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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?»2012-02-27
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    @MarianoSuárez-Alvarez. What I am trying to ask is if we can represent the fundamental group as non-discrete group, then what happens to the quotient? So ,as you said, «what happens if we quotient the upper half plane by a non-discrete group of Moebius transformations?»2012-02-27
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    Then I don't understand the question, I guess!2012-02-27
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    @MarianoSuárez-Alvarez. I modified my question to make it clear.2012-02-27
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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?2012-02-27
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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...2012-02-27
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    @PeteL.Clark You are right, it is not a covering map. But what I am trying to say is 1. Is there anything interesting about the quotient?" I know it is not a surface in general. 2. Can we find some group isomorphic to the fundamental group but non-discrete?2012-02-27
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    Regarding (2): you should be able to find non-discrete copies of $\mathbb Z$, at least!2012-02-27
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    @Mariano Yes, that is what I was trying to ask: what do we get when we mod out by a non-discrete subgroup?2012-02-27
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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.2012-02-27
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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.2012-02-27
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    @Mariano Thanks. Could you give me some refernces, please?2012-02-27
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    Connes wrote a book entitled *Non-commutative geometry*, which is both scary and the best way I know to get the hang of the idea.2012-02-27
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    @yaa09d: Part of my confusion is that the title of your question is "fundamental group representation", and these words also figure prominently in the body of the question. If you understand that quotienting out by a nondiscrete group action has nothing to do with fundamental groups, then...??2012-02-28
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    @PeteL.Clark. I see. I did not write my question clearly (Sorry for the confusion). I edited the title and the question. But part of my question was if there "were" a non-discrete group "isomorphic" to the fundamnetal group, then what would happen?2012-02-29
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    @yaa09rd: If you are asking: are there torsionfree subgroups $\Gamma_1, \Gamma_2$ of $\operatorname{PSL}_2(\mathbb{R})$ which are isomorphic as abstract groups but one is discrete but the other is not, the answer is yes: e.g. there are nondiscrete subgroups isomorphic to $\mathbb{Z}$ (take the subgroup generated by an irrational rotation). The quotient by such a subgroup need not be Hausdorff.2012-02-29
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    @PeteL.Clark In this case, is there anything intresting (worht studying either in geometry or analysis) about the quotient?2012-02-29
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    @yaa09d: Well, I'm certainly not prepared to answer "no" to such a broad question...but nothing positive comes to mind.2012-02-29

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