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Let K be a group (with discrete topology), G be a Lie group. Let $\operatorname{Hom}(K,G)$ be the space of all group homomorphisms from K to G. This is a closed subvariety of the space of all the maps from the generators of K to G, and as such has a topology.

Andre Weil's paper "On Discrete Subgroups of Lie Groups" proves that an important subset $U \subset \operatorname{Hom}(K,G)$ is open. U is defined as the set of all homomorphism $K\to G$ such that the homomorphism is injective, the image is discrete, and the quotient $G/image(K)$ is compact.


  1. What happens if you remove the condition that the quotient is compact?

  2. How often/where is this taught? What kinds of books would it be in, what kind of courses would have it? This looks like a basic result that could be taught anywhere, but it's completely new to me (not that I know much about representation theory). while Weil's paper fortunately seems very readable, I couldn't easily find any other source that would such questions.


In the case where $K=\pi_1 (S)$ is the fundamental group of a surface and $G=PSL_2(\mathbb R)$, the space $\operatorname{Hom}(K,G)$ is very closely related to the Teichmuller space of S. Every Riemann surface is a quotient of $PSL_2(\mathbb R)$ by a discrete subgroup. So, for an element of $\operatorname{Hom}(K,G)$, the quotient $G/image(K)$ corresponds to a Riemann surface and the data of the actual map $K\to G$ gives a marking on it.

Not every homomorphism $K\to G$ corresponds to a point of Teichmuller space. For example, the map that sends all of K to the identity is clearly no good, as the quotient $K/G$ is not topologically the same as the surface S. However, if the map is injective and the image of K in G is discrete, all will be well. So, Weil's theorem basically says that the Teichmuller space of S is an open subset of $\operatorname{Hom}(\pi_1(S),PSL_2(\mathbb R))$.

However, since Weil's theorem requires the quotient to be compact, this won't work if S is a non-compact Riemann surface. I wonder how much more difficult life becomes in this case.

Disclaimer/Another question:

The above has a small lie in it. To get the Teichmuller space, you actually need to look at the quotient $\operatorname{Hom}(K,G)/G$ where G acts on $\operatorname{Hom}(K,G)$ by conjugation of the target. In the case of compact surfaces, this is not supposed to mess up the fact that the subset is open; this seems to be a result of William Goldman but I don't have the exact reference. If you can say anything about this, I'd appreciate it too.

Thank you very much!

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You seem to have confused Andre Weil and Hermann Weyl. – HJRW Jan 22 '10 at 22:11
The "Primer" of Farb and Margalit always deals with first the case of compact surfaces, and makes additional special remarks for punctured surfaces. So I hope along with you that there should be something for noncompact surfaces. +1. – Anweshi Jan 22 '10 at 23:37
I was thinking that it did not sound like something weil would work on. To be fair, if the result was discussed out loud, i can see where the confusion might come from. – Sean Tilson Aug 19 '10 at 22:36
@Henry Wilton: I am quite sure that the OP is correct that the paper is written by Andre Weil, and not Hermann Weyl, Simone Weil, or Andrew Wiles. – Igor Rivin Feb 11 '11 at 6:48

I think that

a) a good place to start is to read pages 60-72 of Misha Kapovich's book "Hyperbolic manifolds and discrete groups".

b) the right context for your question 1 is to consider relative representation varieties (page 62 in above book).

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Thank you for the reference, it does seem like a very good place to start. – Ilya Grigoriev Jan 25 '10 at 19:20

Some tangential comments:

$\textrm{PSL}_{2} \mathbb{R}$ is 3-dimensional; you get a Riemann surface by taking a lattice $\Gamma$ in $\textrm{PSL}_2\mathbb{R}$ and taking the double quotient $\Gamma \backslash \textrm{PSL}_2\mathbb{R} / \textrm{SO}(2)$. (That is, it's $\Gamma \backslash \mathbb{H}^2$ where $\mathbb{H}^2$ is the hyperbolic plane.) But this is compact iff $\Gamma \backslash$ $\textrm{PSL}_2\mathbb{R}$ is, since $\textrm{SO}(2)$ is compact.

If your Riemann surface is non-compact but of finite type, then when you uniformize, the complete hyperbolic metric will have infinite cusps at the punctures. This means that your representation $\Gamma\to$$\textrm{PSL}_2\mathbb{R}$ must send the elements $\gamma\in\Gamma$ corresponding to loops around the punctures to parabolic elements of $\textrm{PSL}_2\mathbb{R}$. Putting this restriction on a representation will force the quotient to have finite volume, and then you have the same theorem that the discrete faithful representations are open in the representation variety.

You might also look at Section 4 of Peter Shalen's paper "Representations of 3-manifold groups". For representations of hyperbolic 3-manifold groups into $\textrm{PSL}_2\mathbb{C}$ we have Mostow rigidity, which says that any two discrete faithful representations are conjugate; thus the appropriate subpace of $\textrm{Hom}(\Gamma,\textrm{PSL}_2\mathbb{C})/\textrm{PSL}_2\mathbb{C}$ is just a point (in stark contrast to the case for surface groups which you attribute to Goldman). But you still have algebraic deformations in the character variety in the case of manifolds with cusps, and these were analyzed by Thurston. In particular, Shalen says that Thurston generalized Weil's results to finite-volume cusped hyperbolic 3-manifolds, by imposing the condition mentioned above that cusp subgroups map to parabolic subgroups of $\textrm{PSL}_2\mathbb{C}$.

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Hey Tom, nice to see you here! Your comments are very useful indeed, Regarding the SO(2) business, I take it that the "common wisdom" that "PSL(2,R) is the group of conformal automorphisms of the unit disk" is wrong. Should, for example, wikipedia's page of SL(2,R) be fixed or am I missing something else now? – Ilya Grigoriev Jan 25 '10 at 19:19
Hi Ilya! No, that's correct -- the point is that Aut(D) = PSL(2,R) is 3-dimensional, but the stabilizer of a point is 1-dimensional (in the unit disk, this is just the rotations). Then since Aut(D) acts transitively on D, the orbit-stabilizer theorem tells us that D = Aut(D) / Stab(point) = PSL(2,R) / SO(2). In particular this implies that D is 3-1=2-dimensional, which is a good check. (You can identify PSL(2,R) with the unit tangent bundle of D, so the full quotient Gamma \ PSL(2,R) can be thought of as the unit tangent bundle of your Riemann surface; this is less frequently useful, though.) – Tom Church Jan 25 '10 at 20:38

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