# Can deformation equivalent Kähler manifolds always be obtained by a deformation where all the fibers are Kähler?

Given compact Kähler manifolds $$X$$ and $$X'$$ deformation equivalent over the unit disk $$\Delta \subset \mathbb{C}$$. More precisely, there is a proper holomorphic surjective map \begin{align*} \pi\colon \mathcal{X}\to \Delta \end{align*} and $$t,t' \in \Delta$$ such that $$X$$ and $$X'$$ are biholomorphic to the fibers $$\pi^{-1}(t), \pi^{-1}(t')$$ respectively. Is there a deformation of $$X$$ and $$X'$$ over $$\Delta$$ such that every fiber is Kähler?

I am specially interested in the case where $$X$$ and $$X'$$ are of hyperkähler type, i.e. irreducible holomorphic symplectic. In other words, simply connected and admitting a unique holomorphic symplectic form. I know that there are (large) deformations where the deformed space is not Kähler.

I don't think this is known. For hyperkahler manifolds, conjecturally, all smooth complex deformations are class C and birational to hyperkahler. If this is true, your conjecture would follow automatically. The only relevant publication that I am aware of is

https://arxiv.org/abs/1703.02001

Arvid Perego

Kählerness of moduli spaces of stable sheaves over non-projective K3 surfaces

We show that a moduli space of slope-stable sheaves over a K3 surface is an irreducible hyperkähler manifold if and only if its second Betti number is the sum of its Hodge numbers h2,0, h1,1 and h0,2.

Perego proves that a (smooth) limit of hyperkahler manifolds is Fujiki class C if $$b_2=h^{2,0}+ h^{1,1} + h^{0,2}$$. This is a bit weaker than what you need, of course.

All the best
Misha