I have seen many references in the (geometric representation theory, symplectic geometry, etc) literature to "infinity local systems".

From what I've been told, given a good cover $\{U_i\}$ of $X$, an infinity local system on a connected space $X$ assigns:

-a chain complex $C_i$ to every contractible open set $U_i$

-a chain morphism $F_{ij}: C_i \to C_j$ to every double intersection $U_i \cap U_j$

-a homotopy of chain morphisms $F_{ik} \sim F_{jk} \circ F_{ij}$ for every triple intersection $U_i \cap U_j \cap U_k$


EDIT: actually, I think that the above definition is rather an "infinity-sheaf". An infinity local system should be a locally constant "infinity-sheaf". My guess is that the Maurer-Cartan condition mentioned below is precisely encoding the "locally-constant" condition.

Some questions:

(1) What are some good sources to learn about these objects (from the perspective of a second year graduate student with limited exposure to these ideas)?

The only paper I have found is Section 2.1 of "A Riemann–Hilbert correspondence for infinity local systems" (https://arxiv.org/pdf/0908.2843.pdf) In this paper, an infinity local system is defined as a set of maps from a simplicial set $K$ to a differential graded category $C$ which satisfy a certain Maurer-Cartan equation. However, it's not clear to me how to make sense of this definition (e.g. what is the role of Maurer-Cartan? how does this definition match up with the intuitive notion of an infinity local system described above?)

(2) The infinity local systems on a topological space $X$ are supposed to form a dg category. I recently heard a talk in which the speaker claimed (as if it was the most natural thing in the world:) )that $hom(k_X, k_X)=C^*(X;k)$, where $k$ is some field, $k_X$ is the locally constant sheaf with stalk $k$ is degree $0$ and $C^*(X;k)$ is the singular cohomology of $X$ with $k$ coefficients.

Is there a good way to understand why $hom(k_X, k_X)=C^*(X;k)$? More generally, how should one understand the dg category structure on $\operatorname{Loc}_{\infty}(X)$?

| cite | improve this question | | | | |
  • 1
    $\begingroup$ Perhaps a more concrete model for the situation you're interested in is just (dg-)modules over $C_*(\Omega X)$? It's noncanonical since it involves choosing a basepoint, but it's at least concrete. Taking chains on the simplicial object $B_{\bullet}\Omega X$ gives a resolution of $k$ over $C_*(\Omega X)$ which yields the computation $hom(k_X, k_X)=C^*(B\Omega X)=C^*(X)$, for example. $\endgroup$ – Dylan Wilson Apr 6 at 18:14
  • 1
    $\begingroup$ For question (2), the point is that $C^*(X;k)$ is the homotopy limit of $k_X$ seen as the constant functor from $X$ to (dg) $k$-modules, therefore the computation you want is just the universal property of the limit. $\endgroup$ – Denis Nardin Apr 6 at 18:24
  • $\begingroup$ Look at proposition 5.3. in the following article www3.nd.edu/~wgd/Dvi/Duality.Algebra.Topology.pdf I guess you do need to assume more finiteness condition on $X$. $\endgroup$ – GSM Apr 6 at 18:30
  • 1
    $\begingroup$ @GSM that proposition is referring to something else (though related). You do not need any finiteness assumptions for the statement I gave, or for what Denis said. $\endgroup$ – Dylan Wilson Apr 6 at 19:08
  • 4
    $\begingroup$ @user142700 Local systems "really" means the functor category $\mathsf{Fun}(X, \mathsf{Ch})$ where one must interpret the space $X$ as yielding an $\infty$-category, and $\mathsf{Ch}$, likewise, as an $\infty$-category. In this setting, $k_X$ means the constant diagram $X \to \bullet \to \mathsf{Ch}$ at the chain complex $k$. The `tensor hom' adjunction tells you that $hom(k_X, -)$ is right adjoint to the functor assigning to a chain complex $C_*$ the constant diagram at $C_*$. On the other hand, the right adjoint to the constant diagram functor is called 'the (homotopy) limit'. $\endgroup$ – Dylan Wilson Apr 6 at 19:56

Your Answer

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

Browse other questions tagged or ask your own question.