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Let X be a topological space, let $\mathcal{U} = \{U_i\}$ be a cover of X, and let $\mathcal{F}$ be a sheaf of abelian groups on X. If X is separated, each $U_i$ is affine, and $\mathcal{F}$ is quasi-coherent, then Cech cohomology computes derived functor cohomology; in general one only gets a spectral sequence $$ H^p(\mathcal{U},\underline{H}^q(\mathcal{F})) \Rightarrow H^{p+q}(X,\mathcal{F}) $$ where $\underline{H}^q(\mathcal{F})$ is the presheaf $U \mapsto H^q(U,\mathcal{F}|_U)$.

Question: For q > 0, $\underline{H}^q(\mathcal{F})$ sheafifies to 0.

For a quasi-coherent sheaf $\mathcal{F}$ this is clear because cohomology vanishes on affines. Is this really true in general? Brian Conrad states this in the introduction to his notes on cohomological descent.

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Yes, this is true in general.

It suffices to show the stalks vanish. Pick $x \in X$ and take an injective resolution $0 \to {\cal F} \to I^0 \to \cdots$. For any open $U$ containing $x$, we get a chain complex

$$0 \to I^0(U) \to I^1(U) \to \cdots$$

whose cohomology groups are $H^p(U,{\cal F}|_U)$.

Taking direct limits of these sections gives the chain complex

$$0 \to I^0_x \to I^1_x \to \cdots$$

of stalks, which has zero cohomology in positive degrees because the original complex was a resolution. However, direct limits are exact and so we find

$$0 = {\rm colim}_{x \in U} H^p(U,{\cal F}|_U) = {\underline H}^p({\cal F})_x$$

as desired.

Generally, cohomology tells you the obstructions to patching local solutions into global solutions, and this says that locally those obstructions vanish.

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Here's another short proof: denote by $I,J$ the inclusion of sheaves on X into presheaves and sheafification respectively, then

$$ \underline{H}^p(\mathcal{F})\cong R^pI(\mathcal{F}). $$

Since $J$ is exact,

$$ J\circ R^pI\cong R^p(J\circ I) $$

and the later vanishes for $p>0$ as $J\circ I=id_{\mathfrak{Ab}(X)}$. So $\underline{H}^p(\mathcal{F})$ sheafifies to 0 for $p>0$.

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  • $\begingroup$ This is an interesting approach, but it relies a lot on understanding derived functors. The longer proof reveals more, so to say. $\endgroup$ – Bombyx mori Apr 3 '17 at 19:04
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    $\begingroup$ @Bombyxmori Depends on point of view. For me this proof reveals the conceptual reason behind the fact, which I could not really see in the longer one. $\endgroup$ – მამუკა ჯიბლაძე Apr 3 '17 at 20:14
  • $\begingroup$ @მამუკაჯიბლაძე: I agree. Upvoted. $\endgroup$ – Bombyx mori Oct 31 '17 at 22:36
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I do not think that sheaves of abelian groups need to be locally acyclic. Let me say what I mean in an example. Take $X=\mathbb{C}^{2}$ with the classical (metric) topology. Let $\mathcal{F} = \mathbb{Z}_{D}$ where $D=\mathbb{C}^{\times} \times \{0\}$.

Then for any arbitrarily small polydisk $U$ containing $(0,0)$ we have $H^{1}(U,\mathbb{Z}_{D}) = H^{1}(U \cap D, \mathbb{Z})$

is not vanishing. It seems in this example that $\underline{H}^{1}(\mathcal{F})_{(0,0)}$ does not vanish.

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  • $\begingroup$ Oh, I could have just used $\mathbb{C}^{\times}$ in $\mathbb{C}$ $\endgroup$ – Oren Ben-Bassat Mar 28 '11 at 10:11
  • $\begingroup$ Just to clarify, the assumptions of locally affine and quasi coherence are crucial. The simple example of the push forward of the constant sheaf $\mathbb{Z}$ under the inclusion of $\mathbb{C}^{\times}$ into $\mathbb{C}$ gives a sheaf of abelian groups which is not locally acyclic. $\endgroup$ – Oren Ben-Bassat Nov 1 '11 at 17:33
  • $\begingroup$ Why should we have $H^{1}(U,\mathbb{Z}_{D}) = H^{1}(U \cap D, \mathbb{Z})$ ? If $X=\mathbb{C}$ and $D=\mathbb{C}^\times$, then $\mathbb{Z}_{D}$ fits into an exact sequence of sheaves on $\mathbb{C}$: $0\rightarrow \mathbb{Z}_{D}\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}_{0}\rightarrow 0$, where the latter two are the constant sheaf resp. the skyscraper sheaf at $0$ in $\mathbb{C}$. The sequence remains exact, when computing global sections on any open disk $U$ containing $0$, and since $H^1(U,\mathbb{Z})=0$, the long exact sequence implies $H^1(U,\mathbb{Z}_D)=0$. Or not? $\endgroup$ – user_1789 Dec 30 '16 at 13:26

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