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For an even integral lattice $L$ (of arbitrary signature, non-degenerate but not necessarily unimodular), consider roots (vectors of square $2$) and their orthogonal hyperplanes.

First, I'd like to know if the collection of root hyperplanes is always locally finite away from the origin. (I'm not actually sure what this means, but I have come across this condition in many papers. At the very least, it should mean that every non-zero point in $L \otimes \mathbb{R}$ has a neighborhood meeting a finite number of the hyperplanes.)

Second, if the above is true, I can take the complement of all of these hyperplanes to be left with a bunch of connected components and I'd like to know if choosing one of them always gives a fundamental domain for the group generated by reflections in the root hyperplanes. I know that this is the case for finite and affine root systems, and am wondering if this is a general fact.

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    $\begingroup$ All of these hyperplanes pass through the origin so as soon as there are an infinite number of roots it is not locally finite. $\endgroup$
    – Torsten Ekedahl
    Commented Aug 7, 2011 at 16:06
  • $\begingroup$ Fair enough. I added the phrase "away from the origin". In the parenthetical remark, I did say I wanted this for every non-zero point. $\endgroup$
    – A. Pascal
    Commented Aug 7, 2011 at 17:01

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Any affine root system in a rank $n$ Lorentzian lattice gives rise to a system of hyperplanes in $\mathbb{R}^{n-1,1}$ that is not locally finite, since all of the root hyperplanes intersect the lightlike line (that is, the line of norm-zero vectors spanned by $\delta$). Perhaps the easiest example is $\widehat{\mathfrak{sl}}_2$, where you have a singular 2-dimensional subspace of $\mathbb{R}^{2,1}$ that is is tangent to the light-cone in a distinguished lightlike line. The real roots of the affine root system make up two chains of vertices in the singular subspace that are parallel to the lightlike line, and that line is perpendicular to everything in the singular subspace, including the real roots. If you have a hyperbolic root system that contains an affine root system, you will get a similar failure of local finiteness.

In the affine case, you will get a fundamental domain in the hyperboloid (and therefore in its cone), but infinitely many domain walls meet at ideal points on the boundary. I have heard that reflection groups in signature $(m,n)$ do not usually admit fundamental domains for $m,n > 1$, but I do not know of a reference or a counterexample.

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  • $\begingroup$ About the last comment on signature $(m,n)$. Shouldn't any discrete group admit some fundamental domain? $\endgroup$
    – A. Pascal
    Commented Aug 8, 2011 at 17:28
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    $\begingroup$ 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. $\endgroup$
    – S. Carnahan ♦
    Commented Aug 9, 2011 at 5:42
  • $\begingroup$ 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. $\endgroup$
    – A. Pascal
    Commented Aug 10, 2011 at 8:46
  • $\begingroup$ 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. $\endgroup$
    – A. Pascal
    Commented Aug 16, 2011 at 9:38

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