Vassilliev invariants of knots and their cables

Question

The following is perhaps a standard question, but i could not find a plain enough answer by simply searching online.

Q: Given a knot $K$ and its $(p,q)$-cable $K_{p,q}$ what is a relation between the Vassiliev invariants of $K$ and $K_{p,q}$?

In particular, I would be happy with a formula for the 2nd coefficient of the Conway polynomial. (One may attempt to solve this via the A-polynomial since it has a simple relationship for satellites, however relating the resulting A-polynomial back to the Conway p. seems nontrivial, ... at least for me. Also it is seems likely that somebody could have already worked this out.)

Answer 1

As I mentioned in a comment, for the degree $2$ invariant $v_2$ which is the coefficient of $z^2$ in the Conway Polynomial, we have that $v_2(K_{p,q})=av_2(K)+b$. If $K$ is the unknot, this implies that $b=v_2(T_{p,q})$, where $T_{p,q}$ is the $(p,q)$-torus knot (assuming here $p,q$ are relatively prime.) Alvarez and Labastida wrote down formulas for Vassiliev invariants of torus knots, and in particular they showed $$v_2(T_{p,q})=\frac{1}{24}(p^2-1)(q^2-1).$$ So that gives you your constant term. Using Ryan's formula for the Alexander polynomial, one should be able to show that $a=p$. This is because when you make the substitution $t\mapsto t^p$, in the conversion to the the Conway polynomial we have $z^2=t+t^{-1}-2$, and so $z^2\mapsto t^p+t^{-p}-2$. It's a lemma that $t^p+t^{-p}=2+pz^2+\cdots$, so the coefficient of $z^2$ will get multiplied by $p$. So the answer will be $$v_2(K_{p,q})=pv_2(K)+\frac{1}{24}(p^2-1)(q^2-1).$$

Answer 2

Let $K$ be a knot, and $\Delta_K$ be the Alexander polynomial of $K$, $\Delta_K \in \mathbb Z[t^\pm]$.

Let's let $K(p,q)$ be the $(p,q)$-cable of $K$. Then

$$ \Delta_{K(p,q)} = \Delta_K(t^{p}) \cdot \Delta_{T_{p,q}}$$

where $\Delta_{T_{p,q}}$ is the Alexander polynomial of the $(p,q)$-torus knot. I believe that's

$$ \Delta_{T_{p,q}} = \frac{ (t^{pq}-1)(t-1) }{(t^p-1)(t^q-1)} $$

The above formulas are fairly classical. It appears at least as early as in Eisenbud and Neumann's book, but it's likely known much earlier.

The type-2 invariant of a knot is given in terms of the Alexander polynomial. In the Conway form it's the coefficient of $z^2$, but in the above Alexander normalization, you'll get it as some kind of linear combination of the first few coefficients. So you just apply whatever that formula is. At present I forget it!