Ok, let me try to understand what’s going on for $G=C^*$. Hopefully, David Ben-Zvi will keep me from leading you too far astray. (I tried G a finite abelian group, but I got very confused.) In this case G is canonically isomorphic to LG. (In general, if M is a free abelian group of finite rank, then the Langlands dual to $Hom(M, C^*)$ is $Hom(Hom(M,Z),C^*)$. But, when M has rank one, there is a canonical isomorphism between M and Hom(M,Z) – the one so that the resulting innner product on M is positive definite.) To avoid confusion with the complex numbers, I am renaming our curve to be X.

For any group G, G local systems are classified by $Hom(\pi_1(X),G)$, modulo conjugation by G. Here G is abelian, so “the moduli stack of G-local systems over X” is just $Hom(pi_1(X),C^*)=Hom(H_1(X), C^*)$. This is an algebraic torus of (complex) dimension 2g; it might be best written as $H^1(X,C)/H^1(X,Z)$. I’ll abbreviate this space by S(X) in the future.

A C^* principle bundle is another name for a line bundle, they are classified by Pic(X). Pic(X) has countably many components, I assume that there is some technical condition telling me to just look at $Pic^0(X)$. This is an abelian variety of (complex) dimension g, it might be best written (by the Abel-Jacobi theorem) as $H^{0,1}(X)/H^1(X,Z)$.

I’m going to go on a crazy tangent here. It seems to me that there is a canonical projection (of complex vector spaces $H^1(X,C) \to H^{0,1}(X)$ and this should induce a map of complex varieties $S(X) \to Pic^0(X)$. The fibers of this map are affine spaces of (complex) dimension g. I don’t see any canonical choice of section, so I’m going to figure this is an affine bundle but not a vector bundle. However, if we took the vector bundle corresponding to the affine bundle $S(X)$, it would be a trivial bundle whose fiber over each point of $Pic^0(X)$ is canonically $H^{1,0}(X)$. Since $H^{1,0}(X)$ is the cotangent space at the origin to $Pic^0(X)$; I’m going to propose identifying this vector bundle with the cotangent bundle $T^*(Pic^0 X)$ to $Pic_0 X$.

Why do I do this? Because we are supposed to relate coherent sheaves on $T(X)$ to D-modules on $Pic^0(X)$. Now, for any smooth space $Y$, there is a close relationship between D-modules on Y and coherent sheaves on $T^* Y$. For simplicity, I’ll sketch this when Y is affine (even though it isn’t in our application). Let Y=Spec A, and let DA be the ring of differential operators on Y. There is a filtration on DA by the order of the operator. The associated graded of this filtration is a commutative ring, call it gr(DA), and Spec gr(DA)=T^*Y. This means that there is a functor from D-modules on Y equipped with a filtration to coherent sheaves on $T^*Y$ equipped with a grading. Exploiting this functor is one of the main ways that people study D-modules, even when the D-modules don’t come with natural filtrations. I don’t know enough here to say anymore.

So, if I understand this correctly, in the special case where $Y=Pic^0(X)$, geometric Langlands says that we can make this correspondence very precise by two tricks (1) passing to the derived category and (2) using $S(X)$ instead of $T^*(Pic^0 X)$.

At this point, I have a nagging feeling that I made a mistake in identifying C^* with its dual, even though there was a canonical way to do it! I suspect that this had the effect of identifying $Pic^0 X$ with its dual. Jacobians of curves are always isomorphic to their duals, but for a general abelian variety, $A$ and $A^{\vee}$ is not isomorphic. I suspect that the right statement here is that, for any abelian variety $A$, there is some canonical affine bundle $E(A)$ over $A$, whose associated vector bundle is $T^*(A)$. And the right statement is probably that the derived category of D-modules on $A^{\vee}$ is equivalent to the derived category of coherent sheaves on $E(A)$.

Anyways, that’s as far as I get. Anyone want to fill in the next steps?

## Re: Generalized Geometric Langlands is False

I assume this is pretty much the same talk of Constantin’s that David Ben-Zvi has put up notes for

here.