# What is a higher derived constructible sheaf

Suppose $X$ is a topological space and $k$ some discrete coefficient field. Let's define the category of "$\infty$-local systems on $X$" to be DG representations of the ring $C_*(\Omega X,k)$ of chains of loops on $X$. (I think this is equivalent, at least when $\text{char}(k)=0$, to locally constant sheaves where "sheaf" is defined in an appropriate higher-topos sense).

I would like some construction of "$\infty$-constructible sheaves" which includes both this category and also constructible complexes of sheaves for some fixed stratification of $X$. Do topologists know of one? Is there an analogue of the Riemann-Hilbert correspondence for $X$ a complex algebraic manifold?

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I think you need X connected for your description of local systems, but the coefficients are arbitrary (i.e., $k$ could be the sphere). But the same category can also be said concretely as complexes of sheaves with locally constant cohomology groups. Then the $\infty$-version of constructible sheaves is just the $\infty$-category of complexes of sheaves with constructible cohomology. And indeed that's derived equivalent to the $\infty$-category of regular holonomic $\D$-modules in the complex analytic setting. –  David Ben-Zvi Apr 24 '13 at 0:43
@David: I think you need to be careful about finiteness issues. For example, local systems on $\mathbb C^\times$ in Dmitry's sense would be representations of $\pi_1 (\mathbb C^\times)$. But you won't be able to see the indecomposible infinite dimensional representations of $\mathbb Z$ using $D$-modules (I think the de Rham homotopy type'' should only be able to the pro-algebraic completion of $\pi_1$, or something along those lines...) –  Sam Gunningham Apr 24 '13 at 1:31
Even in the simply connected case, the two versions of $\infty$-local systems are not quite the same. As I commented below, they correspond to $C_\ast(\Omega X)$ vs $C^\ast(X)$ modules. These categories are closely related (by some version of Koszul duality), but not quite the same. –  Sam Gunningham Apr 24 '13 at 1:41
Thanks Sam! (you keeping tell me this fact enough times I might start to remember it!) OTOH unless you're imposing some such finiteness conditions on your sheaves it's probably a bad idea to call them constructible.. In any case the issue is size (as you say already in the case of the circle), not really anything to do with particularly "infinity" issues. –  David Ben-Zvi Apr 24 '13 at 2:10
@David: I agree about the finiteness issues. I only mention this, as in the question, the $C_\ast(\Omega X)$-version of local systems was talked about. And it happens to be an issue that I am interested in myself! –  Sam Gunningham Apr 24 '13 at 3:59
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In the appendix of "Higher Algebra" (http://www.math.harvard.edu/~lurie/papers/HigherAlgebra.pdf), Jacob Lurie describes constructible sheaves on a stratified space as representations of an exit path $\infty$-category. So one option is to take representations of this exit path $\infty$-category valued in the stable $\infty$-category (or appropriate DG category) of complexes of vector spaces. When the stratification is trivial, this recovers the usual notion of $\infty$-local system (as representations of the fundemental $\infty$-groupoid).
@BZ: Good point. I think the answer should be yes in reasonable cases, but you have to be careful about what you mean by the constructible derived category. For example, a constructible sheaf is usually taken to have finite dimensional stalks. The issue occurs even with no stratification (i.e. local systems). If $X$ is simply connected with no stratification (say), then the I would say that the usual constructible derived category is equivalent to $C^\ast(X)$-mod. This is not (quite) the same as $C_\ast (\Omega X)$-modules (though the two are closely related. –  Sam Gunningham Apr 24 '13 at 1:25
The difference is more clear in the case $X = S^1$; in that case, the usual constructible derived category would decompose in to blocks for each generalized eigenvalue of the monodromy. This sees only the profinite completion of $\pi_1(S^1) = \mathbb Z$, as opposed to all representations. –  Sam Gunningham Apr 24 '13 at 1:28