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The question is in the title. A more precise formulation is:

Let $X$ be a topological space. When does $H^i(X,F) = 0$ for all $i > 0$ and all abelian sheaves $F$ on $X$?

The obvious example is a discrete space. I'd be happy with a characterization of compact Hausdorff topological spaces $X$ satisfying the above property.

Edit: Following Georges Elencwajg's answer, I would like to clarify that these spaces will be quite pathological from the viewpoint of classical topology. Nevertheless, I do not know a single example which does satisfy the above vanishing property and is not discrete. For example, does the Cantor set have this property?

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    $\begingroup$ Before the Cantor set, you might want to look at the one point compactification of $\mathbb N$. $\endgroup$ – André Henriques Apr 13 '13 at 17:59
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    $\begingroup$ If every open cover of $X$ has a refinement consisting of disjoint open sets, then every (locally) surjective map of sheaves on $X$ is surjective on global sections, so that higher sheaf cohomology is trivial. $\endgroup$ – Tom Goodwillie Apr 13 '13 at 19:56
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    $\begingroup$ A compact Hausdorff space has the property you want if and only if it is totally disconnected. See theorem II.16.21 Bredon, "Sheaf theory" (second edition, GTM 170). Examples of compact totally disconnected spaces are the Cantor set and $\mathbb{Z}_p$. $\endgroup$ – user31960 Apr 13 '13 at 21:47
  • $\begingroup$ Thanks! If you write this as an answer, I would be happy to accept it. $\endgroup$ – anon Apr 13 '13 at 23:38
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    $\begingroup$ (The Cantor set is homeomorphic to $\mathbb Z_p$.) $\endgroup$ – Tom Goodwillie Apr 14 '13 at 0:27
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0) I would guess that the compact spaces you are looking for are extremely rare.

1) For example the extremely simple contractible space $I=[0,1]$ is not suitable:
Consider the inclusion $j\colon U=(0,1)\hookrightarrow I $ and take on $I$ the sheaf$j_!(\mathbb Z_U)$, the constant sheaf $\mathbb Z_U$ on $U$ extended to $I$ by zero.
Claim: $H^1(I,F)\cong\mathbb Z$
Proof of claim:
Consider the open embedding $i:F=I\setminus U\hookrightarrow X$ and the short exact sequence of sheaves on $I$ (see Hartshorne's Algebraic Geometry, Exercise I.19, page 68) : $$0\to j_!\mathbb Z_U\to \mathbb Z_I\to i_*\mathbb Z_F\to 0$$ Taking the corresponding long exact sequence in cohomology we get the fragment $$0\to \Gamma(I,j_!\mathbb Z_U) \to \Gamma(I,\mathbb Z_I)\to \Gamma(I,i_*\mathbb Z_F)\to H^1(I,j_!\mathbb Z_U) \to H^1(I,\mathbb Z_I) $$ Since $\Gamma(I,j_!\mathbb Z_U)=0$ and $H^1(I,\mathbb Z_I)=H^1_{\operatorname {singular}}(I,\mathbb Z)=0$ the above fragment becomes $$0\to 0\to \mathbb Z \to \mathbb Z^2\to H^1(I,j_!\mathbb Z_U) \to 0 $$ so that $H^1(I,j_!\mathbb Z_U)\cong \mathbb Z\neq 0$

2) There is a very similar statement in scheme theory saying that $H^1(\mathbb A^1_k,j_!(\mathbb Z_U))=\mathbb Z$, where now $U$ is the complement of two rational points on the affine line $\mathbb A^1_k$: see Hartshorne's Algebraic Geometry, Exercise III 2.1

3) Of course on afffine schemes, quasi-coherent sheaves have zero cohomology in positive degree, but that is not a purely topological statement and as shown in the example 2) above does not apply to arbitrary sheaves of abelian groups: even theorems by Serre necessitate some hypotheses!

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  • $\begingroup$ Thanks for these examples! I should have said that I realize these examples will be "pathological" (and typically very disconnected), and was hoping for some result along the lines that there aren't very many of them besides discrete sets. I will clarify this in the question. $\endgroup$ – anon Apr 13 '13 at 17:30
  • $\begingroup$ I have edited my answer which now contains a complete proof of the claim in 1) $\endgroup$ – Georges Elencwajg Nov 23 '16 at 13:23

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