Jones Polynomial of the trace closure of the fundamental braid - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T02:08:46Z http://mathoverflow.net/feeds/question/71955 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71955/jones-polynomial-of-the-trace-closure-of-the-fundamental-braid Jones Polynomial of the trace closure of the fundamental braid Gorjan Alagic 2011-08-03T02:10:47Z 2011-08-30T16:36:48Z <p>The fundamental braid $\Delta_n \in B_n$ is simply a twist by $\pi$ applied to the entire row of $n$ strands. In terms of Artin generators, it is given by $$\Delta_n = (\sigma_1 \sigma_2 \cdots \sigma_{n-1})(\sigma_1 \sigma_2 \cdots \sigma_{n-2})\cdots (\sigma_1 \sigma_2) \sigma_1~.$$ The square of $\Delta_n$ (i.e., the full $2\pi$ twist) generates the center of $B_n$. </p> <p>I have a rather simple (and quite possibly trivial) question about these braids. What is the Jones polynomial of the trace closure of $\Delta_n$? Do the trace closures of the $\Delta_n$ result in some well-known link family?</p> <p>I have tried computing the J.P. in the obvious way using the Kauffman bracket; some simplifications are possible, but so far nothing sufficient to lead to a general formula. </p> http://mathoverflow.net/questions/71955/jones-polynomial-of-the-trace-closure-of-the-fundamental-braid/74062#74062 Answer by Vaughan Jones for Jones Polynomial of the trace closure of the fundamental braid Vaughan Jones 2011-08-30T13:44:47Z 2011-08-30T13:44:47Z <p>Calculation of the Jones polynomial of this link is a (good) exercise in representation theory. As you have observed by Schur's lemma, in any irreducible representation it is, up to a scalar, a square root of the identity. This scalar can be obtained by a determinant argument. So we reduce to the situation where the eigenvalues are 1 and -1. The trace is then just the difference between the multiplicities. This can be determined by specialisation to the case t=1 where the representation is a symmetric group representation and all such questions are well known. So you have the trace in all irreducible representations going into the Jones representation and you just add them up with their weights. Unfortunately I was persuaded not to include this method in my first paper on the polynomial when I already had the Jones polynomial of torus knots. For Homflypt it is a little more complicated but carried out in detail in my Hecke algebras annals paper, and again in the paper with Marc Rosso where we compute arbitrary quantum invariants of torus knots. It's all a lot simpler for the Jones polynomial itself where there are so few irreducible representations. Have fun, Vaughan Jones</p>