TITLE: Two approaches to compute the signature of a Kaehler manifold
QUESTION [11 upvotes]: Given a compact Kaehler manifold $M$ of complex dimension $2n$, there are essentially two ways to compute its signature $\sigma(M)$, i.e. the index of the intersection form on $H_{2n}(M,\mathbb{R})$:
1.by Hodge index theorem $\sigma(M)=\sum_{p,q}(-1)^p h^{p,q}$, here $h^{p,q}$ stands for the Hodge numbers.
2.by Hirzebruch signature theorem $\sigma(M)=L[M]$, here $L[M]$ stands for the $L$-genus, i.e. the characteristic number of the top $L$-class.This approach is more general since it works on any $4k$ dimensional real manifolds.
My questions are
1.Since these two approaches rest on different levels of cohomology theory, how are they interrelated? 
2.Of course, one possible way to answer Question 1 is to generalize both by the Hirzebruch-Riemann-Roch on Kaehler manifolds, a point already mentioned in Hirzebruch's Neue topologische Methoden. However, I am wondering if someone could relate these two approaches on a more fundamental level. 
To be precise,
Is there a formula to express the Chern numbers/Pontryagin numbers out of the Hodge numbers on a compact Kaehler manifold $M$ of complex dimension $n$? Surely it is the case for $c_n[M]$ interpreted as the Euler characteristic number. 
Or, does anyone know such counterexamples that two Kaehler manifolds(notably, Kaehler surfaces, I guess)  have the same Hodge numbers but different Chern numbers?
Many thanks!

REPLY [5 votes]: "counterexamples that two Kaehler manifolds have the same Hodge numbers but different Chern numbers?"
As you explained above, Chern numbers of surfaces can be expressed in terms of the Euler number and the first Pontrjagin number, so you need dimension at least 3 for a counterexample.
In dimension 3, consider a projective space and a smooth quadric threefold. These two have the same Hodge numbers, same $c_3 = 4$ (Euler number), same $c_1 c_2 = 24$ (by Todd's theorem), but distinct degrees $c_1^3$: for the projective space it equals $64$ and for quadric it is $54$.