2
$\begingroup$

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).

$\endgroup$
1
$\begingroup$

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 $k > n$.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Mike Miller Jan 12 at 23:22

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.