Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Dear all,

my question is: Suppose we have a Fuchsian group G and another group H, containing G with finite index. Must H be a Fuchsian group? Can you give a proof? Or do you know any counter examples?

Best Ali K.

[Edit: Ali K's answer below indicates that he is interested in the case in which $H$ is also a subgroup of $\operatorname{PSL}_2(\mathbb{R})$, so I have answered the question with that additional hypothesis. However the question as stated, though it may not be the OP's intent, is probably more interesting, so I do not wish to officially change it. -- PLC]

share|improve this question

3 Answers 3

Well, if your definition of a Fuchsian group is a finitely generated, discrete subgroup of $PSL_2(\mathbb{R})$, then the answer is 'no'. $H$ could be the direct product of $G$ with an arbitrary finite group $B$, and this isn't in general a Fuchsian group. (If $G$ acts discretely on the hyperbolic plane then the stabiliser of a point would have to contain $B$, but point stabilisers in $PSL_2(\mathbb{R})$ are cyclic or dihedral.)

More interestingly, it is true that Fuchsian groups can be characterised by their actions on their boundaries, as proved by Tukia, Casson--Jungreis and Gabai. In particular, I believe it follows that any group quasi-isometric to a Fuchsian group is virtually a Fuchsian group.

share|improve this answer
    
you have to be slightly careful - one could take e.g. a central extension of a surface group by a finite cyclic group (although maybe this still has a Fuchsian subgroup of finite index). I think the appropriate result is that there is a finite quotient which is a fuchsian group (the kernel of the action on the circle is finite). But maybe that's what you meant by virtually Fuchsian. And this won't arise in Ali's case, since he has a Fuchsian subgroup by hypothesis. –  Ian Agol Oct 5 '10 at 22:30
    
Ian, a finite-by-Fuchsian group is going to be residually finite, and so have a Fuchsian group as a subgroup of finite index, no? –  HJRW Oct 5 '10 at 23:01
    
ok, good point. I was thinking more generally about the definition of quasi-isometric rigidity, but in this case it's equivalent. –  Ian Agol Oct 6 '10 at 1:42

Ali K has further asked how to show that if $G \subset H \subset \operatorname{PSL}_2(\mathbb{R})$ with $G$ Fuchsian and $[H:G] < \infty$, then $H$ is Fuchsian.

Recall that a subgroup of $\operatorname{PSL}_2(\mathbb{R})$ is Fuchsian iff it is discrete in the subgroup topology. Thus the following suffices:

Lemma: Let $H$ be a Hausdorff topological group with a finite index discrete subgroup $G$. Then $H$ is itself discrete.

Proof: $G$ is a locally compact sugroup of the Hausdorff topological group $H$, so $G$ is therefore closed. Since $G$ has finite index in $H$, the complement of $G$ is a finite union of closed cosets of $G$, so $G$ is also open. The trivial subgroup $\{e\}$ is open in $G$ and $G$ is open in $H$, so $\{e\}$ is open in $H$, i.e., $H$ is discrete.

Remark: The Hausdorff condition is necessary here (and obviously satisfied, since $\operatorname{PSL}_2(\mathbb{R})$ is Hausdorff): otherwise let $H$ be a nontrivial finite group endowed with the trivial topology (i.e., the only open subsets are $\varnothing$, $H$) and let $G = \{e\}$.

share|improve this answer

Let us suppose that H is indeed a subgroup of PSL2. How do you then prove the claim?

share|improve this answer
1  
@Ali K: it is probably better if you edit this into the question itself. (You have the ability to do this: just look for a tab near the question that says "edit" and click on it.) –  Pete L. Clark Oct 6 '10 at 14:36
    
@Ali K: I see now that you have two different user IDs. This is a standard pitfall for new users and will inhibit your ability to edit the question (you need to login as the same user that asked it). I took the liberty of editing it myself; I hope that's OK with you. –  Pete L. Clark Oct 6 '10 at 15:19

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.