Let $L$ be a lattice of signature $(1,n)$. Suppose I have a (probably infinite index) subgroup $\Gamma\subset O^+(L)$ of the isometries of $L$ which preserve the positive cone $\mathcal{C}^+\subset L\otimes \mathbb{R}$. Now, suppose that there is an open set $A\subset \mathbb{P}\mathcal{C}^+\cong \mathbb{H}^n$ of hyperbolic space defined by an intersection of (possibly infinitely many) half-spaces such that the following hold:

1) $\Gamma\cdot A=A$

2) There is a finite set of polyhedra $P_i\subset A$ (which may contain up to one cusp of $\mathbb{H}^n$) such that $P_i\rightarrow \Gamma\backslash A$ is injective on the interior of $P_i$ and the images of $P_i$ cover all of $\Gamma\backslash A$.

My question is, is there necessarily a polyhedral fundamental domain for the action of $\Gamma$ on $A$? I know this is a rather specific question, but of course if there is a more general result in this direction, I would love to know. For the purposes of this question, a polyhedron is a subset of $\mathbb{P}(L\otimes \mathbb{R})$ defined by a finite set of linear inequalities. For instance, $\mathbb{P}\mathcal{C}^+$ is NOT a polyhedron.

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    $\begingroup$ Such a fundamental domain can only exist if your group is a discrete subgroup in the whole group of isometries. Moreover, if your polyhedra are assumed to have only finitely many faces, then the group must be geometrically finite. The Wikipedia article en.wikipedia.org/wiki/Geometric_finiteness might be of help. (Its german version has more references to the literature.) $\endgroup$ – ThiKu Dec 3 '15 at 7:32
  • $\begingroup$ Because $\Gamma$ is a subgroup of a discrete group $O^+(L)$ of isometries of lattice, it is discrete. Whether the group $\Gamma$ is geometrically finite may not be that helpful, because I would like to quotient $A$, not $\mathbb{H}^n$, by $\Gamma$. Furthermore, geometric finiteness is not sufficient to conclude the existence of a polyhedral fundamental domain in dimension $4$ and higher. $\endgroup$ – Philip Engel Dec 3 '15 at 7:58
  • $\begingroup$ The answer is no. Any lattice $L$ such that $\mathrm{O}'(L)$ has non-trivial rational $H_1$ will give a counter-example (and there are a lot of them). Sometimes, part of this rational homology comes from the stabilizers of the cusps, and this gives other counter-examples where $A$ is not the whole of $\mathcal H$. You can see this question for explicit examples : mathoverflow.net/questions/141284/… $\endgroup$ – few_reps Dec 3 '15 at 8:00
  • $\begingroup$ Thanks for the reply, though I think I lack the necessary background to see why condition 2) was satisfied in the example described. I will think about it more, but if you could point me in the right direction it would be greatly appreciated. $\endgroup$ – Philip Engel Dec 3 '15 at 8:14
  • $\begingroup$ I sent you an email. $\endgroup$ – few_reps Dec 3 '15 at 10:09

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