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Let $X$ be a compact complex non-Kähler manifold, then what conditions do we need to make it has a Kähler deformation? that is to say it can be deformed to a Kähler manifold.

Obviously not all the compact complex manifolds can be deformed to Kähler ones, for example, the Hopf surface, but certainly, there exist some non-Kähler manifolds which can be deformed to Kähler ones. For example, Hironaka has provided an example that except the central fiber, all the other fibers are projective manifolds, and the central fiber is a non-Kähler Moishezon manifold, so, conversely, for this Moishezon manifold, we can say it has a Kähler (even projective) deformation.

Then, are there any other examples of non-Kähler manifolds which has a Kähler deformation? or even a projective deformation? For example does a $\partial\bar\partial$-manifolds with trivial canonical bundle has a Kähler deformation? Has anyone think about it before? And what's the latest progress on it?

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  • $\begingroup$ By non-Kähler manifold, do you mean (1) a complex manifold which carries no Kähler structure, or (2) a complex manifold with a structure that happens to be non-Kähler? For example, if we consider $\mathbb{P}^1$ equipped with a Hermitian non-Kähler metric, this gives an example of (2), but not (1). Hopf surfaces give examples of (1) but not (2). $\endgroup$
    – AmorFati
    Commented Jul 24, 2021 at 21:53
  • $\begingroup$ @Very Confused, I mean (1), without Kähler structure, so $\mathbb P^1$ is always Kähler no matter which metric it choose. $\endgroup$
    – Tom
    Commented Jul 24, 2021 at 23:46
  • $\begingroup$ A deformation preserves Betti numbers (since all fibres will be diffeomorphic, in particular, homeomorphic). Kähler implies that the odd Betti numbers must be even, so you'll need to restrict your consideration to non-Kähler manifolds whose odd Betti numbers are even. $\endgroup$
    – AmorFati
    Commented Jul 25, 2021 at 2:40

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