2
$\begingroup$

My apologies for the imprecise question(s), it should be clear enough that I´m a complete beginner in this subject.

The (de Rham) Geometric Langlands Conjecture over $\mathbb{C}$ takes as input a smooth proper curve $X$ and reductive group $G$ and relates a Galois side and an automorphic side.

The automorphic side is the dg-category of (all) D-modules on $Bun_{G}X$, and the Galois side is something like $QCoh(Loc_{G^{\vee}}X)$, more precisely ind-coherent sheaves with singular support inside the global nilpotent cone $\mathcal{N}\subset T^{*}Loc_{G^{\vee}}X$. Structure sheaves of points on $Loc_{G^{\vee}}X$ are supposed to correspond to Hecke eigensheaves on $Bun_{G}X$, and this should essentially characterize the equivalence. Note that a basic sanity check is that Hecke eigensheaves with distinct eigenvalues are orthogonal, as they must be if they correspond to structure sheaves of distinct points on the Galois side.

For now let´s pretend that we want to construct a functor $$QCoh(Loc_{G^{\vee}}X)\longrightarrow D(Bun_ {G}X)\cong QCoh((Bun_{G}X)_{dR}).$$ Then a natural place to look for such would be the category $QCoh(Loc_{G^{\vee}}X\times(Bun_{G}X)_{dR}),$ ie complexes of quasi-coherent sheaves on the product $Loc_{G^{\vee}}X\times Bun_{G}X$ with flat connection along the fibres of the projection to $Loc_{G^{\vee}}X$.

A sanity check, in the case of abelian $G$ this is exactly what we do to establish the conjecture. For $GL_{1}$, the integral kernel in this case is simply a universal relative flat line bundle on $A\times A^{fl}$, where $A$ is the abelian variety $Pic^{0}X$ and $A^{fl}$ is the moduli space of flat line bundles on $A$ (a torsor for a vector space over $Pic^{0}(A):=A^{\vee}$.)

Assuming the conjecture, does some suitably fancy version of Morita theory force the existence of such an integral kernel for non-abelian $G$? If not, is there a good reason to assume that such cannot exist? There are other formulations of the categorical Geometric Langlands Conjecture (a Betti version due to Ben-Zvi and Nadler and a Dolbeault one due to Donagi-Pantev), what about a similar question in these cases?

$\endgroup$

0

You must log in to answer this question.