The meaning of knot homology

What I left out of my post on knot homology was, perhaps, the elevator pitch (if you’re in an elevator with a mathematician who already has some background). I’m giving a talk tomorrow which should include this stuff, and so one possibility is to just drop the slides for that on you, and let that speak for itself. Especially recommended are slides 12 and those past 18 (the rest is more standard quantum topology and categorification stuff).

But I’m not so sure that’s a wise plan. So let me try to say something more bloggy:

The knot invariants I’ve constructed are in some sense completely pictorial. They’re entirely defined in terms of algebras much like Khovanov and Lauda’s. What’s different is that where Khovanov and Lauda just had black strands (which in the slides above are actually blue and green), I’ve added red strands. In the categorification of quantum groups picture, these red strands carry highest weights, and a picture where I’ve added red strands for \lambda_1,\dots,\lambda_\ell shows up when I categorify the tensor product representation V_{\lambda_1}\otimes \cdots \otimes V_{\lambda_\ell}.

You should think of these red strands as basically fixed in the place, and the algebra that one uses to categorify the representation (Aaron Lauda would say “the 1-morphisms in the categorification of the representation”) are given by diagrams of black strands which flow around and over these black strands (everything is happening in the plane, and crossings don’t have an over or under strand) with certain relations.

Of course, the philosophy from quantum topology is that you now allow your red stands to not be fixed in place; you should allow them to flow over or under each other (now keeping track of the difference). This gives you a braiding; of course, you have to figure out what to do with the black strands. In principle this is not so complicated. You make the same types of diagrams, and do the same kinds of relations as before. But somehow the presence of a crossing screws things up. Suddenly the first layer of relations isn’t enough; you have to start thinking about relations between relations, and relations between those. Which is my vague and hopefully unscary way of saying that you need to think in the derived category.

But braiding isn’t enough, you have to also have to have cups and caps, that is you have allow a pair of dual representations to appear out of nothing (and it turns out that they must bring with them a particular pattern of black strands. You can see the adjoint \mathfrak{sl}_3 example on slide 24. You also have to have the opposite: these red lines colliding and sucking a great glop of black strands off to nothingness with them. As you might have guessed, both of these also need to be done in a derived way.

The main result of the paper is that this gives you a bigraded vector space which is a knot invariant, and whose Euler characteristic is the corresponding quantum invariant, which in the case of the defining representation of \mathfrak{sl}_n gives you the Mazorchuk-Stroppel-Sussan homology (which in particular matches Khovanov’s homologies when n=2,3).

IFAQ (Infrequently Asked Questions)

I thought Khovanov-Rozansky homology was the knot homology for the defining representation of \mathfrak{sl}_n?

Actually there are (at least) 3 such homologies: Khovanov-Rozansky, Mazorchuk-Stroppel-Sussan, and Cautis-Kamnitzer. For general n, I don’t believe there’s a proof that any pair coincide, though I think everyone would be quite surprised if they didn’t.

What about functoriality?

Well, I have some ideas about that, but there are two hitches: 1. These ideas require the cup and cap functors to be biadjoint. One adjunction is obvious from the definition, the other is hard. I think this is in reach, though. We just need someone to find a clever enough student to sic on the problem. 2. Then somebody has to check the movie moves. This might not be so bad (for example, the isomorphisms for the braid relations are canonical), but might be real ugly.

What about 3-manifold invariants?

Obviously that’s an application I’m very excited about. There are some subtle points in there, though, that I am still quite unclear on how to work out. I have a feeling that if I really knew what was going on for a single 1-framed circle, it would all be clear. But I don’t, so it’s not.

8 thoughts on “The meaning of knot homology

  1. “The invariant of a circle is a seemingly new algebra whose dimension is the dimension of the algebra.”

    This looks like something that was misedited. Is there another formula for the dimension?

    Also, correct “evalution” and “This has lead”.

  2. All those are fixed now.

    I should have said “dimension of the representation.”

    Actually its graded dimension is the quantum dimension (suitably normalized), which is even more interesting.

  3. I should have said “dimension of the representation.”

    Then could the algebra be the one in this paper?

    Title: Opers with irregular singularity and spectra of the shift of argument subalgebra
    Authors: Boris Feigin, Edward Frenkel, Leonid Rybnikov

    … As a byproduct, we obtain the structure of a Gorenstein ring on any such module. …

  4. Then could the algebra be the one in this paper?

    It well could. I think I had heard about the semi-simple version of that algebra, and thus hadn’t thought it was the right sort of thing, but the regular nilpotent version might do the trick.

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