5
$\begingroup$

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.

$\endgroup$

1 Answer 1

1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.