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Let $U(n)$ be the unitary group. From André Weil's paper "On discrete subgroups of Lie groups" it is well known that discrete cocompact subgroups of $U(n)$ have only a finite number of generators and relations.

Does there exist a full classification of discrete subgroups of the unitary group up to isomorphism?

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As David Speyer points out, the question needs a more precise formulation. The indicated Annals papers by Weil (1960, 1962) involve a connected Lie group $G$ (usually having no compact factors) and a discrete cocompact subgroup $\Gamma$ ("uniform lattice"). Here $\Gamma$ is finitely generated, as the fundamental group of a compact manifold. Classification of such discrete groups depends heavily on which Lie group one considers and is closely related to the study of compact manifolds. –  Jim Humphreys Apr 14 '11 at 13:39
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Do you really mean up to isomorphism, or up to conjugation in the unitary group? They are very different questions. If the latter, then the answers to the following question are relevant: mathoverflow.net/questions/17072/the-finite-subgroups-of-sun –  José Figueroa-O'Farrill Apr 14 '11 at 13:45
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$U(n)$ is compact. So any discrete subgroup of $U(n)$ is automatically (i) cocompact and (ii) finite. If you are interested in the classification of finite subgroups of $U(n)$, then the main result is Jordan's theorem: There is an integer $J(n)$ such that any finite subgroup of $U(n)$ has a normal abelian subgroup of index $\leq J(n)$. See Terry Tao's blog for a nice exposition. There is a small industry of improving the bounds for $J(n)$; this paper has some good bounds.

Is this what you wanted to ask?

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Hi David, in this question here mathoverflow.net/questions/91190/… I got the answer that there are infinite and discrete subgroups of $U(n)$. However, here you say the exact opposite, which is correct? –  whistles Mar 16 '12 at 11:57
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There is no infinite subset of $U(n)$ which is discrete in the topology induced by $U(n)$. The answer you link to takes a subgroup of $SL(2, \mathbb{C})$, which is discrete in the topology induced by $SL(2, \mathbb{C})$, and embeds it in $U(n)$. For example, at the end of his answer, Victor embeds the von Dyck group $(2,5,5)$ as a discrete subgroup of $SO(2,1)$, with coefficients in $\mathbb{Q}(\sqrt{5})$. He then observes that applying Galois conjugation takes this representation into $SO(3)$. Galois conjugation is not continuous! The image in $SO(3)$ will have a very different topology. –  David Speyer Mar 16 '12 at 14:47
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Maybe the best way to explain this is to point out that the same issue exists for $U(1)$. There are unitary matrices of infinite order: $( e^{i \pi \alpha})$ for any irrational $\alpha$. However, the group $( e^{i \pi \alpha k})$ is not a discrete subgroup of $U(1)$ -- it is dense in $U(1)$ and, as a subgroup of $U(1)$, has some complicated topology. –  David Speyer Mar 16 '12 at 15:13
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