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Aug 16, 2011 at 9:38 comment added A. Pascal I thought more about the above example with a $\hat{A_{1}}$ inside an $\mathbb{R}^{2,1}$. This is revealing. Thank you. For other readers, and for concreteness, a good model in which to understand the above is the following: $L$ is a lattice with integral basis $u_{1},u_{2},u_{3}$ and pairings $\langle u_{i},u_{i} \rangle=2$, $\langle u_{i},u_{j} \rangle=-2$ if $|i -j|= 1$, and $\langle u_{i},u_{j} \rangle=0$ otherwise. Then the 'singular subspace' is spanned by $u_{1},u_{2}$. Inside there is an $\hat{A_{1}}$. Singular means that the quadratic form restricted to this subspace is degenerate.
Aug 10, 2011 at 8:46 comment added A. Pascal Sorry. I wasn't clear. By 'discrete group' I meant 'group acting discretely', which is not the case for infinite groups acting on compact Hausdorff spaces (like $PGL_{2} (\mathbb{Z})$ on $\mathbb{P}^{1} (\mathbb{R})$, as you point out). In the linear case under question, though, I was considering infinite groups acting on linear (in particular non-compact) spaces, and I'm pretty sure they are acting discretely, being subgroups of $GL_{n}(\mathbb{Z})$. So I think there should be some fundamental domain, even if it is impossible to understand. But maybe I am missing something.
Aug 9, 2011 at 5:42 comment added S. Carnahan Fundamental domains don't necessarily exist when you have non-compact groups of symmetries, because orbits of points can exhibit accumulation. The standard example is the action of $PGL_2(\mathbb{Z})$ on $\mathbb{P}^1 (\mathbb{R})$ by Möbius transformations. In the case at hand, we are considering discrete subgroups of a noncompact orthogonal group acting on a space of substantially lower dimension.
Aug 8, 2011 at 17:28 comment added A. Pascal About the last comment on signature $(m,n)$. Shouldn't any discrete group admit some fundamental domain?
Aug 8, 2011 at 17:27 vote accept A. Pascal
Aug 8, 2011 at 19:04
Aug 8, 2011 at 15:34 history answered S. Carnahan CC BY-SA 3.0