HOMFLY and homology; also superalgebras - MathOverflow

HOMFLY and homology; also superalgebras

2009-10-22T06:48:37Z

My understanding is that an analogy along the following lines is (roughly) true:

"The Alexander polynomial is to knot Floer homology is to gl(1|1)

as the Jones polynomial is to Khovanov homology is to sl(2)

as a-lot-of-other-specializations-of-HOMFLY are to Khovanov-Rozansky homology are to sl(n)."

1) To what extent is it possible to add another line that starts with the (unspecialized) HOMFLY polynomial? I think there is a triply-graded complex that I can put here (and that maybe this is what I should be calling Khovanov-Rozansky homology? or at least is also due to them?), but is there an analogous object to put in place of the Lie (super-)algebras appearing above?

2) Why is gl(1|1) here? That seems weird.

Answer by Geordie Williamson for HOMFLY and homology; also superalgebras

2009-10-22T07:58:22Z

I can clarify your initial questions, but can't help with things super!

Normally, as you point out, Khovanov-Rozansky homology refers to a bigraded theory whose Euler characteristic is sl(n) homology (which is a specialisation of the HOMFLYPT polynomial). This was originally defined by Khovanov and Rozansky in "Matrix factorizations and link homology". In "Matrix factorizations and link homology II" Khovanov and Rozansky define a triply graded theory whose Euler characteristic is the HOMFLYTPT polynomial and Khovanov later showed that theory has a nice construction using using Soergel bimodules. This is often also referred to Khovanov-Rozansky homology, but the correct name seems to be "triply graded link homology".

I think the super situation in general is much less well understood, at least from the representation theoretic perspective.

Answer by Ben Webster for HOMFLY and homology; also superalgebras

2009-10-22T22:44:12Z

In terms of just the knot polynomials, there's a simple explanation for what's going on that makes $\mathfrak{gl}(1|1)$ seem totally in place:

The knot polynomial attached to the defining representation of $\mathfrak{gl}(m|n)$ only depends on m-n (the dimension of that representation in the categorical sense); you just get the specialization of HOMFLY at $t=q^{m-n}$.

Furthermore, nothing much interesting happens at negative values, since they're basically the same as positive ones. So, our current techniques, which work for $\mathfrak{gl}(n)$ knot homology, can't get at dimension 0, which can be minimally described as $\mathfrak{gl}(1|1)$.

Answer by José Figueroa-O'Farrill for HOMFLY and homology; also superalgebras

2009-10-29T15:05:48Z

I can perhaps say something about the second question. I apologise in advance for tooting my own horn, so to speak, as my knowledge derives from a paper I co-wrote years ago with Takashi Kimura and Arkady Vaintrob: http://arxiv.org/abs/q-alg/9602014 .

In the approach to Vassiliev invariants coming from Chern-Simons theory, one constructs a "universal weight system" from some Lie-algebraic data. In the original work of Bar-Natan and of Kontsevich, the data consists of a metric Lie algebra --- i.e., one possessing an ad-invariant inner product --- and a module. However this extends to metric Lie superalgebras and more generally to metric Yang-Baxter Lie algebras, thanks to work of Arkady Vaintrob.

gl(1|1) is a metric Lie superalgebra and in the paper cited above we worked out its universal weight system. It has two parameters (corresponding to the generators of the centre of the universal enveloping algebra) and setting them to specific values one recovers the weight system of the Alexander-Conway polynomial.

We also show en passant that one can obtain the same universal weight system from a solvable four-dimensional metric Lie algebra, so that in fact one does not actually need Lie superalgebras at all in this case.

Answer by Dylan Thurston for HOMFLY and homology; also superalgebras

2010-04-01T08:08:24Z

Lots of good answers above, but the one that seems to be missing is what belongs in the third column for unspecialized HOMFLY. (As Geordie Williamson points out, Khovanov and Rozansky wrote two related papers, one of which does the unspecialized case.) There's a one parameter family of categories that looks like the category of representations of a Lie superalgebra, called $gl(\alpha)$. When $\alpha$ is an integer, you can kill off the representations that have dimension 0 and get the usual category of representations of $gl(n)$. (You can also kill off fewer representations and get the representations of $gl(n+k|k)$ for any $k$.) Many people have written about this, but perhaps the coolest place to look is in Deligne's La série exceptionelle de groupes de Lie, C. R. Acad. Sci. Paris, Série I 322 (1996) 321—326, where he also conjectures a similar structure specializing to all the exceptional groups. (There are also several follow-up papers.)