# Singular and Sheaf Cohomology

Let X = complex manifold of dimension n. Thus, it's a real manifold of dimension 2n.

Now cohomology is a topological concept so it should not depend upon the structure given on a topological space.

We know that k'th Singular cohomology of X is 0 for k > 2n. We can also define a sheaf cohomology on that space using derived functor approach of Grothendieck. Then (by a result of Grothendieck) we know that k'th sheaf cohomology is 0 for k > n.

Now, for constant sheaves [say R], the sheaf cohomology agrees with singular cohomology [with coefficient R]. Does this means that even the k'th singular cohomology of X vanishes for k > n ??

[Edited] I now feel that the result which says that sheaf and singular agrees is actually this that k'th sheaf cohomology [of a complex manifold and constant sheaf] will agree with 2k'th singular cohomology [of the underlying real manifold]. Is this correct?? I would still like others to comment.

Thanks.

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@user1767: I assume you're referring to the theorem of Grothendieck in Hartshorne III.2? That's true for the combinatorial dimension of a noetherian space, which is not the same as the dimension of a manifold (indeed, a manifold is not a noetherian space). –  Akhil Mathew Sep 7 '10 at 11:46
@Akhil, Thanks, that helped. I guess there is no corresponding vanishing theorem for sheaf cohomology on a manifold. –  Amit Sep 7 '10 at 13:13

The cohomological dimension of a real $n$-manifold $M$ is $n$: this means that $H^i(M,\mathscr F)=0$ for each sheaf $\mathscr F$ of abelian groups on $M$ if $i>n$, and that there exist sheaves $\mathcal F$ on $M$ with $H^n(M,\mathscr F)\neq0$. You'll find this proved in Bredon's book on Sheaf theory, §II.16.

It follows that the cohomological dimension of a complex $n$-manifold is $2n$. For example, you reach the maximum, at least for compact ones, for the constant sheaf $\mathbb R$.

The answer to your «Does this means that even the k'th singular cohomology of X vanishes for k > n??» question is No (You can answer it without determining the cohomological dimension: just consider a compact complex $n$-manifold, which is automatically oriented: what is it $2n$-th cohomology group?)

The question in you [Edited] paragraph also has a negative answer. Consider examples to see that it is so.

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Thank you, this clears the picture. Unfortunately, i cant upvote you now [not sufficient reputations]. –  Amit Sep 7 '10 at 14:49
On the other hand, if you have an irreducible variety with the Zariski topology, then sheaf cohomology $H^i$ of the constant sheaf is zero for $i > 0$, because the constant sheaf is flasque. (I'm not sure -- does it make sense to talk about Zariski topology on a (non-algebraic) complex manifold?)
Further remarks: If what you want is to be able to deal with singular cohomology in a more "algebraic" way, you can do so by using the Hodge decomposition $H^n(X;\mathbb{C}) = \bigoplus_{p+q=n}H^q(X,\Omega^p)$ which expresses singular cohomology in terms of sheaf cohomologies of $\Omega^p$'s. This works if your $X$ is a compact Kaehler manifold, e.g., a smooth projective variety. If your $X$ is an algebraic variety, you can also use etale cohomology for an "algebraic" way of dealing with singular cohomology; see Milne's book/notes on etale cohomology for more.
I don't know if this will be helpful but -- another way to express Hodge decomposition for compact Kaehler manifolds is "Every cohomology class (coeffs in $\mathbb{C}$) has a unique $\overline{\partial}$-harmonic representative". There is an analogous statement for compact Riemannian manifolds, namely: "Every cohomology class (coeffs in $\mathbb{R}$) has a unique $d$-harmonic representative", where $d$ is the usual de Rham derivative. –  Kevin H. Lin Sep 8 '10 at 17:16