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Let $X$ be a finite-dimensional compact metrizable space (these properties might partially be irrelevant; on the other hand, the case $X=[0,1]$ is already interesting to me).

Let $\mathcal F$ be a soft sheaf of $\mathbb Q$-vector spaces on the topology $\mathbb O(X)$ (softness means that restriction of global sections to closed subsets is surjective).

Assume that the dimension of $\mathcal F(X)$ is countable. Question: Is $\mathcal F$ necessarily a direct sum of skyscraper sheaves? If $\mathcal F(X)$ is finite-dimensional then this can be proved using induction on the dimension, I think.

Motivation: I have proved a classification result for certain continuous fields of $C^*$-algebras over $X$. The invariant takes values in soft sheaves (actually in flabby cosheaves, but that is "the same"). Now I am wondering about the range of the invariant.

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Are you sure it works when F(X) is finite-dimensional? I feel like if you take a non-principal ultrafilter on X, and define a sheaf by F(U) = Q when U is in the ultrafilter, and 0 otherwise, then F is flabby and hence soft, but is not a skyscraper sheaf. –  Kevin Ventullo Jun 25 '13 at 20:08
    
@KevinVentullo: Thank you for your comment. I thought I had convinced myself that phenomena like what you describe cannot occur. Here is the argument: let $\mathcal F$ be a soft sheaf of $\mathbb Q$-vector spaces with $\mathcal F(X)=\mathbb Q$. Since $\mathcal F\neq 0$, there must be some stalk $\mathcal F_x\neq 0$. Then the map $\mathcal F(X)\to\mathcal F_x$ must be an isomorphism because $\mathbb Q$ is simple. Since $X$ is Hausdorff, $\mathcal F(\{x,y\})\cong\mathcal F_x\oplus\mathcal F_y$ for $x\neq y$. By softness, we therefore must have $\mathcal F_y=0$ for $y\neq x$... –  Rasmus Bentmann Jun 26 '13 at 8:30
    
... Finally, a sheaf with only one non-vanishing stalk must be a skyscraper sheaf. What am I missing? –  Rasmus Bentmann Jun 26 '13 at 8:31
    
Hmm yeah that sounds right. I think what I defined isn't a sheaf, it doesn't satisfying the gluing axiom. –  Kevin Ventullo Jun 27 '13 at 3:45
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1 Answer

This appears to be a counterexample:

Let $Y\subseteq X$ be closed and totally disconnected, such as the Cantor set in $[0,1]$. Set $\mathcal F(U)=C(U\cap Y,\mathbb Q)$ -- locally constant functions on $U\cap Y$ with values in (discrete) $\mathbb Q$.

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