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I am a newcomer to hyperbolic geometry and was trying to understand some of it in the context of dynamics, for reading certain literature.

Let a discrete subgroup $G$ of $SL_2(\mathbb R)$ act on the upper-half plane $\mathbb H$ and consider the quotient $X$ as a hyperbolic space. Fix a point $x \in \mathbb H$. I would like a proof that the following are homeomorphic.

  1. Gromov boundary of the group $G$;

  2. The set of accumulation points of the set $\{gx: g \in G\}$ inside $\mathbb H \cup \mathbb R \cup \infty$.

Also: In each case, supposedly the result is indifferent of the point you start with. How to prove this?

Also, some standard books/articles for such topics will be appreciated. Easily accessible(preferably free) references would also be nice in addition.

Added: In the situation I was looking into, the group $G$ is either: free and convex co-compact, or: co-compact surface group. (Added for the case that this assumption makes things easier.)

In certain other situations, I was reading about similar dynamical results about boundary behavior in the case of more arithmetical groups but stated without notions such as Gromov boundary, and was led to wonder about the contrast. Therefore I am also curious about how different the behavior would be for more arithmetical groups; for example for $SL_2(\mathbb Z)$ and its congruence subgroups, and how would the Gromov boundary look like in those cases.

One more question: In the earlier version, I seem to have confused between the Gromov boundary of a group $G$ and the Gromov boundary of the space $X$. I hope my understanding of the latter is correct: The geodesic boundary in the sense of equivalence class of geodesics starting at $x$, and of bounded distance from each other as you go away from $x$. What exactly is the difference between the two? Some references that might help me in clearing this kind of basic misunderstandings also will be thankfully received.

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I second Theo's recommendation of Beardon's book. It is really wonderful for hyperbolic geometry (and hyperbolic trigonometry, which is deeply beautiful). @George - it would help me if you said a few words about where this question comes from. Is it a class? Research? Is there a paper you are trying to read? Also, this contextual information should be in the question, not in the comments. –  Sam Nead Sep 7 '11 at 19:53
@Theo and George: thanks for the clean-up. –  Willie Wong Sep 7 '11 at 19:57
@Sam Nead: It is research. I had earlier put in a reference to a paper here and later cleared. Now I have added more details into the question itself. It was this paper: maths.warwick.ac.uk/~mpollic/k.pdf .. I didn't permalink it in the question because it is just a preprint on the author's homepage and likely to be taken down soon. –  George Sep 7 '11 at 19:58
You really need convex co-compact - otherwise the homeomorphism need not exist. For example, in my old answer the group $G$ is free on two generators, so has boundary a Cantor set. But the limit set is the entire circle. –  Sam Nead Sep 7 '11 at 20:55
@Sam Nead: But the paper claims that it is true in co-compact case also. –  George Sep 7 '11 at 21:04

3 Answers 3

up vote 2 down vote accepted

New answer.

Notation: $\newcommand{\HH}{\mathbb{H}}$ $\HH$ is the hyperbolic plane. If $X$ is a metric space then write $\partial X$ for the Gromov boundary of $X$. For example $\partial \HH$ is a circle. Now, suppose that $G$ is a finitely generated group. If we think of $G$ as a metric space then we do so by equipping $G$ with the word metric.

Now we suppose that $G$ is a discrete, torsion-free, finitely generated group of isometries of $\HH$. Suppose further that the quotient $X = G\backslash\HH$ has compact convex core. (Equivalently, $G$ has no nontrivial parabolic elements.)

Let $L_G$ be the limit set of $G$ - that is, the set of accumulation points in the Gromov boundary of $\HH$ of the set $G \cdot x$, for any basepoint $x \in \HH$. The referenced paper of Pollicott and Sharp states (on page 8, immediately before Lemma 4.1) that $L_G$ and $\partial G$ are homeomorphic. (More is true: the natural homeomorphism is $G$-equivariant.)

Ok, the proof of this, with details in, is a lengthy "follow-your-nose" argument. But I would advise that you first get the definitions very solid.

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Assuming the action is co-compact, this is fairly easy: Every point in $\mathbb{H}$ is of bounded distance (under the hyperbolic metric) from some point in the orbit of $x$. So you can take any sequence you like approaching an arbitrary boundary point and find a sequence of points $g_ix$ of bounded distance from it - ergo the set of accumulation points is the full bounding circle. (And nothing but the circle, since the action is discrete.)

And Gromov boundary is a quasi-isometry invariant, and since $G$ is acting cocompactly and discretely (therefore properly discontinuously) then the Švarc-Milnor Lemma says that $G$ and $\mathbb{H}$ are quasi-isometric. Thus the Gromov boundary of $G$ is the circle.

If $G$ does not act co-compactly then, as Sam Nead points out, the statement is false, for example when $G$ is a free group on two generators.

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@Invader - It depends on how the free group acts. Some actions of the free group are convex co-compact, and some aren't (as in the example). –  Sam Nead Sep 8 '11 at 9:32
So, it is not the groups themselves that are co-compact, convex co-compact, etc., but the way they act on the universal cover? I was confused by the term "convex co-compact group". Now that I think of it, the term better suited to the action. –  George Sep 8 '11 at 9:43
@Sam - This really confuses me, and after some thought I must be mistaken in my assumption that a discrete subgroup of $SL(2,\mathbb{R})$ must act properly (ie stabilizers are finite and compact sets intersect finitely many points of each orbit). Is that a false assumption? Otherwise, if the group were free and acting co-compactly, it would be quasi-isometric to $\mathbb{H}^2$ by Svarc-Milnor, which it's not. –  MartianInvader Sep 8 '11 at 19:00
@Invader - I'm afraid I don't fully understand your question. But perhaps it is helpful to point out: If a discrete subgroup of isometries of $H^2$ contains parabolics then the action is not co-compact. –  Sam Nead Sep 8 '11 at 21:07
@Sam - Now that I look more carefully, it seems my confusion was that I was simply conflating "cocompact" and "convex cocompact". I believe my answer stands, then, except perhaps "false" could be clarified to "false in general". –  MartianInvader Sep 8 '11 at 21:54

Answer to an old version of the question:

Are these two sets homeomorphic?

It sounds as if you are asking "Is the Gromov boundary of $X$ homeomorphic to the to the limit set of $G$?" This is not generally true. For example, suppose that $G$ is the group generated by the matrices

$$ \left[ \begin{array}{rr} 1 & 2 \\ 0 & 1 \end{array} \right] \quad \mbox{and} \quad \left[ \begin{array}{rr} 1 & 0 \\ 2 & 1 \end{array} \right]. $$ Then $X = H/G$ is a thrice-punctured sphere. The Gromov boundary of $X$ is the discrete space with exactly three points. The limit set of $G$, on the other hand, is the entire Gromov boundary of $H$, and so is homeomorphic to a circle.


You may be interested in the second half of my answer to the following question on MathOverflow.


Briefly, if $G$ is finitely generated then you can classify the possible quasi-isometry types of the resulting surface $X = H/G$. The Gromov boundary will be a disjoint union of circles (one per "funnel") and points (one per "cusp").

As a introduction to hyperbolic geometry I would suggest "A course on geometric group theory" by Brian Bowditch.

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Sam: Sorry, I just saw your comment while I removed mine on Willie's order. The origin of the question is the following passage of a preprint by Pollicott and Sharpe. So the question is how to identify the limit set of a surface group or a convex-cocompact group with its Gromov boundary. –  t.b. Sep 7 '11 at 19:59
Ha-ha! Finally! Ok, what they are saying is that the Gromov boundary of the group (in your notation, $G$) is homeomorphic to the limit set of the group (in their notation and yours, $L_G$). And this is true. –  Sam Nead Sep 7 '11 at 20:19
Yes exactly... :) It took me some time to figure this out... –  t.b. Sep 7 '11 at 20:27

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