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.)

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    $\begingroup$ If $v_k$ is a type $k$ invariant, the invariant $K\mapsto v_k(K_{p,q})$ is also type $k$. In particular, since there are only two invariants for knots of type $\leq 2$, the second coefficient of the Conway polynomial and the constant function, it follows that v_2(K_{p,q})=a v_2+ b for some constants a and b that depend on $p,q$. So you just have to calculate a couple of examples to work out what $a$ and $b$ are. $\endgroup$ – Jim Conant Feb 27 '13 at 22:34
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    $\begingroup$ what's a reference (or argument) for the above statement that cabling preserves $k$? $\endgroup$ – Vivek Shende Feb 27 '13 at 23:53
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    $\begingroup$ @Vivek: it's in section 9.2.2 of "Introduction to Vassiliev Knot Invariants" by Chmutov, Duzhin, and Mostovoy, available at pdmi.ras.ru/~duzhin/papers/cdbook/cdbook.pdf. $\endgroup$ – Steven Sivek Feb 28 '13 at 5:56

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 $$v_2(K_{p,q})=pv_2(K)+\frac{1}{24}(p^2-1)(q^2-1).$$

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  • $\begingroup$ Thanks a lot ! This is a very nice approach for $v_2$, I will try to do my homework and see if this works. What is a reference to Alvarez and Labastida? $\endgroup$ – Roddy Bad Feb 28 '13 at 15:40
  • $\begingroup$ Just Google "Vassiliev invariants for torus knots." You should get their paper. $\endgroup$ – Jim Conant Feb 28 '13 at 15:47
  • $\begingroup$ I accept this answer as the best answer, but more homework needs to be done. $\endgroup$ – Roddy Bad Mar 3 '13 at 21:39
  • $\begingroup$ Good luck. I think there are enough ideas in this and Ryan's answer for you to analyze the general case. $\endgroup$ – Jim Conant Mar 3 '13 at 23:02
  • $\begingroup$ Thanks! For instance the formula should be symmetric w.r.t. p and q, I think. $\endgroup$ – Roddy Bad Mar 4 '13 at 15:58

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!

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  • $\begingroup$ Thanks a lot! the formula for the Alexander polynomial of the (p,q)-cable is useful. $\endgroup$ – Roddy Bad Feb 28 '13 at 15:37
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    $\begingroup$ Yes, much earlier! The formula for satellites (with cabling a very special case) goes back to Seifert, "On the homology invariants of knots", Quart. J. Math. Oxford 2(1950), 23-32. $\endgroup$ – Danny Ruberman Nov 1 '14 at 4:02

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