Let $K_{p,q}$ be a $(p,q)$-cable of the non-trivial knot $K$ in $S^3$.
Is there a closed formula for the Jones polynomial for $K_{p,q}$ as in the case of Alexander polynomial or Seifert matrices?
$\begingroup$
$\endgroup$
2
-
2$\begingroup$ This should be a linear combination of the colored Jones polynomials. en.wikipedia.org/wiki/Jones_polynomial#Colored_Jones_polynomial $\endgroup$– Ian AgolCommented Aug 24, 2020 at 21:15
-
$\begingroup$ As Ian points out, there must be an expression in colored Jones polynomial — this follows eg from the known description of the skein of the annulus. In fact the formula has been written “explicitly”: arxiv.org/abs/0807.2679 $\endgroup$– Vivek ShendeCommented Aug 27, 2020 at 7:45
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
As Ian Agol mentioned, if there were a closed formula for the Jones polynomial $V(K_{p,q})$ in terms of $V(K)$, this would give a closed formula for the colored Jones polynomials $V_n(K)$ in terms of the original Jones polynomial $V(K) = V_2(K)$.
However, this makes me think that there is no such simple formula. If there were, then we'd be able to easily give closed formulas for $V_n(K)$ for arbitrary $n$, but these are usually quite hard to generate. It's one reason why the Volume Conjecture is only known to hold in special cases: the first step of a proof for $K$ [1] is usually to give a closed formula for $V_n(K)$.
[1] Recently there have been proofs for other knots related to "fundamental shadow links" in $\#^k S^2 \times S^1$, which proceed differently. This isn't terribly relevant to your question, but I'm mentioning it for completeness.
answered Aug 25, 2020 at 13:17