# Deformations of K3 surface is again a K3 surface

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!

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.
• 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 arxiv.org/pdf/alg-geom/9705025.pdf remark $2.4$: "...any Kähler deformation of an irreducible symplectic manifold..." May 19, 2020 at 8:41
• 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. May 19, 2020 at 17:30