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Could you give me an example of a compact Kähler manifold which analytically deforms to a non Kähler one?

For example, there is no hope to find a complex structure on a Hopf manifold in order to make it Kähler because of topological obstructions (the second Betti number is zero).

For instance, I think that the Iwasawa manifold should not have topological obstructions.

Of course, the algebraic counterpart of my question has an affirmative answer: complex tori of dimension greater than one give examples of manifolds that can be analytically deformed from an algebraic one to a non algebraic one (but still Kähler).

Thanks in advance!

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Apparently, in this paper ( Hironaka constructs such a thing. I've still not read the paper or worked through the construction, but I keep seeing this given as a reference whenever people say that large deformations of Kahler manifolds may not be Kahler. – Gunnar Þór Magnússon Dec 1 '10 at 21:41
Thanks little brother, I had just found it! I haven't read the paper neither... Apparently he proves more: he constructs a family $V_t$ of 3-dimensional abstract non-singular algebraic varieties such that $V_t$ is projective for $t\ne 0$ but $V_0$ carries a positive $1$-cycle, algebraically equivalent to zero. – diverietti Dec 1 '10 at 22:08
In the same paper Hironaka says that the same question is open for a family of 2-dimensional manifold. By a paper of our big brother Dan Popovici, such an example cannot be of algebraic nature, since he proved that in a holomorphic family where all fibers are projective except at most the central one, all fibers are in fact projective. – diverietti Dec 1 '10 at 22:35
And you want more? Are you looking for a family of non-algebraic Kahler manifolds that deform into a non-Kahler manifold? That might be tricky... Voisin has examples of Kahler manifolds that have the wrong topology to be algebraic, perhaps that's a place to start looking? – Gunnar Þór Magnússon Dec 2 '10 at 7:56
The surface case is known: a compact surface is Kahler iff its first Betti number is even (see f.ex. Buchdahl or Lamari in the Annals de l'Institut Fourier). This is a topological condition, so any deformation of a Kahler surface over a connected base is Kahler. So if you want to find non-algebraic Kahler manifolds that deform into something non-Kahler you have to start looking in dimension $\geq 3$, where we have no idea of the necessary and sufficient conditions to be Kahler. – Gunnar Þór Magnússon Dec 2 '10 at 9:47

(Just so this question has an answer. All manifolds are compact.)

In dimension one every deformation of Kähler manifolds is Kähler because every Riemann surface is Kähler.

In dimension two the same is true but for less trivial reasons. A two-dimensional complex manifold is Kähler if and only if its first Betti number is even, which is purely a topological condition. The result then follows from the fact that the fibres of a deformation are diffeomorphic.

In dimension three, the result is no longer true; that is, there is a deformation of Kähler manifolds such that the central fibre is not Kähler. As Gunnar Magnusson pointed out in the comments, Hironaka gave an example of such a deformation in his paper An Example of a Non-Kählerian Complex-Analytic Deformation of Kählerian Complex Structures. In fact, the construction used gives rise to a whole host of interesting phenomena in dimension three including the existence of a Moishezon manifold which is not projective. See the associated page on Wikipedia for more details.

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