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Let $B_n$ be the braid group on $n$ strands. It has generators $\tau_i$ for $i = 1,\ldots,n-1$ which exchange the $i$th and $(i+1)$st strands, and which satisfy the relations $\tau_i \tau_j = \tau_j \tau_i$ for $|i-j| \ne 1$ and $\tau_i \tau_{i+1} \tau_i = \tau_{i+1} \tau_i \tau_{i+1}$.

$B_n$ maps to the Hecke algebra $H_n$, which has the same generators and the additional relation $(\tau_i-1)(\tau_i + q) = 0$. Composing this map with the Jones-Ocneanu trace yields the HOMFLY polynomial of the closure of a given braid.

In the braid group, there is the half-twist $\Delta$. It exchanges the $i$-th and $(n-i)$th strands by rotating the whole braid counterclockwise. Its square, the full twist, generates the center of the braid group.

One particularly natural choice of basis for the Hecke algebra is the Kazhdan-Lusztig canonical basis.

What is the image of $\Delta^k$ in $H_n$, in the Kazhdan-Lusztig basis?

This is a decategorified version of the question I am really interested in:

What is the complex of Soergel bimodules corresponding to $\Delta^k$ ?

A related question, in case anyone happens to know:

What is the HOMFLY homology of a torus knot?

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  • $\begingroup$ I edited the names, but anyway the link to Wikipedia isn't right. $\endgroup$
    – Jim Humphreys
    Commented Oct 29, 2010 at 19:19
  • $\begingroup$ well, hopefully the link is improved. $\endgroup$
    – Vivek Shende
    Commented Oct 29, 2010 at 19:51
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    $\begingroup$ A quick answer is it's complicated ... expressing T_{w_0} in the KL basis involves inverse Kazhdan-Lusztig polynomials, which are again KL polynomials up to (-1)^{?}. Hence the complex corresponding to the half twist will be very complicated for n large enough (but for, eg n = 3 it is easily described). However, I seem to remember Khovanov mentioning that some nice stabilization happens in the Rouquier complex if one takes high powers of the full twist. I am not sure it has ever been written down. $\endgroup$
    – Geordie Williamson
    Commented Oct 29, 2010 at 22:31
  • $\begingroup$ Thanks Geordie. In fact I spent about the whole day, and not alone either, working out what $\Delta^k$ is for $n=3$, but I guess its description at the end is not all that bad. Do you know what it is for $n=4$? $n=5$? How far does it stay reasonable to write down? Is there any computer software to help me explore this question? $\endgroup$
    – Vivek Shende
    Commented Oct 30, 2010 at 1:00
  • $\begingroup$ I don't know of any software. The thing that makes n = 4 much harder than n = 3 is the fact that there are non-trivial (ie \ne 1) KL polynomials. I once worked out the complex for the half twist (but none of its powers) for n = 4 and remember not finding the answer particularly enlightening! If you do manage to get a conceptual picture of what is going on I'd be very interested to hear about it. $\endgroup$
    – Geordie Williamson
    Commented Oct 30, 2010 at 7:22

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As Geordie says, this is really hard, and will get much harder as n increases (I'd guess $n=3$ is about an order of magnitude harder than $n=2$, and that trend continues).

The relevant papers on this are Stosic:Homological thickness and stability of torus knots and Rozansky:An infinite torus braid yields a categorified Jones-Wenzl projector. Jake Rasmussen had some similar results, but it looks like he never wrote them up (I guess fatherhood has been distracting for him). In fact, you might try emailing Marko; it's possible that the thick/Schur categorification story can actually help here and make things a bit more concrete. I wouldn't hold your breath, though.

Perhaps the most famous fact about $\Delta^k$ is that for the correct power of $q$, the limit $\lim_{k\to\infty}q^?\Delta^k$ is actually the Jones-Wenzl projector. Lev's paper is about how this holds in the categorified picture as well, though he's working in the Temperley-Lieb case, not full Soergel bimodules.

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