Hi. I am studying Hodge theory on Kahler manifolds. I have several questions.

1. Is Hodge number a topological invariant? (I mean, is it independent of the choice of Kahler structure?)

2. If the question 1 is true, then is there any variation formula of Hodge numbers on blowing up(down)? (along Kahler submanifolds) --- please let me know the reference.

3. I read Huybrechts's book "complex geometry". In there, Hodge-index theorem is explained in case of Kahler surfaces. Is there Hodge-index theorem in higher dimensional cases?

4. Is there a symplectic version of Hodge-Riemann bilnear relation?

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@3: The general signature formula is explained one page later in Huybrecht's book. – Heinrich Hartmann Oct 19 '10 at 0:26
Frédéric Campana's text iecn.u-nancy.fr/~gaillard/DIVERS/hodgenumbers.pdf might be related to the question. – Pierre-Yves Gaillard Oct 19 '10 at 7:28
Yes, the link to Greg Kuperberg's answer to a related question below points to the same preprint. – Sándor Kovács Oct 19 '10 at 8:04
Corrected link for Campana's text: iecn.u-nancy.fr/~gaillapy/DIVERS/hodgenumbers.pdf – Gunnar Þór Magnússon Dec 21 '11 at 14:47

Proposition. For compact complex manifolds of dimension $1$ (a.k.a. complex Riemann surfaces) the Hodge numbers are topological invariants.

Proof. The Hodge numbers are determined by the genus, which happens to be a topological invariant.

In a more general setting it is true that

Theorem. Let $f:X\to B$ be a family of complex manifolds and assume that $X_0$ is Kähler for some $0\in B$. Then for $b$ in a neighbourhood of $0$ in $B$, the Hodge numbers of $X_b$ are the same as the Hodge numbers of $X_0$.

Proof. See C.Voisin, Hodge Theory and Complex Algebraic Geometry, I, p.235.

Remark. It actually follows that all nearby fibers are Kähler, but it is not needed in the proof.

From this point I can only speculate: I don't think the Hodge numbers are topological invariants. If they were, then the assumption that one fiber is Kähler is superfluous. So, this suggests that it seems likely that there may be topologically equivalent complex manifolds with different Hodge numbers. I would even expect Kähler ones and perhaps even diffeomorphic ones, but that might be going out on a limb.

Catanese and Manetti gave various examples of orientedly diffeomorphic but not deformation equivalent smooth projective algebraic surfaces of general type. I wonder if those would perhaps have different Hodge numbers. The above theorem definitely does not apply. That in itself, of course, does not prove anything...

EDIT: (Added later) As pointed out by Greg Kuperberg here, this same question had been debated before here.

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Dear Sándor: The Hodge numbers of manifolds in a family are constant in a small neighborhood of the moduli of the given Kahler manifold. In general the Hodge numbers are only upper-semicontinuous as functions of the moduli, so a priori they could jump when deformed too much. I don't have an example at hand (I'll see if I can find one later), but I strongly suspect this falls into Murphy's law of deformation theory: anything that can go wrong, will go wrong in a properly chosen nicely behaved example. – Gunnar Þór Magnússon Oct 19 '10 at 5:50
Dear Gunnar: Yes, of course. Thanks. I was sloppy (and I am not Hodge theorist) and I will edit it. It may not have been clear, but what I was trying to say was that even though there are cases when the Hodge numbers are constant I can't believe they would be topological invariants. After all the miracle about the Hodge decomposition is that it connects topological information (the Betti numbers) to analytic information (the Hodge numbers). Anyway, this just adds to the argument that they are not topological invariants. Thanks! – Sándor Kovács Oct 19 '10 at 6:22
Dear Gunnar, it seems that it still follows that if all fibers of a family are Kähler, then the Hodge numbers are constant and a jump can happen only at non-Kähler points in the moduli space. Right? – Sándor Kovács Oct 19 '10 at 6:47
Sándor, I agree with you. If the base is connected, a locally constant function is constant. So for an everywhere Kähler family, there should be no problem. – Donu Arapura Oct 19 '10 at 8:44
@YCho: the upper semicontinuity of Hodge numbers is in for example C.Voisin, Hodge Theory and Complex Algebraic Geometry, I. I don't have the book with me right now, but it is at around p.230. I think there may even be a section named after this. As for being a topological invariant, I am not entirely sure how your proof would go, but it is not true. It is not even a differentiable invariant. See this post: mathoverflow.net/questions/42744/… – Sándor Kovács Oct 21 '10 at 6:54

For a Kaehler surface, the Hodge numbers are topological invariants. By Hodge Index Theorem, the signature of the Poincare pairing is equal to 2h^{2,0} +2 - h^{1,1}, hence h^{2,0} is a topological invariant, and for the rest of the numbers it is obvious.

For dimension 3, I think there are counterexamples.

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It depends on how you define Hodge Number, as far as I know you don't even need Kahler structure to define Hodge Number. See Griffiths-Harris pp 105, there the Hodge number is defined as the dimension of qth Cech Cohomology group of the sheaf of Holomorphic p-forms over the complex manifold M. However it is not dependent on the complex structure over M.

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Right, you don't need a K\"ahler structure just a comples structure to define Hodge numbers. Perhaps, the "not" in your last sentence is a typo. – Donu Arapura Oct 19 '10 at 8:37