Let $G$ be a finite subgroup of the group of automorphsims of a $K3$ surface $S$. Consider the quotient $S/G$.

I am interested in the collection of such qutients:

$$\{ S/G \mid S\text{ is a K3 surface, }G\text{ is a finite subgroup of }Aut(S) \}$$

Is there any classification result for the collection of quotients?

What if I assume that $G$ contains an involution that acts on $H^{2,0}(S)$ as multiplication by $-1$? Is the collection finite up to deformation?

  • 4
    $\begingroup$ Up to deformation there are only a finite number of possible group actions, so a finite number of quotients. There is a huge literature about this, see for instance the work of Mukai. $\endgroup$ – abx Oct 15 '19 at 13:44
  • $\begingroup$ One starting point for a classification is to work birationally and to note that the resolutions of $S/G$ are all either K3 surfaces or rational surfaces (because K3 surfaces do not admit dominant rational maps into other types of surfaces). Then one would ask which K3s admit a generically finite (Galois) cover by another K3, and there is some work on this but probably no classification. $\endgroup$ – Evgeny Shinder Oct 22 '19 at 22:10
  • $\begingroup$ Sorry, I forgot about other surfaces of Kodaira dimension zero: K3s definitely cover Enriques surfaces, and I do not know about whether K3 surfaces admit rational dominant maps to bi-elliptic surfaces. $\endgroup$ – Evgeny Shinder Oct 23 '19 at 16:48
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    $\begingroup$ @EvgenyShinder Complex K3s do not admit rational dominant maps to bi-elliptic surfaces. One could show this as follows: Resolving the map would give a regular map from a simply connected surface (a blow-up of a K3 surface a finite # of times) to a bi-elliptic surface whose universal cover is $\mathbb{C}^2$. Lifting the map to the universal cover shows the map is constant. $\endgroup$ – Yosemite Stan Oct 30 '19 at 21:20

This question can be divided into two: what is the classification result for finite group actions on $K3$ surfaces, and what is the quotient space of $K3$ surface by a finite group action. As @abx said, there is a huge literature about these, and I'll try to list some of them.

Since we have global Torelli theorem for $K3$ surfaces, the classification problem for automorphism groups can be translated into lattice theoretic problem. V.V. Nikulin (1979) first systematically studied this. Mukai (1988) classified the maximal finite groups acting symplectically on $K3$. There are $11$ such groups, and they are related to the Mathieu group. The quotient of a $K3$ by a finite symplectic group action is still a $K3$ (with ADE singularities). By looking at the configurations of those singularities, Gang Xiao (1996) gave another (more geometric) approach for the classification by Mukai. Kondo (1998) simplified Mukai's approach via Niemeier lattices.

A finite group $G$ acting on a $K3$ fits into a short exact sequence: $$1\longrightarrow G_s\longrightarrow G\longrightarrow \mu_n\longrightarrow 1$$ where $G_s$ is the subgroup of symplectic automorphisms in $G$, and $\mu_n$ is a cyclic group. To understand the quotient, we first quotient by $G_s$ and obtain an ADE $K3$, then quotient by the non-symplectic action of $\mu_n$.

A $K3$ surface with a non-symplectic involution is called a $2$-elementary $K3$. Such $K3$ are classified by Nikulin (1981, factor groups of groups of automorphisms of hyperbolic forms with respect to subgroups generated by 2-reflections), and the fix locus of the involution in each case is described there (Theorem 4.2.2). It turns out that there is only one case when the involution has no fixed point, then the quotient is Enriques surface; for other cases, the fixed loci are nonsingular curves, and the quotients are smooth rational surfaces (Remark 4.5.4).

If a $K3$ $S$ has group action by $G$ with non-symplectic part $\mu_n$ has order at least $3$, then the quotient $S/G$ must be rational. For proof of this, I recommend G. Xiao's paper https://arxiv.org/abs/alg-geom/9512007.

The action of $\mu_n$ on the trascendental lattice is free, and this implies that $\varphi(n)\le 21$. Such $n$ is at most $66$. G. Xiao proved in the above paper that $n=60$ can not be realized. All other $n$ with $\varphi(n)\le 21$ can be realized, as constructed mainly by Kondo (1992, Automorphisms of algebraic K3 surfaces which act trivially on Picard groups).

Therefore, the symplectic part and non-symplectic part have only finitely many possibilities. Of course there are different ways to combine the two parts to get the group $G$, and I think a classification in general is still widely open. However, we at least know that there are only finite possibilities for the group $G$, and not hard to prove that there are only finitely many deformation types of actions of $G$ on $K3$.


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