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If L and M are two local systems on a space X, what can we say about the cohomology groups $H^i(X,L\otimes M)$ in terms of the cohomology of L and M? For example, can we determine their dimensions. You can assume X to be an affine connected smooth curve, if that helps.

For the same question for coherent locally free sheaves, it seems that Riemann-Roch is helpful sometimes. But I don't know if there is a Riemann-Roch type theorem in l-adic cohomology?

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    $\begingroup$ I'm not sure there's much useful to say; I feel like one should be able to come up with at least a spectral sequence of some sort, but even that I'm not coming up with. $\endgroup$
    – Ben Webster
    Dec 14, 2009 at 2:21

2 Answers 2

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If $X$ can just be a topological space, there are examples in which $H^i(X;L)$ and $H^i(X;M)$ vanish entirely, but $H^i(X,;L \otimes M)$ does not. For example, in cohomology with real coefficients, maybe $X = \mathbb{R}P^3$ and $L = M$ is the local system corresponding to the tautological line bundle, or if you like the non-trivial representation of $\pi_1(\mathbb{R}P^3) = \mathbb{Z}/2$. Then $H^i(X,L)$ vanishes, but $H^i(X;L \otimes L)$ is the usual cohomology of $\mathbb{R}P^3$ and does not vanish in degrees $0$ and $3$.

One thing that exists (again for cohomology in the setting of topological spaces) is a well-defined cup product map: $$\cup:H^i(X;L) \otimes H^j(X;M) \to H^{i+j}(X;L \otimes M).$$ This map is not a direct statement about dimensions of anything, but it is an important map in the theory of cohomology with local coefficients. For instance, if $L$ and $M$ are inverse line bundles and $X$ is a closed manifold, then this cup product is the right way to set up Poincare duality. (To review, they have to be line bundles with flat connections to give you local coefficients.)

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My example is similar to Greg's, and shows that $H^i(X,M), H^i(X,L)$ can vanish while $H^i(X, M \otimes L)$ doesn't have to for $X$ the affine line over a finite field. Let $\psi \colon \mathbb{F}_q \to \overline{\mathbb{Q}}_\ell^\times$ be a non trivial character. The Artin-Schreier cover of the affine line $\mathbb{A}^1$ over $\mathbb{F}_q$ is given by $$ p \colon \mathbb{A}^1 \to \mathbb{A}^1, \quad x \mapsto x^q - x. $$ Hence there is a surjection $\pi_1^\text{et}(\mathbb{A}^1) \to \mathrm{Aut}(p) \simeq \mathbb{F}_q$. Let $\mathcal{L}(\psi)$ be the lisse sheaf associated to the representation $$ \pi_1^\text{et}(\mathbb{A}^1) \to \mathbb{F}_q \xrightarrow{\psi} \overline{\mathbb{Q}}_\ell^\times $$ If $x \in \mathbb{F}_q$ write $\psi_x$ for the character $y \mapsto \psi(xy)$. Then $p_\ast \overline{\mathbb{Q}}_\ell \simeq \oplus_{x \in \mathbb{F}_q} \mathcal{L}(\psi_x)$. From the Leray spectral sequence, we get that $H^i(\mathbb{A}^1, \mathcal{L}(\psi_x)) = 0$ unless $i = 0$ and $x = 0$. Now given $x \in \mathbb{F}_q^\times$ we have $\mathcal{L}(\psi_x) \otimes \mathcal{L}(\psi_{-x}) \simeq \overline{\mathbb{Q}}_\ell$.

So summing up, for any $x \in \mathbb{F}_q^\times$ we have $R\Gamma(\mathbb{A}^1, \mathcal{L}(\psi_x)) = 0$ but $R\Gamma(\mathbb{A}^1, \mathcal{L}(\psi_{x}) \otimes \mathcal{L}(\psi_{-x})) = \overline{\mathbb{Q}}_\ell$. A similar argument shows also that $R\Gamma_c(\mathbb{A}^1, \mathcal{L}(\psi_x)) = 0$ but $R\Gamma_c(\mathbb{A}^1, \mathcal{L}(\psi_{x}) \otimes \mathcal{L}(\psi_{-x})) = \overline{\mathbb{Q}}_\ell[2]$.

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