# Non-zero sheaf cohomology

Let $\mathbb{R}$ denote the real line with its usual topology. Does there exist a sheaf $F$ of abelian groups on $\mathbb{R}$ whose second cohomology group $H^{2}\left(\mathbb{R},F\right)$ is non-zero? What about $H^{j}\left(\mathbb{R},F\right)$ for integers $j\ge 2$ ?

(Here cohomology means derived functor cohomology as in, say, Hartshorne or EGA. Anyway this cohomology coincides with Cech cohomology since $\mathbb{R}$ is paracompact.)

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What is R here? –  Ben Webster Oct 11 '09 at 13:28
@Ben- My guess: R is a commutative ring and the sheaf is on Spec(R). –  Anton Geraschenko Oct 11 '09 at 14:23
Your question would certainly benefit from more info – what is R and what type of sheaf cohomology you are considering? –  Ilya Nikokoshev Oct 11 '09 at 14:38
Now I agree with ilya, R is probably the real line. –  Anton Geraschenko Oct 11 '09 at 14:43
Yes, R is the real line with its usual topology. Cohomology is derived functor cohomology for the functor "Global Sections" (the cohomology used in Hartshorne or EGA, defined via injective resolutions) . It coincides with Cech cohomology since R is paracompact. @ilya n. Could you please explain how the vanishing of cohomology in dimensions 2 and larger is a simple consequence of the definition of Cech cohomology ? Of course I do not assume that my sheaf is constant (in which case even the first cohomology group would vanish). Thanks to all for your interest. –  Georges Elencwajg Oct 11 '09 at 18:01

This holds true in much more generality actually: One has $H^i(X,F) = 0$ for $i > n$ where $n$ is the Lebesgue covering dimension of $X$, for every paracompact Hausdorff space $X$. For a proof see the book of Godement, II.5.12 –  Nicolas Schmidt May 1 at 19:40
Since now we know that R in your question refers to real line equipped with standard topology, sheaf cohomology will always have H^i(F) = 0 for i>1 — depending on how you define sheaf cohomology this is a theorem of different difficulty.