5

I define a $K3$ surface as a smooth complex manifold of dimension two which is simply-connected and such that the canonical bundle is trivial.

I know that two $K3$ surfaces are always deformation equivalents and I know that $K3$ surfaces are Kähler. Conversely, if a deformation $X$ of a $K3$ surface is Kähler, then Hodge structure is preserved so $X$ is again simply-connected, and the symplectic form on the $K3$ surface extends to $X$ so the canonical bundle of $X$ is trivial too. Hence $X$ is a $K3$ surface.

How about non-Kähler deformations of a $K3$ surface? If they exists they aren't $K3$ surfaces, but do they exists?

Thank you!

Nutella Warrior
  • 265

1 Answers1

8

A deformation of a compact Kähler manifold of dimension 2 is always Kähler. Indeed, it is a theorem of Kodaira and Siu that a compact complex surface is Kähler if and only if $b_1(X)$ is even. Since a deformation of the complex structure preserves the underlying topology (by Ehresmann's theorem), the property of a compact complex surface being Kähler is invariant under deformation.

MarkM
  • 826
  • Thank you! I am missing something, or this argument works in every dimension? I ask because I often see that a word of caution is needed when deforming Kähler manifolds, see for example https://arxiv.org/pdf/alg-geom/9705025.pdf remark $2.4$: "...any Kähler deformation of an irreducible symplectic manifold..." – Nutella Warrior May 19 '20 at 08:41
  • 3
    No, it only works in dimension 2. The theorem that I mentioned states that a compact complex surface (i.e. dimension 2) is Kähler if and only if $b_1(X)$ is even. It is not true in higher dimensions - being Kähler is not in general a topological condition. Check out ‘Hironaka’s example’ for a deformation of a Kähler threefold into one which is not Kähler. – MarkM May 19 '20 at 17:30
  • Ok, I misremembered it. Thank you for Hironaka example too! – Nutella Warrior May 20 '20 at 07:08