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 !


  • $\begingroup$ What do you mean by 'can one tell me the algebraic structure of $\Gamma$'? What do you want to know about it? $\endgroup$
    – HJRW
    Jun 4, 2013 at 12:09
  • $\begingroup$ The generator, the relations. And if one could find matrixes which are generators, it would be great. $\endgroup$
    – Selim G
    Jun 4, 2013 at 12:15
  • $\begingroup$ 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? $\endgroup$
    – HJRW
    Jun 4, 2013 at 12:17

1 Answer 1


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.

You can find a brief discussion of a similar procedure in the complex-hyperbolic setting in this paper: D. Cartwright, T. Steger, Enumeration of the 50 fake projective planes, C. R. Acad. Sci. Paris, Ser. I 348 (2010), 11-13.

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 $\ge i$.

  • $\begingroup$ Thanks Misha! This is not just an answer but also a few important problems. $\endgroup$ Oct 6, 2013 at 22:55
  • 3
    $\begingroup$ 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. $\endgroup$
    – Ian Agol
    Oct 7, 2013 at 0:18
  • $\begingroup$ @IanAgol: Very true. To be honest, I do not know in details how the algorithm of Cartwright and Steger works. They first find a Siegel domain (using reduction theory) which contains the actual fundamental domain, but is typically larger. Then they somehow shop off some of its pieces until they get the actual fundamental domain. $\endgroup$
    – Misha
    Oct 7, 2013 at 8:43

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