$\dim_k H^1(X, \Omega_X)$, $X \subset \mathbb{P}^3$ a projective surface Is it possible to calculate $\dim_k H^1(X, \Omega_X)$, for $X \subseteq \mathbb{P}_\mathbb{C}^3$ a smooth projective surface of degree $d$, without using Chern classes?
I've been trying to do this by playing around with the long exact sequences in cohomology associated to the Euler sequence and the conormal sequence
$$0 \to \mathcal{O}_X(-d) \to i^\ast\Omega_{\mathbb{P}^3} \to \Omega_X \to 0$$
 for $i: X \hookrightarrow \mathbb{P}^3$.  I don't really understand Chern classes very well.  Mainly I guess I want to know whether I need them to solve this problem.
(This is homework: I was assigned to calculate the Hodge diamond of a complex projective surface, and I have the other Hodge numbers.)
 A: If you know the other Hodge numbers, it suffices to compute the Euler characteristic. You can compute this using a mild generalization of Riemann-Hurwitz (which unfortunately is slightly annoying to state and which I don't know how to prove quickly) and looking at a suitable projection. 
Edit: Some details. Recall that all projective hypersurfaces in $\mathbb{P}^3$ of the same degree are diffeomorphic. Hence it suffices to compute the Euler characteristic of the Fermat hypersurface $X^d + Y^d + Z^d + W^d = 0$. Project onto the first two coordinates (you need to delete some points for this to be well-defined but this has a predictable effect on the Euler characteristic; here and possibly at other points in this argument I'm secretly using the fact that Euler characteristic and compactly supported Euler characteristic agree for complex varieties). This gives us a map to $\mathbb{P}^1$ with generic fiber curves of the form $(1 + \lambda) X^d + Z^d + W^d = 0$ minus $d$ points, whose Euler characteristic you can compute from the genus-degree formula. At $d$ points $(X : Y) = (1 : \zeta_d^i)$ there are special fibers, and as I said above, a mild generalization of Riemann-Hurwitz will help you deal with these (cut stuff up and glue it back together appropriately). The final Euler characteristic you get should be
$$\chi = d^3 - 4d^2 + 6d.$$
