# Diffeomorphic Kähler manifolds with different Hodge numbers

This question made me wonder about the following:

Are there orientedly diffeomorphic Kähler manifolds with different Hodge numbers?

It seems that this would require that those manifolds are not deformation equivalent. However, there are examples by Catanese and Manetti that that happens already for smooth projective surfaces.

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"deformation invariant" = "deformation equivalent"? –  Kevin H. Lin Oct 19 '10 at 7:26
Kevin: yes, of course. Thanks for catching that. –  Sándor Kovács Oct 19 '10 at 7:29

This question was debated in another forum a few years ago. The result was a note by Frédéric Campana in which he describes a counterexample as a corollary of another construction. In 1986 Gang Xiao (An example of hyperelliptic surfaces with positive index Northeast. Math. J. 2 (1986), no. 3, 255–257.) found two simply connected complex surfaces $S$ and $S'$ (that is, complex dimension 2), with different Hodge numbers, that are homeomorphic by Freedman's classification. The homeomorphism has to be orientation-reversing, but $S \times S$ and $S' \times S'$ are orientedly diffeomorphic and of course still have different Hodge numbers. Freedman's difficult classification is not essential to the argument, because in 8 real dimensions you can use standard surgery theory to establish the diffeomorphism.

Campana also explains that Borel and Hirzebruch found the first counterexample in 1959, in 5 complex dimensions.

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Greg, thanks for the great answer. Did you mean to say "still have different Hodge numbers" above instead of "equal"? –  Sándor Kovács Oct 19 '10 at 7:28
Yes, you are right, they still have different Hodge numbers. –  Greg Kuperberg Oct 19 '10 at 7:30
S and S' are homeomorphic but not diffeomorphic? –  Kevin H. Lin Oct 19 '10 at 7:38
No because Campana's point is that by taking products things become (mysteriously) diffeomorphic, Note that for surfaces, Hodge numbers can be recovered from the first Betti number and first Pontryagin class by Hirzebruch. So there is no counterexample in dimension < 3. –  Donu Arapura Oct 19 '10 at 8:10
Donu, what's mysterious is that $S$ and $S'$ are not diffeomorphic. There is no obstruction to a diffeomorphism between them using ordinary topological invariants. They fail to be diffeomorphic because 4 dimensions is too low to create the diffeomorphism, but on the other hand high enough to create a problem. –  Greg Kuperberg Oct 19 '10 at 8:14