# Sheaf Cohomology on a Stone Space

Let $X$ be a stone space, i.e. a compact, totally disconnected hausdorff space. Then $H^1(X,\mathbb{Z}/2)=0$. Here's one way of proving this: $X$ with $\mathbb{Z}/2$ (the constant sheaf) is an affine scheme, now use the vanishing result for quasicoherent sheaves on affine schemes.

What happens if we also allow locally compact, totally disconnected hausdorff spaces? Then $X$ with $\mathbb{Z}/2$ is again a scheme, but it is not affine (unless $X$ is compact). A typical example would be an open subset of a stone space (actually this is generic), or the underlying topological space of a local field.

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Consider Aleph_1 with its order topology. Do you know how to show that H^1(Aleph_1,Z/2)=0? I would actually bet it's non-trivial. – André Henriques Jun 4 '10 at 13:47
I think $\aleph_1$ is a disjoint union of compact open subsets. Thus the cohomology vanishes. – Martin Brandenburg Jun 9 '10 at 23:47

A search brought up this paper by R. Wiegand. This article by the same author might also be of interest.

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Theorem 4.1 in the first paper is very useful. And in the second paper, the remark after Prop. 5.3 provides a counterexample for vanishing $H^1(X;\mathbb{Z}/2)$. – Martin Brandenburg Jun 4 '10 at 14:31

The vanishing theorem holds under reasonable topological hypetheses, say when we have a totally disconnected topological space $X$ that is paracompact, Hausdorff, locally compact and has a countable base (e.g. the $p$-adic rationals).

In that case the space is regular (since it is Hausdorff and locally compact) and hence satisfies the Lindelof condition. Then for each closed $A$ and $B$ with empty intersection there is a clopen $U$ such that $A\subset U\subset X\setminus B$ (see e.g. theorem 6.2.7. in Engelking, General topology). So the constant sheaf with stalk $\mathbf{Z}$ or $\mathbf{Z}/2$ is soft and (using the paracompactness again) its higher cohomology vanishes.

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interesting. what are the ring-theoretic properties of the corresponding boolean ring $C_0(X,\mathbb{Z}/2)$ (without unit), corresponding to your topological assumptions? – Martin Brandenburg Jun 5 '10 at 6:46
Martin -- no idea, sorry! This came up as a weird example in a course I took a long time ago and luckily I still have the notes. – algori Jun 5 '10 at 14:30
Ok I will ask a new question. – Martin Brandenburg Jun 9 '10 at 15:52