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

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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.

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The surface $X/G$ may be singular, so you need to be a bit careful. –  rita Oct 2 '12 at 16:35
3  
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 arxiv.org/abs/0902.3761, in particular Theorem 1.8 page 13. –  Francesco Polizzi Oct 2 '12 at 17:23

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