Given a sheaf on say a manifold, such that all its stalks are isomorphic to a fixed finite dimensional vector-space. Is it true that the sheaf is a local system?
Criterion for local systems
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sheaf-theory
1 Answers
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No. Let $M$ be your manifold and $p\in M$. Let $V$ be a finite-dimensional vector space. The sheaf $S$ is defined as follows: for any connected open subset $U\subset M$, $S(U)=V$. (for not-connected $U$ we would then get a direct product of copies of $V$ - one for each component). If U'\subset U is a connected open subset of a connected open $U$, the restriction map is as follows: if $p\in U$ and p\notin U' then S(U)\to S(U') is the zero map, otherwise it's the identity.
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0Thanks, your sheaf is even constructible right? – 2011-05-05