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Let $n$ be a large integer.

I am looking for a cocompact properly discontinuous isometric action on $n$-dimensional Lobachevky space which is generated by elements of finite order.

Or equivalently, I need a cocompact properly discontinuous isometric action $\Gamma\curvearrowright\mathbb{H}^n$ such that $\mathbb{H}^n/\Gamma$ is simply connected, here $\mathbb{H}^n/\Gamma$ stays for quotient space (NOT for orbifold).

Comments.

  • I am aware that for large $n$ these is no cocompact action generated by reflections (Vinberg, 1984).
  • Consider the group of matrices with integer coefficients from $\mathbb{Q}[\sqrt{5}]$ which preserve the quadratic form $$\tfrac{1+\sqrt{5}}{2}\cdot x_0^2-x_1^2-\cdots-x_n^2.$$ This gives cocompact properly discontinuous isometric action, say $\Gamma\curvearrowright\mathbb{H}^n$. (For $n=2$, the group $\Gamma$ contains the Coxeter group of regular right-angled pentagon.)
    I would be very happy if the subgroup generated by the elements of finite order in $\Gamma$ would have finite index. Or, equivalently, if $$|\pi_1(\mathbb{H}^n/\Gamma)|<\infty.$$

P.S. It seems that examples of such actions are not known. (In addition to Agol's comment, I've got a letter from Vinberg stating this.)

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I suspect this is an open question, but I'm not aware of a reference explicitly stating it. –  Ian Agol Jul 19 '12 at 16:40
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At one point, I checked that all of the groups $O(n,1;Z)$ are generated by rank one involutions (reflections through planes and points), up to the dimensions that Vinberg had computed them. In fact, the even unimodular Lorentzian lattice in dimension 25 is also generated by such involutions (Daniel Allcock checked this using properties of the Leech lattice and Conway groups). I don't see why this might not be true in all dimensions, but I would have no clue how to approach proving it. Are you interested in non-uniform lattices too? –  Ian Agol Jul 20 '12 at 19:19
    
@Agol; What do you mean by "up to the dimensions that Vinberg had computed them"? Did Vinberg gave a list of generators of $O(n,1;\mathbb{Z})$ for small $n$? –  Anton Petrunin Jul 20 '12 at 20:58
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@ Anton: I mean he computed the reflection subgroup for $n$ such that the subgroup is finitely generated. The full group is then generated by the reflection subgroup together with the symmetries of its fundamental domain, so in some sense he computed $O(n,1;Z)$. All that I'm pointing out is that these are actually generated by involutions. –  Ian Agol Jul 23 '12 at 17:41
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