Given a hyperbolic reflection group $G$ acting on hyperbolic space $\mathbb{H}_n$ by, well, reflections in hyperplanes.

Does a neat subgroup of $G$ act fixed-point free on $\mathbb{H}_n$? If not, what are sufficient conditions for this?

Some thoughts on this: The stabilizer of a point $x\in\mathbb{H}_n$ in some fundamental chamber of the action of $G$ is generated by the reflections in the walls containing $x$. If this is a finite reflection group, its intersection with any neat subgroup of $G$ is empty and this subgroup acts fixed-point free.

What if the stabilizer is infinite? When does its intersection with a neat subgroup vanish?

(Remember: A subgroup of $G(\mathbb{Q})$ of a linear algebraic group over $\mathbb{Q}$ is called neat if the image of $G$ under some faithful represenation $G\to \textrm{GL}_n(\mathbb{Q})$ is neat, the latter neat here meaning that the subgroup of $\mathbb{C}^\ast$ generated by the eigenvalues of the elements in the subgroup of $\textrm{GL}_n(\mathbb{Q})$ is torsion free. Think of this as "super-torsion free": Not only the matrices are torsion free but every entry of their diagonalization.)

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    $\begingroup$ A group cannot act by reflections. It can be generated by reflections; I don't know if this is what you mean. It then acts by isometries. $\endgroup$ – YCor Mar 17 '18 at 21:28
  • $\begingroup$ "Does a neat subgroup": what do you mean? "Does some neat subgroup"? if so you need more assumptions to make the questions nontrivial. Or you mean "Does every neat subgroup?" It's energy-consuming to try every possible interpretation to guess which one is more interesting and is likely to be the intended question... $\endgroup$ – YCor Mar 17 '18 at 21:31
  • $\begingroup$ "What if the stabilizer is infinite?" which stabilizer? $\endgroup$ – YCor Mar 17 '18 at 21:32
  • $\begingroup$ @YCor: You're right, I should have been more precise. Sorry! Your interpretation is right, the group shall be generated by reflection and acting via isometries. Regarding your second comment: I meant "every neat subgroup" (and didn't thought about the other interpretation of my wording). For your last comment: This refers to the section before: the stabilizer of an arbitrary point. If there is no point with infinite stabilizer (or, weaker, no point with an element of infinite order in the stabilizer), I can answer my question myself, but if there is such a point, I don't know how to proceed. $\endgroup$ – mathmax Mar 19 '18 at 8:24

Take two distinct intersecting lines in $\mathbb{H}^2$, whose angle of incidence is not in $\pi\mathbf{Q}$. Let $r,s$ be the corresponding reflections. Then the cyclic subgroup generated by $rs$ in $\langle r,s\rangle$ has index 2, and is generated by somme irrational rotation. In particular, it is neat (viewed in $\mathrm{SL}_2(\mathbf{R})$). But not fixed-point-free (it fixes the intersection of the two lines).

However I still see ambiguity in the question, because being neat a priori depends on the choice of a linear representation of the isometry group (the 2-dimensional one is not really a linear representation since it's not faithful).

Still, for every linear representation of the isometry group, it is true that $\langle rs\rangle$ is neat. This is because its closure in the isometry group is connected.


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