6
$\begingroup$

It is known that smooth complex hypersurfaces with degree bigger than 2 and dimension bigger than 1 have finite automorphism groups, except for K3 surfaces.

But the group of polarised automorphisms over a K3 surface is finite. Do we have a simple argument for it?

Improve this question
$\endgroup$
1
  • 6
    $\begingroup$ Yes. The group of polarized automorphisms of any variety $X$ is an algebraic group. Therefore if its Lie algebra $H^0(X,T_X)$ is zero, it is finite. But for a K3 surface $H^0(X,T_X)=H^0(X,\Omega ^1_X)=0$. $\endgroup$
    – abx
    Jul 1, 2022 at 16:50

1 Answer 1

Reset to default
5
$\begingroup$

Calabi-Yau theorem implies that any diffeomorphism of a Calabi-Yau manifold which preserves the complex structure and the Kahler class also preserves the Calabi-Yau metric. However, the group of isometries of a compact metric space is compact. The group of isometries of a Riemannian manifold is a Lie group by Meyers-Steenrod. To prove that the group of polarized isometries of K3 is finite, it remains to show that it is 0-dimensional. This would follow if we prove that there are no holomorphic vector fields. However, K3 is symplectic, hence holomorphic vector fields are the same as holomorphic 1-forms, and they all vanish because $b_1(M)=0$.

For an algebro-geometric argument, please see the comment of abx above this.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.