5
$\begingroup$

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!

$\endgroup$

1 Answer 1

7
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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..." $\endgroup$ May 19, 2020 at 8:41
  • 3
    $\begingroup$ 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. $\endgroup$
    – MarkM
    May 19, 2020 at 17:30
  • $\begingroup$ Ok, I misremembered it. Thank you for Hironaka example too! $\endgroup$ May 20, 2020 at 7:08

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.