# Why is there no stack of $\ell$-adic sheaves on a curve?

One of the main players in the categorical geometric langlands correspondence is the moduli stack of rank n integrable connections on a complex curve. The reason for considering such objects is that they are the de Rham analogue of lisse $\ell$-adic sheaves, which have an obvious Galois interpretation. I have read in many places that there is no moduli stack of lisse $\ell$-adic sheaves on a curve over a finite field, making the naive $\ell$-adic formulation of the categorical geometric Langlands correspondence problematic. I have two questions

1. Why do $\ell$-adic sheaves not form a stack?

2. Is it possible that they form some kind of higher stack?

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Let's work over the complex numbers for simplicity. Let $X$ be a Riemann surface, fix a base point $x \in X$, and let $\Gamma$ be the fundamental group $\pi_{1}(X,x)$.

The data of an $n$-dimensional local system on $X$, together with a trivialization at the point $x$, is equivalent to the data of an $n$-dimensional representation of $\Gamma$. This is true in either the $\ell$-adic setting (in which case you'd study representations into $GL_n( \mathbf{Z}_{\ell} )$) or in the setting of "usual" local systems (where you'd study representations into $GL_n( \mathbf{C})$).

In particular, you can topologize the set of representations by picking a set of generators for $\Gamma$ and thereby embedding into some power of $GL_n(k)$. In the case $k = \mathbf{C}$, you can do even better: $GL_n( \mathbf{C} )$ is again an algebraic variety over $\mathbf{C}$, so that the collection of representations of $\Gamma$ into $GL_n( \mathbf{C} )$ inherits the structure of an algebraic variety. (In fact, it has multiple inequivalent "algebraic structures"; the one that I've described is the "Betti version", which is different from the one used in geometric Langlands).

Working with $\ell$-adic coefficients, you can still talk about whether or not two representations are "close" to one another, but the space of representations no longer bears any resemblance to an algebraic variety defined over $\mathbf{C}$ (for example, it's totally disconnected). This has nothing to do with stacks versus higher stacks: it's a disconnect between the "coefficient field" for your local systems and the field that your variety is defined over.

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Does this mean that the correct replacement in the $\ell$-adic setting should be a rigid analytic stack? Perhaps something akin to Galois deformation spaces in the sense of Mazur? – anon Jun 19 '13 at 21:48
If $X$ is a curve over $\mathbb{C}$ can't you build the moduli of rank $n$ de Rham local systems on $X$ in a straightforward way using $(X,x)^{DR}$, the Toen de Rham homotopy type of $X$. Namely $Loc^{DR}_n(X) = Map( (X,x)^{DR}, BGL_n )$, where we're in the $\infty$-topos of (higher) stacks over $\mathbb{C}$. If $X$ is a curve over a finite field (of characteristic not equal to $\ell$), couldn't we just define $Loc^{\ell}_n(X) = Map ( (X^{et}\otimes \mathbb{Q}_{\ell})^{sch} , BGL_n)$, now in the topos of (higher) stacks over $\mathbb{Q}_{\ell}$. – anon Jun 19 '13 at 23:53
where $(X^{et}\otimes \mathbb{Q}_{\ell})^{sch}$ Toen's $\ell$-adic schematic homotopy type of $X$? – anon Jun 20 '13 at 0:01
You could make that definition. However, the business about $\infty$-topoi and higher stacks is a red herring, because $BGL_n$ is an ordinary stack. You also don't get very much structure this way: for example, any $\ell$-adic sheaf $F$ with Zariski-dense monodromy will have an open neighborhood in your $Loc^{\ell}_{n}(X)$ which is isomorphic to BG_m. In particular, $Loc^{\ell}_{n}(X)$ doesn't allow you to "move continuously" between nonisomorphic sheaves with dense monodromy. – Jacob Lurie Jun 20 '13 at 4:14
I suppose my original answer is misleading: the "Betti" moduli space is defined over the rational numbers (even over the integers), so you could base change it to any field you like. And when you base change it to the $\ell$-adics, you get something closely related to your $\Loc_{n}^{\ell}(X)$. However, the Betti moduli space depends only on the topology of $X$, not on its complex structure. Geometric Langlands is about the deRham moduli spaces, where there is a connection between the algebraic structure on $X$ and on the moduli space. – Jacob Lurie Jun 20 '13 at 4:20