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Let $X$ be a compact complex manifold. Its Hodge-Deligne polynomial is then defined to be $\sum_{p, q \geq 0} (-1)^{p+q} h^{p, q}(X)$ where $h^{p, q}(X):= \mbox{dim}_{\mathbb{C}}H^{p, q}(X)$.

The question now is: why is this a polynomial? More concretely I want to know the following:

  • Why are the numbers $h^{p, q}(X)$ always finite, or equivalently the vector spaces $H^{p, q}(X)$ finite-dimensional.

  • Why are the values $h^{p, q}(X)$ equal to $0$ if $p+q$ is large enough?

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    I am not even sure whether the OP $i$s st$i$ll around, since he was last seen on the day when he asked this question, he never commented on any of the answers, nor has he accepted an answer to either of his questions.2011-08-28

2 Answers 2

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Let $X$ be a compact complex manifold of dimension $n$. I won't assume $X$ Kähler since you don't (and also, I must confess that I like general theorems!). By definition $H^{p,q}(X)=H^q(X,\Omega ^p_X)$ and you can reason as follows.

a) Since $ \Omega ^p _X $ is the zero sheaf for $p\gt n$, you then obviously have in that case $H^q(X,\Omega ^p_X)=0 \:$ and so

$h^{p,q}(X)=0 \quad \text {for} \quad p\gt n$

b) Cartan-Serre proved in 1953 that the cohomology spaces $H^q(X,\mathcal F)$ of a coherent sheaf $\mathcal F$ on a compact manifold are all finite-dimensional, by putting the structure of a Fréchet space on the spaces $H^q(V,\mathcal F)$ for certain Stein open subsets $V\subset X$. Let me emphasize again that $X$ needn't be Kähler in their theorem: no Hodge theory is involved. So we have $ h^{p,q}(X)\lt \infty \quad \text {for} \quad p,q\geq 0 $

c) And now for the sting: Andreotti and Grauert proved in 1962 the incredible theorem that for a complex manifold $Y$ of dimension $n$, compact or not, the cohomology groups vanish above $n$ for all coherent sheaves: $H^q(Y,\mathcal F)=0$ for $q\gt n$. And actually it is even better if $Y$ is noncompact because then you also have $H^n(Y,\mathcal F)=0$ ! Anyway we have:

$ h^{p,q}(X)=0 \quad \text {for} \quad q\gt n $

[Beware that this does not follow from a) if $X$ is not assumed Kähler because then you don't necessarily have $ h^{p,q}(X)=h^{q,p}(X)$]

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The second part is easy: if the manifold is $n$-dimensional, then there are no non-zero differential forms of degree higher than $n$ (this follows almost immediately from the definition of a differential form). So the complex that defines the de Rham cohomology is just 0 beyond the $n$-th term.

The first part is more subtle. One way of proving finite-dimensionality for compact manifolds goes roughly as follows: one defines certain operators $\Delta$ on $\Omega^k(X)$ such that their kernel $\mathcal H_\Delta^k(X)=\{\alpha\in\Omega^k(X)\mid\Delta\alpha=0\}$ is naturally isomorphic to $H^k(X)$ (this isomorphism was proved by Hodge). Now, these operators $\Delta$ are elliptic and kernels of elliptic operators on compact manifolds are finite dimensional.