## Questions about the MOY relations				August 9, 2007

Posted by Joel Kamnitzer in link homology, Pictorial Algebra, quantum groups.
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I have some rather specific questions about the MOY knot invariant which have come up as Sabin Cautis and I have been thinking about Khovanov-Rozansky homology. I’ll start by explaining the following “theorem” and then I’ll ask some questions about it. Hopefully someone (for eg. Scott) will be able to answer them.

Consider closed crossingless Reshetikhin-Turaev diagrams for sl(n) where all strands are labelled by 1 or 2. This means that we look at oriented planar graphs with all edges either labelled 1 or 2, and each vertex trivalent with either two single edges and one double edge and matching orientations.

We can think of such a graph as being related to the representation theory of (quantum) sl(n). The single strands correspond to the standard representation $\mathbb{C}^n$ and the double strands to $\Lambda^2 \mathbb{C}^n$ The vertices correspond to morphisms of representations (for example $\mathbb{C}^n \otimes \mathbb{C}^n \rightarrow \Lambda^2 \mathbb{C}^n$). Within this representation theory context, each graph G can be evaluated to a Laurent polynomial c(G) (an element of the trivial representation of quantum sl(n)).

Now, ignore the representation theory interpretation for a moment and just consider the set of all possible graphs. Let us impose the “MOY relations” on such graphs, by which I mean the relations founds on page 4 of Khovanov-Rozansky’s link homology paper.

Theorem.
Modulo these MOY relations, any graph G is equivalent to c(G). Namely, c(G) is the only Laurent polynomial one can assign to each graph which is consistent with the MOY relations.

My first question is whether this theorem is correct as stated and if so where can I find a proof. It does not seem to be stated (or proven) in the MOY paper. It is stated on page 3 of the above Khovanov-Rozansky paper.

Now, come some more substantial questions:
1. Is there any theory to explain how such a graph is equivalent to c(G)? Is there a sequence of moves that one can do in order to reduce any graph G to c(G) and to what extent is that sequence of moves unique? The last move involving sums on both sides is the most confusing in this way of thinking. I guess I am asking for a theory of “invertible foams”, which should be the “simple” part of the theory being developed by Mackaay, et al. (and Scott).

2. Is there an extension of the above theorem to “open graphs” (ie graphs with open edges which represent invariants in a tensor product)? I am mostly interested in the case where there are just 4 open edges, all of them single with opposite orientations, so that in this case I know a basis for the corresponding invariant space $Inv(V \otimes V \otimes V^\star \otimes V^\star)$ (it is just two dimensional).

## Comments»

1. Ben Webster - August 9, 2007

The theorem is true. You can find most of the proof in Jake Rasmussen’s paper “Some Differentials…” That only deals with braid-like link diagrams, but applying Vogel’s algorithm to an arbitrary diagram shows that it is congruent to a sum of braid-like ones (in MOY language, this is using the 3rd line in Khovanov and Rozansky’s list to remove badly oriented Seifert circles). Once in braid-land, we can induct on the braid index and the number of intersection points with the inner-most circle, using the 4th line to move intersection points outward, and the relations of the second line to reduce the number of intersections on the inner circle to one, and then make it disappear.

I can explain this is person if you’re coming to berkeley some time in the next week or so.

2. Ben Webster - August 9, 2007

Oh, and for your second question, the braids modulo these moves are just the Hecke algebra. so if your question was whether it surjects onto the invariant space, the answer is yes.

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