Why is the Hodge-Deligne polynomial a polynomial? 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?
 A: 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)$]
A: 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.
