# When do reflection groups act discretely on quadrics in indefinite/semi-Riemannian situation?

The hyperbolic case seems to be well understood after work of Vinberg. Given a lattice $L$ with quadratic form $Q$ of signature $(1,n)$, the orthogonal group of $L$ acts discretely on the affine quadric $Q=1$ in $L \otimes \mathbb{R}$ and the subgroup generated by reflections has a convex polyhedral domain, the walls of this domain meet in non-obtuse angles, and the corresponding reflections give a Coxeter group.

Note that if the quadratic form has signature $(n,1)$, then we should take the quadric $Q=-1$, since we would like the stabilizer in $O(n,1)$ of a point on the quadric to be compact.

My question is to what extent the above situation extends to signature $(p,q)$.

I've learned from some papers of Toshiyuki Kobayashi that there is no chance for an infinite group to act discretely if you have chosen the wrong sign for the quadric (see his papers on Discrete groups on pseudo-Riemannian homogeneous spaces), but it seems that if you choose $Q=1$ when $p \leq q$ and $Q=-1$ when $p \geq q$, then there is still a chance that the group generated by reflections on the lattice $L$, or even the whole orthogonal group of $L$, could act discretely on the appropriate quadric.

This question arose after reading some Vinberg references suggested by Agol in the answer to this question Fundamental domain for group generated by reflections in -2 curves and (EDIT) after the comment by Scott Carnahan that he had heard there were reflection groups with no fundamental domains When are root hyperplanes locally finite?.

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I suspect that in general the reflection group of the lattice will not act discretely on any of the quadrics. For example, let $L=I_{p,q}$, the unique-up-to-isometry unimodular lattice of signature $(p,q)$, with diagonal basis $e_{1},...,e_{p}, f_{1},...,f_{q}$. Now consider root hyperplanes containing $f_{q}$ (if $Q=-1$). So long as $q>1$, we will have a whole bunch of those, namely one for each root in $I_{p,q-1}$. Likewise, if $Q=1$, take root hyperplanes through $e_{1}$. We'll have one for each root in $I_{p-1,q}$. Thus the reflection group of the lattice will have infinite stabilizer at the points $f_{q}$ or $e_{1}$.