Fuchsian groups and topological isomorphism

I have a (finite) presentation of a group and I am wanting to prove that it is not Fuchsian. Because it is given by a presentation, a neat, algebraic description of Fuschian groups would be nice. This exists for Fuchsian groups of the first kind, as Poincaré gave a presentation for these groups. Using this description, I can show that my group is not a Fuchsian group of the first kind. So, I am wondering if such a description exists for Fuschian groups of the second kind. This is the background to my question.

According to the "encyclopedia of maths", every finitely-generated Fuchsian group of the second kind is "topologically isomorphic (as a group of the disc) to a finitely-generated Fuchsian group of the first kind". Now, this makes absolutely no sense to me. The disc is referring to viewing Fuchsian groups as transformations of the unit disc onto the complex plane. However, I do not understand what the "topologically isomorphic" means. The groups are discrete and have no topology. No? (And does topologically isomorphic imply algebraically isomorphic? If so, I am happy because then I am done...no?)

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My guess would be that "topologically isomorphic" means the groups are isomorphic, and one can find a homeomorphism $h$ from the open unit disc $D$ to itself such that the action of one group on $D$ is exactly the same and the action of the other on $h(D)$. This would mean that you're done. – MartianInvader Aug 22 '13 at 20:46

Instead of "topologically isomorphic" people usually say "topologically conjugate" exactly as in the comment by @MartianInvader. In general, given a group presentation it is hard if not impossible to say anything about the group (one cannot even decide if the group is trivial!), since the same group can have two completely different presentations. However, if, by some magic, you can prove that your group is not isomorphic to a Fuchsian group of the 1st kind, then, with few exceptions, it is also not isomorphic to a Fuchsian group of the 2nd kind since, algebraically speaking, there is no difference between the two. The exceptions are groups which are either finite cyclic, or finite dihedral, or infinite cyclic or infinite dihedral. Thus, you should test your magic powers and see if you can prove, say, that your group is not trivial. If you can, then repeat the same for cyclic and dihedral groups.

Here are some details on the proof you requested. I will do so under the assumption that the Fuchsian groups are finitely generated and orientation-preserving (it is also true in general but more tedious/hideous).

First, some terminology: A discrete subgroup of $PSL(2,R)$ is called elementary if it is either finite or infinite cyclic. I will call finitely generated discrete subgroups of $PSL(2,R)$ "Fuchsian" whether they are elementary or not (it seems that different people use different terminology here). With every Fuchsian subgroup $\Gamma$ one associates the quotient orbifold $O=D/\Gamma$, where $D$ is the open unit disk. The best introduction to orbifolds I know is the article of Peter Scott "Geometries of 3-manifolds". I also found Thurstons treatment of 2-dimensional orbifolds in Chapter 13 of his Princeton Lecture Notes very illuminating.

Such orbifolds $O$ always admit a compactification by adding some "boundary circles" so that the result is a compact orbifold with boundary $\overline{O}$.

One then defines the Euler characteristic $\chi$ of orbifolds, so that $\chi(O)=\chi(\overline{O})$. This Euler characteristic (like the usual one) is multiplicative under finite coverings. The orbifolds $O$ (and $\overline{O}$) which appear in this context are good and, moreover, they admit finite coverings by surfaces.

One further verifies that $F=\Gamma$ is elementary if and only if $\chi(O)=0$; this is easy in the case when $F$ is torsion-free and then proven in general by taking a finite index torsion-free subgroup. The same argument shows that $\chi(O)<0$ once $F$ is nonelementary.

The key statement then is that if $O$ is an orbifold without boundary which has negative Euler characteristic and "finite topology" (finitely many singular points and finitely many handles and holes) then $O$ admits a complete hyperbolic metric of finite area, i.e., is isometric to the quotient orbifold $D/F'$, where $F'$ is a Fuchsian group (necessarily of the 1st kind) and with $Area(D/F')<\infty$. You can find proofs in Scott's paper (he treats only compact case, but, considering orbifolds $\overline{O}$, one can repeat his proof in the noncompact case) and by Thurston. Here is the proof in a nutshell: Every boundaryless orbifold (as above) with $\chi(O)<0$ admits a topological "pair of pants" decomposition, along simple loops, where the "pairs of pants" are also allowed to be orbifolds which become pairs of pants after singular points are removed.

For every such "pair of pants" $P$ one designates some boundary components to be "cusps" (where the hyperbolic metric is complete and of finite area) or "waists", where the hyperbolic metric completes by a closed geodesic of certain positive length. One then considers various hyperbolic metrics on $P$ and one shows that one can choose such a metric to have prescribed length of "waists" (say, 1 unit). Given this information, one then assembles a hyperbolic metric on $O$ by isometrically gluing "pairs of pants" along "waists", which is possible since they all have the same length. I will denote the new hyperbolic orbifold by $O'$. Then, as in the case of hyperbolic metrics on surfaces, the universal cover of $O'$ is isometric to the hyperbolic plane (unit disk $D$) and the group of covering transformations acts as a Fuchsian group $F'$ on $D$. Finiteness of the area of $O'$ forces $F'$ to be of the first kind.

Now given two hyperbolic orbifolds $O=D/F, O'=D/F'$, where $F, F'$ are Fuchsian groups (need not be of 1st kind), if $f: O\to O'$ is a homeomorphism of orbifolds, it lifts to a homeomorphism of the universal covers $\tilde{f}: D\to D$ which topologically conjugates the groups $F$ and $F'$: $$\tilde{f} F (\tilde{f})^{-1}=F',$$ this is the same as in the case of the usual covering spaces theory.

I hope, it helps.

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Thanks for your answer. Do you have a reference? (Or even a quick proof?) Also, I had meant to say this in my question: One of the reasons I was wondering this was because I was reading Fine and Rosenberger's book "Algebraic generalisations of discrete groups", and they seem to only be generalising groups of the first kind and not those of the second kind. But, if they are algebraically equivalent, why make this distinction? You wouldn't happen to know, would you?... – user1729 Aug 26 '13 at 8:13
@user1729: I will write a proof when I have time. I do not know their book, but I have two guesses: Either by 1st kind they mean "cocompact" (which is not quire right, of course) or they want to avoid dealing with cyclic and dihedral groups (Vinberg does not even regard them as Fuchsian). – studiosus Aug 26 '13 at 13:17
I would just like to remark that I believe the most common usage of "elementary" in this context is for groups which are virtually cyclic (so including dihedral groups as well). – user641 Aug 31 '13 at 3:06