How do you find the Alexander polynomial of the closure of the following braid,
$(\sigma_1^{-2}\sigma_2^{-1}\sigma_3^{-1}\sigma_4^{-1}\sigma_5^{-1}...\sigma_{A-1}^{-1})^B$ where $A$ and $B$ are positive integers?
I have found the general form of the Seifert matrix of link, and then tried to use the formula
$$\Delta(t) = \det (V-tV^T) $$ where $V$ is the Seifert matrix and $\Delta(t)$ is the Alexander polynomial.

I therefore tried using MATLAB to calculate the Alexander polynomials of the knots for $A$ and $B$ running from $1$ to $10$. From the results I guessed the general form of the Alexander polynomial, but I do not end up in a rigorous proof.

I am not quite familiar with the computation of Alexander polynomials using other techniques such as the knot group and Jacobian, so I am not sure if those routes would give a good way in attacking the problem.

PS. I did calculate the Jones polynomial quite quickly, but I don't see any help from that.


1 Answer 1


The braid you mention is an element of $B_A$ I assume? It doesn't really matter if you add strings after this, as taking the braid closure will just add a new unlink for every new string.

In any case, it seems like you would be interested in the Burau representation of the braid group. The Alexander polynomial of a braid closure $\hat{\beta}$ is given by the determinant of the matrix $I-\beta_*$ where $\beta_*$ is the reduced Burau representation of the braid $\beta$. There is a large amount of literature on the Burau representation of a braid going back to the 60s and 70s. I would strongly suggest picking up Birman's book1 for a good introduction to theory behind the representations of the braid groups and how to calculate such maps.

1J. S. Birman Braids, Links and Mapping Class Groups, Annals of Mathematical Studies 82, Princeton University Press, 1975.

  • $\begingroup$ I read your reference book, and tried to get to the answer using the theory of mapping class groups, but I still get to the problem: I have to find the determinant of a matrix, but it is so complicated and it is hard to resolve it. $\endgroup$
    – wilsonw
    Jun 15, 2013 at 6:22

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