Let $X$ be a manifold such that $dim(X)=n$. It is well-know that if $\mathcal{F}$ is a coherent sheaf $H^m(X,\mathcal{F})=0$ for all $m >n$ (where I denote with $H(-)$ Cech cohomology). But is there a sheaf such that $H^m(X,\mathcal{F}) \ne 0$ for $m>n$?
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I'm not sure what you mean by "coherent sheaf", as that term is usually only used in the presence of something like a complex structure. But by this answer, the cohomology of any sheaf vanishes in degrees above $n$ on any topological $n$-manifold. Essentially, it can be shown that any open cover of an $n$-manifold admits a refinement for which all $(n+2)$-fold intersections are empty, so the Cech cohomology automatically vanishes above $n$.
answered Dec 3, 2013 at 22:14
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$\begingroup$ Dear @EricWofsey, How can I prove that any open cover of an $n$-manifold admits a refinement for which all $(n+2)$-fold intersections are empty? $\endgroup$ Commented Dec 4, 2013 at 10:19
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$\begingroup$ I don't know how the proof goes off the top of my head, but if you look for a reference on "covering dimension" you should be able to find it. $\endgroup$ Commented Dec 4, 2013 at 17:29