MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\sigma_1$ and $\sigma_2$ be two braids with $n$-strings.

Are there any formulas relating $J_{\widehat{\sigma_1\sigma_2}}(q)$, $J_{\hat{\sigma_1}}(q)$, and $J_{\hat{\sigma_2}}(q)$?

Here, $J_L(q)$ is the jones polynomial of a link $L$ and $\hat{\sigma}$ stands for the closure of a braid $\sigma$.

share|cite|improve this question
Maybe there are, in some special cases. But, for example, if you take $\sigma_1 = \sigma_2$ to be 1/n-th of a twist, then $\widehat{\sigma}_1 = \widehat{\sigma}_2 = T_{1,n}$ are both unknotted, but the closure of their concatenation is $T_{2,n}$, which has nontrivial Jones polynomial. – Marco Golla May 19 '14 at 9:49
Even better, if you take $\sigma_1 = \sigma_2^{-1}$, then you need to find a relationship between the trivial polynomial, and $V(q), V(q^{-1})$ for some Jones polynomial $V$. – Marco Golla May 19 '14 at 9:55
It may be worth mentioning that there is a map on Khovanov homology relating the closures of ${\sigma}_1, {\sigma}_2$, and ${\sigma}_1{\sigma}_2$. The map corresponds to the cobordism which consists of $n$ saddles joining the respective strands of ${\sigma}_1, {\sigma}_2$. This doesn't give a formula relating the Jones polynomial (=graded Euler characteristic) of the three links, but this does provide some explanation for the various special cases mentioned in a few other comments and answers. – Slava Krushkal May 19 '14 at 20:13
up vote 15 down vote accepted

Here is one reason not to expect such a relationship (although I'm not sure if it can be completed to a proof). The Jones polynomial $J_\sigma$ (roughly) comes from taking the trace of a linear map $A_\sigma$ associated to the braid $\sigma$, so the question (roughly) asks about relations between $Tr(A_\sigma)$, $Tr(A_\tau)$, and $Tr(A_\sigma A_\tau)$.

Let $a,b$ generate a free group $\pi$, and let $Chr$ be the $SL_2$ character scheme of $\pi$ (i.e. the space of pairs of $SL_2$ matrices considered up to simultaneous conjugation). Define $Tr(g):Chr \to \mathbb C$ to be the function $\rho \mapsto Tr(\rho(g))$. Then the functions $Tr(a)$, $Tr(b)$, and $Tr(ab)$ are algebraically independent. So there shouldn't be any universal formula relating these three traces, whether $a,b \in SL_2$ or are bigger matrices.

share|cite|improve this answer

In the very special case where $\sigma_2$ is a full twist of all $n$ strands, there is some information. This is essentially because full twists are in the center of the braid group. See this paper by Champanerkar and Kofman for details:

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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