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Jul 25, 2021 at 2:40 comment added AmorFati 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.
Jul 24, 2021 at 23:46 comment added Tom @Very Confused, I mean (1), without Kähler structure, so $\mathbb P^1$ is always Kähler no matter which metric it choose.
Jul 24, 2021 at 21:53 comment added AmorFati 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).
Jun 29, 2021 at 12:37 history edited Tom CC BY-SA 4.0
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Jun 28, 2021 at 17:09 history asked Tom CC BY-SA 4.0