Basic question: irregularity and geometric genus in Hodge theory Please could you give me a reference to the theorems that give rise to the following equations which are taken from page 8 and 9 of Friedman' book on holomorphic vector bundles and algebraic surfaces (${X}$ is a compact complex surface):
$b_1(X)=2q(X)$
$b_2(X)=2p_g(X)+h^{1,1}(X)$
$b_2^+(X)=2p_g(X)+1$
where
$q(X)=dim_{\mathbb{C}}H^0(X;\Omega_X^1)=dim_{\mathcal{C}}H^1(X;\mathcal{O}_X)$
$p_g(X)=dim_{\mathbb{C}}H^0(X;\Omega_X^2)=dim_{\mathbb{C}}H^2(X;\Omega_X)$
Can I ask also for a place to read about it please? Thank you very much.
 A: A good place to read about everything I'm about to say is in Huybrechts, citation 2 at the link.
It's important to note that $X$ here is not just a compact complex surface: these identities need not hold in that generality. $X$ is, rather, an algebraic surface, that is, one which embeds into projective space. In particular, $X$ receives a Kähler structure from this embedding, which is the only reason we can apply Hodge theory as we are about to do.
The sheaf or Čech cohomology groups $H^p(X;\Omega^q_X)$ are isomorphic to the Dolbeault cohomology groups $H^{p,q}(X)$, where $H^{p,q}$ is the $q$th cohomology of the complex of complex-valued differential forms of degree $(p,q)$ with differential $\bar\partial$. This is Dolbeault's theorem.
Real Hodge theory gives a unique harmonic form for each de Rham cohomology class; Kähler Hodge theory gives a unique harmonic form for each Dolbeault cohomology class; and so on a Kähler manifold, such as our $X$, there's a decomposition of singular cohomology $H^k(X)$ into the direct sum of Dolbeault cohomology, $\oplus_{p+q=k} H^{p,q}(X)$ (which is independent of the Kähler metric.) 
In particular $H^1(X)=H^{0,1}(X)\oplus H^{1,0}(X)$. Serre duality also implies that $H^{p,q}(X)\cong H^{q,p}(X) $, which yields your first identity. The second identity is proved in the same way, via the decomposition $H^2=H^{0,2}\oplus H^{1,1}\oplus H^{2,0}$.
The last identity follows from the Hodge index theorem, although it's not the statement from general algebraic geometry you'll most often see-it's likely you'll find this version described in Huybrechts.
