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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?

t.b.
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phil
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  • If you want a line break, simply hit return twice, or use the html <br> for that. Only stuff enclosed in $-signs counts as LaTeX here. For the rest you have html or markdown. – t.b. Aug 24 '11 at 14:02
  • Polynomial in which variables? – unkkk Aug 24 '11 at 15:35
  • @unkkk: I have converted your answer to a comment. Answers should be reserved for posts that answer the question. But because you do not have 50 reputation points yet, you can only comment on your own questions and answers. So, you didn't do anything wrong; the "add comment" button will only appear for you once you gain 50 points. Here is an explanation of reputation points. – Zev Chonoles Aug 24 '11 at 22:05
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    What you've written to me looks like a number, not a polynomial. – Qiaochu Yuan Aug 28 '11 at 04:48
  • @Qiaochu I think phil meant $\sum_{p,q}(-1)^{p+q}h^{p,q}(X)u^pv^q$ as a polynomial in $u$ and $v$. Anyway, the question can be easily stated (and essentially is stated) without ever referring to the Hodge-Deligne polynomial. – Alex B. Aug 28 '11 at 06:14
  • I am not even sure whether the OP is still 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. – Alex B. Aug 28 '11 at 06:16

2 Answers2

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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)$]

Georges Elencwajg
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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.

Alex B.
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