In http://arxiv.org/pdf/1410.6240.pdf M. McBreen and N. Proudfoot conjectured a precise relationship between the quantum cohomology of a symplectic resolution and the intersection cohomology of the cone which it is resolving. However, I could not find the motivation for this conjecture in the above article. Is there a heuristic reason why the two sides should be related?

1$\begingroup$ Nicholas Proudfoot is a user here, by the way. $\endgroup$– Sam HopkinsFeb 9, 2016 at 23:00

5$\begingroup$ @SamHopkins Seems to me that's all the more reason to ask this question :) OTOH, there are random people like me who will be interested in the answer, and there might be random people who know the answer, so I'm glad that OP asked the question here rather than over email. $\endgroup$– Theo JohnsonFreydFeb 10, 2016 at 1:17

$\begingroup$ Is there a cartoon if you saw this cone as a Coulomb or a Higgs branch? $\endgroup$– AHusainFeb 11, 2016 at 3:38
1 Answer
I'll try to explain the genesis of the paper, then some intuition we developed later.
The original motivation was a coincidence. For $X$ hypertoric, Tom Braden and Nick Proudfoot had defined a ring structure on $IH^*(X^{aff})$ whereas Daniel Shenfeld and I found a generators & relations presentation of $QH_{\mathbb{C}^*}(X)$. When the quantum parameter $q$ is specialized to the 'most singular value' $q_0$, our two rings coincide.
We ended up finding something much stronger. Namely, the decomposition theorem tells us $H^*(X)= IH^*(X^{aff}) \oplus K$ for some perverse sheaf $K$, and hence gives an inclusion $IH^*(X^{aff}) \to H^*(X)$. The latter maps to a `polynomial subbring' $QH^*_{\mathbb{C}^*}(X)_{pol} \subset QH^*_{\mathbb{C}^*}(X)$, which admits a specialization $QH^*_{\mathbb{C}^*}(X)_{pol} \to R'$ at $q=q_0$.
In our paper, we show that for $X$ hypertoric, the composition $$ IH^*(X^{aff}) \to H^*(X) \to QH^*_{\mathbb{C}^*}(X)_{pol} \to R' $$
is an isomorphism. This is (to me) still rather surprising! Neither surjectivity nor injectivity are given to you for free, and the fact that they hold for all hypertorics is very suggestive.
There are nonetheless some general heuristics. For $X$ a symplectic resolution, the Steinberg algebra $A = H^{top}(X \times_{X^{aff}} X)$ acts on $H^*(X)$, and the trivial isotypical component of this module is exactly $IH^*(X^{aff})$. In fact, the Steinberg algebra decomposes as $\mathbb{C} id \oplus A^+$, where $id$ is the diagonal correspondence, and in the decomposition $$ H^*(X) = IH^*(X^{aff}) \oplus K $$ we have $K = image(A^+)$.
The conjecture can be roughly reformulated as `specializing $q=q_0$ kills the image of $A^+$'.
Now, let $u \in QH^2_{\mathbb{C}^*}(X)$. Then [BravermanMaulikOkounkov] tells us that the operator of quantum product by $u$, viewed as a function of the quantum parameter $q \in H^2(X,\mathbb{C}^*)$, has poles along certain divisors $D_{\alpha} \subset H^2(X,\mathbb{C}^*)$. The residues are Steinberg operators $Z_{\alpha} \in A^+$.
For many symplectic resolutions, the $D_{\alpha}$ intersect in a 'maximally singular point $q_0$' (when this is not the case, our conjecture runs into some trouble and needs to be corrected).
One would like to say: specializing $q$ to $D_{\alpha}$ kills the image of $Z_{\alpha}$ in $H^*(X)$; specializing to $q_0$ kills the image of all $Z_{\alpha}$. In fact, modulo some technical difficulties it seems likely one can prove such a statement.
Hence, very roughly, our conjecture ressembles the claim that the $Z_{\alpha}$ generate $A^+ \subset A$, or at least the set of idempotents in $A^+$. I do not know of a general reason why the $Z_{\alpha}$ should generate, but perhaps you prefer this mystery to the one we started with.
There are also some motivations coming from representation theory, which I won't go into here as they require a lot of setup.
More speculatively, the limit $q \to q_0$ should correspond to a singular degeneration $Y \to Y_0$ of the mirror variety of $X$. Since symplectic resolutions are almost 'selfmirror', one may hope that the cone singularity of $X^{aff}$ appears in $Y_0$. Perhaps one can directly identify $R'$ with the intersection cohomology of this singularity. Of course, for now there are too many difficulties to make this precise.

$\begingroup$ Thank you very much, this is really cool! Do you think you could briefly mention or just name (some) representationtheoretic motivations? $\endgroup$ Feb 11, 2016 at 6:57