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Let $X$ be a 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 $k \gt 2n$. 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 $k \gt n$.

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 $k \gt n$?

[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.


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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
up vote 14 down vote accepted

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

The statement is the following: If you take a variety with the analytic topology (or a complex manifold), then yes, sheaf cohomology of the constant sheaf agrees with singular cohomology. More generally, sheaf cohomology of a constant sheaf on a locally contractible space agrees with singular cohomology of that space. I think there's an explanation of this somewhere in Warner's book "Foundations of Differentiable Manifolds and Lie Groups".

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.

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 Kähler 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.

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Thanks Kevin. Hodge decomposition gives rise to another question. We are using complex coefficients for singular cohomology [on the left side of the expression] because X is a complex manifold. Right? Does similar decomposition exists for say real manifold [where we can use R coefficients for singular cohomology] ?? – Amit Sep 8 '10 at 9:12
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
It is helpful, thanks again. – Amit Sep 11 '10 at 11:07
Doesn't one need to add paracompactness for the claim in your first paragraph? (The long line is a counterexample otherwise, iirc) – Mariano Suárez-Alvarez Jun 13 '12 at 20:01

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