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Given a proper algebraic surface $S$, the Picard group $Pic(S)$ is endowed with the (symmetric) intersection form. We can therefore talk about reflections in the classes of $-2$-curves. These will preserve the intersection form and therefore the positive cone inside the lattice $Pic(S)$. Many times I have come across the claim that if $S$ is a smooth K3, then the action of this group generated by reflections is properly discontinuous on the positive cone inside $Pic(S) \otimes \mathbb{R}$ and has a (maybe infinite) polyhedral fundamental domain with faces defined by the hyperplanes orthogonal to the classes of effective $-2$-classes. Why is this? And is there a good reference for the proof of this fact?

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Did you already have a look at Chapter VIII of "Compact Complex Surfaces" by Barth, Hulek, Peters, van de Ven; as well as S. Kovacs' paper "The cone of curves of a K3 surface"? – Christian Liedtke Aug 8 '11 at 14:52
I just took a look at both sources, which don't seem to address this question. Any other suggestions? – A. Pascal Aug 8 '11 at 15:23
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This is a general fact about groups generated by reflections acting on lattices. The reflections in $-2$ curves generates an action on $H^2(S;\mathbb{Z})$ by reflections, with quadratic form given by the cup product. Restricting this to $H^{1,1}(S)\cap H^2(S;\mathbb{Z})$, one gets an action on an integral lattice of signature $(1,19)$ (check out McMullen). For any group preserving a Lorentzian lattice, the subgroup generated by reflections will have fundamental domain bounded by hyperplanes fixed by a subset of reflections. The point is that the set of all hyperplanes which are fixed by reflections is equivariant with respect to the group action, since the conjugate of a reflection is a reflection. Thus, they cut out polyhedra in the positive cone (the hyperplanes must be locally finite since the group is discrete as it preserves a lattice). The fact that the quadratic form is Lorentzian is used here, since otherwise none of the complementary pieces of the null cone are convex. Given any two points in the cone, one takes a generic path between them, which will cross finitely many hyperplanes. The sequence of reflections fixing the sequence of hyperplanes crossed by the path will send the polyhedron containing one point to the polyhedron containing the other. Thus, the polyhedra are all equivalent by the group action, and therefore form a fundamental domain for the subgroup generated by reflections (one must also check that no reflection fixes one of the polyhedra, which is not hard to show geometrically as well). In fact, Vinberg has a nice algorithm which will compute the fundamental domain inductively, and I believe this kind of argument is discussed in his book.

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I'd like to accept this as the answer, but don't completely understand yet. In the above argument, do we really use the Lorentzian condition? This is similar to another question I asked about locally finite collections of root hyperplanes for an arbitrary even symmetric bilinear form. Also, which book of Vinberg might you have in mind? I took a look at the linked paper, which gives an algorithm, but he doesn't really explain the existence of the chamber decomposition. – A. Pascal Aug 9 '11 at 5:27
A. Pascal: I put a link to the Mathscinet review of his book,although I don't have access to it now so I haven't checked if it contains the argument. In his papers, he just says that the argument is exactly as in Coxeter's proof of the existence of a fundamental domain for a group generated by reflections in the Euclidean case. The fact that the quadratic form is Lorentzian is used because the positive cone is convex - this is not true in other signatures. In some cases, one can find a convex fundamental domain for a reflection group preserving a quadratic form of other signatures(Tits cone). – Ian Agol Aug 9 '11 at 16:59
Thanks. This cleared up some things and the Shvartsman-Vinberg book is very useful. – A. Pascal Aug 10 '11 at 8:41
Great answer, I needed this exactly. – Philip Engel Jun 5 '13 at 4:01

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