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Let $X$ be a compact complex Kahler manifold with first real Chern class $c_1 = 0$. Consider a family $\pi : \mathcal X \to \Delta$ over the unit disc in $\mathbb C$, where the fibers $X_s$ are compact Kahler with $c_1(X_s) = 0$ for $s \not= 0$. Do we know that the central fiber $X_0$ is Kahler with $c_1 = 0$?

Some remarks:

a) The condition on the Chern class is topological, so the question is really if the central fiber is Kahler.

b) This is true for complex tori and K3 surfaces, though for K3 surfaces being Kahler is a consequence of the topological condition of having even first Betti number.

c) Kuranishi constructed an example of non-Kahler deformations of projective manifolds, but the manifolds in question were not Kahler-Einstein so that example does not apply here.

d) There exist non-Kahler compact complex manifolds with $c_1 = 0$, like the Iwasawa manifold. The Iwasawa manifold does not have the right first Betti number to provide a counterexample to the question. However, I've heard physicists have found many examples of non-Kahler manifolds of Calabi-Yau type, and maybe one of those does at least not have topological obstructions to being a counterexample?

[edit] Two more remarks:

e) For the special case of Calabi-Yau manifolds, Popovici gives that the central fiber is Moishezon. One could hope that a Moishezon manifold with the Hodge numbers of a Calabi-Yau manifold is Kahler, but this is false by an example of Oguiso. His example is however not homeomorphic to a projective manifold, leaving the question open.

f) One could try looking at Calabi-Yau threefolds, of which many examples are apparently known (I only know of complete intersections of appropriate degree in projective space). The case of a Calabi-Yau hypersurface in $\mathbb P^4$ is uninteresting, as they are rigid, so the central fiber is isomorphic to the general fiber.

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1 Answer 1

up vote 11 down vote accepted

There are counterexamples: a Moishezon manifold, which has trivial canonical class and is birationally equivalent to a hyperkahler manifold, is also deformationally equivalent to a hyperkaehler manifold (this is a result of Huybrechts, I am not sure if he stated it in this generality, but his proof certainly works). Such Moishezon manifolds can be non-Kaehler (see e.g. Periods of Enriques Manifolds, Keiji Oguiso, Stefan Schroeer, http://arxiv.org/abs/1010.0820, Proposition 6.2).

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Excellent! (Well, not really, but you know...) –  Gunnar Magnusson Mar 3 '11 at 18:58

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