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Let $\Gamma_1,\Gamma_2\leq SL_2(R)$ be two Fuchsian subgroups. Take any $S_1\in\Gamma_1,S_2\in\Gamma_2$. Clearly $<S_i>\leq\Gamma_i$ and they are discrete.

Now consider $G=<S_1,S_2>\leq SL_2(R)$.

Q1: How do I show this group $G$ is discrete? Is this even a Fuchsian subgroup? I hope it is but I do not know how to prove it.

Q2: How do I know $G$ has no elliptic elements?

The purpose is to ask if $\Gamma_i$ are fuchsian subgroups, then $<\Gamma_1,\Gamma_2>$ is fuchsian as well. This is implied by pants gluing procedure.

user45765
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    Think in terms of fixed points. What happened if two of them has a common fixed point in the infinity boundary?? – Anubhav Mukherjee Feb 26 '18 at 03:16
  • @AnubhavMukherjee I am not sure what you are trying to imply here. I assume you are saying that there is at least one of them being hyperbolic. Thus if they have a common fixed point, they will have the same fixed point. Hence it is going to be torus. However, it is possible both $S_i$ are elliptic. Any elliptic elements sitting inside Fuchsian is finite order. The group generated by two elliptic might not be finite order at all. I am not sure whether one can keep deducing this way via fixed points. – user45765 Feb 26 '18 at 04:43
  • @AnubhavMukherjee In particular, I am not sure how to deduce that if both $S_i$ are elliptic, $G$ has to be discrete. – user45765 Feb 26 '18 at 04:46
  • See the following for an answer to a slightly different question which also answers yours. Short answer: it's a difficult computational problem to determine whether $\langle S_1,S_2 \rangle$ is discrete, but there is an algorithm to do it. https://math.stackexchange.com/questions/2651962/how-to-draw-fundamental-domain-for-a-group-generated-by-2-elements-acting-on-upp/2652089#2652089 – Lee Mosher Feb 26 '18 at 14:36
  • What is your definition of a Fuchsian group? – Moishe Kohan Feb 27 '18 at 00:48
  • @MoisheCohen Discrete subgroups of $SL_2(R)$ are by definition Fuchsian groups. – user45765 Feb 27 '18 at 00:56
  • So you do not care if they are first kind or second kind, right? In any case, there are "combination theorems" which provide sufficient conditions to ensure that $\langle \Gamma_1, \Gamma_2\rangle$ is discrete. You can find them, for instance, in Maskit's book "Kleinian Groups". – Moishe Kohan Feb 27 '18 at 00:59
  • @MoisheCohen I do not think the author's statement depends upon whether limit points of infinities are dense or not. Is this fact obvious or any easier way to interpret for discreteness? – user45765 Feb 27 '18 at 01:02
  • Some people only consider those discrete subgroups $\Gamma< PSL(2,R)$ to be Fuchsian for which $H^2/\Gamma$ has finite area. In some sense, this case is the most important one. If you have two such subgroups then $\langle\Gamma_1, \Gamma_2\rangle$ is discrete if and only if $\Gamma_1, \Gamma_2$ are "commensurable", i.e. their intersection has finite index in both. In other words, $\langle\Gamma_1, \Gamma_2\rangle$ is "almost never discrete" in this situation. Who is the "author" you are talking about? – Moishe Kohan Feb 27 '18 at 01:06
  • @MoisheCohen Imayoshi's Introduction to Teichmuller space. I do not think the fuchsian group in the textbook means finiteness of area in general. Especially, I poke a hole on the surface. – user45765 Feb 27 '18 at 01:19
  • It's a nice book. Are you trying to understand a particular proof? (Which one?) Otherwise, your question does not quite make sense. In general, a subgroup generated by two Fuchsian subgroups will not be discrete and even if it is discrete, it can contain elliptic elements even if the original subgroups did not. – Moishe Kohan Feb 27 '18 at 03:39
  • @MoisheCohen I am trying to understand given two pants $P_1,P_2$. Suppose I can glue them along one boundary through Dehn's twist. Then I need to join two fuchsian groups associated to each $P_i$ to form a larger group to glue the pants. So I do expect quotienting out the join of two groups will result in orbifold as elliptic element may fix around an element of finite order? – user45765 Feb 27 '18 at 14:12
  • If you do this correctly, the result is discrete and has no elliptic elements. This is a special case of a "Maskit combination theorem." See his book "Kleinian Groups". – Moishe Kohan Feb 27 '18 at 23:16

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