7
$\begingroup$

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.)

$\endgroup$
3
  • 1
    $\begingroup$ 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, 2013 at 22:34
  • 1
    $\begingroup$ what's a reference (or argument) for the above statement that cabling preserves $k$? $\endgroup$ Feb 27, 2013 at 23:53
  • 2
    $\begingroup$ @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. $\endgroup$ Feb 28, 2013 at 5:56

2 Answers 2

9
$\begingroup$

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

$\endgroup$
6
  • $\begingroup$ 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, 2013 at 15:40
  • $\begingroup$ Just Google "Vassiliev invariants for torus knots." You should get their paper. $\endgroup$
    – Jim Conant
    Feb 28, 2013 at 15:47
  • $\begingroup$ I accept this answer as the best answer, but more homework needs to be done. $\endgroup$
    – Roddy Bad
    Mar 3, 2013 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, 2013 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, 2013 at 15:58
6
$\begingroup$

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!

$\endgroup$
2
  • $\begingroup$ Thanks a lot! the formula for the Alexander polynomial of the (p,q)-cable is useful. $\endgroup$
    – Roddy Bad
    Feb 28, 2013 at 15:37
  • 1
    $\begingroup$ 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. $\endgroup$ Nov 1, 2014 at 4:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.