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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 $2n$? n$? Surely it is the case for$c_n[M]$interpreted as the Euler characteristic number. Or, does anyone know such counterexamples that two Kaehler manifolds(notably, Kaehler surfaces, I guess) have the same Hodge numbers but different Chern numbers? Many thanks! 9 added 88 characters in body; [made Community Wiki] Given a compact Kaehler manifold$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$2n$? 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! 8 edited tags; added 7 characters in body 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$2n\$?

Or, does anyone know such examples counterexamples that two Kaehler manifolds(notably, Kaehler surfaces, I guess) have same Hodge numbers but different Chern numbers?

Many thanks!

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