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How can one construct families of cocompact discrete subgroups of the topological group $\text{SL}_2(\mathbb{C})$?

Here quaternion algebra's might help, I believe, but I have some difficulties with the construction.

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up vote 10 down vote accepted

Let $k$ be a number field with one complex place and let $B$ be a quaternion algebra defined over $k$ which ramifies at every real place of $k$ (and perhaps some finite places as well - Hilbert reciprocity implies that the total number of ramified places must be even). Let $\mathcal{O}$ be an order of $B$ and $\mathcal{O}^1$ its elements of reduced norm $1$. Choosing an embedding $k\hookrightarrow \mathbb{C}$ induces an embedding $B\hookrightarrow M_2(\mathbb{C})$, which in turn restricts to an embedding $\Psi: \mathcal{O}^1\hookrightarrow SL_2(\mathbb{C})$. A group which is commensurable with $\Psi(\mathcal{O}^1)$ is called arithmetic. An arithmetic group is cocompact if and only if the quaternion algebra $B$ is a division algebra (by Wedderburn's Theorem this is equivalent to saying that the set of places of $k$ which ramify in $B$ is nonempty).

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This is very clear, thank you. – Wanderer Feb 12 '10 at 10:53

For the arithmetic point of view you mention, Maclachlan and Reid's book "The arithmetic of hyperbolic 3-manifolds" is a great reference.

In case you're interested, there are also many geometric ways of doing this too (though you might object that some of them don't explicitly give you the subgroup).

You can explicitly build closed hyperbolic 3-manifolds out of polyhedra.

You can take a finite covolume subgroup and produce a cocompact subgroup via hyperbolic dehn surgery.

By a theorem of R. Brooks ("Circle packings and co-compact extensions of Kleinian groups", Invent. math. 86, 461-469 (1986)), you can take any cocompact subgroup of $\mathrm{SL}_2(\mathbb{R})$, and after a small quasiconformal conjugacy, it will lie in a cocompact subgroup of $\mathrm{SL}_2(\mathbb{C})$.

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Thanks also for the reference! It looks a bit heavy, but I'll take a closer look at it when I have more time. – Wanderer Feb 12 '10 at 10:47

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