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 $B_n$.

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?

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.

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    $\begingroup$ If you were to follow the nth Jones-wenzl idempotent by that braid, close it up and take the bracket, you could compute that easily. $\endgroup$ – Charlie Frohman Aug 3 '11 at 2:28
  • $\begingroup$ For $n$ even, it is a torus link of $n/2$ components, and for $n$ odd it is a cable link. You can convince yourself easily that the braid closure may be drawn on a standard Mobius strip, which sits nicely inside a solid torus with boundary a $(1,2)$ curve on the boundary torus. When $n$ is even, the strands are parallel, and may be pushed onto parallel curves on the boundary torus. When $n$ is odd, one may do this for all but one strand, which is isotopic to the core of the Mobius strip, and therefore it is cabled. $\endgroup$ – Ian Agol Aug 30 '11 at 14:40

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

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