cohomology groups of tensor product of sheaves

2009-12-14T01:11:27Z

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 cohomologies of L and M? Like 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.

Answer by Greg Kuperberg for cohomology groups of tensor product of sheaves

2009-12-14T07:35:14Z

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.)

cohomology groups of tensor product of sheaves - MathOverflow

cohomology groups of tensor product of sheaves

Answer by Greg Kuperberg for cohomology groups of tensor product of sheaves