# Quantum equivariant $K$-theory and DAHA.

Theorem 3.2 of the paper "Quantum cohomology of the Springer resolution" by Braverman, Maulik and Okounkov relates equivariant quantum cohomology of the cotangent bundle of $G/B$ to the trigonometric Dunkl operators. A naive guess (or maybe WAG?) is that replacing cohomology with $K$-theory gives the Dunkl representation of the DAHA. An explanation (or reference to an explanation) of what (if any) significant obstacles remain for doing the $K$-theoretic analog of BMO will be greatly appreciated! Is it the sort of thing that should be straightforward (though perhaps technical) at this point? I'd prefer an answer that addresses these particular varieties, rather than general foundational problems.

-
With this question I continue the tradition of only posting questions and answers into which the acronym DAHA can reasonably be shoehorned. –  GS Jan 12 '10 at 14:28

My best guess is that either

1. this is true for $\mathbb{CP}^1$, and it's pretty easy to generalize that given what's already in that paper, or
2. this is false for $\mathbb{CP}^1$, and you're hosed.

The bit one has to understand is the map from the 2 point genus 0 moduli space to the Steinberg variety. BMO get away with just noting that the two spaces have the same dimension, so the pushforward of the fundamental class of the moduli space has to be a sum of fundamental classes of components of Steinberg, whose coefficients they work out by deforming to an almost generic situation and doing the calculation for $\mathbb{CP}^1$.

I think by looking at the pushforward of the structure sheaf on the 2-point moduli space, you'll find that quantum correction is some K-class on the Steinberg variety and thus something in the affine Hecke algebra, and I think it should be the sum of SL(2) contributions for each root by the same deformation arguments that BMO use.

I just spoke to Davesh Maulik about this, and it seems my intuition has failed me: he claims it is just hard, and the techniques of that paper will not work.

-
Thanks; exactly the type of answer I was asking for. –  GS Jan 13 '10 at 11:32