# Fundamental group of an hyperbolic $4$-manifold

Good afternoon everyone,

I have a very general question about hyperbolic manifolds and their fundamental groups in high dimension (at least $4$). If the theory of surfaces and $3$-manifolds provide a lot of constructive examples of compact hyperbolic manifolds, it is not that obvious that in higher dimensions, such objects exists.

A way to construct a $4$ dimensional hyperbolic manifold is to find a cocompact lattice in $\text{O}(n,1)$. Let $\varphi = x_1^2 + ... + x_n^2 - \sqrt{p} x_{n+1}^2$ , and $\Gamma = \text{Aut}(\varphi) \cap \text{Gl}(n+1, \mathbb{Z}[\sqrt{p} ])$. A theorem by Borel ensures us that $\Gamma$ is a cocompact lattice in $\text{Aut}(\varphi) \simeq \text{O}(n,1)$. By Selberg's lemma, we can find a finite index subgroup of $\gamma$ which is torsion free, and this rises to the construction of a compact hyperbolic $4$-manifold.

My question is : can one tell me the algebraic structure of $\Gamma = \text{Aut}(\varphi) \cap \text{Gl}(n+1, \mathbb{Z}[\sqrt{p} ])$ ? Is there a known method to compute it ? Is there a general method for all quadratic forms with coefficient in a number field ?

Thank you very much for your attention !

Selim

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What do you mean by 'can one tell me the algebraic structure of $\Gamma$'? What do you want to know about it? –  HJRW Jun 4 '13 at 12:09
The generator, the relations. And if one could find matrixes which are generators, it would be great. –  Selim G Jun 4 '13 at 12:15
Although they only deal with the 3-dimensional case, there's a lot of information about arithmetic hyperbolic manifolds in Machlachlan and Reid's book The arithmetic of hyperbolic 3-manifolds. Have you tried looking there? –  HJRW Jun 4 '13 at 12:17

For general $p$, the only known method is to construct a Dirichlet fundamental domain and read off the group presentation from it. The procedure for computation of a fundamental domain is called "Jorgenesen's algorithm: List elements of $\Gamma$ (using the embedding to $GL(2(n+1),Z)$ via restriction of scalars). For each $\gamma\in \Gamma$ construct the bisector of $o, \gamma(o)$ in the unit ball $H^4$. On each step, check if conditions of Poincare's fundamental domain theorem hold. If it they do, the process stops and you got your fundamental domain $D$. Now, generators of $\Gamma$ are face-pairing transformations for $D$. The relators correspond to cycles of 2-dimensional faces of $D$.
For small $p$'s one can do better: If $p=1$ (and $n=4$), the group is a reflection group (whose fundamental domain is a simplex), so you get an explicit presentation.
To indicate limits of our present knowledge of higher-dimensional hyperbolic manifolds/orbifolds of finite volume: There is not a single known lattice in $O(n,1)$, $n\ge 1000$, for which we know the abelianization. All what we know that for each $i$, every such arithmetic lattice (note that $n> 7$) contains a congruence subgroup $\Gamma_i$ so that abelianization of $\Gamma_i$ has rank $\ge i$.
One comment, Misha. If you choose an ordering on the elements, and find a finite-sided polyhedron satisfying Poincare's polyhedron theorem, then you don't necessarily know that those elements generate $\Gamma$, just a finite-index subgroup. Since the question asks about congruence arithmetic groups, in principle one could use number-theoretic formulae for the covolume of $\Gamma$ to find the index of the subgroup (one need not compute the covolume, just the Euler characteristic in dim 4). Then you keep going until you find a subgroup of index 1. –  Ian Agol Oct 7 '13 at 0:18