Let $G \subset SL_2(\mathbb R)$ be a free subgroup generated by a symmetric set of generators $\{ a_1^{\pm 1},\ldots,a_n^{\pm 1} \}$ such that the action of $G$ on the upper-half plane $\mathbb H$ in the usual way via Möbius transformations, is convex co-compact. Let $X$ be the quotient of this action.

Let $x \in X$ be the image of some point $\tilde x \in \mathbb H$. How to find the fundamental group $\pi_1(X,x)$? What relation does it have with $G$?

Denote the limit set of $L_G$ to be the set of limit points of $G \tilde x$. In analogy with the compactification of the quotient $SL_2(\mathbb Z) \backslash \mathbb H$ by adding a cusp at $\infty$, and similarly for subgroups of $SL_2(\mathbb Z)$, can we say that the compactification of $G \backslash \mathbb H$ is missing some points from the boundary? If so, what points in the boundary $L_G$ are missing while compactifying $G \backslash \mathbb H$ ?

  • $\begingroup$ In the second paragraph - what is $X$? If $X$ is the quotient $G \backslash \mathbb H$ then in the third paragraph what is $G \backslash X$? $\endgroup$ – Sam Nead Oct 4 '11 at 17:43
  • $\begingroup$ It would be helpful to have the context for the question; eg your background and where the question is coming from. Is this from a class? Self-study? Is there a particular book or article you are reading? $\endgroup$ – Sam Nead Oct 4 '11 at 17:47
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    $\begingroup$ George - by making your question clearer, and using standard notation, you will make it easier for others to help you. I have no intention to give you, personally, a hard time. $\endgroup$ – Sam Nead Oct 23 '11 at 11:41
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    $\begingroup$ @George: Respectfully, I think you are overreacting. It is common practice on this site to ask where a question comes from, or what motivated it, as well as the general background of the asker; this helps readers (your potential answerers) understand your question better, avoid telling you things you already know, and explaining the context of a question often means that you will get information you had not known to ask for, but that people knew to mention to you because they knew the context. $\endgroup$ – Zev Chonoles Oct 23 '11 at 15:36
  • $\begingroup$ @Sam Nead : I am sorry for the touchy display and I apologize. I have deleted the comments. Thank you very much for your answer. Zev : Thanks for the explanation. $\endgroup$ – George Oct 24 '11 at 10:36

$\newcommand{\HH}{\mathbb H}$ $\newcommand{\RR}{\mathbb R}$ $\newcommand{\ZZ}{\mathbb Z}$ $\newcommand{\SL}{\mathop {\rm SL}}$

In your question $X = G \backslash \HH$ is given as a quotient. Also, $G$ is a finitely generated free group and a subgroup of $\SL(2,\RR)$. As $G$ is convex co-compact it follows that $G$ is discrete. In this case $\HH$ is the universal cover of $X$, the group $G$ is the so-called "deck-group", and there is a non-canonical isomorphism of $G$ with $\pi_1(X)$. (Choosing basepoints doesn't make the isomorphism canonical.)

The last paragraph of your question doesn't make much sense. The usual action of $\SL(2,\ZZ)$ on $\HH$ is not convex co-compact. So it is not a good example to think about. Here is a better example which fits your situation very tightly:

Suppose that $G$ is the free group of rank one -- that is, $G \cong \ZZ$. Suppose that $G$ is generated by a hyperbolic isometry $\gamma$. Let $A_\gamma \subset \HH$ be the axis of $\gamma$ acting on $\HH$. So $A_\gamma$ is a copy of $\RR$, topologically, and $G$ acts on $A_\gamma$ as $\ZZ$ acts on $\RR$, by translation. So the quotient $g = G \backslash A_\gamma$ is a circle. Also $X = G \backslash \HH$ is a hyperbolic annulus with two "flaring ends". To make this precise, note that $g \subset X$ is an essential loop in the annulus $X$. Note that $X - g$ has two components $L$ and $R$ (coming from the left and right sides of $A_\gamma$ in $\HH$). $L$ is again an annulus with one boundary on $g$. Also, $L$ has an exponentially flaring metric (as you move away from $g$) so the other boundary of $L$ is "at infinity".

Returning to $\HH$, note that the limit set of $G$, $\Lambda_G$, is exactly two points: the endpoints of $A_\gamma$. Let $\Omega_G = \partial_\infty \HH - \Lambda_G$ be the domain of discontinuity of $G$. Then $\Omega_G$ is two open sub-arcs of $\partial_\infty \HH$. As before $G$ acts on each of these by translation. The quotients are the circles at infinity for $L$ and $R$ respectively. These two circles at infinity are the Gromov boundary of the hyperbolic annulus $X$.

Finally, in this example the convex core of $X = G \backslash \HH$ is exactly the circle $g = G \backslash A_\gamma$ that we started with.

This discussion generalizes. If $G \subset \SL(2,\RR)$ is any (torsion free) group acting convex co-compactly on $\HH$ then the fundamental group of $X = G \backslash \HH$ is isomorphic to $G$. Also, the Gromov boundary of $X$, $\partial_\infty X$, is homeomorphic to $G \backslash \Omega(G)$. Each component of $\partial_\infty X$ is a circle - there is one for each flaring annular end of $X$. All ends are of this type - convex co-compactness rules out cusps.

  • $\begingroup$ Thanks a lot. This is very clear and informative. $\endgroup$ – George Oct 24 '11 at 10:37
  • $\begingroup$ Sorry about blowing up. I really don't have much knowledge on topology and I was just trying to understand a paper on dynamics, in which the topological matters are not elaborated in detail. Thanks a lot for the detailed answer. It helped a lot. $\endgroup$ – George Oct 24 '11 at 11:12

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