# Is there non-discrete group isomorphic to the fundamental group, what about the quotient?

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
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
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
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
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