Let $X$ be an orientable surface (real manifold of dim 2). Let $F$ be a local system (a locally free sheaf of abelian groups) on $X$. Is is true that $H^n(X,F)=0$ for $n>2$? And that $H^n(X,F)=0$ for $n \geq 2$ if $X$ is non-compact?

(Here the cohomology used is either Grothendieck's sheaf cohomology or the standard singular cohomology with local systems -- they amount to the same thing).


The analagous statement is true for any $n$-dimensional CW complex $X$ equipped with a local system $F$. I am sure there is a sheafy proof which works in much more generality but I am less comfortable language so I won't try to find one.

Represent $F$ by a homomorphism $\rho_F: \pi_1 X \to \text{Aut}(A)$, where $A$ is some abelian group. Then the cohomology groups you are interested are given by taking your favorite model for $C_*(\widetilde X;\Bbb Z)$ as a free $\Bbb Z[\pi_1(X)]$-module, and taking the homology groups of $$\text{Hom}_{\pi_1 X}(C_*(\widetilde X;\Bbb Z), A).$$ In particular, if you take the cellular chain complex of $X$, one may use the corresponding cellular chain complex for the cell decomposition of the universal cover. In particular, in this model $$C^k(X;F) = \text{Hom}_{\pi_1 X}(C_k(\widetilde X;\Bbb Z), A) = 0 \;\;\;\;\;\; \text{for } k>n.$$

Therefore the cohomology groups $H^k(X;F)$ vanish for $k > n$.

  • $\begingroup$ I think a Cech-to-sheaf cohomology spectral sequence for a sufficiently well-chosen open cover proves the desired result as well, so long as the cohomology of $\mathcal F$ vanishes in positive degrees on sufficiently small contractible neighborhoods of any point. This is not true for arbitrary sheaves, and is for locally constant sheaves. Maybe constructible is enough? $\endgroup$ – user98602 Jan 12 at 23:22

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