cohomology of T_X on K3 surfaces

I'm starting to study K3 surfaces, do you know some books on the subject which i could read?

In particular i need the proofs that for a complex k3 surface $X$ it is $H^0(X,T_X)=H^2(X,T_X)=0$ and $h^1(X,T_X)=20$ (i consider the Dolbeault cohomology groups)

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There are proofs that $H^0(X, T_X)=0$ directly, but I think they are all fairly major theorems. For example, see Nygaard's paper " A $p$-adic proof of the non-existence of vector fields on K3 surfaces." One reason for starting with this, is that having enough $0$'s in the $E_1$-page of the Hodge-de Rham spectral sequence makes it obvious that it degenerates, and then you can read off the other Hodge numbers.

On the other hand, as long as your definition of K3 surface involves being a smooth, proper, algebraic variety over $\mathbb{C}$, then we can start with the fact that the Hodge-de Rham spectral sequence degenerates at $E_1$. You could either take this as a result from Hodge theory, or (better in my opinion) the result from Illusie.

Since $\omega\simeq \mathcal{O}_X$, the Hodge numbers you care about are $h^{1,0}=h^{1,2}=0$ and $h^{1,1}=20$. By assumption $H^1(X,\mathcal{O}_X)=0$, so by Hodge symmetry $h^{1,0}=h^{0,1}=0$ and you get the first one. Again, by Hodge symmetry $h^{1,2}=h^{2,1}=h^1(X, \omega)=h^1(X, \mathcal{O}_X)=0$ and you get the second one.

The $H^2_{dR}(X/\mathbb{C})$ part is a little more subtle. At this point we can fill in the entire Hodge diamond except $s:=h^{1,1}$ (use the above + Serre duality + connected):

$\begin{matrix} & & 1 & & \\ & 0 & & 0 & \\ 1 & & s & & 1 \\ & 0 & & 0 & \\ & & 1 & & \\ \end{matrix}$

By Noether's formula $c_1^2+c_2=12\chi$. But $c_1^2=K_X\cdot K_X=0$ and $\chi=2$, so $c_2=24$

We now use that $c_2=\sum (-1)^nh^n_{dR}(X/\mathbb{C})$. Thus $24=s+4$ and hence $s=20$.

(Some of this can be skipped if you use Example 15.2.2 in Fulton's Intersection Theory since it is written for surfaces $\chi (\mathcal{O}_X)=1/12(K\cdot K+\chi)$ without needing to worry about interpreting Chern classes).

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