials $P_{L,n}(t)$, where $n$ is which quantum $\mathfrak{sl}(n)$ is used for the adjoint representation. As it turns out, this is a (Laurent) polynomial in both $n$ and $t$. They substitute $t=e^x$ again and show that each coefficient of the power series expansions for each $n$ is a finite-type invariant.

Theorem 4.8 is for the Kauffman polynomial, and they do a similar substitution.

That said, I haven't read the paper and don't know how the argument actually works, but hopefully the reference helps.