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Suppose we have a reasonable topological space $X$ (i.e. a complex algebraic variety or a manifold) whose integral singular cohomology and fundamental group we understand well.

Suppose that we are given a monodromy representation $\rho: \pi_1(X,x) \longrightarrow \text{GL}(V)$.

How does one compute the cohomology $H^i(X,L)$ of the local system $L$ associated to $\rho$?

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3 Answers 3

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Suppose $X$ is a connected CW complex with fundamental group $\pi:=\pi_1(X,x)$. Then the cellular chain complex $C_*(\widetilde{X})$ of the universal cover is a chain complex of free $\mathbb{Z}\pi$-modules, where $C_i(\widetilde{X})$ has rank equal to the number of $i$-cells of $X$. We can also regard $L$ as a $\mathbb{Z}\pi$-module. By definition $H^i(X;L)$ is the $i$-th homology of the cochain complex $\operatorname{Hom}_{\mathbb{Z}\pi}(C_*(\widetilde{X}),L)$. This can sometimes be computed directly, if the cell structure of $X$ and the boundary maps are well enough understood.

Then there are various tricks involving long exact sequences associated to short exact sequences of coefficient systems, transfer arguments, Shapiro's lemma, spectral sequences, and so on. For manifolds, there is Poincaré duality with local coefficients, which can sometimes inform computations. These methods are descirbed in Brown's "Cohomolgy of Groups" for $X=K(\pi,1)$, but mostly they apply more generally.

It really does depend on the space and the coefficient system as to which of these methods works best.

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  • $\begingroup$ When you say that "$H^i(X; L)$ is by definition the $i$-th cohomology of $\mathrm{Hom}_{\mathbb{Z} \pi}(C_*(\tilde{X}), L)$" what do you mean? Do we get an injective resolution for $L$ in the category of sheaves? I don't see how. $\endgroup$
    – Eduardo de Lorenzo
    Commented Oct 22, 2021 at 10:02
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    $\begingroup$ @EduardodeLorenzo: I mean that this is the classical definition of cohomology of a space $X$ with coefficients in a $\mathbb{Z}\pi_1(X)$-module, as given by Steenrod and others. I don't think it can really be thought of as an $\operatorname{Ext}$-group, unless the universal cover $\tilde{X}$ is acyclic, in which case $C_*(\tilde{X})$ is a projective resolution of the trivial module $\mathbb{Z}$. $\endgroup$
    – Mark Grant
    Commented Oct 22, 2021 at 10:31
  • $\begingroup$ Even in the case where the universal cover is contractible, couldn't it happen that an injective resolution of $L$ in the category of local systems/$\mathbb{Z}\pi$-modules is not injective when considered in the (larger) category of sheaves? If I'm not mistaken, this is the category you need to take so that cohomology of constant sheaves coincides with singular cohomology, which is the origin of my question. $\endgroup$
    – Eduardo de Lorenzo
    Commented Oct 22, 2021 at 11:54
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    $\begingroup$ @EduardodeLorenzo: Maybe we're using different definitions of "local coefficent system"? To see that what I wrote coincides with singular cohomology when $X$ is a CW-complex and $L$ is trivial, use the equivalence of singular with cellular, and the fact that $\operatorname{Hom}_{\mathbb{Z}\pi}(C_*(\tilde{X}),L)\cong \operatorname{Hom}_{\mathbb{Z}}(C_*(X),L)$. $\endgroup$
    – Mark Grant
    Commented Oct 23, 2021 at 12:11
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In general you should have $H^1(\pi_1(X,x),V)\cong H^1(X,L)\,,$ where on the left we have the group cohomology of $\pi_1(X,x)$ acting on $V$ according to the representation. You will also have an embedding $H^2(\pi_1(X,x),V)\hookrightarrow H^2(X,L)\,,$ that will be an isomorphism if $X$ has $\pi_2(X)=0\,.$ In general you will have, if $\pi_k(X)=0$ for all $2\le k\le n\,,$ that $H^n(\pi_1(X,x),V)\cong H^n(X,L)\,,$ and you will have an embedding $H^{n+1}(\pi_1(X,x),V)\hookrightarrow H^{n+1}(X,L).$

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If you're looking for something very concrete in terms of things like explicit boundary map computations, you might be interested in looking at Section 31 of Steenrod's Topology of Fibre Bundles. As in the other answers, though, it sort of depends on what sort of information you already possess or can get at for your space.

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