Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I have read somewhere that "$K3$ surfaces admitting finite non-symplectic group actions are projective". Could someone remind me of a proof?

share|cite|improve this question

1 Answer 1

A sketch of the proof goes as follows.

Let $G$ be a finite group of non-symplectic automorphisms on a $K3$ surface $X$. Since $G$ is non-symplectic, there exists $g \in G$ such that $g \omega \neq \omega$, where $\omega$ is the holomorphic $2$-form on $X$.

Then, setting $Y:=X/G$, one has $q(Y)=p_g(Y)=0$, since $q(X)=0$ and by the previous remark the holomorphic $2$-form $\omega$ does not descend to the quotient.

From this, one proves that either $Y$ is rational (i.e, bimeromorphic to $\mathbb{P}^2$) or the minimal desingularization of $Y$ is an Enriques surface. In any case, $Y$ is a compact algebraic surface, so there exists an ample divisor $H$ on $Y$.

Finally, the quotient $\pi \colon X \longrightarrow Y=X/G$ is a finite holomorphic map since $G$ is a finite group. It follows that $\pi^*H$ is an ample divisor on $X$, hence $X$ is projective.

Remark 1. As pointed out by Rita in her comment, one must be a bit careful since $Y$ can be singular. I skip the details, which can be found in Frantzen's dissertation.

Remark 2. One also proves that $G$ is a symplectic grup of automorphism if and only if the minimal desingularization of $X/G$ is again a $K3$ surface. The reason is that in this case $g \omega=\omega$ for all $g \in G$, so $\omega$ descends to the quotient.

share|cite|improve this answer
The surface $X/G$ may be singular, so you need to be a bit careful. – rita Oct 2 '12 at 16:35
You are right, but this was only a sketch of the proof. For the details, one can look for instance at Kristina Frantzen's Phd Thesis "K3-surfaces with special symmetry", see, in particular Theorem 1.8 page 13. – Francesco Polizzi Oct 2 '12 at 17:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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