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It is known that (uniformization theorem) any Riemann surface can be written as the quotient of its universal cover by a discrete group (of Möbius transformations). This group is isomorphic to the fundamental group of the surface. My question is:

Can we choose a non-discrete group that is isomorphic to the fundamental group of the surface? What happens if we consider the quotient? (I know that it is not a surface in general).

In other words:

What happens if we quotient the upper-half plane by a non-discrete group?

Is it interesting to study such quotients? If so, why?

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I don't under why you are talking about representations in your question. Aren't you trying to ask «what happens if we quotient the upper half plane by a non-discrete group of Moebius transformations?» – Mariano Suárez-Alvarez Feb 27 '12 at 2:38
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I don't really get the question. Certainly non-discrete subgroups act on the upper half plane -- indeed $\operatorname{PSL}_2(\mathbb{R})$ acts on the upper half plane, is not discrete and has plenty of nondiscrete subgroups. But a nondiscrete group $G$ cannot act properly discontinuously on a space $X$ and therefore the quotient $X \rightarrow G \backslash X$ cannot be a covering map, and thus the connection to fundamental groups is lost. Maybe I'm interpreting what you're saying too narrowly (or, simply, incorrectly): if so, how? – Pete L. Clark Feb 27 '12 at 3:55
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One way to interpret this is: «what do we gt when we mod out by a non-discrete subgroup?» For example, if we mod out by a subgroup which is virtually discrete, so that it acts with finite stabilizers, we can make sense of the result as an orbifold, I guess... – Mariano Suárez-Alvarez Feb 27 '12 at 4:06
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Maybe you should think about a simpler example of this sort of phenomenon. You can get a nondiscrete copy of $\mathbb Z$ to act on the circle as rotations by an irrational angle. Take a look at what you get this way; it's not pretty. You leave pretty much every nice topological feature behind, and get into some nasty bits of analysis really quickly; you're not too far from the construction of a non-measurable set, for instance. – NKS Feb 27 '12 at 4:25
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To analyse such situations is one of Connes's motivations for his non-commutative geometry. He avoids the mess NKS mentions by replacing the quotient by, say, $C^\infty(H)\rtimes G$, the algebra obtained as the cross-product of the ring of smooth functions on the upper halfplane, by the group you want to mod out. When $G$ acts nicely, this algebra is Morita equivalent to the algebra $C^\infty(H/G)$; so when $G$ does not act nicely, we can use it as a replacement for the quotient. – Mariano Suárez-Alvarez Feb 27 '12 at 4:31

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