11
$\begingroup$

Let $G$ a complex reductive algebraic group, $X$ be a smooth compact complex curve. The moduli stack $Bun_G$ of principal $G$-bundles on $X$ is generally speaking not quasi-compact (so we do not have Verdier duality). However, Drinfeld has conjectured that a different functor, $Ps\text{-}Id_{Bun_G, !}$, gives us an equivalence between the category of D-modules on $Bun_G$ and its dual $$ Ps\text{-}Id_{Bun_G, !}\colon (D\text{-}mod(Bun_G))^\vee\rightarrow D\text{-}mod(Bun_G). $$ This conjecture has been proved by Gaitsgory. As a by-product of the proof, Gaitsgory has found that there is a canonical isomorphism of functors $D\text{-}mod(Bun_M)\rightarrow D\text{-}mod(Bun_G)$ $$ Eis_!^-\cdot Ps\text{-}Id_{Bun_M, !}\simeq Ps\text{-}Id_{Bun_G, !}\cdot (CT_*)^\vee, $$ where $M$ is the Levi quotient for a parabolic $P \subset G$, $Eis_!^-$ is the geometric Eisenstein series functor for the opposite parabolic $P^-$ and $CT_*^\vee$ is the dual of geometric constant term functor. He dubbed this isomorphism 'strange functional equation'.

My question is: have decategorifications of this functional equation been studied? Does it descend to an interesting relation on the level of Hochschild cohomologies, for example?

$\endgroup$
  • 1
    $\begingroup$ Do you mean $D\text{-}Mod$ and $Ps\text{-}Id$? Note that a hyphen in math mode behaves like a minus and gets lots of spacing. Also, do you have a reference for the proof? $\endgroup$ – David Roberts Jun 20 '18 at 21:50
  • $\begingroup$ @DavidRoberts I do actually. How should I change the tex code? $\endgroup$ – user74900 Jun 20 '18 at 22:39
  • 1
    $\begingroup$ Use \text{-} instead of just -. I did it for you. $\endgroup$ – David Roberts Jun 20 '18 at 23:01
1
$\begingroup$

This is essentially the question studied in https://arxiv.org/abs/1503.04705v4.

A (possibly wrong) comment: Unlike $\operatorname{Eis}_*$, which decategorifes (under Drinfeld-Wang's conventions) to the function-theoretic Eisenstein series, there is no obvious decategorification of $ \operatorname{Eis}_!.$ The strange functional equation, rather than giving some classical identity for Eisenstein series, tells us you how to decategorify $\operatorname{Eis}_!.$

The decategorification is via the fonction-faisceaux correspondence (see e.g., fonctions-faisceaux correspondence ). Your Eisenstein $*$ and $!$ functors becomes maps from the space of functions on $\operatorname{Bun}_T(\mathbb{F}_p)$ to the space of functions on $\operatorname{Bun}_G(\mathbb{F}_p)$.

In the usual fonction-faisceaux correspondence, $!$-pushforward corresponds to "integration", i.e., summing over points in fibers. The Drinfeld-Wang conventions are actually such that instead $*$-pushforward plays this role, and it is $!$-pushforward that admits no classical analogue. This in turn leads to $\operatorname{Eis}_*$ being the categorical counterpart of classical Eisenstein series, while $\operatorname{Eis}_!$ a priori doesn't necessarily have a function-theoretic analogue.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ And what would be the result of such a decategorification, in more concrete terms? $\endgroup$ – Qfwfq Jun 20 '18 at 22:11
  • $\begingroup$ @Qfwfq: I added some more details to my answer. Does this answer your question satisfactorily? $\endgroup$ – dhy Jun 21 '18 at 3:23
  • $\begingroup$ Thank you! Actually, I hoped to see an actual functional equation satisfied by an analytic function and corresponding to the higher categorical relations you mentioned, but I realize it's just because I don't know the subject at all and probably my expectation was very naive! :) $\endgroup$ – Qfwfq Jun 21 '18 at 21:13
  • 1
    $\begingroup$ @Qfwfq: Well, there is such a functional equation, (A.22) in Drinfeld-Wang, which gives a formula for the decategorification of $\operatorname{Eis}_!$. But I don't know any way to express the decategorification of $\operatorname{Eis}_!$ in terms of standard automorphic form operations, so I'm not sure this is really useful for anything... $\endgroup$ – dhy Jun 22 '18 at 7:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy