# Deform a compact Kähler manifold to a non Kähler one

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).

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Apparently, in this paper (jstor.org/pss/1970426) 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