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I know that all fibers in a analytic fibration (proper, holomorpic) are homeomorphic, and if the fibers are Kählerian manifolds, then they have equal Hodge numbers.

Could it happen however that a manifold admits different Kählerian structures for which the Hodge numbers differ?

doetoe
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1 Answers1

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An Explicit Example. Let $\displaystyle\Sigma\equiv\mathbb{CP}^2\#8\overline{\mathbb{CP}^2}$ be the blow-up of complex projective plane $\mathbb{CP}^2$ at $8$ points in general position, this is a Del Pezzo surface (proof and definition), so it is an algebraic complex surface; in particular, it is a compact Kähler surface. By theorem 1 in D. Kotschick (1989) - On manifolds homeomorphic to $\displaystyle\mathbb{CP}^2\#8\overline{\mathbb{CP}^2}$, Invent. Math., 95 3, 591–600, $\Sigma$ is homeomorphic but not diffeomorphic to the Barlow surface, which is another compact Kähler surface.

Armando j18eos
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  • Thanks, this is interesting. However, if I'm not mistaken, they both have the same Hodge numbers... – doetoe Sep 20 '18 at 16:00
  • Yes, it is; but I don't know neither a proof nor a source for a proof. – Armando j18eos Sep 21 '18 at 07:53
  • They are both rational, so they can only differ in their $h^{1,1}$. Since they are homeomorphic, their Betti numbers $b_i$ are equal, and $b_2 = h^{0,2} + h^{1,1} + h^{2,0}$. – doetoe Sep 21 '18 at 09:19
  • Sorry, I misunderstood your question; I meant: I know the common Hodge diamond of these varieties, but I don't know neither a proof nor a source for a proof. – Armando j18eos Sep 21 '18 at 09:33